text:
\section*{Introduction}
Problems focused around the issue of calculating the essential norm\footnote{We kindly refer the reader to the \enquote{Toolbox},
that is, Section~\ref{SECTION: Toolbox} below, for all unexplained terms we use in this introduction.}, that is, the quotient norm modulo the compact operators,
of multiplication and composition operators (or, a mix of both, the so-called weighted composition operators) has a long history
and a well-established reputation (see, for example, the papers of Axler \cite{Axl82}, Axler, Jawell and Shields \cite{AJS80},
Bonet, Doma{\' n}ski and Lindstr{\" o}m \cite{BDL99, BDL99b}, Buckler, Ramanujan and Vuktoi{\' c} \cite{BRV99}, Edmunds and Shargorodsky \cite{ES05},
Lin and Rochberg \cite{LR93}, de Pagter and Schep \cite{PS88}, Saksman and Tylli \cite{ST94}, Schep \cite{Sch80, Sch89} and Shapiro \cite{Sha87}).
Nevertheless, even more recently, this subject attracts the attention of many researchers and seems to be still an inexhaustible
source of inspirations (to name just a few, let us mention the work of Albanese, Bonet and Ricker \cite{ABR18}, Bonet and Ricker \cite{BR20}, Contreras and Hern{\' a}ndez-D{\' i}az
\cite{CHD03}, {\u C}u{\u c}kovi{\' c} and Zhao \cite{CZ07}, Douglas, Sun and Zheng \cite{DSZ11}, Goldstein, Jarchow and Labuschagne \cite{GJL06},
Guo, Sun, Zheng and Zhong \cite{GSZZ09}, Hyv{\" a}rinen, Kemppainen, Lindstr{\" o}m, Rautio and Saukko \cite{HKLRS12}, Izuchi, Izuchi and Ohno \cite{IIO11},
De Jager and Labuschagne \cite{DJL19}, Lang and Maz'ya \cite{LM08}, Lef{\` e}vre \cite{Lef09}, Le{\' s}nik and Mleczko \cite{LM21},
Mleczko \cite{Mle09} and Schep \cite{Sch22}).

What sets the overwhelming majority of articles dealing with multiplication operators apart is the fact that they focus almost exclusively on the case
where the operator acts from one to the same space (perhaps modified by different weights); for example, if $1 \leqslant p \leqslant \infty$,
then it is part of the folklore that
\begin{equation} \label{EQ: ess norm lp -> lp}
	\norm{M_{\lambda} \colon \ell_p \rightarrow \ell_p}_{ess} = \limsup_{n \rightarrow \infty} \abs{\lambda_n}
\end{equation}
and
\begin{equation*}
	\norm{M_{\lambda} \colon L_p(\Omega) \rightarrow L_p(\Omega)}_{ess} = \norm{M_{\lambda} \colon L_p(\Omega) \rightarrow L_p(\Omega)} = \norm{\lambda}_{L_{\infty}(\Omega)},
\end{equation*}
where $\Omega = (\Omega,\Sigma,\mu)$ is a non-atomic and complete measure space.

Our object in studying essential norms of the pointwise multipliers acting between two K{\" o}the spaces is basically twofold.

First, and this seems to be the main novelty here, we will consider the {\it non-algebraic case}\footnote{The idea of considering non-algebraic case seems
to be in the air for some time (see, e.g., \cite{ABR18}, \cite{CHD03}, \cite{CZ07}, \cite{ES05} and \cite{LM21}), but in the context of K{\" o}the
spaces and essential norms of multiplication operators acting on them it seems to be a blank page. However, the reason for this, it seems, is not
a lack of interest in this subject, but rather a lack of available techniques or a lack of discernment in the current results. In fact,
our results answer a rather old Djakov and Ramanujan's question from \cite{DR00}, five of six questions posed quite recently by Ramos-Fern{\' a}ndez in \cite{RF16}
and the question asked by Ramos-Fern{\' a}ndez, Rivera-Sarmiento and Salas-Brown in \cite{RFRSSB19}.}, that is, a situation in which the multiplication
operator $M_{\lambda}$ acts between two \enquote{distinct} K{\" o}the spaces (both, however, defined over the same complete and $\sigma$-finite measure space).
In other words, we are interested in a precise estimation of
\begin{equation*}
	\norm{M_{\lambda} \colon X \rightarrow Y}_{ess} \coloneqq \inf \left\{ \norm{M_{\lambda} - K}_{X \rightarrow Y} \colon K \colon X \rightarrow Y \text{ is compact} \right\},
\end{equation*}
where $X$ and $Y$ are two K{\" o}the spaces. Let us observe that in this situation the symbol $\lambda$ of the multiplication operator $M_{\lambda} \colon X \rightarrow Y$
belongs to the vector space
\begin{equation*}
	M(X,Y) \coloneqq \left\{ g \in L_0 \colon fg \in Y \text{ for all } f \in X \right\}
\end{equation*}
which, when equipped with the natural operator norm
\begin{equation*}
	\norm{g}_{M(X,Y)} \coloneqq \norm{M_g \colon f \mapsto fg}_{X \rightarrow Y} = \sup_{\norm{f}_X \leqslant 1} \norm{fg}_Y,
\end{equation*}
becomes a K{\" o}the space itself commonly called the {\it space of pointwise multipliers} between $X$ and $Y$. Now,
in stark contrast to the algebraic case, when simply the space $M(X,X)$ coincide up to the equality of norms with $L_{\infty}(\Omega)$,
where now $\Omega = (\Omega,\Sigma,\mu)$ is a complete and $\sigma$-finite measure space, the space $M(X,Y)$ can be significantly
different than $L_{\infty}(\Omega)$; for example, if $1 \leqslant q < p \leqslant \infty$, then it is rather straightforward to see that
\begin{equation*}
	M(L_p(\Omega), L_q(\Omega)) = L_r(\Omega)
\end{equation*}
with $1/r = 1/q - 1/p$. Moreover, since bounded multiplication operators $M_{\lambda} \colon X \rightarrow Y$ are
in correspondence with the K{\" o}the space $M(X,Y)$, so it is natural to expect\footnote{Directly from (\ref{EQ: ess norm lp -> lp}) it follows
that $M_{\lambda} \colon \ell_p \rightarrow \ell_p$ is compact if, and only if, $\lambda = \{\lambda_n\}_{n=1}^{\infty} \in c_0$. On the other hand,
the separable part of the space $M(\ell_p,\ell_p) \equiv \ell_{\infty}$, that is $M(\ell_p,\ell_p)_o$, coincide with $c_0$.} that compact
multiplication operators $M_{\lambda} \colon X \rightarrow Y$
somehow correspond to the separable part of the space $M(X,Y)$, that is, the ideal $M(X,Y)_o$. Ideologically, it is a bit reminiscent of Calkin's
correspondence, which states, that there is a correspondence between the operator ideal of bounded (respectively, compact) operators on
a separable Hilbert space and a K{\" o}the sequence space $\ell_{\infty}$ (respectively, $c_0$).

Secondly, and this seems to be the most pragmatic here, since the class of K{\" o}the spaces often serve as a basic skeleton for building other spaces
(for example, Bergman spaces, Bloch spaces, Ces{\` a}ro spaces, Hardy spaces, Sobolev spaces and Tandori spaces can be build upon Orlicz, Lorentz, Marcinkiewicz,
Calder{\' o}n--Lozanovski{\u \i}, Musielak--Orlicz or, even more generally, rearrangement invariant and K{\" o}the spaces), so it can be expected\footnote{The
reason for this is actually the result of Anderson and Shields saying that if $E$ is a K{\" o}the sequence space and $V$ is a vector space then
$M(V,E) = M(S(V),E)$, where $S(V)$ is the {\it solid envelope} of $V$, that is to say, the intersection of all vector spaces with the ideal property that contain $V$
as a subspace (see \cite[Lemma~2 and Lemma~3(i)]{AS76}).}
that the techniques developed by us will find some applications in certain classes of function spaces (sometimes not belonging to the family of K{\" o}the spaces).
We make here the first step in this direction by improving Mleczko's \cite{Mle09} result on compact Fourier multipliers acting on Banach spaces of
analytic functions on the unit disc (in particular, acting on abstract Hardy spaces), and Albanese, Bonet and Ricker's
\cite{ABR18} and Bennett's \cite{Ben96} (see also \cite{BR20}) results on compact multipliers acting between classical Ces{\` a}ro and Tandori spaces.

What also seems to be noteworthy is that the non-algebraic case is not a simple generalization of what is currently known.
On the one hand, it requires a fresh approach avoiding, for example, the spectral theory techniques (after all, $M(X,X) \equiv L_{\infty}(\Omega)$
is a Banach algebra, which cannot be said about $M(X,Y)$ in general). On the other, because the description of compact multipliers between
$X$ and $Y$ is inextricably linked with the description of a separable part of the space $M(X,Y)$, some more advanced theory related
to the so-called {\it arithmetic of K{\" o}the spaces}\footnote{Arithmetic of K{\" o}the spaces, broadly speaking, can be described as a certain collection
of techniques dealing with constructions such as the pointwise product, the pointwise multipliers and the symmetrization (we refer to \cite{KLM19} and Section~\ref{SECTION: examples}
for some illustrations of these ideas and examples; cf. also \cite{CS14}, \cite{CDSP08}, \cite{KLM12, KLM14} and \cite{Sch10}).}
and {\it local properties}\footnote{The {\it order continuity} property will play the main role here, because \enquote{globally} it is an analogue
of the separability, but can also be \enquote{localized} at a single point, that is, it make perfect sense to ask if a function in a K{\" o}the space
is order continuous or not.} of K{\" o}the spaces, comes into play.

After all this propaganda, let us only add that the proofs we propose seems to lie much closer to the methods of functional analysis, in particular,
to the theory of function spaces and to the operator theory, than to rely on some {\it ad hoc} tricks. As usual, more sophisticated methods put certain
phenomena in a more complete and interesting context and, we hope, that this is the case here.

\subsection*{Neverending story} We have one more, slightly less official, but short, story to tell. Initially, the reason for our work was Voigt's
preprint \cite{Voi22} posted on {\tt arXiv.org} in late March, where the essential norm of the multiplication operators acting from $L_p$ into itself is calculated.
Already a superficial reading of this work, as well as the works of authors cited therein, almost immediately convinced us that everything should work
in much more general framework. However, we also could not shake off the impression that the problem raised in \cite{Voi22} is too classical not to have
a solution in the existing literature.
To our surprise, just a few weeks later, a commentary on Voigt's article in the form of Schep's preprint \cite{Sch22} appeared. In introduction to his paper, Schep writes
\begin{quote}
	\begin{small}
		\enquote{We refer to a recent preprint of J. Voigt \cite{Voi22} for some references. It seems that these authors are not aware that these multiplication
		operators are examples of order bounded band preserving operators (or orthomorphisms). (...) The current paper was prompted by the preprint \cite{Voi22},
		which didn't use our results form \cite{Sch80}. By using the term abstract multiplication operator in our title we hope that future duplication
		of our results by authors working on these questions for concrete function spaces will be avoided by a web search of MathSciNet search. (...) The current
		paper describes completely the essential spectrum and norm of an operator $T \in Z(E)$ for an arbitrary Banach lattice $E$.}
	\end{small}
\end{quote}
Although we don't want to breach open doors here, we feel that the last word has not yet been given, because as we have already tried to explain above
an analogous questions for something, what we call here the {\it non-algebraic case}, seems to be almost completely untouched.

\subsection*{Outline}

Now, we describe the organization of this work and mention some other results contained herein.

We begin in Section~1 with rather rapid recollection of required definitions and notation that will be used through this paper.
As it seems to us, the only slightly unconventional approach is presented here in the context of Rademacher functions (Definition~\ref{DEF: Abstract Rademachers}).

Section~2 is the main part of the paper. At the beginning we compute the essential norms of multiplication operators acting between two
distinct K{\" o}the function spaces (Theorem~\ref{THM: essnorm multipliers function}) or two distinct K{\" o}the sequence spaces
(Theorem~\ref{THM: Essential norm Kothe sequence}) culminating in Theorem~\ref{THM: ess norm func + seq}, where the essential norm
of the multiplication operator acting between two distinct K{\" o}the spaces is calculated.
We then explain how these results can be viewed through the prism of Pitt's theorem (Proposition~\ref{PROP: Pitt's multiplication}) and,
using this point of view, we answer Djakov and Ramanujan's question from \cite{DR00} (Remark~\ref{REMARK: Djakov-Ramanujan}).
In the following, we will focus our attention on the weak essential norms of multiplications operators acting between two distinct K{\" o}the
function spaces (Theorem~\ref{THM: weak ess norm}). {\it En route} to the proof on this result, we will use Flores, Tradacete and Troitsky's
notion of disjointly homogeneous spaces in tandem with some recent results from \cite{LTM22} about the Dunford--Pettis criterion in K{\" o}the
function spaces. Closing this thread, we will give an example of a multiplication operator acting between two Marcinkiewicz spaces which
is not weakly compact (Example~\ref{EXP: not weakly compact between Marcinkiewicz}), and of a weakly compact multiplication operator
acting between Lorentz spaces with a symbol which is not order continuous (Example~\ref{EXP: Lorentz symbol not OC but WC}).
Finally, the last theme that we will devote our attention is the interplay between multiplier and its symbol.
More precisely, relying on certain ideas from \cite{KKT22} we will show an elegant result (Proposition~\ref{Proposition: odleglosci = odleglosci})
which equates the (weak) essential norm of the multiplication operator $M_{\lambda} \colon X \rightarrow Y$ with the distance from its symbol
$\lambda$ to the ideal $M(X,Y)_o$ of all order continuous functions in $M(X,Y)$.
We also propose aesthetically pleasing expression of the weak essential norm of multipliers acting between two rearrangement invariant function
spaces (Corollary~\ref{COR: ess r.i. function}).

Then, in Section~3, we will harness our techniques and results developed so far to work. To begin with, by calculating the essential norms of
multiplication operators acting between two distinct abstract Ces{\` a}ro sequence spaces or two distinct abstract Tandori sequence spaces
(Theorem~\ref{THM: ess norm Cezaro and Tandori function}), we answer Ramos-Fern{\' a}ndez, Rivera-Sarmiento and Salas-Brown's question from \cite{RFRSSB19}.
To illustrate our, as nomenclature itself suggests, \enquote{abstract} approach, we will consider (in Remark~\ref{Remark: ces_p -> ces_q [ABR18]})
pointwise multipliers acting between the classical Ces{\` a}ro sequence $ces_p$ spaces, and, their duals, the so-called Tandori sequence
spaces $\widetilde{\ell_p}$. This way we will recreate Albanese, Bonet and Ricker's result from \cite{ABR18} and some of Bennett's results from \cite{Ben96}.
Next, we do more or less the same in the case of abstract Ces{\` a}ro function spaces (Theorem~\ref{THM: ess norm Cezaro and Tandori function}),
but we also explain why our results on weak essential norms cannot be easily transferred to the ground of operators acting between Ces{\` a}ro
function spaces.
We will close this section by computing the essential norms of Fourier multipliers acting from a Banach space of analytic functions on the unit disc
$\mathbb{D}$, in particular, from an abstract Hardy space built upon a K{\" o}the function space defined the unit circle $\mathbb{T}$, into some
K{\" o}the sequence space (Theorem~\ref{THM: ess norm holomorphic}). This improves Mleczko's main result from \cite{Mle09}.

Last Section 4 is devoted to examples. Here, we calculate the space $M(X,Y)$ for the different configurations of spaces $X$ and $Y$, coming from
some more concrete classes of K{\" o}the sequence spaces, like Musielak--Orlicz spaces (including Nakano spaces and, most importantly, Orlicz spaces;
see Proposition~\ref{PROPOSITION: Compact multipliers between Nakano}) or Lorentz and Marcinkiewicz spaces (Proposition~\ref{Prop: computing multipliers from m and to lambda}).
This can be also thought of as a tutorial to the arithmetic of K{\" o}the spaces having their roots in \cite{KLM12}, \cite{KLM14}, \cite{KLM19} and \cite{Sch10}
(Lemma~\ref{LEMMA: a'la KLM12/14}), and as a reincarnation of some Kolwicz's ideas about symmetrization construction from \cite{Kol16} (Lemma~\ref{LEMMA: a'la Kolwicz OC}).

All sections are sprinkled with examples and open problems.

Finally, in Appendix~\ref{Appendix B} we present an example (Proposition~\ref{PROPOSITION: Appendix A - counterexample factorization})
of a certain Orlicz space $\ell_{\widetilde{M}}$ with the property that
\begin{equation*}
	\ell_{\widetilde{M}} \odot M(\ell_{\widetilde{M}},\ell_2) \neq \ell_2,
\end{equation*}
that is, $\ell_{\widetilde{M}}$ does {\it not} factorize through a Hilbert space $\ell_2$, but $M(\ell_{\widetilde{M}}, \ell_2) \neq \ell_{\infty}$.
To the best of our knowledge, this is the first example of such a space.

\subsection*{Acknowledgements}

We are very grateful to Lech Maligranda for careful reading of an early draft for our work and all his helpful advise, and to Pawe{\l} Kolwicz
for all the hints that led us back on the right track. Moreover, both authors would like to express their thanks to Damian Kubiak for
sharing his paper with them. We also thank Karol Le{\' s}nik for some useful discussions. Last but not least, we feel obligated
to our dear friends Przemys{\l}aw Trela and Mariusz {\. Z}yluk for some, as usual, interesting comments about our work.

Our research was carried out at the Pozna{\' n} University of Technology as a part of the Rector's grant.

\section{Toolbox: Preliminary definitions and prerequisite concepts} \label{SECTION: Toolbox}

Our notation is rather standard and is, more or less, a mix of notations used in Bennett--Sharpley \cite{BS88}, Kre{\u \i}n--Petunin--Semenov\cite{KPS82},
Lindenstrauss--Tzafriri \cite{LT79} and Pietsch \cite{Pie80}.
Below, for the reader's convenience, we provide in detail the most important definitions and the necessary facts.
However, the seasoned reader may skip this part without any harm (probably a glance should be given to the Section~\ref{SUBSEC: Rademachers}
devoted to Rademacher functions). Other more specific terminology will be introduced in context when needed.

\subsection{Function spaces}
Let $(\Omega, \Sigma, \mu)$ be a complete\footnote{This means that any subset of a set of measure zero is measurable.} and
$\sigma$-finite measure space. Let, moreover, $L_0(\Omega,\Sigma,\mu)$ (briefly, $L_0(\Omega)$, or even $L_0$, as long as it does
not lead to confusion) be the set consisting of all equivalence classes, modulo equality almost everywhere, of locally
integrable\footnote{We will say, that a function $f$ is {\it locally integrable} if it is measurable and for every $\sigma \in \Sigma$
with $\mu(\sigma) < \infty$ we have that $\int_{\sigma} \abs{f(\omega)} d\mu(\omega) < \infty$} real or complex valued functions on $\Sigma$.
A Banach space $X$ is called a {\it K{\" o}the space} (also a {\it Banach ideal space} or a {\it Banach function space}) if the following two conditions hold:
\begin{itemize}
	\item[\ding{228}] $X$ is a linear subspace of $L_0(\Omega,\Sigma,\mu)$,
	\item[\ding{228}] if $\abs{f(\omega)} \leqslant \abs{g(\omega)}$ almost everywhere on $\Omega$, with $f$ measurable
	and $g \in X$, then $f \in X$ and $\norm{f}_X \leqslant \norm{g}_X$;
\end{itemize}
cf. \cite[pp.~1--3]{BS88}, \cite[Definition~1.1, p.~40]{KPS82}, \cite[Definition~1.b.17, p.~28]{LT79} or \cite[p.~161]{Ma89}.
However, unless we write otherwise, we will also assume that
\begin{itemize}
	\item[\ding{228}] for every $\sigma \in \Sigma$ with $\mu(\sigma) < \infty$ the characteristic function $\chi_{\sigma}$ belongs to $X$.
\end{itemize}
Hereafter, whenever we use the term {\it K{\" o}the sequence space} we mean a K{\" o}the space over a purely atomic measure space, while, the name
{\it K{\" o}the function space} is reserved for a K{\" o}the space over a non-atomic measure space (the choice of vocabulary
seems, perhaps, a bit misplaced here, because formally every sequence is also a function, but this nomenclature seems to be
well-rooted in the literature). It follows from Maharam's theorem that every K{\" o}the space is, roughly speaking, a mixture
of a certain K{\" o}the sequence and K{\" o}the function space. Moreover, since we only consider $\sigma$-finite measure spaces,
so without losing generality we can, and we will, always assume that K{\" o}the sequence spaces are defined over $(\mathbb{N}, 2^{\mathbb{N}}, \mu)$.

Due to the closed graph theorem, an inclusion of two K{\" o}the spaces, say $X$ and $Y$, is bounded and to emphasize this
fact we will write $X \hookrightarrow Y$ instead of just $X \subset Y$ (in some situations we will also use the symbol $X \overset{C}{\hookrightarrow} Y$
to indicate that the norm of the inclusion $X \subset Y$ is not greater than $C > 0$). Moreover, $X = Y$ ($X \equiv Y$) means that the
spaces $X$ and $Y$ are the same as sets and their norms are equivalent (respectively, equal).

Given a K{\" o}the space $X$ on $(\Omega,\Sigma,\mu)$ the {\it K{\" o}the dual space $X^{\times}$}
(or the {\it associated space}) of the space $X$ is defined as a vector space
\begin{equation*}
	X^{\times} \coloneqq \left\{f \in L_0(\Omega) \colon \int_{\Omega} \abs{f(\omega) g(\omega)} d\mu(\omega) < \infty \text{ for all } g \in X \right\}
\end{equation*}
with the norm $\norm{f}_{X^{\times}} \coloneqq \sup_{\norm{g}_X = 1} \norm{fg}_{L_1}$. Recall that $X \overset{1}{\hookrightarrow} X^{\times \times}$
and we have the equality $X \equiv X^{\times \times}$ if, and only if, the norm in $X$ has the so-called {\it Fatou property}, that is, the closed
unit ball in $X$ is also closed in $L_0(\Omega)$ with respect to the topology of local convergence in measure. Sometimes it will be convenient to use
the symbol $\langle \cdot, \cdot \rangle$ for the natural pairing between the K{\" o}the space $X$ and its topological dual $X^*$
or its K{\" o}the dual $X^{\times}$.

Moreover, for a K{\" o}the space $X$ on $(\Omega,\Sigma,\mu)$ and a positive and locally integrable function $w \colon \Omega \rightarrow (0,\infty)$
(briefly, a {\it weight} on $\Omega$) by a {\it weighted K{\" o}the space $X(w)$} we will understand a vector space
\begin{equation*}
	X(w) \coloneqq \{f \in L_0(\Omega) \colon fw \in X\}
\end{equation*}
together with the norm $\norm{f}_{X(w)} \coloneqq \norm{fw}_X$. There should be no doubt that the space $X(w)$ is a K{\" o}the space itself.
It is also clear that
\begin{equation*}
	\left[ X(w) \right]^{\times} \equiv X^{\times}\left( \frac{1}{w} \right).
\end{equation*}

We will say that a function, say $f$, in a K{\" o}the space $X$ is {\it order continuous} (or it has an {\it order continuous norm})
if for any sequence $\{f_n\}_{n=1}^\infty$ in $X$ with $0 \leqslant f_n \leqslant \abs{f}$ and $f_n \downarrow 0$ point-wise we have
that $\norm{f_n}_X \downarrow 0$.
By $X_o$ we denote the subspace of all order continuous elements of $X$. It turns out that $X_o$ is an {\it order ideal} in $X$, that is to say,
$X_o$ is a closed linear subspace of $X$ with the ideal property\footnote{This means that if $f \in X_o$, $g \in X$ and $\abs{g(\omega)} \leqslant \abs{f(\omega)}$,
then also $g \in X_o$.} (see \cite[Theorem~3.8, p.~16]{BS88}).
Whenever $X = X_o$, that is, every element of $X$ is order continuous, the space $X$ is said to be {\it order continuous} (pedantically, {\it $\sigma$-order continuous})
In this situation, we will occasionally write $X \in (OC)$. Notice that this is in fact equivalent to the separability of $X$ (see \cite[Theorem~5.5, p.~27]{BS88}).
In general, we refer to \cite{Wn99} for much more information about order continuity property. Let's add that we have two useful characterizations:
\begin{itemize}[leftmargin=0.5cm]
	\item The dual space $X^*$ of a K{\" o}the space $X$ can be naturally identified with the the K{\" o}the dual space $X^{\times}$ if, and only if,
		the space $X$ is order continuous (see \cite[Corollary~4.3, p.~23]{BS88});
	\item A K{\" o}the space $X$ with the Fatou property is reflexive if, and only if, the space $X$ itself and its K{\" o}the dual $X^{\times}$
		are both order continuous (see \cite[Corollary~4.4, p.~23]{BS88}).
\end{itemize}
In practice, it is also useful to observe that a function $f$ in a K{\" o}the space $X$ is order continuous if, and only if,
$\lim_{n \rightarrow 0} \norm{f\chi_{\sigma_n}}_X = 0$ for every decreasing sequence $\{\sigma_n\}_{n=1}^{\infty}$ of measurable
sets with the intersection of measure zero (cf. \cite[Proposition~3.5, p.~15]{BS88}). Consequently, in the particular case when $X$
is a K{\" o}the sequence space, a sequence $x = \{x_n\}_{n=1}^{\infty}$ from $X$ is order continuous if, and only if, $\lim_{n \rightarrow \infty} \norm{x\chi_{\{n,n+1,...\}}}_X = 0$.

Now, let $(\Omega, \Sigma, \mu)$ be a {\it resonant measure space}, that is, non-atomic or purely atomic with all atoms having equal measure,
complete and $\sigma$-finite measure space (see, e.g., \cite[Theorem~2.7, p.~51]{BS88}).
A K{\" o}the space $X$ of measurable functions on $\Omega$ is called {\it rearrangement invariant} (or {\it symmetric}; sometimes we will
also write briefly {\it r.i.}) if it satisfies the following condition:
\begin{itemize}
	\item[\ding{228}] if $f$ and $g$ are equi-measurable and $g \in X$, then $f \in X$ and $\norm{f}_X = \norm{g}_X$;
\end{itemize}
cf. \cite[Definition~4.1, p.~59]{BS88}, \cite[Definition~2.a.1, pp.~117--118]{LT79} or \cite[p.~90]{KPS82}. Equivalently,
\begin{itemize}
	\item[\ding{228}] if $f^*(t) \leqslant g^*(t)$ for almost all $t > 0$ and $g \in X$, then $f \in X$ and $\norm{f}_X \leqslant \norm{g}_X$,
\end{itemize}
where $f^*$ denote the {\it non-decreasing rearrangement} of $f$ (see \cite[p.~90]{KPS82}; cf. \cite[Definition~1.5, p.~39]{BS88} and \cite[p.~117]{LT79}).
Every K{\" o}the space is a Banach lattice in a sense of Meyer-Nieberg \cite{MN91}. Moreover, every order continuous Banach lattice with the weak
unit is order isometric to a K{\" o}the space (see \cite[Theorem~1.b.14, p.~25]{LT79} and references given there).
It follows from the {\it Luxemburg representation theorem} (see \cite[Theorem~4.10, p.~62]{BS88}) that when considering rearrangement invariant
spaces it is enough to limit oneself to the case of one of the three separable measure spaces:
\begin{itemize}
	\item[\ding{228}] the natural numbers $\mathbb{N} = \{1,2,...\}$ with the counting measure $\#$,
	\item[\ding{228}] the unit interval $(0,1) $ with the usual Lebesgue measure $m$,
	\item[\ding{228}] the semi-axis $(0,\infty)$ with the usual Lebesgue measure $m$;
\end{itemize}
cf. \cite[pp.~114--115]{LT79} (note, however, that in \cite{KPS82} one consider only r.i. spaces defined on the semi-axis and in \cite{BS88}
the Fatou property is essentially included in their definition; see~\cite[p.~90]{KPS82} and \cite[Definition~1.1, p.~2]{BS88}). Again, we take
here the convention that whenever a rearrangement space $X$ is defined over $(\mathbb{N}, 2^{\mathbb{N}}, \#)$ we will call it a {\it rearrangement
invariant sequence space}, but in the other two cases, we will use the name {\it rearrangement invariant function space}.

The class of rearrangement invariant spaces includes many important and classical examples of spaces, like
\begin{itemize}[leftmargin=0.5cm]
	\item Lebesgue--Riesz spaces $L_p$,
	\item Orlicz spaces $L_M$ (see \cite[\S4.8]{BS88}, \cite[\S3]{Mal83}),
	\item Lorentz spaces $\Lambda_{\varphi}$ (see \cite[Definition~5.12, pp.~71--72]{BS88}, \cite[p.~107]{KPS82}),
	\item Lorentz spaces $L_{p,q}$ (see \cite[Definition~2.b.8, p.~142]{LT79}; cf. also Section~\ref{SECTION: ESS NORMS}),
	\item Marcinkiewicz spaces $M_{\varphi}$ (see \cite[Definition~5.7, p.~69]{BS88}, \cite[p.~114]{KPS82}),
	\item Lorentz--Zygmund spaces $L_{p,q}(\log L)^{\alpha}$ (see \cite[Definition~6.13, p.~253]{BS88}),
	\item Grand Lebesgue spaces $L_{p)}$ and, their duals, small Lebesgue spaces $L_{(p}$,
	\item and some combined forms of these spaces, like, for example, Orlicz--Lorentz spaces $\Lambda_{M,w}$.
\end{itemize}
For general properties of K{\" o}the and rearrangement invariant spaces we refer to Bennett--Sharpley \cite{BS88},
Kre{\u \i}n--Petunin--Semenov \cite{KPS82}, Lindenstrauss--Tzafriri \cite{LT77} and \cite{LT79}, Luxemburg \cite{Lux67}, Maligranda \cite{Ma89}
and Zaanen \cite[Chapter~15]{Zaa67}.
More information on rearrangement invariant spaces defined {\it not} necessary on a $\sigma$-finite measure space can be found, for example, in Folland \cite{Fol99}.
For much more information about Banach lattices and, even more generally, Riesz spaces see, for example, Aliprantis--Burkinshaw \cite{AB85},
Kantorovich--Akilov \cite{KA82}, Meyer-Nieberg \cite{MN91} and Zaanen \cite{Za83}.

\subsection{Essential norms and operator ideals}
Denote by $\mathscr{L}$ the class of all operators, which we {\it always} assumed to be linear and bounded, between all Banach spaces.
For two Banach spaces, say $X$ and $Y$, let $\mathscr{L}(X,Y)$ be the set of all operators from $X$ into $Y$. More generally, for
$\mathscr{I} \subset \mathscr{L}$ any two Banach spaces $X$ and $Y$ we set $\mathscr{I}(X,Y) \coloneqq \mathscr{I} \cap \mathscr{L}(X,Y)$.
If the need arises, we will shorten the symbols $\mathscr{L}(X,X)$ and $\mathscr{I}(X,X)$ to $\mathscr{L}(X)$ and $\mathscr{I}(X)$, respectively.
Following Pietsch \cite{Pie80} we recall that a class $\mathscr{I}$ is called a {\it operator ideal} if the following conditions are satisfied:
\begin{itemize}
	\item[\ding{228}] for any one-dimensional Banach space $X$ the identity operator $\text{id}_X \colon X \rightarrow X$ belongs to $\mathscr{I}$,
	\item[\ding{228}] for all Banach spaces, say $X$ and $Y$, the set $\mathscr{I}(X,Y)$ is a linear subspace of $\mathscr{L}(X,Y)$,
	\item[\ding{228}] for any four Banach spaces, say $W$, $X$, $Y$ and $Z$, the composite operator $UTS$ belongs to $\mathscr{I}(W,Z)$
	whenever $S$ belongs to $\mathscr{L}(Y,Z)$, $T$ to $\mathscr{I}(X,Y)$, and $U$ to $\mathscr{L}(W,X)$.
\end{itemize}
An operator ideal $\mathscr{I}$ is said to be a {\it closed operator ideal} provided $\mathscr{I}(X,Y)$ is closed in $\mathscr{L}(X,Y)$ for any two Banach
spaces $X$ and $Y$. Some very well-known classes of operator ideals, like compact operators, weakly compact operators, strictly singular operators
or completely continuous operators are examples of closed operator ideals.

For any closed operator ideal $\mathscr{I}$ and any operator, say $T$, acting between two Banach spaces $X$ and $Y$ by the {\it essential
norm of $T \colon X \rightarrow Y$ with respect to an operator ideal $\mathscr{I}$} we mean
\begin{equation}\label{def: general essnorm}
	\norm{T \colon X \rightarrow Y}_{{ess}_{\mathscr{I}}} \coloneqq \norm{T \colon X \rightarrow Y}_{{\mathscr L}(X,Y)/\mathscr{I}(X,Y)}
		= \inf\{\norm{T - J}_{X \rightarrow Y} \colon J \in \mathscr{I}(X,Y)\};
\end{equation}
cf. with the definitions given in \cite[p.~96]{Tyl95} and \cite[Definition~0.3]{Lef09}. The last expression on the right in (\ref{def: general essnorm})
is, of course, nothing else but a natural quotient norm on the space ${\mathscr L}(X,Y)/\mathscr{I}(X,Y)$, that is to say,
\begin{equation*}
	\norm{T \colon X \rightarrow Y}_{{\mathscr L}(X,Y)/\mathscr{I}(X,Y)} = \text{dist}_{\mathscr{L}(X,Y)}(T \colon X \rightarrow Y,\mathscr{I}(X,Y)).
\end{equation*}
The above definition, in its two most common versions, looks like this:
\begin{itemize}[leftmargin=0.5cm]
	\item the quotient algebra ${\mathscr L}(X,Y)/{\mathscr K}(X,Y)$, where ${\mathscr K}(X,Y)$ denote the ideal of compact operators from $X$ to $Y$,
		is called the {\it Calkin algebra} and the corresponding quotient norm $\norm{T \colon X \rightarrow Y}_{{\mathscr L}(X,Y)/\mathscr{K}(X,Y)}$
		(or, a bit more concisely, $\norm{T \colon X \rightarrow Y}_{ess}$) is called the {\it essential norm} of $T$;
	\item alike, the quotient algebra ${\mathscr L}(X,Y)/{\mathscr W}(X,Y)$, where ${\mathscr W}(X,Y)$ denote the ideal of weakly compact operators
		acting between $X$ and $Y$, is called the {\it weak Calkin algebra} and the corresponding quotient norm $\norm{T \colon X \rightarrow Y}_{{\mathscr L}(X,Y)/\mathscr{W}(X,Y)}$
		(briefly, just $\norm{T \colon X \rightarrow Y}_w$) is called the {\it weak essential norm} of $T$.
\end{itemize}
Clearly, an operator $T$ acting between $X$ and $Y$ belongs to the closed operator ideal $\mathscr{I}(X,Y)$ if, and only if,
\begin{equation*}
	\norm{T \colon X \rightarrow Y}_{{\mathscr L}(X,Y)/\mathscr{I}(X,Y)} = 0.
\end{equation*}
However, the concept of the essential norm has in general much more to offer (see, e.g., \cite{Tyl95} and references given there).

\subsection{Pointwise multipliers}
Let $X$ and $Y$ be two K{\" o}the spaces defined on the same complete and $\sigma$-finite measure space $(\Omega,\Sigma,\mu)$.
The space of {\it pointwise multipliers $M(X,Y)$} is defined as a vector space
\begin{equation*}
	M(X,Y) \coloneqq \{f \in L_0(\Omega) \colon fg \in Y \text{ for all } g \in X\}
\end{equation*}
equipped with the natural operator norm $\norm{f}_{M(X,Y)} \coloneqq \sup_{\norm{g}_X = 1} \norm{fg}_Y$.

Note that the space $M(X,L_1)$ is nothing else but the K{\" o}the dual space $X^{\times}$ of the space $X$, so we can naturally
think of the construction $M(X,Y)$ as a generalized K{\" o}the dual of $X$ (cf. \cite{CDSP08} and \cite{MP89}). Moreover,
looking from a different perspective, the space $M(X,Y)$ can naturally be understood as a \enquote{pointwise quotient} of the
space $Y$ by $X$.

Some properties and examples of this construction can be found, for example, in \cite{CS14}, \cite{CDSP08}, \cite{KLM12, KLM14, KLM19},
\cite{LT21}, \cite{LTM22}, \cite{MP89} and \cite{Sch10} (see also Bennett's memoir \cite[\S14]{Ben96}, Grosse-Erdmann's book \cite{GE98} and references given there).
We will collect here only a few of the most important from the point of view of our future needs (see \cite{AS76}, \cite{KLM12} and \cite{Sch10}).
Hereinafter, unless stated otherwise, $X$, $Y$ and $Z$ will denote three K{\" o}the spaces defined over the same measure space.
\begin{itemize}[leftmargin=0.5cm]
	\item If $X \hookrightarrow Y$, then $M(Y,Z) \hookrightarrow M(X,Z)$. Moreover, if $Y \hookrightarrow Z$, then $M(X,Y) \hookrightarrow M(X,Z)$.
	\item $L_{\infty} \hookrightarrow M(X,Y)$ if, and only if, $X \hookrightarrow Y$. In particular, $M(X,X) \equiv L_{\infty}$.
	\item $M(X,Y) \hookrightarrow M(Y^{\times},X^{\times}) = M(X^{\times \times}, Y^{\times \times})$. Moreover, if $Y$ has the Fatou property,
		then $M(X,Y) = M(Y^{\times},X^{\times})$.
	\item If $Y$ has the Fatou property, then $M(X,Y)$ share this property too.
	\item If $X$ and $Y$ are two r.i. function spaces, then $M(X,Y)$ is also a r.i. function space (the same is true for r.i. sequence spaces;
		see the first part of the proof of Lemma~\ref{LEMMA: a'la KLM12/14}).
\end{itemize}
Note that for $1 \leqslant q < p < \infty$ we have the following identifications $M(L_p,L_q) = L_r$ and $M(\ell_p,\ell_q) = \ell_r$,
where $1/r = 1/q - 1/p$. On the other hand, if $1 \leqslant p < q < \infty$, then $M(L_p,L_q) = \{0\}$, but $M(\ell_p,\ell_q) = \ell_{\infty}$.

The space of pointwise multipliers $M(X,Y)$ and the space of all {\it multiplication operators $M_{\lambda}$} (or {\it diagonal operators}; cf. \cite{OT72})
acting between $X$ and $Y$ are isometrically isomorphic via the mapping
\begin{equation*}
	\Upsilon \colon \lambda \mapsto (M_{\lambda} \colon f \mapsto \lambda f),
\end{equation*}
where $\lambda \in M(X,Y)$ and $f \in X$. Function $\lambda$ defining the multiplication operator $M_{\lambda}$ will be called its {\it symbol}.

\subsection{Symmetrizations of K{\" o}the spaces} \label{SUBSEC: Symmetrization}
Let $X$ be a K{\" o}the space defined on complete and $\sigma$-finite measure space $(\Omega,\Sigma,\mu)$. By the {\it symmetrization $X^{(*)}$}
of $X$ we will understand the set
\begin{equation*}
	X^{(*)} \coloneqq \left\{ x \in L_0(\Omega) \colon x^* \in X \right\}
\end{equation*}
together with the functional $\norm{x}_{X^{(*)}} \coloneqq \norm{x^*}_X$.

Many classical rearrangement invariant spaces, like Lorentz spaces $\Lambda_{\varphi}$ and $L_{p,q}$, Marcinkiewicz spaces $M_{\varphi}$, and Orlicz--Lorentz
spaces $\Lambda_{M,w}$ are particular cases of this construction, namely,
\begin{equation*}
	\Lambda_{\varphi} = \left[ L_1(\varphi') \right]^{(*)}, L_{p,q} = \left[ L_q\left( t^{\frac{1}{p} - \frac{1}{q}} \right) \right]^{(*)},
		M_{\varphi} = \left[ L_{\infty}(\varphi) \right]^{(*)} \text{ and } \Lambda_{M,w} = (L_{\Phi})^{(*)},
\end{equation*}
where $L_{\Phi}$ is a Musielak--Orlicz space generated by the Musielak--Orlicz function $\Phi(u,t) = M(u)w(t)$ and $M$ is a Young function.

In general, however, even the fact that the symmetrization $X^{(*)}$ of a K{\" o}the space $X$ is a linear space is not so obvious
(see \cite[Example~3.1]{CKMP04} for a counter-example). In fact, it was proved by Kami{\' n}ska and Raynaud \cite[Lemma~1.4]{KR09}
(and later supplemented by Kolwicz, Le{\' s}nik and Maligranda in \cite[Corollary~1]{KLM19}) that the symmetrization $X^{(*)}$ of
a K{\" o}the space $X$ is a linear spaces if, and only if, the dilation operator $D_2$ is bounded on the cone of non-negative and
non-increasing functions from $X$. Even then, the functional $\norm{\cdot}_{X^{(*)}}$ may be only a quasi-norm\footnote{In quasi-Banach
spaces, the triangle inequality is allowed to hold approximately, that is, $\norm{f + g}_X \leqslant \Delta \left( \norm{f}_X + \norm{g}_X \right)$
for some constant $\Delta \geqslant 1$ (see, for example, \cite{Kal03} for more).}.

This construction enjoys some interest nowadays (see \cite{Kol16}, \cite{KLM19}, \cite{KR09} and references given there).
For us, as we will see later, its usefulness stems primarily from the fact that it often allows us to reduce some rather difficult
questions about the structure of rearrangement invariant spaces to simpler questions about the structure of weighted $L_p$-spaces;
for example, it takes little skill to notice that
\begin{equation*}
	M(L_{\infty}(w), F) = F\left( \frac{1}{w} \right),
\end{equation*}
where $F$ is a rearrangment invariant space. At the same time, it is not entirely clear how to identify the space
$M(M_{\varphi},F)$, until we notice that the following equalities should be true
\begin{equation*}
	M(M_{\varphi},F) = M\left( [L_{\infty}(\varphi)]^{(*)}, F^{(*)} \right) \overset{\textdbend}{=} \left[ M(L_{\infty}(\varphi), F) \right]^{(*)}
		= \left[ F\left( \frac{1}{w} \right) \right]^{(*)}.
\end{equation*}
The point is, that we can make the second step somehow rigorous (see Section~\ref{SECTION: examples}; cf. \cite{KLM14}).

\subsection{Rademachers on general measure spaces} \label{SUBSEC: Rademachers}
Usually, without unnecessary confusion, the {\it Rademacher system $\left\{ r_n \right\}_{n=1}^{\infty}$} is understood as the Bernoulli
sequence of independent identically and symmetrically distributed random variables on the unit interval taking the values $\pm 1$. In more
familiar form the {\it $n^{\text{th}}$ standard Rademacher $r_n$}, where $n \in \mathbb{N}$, is just a function
\begin{equation} \label{EQ: n-th standard Rademachers}
	r_n(t) \coloneqq \sign(\sin 2^n\pi t) \text{ for } 0 < t < 1
\end{equation}
(see \cite[Definition~6.2.1, p.~132]{AK06} and \cite[p.~24]{LT77}; Astashkin's book \cite{Ast20} seems to be an
inexhaustible source of information about Rademacher's functions).

However, for some reasons, which will become clear later, the object we really want to construct here is a family of functions that shares
the most important properties of Rademacher system, but can be defined on {\it any} $\sigma$-finite measure space.
Since on a finite measure space $(\Omega,\Sigma,\mu)$ Rademachers are nothing else but a sequence of independent random variables
each of which takes values $\pm 1$ with probability $\mu(\Omega)/2$, so the construction of Rademachers on any finite and
non-atomic measure space should present no difficulties (cf. \cite[p.~224]{JL0506}). With a little patience while gluing the appropriate
fragments together we can do the same for every $\sigma$-finite measure space. Unfortunately, handling with such a construction turned out
to be quite difficult to us. Therefore, we propose the following, easier-to-work, one.

\begin{definition}[Abstract Rademachers $\left\{ r^{\Omega}_n \right\}_{n=1}^{\infty}$] \label{DEF: Abstract Rademachers}
	Let $(\Omega,\Sigma,\mu)$ be a non-atomic complete and $\sigma$-finite measure space and let $\Omega = \{\Omega_{N}\}_{N=1}^{\infty}$
	be a decomposition of $\Omega$ into countable family of pairwise disjoint sets of measure not exceeding one. Further,
	for every $N \in \mathbb{N}$, let $\varpi_{N} \colon \Omega_{N} \rightarrow (0,\mu(\Omega_N))$ be a measure-preserving transformation
	the existence of which is guaranteed by Ryff's theorem (see \cite[Theorem~7.5, p.~82]{BS88}). Put
	\begin{equation*}
		r^{\Omega_{N}}_n(\omega) \coloneqq (r_n \circ \varpi_{N})(\omega) \text{ for } \omega \in \Omega_N,
	\end{equation*}
	where, as usual, the symbol $r_n$ stands for the $n^{\text{th}}$ standard Rademacher (like in (\ref{EQ: n-th standard Rademachers}) above).
	Then, for $n \in \mathbb{N}$, by the $n^{\text{th}}$ {\it abstract Rademacher on $\Omega$} we understand a function $r^{\Omega}_n \colon \Omega \rightarrow \{\pm 1\}$
	defined as
	\begin{equation*}
		r^{\Omega}_n(\omega) = \sum_{N=1}^{\infty} r^{\Omega_{N}}_n(\omega) \text{ for } \omega \in \Omega.
	\end{equation*}
\end{definition}

With such a definition at hand, it is not difficult to prove the following, essentially technical, lemma which seems to be well-known
in its basic version (that is, when $\Omega$ is just a unit interval or a semi-axis with the Lebesgue measure; see, for example,
\cite[Lemma~3.2]{Kub14}). However, this lemma will prove its usefulness in the proof of Theorem~\ref{THM: essnorm multipliers function}.

\vspace{5pt} \noindent {\bf Lemma~$\mathscr{R}$.}
{\it Let $(\Omega,\Sigma,\mu)$ be a non-atomic complete and $\sigma$-finite measure space. Then for any function, say $f$,
	from $L_1(\Omega,\Sigma,\mu)$ we have}
	\begin{equation*}
		\lim\limits_{n \rightarrow \infty} \int_{\Omega} f(\omega) r^{\Omega}_n(\omega) d\mu(\omega) = 0.
	\end{equation*}
\begin{proof}
	Take $f \in L_1(\Omega,\Sigma,\mu)$ and put $\Omega = \{\Omega_{N}\}_{N=1}^{\infty}$, where $\Omega_{N}$'s are like
	in the Definition~\ref{DEF: Abstract Rademachers}.

	First, we claim that for any $N \in \mathbb{N}$ we have
	\begin{equation} \label{rademacherydozera}
		\lim\limits_{n \rightarrow \infty} \int_{\Omega} f(\omega) r^{\Omega_N}_n(\omega) d\mu(\omega) = 0.
	\end{equation}
	Indeed,
	\begin{align*}
		\lim\limits_{n \rightarrow \infty} \int_{\Omega} f(\omega) r^{\Omega_N}_n(\omega) d\mu(\omega) &
		=\lim\limits_{n \rightarrow \infty} \int_{\Omega_N} f(\omega) \left( r_n \circ \varpi_{N} \right)(\omega) d\mu(\omega) \\
		& = \lim\limits_{n \rightarrow \infty} \int_{0}^{\mu\left( \Omega_N \right)} \left( f \circ \varpi_N^{-1} \right)(t) r_n(t) dm(t)  = 0,
	\end{align*}
	where the last equality follows from, for example, \cite[Lemma~3.2]{Kub14}. This proves the claim. Next, fix $\varepsilon>0$. Since the space $L_1(\Omega,\Sigma,\mu)$
	is order continuous (see \cite[Theorem~5.5, p.~27]{BS88}), so there is a finite sequence of natural numbers, say $\{N_j\}_{j=1}^{M}$,
	such that
	\begin{equation*}
		\int_{\Omega\setminus \bigcup\limits_{j=0}^M\Omega_{N_j}}\abs{ f(\omega)} d\mu(\omega) < \frac{\varepsilon}{2}.
	\end{equation*}

	Next, using (\ref{rademacherydozera}), for every $j = 1, 2, ..., M$ we can find $n_j \in \mathbb{N}$ satisfying
	\begin{equation*}
		\abs{\int_{\Omega} f(\omega)r^{\Omega_{N_j}}_n(\omega) d\mu(\omega)} < \frac{\varepsilon}{2M} \text{ for } n \geqslant n_j.
	\end{equation*}
	Finally, for all $n \geqslant \max\{n_1,n_2,... , n_M\}$, we have
	\begin{align*}
		\abs{\int_{\Omega} f(\omega) r^{\Omega}_n(\omega) d\mu(\omega)}
		&=\abs{\sum\limits_{j=0}^M\int_{\Omega_{N_j}} f(\omega) r^{\Omega}_n(\omega) d\mu(\omega)
			+\int_{\Omega\setminus \bigcup\limits_{j=0}^M\Omega_{N_j}} f(\omega)r^{\Omega}_n(\omega) d\mu(\omega)}\\
		&\leqslant\sum\limits_{j=0}^M\abs{\int_{\Omega_{N_j}} f(\omega) r^{\Omega_{N_j}}_n(\omega) d\mu(\omega)}
		+ \int_{\Omega\setminus \bigcup\limits_{j=0}^M\Omega_{N_j}}\abs{ f(\omega)} d\mu(\omega)
		< \varepsilon.
	\end{align*}
	In view of $\varepsilon$'s arbitrariness this finishes the proof.
\end{proof}

We end this section with one more remark about notation. First, given two quantities, say $A$ and $B$, depending (maybe) on certain
parameters, we will write $A \approx B$ understanding that there exist absolute constants (that is, independent of all involved parameters),
say $c, C > 0$, such that $cA \leqslant B \leqslant CA$. Second, we will sometimes use the symbol $T \colon X \circlearrowleft$, which will
mean exactly the same as $T \colon X \rightarrow X$.

\section{Main results: On essential norms of multiplication operators} \label{SECTION: ESS NORMS}

This section is a main part of our work. Since the title of this paragraph is basically self-explanatory, so let's stop at the following remark,
which we will use repeatedly without making any explicit reference to it: Since
\begin{equation*}
	M_{\lambda} = M_{e^{i\varphi} \abs{\lambda}} = M_{e^{i\varphi}}M_{\abs{\lambda}},
\end{equation*}
so the multiplication operator $M_{\lambda}$ belongs to some operator ideal, say $\mathscr{I}$, if, and only if, the same can be said about the
multiplication operator $M_{\abs{\lambda}}$.

\subsection{Compact multiplication operators} For the sake of completeness, let us remind that an operator acting from one Banach space to another is
called {\it compact} provided it carries bounded sets to relatively compact one.

Intuition tells us that there should be no non-trivial compact multiplication operators between K{\" o}the function spaces.
In fact, the point is that \enquote{locally} the multiplication operator behaves similarly to the identity. More precisely,
let $M_{\lambda} \colon X \rightarrow Y$ be the multiplication operator between two K{\" o}the function spaces both defined on
the same complete and $\sigma$-finite measure space $(\Omega,\Sigma,\mu)$. Since the symbol $\lambda \in M(X,Y)$ is not identically
zero (otherwise it's all trivial), so there is a set $\sigma \in \Sigma$ with $\mu(\sigma) > 0$ such that
\begin{equation*}
	\lambda\chi_{\sigma} \approx \chi_{\sigma}.
\end{equation*}
However, since the operator $\text{id}_{\sigma} \colon X \rightarrow Y$ defined as
\begin{equation*}
	\text{id}_{\sigma}(x) \coloneqq x\chi_{\sigma},
\end{equation*}
is an isomorphic embedding, so know from Riesz that it cannot be compact. Therefore, $M_{\lambda} \colon X \rightarrow Y$ is not compact as well.

What we want to do now is to calculate the essential norm of the multiplication operator which acts between two K{\" o}the function spaces.
Roughly speaking, we will show that it is as far from being compact as it can be.

\begin{theorem}[Essential norm of multiplication operators between K{\" o}the function spaces] \label{THM: essnorm multipliers function}
	{\it Let $X$ and $Y$ be two K{\" o}the function spaces with the Fatou property both defined on the same complete and $\sigma$-finite
		measure space $(\Omega,\Sigma,\mu)$. Suppose that either the space $X$ is order continuous or the space $Y$ is reflexive.
		Then
		\begin{equation*}
			\norm{M_{\lambda} \colon X \rightarrow Y}_{ess} = \norm{M_{\lambda} \colon X \rightarrow Y} = \norm{\lambda}_{M(X,Y)}.
		\end{equation*}
		In particular, there are no non-trivial compact multiplication operators between $X$ and $Y$.}
\end{theorem}
\begin{proof}
	Of course, we can immediately assume that the space $M(X,Y)$ is non-trivial, because otherwise there is nothing to prove. What's more,
	the inequality
	\begin{equation*}
		\norm{M_{\lambda} \colon X \rightarrow Y}_{ess} \leqslant \norm{M_{\lambda}}_{X \rightarrow Y} = \norm{\lambda}_{M(X,Y)}
	\end{equation*}
	is true for obvious reasons, so there is noting else but to turn to the proof of a more interesting inequality.

	Let's start with the case when the space $X$ is order continuous. Fix $\varepsilon > 0$ and take $\lambda \in M(X,Y)$. It follows from the definition
	of the norm in the space $M(X,Y)$ that there exists $f \in \text{Ball}(X)$ with $\norm{\lambda f}_Y \geqslant \norm{\lambda}_{M(X,Y)} - \varepsilon$.
	Moreover, it is also clear that
	\begin{equation*}
		\norm{\lambda f r^{\Omega}_n}_Y = \norm{\lambda f}_Y \geqslant \norm{\lambda}_{M(X,Y)} - \varepsilon,
	\end{equation*}
	where $r^{\Omega}_n$ denote the $n^{\text{th}}$-abstract Rademachers on $\Omega$ (cf. Definition~\ref{DEF: Abstract Rademachers}). Now,
	for any compact operator, say $K$, acting from $X$ to $Y$ we have that
	\begin{equation*}
		\norm{M_{\lambda} - K}_{X \rightarrow Y} \geqslant \norm{(M_{\lambda} - K)(f r^{\Omega}_n)}_Y \geqslant \abs{\norm{\lambda f r^{\Omega}_n}_Y - \norm{K(f r^{\Omega}_n)}_Y},
	\end{equation*}
	We claim that the sequence $\left\{ f r^{\Omega}_n \right\}_{n=1}^{\infty}$ is weakly null. To see this, take $g \in X^{\times}$ (due to our assumptions
	the space $X$ is order continuous, thus we have the equality $X^* = X^{\times}$; see, e.g., \cite[Theorem~4.1, p.~20]{BS88}). Since
	$fg \in L_1(\Omega,\Sigma,\mu)$, so it is enough to apply Lemma~$\mathscr{R}$ from Section~\ref{SUBSEC: Rademachers} to conclude that
	\begin{equation*}
		\lim\limits_{n \rightarrow \infty} \abs{ \langle f r^{\Omega}_n, g \rangle } = \lim\limits_{n \rightarrow \infty} \abs{\int_{\Omega} (f g) r^{\Omega}_n d\mu} = 0
	\end{equation*}
	and this proves our claim. Consequently, $\lim_{n \rightarrow \infty} \norm{K(f r^{\Omega}_n)}_Y = 0$, because every compact operator
	is {\it a fortiori} completely continuous. But now, using the above inequalities, we get that
	\begin{equation*}
		\norm{M_{\lambda} - K}_{X \rightarrow Y} \geqslant \norm{\lambda}_{M(X,Y)} - \varepsilon.
	\end{equation*}
	This concludes the proof in the first case.

	Now, let us move on to the second case when the space $Y$ is reflexive. It follows from Schauder's theorem that an operator, say $T$, acting
	between Banach spaces is compact if, and only if, its adjoint $T^*$ is compact too (see., e.g., \cite[A.4, p.~329]{AK06}). Hence, we can
	conclude\footnote{Note that in general there is no hope for an equality in
	(\ref{inequality: ess norm of T and adjoint T}) (see \cite[Example~1]{AJS80}; cf. \cite[p.~160]{AJS80}).} that
	\begin{equation} \label{inequality: ess norm of T and adjoint T}
		\norm{M_{\lambda} \colon X \rightarrow Y}_{ess} \geqslant \norm{M_{\lambda}^* \colon Y^* \rightarrow X^*}_{ess},
	\end{equation}
	where $M_{\lambda}^*$ is the adjoint of the multiplication operator $M_{\lambda}$.
	Next, we will need the following observation.

	\vspace{5pt} \noindent {\bf Claim $\spadesuit$.}
	{\it Let $W$ and $Z$ be two K{\" o}the function space both defined on the same complete and $\sigma$-finite measure space $(\Omega,\Sigma,\mu)$.
		Let, moreover, $M_{\lambda} \colon W \rightarrow Z$ be the multiplication operator with the symbol $\lambda \in M(W,Z)$. Suppose
		the space $Z$ is order continuous. Then the following diagram commutes}
		\begin{equation*}
			\xymatrix{
				Z^* \ar[r]^{M_{\lambda}^*} & W^*\\
				Z^{\times} \ar[u]^{L} \ar[r]^{M_{\lambda}^{\times}} & W^{\times} \ar[u]_{\text{id}}}
		\end{equation*}
	{\it where $L \colon Z^{\times} \rightarrow Z^*$ is the natural isometric isomorphism between the K{\" o}the dual $Z^{\times}$
		and the topological dual $Z^*$ (see below for details), and $M_{\lambda}^{\times} \colon Z^{\times} \rightarrow W^{\times}$ is the K{\" o}the
		adjoint\footnote{Recall that for a bounded operator $T$ acting between two K{\" o}the spaces, say $X$ and $Y$, defined on possibly
		different measure spaces its {\it K{\" o}the adjoint} (or {\it K{\" o}the dual}) operator $T^{\times} \colon Y^{\times} \rightarrow X^{\times}$
		is defined by
		\begin{equation*}
			\langle x,T^{\times}(y^{*}) \rangle = \langle T(x),y^{*} \rangle \text{ for all } x \in X \text{ and } y^* \in Y^{\times}.
		\end{equation*}
		} of $M_{\lambda} \colon W \rightarrow Z$.}

	\vspace{5pt} \noindent
	{\it Proof of Claim $\spadesuit$.}
	It is clear that each function $g \in Z^{\times}$ induces a bounded linear functional
	\begin{equation*}
		L_g \colon f \mapsto L_g(f) \coloneqq \int_{\Omega} f g d\mu
	\end{equation*}
	on $Z$ and we can think of $L_g$ as a member of the space $Z^*$. Moreover, since the space $Z$ is order continuous, it turns out that the mapping
	$L \colon Z^{\times} \rightarrow Z^*$ given by
	\begin{equation*}
		L \colon g \mapsto \left( f \mapsto L_g(f) \right) \text{ for } f \in Z
	\end{equation*}
	is an isometric isomorphism (see, e.g., \cite[Theorem~4.1, p.~20]{BS88}). Therefore, for $f \in W$ and $g \in Z^{\times}$, we have
	\begin{equation*}
		\langle f, M_{\lambda}^{*}(L(g)) \rangle = \langle M_{\lambda}(f), L(g) \rangle
			= \int_{\Omega} M_{\lambda}(f) g d\mu = \int_{\Omega} f M_{\lambda}^{\times}(g) d\mu
			= \langle f, M_{\lambda}^{\times}(g) \rangle.
	\end{equation*}
	This ends the proof of Claim $\spadesuit$.

	\vspace{5pt}
	Now, it follows from the preceding Claim $\spadesuit$ that the adjoint operator $M_{\lambda}^* \colon Y^* \rightarrow X^*$ can be naturally identified
	with the multiplication operator $M_{\lambda}^{\times} \colon Y^{\times} \rightarrow X^{\times}$ and, in consequence, we have that
	\begin{equation*}
		\norm{M_{\lambda} \colon X \rightarrow Y}_{ess} \geqslant \norm{M_{\lambda}^{\times} \colon Y^{\times} \rightarrow X^{\times}}_{ess}
			= \norm{\lambda}_{M(Y^{\times},X^{\times})} = \norm{\lambda}_{M(X^{\times\times},Y^{\times\times})} = \norm{\lambda}_{M(X,Y)},
	\end{equation*}
	where the above inequality is a consequence of (\ref{inequality: ess norm of T and adjoint T}), the first equality results
	from the fact that under our assumption that the space $Y$ is reflexive, the space $Y^{\times}$ is order continuous (see \cite[Corollary~4.4, p.~23]{BS88})
	and the first part of this proof, while, finally, the last equality follows from the fact that $X^{\times\times}$ (resp. $Y^{\times\times}$)
	and $X$ (resp. $Y$) are isometrically isomorphic whenever the space $X$ (resp. $Y$) has the Fatou property (see \cite[Theorem~2.7, p.~10]{BS88}).
	We are done.
\end{proof}

We immediately get the following

\begin{corollary}[Compact multiplication operators from a K{\" o}the function space to itself] \label{COR: compact function X to X}
	{\it Let $X$ be an order continuous K{\" o}the function space with the Fatou property. Then
	\begin{equation*}
		\norm{M_{\lambda} \colon X \circlearrowleft}_{ess} = \norm{M_{\lambda} \colon X \circlearrowleft} = \norm{\lambda}_{L_{\infty}}.
	\end{equation*}
	In particular, there are no non-trivial compact multiplication operators on $X$.}
\end{corollary}
\begin{proof}
	Just observe that $M(X,X) \equiv L_\infty$ and use Theorem~\ref{THM: essnorm multipliers function}.
\end{proof}

Let us now turn to the case of K{\" o}the spaces defined over purely atomic measure spaces. Now the situation is completely different, because,
for example, any sequence from $c_{00}$ (that is, a sequence which is eventually zero) induces a compact multiplication operator. Precisely,
we have the following

\begin{theorem}[Essential norm of multiplication operators between K{\" o}the sequence spaces] \label{THM: Essential norm Kothe sequence}
	{\it Let $X$ and $Y$ be two K{\" o}the sequence spaces with the Fatou property. Suppose that either the space $X$ is reflexive
		or the space $Y$ is order continuous. Then
	\begin{equation*}
		\norm{M_{\lambda} \colon X \rightarrow Y}_{ess} = \lim\limits_{n \rightarrow \infty} \norm{\lambda \chi_{\{n, n+1, \ldots\}}}_{M(X,Y)}.
	\end{equation*}
		In particular, the multiplication operator $M_{\lambda} \colon X \rightarrow Y$ is compact if, and only if, $\lambda \in M(X,Y)_o$.}
\end{theorem}

Note that, as the first part of the proof presented below shows, the inequality
\begin{equation*}
	\norm{M_{\lambda} \colon X \rightarrow Y}_{ess} \leqslant \lim\limits_{n \rightarrow \infty} \norm{\lambda \chi_{\{n, n+1, \ldots\}}}_{M(X,Y)}
\end{equation*}
holds without any additional assumptions on $X$ and $Y$.

\begin{proof}
	In order to show the first of the two inequalities, take $\lambda = \{\lambda_n\}_{n=1}^\infty \in M(X,Y)$ and notice that
	\begin{align*}
		\norm{M_{\lambda} \colon X \rightarrow Y}_{ess} & = \inf \left\{ \norm{M_{\lambda} - K}_{X \rightarrow Y} \colon K \in {\mathscr K}(X,Y) \right\} \\
				& \leqslant \norm{M_{\lambda} - M_{\lambda \chi_{\{1, \ldots, n - 1\}}}}_{X \rightarrow Y} = \norm{\lambda \chi_{\{n, n+1, \ldots\}}}_{M(X,Y)},
	\end{align*}
	where to get the inequality we have used the plain fact that finite rank operators acting between Banach spaces are always compact.
	In consequence, we conclude that
	\begin{equation*}
		\norm{M_{\lambda} \colon X \rightarrow Y}_{ess} \leqslant \lim_{n \rightarrow \infty} \norm{\lambda \chi_{\{n, n+1, ...\}}}_{M(X,Y)}.
	\end{equation*}

	Now, let us turn to the proof of the second inequality. We will consider both situations mentioned in the formulation of the theorem separately.

	$1^{\text{st}}$. Let's start with the case when the space $X$ is reflexive. Fix $\varepsilon > 0$. It is clear from the definition of the norm in the space $M(X,Y)$
	that for any $n \in \mathbb{N}$ we can find $x^{(n)} \in \text{Ball}(X)$ with
	\begin{equation*}
		\norm{\lambda \chi_{\{n, n+1, \ldots\}} x^{(n)}}_Y \geqslant \norm{\lambda \chi_{\{n, n+1, \ldots\}}}_{M(X,Y)} - \varepsilon.
	\end{equation*}
	It should be also clear that without losing generality we can assume that $\supp\left( x^{(n)} \right) \subset \{n, n+1, ...\}$ for $n \in \mathbb{N}$.
	This, however, gives us that
	\begin{equation} \label{inequality: seq multip}
		\norm{\lambda x^{(n)}}_Y \geqslant \norm{\lambda \chi_{\{n, n+1, \ldots\}}}_{M(X,Y)} - \varepsilon \text{ for every } n \in \mathbb{N}.
	\end{equation}
	Next, observe that the sequence $\left\{ x^{(n)} \right\}_{n=1}^{\infty} \subset \text{Ball}(X)$ is weakly null. Indeed, since the space $X$ is reflexive,
	so the space $X^{\times}$ is order continuous and $X^* = X^{\times}$ (keeping in mind that the space $X$ has the Fatou property we refer to \cite[Corollaries 4.3 and 4.4, p.~23]{BS88}).
	Therefore, for any $x^* \in X^{\times}$, we have that
	\begin{equation*}
		\abs{\langle x^{(n)}, x^* \rangle} = \abs{\sum_{j=1}^{\infty} x^{(n)}(j) x^*(j)} = \abs{\sum_{j=n}^{\infty} x^{(n)}(j) x^*(j)}
			\leqslant \norm{x^{(n)}}_X \norm{x^*\chi_{\{n, n+1, \ldots\}}}_{X^{\times}} \xrightarrow{n \rightarrow \infty} 0.
	\end{equation*}
	Now, for any compact operator $K$ between $X$ and $Y$, we have
	\begin{equation*}
		\norm{M_{\lambda} - K}_{X \rightarrow Y} \geqslant \norm{(M_{\lambda} - K) \left( x^{(n)} \right) }_Y \geqslant \norm{\lambda x^{(n)}}_Y - \norm{K \left( x^{(n)} \right)}_Y,
	\end{equation*}
	where the last inequality follows from the reverse triangle inequality. Moreover, remembering about inequality (\ref{inequality: seq multip}),
	we get that actually
	\begin{equation*}
		\norm{M_{\lambda} - K}_{X \rightarrow Y} \geqslant \norm{\lambda \chi_{\{n, n+1, \ldots\}}}_{M(X,Y)} -  \norm{K \left( x^{(n)} \right)}_Y - \varepsilon.
	\end{equation*}
	Going further, since every compact operator is {\it a fortiori} completely continuous and, as we have shown above, the sequence
	$\left\{ x^{(n)} \right\}_{n=1}^{\infty} \subset \text{Ball}(X)$ is weakly null, so $\norm{K \left( x^{(n)} \right)}_Y \rightarrow 0$ as $n \rightarrow \infty$.
	Consequently,
	\begin{equation*}
		\norm{M_{\lambda} - K}_{X \rightarrow Y} \geqslant \lim\limits_{n \rightarrow \infty} \norm{\lambda \chi_{\{n, n+1, \ldots\}}}_{M(X,Y)} - \varepsilon
	\end{equation*}
	and this, due to $\varepsilon$'s arbitrariness, ends the proof in the first case.

	$2^{\text{nd}}$. Next, suppose that the space $Y$ is order continuous. Again, fix $\varepsilon > 0$. Let $\left\{ x^{(n)} \right\}_{n=1}^{\infty} \subset \text{Ball}(X)$
	be the sequence constructed above. Moreover, let $K \colon X \rightarrow Y$ be a compact operator. Since the set $\left\{ K\left( x^{(n)} \right) \right\}_{n=1}^{\infty}$
	is relatively compact in $Y$, so it has an accumulation point, say $y_a \in Y$. This means that there is a subsequence $\left\{ x^{(n_k)} \right\}_{k=1}^{\infty}$
	of the sequence $\left\{ x^{(n)} \right\}_{n=1}^{\infty}$ with
	\begin{equation}\label{inequality: podciag zbiezny}
		\norm{K\left( x^{(n_k)} \right) - \widetilde{y}}_Y < \varepsilon \text{ for all } k \text{ big enough}.
	\end{equation}
	Therefore, using again the reverse triangle inequality,
	\begin{equation*}
		\norm{M_{\lambda} - K}_{X \rightarrow Y} \geqslant \norm{(M_{\lambda} - K)\left( x^{(n_k)} \right)}_Y
			\geqslant \norm{M_{\lambda}\left( x^{(n_k)} \right) - y_a}_Y - \norm{K\left( x^{(n_k)} \right) - y_a}_Y.
	\end{equation*}
	But, the above inequality in tandem with (\ref{inequality: podciag zbiezny}) implies that for sufficiently big $k$ we have
	\begin{equation} \label{inequality: nierownosc ess seq}
		\norm{M_{\lambda} - K}_{X \rightarrow Y} \geqslant \norm{M_{\lambda}\left( x^{(n_k)} \right) - y_a}_Y - \varepsilon
	\end{equation}
	On the other hand, recalling that $\supp\left( x^{(n)} \right) \subset \{n, n+1, ...\}$ for $n \in \mathbb{N}$, we get that
	\begin{align*}
		\norm{M_{\lambda}\left( x^{(n_k)} \right) - \widetilde{y}}_Y & = \norm{\lambda x^{(n_k)} - y_a}_Y \\
			& \geqslant \norm{\left( \lambda x^{(n_k)} - y_a \right)\chi_{\supp\left( x^{(n_k)} \right)}}_Y \\
			& \geqslant \norm{\lambda x^{(n_k)}}_Y - \norm{y_a\chi_{\supp\left( x^{(n_k)} \right)}}_Y \\
			& \geqslant \norm{\lambda \chi_{\{n_k, n_{k+1}, \ldots\}}}_{M(X,Y)} - 2\varepsilon,
	\end{align*}
	where the last inequality follows from inequality (\ref{inequality: seq multip}) together with our assumption that the space $Y$ is order continuous.
	Finally, by combining the just obtained inequality with (\ref{inequality: nierownosc ess seq}), we get immediately that
	\begin{equation*}
		\norm{M_{\lambda} - K}_{X \rightarrow Y} \geqslant \norm{\lambda \chi_{\{n_k, n_{k+1}, \ldots\}}}_{M(X,Y)} - 3\varepsilon,
	\end{equation*}
	which due to the arbitrariness of $K$ and the ideal property of the space $M(X,Y)$ gives
	\begin{equation*}
		\norm{M_{\lambda} \colon X \rightarrow Y}_{ess} \geqslant \lim\limits_{n \rightarrow \infty} \norm{\lambda \chi_{\{n, n+1, ...\}}}_{M(X,Y)} - 3\varepsilon.
	\end{equation*}
	This proves what we wanted.
\end{proof}

The results consisting in the estimation of the quantity $\norm{M_{\lambda} \colon X \circlearrowleft}_{\textit{ess}}$ in various concrete classes of K{\" o}the
sequence spaces, like Musielak--Orlicz spaces, Lorentz spaces, Marcinkiewicz spaces or Ces{\` a}ro spaces, are scattered throughout the existing literature
(we refer to \cite{RF16}, \cite{RFRSSB19} and references given there). Below, we propose the following consequence of Theorem~\ref{THM: Essential norm Kothe sequence},
which is a general supplement to the results of this type. Later, in Section~\ref{SECTION: examples}, we will pick up this thread again with much more due attention.

\begin{corollary}[Compact multiplication operators from a K{\" o}the sequence space to itself] \label{COR: compact sequence X to X}
	{\it Let $X$ be an order continuous K{\" o}the sequence space with the Fatou property. Then
		\begin{equation*}
			\norm{M_{\lambda} \colon X \circlearrowleft}_{ess} = \limsup\limits_{n \rightarrow \infty} \abs{\lambda_n}.
		\end{equation*}
		In particular, the multiplication operator $M_{\lambda} \colon X \circlearrowleft$ is compact if, and only if, $\lambda \in c_0$.}
\end{corollary}
\begin{proof}
	Since $M(X,X) \equiv \ell_\infty$, so using Theorem~\ref{THM: Essential norm Kothe sequence} we immediately obtain that
	\begin{equation*}
		\norm{M_{\lambda} \colon X \circlearrowleft}_{ess} = \lim\limits_{n \rightarrow \infty} \norm{\lambda \chi_{\{n, n+1, ...\}}}_{\ell_\infty}
			= \lim\limits_{n \rightarrow \infty} \left( \sup\limits_{m \geqslant n} \abs{\lambda_m} \right) = \limsup\limits_{n \rightarrow \infty} \abs{\lambda_n}.
	\end{equation*}
\end{proof}

Let us admit, however, that very recently Schep in \cite{Sch22} obtained a better result avoiding\footnote{He was also able to avoid the assumption
about $\sigma$-finiteness of the involved measure space and he considered a more general class of operators, namely, the {\it band preserving operators}.}
the assumption about order continuity of the space $X$. On the other hand, his argument is based on the spectral theory and does not work in the non-algebraic
case.

Finally, we can combine an atomic case (Theorem~\ref{THM: Essential norm Kothe sequence}) together with a non-atomic case (Theorem~\ref{THM: essnorm multipliers function})
to describe the essential norm of multiplication operators acting between two K{\" o}the spaces.

\begin{theorem}[Essential norm of multiplication operators between K{\" o}the spaces] \label{THM: ess norm func + seq}
	{\it Let $X$ and $Y$ be two K{\" o}the spaces with the Fatou property both defined over the same complete and $\sigma$-finite measure space $(\Omega,\Sigma,\mu)$.
		Let, moreover, $\Omega_c$ and $\Omega_a$ denote the non-atomic and the purely atomic part of $(\Omega,\Sigma,\mu)$, respectively. Suppose that one of the
		spaces $X$ or $Y$ is reflexive. Then}
		\begin{equation*}
			\norm{M_{\lambda} \colon X \rightarrow Y}_{ess} \approx
				\max \left\{ \norm{M_{\lambda} \colon X|_{\Omega_c} \rightarrow Y|_{\Omega_c}}_{ess}, \norm{M_{\lambda} \colon X|_{\Omega_a} \rightarrow Y|_{\Omega_a}}_{ess} \right\}
		\end{equation*}
	{\it with an equivalence involving universal constants only.}
\end{theorem}
\begin{proof}
	Without loss of generality, we can assume that the space $M(X,Y)$ is non-trivial. Take $\lambda \in M(X,Y)$.

	To show the first inequality let $\{K_n\}_{n=1}^{\infty}$ be a sequence of compact operators from $X$ into $Y$ defined as
	\begin{equation*}
		K_n(x) = \lambda x \chi_{\{1,2,..., n-1\}}
	\end{equation*}
	for $x \in X$ and $n = 2, 3, ...$. We have
	\begin{align*}
		\norm{M_{\lambda} \colon X \rightarrow Y}_{ess} & \leqslant \norm{M_{\lambda} - K_n}_{X \rightarrow Y} \\
			& \leqslant \norm{M_{\lambda} - K_n}_{X|_{\Omega_c} \rightarrow Y|_{\Omega_c}} + \norm{M_{\lambda} - K_n}_{X|_{\Omega_a} \rightarrow Y|_{\Omega_a}} \\
			& = \norm{M_{\lambda}}_{X|_{\Omega_c} \rightarrow Y|_{\Omega_c}} + \norm{M_{\lambda\chi_{\{n,n+1,...\}}}}_{X|_{\Omega_a} \rightarrow Y|_{\Omega_a}}.
	\end{align*}
	Now, it follows from Theorem~\ref{THM: essnorm multipliers function} and Theorem~\ref{THM: Essential norm Kothe sequence} that
	\begin{equation*}
		\norm{M_{\lambda} \colon X|_{\Omega_c} \rightarrow Y|_{\Omega_c}}_{ess} = \norm{M_{\lambda}}_{X|_{\Omega_c} \rightarrow Y|_{\Omega_c}},
	\end{equation*}
	and, respectively,
	\begin{equation*}
		\norm{M_{\lambda} \colon X|_{\Omega_a} \rightarrow Y|_{\Omega_a}}_{ess} = \lim\limits_{n \rightarrow \infty} \norm{\lambda\chi_{\{n,n+1,...\}}}_{M(X|_{\Omega_a},Y|_{\Omega_a})}
			= \lim\limits_{n \rightarrow \infty} \norm{M_{\lambda\chi_{\{n,n+1,...\}}}}_{X|_{\Omega_a} \rightarrow Y|_{\Omega_a}}.
	\end{equation*}
	In consequence, we have
	\begin{equation*}
		\norm{M_{\lambda} \colon X \rightarrow Y}_{ess}
			\leqslant 2\max \left\{ \norm{M_{\lambda} \colon X|_{\Omega_c} \rightarrow Y|_{\Omega_c}}_{ess}, \norm{M_{\lambda} \colon X|_{\Omega_a} \rightarrow Y|_{\Omega_a}}_{ess} \right\}
	\end{equation*}

	To prove the second inequality, let's first define two operators $P_{\square} \colon X \rightarrow X|_{\Omega_{\square}}$
	and $R_{\square} \colon Y \rightarrow Y|_{\Omega_{\square}}$ in the following way
	\begin{equation*}
		P_{\square}(x) = x\chi_{\Omega_{\square}} \text{ for } x \in X
	\end{equation*}
	and
	\begin{equation*}
		R_{\square}(y) = y\chi_{\Omega_{\square}} \text{ for } y \in Y.
	\end{equation*}
	where $\square \in \{a, c\}$. Clearly, $P_{\square}$ and $R_{\square}$ are norm one projections onto $X|_{\Omega_{\square}}$ and, respectively, $Y|_{\Omega_{\square}}$.
	Now, observe that for $\square \in \{a, c\}$ we have
	\begin{align*}
		\norm{M_{\lambda} \colon X|_{\Omega_{\square}} \rightarrow Y|_{\Omega_{\square}}}_{ess}
			& = \inf \left\{ \norm{M_{\lambda} - K}_{X|_{\Omega_{\square}} \rightarrow Y|_{\Omega_{\square}}} \colon K \in \mathscr{K}(X|_{\Omega_{\square}},Y|_{\Omega_{\square}}) \right\} \\
			& = \inf \left\{ \norm{R_{\square}(M_{\lambda} - K)P_{\square}}_{X \rightarrow Y} \colon K \in \mathscr{K}(X,Y) \right\} \\
			& \leqslant \inf \left\{ \norm{M_{\lambda} - K}_{X \rightarrow Y} \colon K \in \mathscr{K}(X,Y) \right\} \\
			& = \norm{M_{\lambda} \colon X \rightarrow Y}_{ess}.
	\end{align*}
	Therefore,
	\begin{equation*}
		\max \left\{ \norm{M_{\lambda} \colon X|_{\Omega_c} \rightarrow Y|_{\Omega_c}}_{ess}, \norm{M_{\lambda} \colon X|_{\Omega_a} \rightarrow Y|_{\Omega_a}}_{ess} \right\}
			\leqslant \norm{M_{\lambda} \colon X \rightarrow Y}_{ess},
	\end{equation*}
	and this ends the proof.
\end{proof}

Although Theorem~\ref{THM: essnorm multipliers function} and Theorem~\ref{THM: Essential norm Kothe sequence} are isometric in nature, the above result
unfortunately loses this value. Therefore, in this context, it is natural to ask the following question.

\begin{question}
	{\it Let $X$ and $Y$ be two K{\" o}the spaces with the Fatou property both defined over the same complete and $\sigma$-finite measure space $(\Omega,\Sigma,\mu)$.
		Suppose they can be represented up to the equality of norms as a direct sums of their atomic and non-atomic parts, that is to say, there are two two-dimensional
		K{\" o}the spaces, say $E$ and $F$, such that}
		\begin{equation*}
			X \equiv X|_{\Omega_a} \oplus_E X|_{\Omega_c} \textit{ and } Y \equiv Y|_{\Omega_a} \oplus_F Y|_{\Omega_c}.
		\end{equation*}
	{\it Do we then have the equality}
		\begin{equation*}
			\norm{M_{\lambda} \colon X \rightarrow Y}_{ess}
				= \norm{ \left\{ \norm{M_{\lambda} \colon X|_{\Omega_{\square}} \rightarrow Y|_{\Omega_{\square}}}_{ess} \right\}_{\square \in \{a, c\}} }_{M(E,F)}?
		\end{equation*}
\end{question}

We will close this subsection by taking a closer look to the landscape surrounding results presented so far. This will show Theorem~\ref{THM: Essential norm Kothe sequence}
in a slightly richer light.

\begin{remark}[On Pitt's theorem] \label{Remark: Pitt's theorem for multipliers}
	We know from Pitt \cite{P79} (cf. also \cite[Proposition~2.c.3, p.~76]{LT77} or \cite[Theorem~2.1.4, p.~32]{AK06}) that every operator from $\ell_p$
	into $\ell_q$ is compact if, and only if, $1 \leqslant q < p \leqslant \infty$ (with the convention that whenever $p = \infty$, then we are working
	with $c_0$ instead of $\ell_{\infty}$), or, pictographically,
	\begin{equation*}
		{\mathscr L}(\ell_p,\ell_q) = {\mathscr K}(\ell_p,\ell_q)
	\end{equation*}
	(have a look at \cite[Theorem~A2]{Ros69} for even more general result). On the other hand, the natural inclusion $\ell_q \hookrightarrow \ell_p$,
	where again $1 \leqslant q < p \leqslant \infty$, is not compact\footnote{But it is \enquote{nearly} compact, because it is strictly singular.}.

	Many generalizations of this classical result are also available. For example, let $M$ and $N$ be two Orlicz functions both satisfying
	the $\Delta_2$-condition at zero. Then every operator from $\ell_M$ into $\ell_N$ is compact if, and only if, $\alpha_M > \beta_N$, where $\alpha_M$ and
	$\beta_N$ are the so-called Matuszewska--Orlicz indices (see \cite[p.~149]{LT77}; cf. \cite[Theorem~4.a.9, p.~143]{LT77} and \cite[Chapter~11]{Ma89}).

	For the multiplication operators Pitt's theorem corresponds - via Theorem~\ref{THM: Essential norm Kothe sequence} - with the fact that the space
	$M(\ell_p,\ell_q) = \ell_r$ with $1/r = 1/q - 1/p$ is separable. In general, we can prove the following result (which should probably be compared
	with \cite[Proposition~3.1]{Cro69}).

	\begin{proposition}[One more analogue of Pitt's theorem] \label{PROP: Pitt's multiplication}
		{\it Let $X$ and $Y$ be two K{\" o}the sequence spaces with the Fatou property. Then every multiplication operator $M_{\lambda}$
			acting from $X$ into $Y$ is compact if, and only if, the space $M(X,Y)$ is order continuous.}
	\end{proposition}
	\begin{proof}
		One way, this is a direct consequence of the first part of the proof of Theorem~\ref{THM: Essential norm Kothe sequence}.

		Next, suppose the space $M(X,Y)$ is not order continuous. Take $\lambda_0 \in M(X,Y) \setminus M(X,Y)_o$. Then, due to \cite[Lemma~5]{HN04},
		there exists $\delta > 0$ and a strictly increasing sequence, say $\{n_k\}_{k=1}^{\infty}$, of natural numbers such that
		\begin{equation} \label{eq: 1}
			\norm{\lambda_0 \chi_{\left\{ n_k, n_k +1, ..., n_{k+1} \right\}}}_{M(X,Y)} > \delta \text{ for } k \in \mathbb{N}.
		\end{equation}
		Moreover, it follows from the definition of the norm in the space $M(X,Y)$, there exists a sequence $\left\{ x^{(k)} \right\}_{k=1}^{\infty} \subset \text{Ball}(X)$
		with $\text{supp}\left( x^{(k)} \right) \subset \left\{ n_k, n_k +1, ..., n_{k+1} \right\}$ such that
		\begin{equation} \label{eq: 2}
			\norm{\lambda_0\chi_{\left\{ n_k, n_k +1, ..., n_{k+1} \right\}} x^{(k)}}_Y
				\geqslant \norm{\lambda_0 \chi_{\left\{ n_k, n_k +1, ..., n_{k+1} \right\}}}_{M(X,Y)} - \frac{\delta}{2} \text{ for } k \in \mathbb{N}.
		\end{equation}
		Now, combining (\ref{eq: 1}) together with (\ref{eq: 2}) we get
		\begin{equation*}
			\norm{\lambda_0 x^{(k)}}_Y = \norm{\lambda_0\chi_{\left\{ n_k, n_k +1, ..., n_{k+1} \right\}} x^{(k)}}_Y > \frac{\delta}{2} \text{ for } k \in \mathbb{N}.
		\end{equation*}
		However, since $M_{\lambda_0}\left( \left\{ x^{(k)} \right\}_{k=1}^{\infty} \right) = \left\{ \lambda_0 x^{(k)} \right\}_{k=1}^{\infty}$,
		so this means that the image of a bounded sequence $\left\{ x^{(k)} \right\}_{k=1}^{\infty}$ through the multiplication operator $M_{\lambda_0}$,
		is not relatively compact in $Y$. In consequence, the operator $M_{\lambda_0} \colon  X \rightarrow Y$ is not compact and this completes the proof.
	\end{proof}

	Let us mention one more thing here. If $X \hookrightarrow Y$ (in particular, if $\ell_q \hookrightarrow \ell_p$, that is, $q \leqslant p$),
	then $M(X,Y) = \ell_{\infty}$ and, consequently, there are some non-compact multiplication operators acting from $X$ into $Y$
	(just take $\lambda \in \ell_{\infty} \setminus c_0$). Nevertheless, even if $Y \hookrightarrow X$, but the spaces $X$ and $Y$ are too \enquote{close},
	the space $M(X,Y)$ can be not order continuous and then there will be some non-compact multiplication operators from $X$ into $Y$.
	An example illustrating this situation is given in Appendix~\ref{Appendix B}.
\end{remark}

\begin{remark}[On Djakov and Ramanujan's question] \label{REMARK: Djakov-Ramanujan}
	In \cite{DR00} Djakov and Ramanujan ask the following question
	\begin{quote}
		\enquote{Is it true that every compact multiplier between the spaces $\ell_M$ and $\ell_N$ is a limit of finitely-supported multipliers?}
	\end{quote}
	(precisely, see \cite[p.~318]{DR00}). Thanks to our Proposition~\ref{PROP: Pitt's multiplication} we can answer this question in the affirmative.
	In fact, to see this, just note that the ideal $M(X,Y)_o$ is contained in the closure of finitely-supported functions living in $M(X,Y)$ (see \cite[Theorem~3.11, p.~18]{BS88}).
	Even more is true, because - as it follows from Proposition~\ref{PROP: Pitt's multiplication} - actually {\it every} compact multiplication operator
	acting between two K{\" o}the sequence spaces is a limit of finitely-supported multipliers.
\end{remark}

\begin{remark}[On the approximation numbers]
	Let us recall, at the beginning, that the $n^{\text{th}}$ {\it approximation number $a_n(T \colon X \rightarrow Y)$} of an operator $T$ acting
	between two Banach spaces $X$ and $Y$ is defined in the following way
	\begin{equation*}
		a_n(T \colon X \rightarrow Y) \coloneqq \inf \left\{ \norm{T - T_n}_{X \rightarrow Y} \colon \text{rank}(T_n) < n \right\},
	\end{equation*}
	where $n = 1, 2, ...$.
	It was proved by Pietsch that the operator $T$ is compact if, and only if, $\lim_{n \rightarrow \infty} a_n(T \colon X \rightarrow Y) = 0$.
	Moreover,
	\begin{equation*}
		\lim\limits_{n \rightarrow \infty} a_n(T \colon X \rightarrow Y) = \norm{T \colon X \rightarrow Y}_{ess}.
	\end{equation*}
	Now, it is not very hard to prove that for $1 < p < q < \infty$ we have the following formula
	\begin{equation*}
		a_n(M_{\lambda} \colon \ell_q \rightarrow \ell_p) = \left( \sum_{k=n}^{\infty} \left( \lambda_k^* \right)^r \right)^{1/r} = \norm{\lambda^* \chi_{\{n,n+1,...\}}}_{\ell_r},
	\end{equation*}
	where $1/r = 1/q - 1/p$ and $n \in \mathbb{N}$. On the other hand, we don't even know how to express $a_n(M_{\lambda} \colon \ell_p \rightarrow \ell_q)$.
	The most obvious guess that $a_n(M_{\lambda} \colon \ell_p \rightarrow \ell_q) = \lambda_{n+1}^*$ doesn't seem to work.
	Hence, in connection with this, we would like to ask the following question.
	\begin{question}
		{\it For which pairs of K{\" o}the sequence spaces $X$ and $Y$ we have the following equality}
		\begin{equation*}
			a_n(M_{\lambda} \colon X \rightarrow Y) = \norm{\lambda^* \chi_{\{n,n+1,...\}}}_{M(X,Y)}?
		\end{equation*}
		{\it What can then be said about $a_n(M_{\lambda} \colon Y \rightarrow X)$?}
	\end{question}
\end{remark}

The assumptions that appear in Theorem~\ref{THM: essnorm multipliers function} and Theorem~\ref{THM: Essential norm Kothe sequence} seem
to be only technical artifacts for us. We believe (highly motivated by \cite{Sch22}) that all major results from this section should work
in full generality. Let us express our expectations in the following question.

\begin{question}
	{\it How to get rid of the additional assumptions in Theorem~\ref{THM: essnorm multipliers function} and Theorem~\ref{THM: Essential norm Kothe sequence}?}
\end{question}

\subsection{Weakly compact multiplication operators} Recall that an operator acting between two Banach spaces is {\it weakly compact} if it takes
bounded subset to relatively weakly compact one. Equivalently, thanks to the Eberlein--{\v S}mulian theorem, an operator is weakly compact if, and
only if, every bounded sequence admits a subsequence whose image is weakly convergent.

As we saw earlier (see Theorem~\ref{THM: essnorm multipliers function}), there are basically no compact multiplication operators between K{\" o}the
function spaces. It is natural, then, to weaken this notion and ask, for example, about weakly compact multipliers.

But, before we proceed to the formulation of the next result, let us remind of a concept that, in a sense, will play one of the main roles here.
Precisely, a K{\" o}the space (in general, nothing prevents considering Banach lattices) has the {\it positive Schur property} if every weakly null
sequence\footnote{As it turns out, it is enough to consider disjoint sequences only (see~\cite[Corollary~2.3.5]{MN91}).} with positive terms is norm
convergent. This property was investigated by Wnuk (see \cite{Wnu93} and \cite{Wnu89}). Due to Rosenthal's $\ell_1$-theorem, a K{\" o}the spaces having
the positive Schur property are exactly those spaces which are {\it 1-disjointly homogeneous} (briefly, 1-DH), that is to say, in which every normalized
disjoint sequence has a subsequence equivalent to unit vector basis of $\ell_1$ (see~\cite[Theorem~7]{Wnu93}; cf.~\cite[Proposition~4.9]{FHSTT14}).
This class of spaces was introduced by Flores, Tradacete and Troitsky in \cite{FTT09} and attracted much attention in recent times (see, e.g., \cite{FHSTT14} and
\cite{LTM22} and references therein). The class of 1-disjointly homogeneous spaces significantly exceeds the family of $L_1$-spaces, because,
for example,
\begin{itemize}[leftmargin=0.5cm]
	\item any Orlicz space $L_M$ generated by Young's function $M$ such that $M$ satisfies the $\Delta_2$-condition for \enquote{large} arguments,
		Young conjugate $M^*$ of $M$ satisfies the $\Delta_0$-condition and $M$ is equivalent to the linear function for \enquote{small} arguments
		is 1-DH (clearly, if we narrow our considerations only to Orlicz spaces defined on $(0,1)$, then the last condition may be ignored; see \cite{LTM22} for more details);
	\item any Lorentz space $\Lambda_{\varphi}(0,1)$ is 1-DH provided the following three conditions are satisfied, namely,
		  $1^{\circ}$. $\lim_{t \rightarrow 0^{+}} \varphi(t) = 0$; $2^{\circ}$. $\varphi(1) > 0$; $3^{\circ}$. $\int_0^1 \varphi(t)dt = 1$
		  (in particular, Lorentz spaces $L_{p,1}(0,1)$ are 1-DH for all $1 \leqslant p < \infty$);
	\item for each $1 < p < \infty$ the space $L_{p,1}(0,\infty) \cap L_1(0,\infty)$ is 1-DH (see~\cite[Theorem~8]{LTM22}).
\end{itemize}
Note, however, that spaces with the positive Schur property, or, which is one thing, 1-DH spaces, are in general not stable under isomorphism.

We are now ready to present the main result here.

\begin{theorem}[Weak essential norm of multiplication operators acting between K{\" o}the function spaces] \label{THM: weak ess norm}
	{\it Let $X$ and $Y$ be two K{\" o}the function spaces with the Fatou property both defined over the same complete and $\sigma$-finite measure space
		$(\Omega,\Sigma,\mu)$. Suppose that the space $Y$ is} $1$-DH {\it and that every simple function\footnote{By a {\it simple function}
		we understand here any function defined on $\Omega$ which is a finite linear combination of characteristic functions of sets of
		finite measure.} in the space $M(X,Y)$ has an absolutely continuous norm. Then}
		\begin{equation*}
			\norm{M_{\lambda} \colon X \rightarrow Y}_{w} = \lim_{n \rightarrow \infty} \norm{\lambda\chi_{\mathcal{U}_n(\lambda)}}_{M(X,Y)}.
		\end{equation*}
		{\it where\footnote{It is clear that if the measure space $(\Omega,\Sigma,\mu)$ is finite, then it is enough to put
			$\mathcal{U}_n(\lambda) \coloneqq \left\{ \omega \in \supp(\lambda) \colon \abs{\lambda(\omega)} > n \right\}$ for $n \in \mathbb{N}$.}}
		\begin{equation*}
			\mathcal{U}_n(\lambda) \coloneqq \bigcup\limits_{N\geqslant n}\Omega_N
				\cup \left\{ \omega \in \supp(\lambda) \colon \abs{\lambda(\omega)} > n \right\} \text{ for } n \in \mathbb{N}
		\end{equation*}
		{\it and $\{\Omega_N\}_{N=1}^{\infty}$ is a decomposition of $\Omega$ into a pairwise disjoint family of sets of finite measure.
		In particular, the multiplication operator $M_{\lambda} \colon X \rightarrow Y$ is weakly compact if, and only if, $\lambda \in M(X,Y)_o$.}
\end{theorem}
\begin{proof}
	Take $\lambda \in M(X,Y)$ and note that $\{\mathcal{U}_n(\lambda)\}_{n=1}^{\infty}$ is a decreasing family of measurable sets with
	$\mu\left( \bigcap_{n=1}^{\infty} \mathcal{U}_n(\lambda) \right) = 0$.

	To demonstrate the first inequality observe that all sets
	\begin{equation*}
		\mathcal{U}_n(\lambda)^{\complement} \coloneqq \supp(\lambda) \setminus \mathcal{U}_n(\lambda)
	\end{equation*}
	have a finite measure and that the symbol $\lambda$ is a bounded function when restricted to the set $\mathcal{U}_n(\lambda)^{\complement}$ for all $n \in \mathbb{N}$.
	Therefore, thanks to our assumptions imposed on the space $M(X,Y)$, we immediately see that $\lambda\chi_{\mathcal{U}_n(\lambda)^{\complement}} \in M(X,Y)_o$.
	To make our life easier, put $\widetilde{\lambda}_n \coloneqq \lambda\chi_{\mathcal{U}_n(\lambda)^{\complement}}$. Now, for each $n \in \mathbb{N}$, let
	$\left\{ \sigma^n_k \right\}_{k=1}^{\infty} \subset \Sigma$ be a decreasing family of measurable subsets with $\mu\left( \bigcap_{k=1}^{\infty} \sigma^n_k \right) = 0$.
	Since $\widetilde{\lambda}_n \in M(X,Y)_o$, so it follows that for all $\varepsilon > 0$ there is $K \in \mathbb{N}$ such that for all $k > K$ we have
	\begin{equation*}
		\sup\limits_{g \in M_{\widetilde{\lambda}_n}(\text{Ball}(X))} \norm{g\chi_{\sigma^n_k}}_Y
			= \sup\limits_{h \in \text{Ball}(X)} \norm{\widetilde{\lambda}_n h\chi_{\sigma^n_k}}_Y
			= \norm{\widetilde{\lambda}_n \chi_{\sigma^n_k}}_{M(X,Y)} < \varepsilon.
	\end{equation*}
	However, this exactly means that the set $M_{\widetilde{\lambda}_n}(B(X))$ is $Y$-equicontinuous (or, using a different nomenclature, $L$-weakly compact;
	see \cite[Definition~3.6.1, p.~212]{MN91}), so applying \cite[Proposition~3.6.5, p.~212]{MN91}, also relatively weakly compact. Consequently, all operators
	$M_{\widetilde{\lambda}_n}$ for $n \in \mathbb{N}$ are weakly compact, and we have the inequalities
	\begin{align*}
		\norm{M_{\lambda} \colon X \rightarrow Y}_{w} & \leqslant \inf\{\norm{M_{\lambda} - W}_{X \rightarrow Y} \colon W \in \mathscr{W}(X,Y)\} \\
		& \leqslant \norm{M_{\lambda} - M_{\widetilde{\lambda}_n}}_{X \rightarrow Y} = \norm{\lambda - \widetilde{\lambda}_n}_{M(X,Y)}
		= \norm{\lambda\chi_{\mathcal{U}_n}}_{M(X,Y)}.
	\end{align*}
	But this clearly implies that
	\begin{equation*}
		\norm{M_{\lambda} \colon X \rightarrow Y}_{w} \leqslant \lim\limits_{n \rightarrow \infty} \norm{\lambda\chi_{\mathcal{U}_n}}_{M(X,Y)}
	\end{equation*}
	and we are done.

	Let us pass to the proof of the second inequality. Fix $\varepsilon > 0$ and take $\lambda \in M(X,Y)$. Still, let $\left\{ \mathcal{U}_n(\lambda) \right\}_{n=1}^{\infty}$
	be as in the formulation of the theorem. For any $n \in \mathbb{N}$ we can find $f_n \in \text{Ball}(X)$ with $\supp(f_n) \subset \mathcal{U}_n(\lambda)$
	such that
	\begin{equation} \label{inequality: wydobywanie normy weakly compact}
		\norm{\lambda f_n}_Y = \norm{\lambda \chi_{\mathcal{U}_n(\lambda)} f_n}_Y \geqslant \norm{\lambda \chi_{\mathcal{U}_n(\lambda)}}_{M(X,Y)} - \varepsilon.
	\end{equation}
	Take any weakly compact operator, say $W$, acting between $X$ and $Y$. Since the space $Y$ is $1$-disjointly homogeneous, so due to
	\cite[Theorem~3.6]{LTM22} the space $Y$ satisfies\footnote{In fact, in the class of K{\" o}the function spaces, both conditions are equivalent.}
	the so-called Dunford--Pettis criterion, that is to say, every bounded and relatively weakly compact subset of $Y$ is $Y$-equicontinuous. However, this means
	that the set $\{W(f_n)\}_{n=1}^{\infty}$ is actually $Y$-equicontinuous and
	\begin{equation} \label{inequality: equi-continuity}
		\norm{W(f_n)\chi_{\mathcal{U}_n(\lambda)}}_Y < \varepsilon
	\end{equation}
	for all sufficiently large $n$. Therefore,
	\begin{equation*}
		\norm{M_{\lambda} - W}_{X \rightarrow Y} \geqslant \norm{(M_{\lambda} - W)(f_n)}_Y = \norm{(\lambda f_n - W(f_n))\chi_{\mathcal{U}_n(\lambda)}}_Y
		\geqslant \norm{\lambda f_n}_Y - \norm{W(f_n)\chi_{\mathcal{U}_n(\lambda)}}_Y,
	\end{equation*}
	and, using (\ref{inequality: wydobywanie normy weakly compact}) along with (\ref{inequality: equi-continuity}), we get that for all
	sufficiently big $n$ we have
	\begin{equation*}
		\norm{M_{\lambda} - W}_{X \rightarrow Y} \geqslant \norm{\lambda \chi_{\mathcal{U}_n(\lambda)}}_{M(X,Y)} - 2 \varepsilon.
	\end{equation*}
	This is enough to conclude the proof.
\end{proof}

Next two examples explain the role of the assumption imposed on the space $Y$ in the above result. Roughly speaking,
there is a chance for an analogue of Theorem~\ref{THM: weak ess norm}, but only replacing the $1$-DH property with \enquote{something}
else. But, before we dive deeper into details, we need to define the objects of our considerations.

Let $\varphi$ be an increasing concave function on $(0,1)$ with $\varphi(0^+) = 0$. Recall that the {\it Lorentz space
$\Lambda_{{\varphi},q}(0,1)$}, where $1 \leqslant q < \infty$, (resp. {\it Marcinkiewicz space
$M_{\varphi}(0,1)$}) consists of all equivalence classes, modulo equality almost everywhere, of real valued
Lebesgue measurable functions defined on the unit interval $(0,1)$ with
\begin{equation} \label{EQ: Lorentz function}
	\norm{f}_{\Lambda_{{\varphi},q}} \coloneqq \left( \int_0^1 \left( \varphi(t)f^{*}(t) \right)^q \frac{dt}{t} \right)^{1/q} < \infty,
\end{equation}
respectively,
\begin{equation} \label{EQ: Marcinkiewicz function}
	\norm{f}_{M_{\varphi}} \coloneqq \sup\limits_{0 < t < 1} \frac{\varphi(t)}{t} \int_0^t f^{*}(s)ds  < \infty.
\end{equation}
In particular, if the Lorentz space  $\Lambda_{{\varphi},q}$ (resp. Marcinkiewicz space $M_{\varphi}$) is generated by
the power function $\varphi(t) = t^{1/p}$, where $1 < p < \infty$, then it is denoted by $L_{p,q}$ (respectively, $L_{p,\infty}$).
Of course, $L_{p,p} \equiv L_p$. Moreover, the spaces $L_{p,\infty}$ are sometimes also called the {\it weak $L_p$-spaces}.

\begin{example}[Multipliers between weak $L_p$-spaces] \label{EXP: not weakly compact between Marcinkiewicz}
	Let $1 \leqslant p < q < \infty$. Recall that
	\begin{equation*}
		M\left( L_{p,\infty}(0,1), L_{q,\infty}(0,1) \right) = L_{r,\infty}(0,1),
	\end{equation*}
	where $1/r = 1/p - 1/q$ (see, e.g., \cite[Theorem~4]{KLM19} and references given there). Note also that the space $L_{q,\infty}$ is not\footnote{This space
	is not even {\it disjointly homogeneous} (in particular, it is not $p$-DH for any $1 \leqslant p \leqslant \infty$), because inside $L_{q,\infty}$ we can find a disjoint
	sequence with a subsequence equivalent to unit vector basis of $c_0$, as well as another disjoint sequence spanning $\ell_q$ (see, e.g., \cite[Example, p.~659]{FTT09}
	for more details). On the other hand, the space $(L_{q,\infty})_o$ is $\infty$-DH.}	$1$-DH.

	Take the function $\lambda_0(t) = t^{-1/r}$ from $M(L_{p,\infty}(0,1), L_{q,\infty}(0,1))$. Since for all $\varepsilon > 0$ we have that
	\begin{equation*}
		\norm{\lambda_0\chi_{(0,\varepsilon)}}_{L_{r,\infty}} = 1,
	\end{equation*}
	so $\lambda_0$ is not order continuous.

	Next, we will show that the multiplication operator $M_{\lambda_0} \colon L_{p,\infty}(0,1) \rightarrow L_{q,\infty}(0,1)$ is not weakly compact.
	For this, take $g(t) = t^{-1/p} \in L_{p,\infty}(0,1)$. Observe that to achieve the intended goal it is enough to show that
	\begin{equation} \label{functional yes}
		\lim\limits_{n \rightarrow \infty} \langle \lambda_0 g r_n, \varphi \rangle = 0 \text{ for every } \varphi \in (L_{q,\infty}(0,1))^{\times}
	\end{equation}
	and, moreover, to construct a linear functional, say $\psi$, in $(L_{q,\infty}(0,1))^*$ such that
	\begin{equation} \label{functional no}
		\langle \lambda_0 g r_n, \psi \rangle = \frac{q}{q-1} \text{ for all } n \in \mathbb{N},
	\end{equation}
	where $\{r_n\}_{n=1}^{\infty}$ is the sequence of Rademacher functions.
	In fact, if this can be done, then the image of the sequence $\{g r_n\}_{n=1}^{\infty} \subset L_{p,\infty}$ through the operator $M_{\lambda_0}$, that is,
	the sequence $\{\lambda_0 g r_n\}_{n=1}^{\infty} \subset L_{q,\infty}$, does not possess a weakly convergent subsequence (if it did, then because the intersection
	of kernels of all functionals from $(L_{q,\infty}(0,1))^{\times}$ is zero, this hypothetical subsequence would have to converge weakly to zero, which in view of
	(\ref{functional no}) is just impossible). But this would exactly mean that the operator $M_{\lambda_0} \colon L_{p,\infty}(0,1) \rightarrow L_{q,\infty}(0,1)$
	is not weakly compact.

	First, take $h \in (L_{q,\infty}(0,1))^{\times}$ and put
	\begin{equation*}
		\langle f, \varphi \rangle \coloneqq \int_0^1 f(t)h(t)dt \text{ for } f \in L_{q,\infty}(0,1).
	\end{equation*}
	Because of Lemma~$\mathscr{R}$ (from Section~\ref{SUBSEC: Rademachers}) and the fact that $\lambda_0(t)g(t) = t^{-1/q} \in L_{q,\infty}(0,1)$ we infer that
	\begin{equation*}
		\langle \lambda_0 g r_n, \varphi \rangle = \int_0^1 \underbrace{\lambda_0(t)g(t)}_{\in L_{q,\infty}} r_n(t) \underbrace{h(t)}_{\in (L_{q,\infty})^{\times}}dt
			\xrightarrow{n \rightarrow \infty} 0.
	\end{equation*}
	This proves our first claim (\ref{functional yes}).

	Second, let $\mathscr{U}$ be a free ultrafilter on $\mathbb{N}$. For $f \in L_{q,\infty}(0,1)$ put
	\begin{equation*}
		\langle f, \psi \rangle \coloneqq \lim\limits_{\mathscr{U}} \left[ 2^{n\left( 1-\frac{1}{q} \right)} \int_0^{2^{-n}} f(t)dt \right].
	\end{equation*}
	It is not hard to see that $\psi$ is bounded linear functional on $L_{q,\infty}(0,1)$. Indeed, it follows from Hardy--Littlewood's inequality
	(see, e.g., \cite[Theorem~2.2, p.~44]{BS88}) that
	\begin{align*}
		\abs{2^{n\left( 1-\frac{1}{q} \right)} \int_0^{2^{-n}} f(t)dt} \leqslant 2^{n\left( 1-\frac{1}{q} \right)} \int_0^{2^{-n}} f^*(t)dt
			\leqslant \norm{f}_{L_{q,\infty}} 2^{n\left( 1-\frac{1}{q} \right)} \int_0^{2^{-n}} \frac{dt}{t^{\frac{1}{q}}} \leqslant \frac{\norm{f}_{L_{q,\infty}}}{1 - \frac{1}{q}},
	\end{align*}
	where the second inequality is due to the fact that $f^*(t) \leqslant \norm{f}_{L_{q,\infty}} t^{-1/q}$ for all $f \in L_{q,\infty}(0,1)$.
	Now, it remains only to note that for all $n \in \mathbb{N}$ there exists a big enough $N \in \mathbb{N}$ with
	\begin{equation*}
		2^{N\left( 1-\frac{1}{q} \right)} \int_0^{2^{-N}} \lambda_0(t)g(t)r_n(t)dt = 2^{N\left( 1-\frac{1}{q} \right)} \int_0^{2^{-N}} t^{-\frac{1}{q}}r_n(t)dt
			= 2^{N\left( 1-\frac{1}{q} \right)} \int_0^{2^{-N}} t^{-\frac{1}{q}}dt = \frac{q}{q-1}.
	\end{equation*}
	Hence, as a consequence of the definition of functional $\psi$,
	\begin{equation*}
		\langle \lambda_0 g r_n, \psi \rangle = \frac{q}{q-1} \text{ for all } n \in \mathbb{N}.
	\end{equation*}
	That's all we needed.
\end{example}

\begin{example}[Multipliers between Lorentz spaces $L_{p,q}$] \label{EXP: Lorentz symbol not OC but WC}
	Let $1 < p < q < \infty$. Again, in view of \cite[Theorem~4]{KLM19}, we have
	\begin{equation*}
		M(L_p(0,1),L_{p,q}(0,1)) = L_{\infty}(0,1).
	\end{equation*}
	Moreover, it follows from \cite[Proposition~5.1]{FJT75} that Lorentz function spaces $L_{p,q}$ are $q$-DH, but since $q > 1$, they are not $1$-DH.
	Of course, the ideal $M(L_p(0,1),L_{p,q}(0,1))_o$ is trivial. On the other hand, due to the reflexivity of the space $L_p(0,1)$, each multiplication
	operator $M_{\lambda} \colon L_p(0,1) \rightarrow L_{p,q}(0,1)$ is weakly compact (see, e.g., \cite[p.~348]{AK06}). Therefore, in general, the multiplication
	operator can be weakly compact even though its symbol is not order continuous.
\end{example}

It should also be added that the class of disjointly homogeneous K{\" o}the {\it sequence} spaces is not particularly rich, because, for example (and in
opposition to the function case presented above), Orlicz spaces $\ell_M$ and Lorentz spaces $\ell_{p,q}$ are, excluding trivial cases,
not disjointly homogeneous at all. This is because basic sequences in K{\" o}the sequence spaces are usually equivalent to disjoint sequences.
The culprit for this are the echoes of the Bessaga--Pe{\l}czy{\' n}ski selection principle (cf. \cite[Proposition~3.1.10, p.~14]{AK06}).

\subsection{Interplay between the multiplication operator and its symbol}
Typically, (some) operator theoretic aspects of the multiplication operator $M_{\lambda} \colon X \rightarrow Y$ (that is, treating $M_{\lambda}$
as a member of $\mathscr{L}(X,Y)$) are reflected in function theoretic properties of its symbol $\lambda$ (which, of course,
live a fully-fledged life in $M(X,Y)$). Here, we aim to study this symmetry in the context specified by the next result.

\begin{proposition} \label{Proposition: odleglosci = odleglosci}
	{\it Let $X$ and $Y$ be two K{\" o}the spaces with the Fatou property both defined on the same complete and $\sigma$-finite measure
		space $(\Omega,\Sigma,\mu)$.}
	\begin{enumerate}
		\item [(i)] {\it Suppose $(\Omega,\Sigma,\mu)$ is purely atomic. Assume further that either the space $X$ is reflexive
			or the space $Y$ is order continuous. Then}
		\begin{equation*}
			\text{dist}_{\mathscr{L}(X,Y)}(M_{\lambda} \colon X \rightarrow Y, {\mathscr K}(X,Y)) = \text{dist}_{M(X,Y)}(\lambda, M(X,Y)_o).
		\end{equation*}
		\item [(ii)] {\it Suppose $(\Omega,\Sigma,\mu)$ is non-atomic. Assume further that the space $Y$ is} $1$-DH {\it and every simple
			function in $M(X,Y)$ has an absolutely continuous norm. Then}
	\end{enumerate}
	\begin{equation*}
		\text{dist}_{\mathscr{L}(X,Y)}(M_{\lambda} \colon X \rightarrow Y, {\mathscr W}(X,Y)) = \text{dist}_{M(X,Y)}(\lambda, M(X,Y)_o).
	\end{equation*}
\end{proposition}
\begin{proof}
	(i) First of all, there is no doubt in assuming that $M(X,Y)_o \neq M(X,Y)$, because if both spaces $M(X,Y)_o$
	and $M(X,Y)$ are equal, then it follows from Theorem~\ref{THM: Essential norm Kothe sequence} that every
	multiplication operator $M_{\lambda}$ acting between $X$ and $Y$ is compact and, consequently,
	\begin{equation*}
		\text{dist}_{\mathscr{L}(X,Y)}(M_{\lambda} \colon X \rightarrow Y, {\mathscr K}(X,Y)) = 0 = \text{dist}_{M(X,Y)}(\lambda, M(X,Y)_o).
	\end{equation*}

	Now, let us begin the proof with the following observation\footnote{This result can be seen as a spin-off from Theorem~3.2 in \cite{KKT22}.
	However, in the mentioned article we were more interested in the rearrangement invariant structure. Morally speaking, it may be instructive
	to compare our Theorem~\ref{THM: Essential norm Kothe sequence} and Theorem~\ref{THM: weak ess norm} with Theorem~3.4 from \cite{KKT22}.}
	which seems quite interesting in itself.

	\vspace{5pt} \noindent {\bf Claim~$\clubsuit$.}
	{\it Let $Z$ be a K{\" o}the sequence space. Suppose that the ideal $Z_o$ is non-trivial. Then}
	\begin{equation}
		\text{dist}_Z(z,Z_o) = \lim\limits_{n \rightarrow \infty} \norm{z \chi_{\{n, n+1, ...\}}}_Z.
	\end{equation}
	\begin{proof}[Proof of Claim $\clubsuit$]
		Using the ideal property of the space $Z$ it is straightforward to see that
		\begin{equation*}
			\norm{z - y}_Z \geqslant \lim\limits_{n \rightarrow \infty} \norm{(z - y)\chi_{\{n, n+1, ...\}}}_Z
			\geqslant \lim\limits_{n \rightarrow \infty} \abs{\norm{z\chi_{\{n, n+1, ...\}}}_Z - \norm{y\chi_{\{n, n+1, ...\}}}_Z},
		\end{equation*}
		where the second inequality follows from the reverse triangle inequality. Consequently, for $y \in Z_o$,
		we have that $\lim_{n \rightarrow \infty} \norm{y\chi_{\{n, n+1, ...\}}}_Z = 0$, which means that
		\begin{equation*}
			\text{dist}_Z(z,Z_o) = \inf\{\norm{z - y}_Z \colon y \in Z_o\} \geqslant \lim\limits_{n \rightarrow \infty} \norm{z \chi_{\{n, n+1, ...\}}}_Z.
		\end{equation*}

		To prove the opposite inequality just note that
		\begin{equation*}
			\text{dist}_Z(z,Z_o) = \inf\{\norm{z - y}_Z \colon y \in Z_o\}
			\leqslant \lim\limits_{n \rightarrow \infty} \norm{z - z\chi_{\{1, 2, ..., n-1\}}}_Z
			= \lim\limits_{n \rightarrow \infty} \norm{z \chi_{\{n, n+1, ...\}}}_Z,
		\end{equation*}
		where the first inequality is due to the fact\footnote{Even a slightly more general statement is true, namely, if $Z$ is a K{\" o}the sequence space
		on $(\Omega,\Sigma,\mu)$ and $\sigma \in \Sigma$ has finite measure, then $z\chi_{\sigma} \in Z_o$ for every $z \in Z$. In other words, $c_{00} \subset Z_o$.}
		that $x\chi_{\{1, 2, ..., n-1\}} \in Z_o$. Thus we have completed the proof of Claim $\clubsuit$.
	\end{proof}

	Let's go back to the proof of the theorem. Using Theorem~\ref{THM: Essential norm Kothe sequence}
	once again together with the above Claim $\clubsuit$, but for the space  $M(X,Y)$ instead of $Z$, we see that
	\begin{align*}
		\text{dist}_{\mathscr{L}(X,Y)}({M_{\lambda} \colon X \rightarrow Y, {\mathscr K}(X,Y)}) & = \norm{M_{\lambda} \colon X \rightarrow Y}_{ess} \\
		& = \lim\limits_{n \rightarrow \infty} \norm{\lambda \chi_{\{n, n+1, \ldots\}}}_{M(X,Y)} \\
		& = \text{dist}_{M(X,Y)}({\lambda,(M(X,Y))_o}).
	\end{align*}
	That's all.

	(ii) The proof of the second part is very similar. It requires two ingredients. First, if $M(X,Y)_o = M(X,Y)$,
	then, as Theorem~\ref{THM: weak ess norm} teaches us, every multiplication operator $M_{\lambda}$ between $X$ and $Y$ is weakly compact,
	so we have the equality
	\begin{equation*}
		\text{dist}_{\mathscr{L}(X,Y)}(M_{\lambda} \colon X \rightarrow Y, {\mathscr W}(X,Y)) = 0 = \text{dist}_{M(X,Y)}(\lambda, M(X,Y)_o).
	\end{equation*}
	Therefore, without loss of the generality we can assume that $M(X,Y)_o \neq M(X,Y)$. Next, we need an analogue of Claim~$\clubsuit$
	in the following form

	\vspace{5pt} \noindent {\bf Claim~$\heartsuit$.}
	{\it Let $Z$ be a K{\" o}the function space defined on a complete and $\sigma$-finite measure space $(\Omega,\Sigma,\mu)$.
		Suppose that every simple function in $Z$ has an absolutely continuous norm. Then}
		\begin{equation}
			\text{dist}_Z(z,Z_o) = \lim\limits_{n \rightarrow \infty} \norm{z \chi_{\mathcal{U}_n(\lambda)}}_Z,
		\end{equation}
	{\it where, as in Theorem~\ref{THM: weak ess norm},}
		\begin{equation*}
			\mathcal{U}_n(\lambda) \coloneqq \bigcup\limits_{N\geqslant n}\Omega_N
				\cup \left\{ \omega \in \supp(\lambda) \colon \abs{\lambda(\omega)} > n \right\} \text{\it for } n \in \mathbb{N}
		\end{equation*}
	{\it and $\{\Omega_N\}_{N=1}^{\infty}$ is a decomposition of $\Omega$ into a pairwise disjoint family of sets of finite measure.}

	\vspace{5pt} We will omit the proof of this fact, as it does not differ significantly from that of Claim~$\clubsuit$. Finally, using
	Theorem~\ref{THM: weak ess norm} in combination with Claim $\heartsuit$ we complete the proof.
\end{proof}

With the help of the above result we can express the quantity $\lim\limits_{n \rightarrow \infty} \norm{\lambda \chi_{\mathcal{U}_n(\lambda)}}_{M(X,Y)}$
from Theorem~\ref{THM: weak ess norm} which, due to the quite ephemeral character of the family $\left\{ \mathcal{U}_n(\lambda) \right\}_{n=1}^{\infty}$,
seems not suitable for calculations, in a fairly reasonable way. The price we have to pay for this is to confine our consideration to rearrangement invariant
spaces.

\begin{corollary}[Weak essential norm of multiplication operator between rearrangement invariant function spaces] \label{COR: ess r.i. function}
	{\it Let $X$ and $Y$ be two rearrangement invariant function spaces with the Fatou property. Assume that the space $Y$ is} $1$-DH
	{\it and that the ideal $M(X,Y)_o$ is non-trivial. Then}
	\begin{equation} \label{EQ: weak ess norm in r.i.}
		\norm{M_{\lambda} \colon X \rightarrow Y}_{w} = \lim\limits_{n \rightarrow \infty} \norm{\lambda^* \chi_{\left( 0,\frac{1}{n} \right) \cup (n, \infty)}}_{M(X,Y)}.
	\end{equation}
\end{corollary}

It is clear that this result in only a special case of Theorem~\ref{THM: weak ess norm}, so, somehow, the set $\mathcal{U}_n(\lambda)$
has to be hidden in the formula (\ref{EQ: weak ess norm in r.i.}). Intuitively, the first part of the set $\mathcal{U}_n(\lambda)$, that is,
$\bigcup_{N \geqslant n}^{\infty} \Omega_N$ corresponds to the set $(n, \infty)$, and the second part, that is,
$\left\{ \omega \in \supp(\lambda) \colon \abs{\lambda(\omega)} > n \right\}$, corresponds to the set $\left( 0,\frac{1}{n} \right)$.
However, what is completely not obvious here, is the fact that instead of the symbol $\lambda$ we can consider its non-increasing
rearrangement $\lambda^*$. Fortunately, such questions were the subject of our considerations in \cite{KKT22}.

\begin{proof}
	First of all, in the class of rearrangement invariant spaces we essentially have only two possibilities, namely, either the ideal of order continuous
	functions is trivial or it is equal to the closure of simple functions (see \cite[Theorem~5.5, pp.~67--68]{BS88}; cf. \cite[Theorem~B]{KT17}). Hence,
	since the space $M(X,Y)$ is also rearrangement invariant (see \cite[Theorem~2.2(i)]{KLM12}) and we assumed that the ideal $M(X,Y)_o$ is non-trivial,
	so every simple function in $M(X,Y)$ has an absolutely continuous norm. Consequently, the use of Proposition~\ref{Proposition: odleglosci = odleglosci}(ii)
	is justified, and we have the following identifications
	\begin{equation*}
		\norm{M_{\lambda} \colon X \rightarrow Y}_{w} = \text{dist}_{\mathscr{L}(X,Y)}(M_{\lambda} \colon X \rightarrow Y, {\mathscr W}(X,Y))
			= \text{dist}_{M(X,Y)}(\lambda, M(X,Y)_o).
	\end{equation*}
	On the other hand, it follows from \cite[Corollary~3.7]{KKT22} that
	\begin{equation*}
		\text{dist}_{M(X,Y)}(\lambda, M(X,Y)_o) = \lim\limits_{n \rightarrow \infty} \norm{\lambda^* \chi_{\left( 0,\frac{1}{n} \right) \cup (n, \infty)}}_{M(X,Y)}
	\end{equation*}
	and this ends the proof.
\end{proof}

\section{Applications: Ces{\` a}ro, Tandori and Hardy spaces}

Because both Ces{\` a}ro spaces $CX$ and Tandori spaces $\widetilde{X}$ are usually defined as certain spaces of real valued functions,
so in order not to complicate things unnecessarily, let's agree that whenever we allow these spaces into our considerations, then
we will tacitly assume that the underlying K{\" o}the spaces $X$ are defined over the field of real numbers. If the need arose,
we can always consider the complexifications of these spaces.

\subsection{Abstract Ces{\` a}ro and Tandori sequence spaces}
For a K{\" o}the sequence space $X$ we define the {\it abstract Ces\`aro sequence space $CX$} as a vector space
\begin{equation*}
	CX \coloneqq \left\{ x = \{x_n\}_{n=1}^{\infty} \in \omega \colon \mathscr{H}(\abs{x}) \in X \right\}
\end{equation*}
with the norm $\norm{x}_{CX} \coloneqq \norm{\mathscr{H}(\abs{x})}_X$, where the symbol $\mathscr{H}$ will stand for the {\it discrete Hardy operator}
(or the {\it discrete Ces{\` a}ro operator}) which is defined as
\begin{equation*}
	\mathscr{H} \colon \{x_n\}_{n=1}^\infty \mapsto \left\{ \frac{1}{n} \sum_{k=1}^n x_k \right\}_{n=1}^\infty.
\end{equation*}
These spaces can be viewed, on the one hand, as a generalization of the {\it classical Ces{\` a}ro sequence spaces $ces_p \coloneqq C\ell_p$}
(see \cite{AM14} and references given there; see also Bennett's memoir \cite{Ben96} and Grosse-Erdmann book \cite{GE98}) and, on the other hand,
as the so-called {\it optimal domains} of Hardy's operator $\mathscr{H}$ for the space $X$, because from the very definition the operator
$\mathscr{H} \colon CX \rightarrow X$ is bounded and the space $CX$ is the largest K{\" o}the sequence space with this property (see \cite{DS07},
\cite{CR16} and \cite{NP10}; cf. Okada, Ricker and S{\' a}nchez P{\' e}rez book \cite{ORS08} and references therein); diagrammatically,
\begin{equation*}
	\xymatrix{
		CX \ar[r]^{\mathscr{H}} & X\\
		X \ar[u]^{\text{id}} \ar[r]^{\mathscr{H}} & X \ar@{=}[u]}
\end{equation*}
Furthermore, the Ces{\` a}ro spaces are also closely related to Sinnamon's {\it down spaces $X^{\downarrow}$} (see \cite{KMS07}, \cite{MS06} and \cite{Si94}).
They owe the nickname \enquote{abstract} due to Le{\' s}nik and Maligranda, who proposed this nomenclature in \cite{LM15a}.
Nowadays, the structure of these spaces, especially in its most general form, enjoys unflagging interest (see \cite{ALM19} and \cite{KKT22}
and their references).

In a sense, a dual construction to the Ces{\` a}ro spaces are the so-called Tandori spaces. Precisely, for a K{\" o}the sequence space $X$ by the
corresponding {\it abstract Tandori sequence space $\widetilde{X}$} we understand a vector space
\begin{equation*}
	\widetilde{X} \coloneqq \left\{ x = \{x_n\}_{n=1}^{\infty} \in \omega \colon \widetilde{x} \in X \right\}
\end{equation*}
with the norm $\norm{x}_{\widetilde{X}} \coloneqq \norm{\widetilde{x}}_X$, where $\widetilde{x}$ is the {\it non-increasing majorant}
of a given sequence $x = \{x_n\}_{n=1}^{\infty}$ defined as
\begin{equation*}
	\widetilde{x} = \left\{ \widetilde{x_n} \right\}_{n=1}^{\infty} \coloneqq \left\{ \sup\limits_{m \geqslant n} \abs{x_m} \right\}_{n=1}^{\infty}.
\end{equation*}
The name of these spaces was proposed in \cite[p.~767]{LM16}, where the authors refer to Tandori's paper \cite{Tan55} from 1954. However,
despite being nameless, these spaces were investigated even earlier (see \cite{LM15a} and \cite{AM14}). Their significance stems mostly from
the fact that they are K{\" o}the duals of the Ces{\` a}ro spaces (see~\cite{LM15a}; cf. \cite{Ben96} and \cite{GE98}), but some questions about their
isomorphic structure or interpolation properties are interesting in themselves (see, for example, \cite{ALM19} and \cite{LM16}).

In \cite{RFRSSB19} Ramos-Fern{\' a}ndez, Rivera-Sarmiento and Salas-Brown computed the essential norm of the multiplication operator $M_{\lambda}$
acting from the classical Ces{\` a}ro sequence spaces $ces_p$ into itself. They showed that
\begin{equation*}
	\norm{M_{\lambda} \colon ces_p \rightarrow ces_p}_{ess} = \limsup\limits_{n \rightarrow \infty} \abs{\lambda_n},
\end{equation*}
where $\lambda = \{\lambda_n\}_{n=1}^{\infty} \in \ell_{\infty}$. In addition, they write the following
\begin{quote}
	\enquote{Thus, it is natural to ask if the result obtained in this article is also valid for multiplication operator acting on any K{\" o}the sequence space $X$.}
\end{quote}
Here, we are interested not only in answering the above question, but exactly as we have so far, we want to consider the non-algebraic case,
that is to say, our goal is to evaluate
\begin{equation*}
	\norm{M_{\lambda} \colon CX \rightarrow CY}_{ess} \text{ and also } \norm{M_{\lambda} \colon \widetilde{X} \rightarrow \widetilde{Y}}_{ess}.
\end{equation*}

To do this, let's start with the following preparatory result.

\begin{lemma}[Separable part of Tandori sequence spaces] \label{LEMMA: OC w Tandori}
	{\it Let $X$ be a K{\" o}the sequence space. Then}
	\begin{equation*}
		\left( \widetilde{X} \right)_o = \widetilde{X_o}.
	\end{equation*}
\end{lemma}
\begin{proof}
	Let us start by recalling the well-known fact: {\it Let $Z$ be a K{\" o}the sequence space. Then}
	\begin{equation} \tag{$\spadesuit$}
		z = \{z_n\}_{n=1}^{\infty} \in Z \textit{ belongs to the ideal } Z_o \textit{ if, and only if, } \lim\limits_{n \rightarrow \infty} \norm{z\chi_{\{n, n+1, ...\}}}_Z = 0.
	\end{equation}
	If we keep in mind ($\spadesuit$), the rest of the proof basically follows from the following simple-to-see inequality, namely,
	\begin{equation} \label{INEQ: tandori X o = X o tandori}
		\stackon[-8pt]{$x\chi_{\{n, n+1, ...\}}$}{\vstretch{1.5}{\hstretch{2.4}{\widetilde{\phantom{\;\;\;\;\;\;\;\;}}}}} = \widetilde{x}\chi_{\{n, n+1, ...\}} \text{ for all } n \in \mathbb{N}.
	\end{equation}
	In fact, after imposing the norm $\norm{\cdot}_X$ on both sides of (\ref{INEQ: tandori X o = X o tandori}) we get
	\begin{equation} \label{EQ: 1}
		\norm{x\chi_{\{n, n+1, ...\}}}_{\widetilde{X}} = \norm{\stackon[-8pt]{$x\chi_{\{n, n+1, ...\}}$}{\vstretch{1.5}{\hstretch{2.4}{\widetilde{\phantom{\;\;\;\;\;\;\;\;}}}}}}_X
			= \norm{\widetilde{x}\chi_{\{n, n+1, ...\}}}_X \text{ for all } n \in \mathbb{N}.
	\end{equation}
	Now, if $x \in \left( \widetilde{X} \right)_o$, then it follows from ($\spadesuit$) that $\lim_{n \rightarrow \infty} \norm{x\chi_{\{n, n+1, ...\}}}_{\widetilde{X}} = 0$.
	Therefore, in view of (\ref{EQ: 1}), also $\lim_{n \rightarrow \infty} \norm{\widetilde{x}\chi_{\{n, n+1, ...\}}}_X = 0$.
	But this exactly means that $\widetilde{x} \in X_o$. Consequently, $x \in \widetilde{X_o}$ and we have the inclusion $\left( \widetilde{X} \right)_o \subset \widetilde{X_o}$.
	On the other hand, take $x \in \widetilde{X_o}$. Then $\widetilde{x} \in X_o$, so  $\lim_{n \rightarrow \infty} \norm{\widetilde{x}\chi_{\{n, n+1, ...\}}}_X = 0$.
	However, using again (\ref{EQ: 1}), we see that $\lim_{n \rightarrow \infty} \norm{x\chi_{\{n, n+1, ...\}}}_{\widetilde{X}} = 0$ as well.
	This gives us that $x \in \left( \widetilde{X} \right)_o$ and, in consequence, $\left( \widetilde{X} \right)_o \supset \widetilde{X_o}$.
	The proof is complete.
\end{proof}

Now we are ready to present and also prove the main result of this section.

\begin{theorem}[Essential norm of multipliers acting between abstract Ces{\` a}ro and Tandori sequence spaces] \label{THM: multipliers Cezaro i Tandori}
	{\it Let $X$ and $Y$ be two rearrangement invariant sequence spaces with the Fatou property. Suppose that either the space $X$ is order
		continuous or the space $Y$ is reflexive. Further, suppose that $Y$ factorizes\footnote{More precisely, for two K{\" o}the spaces, say $E$ and $F$,
		both defined over the same complete and $\sigma$-finite measure space $(\Omega,\Sigma,\mu)$ by the {\it pointwise product space $E \odot F$} we understand
		a vector space $\{gh \colon g \in E \text{ and } h \in F\}$ equipped with the quasi-norm $\norm{f}_{E \odot F} \coloneqq \inf\{ \norm{g}_E\norm{h}_F \colon f = gh,\ g \in E,\ h \in F\}$.
		Now, we will say that $E$ {\it factorizes} though $F$ provided $E \odot M(E,F) = F$. For much more information about pointwise products and factorization
		we refer to \cite{CS14}, \cite{KLM14}, \cite{KLM19} and \cite{Sch10}.} though $X$, that is to say,
		\begin{equation*}
			X \odot M(X,Y) = Y.
		\end{equation*}
		Then
		\begin{equation*}
			\norm{M_{\lambda} \colon \widetilde{X} \rightarrow \widetilde{Y}}_{ess} \approx \lim\limits_{n \rightarrow \infty} \norm{\widetilde{\lambda}\chi_{\{n, n+1, ...\}}}_{M(X,Y)}.
		\end{equation*}
		Moreover, if the Hardy operator $\mathscr{H}$ is bounded on both spaces $X$ and $Y$, then also
		\begin{equation*}
			\norm{M_{\lambda} \colon CX \rightarrow CY}_{ess} \approx \lim\limits_{n \rightarrow \infty} \norm{\widetilde{\lambda}\chi_{\{n, n+1, ...\}}}_{M(X,Y)}.
		\end{equation*}
		Both of the above estimates depend on $X$ and $Y$ only.
		In particular, the multiplication operator $M_{\lambda} \colon \begin{array}{l} CX \rightarrow CY \\ \;\;\: \widetilde{X} \rightarrow \widetilde{Y} \end{array}$
		is compact if, and only if, the function $\widetilde{\lambda} = \left\{ \widetilde{\lambda_n} \right\}_{n=1}^{\infty}$ belongs to the ideal $M(X,Y)_o$.}
\end{theorem}

\begin{proof}
As a warm-up, let's explain the following, not very difficult issues, that will be useful later.

	(a) Due to our assumption about the boundedness of Hardy's operator $\mathscr{H}$ on $X$ and $Y$, both spaces $CX$ and $CY$ are non-trivial
	(this easily follows from the definition of the Ces{\` a}ro space construction, however, this statement is also a form of Hardy's inequality,
	which in its qualitative version can be written as an inclusion; cf.~\cite[p.~1]{Ben96} and \cite[pp.~19--21 and 147--159]{KMP07}).

	(b) If a K{\" o}the sequence space, say $Z$, has the Fatou property then the Ces{\` a}ro space $CZ$ as well as the Tandori space
	$\widetilde{Z}$ have the Fatou property (see, e.g., \cite[Theorem~1(d)]{LM15a}; although, formally, the proof given in \cite{LM15a}
	covers only K{\" o}the function spaces, it works just as well for their sequence counterparts).

	(c) Both spaces $\widetilde{Z}$ and $CZ$ are order continuous provided $Z$ is an order continuous K{\" o}the sequence space
	(this is a direct consequence of Lemma~\ref{LEMMA: OC w Tandori} and a nearly direct consequence of Lemma~11 in \cite{KT17};
	cf. the proof of Proposition~5.1 in \cite{KKT22}).

	(d) Lastly, we claim that

	\vspace{5pt} \noindent {\bf Claim $\diamondsuit$.}
	{\it Let $Z$ be a reflexive rearrangement invariant sequence space with the Fatou property such that the Hardy's operator $\mathscr{H}$ is bounded on $Z$.
		Then both spaces $\widetilde{Z}$ and $CZ$ are reflexive.}
	\begin{proof}[Proof of Claim $\diamondsuit$]
		Both spaces are related to each other through duality, so let's stop at the proof for the Ces{\` a}ro space construction.

		Since the space $Z$ is reflexive and has the Fatou property, so it follows from \cite[Corollary~4.4, p.~23]{BS88} that both spaces $Z$ and $Z^{\times}$
		are order continuous. Because of (b) and, one more time, \cite[Corollary~4.4, p.~23]{BS88}, to complete the proof it is enough to show that the spaces
		$CZ$ and $(CZ)^{\times}$ are order continuous. Now, in the first case, that is, when we consider $CZ$, it results immediately from (c),
		while in the second case, that is, when we consider $(CZ)^{\times}$, things are a bit more complicated. Namely, using the Le{\' s}nik
		and Maligranda duality results from \cite{LM15a} we get the equality
		\begin{equation*}
			(CX)^{\times} = \widetilde{X^{\times}}
		\end{equation*}
		(see \cite[Theorem~6]{LM15a}). But, with Lemma~\ref{LEMMA: OC w Tandori} in mind, we also have
		\begin{equation*}
			\left( (CX)^{\times} \right)_o = \left( \widetilde{X^{\times}} \right)_o = \widetilde{(X^{\times})_o} = \widetilde{X^{\times}} = (CX)^{\times}.
		\end{equation*}
		This means that also the space $(CX)^{\times}$ is order continuous. And that is all we needed to finish the proof of Claim~$\diamondsuit$.
	\end{proof}

	Now, the essential part of the proof is based on the following three steps:

	$1^{\text{st}}$. In view of (a), (b), (c) and (d), there is noting to prevent us from using Theorem~\ref{THM: Essential norm Kothe sequence}.
	Consequently, we get the formula
	\begin{equation*}
		\norm{M_{\lambda} \colon \begin{array}{l} CX \rightarrow CY \\ \;\;\: \widetilde{X} \rightarrow \widetilde{Y} \end{array}}_{ess}
			= \lim\limits_{n \rightarrow \infty} \norm{\lambda\chi_{\{n, n+1, ...\}}}_{M(CX,CY)}.
	\end{equation*}
	Moreover, it follows that the multiplication operator $M_{\lambda} \colon \begin{array}{l} CX \rightarrow CY \\ \;\;\: \widetilde{X} \rightarrow \widetilde{Y} \end{array}$
	is compact if, and only if, $\lambda \in \left( \widetilde{M(X,Y)} \right)_o$.
	However, as shown in Lemma~\ref{LEMMA: OC w Tandori}, $\left( \widetilde{M(X,Y)} \right)_o = \widetilde{M(X,Y)_o}$. Therefore, the condition
	\enquote{$\lambda \in \left( \widetilde{M(X,Y)} \right)_o$} can be equivalently replaced by the condition \enquote{$\lambda \in \widetilde{M(X,Y)_o}$}.
	But the latter means that $\widetilde{\lambda} \in M(X,Y)_o$.

	$2^{\text{nd}}$. Next, due to our factorization assumption that $X \odot M(X,Y) = Y$ and \cite[Theorem~6]{KLM19} (cf. \cite[pp. 67--77]{Ben96}), we get
	\begin{equation} \label{EQ: multiplikatory ktorych nie ma}
		M(CX,CY) = \widetilde{M(X,Y)}.
	\end{equation}
	Moreover, by duality (see \cite[Theorem~6]{LM15a}),
	\begin{align*}
		M\left( \widetilde{X},\widetilde{Y} \right) & = M\left( \left[ C\left( X^{\times} \right) \right]^{\times}, \left[ C\left( Y^{\times} \right) \right]^{\times} \right) \\
			& = M\left( C \left( Y^{\times} \right), C\left( X^{\times} \right) \right)
			= \widetilde{M\left( Y^{\times}, X^{\times} \right)} = \widetilde{M(X,Y)}.
	\end{align*}
	In consequence, we actually have that
	\begin{equation*}
		\norm{M_{\lambda} \colon \begin{array}{l} CX \rightarrow CY \\ \;\;\: \widetilde{X} \rightarrow \widetilde{Y} \end{array}}_{ess}
			\approx \lim\limits_{n \rightarrow \infty} \norm{\lambda\chi_{\{n, n+1, ...\}}}_{\widetilde{M(X,Y)}}.
	\end{equation*}
	The only thing that remains to be clarified is (\ref{EQ: multiplikatory ktorych nie ma}). But this is a consequence of \cite[Theorem~6]{KLM19}
	and duality results contained in \cite{LM15a} (see also \cite[pp. 67--77]{Ben96}).

	$3^{\text{rd}}$. Finally, using (\ref{EQ: 1}) form the proof of Lemma~\ref{LEMMA: OC w Tandori}, we see that
	\begin{equation*}
		\norm{\lambda\chi_{\{n, n+1, ...\}}}_{\widetilde{M(X,Y)}}
			= \norm{\stackon[-8pt]{$\lambda \chi_{\{n, n+1, ...\}}$}{\vstretch{1.5}{\hstretch{2.4}{\widetilde{\phantom{\;\;\;\;\;\;\;\;}}}}}}_{M(X,Y)}
			= \norm{\widetilde{\lambda} \chi_{\{n, n+1, ...\}}}_{M(X,Y)}.
	\end{equation*}
	Putting it all together ends the proof.
\end{proof}

Algebraic situation is much simpler.

\begin{corollary}[Compact multipliers form abstract Ces{\` a}ro or Tandori sequence space to itself]
	{\it Let $X$ be an order continuous rearrangement invariant sequence space with the Fatou property. Then}
	\begin{equation*}
		\norm{M_{\lambda} \colon \begin{array}{l} CX \circlearrowleft \\ \;\, \widetilde{X} \circlearrowleft \end{array}}_{ess}
			= \limsup\limits_{n \rightarrow \infty} \abs{\lambda_n}.
	\end{equation*}
	{\it In particular, the multiplication operators $M_{\lambda} \colon \begin{array}{l} CX \circlearrowleft \\ \;\, \widetilde{X} \circlearrowleft \end{array}$
		are compact if, and only if, $\lambda \in c_0$.}
\end{corollary}
\begin{proof}
	This result is an easy consequence of Corollary~\ref{COR: compact sequence X to X} and remark (c) from the proof of Theorem~\ref{THM: multipliers Cezaro i Tandori}.
	Indeed, since $M(CX,CX) \equiv \ell_{\infty}$ and the space $CX$ is order continuous (this follows from our assumptions together with \cite[Lemma~1]{LM16};
	cf. \cite[Proposition~5.1]{KKT22}), so
	\begin{equation*}
		\norm{M_{\lambda} \colon \begin{array}{l} CX \circlearrowleft \\ \;\, \widetilde{X} \circlearrowleft \end{array}}_{ess}
			= \lim\limits_{n \rightarrow \infty} \norm{\lambda \chi_{\{n, n+1, ...\}}}_{\ell_{\infty}}
			= \lim\limits_{n \rightarrow \infty} \left( \sup\limits_{m \geqslant n} \abs{\lambda_m} \right)
			= \limsup\limits_{n \rightarrow \infty} \abs{\lambda_n}.
	\end{equation*}
\end{proof}

Let us close this section with the touch of classics.

\begin{remark}[Compact multipliers between classical Ces{\` a}ro and Tandori sequence spaces] \label{Remark: ces_p -> ces_q [ABR18]}
		Assume that $1 < p < q < \infty$. With the help of technology we have developed, let us now try to answer
		the following question: {\it When are the multiplication operators}
		\begin{equation*}
			M_{\lambda} \colon \begin{array}{l} ces_q \rightarrow ces_p \\ ces_p \rightarrow ces_q \end{array}
				\textit{ and }
			M_{\lambda} \colon \begin{array}{l} \widetilde{\ell_q} \rightarrow \widetilde{\ell_p} \\ \widetilde{\ell_p} \rightarrow \widetilde{\ell_q} \end{array}
		\end{equation*}
		{\it compact?}.

		It was proved by Bennett \cite{Ben96} (cf. Grosse-Erdmann \cite{GE98}) that
		\begin{equation} \tag{A} \label{first situation}
			M(ces_q, ces_p) = \widetilde{M(\ell_q, \ell_p)} = \widetilde{\ell_r},
		\end{equation}
		where $1/r = 1/p - 1/q$, and
		\begin{equation} \tag{B} \label{second situation}
			M(ces_p, ces_q) = \ell_{\infty}\left( n^{\frac{1}{q} - \frac{1}{p}} \right)
			= \left\{ x = \{x_n\}_{n=1}^{\infty} \colon \sup\limits_{n \in \mathbb{N}} \left( n^{\frac{1}{q} - \frac{1}{p}} \abs{x_n} \right) < \infty \right\}.
		\end{equation}

		The first situation (\ref{first situation}) is not particularly complicated as it fits exactly Theorem~\ref{THM: multipliers Cezaro i Tandori}.
		Indeed, since we have the factorization
		\begin{equation*}
			\ell_q \odot M(\ell_q, \ell_p) = \ell_q \odot \ell_r = \ell_p
		\end{equation*}
		and both spaces $ces_p$ and $ces_q$ are even reflexive, so the multiplication operator $M_{\lambda} \colon ces_q \rightarrow ces_p$
		is compact if, and only if, $\lambda \in \widetilde{(\ell_r)_o}  = \widetilde{\ell_r}$.

		The second situation (\ref{second situation}) is much more interesting, because now
		\begin{equation*}
			\ell_p \odot M(\ell_p, \ell_q) = \ell_p \odot \ell_{\infty} = \ell_p \neq \ell_q
		\end{equation*}
		and we {\it can't} use Theorem~\ref{THM: multipliers Cezaro i Tandori} directly. Fortunately, Theorem~\ref{THM: Essential norm Kothe sequence}
		remains at our disposal all the time, so the multiplication operator $M_{\lambda} \colon ces_p \rightarrow ces_q$ is compact if, and only if, the
		symbol $\lambda = \{\lambda_n\}_{n=1}^{\infty}$ belongs to the space
		\begin{equation*}
			\left( M(ces_p,ces_q) \right)_o = \left( \ell_{\infty}\left( n^{\frac{1}{q} - \frac{1}{p}} \right) \right)_o = c_0\left( n^{\frac{1}{q} - \frac{1}{p}} \right),
		\end{equation*}
		that is, $\left\{ \lambda_n n^{1/q-1/p} \right\}_{n=1}^{\infty} \in c_0$. The above equality is probably interesting in its own way,
		because $M(\ell_p,\ell_q) = \ell_{\infty}$, so the fact that the space $M(ces_p,ces_q)$ is not just $\ell_{\infty}$ may come as a bit of surprise.
		Anyhow, in this way, we recreated the Albanese, Bonet and Ricker result from \cite{ABR18} (precisely, see \cite[Proposition~2.2]{ABR18}).

		The analogous answer in the case of Tandori spaces coincides with what is contained in Table~6 on page 69 and Table~22 on page 70 in Bennett memoir
		\cite{Ben96} (see also \cite{BR20}).
\end{remark}

\subsection{Abstract Ces{\` a}ro and Tandori function spaces: Open ends}
In the function case, the {\it continuous Hardy operator} (or the {\it continuous Ces{\` a}ro operator}) $\mathscr{H}$ is defined by
$\mathscr{H}(f)(x) \coloneqq \frac{1}{x}\int_0^x f(t)dt$ for $x > 0$ and the corresponding {\it abstract Ces{\` a}ro function space $CX$},
where $X$ is a K{\" o}the function space defined on the unit interval $(0,1)$ or the semi-axis $(0,\infty)$ (both equipped with the usual Lebesgue
measure $m$), is defined - similarly to the sequence case considered in the previous subsection - as the set of all real-valued measurable
functions, say $f$, such that $\mathscr{H}(\abs{f}) \in X$, equipped with the norm $\norm{f}_{CX} \coloneqq \norm{\mathscr{H}(\abs{f})}_X$.

Without any surprises, for a K{\" o}the function space $X$ exactly as above, by the {\it abstract Tandori function space $\widetilde{X}$} we mean
a set of all real-valued measurable functions, say $f$, with $\widetilde{f} \in X$, where $\widetilde{f}$ is the {\it decreasing majorant} of a given function $f$,
that is to say, $\widetilde{f}(x) \coloneqq \esssup_{t \geqslant x} \abs{f(t)}$, with the norm $\norm{f}_{\widetilde{X}} \coloneqq \norm{\widetilde{f}}_X$.

Here, we don't really need more than the above two definitions, but the isomorphic and isometric structure of the Ces{\` a}ro and Tandori
function spaces has been the subject of extensive research and has many non-trivial results to offer (we refer to \cite{ALM19}, \cite{CR16},
\cite{DS07}, \cite{KKM21}, \cite{KKT22}, \cite{KT17}, \cite{Kub14}, \cite{LM15a, LM16} and \cite{MS06}; see also \cite{AM14} and references therein).

Let's get to the point.

\begin{theorem}[Essential norm of multiplication operators between abstract Ces{\` a}ro function spaces]\label{THM: ess norm Cezaro and Tandori function}
	{\it Let $X$ and $Y$ be two K{\" o}the function spaces with the Fatou property both defined on the semi-axis $(0,\infty)$ equipped with the Lebesgue measure.
		Suppose that both operators\footnote{Here, the symbol $\mathscr{H}^{\times}$ stands for the K{\" o}the adjoint operator to the Hardy operator
		$\mathscr{H}$, the so-called {\it continuous Copson operator}, which can be explicitly defined via
		\begin{equation*}
			\mathscr{H}^{\times} \colon f \mapsto \mathscr{H}^{\times}(f)(x) \coloneqq \int_{x}^{\infty} \frac{f(t)}{t}dt \text{ for } x > 0.
		\end{equation*}
		}
		$\mathscr{H}$ and $\mathscr{H}^{\times}$ as well as the dilation operator $D_{\tau}$, for some $0 < \tau < 1$, are bounded on $X$ and on $Y$.
		Suppose further that the space $Y$ is order continuous. Assume also that $Y$ factorizes though $X$, that is to say,
		\begin{equation*}
			X \odot M(X,Y) = Y.
		\end{equation*}
		Then
		\begin{equation*}
				\norm{M_{\lambda} \colon CX \rightarrow CY}_{ess} = \norm{M_{\lambda} \colon CX \rightarrow CY} \approx \norm{\lambda}_{\widetilde{M(X,Y)}}
		\end{equation*}
		and the above estimate depend on $X$ and $Y$ only.
		Therefore, in particular, there is a lack of compact multiplication operators acting between two abstract Ces{\` a}ro function spaces.}
\end{theorem}
\begin{proof}
	The proof of this fact is to mimic the argument used in the proof of Theorem~\ref{THM: multipliers Cezaro i Tandori}. And so:

		($\alpha$) Since the Hardy's operator $\mathscr{H}$ is bounded on both spaces $X$ and $Y$, so it follows from the optimal domain' point of view on
		the Ces{\` a}ro space construction that the spaces $CX$ and $CY$ are both non-trivial. Note, however, that the boundedness of Hardy's
		operator $\mathscr{H}$ is in general not necessary for a non-triviality of the Ces{\` a}ro construction (see Example~3.3(d) and Example~~3.3(f)
		in \cite{KKM21}).

		($\beta$) As in the sequence case, the Fatou property and order continuity property are inherited by the Ces{\` a}ro space construction
		(here we need \cite[Lemma~1(d)]{LM15a} and \cite[Lemma~11]{KT17}, respectively). Note also that the abstract Ces{\` a}ro function spaces
		are {\it never} reflexive, because, for example, they contain a complemented copy of $L_1(0,1)$ (see \cite[Lemma~4.1]{KKM21}).

		($\gamma$) In view of the two previous steps, we can now use Theorem~\ref{THM: essnorm multipliers function} to conclude that
		\begin{equation*}
			\norm{M_{\lambda} \colon CX \rightarrow CY}_{ess} = \norm{M_{\lambda} \colon CX \rightarrow CY} \approx \norm{\lambda}_{M(CX,CY)}.
		\end{equation*}

		($\delta$) Finally, due to our assumptions and \cite[Theorem~6]{KLM19} (together with \cite[Remark~9]{KLM19}) we have
		\begin{equation*}
			M(CX,CY) = \widetilde{M(X,Y)}.
		\end{equation*}

		All this completes the proof.
\end{proof}

For the sake of completeness, let us also formulate the following result, which for the classical Ces{\` a}ro function spaces $Ces_p(0,1) \coloneqq CL_p(0,1)$
was proved, for example, in \cite{AGHLM18} (see \cite[Theorem~5.2]{AGHLM18}).

\begin{corollary}[Compact multipliers form abstract Ces{\` a}ro function space to itself]
	{\it Let $X$ be an order continuous K{\" o}the function spaces with the Fatou property defined either on the semi-axis $(0,\infty)$
		or the unit interval $(0,1)$ equipped with the Lebesgue measure. Then}
		\begin{equation*}
			\norm{M_{\lambda} \colon CX \circlearrowleft}_{ess} = \norm{M_{\lambda} \colon CX \circlearrowleft} = \norm{\lambda}_{L_{\infty}}.
		\end{equation*}
		{\it In particular, there are no non-trivial compact multiplication operators on $CX$.}
\end{corollary}
\begin{proof}
	To begin with, recall that $M(CX,CX) \equiv L_{\infty}$. Moreover, it follows from \cite[Lemma~11]{KT17} and, respectively, \cite[Lemma~1(d)]{LM15a}
	that if $X$ is order continuous or, respectively, has the Fatou property, then the same can be said about the space $CX$. Now, this result is an immediate
	consequence of Corollary~\ref{COR: compact function X to X}.
\end{proof}

This is, unfortunately, the moment when things stop working as we would like them to. Let us explain exactly why.

First of all, it is know that the abstract Tandori function spaces $\widetilde{X}$ are {\it never} order continuous (see, for example,
\cite[Proposition~4.8]{KKM21}). Even worse, the ideal $\left( \widetilde{X} \right)_o$ is always trivial. Therefore, in particular, they are
also not reflexive and we cannot use Theorem~\ref{THM: essnorm multipliers function} directly to estimate the quantity
\begin{equation*}
	\norm{M_{\lambda} \colon \widetilde{X} \rightarrow \widetilde{Y}}_{ess}
\end{equation*}
in this case.

Next, if we try to calculate the weak essential norms, that is,
\begin{equation*}
	\norm{M_{\lambda} \colon CX \rightarrow CY}_w \quad (\text{or, maybe, even } \norm{M_{\lambda} \colon \widetilde{X} \rightarrow \widetilde{Y}}_{w})
\end{equation*}
using Theorem~\ref{THM: weak ess norm}, we run into another problem, because we need to know which Ces{\` a}ro function spaces are 1-disjointly homogeneous.
In our experience, this should happen sporadically\footnote{For example, let $X$ be the weighted $L_{\infty}$-space, where the weight is defined
as $w(t) = t$, for $t > 0$. Then the Ces{\` a}ro space $CX$ is 1-disjointly homogeneous, because $CX \equiv L_1$. We will even risk the hypothesis
that for $X$ being a r.i. function space this situation will never occurs.}, but we don't know for sure.

Finally, due to some annoying technicalities (see the paragraph after Problem~2 in \cite{KLM19} for more details),
we do not have a satisfactory description of the space
\begin{equation*}
	M(CX, CY) \quad (\text{so, also } M\left( \widetilde{X}, \widetilde{Y} \right))
\end{equation*}
when both spaces $X$ and $Y$ are defined on the unit interval $(0,1)$ equipped with the usual Lebesgue measure.

It seems like the best we can do for now is to ask the following three questions.

\begin{question} {\it Let $X$ and $Y$ be two K{\" o}the function spaces with the Fatou property.}
	\begin{enumerate}
		\item [(i)] {\it How to estimate the quantity $\norm{M_{\lambda} \colon \widetilde{X} \rightarrow \widetilde{Y}}_{ess}$?}
		\item [(ii)] {\it When the abstract Ces{\` a}ro spaces are $1$-disjointly homogeneous?}
		\item [(iii)] {\it How to describe the space $M(CX,CY)$ in general?}
	\end{enumerate}
\end{question}

\subsection{Banach spaces of analytic functions}
In this subsection, we will improve and complete the main result from Mleczko's paper \cite{Mle09} (precisely, see \cite[Theorem~2.5]{Mle09}).
However, before we do that, we need short preliminaries to clarify our notation.

Let $\mathscr{H}(\mathbb{D})$ be the space of all analytic functions on the unit disc $\mathbb{D} \coloneqq \{z \in \mathbb{C} \colon \abs{z} < 1\}$.
Further, let $E$ be a K{\" o}the sequence space. With a linear subspace, say $H(\mathbb{D})$, of $\mathscr{H}(\mathbb{D})$ we can associate
the space
\begin{equation*}
	\widehat{H}(\mathbb{D}) \coloneqq \left\{ \left\{ \widehat{f}(n) \right\}_{n=0}^{\infty} \colon \sum_{n=0}^{\infty} \widehat{f}(n)\chi_n \in H(\mathbb{D}) \right\}
\end{equation*}
of Taylor's coefficients of functions from $H(\mathbb{D})$, where $\chi_n(z) = z^n$ for $z \in \mathbb{D}$ and $n = 0, 1, 2, ...$. Then with
any sequence $\lambda = \{\lambda_n\}_{n=0}^{\infty}$ from $M(\widehat{H}(\mathbb{D}),E)$ we can associate its {\it Fourier multiplier operator}
$\mathscr{M}_{\lambda} \colon H(\mathbb{D}) \rightarrow E$ defined by
\begin{equation*}
	\mathscr{M}_{\lambda} \colon f \mapsto \left\{ \lambda_n \widehat{f}(n) \right\}_{n=0}^{\infty},
\end{equation*}
where $f = \sum_{n=0}^{\infty} \widehat{f}(n)\chi_n \in H(\mathbb{D})$ for $n = 0, 1, 2, ...$.

Moreover, let $X$ be a K{\" o}the space on the unit circle $\mathbb{T} \coloneqq \{z \in \mathbb{C} \colon \abs{z} = 1\}$ equipped with the normalized
Lebesgue measure $dm(t) = \abs{dt}/2\pi$. Then the {\it abstract Hardy space $H[X]$} is defined as a vector space
\begin{equation*}
	H[X] \coloneqq \left\{ f \in X \colon \widehat{f}(n) = 0 \text{ for all } n < 0 \right\}
\end{equation*}
equipped with the norm inherited from $X$ (cf. \cite{Mle09}). Equivalently, via convolution with the Poisson kernel, we can treat the Hardy space $H[X]$
as a space of analytic functions on the unit disc $\mathbb{D}$. Of course, $H[L_p]$, where $1 \leqslant p \leqslant \infty$, is nothing
else, but a {\it classical Hardy space} $H_p$ (for more information we refer to \cite{LM21}, \cite{Mle09} and references given there).

Now we are ready to prove the promised result.

\begin{theorem} \label{THM: ess norm holomorphic}
	{\it Let $H(\mathbb{D})$ be a Banach space of analytic function on the unit disc $\mathbb{D}$ intermediate between $H_{\infty}(\mathbb{D})$
		and $H_2(\mathbb{D})$. Let, moreover, $E$ be an order continuous K{\" o}the sequence space with the Fatou property, Then}
		\begin{equation*}
			\norm{\mathscr{M}_{\lambda} \colon H(\mathbb{D}) \rightarrow E}_{ess} = \lim\limits_{n \rightarrow \infty} \norm{\lambda\chi_{\{n, n+1, ...\}}}_{M(\ell_2,E)}.
		\end{equation*}
		{\it In particular, the Fourier multiplier $\mathscr{M}_{\lambda} \colon H(\mathbb{D}) \rightarrow E$ is compact if, and only if, $\lambda \in M(\ell_2,E)_o$.}
\end{theorem}

Clearly, if $X$ is a K{\" o}the space on $\mathbb{T}$ such that $L_{\infty}(\mathbb{T}) \hookrightarrow X \hookrightarrow L_2(\mathbb{T})$,
then $H_{\infty}(\mathbb{D}) \hookrightarrow H[X] \hookrightarrow H_2(\mathbb{D})$ and, consequently, the corresponding Hardy space $H[X]$
may serve here as a main illustration for this result.

\begin{proof}[Sketch of the proof]
	The proof basically consists of two steps.

	$1^{\text{st}}$. First, we need to transfer the situation to our \enquote{ideal} word. This can be done using the so-called solid envelopes
	together with Anderson and Shield's results from \cite{AS76}. More precisely, thanks to our assumption that
	\begin{equation*}
		H_{\infty}(\mathbb{D}) \hookrightarrow H(\mathbb{D}) \hookrightarrow H_2(\mathbb{D})
	\end{equation*}
	and knowing that the solid envelope of $H_{\infty}(\mathbb{D})$ and $H_2(\mathbb{D})$ is $\ell_2$ (this result is essentially due to S. Kisliakov),
	it is not hard to see that the following diagram commutes
	\begin{equation*}
		\xymatrix{
			H(\mathbb{D}) \ar[d]_{\text{id}} \ar[r]^{\mathscr{M}_{\lambda}} & E\\
			H_2 \ar[r]^{\mathscr{F}} & \ell_2 \ar[u]_{M_{\lambda}}}
	\end{equation*}
	where $\lambda \in \widehat{H}(\mathbb{D})$ and $\mathscr{F}(f) = \left\{ \widehat{f}(n) \right\}_{n=0}^{\infty}$ for $f \in H(\mathbb{D})$
	(cf. Proposition~1.1 and Corollary~1.2 in \cite{Mle09}). But this means that
	\begin{equation*}
		\mathscr{M}(H(\mathbb{D}),E) = M(\ell_2,E).
	\end{equation*}

	$2^{\text{nd}}$. Now, because the space $E$ is order continuous, so it is enough to apply our Theorem~\ref{THM: Essential norm Kothe sequence}.

	That is all.
\end{proof}

How to get back to Mleczko's result from here? Well, with the extra assumption that $\ell_2 \hookrightarrow E$, we get that
$M(\ell_2,E) = \ell_{\infty}$ and, consequently,
\begin{equation*}
	\lim\limits_{n \rightarrow \infty} \norm{\lambda\chi_{\{n, n+1, ...\}}}_{M(\ell_2,E)}
		= \lim\limits_{n \rightarrow \infty} \norm{\lambda\chi_{\{n, n+1, ...\}}}_{\ell_{\infty}}
		= \lim\limits_{n \rightarrow \infty} \left( \sup_{m \geqslant n} \abs{\lambda_m} \right)
		= \limsup_{n \rightarrow \infty} \abs{\lambda_n}.
\end{equation*}
This gives us Theorem~2.5 from \cite{Mle09}.

\begin{corollary}[Mleczko, 2009]
	{\it Let $H(\mathbb{D})$ be a Banach space of analytic function on the unit disc $\mathbb{D}$ intermediate between $H_{\infty}(\mathbb{D})$
		and $H_2(\mathbb{D})$. Moreover, let $E$ be an order continuous K{\" o}the sequence space with the Fatou property intermediate between
		$\ell_2$ and $\ell_{\infty}$. Then the Fourier multiplier $\mathscr{M}_{\lambda} \colon H(\mathbb{D}) \rightarrow E$ is compact if, and
		only if,
		\begin{equation*}
			\limsup_{n \rightarrow \infty} \abs{\lambda_n} = 0.
		\end{equation*}
	}
\end{corollary}

Let us close this section by noting that the space $M(\ell_2,E)_o$ can be precisely identified in many situations.

\begin{remark}
	\hfill

	(a) Let $1 \leqslant q < 2$. Then it is straightforward to see that $M(\ell_2,\ell_q) = \ell_r$, where $1/r = 1/q - 1/2$.
	Now, let $2 \leqslant q < \infty$. In this case $\ell_2 \hookrightarrow \ell_q$, so $M(\ell_2, \ell_q) = \ell_{\infty}$.
	Therefore, due to Theorem~\ref{THM: ess norm holomorphic}, the Fourier multiplier $\mathscr{M}_{\lambda} \colon H_p \rightarrow \ell_q$
	is compact if, and only if, its symbol $\lambda = \{\lambda_n\}_{n=1}^{\infty}$ belongs either to $\ell_r$ with $1/r = 1/q - 1/2$, provided
	$1 \leqslant q < p < \infty$, or to $c_0$, provided $1 \leqslant p \leqslant q < \infty$.

	(b) In general, if a K{\" o}the sequence space $E$ is $2$-concave\footnote{Recall that a K{\" o}the space $X$ is {\it $p$-concave} (for some $1 \leqslant p < \infty$)
	if there exists a constant, say $C_p > 0$, such that, for every choice of elements $\{x_i\}_{i=1}^n$ in $X$, we have
	\begin{equation*}
		\left( \sum_{i=1}^{n} \norm{x_k}_X^p \right)^{1/p} \leqslant C_p \norm{\left( \sum_{i=1}^{n} \abs{x_k}^p \right)^{1/p}}_X.
	\end{equation*}
	Note that for $1 \leqslant p < \infty$ the space $\ell_p$ is $p$-concave with constant $C_p$ equal to $1$.}, then
	\begin{equation*}
		M(\ell_2,E) = M(E^{\times},\ell_2) = M\left( \left( E^{\times} \right)^{(1/2)},\ell_1 \right)^{(2)}
			= \left( \left( \left( E^{\times} \right)^{(1/2)} \right)^{\times} \right)^{(2)},
	\end{equation*}
	where the symbol $X^{(p)}$, for $0 < p < \infty$ and a K{\" o}the space $X$, denote a K{\" o}the space given by the (quasi-)norm $\norm{x}_{X^{(p)}} \coloneqq \norm{\abs{x}^p}_X^{1/p}$.
	Unfortunately, the description of $M(\ell_2,E)_o$ seems out of reach in this case.
\end{remark}

\section{Examples: On compact multipliers between some concrete classes of K{\" o}the spaces} \label{SECTION: examples}

The space $M(X,Y)_o$, that is, a separable part of the space of pointwise multipliers between $X$ and $Y$ plays a fundamental role in virtually
all of our results regarding the problem of description of compact multipliers acting between two K{\" o}the spaces (see Section~\ref{SECTION: ESS NORMS};
especially Theorem~\ref{THM: Essential norm Kothe sequence} and Theorem~\ref{THM: weak ess norm}).
In many specific situations the space $M(X,Y)$ can be explicitly\footnote{Almost always up to the {\it equivalence}, but not equality, of norms.}
calculated and, on the other hand, the points of order continuity of certain classes of K{\" o}the spaces can be characterized in a fairly satisfactory manner.
In fact, these two problems have been the leitmotif of many articles (to mention a few, see \cite{CS14}, \cite{CDSP08}, \cite{DR00}, \cite{Kol16},
\cite{KLM12, KLM14, KLM19}, \cite{KP12}, \cite{LT21} and \cite{MP89}; cf. \cite{Ben96} and \cite{GE98}).
Unfortunately, there seems to be no overall satisfactory\footnote{In some sense, an exact description of the ideal $M(X,Y)_o$ in the language
of the properties of $X$ and $Y$ should not be possible in general, because the properties of the space $M(X,Y)$ also depend on the relative
position of the spaces $X$ and $Y$ on the scale of K{\" o}the spaces.} and general description of the points of order continuity in the space
of pointwise multipliers $M(X,Y)$, where $X$ and $Y$ are two K{\" o}the sequence spaces (this fact should be compared with \cite[Corollary~6.3]{LTM22}).
That is why we have chosen to focus here on more specific pairs of K{\" o}the spaces for which we can actually calculate the space $M(X,Y)$ and describe
the ideal $M(X,Y)_o$ in some sensible way. To this end, we will sometimes use existing results, but anyway, \cite{CS14}, \cite{KLM12, KLM14, KLM19},
\cite{Kol16}, \cite{KP12}, \cite{LT21} and \cite{Sch10} will be the main sources of inspiration.

\subsection{Musielak--Orlicz sequence spaces} \label{SUBSEC: Musielak-Orlicz sequence}
	Let $\{M_n\}_{n=1}^{\infty}$ be a sequence of Orlicz functions\footnote{By an {\it Orlicz function $M \colon [0,\infty) \rightarrow [0,\infty)$}
	we understand a convex and strictly increasing function vanishing at zero. We decided to use only the Orlicz functions here instead of the more
	general Young functions (which can \enquote{jump} to infinity) to avoid some unnecessary complications. However, formally, all the results presented
	here work in full generality (actually, everything work in even more general setting of the so-called {\it generalized Calder{\' o}n--Lozanovski{\u \i}
	sequence spaces $E_{\{M_n\}}$}, where $E$ is a K{\" o}the sequence space and $\{M_n\}_{n=1}^{\infty}$ is a sequence of Young functions; however,
	we will not elaborate on this topic here, because we feel that such a generality would disturb the very shaky balance between applicability and technicality,
	to the detriment of the first).}.
	The {\it Musielak--Orlicz sequence space} $\ell_{\{M_n\}}$ is a Banach space
	of all sequences $x = \{x_n\}_{n=1}^{\infty}$ of real numbers with $\sum_{n=1}^{\infty} M_n(\abs{x_n}/\varrho) < \infty$ for some $\varrho > 0$,
	equipped with the so-called {\it Luxemburg--Nakano norm}
	\begin{equation*}
		\norm{x}_{\ell_{\{M_n\}}} \coloneqq \inf \left\{ \varrho > 0 \colon \sum_{n=1}^{\infty} M_n\left( \frac{x_n}{\varrho} \right) \leqslant 1 \right\}.
	\end{equation*}

	This general construction includes two very important types of spaces: On the one hand, {\it Nakano sequence spaces $\ell_{\{p_n\}}$} (or {\it variable
	exponent Lebesgue--Riesz spaces}) which are nothing else but Musielak--Orlicz sequence spaces $\ell_{\{M_n\}}$ with all generating functions $M_n$
	being just power functions, that is, $M_n(t) = t^{p_n}$ for some $p_n \geqslant 1$; On the other, classical {\it Orlicz sequence spaces $\ell_M$}
	which are Musielak--Orlicz sequence spaces $\ell_{\{M_n\}}$ with \enquote{constant} sequence of Orlicz functions $\{M_n\}_{n=1}^{\infty}$, that is,
	with $M_n = M$ for all $n \in \mathbb{N}$ (see \cite[Chapter~4]{LT77} and \cite{Ma89} for much more information about Musielak--Orlicz spaces and,
	respectively, Orlicz spaces).

	Recently, Le{\' s}nik and the second named author in \cite{LT21} proved that the space of pointwise multipliers between two Musielak--Orlicz sequence spaces,
	say $\ell_{\{M_n\}}$ and $\ell_{\{N_n\}}$, is another Musielak--Orlicz sequence space. More precisely, it follows from Theorem~9 in \cite{LT21} that
	\begin{equation*}
		M\left( \ell_{\{M_n\}}, \ell_{\{N_n\}} \right) = \ell_{\{N_n \ominus M_n\}},
	\end{equation*}
	where
	\begin{equation*}
		(N_n \ominus M_n)(t) \coloneqq \sup\limits_{0 \leqslant u \leqslant 1} [N_n(tu) - M_n(u)] \text{ for } n \in \mathbb{N}
	\end{equation*}
	is the so-called {\it generalized Young conjugate} of $M_n$ with respect to $N_n$.

	Moreover, Kolwicz and Panfil in \cite{KP12} gave complete criteria describing the points of order continuity in Musielak--Orlicz spaces. Precisely,
	it can be deduced from Theorem~11 in \cite{KP12} that a function, say $x$, is an order continuous element of the  Musielak--Orlicz sequence space
	$\ell_{\{M_n\}}$ if, and only if, the sequence of Orlicz functions $\{M_n\}_{n=1}^{\infty}$ satisfies the so-called {\it local $\Delta_2$-condition
	with respect to $x$} (see \cite[Definition~1]{KP12}).

 	Without going into technical details, the aforementioned results in tandem with our Theorem~\ref{THM: Essential norm Kothe sequence} make it possible
 	to describe the compact multipliers between two Musielak--Orlicz sequence spaces. However, sacrificing a little generality for the sake of clarity,
 	we will now describe the compact multipliers between two Nakano sequence spaces.

	\begin{proposition}[Compact multipliers between Nakano sequence spaces] \label{PROPOSITION: Compact multipliers between Nakano}
		{\it Let $\{p_n\}_{n=1}^{\infty}$ and $\{q_n\}_{n=1}^{\infty}$ be two sequences of real numbers with $1 \leqslant p_n \leqslant q_n < \infty$
			for all $n \in \mathbb{N}$. Suppose that either
			\begin{equation*}
				1 < \inf\limits_{n \in \mathbb{N}} p_n \leqslant \sup\limits_{n \in \mathbb{N}} p_n < \infty
			\end{equation*}
			or
			\begin{equation*}
				\sup\limits_{n \in \mathbb{N}} q_n < \infty.
			\end{equation*}
			Then the multiplication operator $M_{\lambda} \colon \ell_{\{p_n\}} \rightarrow \ell_{\{q_n\}}$ is compact if, and only if,}
		\begin{equation*}
			\forall \ell > 1 \colon \lim\limits_{k \rightarrow \infty} \sum_{\left\{ n \in \supp(\lambda) \colon r_n > \frac{\log k}{\log \ell} \right\}}
				\left( \abs{\lambda_n} \ell \right)^{r_n} = 0,
		\end{equation*}
		{\it where $1/r_n = 1/q_n - 1/p_n$ for $n \in \mathbb{N}$.}
	\end{proposition}
	\begin{proof}
		We will prove this result in four steps.

		$1^{\text{st}}$. It is part of folklore that the Nakano space $\ell_{\{s_n\}}$, where $\{s_n\}_{n=1}^{\infty}$ is from $[1,\infty)$,
		is order continuous if, and only if, $p_n < \infty$ for all $n \in \mathbb{N}$. In consequence, using duality and \cite[Corollary~4.4, p.~23]{BS88},
		the space $\ell_{\{s_n\}}$ is reflexive if, and only if, $\inf_{n \in \mathbb{N}} s_n > 1$ and $\sup_{n \in \mathbb{N}} s_n < \infty$.

		$2^{\text{nd}}$. Suppose $1 \leqslant p_n \leqslant q_n < \infty$ for all $n \in \mathbb{N}$. Then it follows from \cite[Theorem~9]{LT21}
		that
		\begin{equation*}
			M\left( \ell_{\{p_n\}}, \ell_{\{q_n\}} \right) = \ell_{\{q_n \ominus p_n\}} = \ell_{\{r_n\}},
		\end{equation*}
		where $1/r_n = 1/q_n + 1/p_n$ for $n \in \mathbb{N}$.

		$3^{\text{rd}}$. Next, it follows from \cite[Theorem~11]{KP12} that the sequence $x = \{x_n\}_{n=1}^{\infty}$ belongs to the separable part of the
		Musielak--Orlicz sequence space $\left( \ell_{\{M_n\}} \right)_o$, where $\{M_n\}_{n=1}^{\infty}$ is a sequence of Orlicz functions if, and only
		if, for each $\ell > 1$ we have that
		\begin{equation*}
			\lim_{k \rightarrow \infty} \norm{ \left\{ M_n\left( \ell x_n\chi_{A_k^{\ell}} \right) \right\}_{n=1}^{\infty} }_{\ell_1} = 0,
		\end{equation*}
		where $A_k^{\ell} = \left\{ n \in \supp(x) \colon M_n(\ell x_n) > k M_n(x_n) \right\}$.
		However, in our situation, we need to apply the above criterion to the Nakano sequence space $\ell_{\{r_n\}}$. After some simple calculations
		it is easy to see that the sequence $x = \{x_n\}_{n=1}^{\infty}$ belongs to $(\ell_{r_n})_o$ if, and only if,
		\begin{equation*}
			\lim_{k \rightarrow \infty} \norm{ \left\{ \left( \ell x_n \right)^{r_n} \chi_{A_k^{\ell}} \right\}_{n=1}^{\infty} }_{\ell_1} = 0,
		\end{equation*}
		where
		\begin{equation*}
			A_k^{\ell} = \left\{ n \in \supp(x) \colon r_n > \frac{\log k}{\log \ell} \right\}.
		\end{equation*}

		$4^{\text{th}}$. Now, to complete the proof, it is enough to use the previous steps and Theorem~\ref{THM: Essential norm Kothe sequence}.
	\end{proof}

\subsection{Lorentz and Marcinkiewicz sequence spaces}
	Let $\varphi$ be a positive, increasing and concave function on $(0,\infty)$.
	By the {\it Lorentz sequence space $\lambda_{\varphi}$}, respectively, the {\it Marcinkiewicz sequence space $m_{\varphi}$}, we understand
	the space of all sequences $x = \{x_n\}_{n=1}^{\infty}$ for which the following norm
	\begin{equation*}
		\norm{x}_{\lambda_{\varphi}} \coloneqq \sum_{n=1}^{\infty} x_n^* (\varphi(n+1) - \varphi(n)),
	\end{equation*}
	respectively,
	\begin{equation*}
		\norm{x}_{m_{\varphi}} \coloneqq \sup\limits_{n \in \mathbb{N}} \frac{\varphi(n)}{n} \sum_{k=1}^n x_k^*,
	\end{equation*}
	is finite. Here, $x^* = \{x_n^*\}_{n=1}^{\infty}$ is the non-increasing rearrangement of the sequence $\abs{x} = \{\abs{x_n}\}_{n=1}^{\infty}$.
	In the case when $\varphi(n) = n^{1/p}$ for $1 < p < \infty$, the corresponding Marcinkiewicz sequence space $m_{\varphi}$
	is denoted by $\ell_{p,\infty}$ and is sometimes called the {\it weak $\ell_p$-space}.

	Let us now collect some important facts about Lorentz and Marcinkiewicz spaces.
	\begin{itemize}[leftmargin=0.5cm]
		\item For any rearrangement invariant sequence space $X$ with the fundamental function\footnote{Recall, that the {\it fundamental
			function $\varphi_X$} of a rearrangement invariant sequence space $X$ is the function $\varphi_X(n) \coloneqq \norm{\sum_{k=1}^{n} e_k}_X$,
			where $n \in \mathbb{N}$ and $\{e_n\}_{n=1}^{\infty}$ is the canonical basic sequence of $X$. Note that every such a function is equivalent
			to a concave function (see \cite[Proposition~5.11, p.~71]{BS88}).} $\varphi_X$ we have the following inclusions
			\begin{equation*}
				\ell_1 \hookrightarrow \lambda_{\varphi_X} \hookrightarrow X \hookrightarrow m_{\varphi_X} \hookrightarrow \ell_{\infty},
			\end{equation*}
			that is to say, Lorentz spaces $\lambda_{\varphi_X}$ are the \enquote{smallest} and Marcinkiewicz spaces $m_{\varphi_X}$ are the
			\enquote{largest} spaces among all rearrangement invariant spaces with a given fundamental function $\varphi_X$
			(see \cite[Theorem~5.13, p.~72]{BS88}, \cite[Theorem~5.5, p.~118 and Theorem~5.7, p.~119]{KPS82}).
		\item It is well-known that Lorentz space $\lambda_{\varphi}$ is separable if, and only if, $\varphi(\infty) \coloneqq \lim_{t \rightarrow \infty} \varphi(t) = \infty$
			(see \cite[Lemma~5.1, p.~110]{KPS82}; cf. also \cite[Corollary~4.20]{Kol16} together with the note on p.~820 about the \enquote{sequence case}).
			Note also that if $\varphi(\infty) < \infty$, then $\lambda_{\varphi} = \ell_{\infty}$ and, in consequence, $(\lambda_{\varphi})_o = c_0$.
			On the other hand, Marcinkiewicz spaces $m_{\varphi}$ are never separable (see, e.g., \cite[Remark~4.4(ii)]{Kol16}). However, the separable part
			$(m_{\varphi})_o$ of $m_{\varphi}$ can be described precisely in the following way
			\begin{equation*}
				(m_{\varphi})_o = \left\{ x = \{x_n\}_{n=1}^{\infty} \in m_{\varphi} \colon \lim\limits_{n \rightarrow \infty} \frac{\varphi(n)}{n} \sum_{k=1}^n x_k^* = 0 \right\}
			\end{equation*}
			(see \cite[pp.~115--116]{KPS82}).
		\item It follows from the very definition that Lorentz spaces $\lambda_{\varphi}$ are (up to the equality of norms) symmetrizations
			of weighted $\ell_1$-spaces; more precisely,
			\begin{equation*}
				\lambda_{\varphi} \equiv \left[ \ell_1(\Delta \varphi) \right]^{(*)},
			\end{equation*}
			where $\Delta \varphi(n) = \varphi(n+1) - \varphi(n)$ for $n \in \mathbb{N}$.
		\item Dually, down to one technical detail, Marcinkiewicz spaces $m_{\varphi}$ are (but only up to the equivalence of norms) symmetrizations
			of weighted $\ell_{\infty}$-spaces, namely,
			\begin{equation*}
				m_{\varphi} = \left[ \ell_{\infty}(\varphi) \right]^{(*)}.
			\end{equation*}
			The detail is the assumption that the so-called lower index\footnote{For an increasing concave function $\varphi$ defined on $[1,\infty)$
			the corresponding {\it lower index $p_{\varphi}^{\infty}$} and {\it upper index $q_{\varphi}^{\infty}$} are the numbers defined as
			\begin{equation*}
				p_{\varphi}^{\infty} = \lim\limits_{s \rightarrow 0^{+}} \frac{\log s}{\log \overline{\varphi}^{\infty}(s)} \text{, }
				q_{\varphi}^{\infty} = \lim\limits_{s \rightarrow \infty} \frac{\log s}{\log \overline{\varphi}^{\infty}(s)},
				\text{ where } \overline{\varphi}^{\infty}(s) = \sup\limits_{t \geqslant \max(1, 1/s)} \frac{\varphi(st)}{\varphi(t)}.
			\end{equation*}
			It is known that $1 \leqslant p_{\varphi}^{\infty} \leqslant q_{\varphi}^{\infty} \leqslant \infty$.}
			$p_{\varphi}^{\infty}$ of the function $\varphi$ is strictly greater than one.
	\end{itemize}

	After this short reminder we are in the position to show the following result describing the compact multipliers
	acting either from the Marcinkiewicz sequence space into a rearrangement invariant sequence space or
	from a rearrangement invariant sequence space into the Lorentz sequence space.

	\begin{proposition}[Compact multipliers between Lorentz and Marcinkiewicz sequence spaces] \label{Prop: computing multipliers from m and to lambda}
		{\it Let $\varphi$ and $\psi$ be two positive, increasing and concave functions defined on $(0,\infty)$.
			Let, moreover, $X$, $Y$ and $Z$ be three rearrangement invariant sequence spaces with the Fatou property.}
		\begin{enumerate}
			\item {\it Suppose $X$ is reflexive and $q_{\varphi}^{\infty} < \infty$. Then the multiplication operator
				$M_{\lambda} \colon X \rightarrow \lambda_{\varphi}$ is compact if, and only if,
				either $\left\{ \frac{\varphi(n)}{n}\lambda_n^* \right\}_{n=1}^{\infty} \in X^{\times}$, provided
				$\left\{ \frac{\varphi(n)}{n} \right\}_{n=1}^{\infty} \notin X^{\times}$, or $\{\lambda_n\}_{n=1}^{\infty} \in c_0$,
				provided $\left\{ \frac{\varphi(n)}{n}\lambda_n^* \right\}_{n=1}^{\infty} \in X^{\times}$.}
			\item {\it Suppose $Y$ is order continuous and $1 < p_{\varphi}^{\infty}$. Then the multiplication operator
				$M_{\lambda} \colon m_{\varphi} \rightarrow Y$ is compact if, and only if, either
				$\left\{ \frac{1}{\varphi(n)}\lambda_n^{*} \right\}_{n=1}^{\infty} \in Y$, provided
				$\left\{ \frac{1}{\varphi(n)} \right\}_{n=1}^{\infty} \notin Y$, or $\{\lambda_n\}_{n=1}^{\infty} \in c_0$,
				provided $\left\{ \frac{\varphi(n)}{n}\lambda_n^* \right\}_{n=1}^{\infty} \in Y$.}
			\item {\it Suppose $q_{\varphi}^{\infty} < \infty$. Then the multiplication operator
				$M_{\lambda} \colon Z \rightarrow \lambda_{\varphi}$ is compact if, and only if, either
				$\left\{ \frac{\varphi(n)}{n} \lambda_n^* \right\}_{n=1}^{\infty} \in (Z^{\times})_o$, provided
				$\left\{ \frac{\varphi(n)}{n} \right\}_{n=1}^{\infty} \notin Z^{\times}$, or $\{\lambda_n\}_{n=1}^{\infty} \in c_0$,
				provided $\left\{ \frac{\varphi(n)}{n} \right\}_{n=1}^{\infty} \in Z^{\times}$.}
		\end{enumerate}
		{\it In particular,}
		\begin{enumerate}
			\item[(iv)] {\it if $\limsup_{n \rightarrow \infty} \frac{\psi(n)}{\varphi(n)} = \infty$
				and $1 < p_{\varphi}^{\infty}$, then the multiplication operator $M_{\lambda} \colon m_{\varphi} \rightarrow \lambda_{\psi}$
				is compact if, and only if, $\left\{ \frac{\psi(n+1) - \psi(n)}{\varphi(n)} \lambda_n^* \right\}_{n=1}^{\infty} \in \ell_1$.}
			\item[(v)] {\it if $\limsup_{n \rightarrow \infty} \frac{\varphi(n)}{\psi(n)} = \infty$ and $q_{\varphi}^{\infty} < \infty$,
				then the multiplication operator $M_{\lambda} \colon \lambda_{\psi} \rightarrow \lambda_{\varphi}$
				is compact if, and only if, $\left\{ \frac{\varphi(n)}{\psi(n)} \lambda_n^* \right\}_{n=1}^{\infty} \in c_0$.}
		\end{enumerate}
	\end{proposition}

	Note that (i) and (ii) means exactly that the operators $M_{\lambda} \colon X \rightarrow \lambda_{\varphi}$ and $M_{\lambda} \colon m_{\varphi} \rightarrow Y$
	are compact if, and only if, they are bounded.

	The proof of Proposition~\ref{Prop: computing multipliers from m and to lambda} will essentially consists of two parts:
	we compute the space of pointwise multipliers first, and then we try to characterize the points of order continuity in this space.
	However, since the whole argument seems quite long, we decided to include its most important parts in two auxiliary lemmas.

	We start with the following one, which can be viewed as a discrete version of some of the Le{\' s}nik, Kolwicz and Maligranda results from \cite{KLM14}.

		\begin{lemma} \label{LEMMA: a'la KLM12/14}
			{\it Let $X$ and $Y$ be two rearrangement invariant sequence spaces with the Fatou property. Let, moreover, $\varphi$ be a positive, increasing
				and concave function on $(0,\infty)$.}
			\begin{enumerate}
				\item {\it Suppose $1 < p_{\varphi}^{\infty}$. Then $M(m_{\varphi},Y) = \left[ Y\left( \frac{1}{\varphi} \right) \right]^{(*)}$.}
				\item {\it Suppose $q_{\varphi}^{\infty} < \infty$. Then $M(X,\lambda_{\varphi}) = \left[ X^{\times}\left( \frac{\varphi(t)}{t} \right) \right]^{(*)}$.}
			\end{enumerate}
		\end{lemma}
		\begin{proof}
			The proof presented here will abound in many small technical details and facts that are either certainly known to some specialists
			or are based on earlier results scattered over at least a few papers (mainly \cite{KLM12} and \cite{KLM14}). However, in order to relieve
			the reader from the tedious and time-consuming task of checking and converting all arguments by himself, and, because we feel that by simplifying
			some of the proofs we are able to present the material in a clearer and a bit more complete manner, we decided to present all the details.
			Simultaneously, the proofs given here can serve as a tutorial to the arithmetic of K{\" o}the spaces.

			Let's start with explanation that spaces $M(m_{\varphi},Y)$ and $M(X,\lambda_{\varphi})$ are actually rearrangement
			invariant. We only need to show that the space $M(E,F)$, where $E$ and $F$ are two rearrangement
			invariant sequence spaces, is rearrangement invariant as well (for a pair of rearrangement invariant function spaces this fact is known;
			see \cite[Theorem~2.2]{KLM12}). To see this, take $x = \{x_n\}_{n=1}^{\infty} \in M(E,F)$ and let's consider two cases.

			Firstly, suppose that $x^{*}(\infty) \coloneqq \lim_{n \rightarrow \infty} x_n^* > 0$.
			This implies that $\ell_{\infty} \hookrightarrow M(E,F)$. Moreover, if $F \hookrightarrow G$, then it is straightforward to
			see that $M(E,F) \hookrightarrow M(E,G)$. Note also that $M(E,\ell_{\infty}) \hookrightarrow \ell_{\infty}$. Indeed, if an unbounded
			function, say $z = \{z_n\}_{n=1}^{\infty}$, belonged to $M(E,\ell_{\infty})$, then $z_n = e_n z \in \ell_{\infty}$ for all $n \in \mathbb{N}$,
			which is impossible. Thus, combining all these three facts together and remembering that $F \hookrightarrow \ell_{\infty}$
			(see \cite[Corollary~6.7, p.~78]{BS88}), we get
			\begin{equation*}
				\ell_{\infty} \hookrightarrow M(E,F) \hookrightarrow M(E,\ell_{\infty}) \hookrightarrow \ell_{\infty}.
			\end{equation*}
			In consequence, $M(E,F) = \ell_{\infty}$.

			Secondly, suppose that $x^*(\infty) = 0$. Here we will, more or less, follow the argument given in \cite[Theorem~2.2]{KLM12}.
			Let $y = \{y_n\}_{n=1}^{\infty}$ be any function that is equi-measurable with $x$. Then it follows from \cite[Lemma~2.1, p.~60]{KPS82}
			that for any $\varepsilon > 0$ there exists a measure-preserving transformation, say $\varpi_{\varepsilon}$, such that
			\begin{equation*}
				\norm{x - y(\varpi_{\varepsilon})}_{\ell_1} < \varepsilon.
			\end{equation*}
			Consequently, since $\ell_1 \overset{\varphi_F(1)}{\hookrightarrow} F$ and $E \overset{1/\varphi_E(1)}{\hookrightarrow} \ell_{\infty}$
			(see \cite[pp.~77--78]{BS88}), so for all $z = \{z_n\}_{n=1}^{\infty} \in E$ with $\norm{z}_E \leqslant 1$ we have
			\begin{align*}
				\norm{yz}_F & = \norm{y(\varpi_{\varepsilon})z(\varpi_{\varepsilon})}_F \\
					& \leqslant \norm{xz(\varpi_{\varepsilon})}_F + \norm{\left( y(\varpi_{\varepsilon}) - x \right)z(\varpi_{\varepsilon})}_F \\
					& \leqslant \norm{x}_{M(E,F)} + \norm{\left( y(\varpi_{\varepsilon} \right) - x)z(\varpi_{\varepsilon})}_F \\
					& \leqslant \norm{x}_{M(E,F)} + \norm{z(\varpi_{\varepsilon})}_{\ell_{\infty}} \norm{y(\varpi_{\varepsilon}) - x}_F \\
					& \leqslant \norm{x}_{M(E,F)} + \frac{\varphi_F(1)}{\varphi_E(1)}\norm{z}_{E} \norm{y(\varpi_{\varepsilon}) - x}_{\ell_1}
					< \norm{x}_{M(E,F)} + \frac{\varphi_F(1)}{\varphi_E(1)}\varepsilon.
			\end{align*}
			Due to $\varepsilon$'s arbitrariness we conclude that
			\begin{equation*}
				\norm{y}_{M(E,F)} \leqslant \norm{x}_{M(E,F)}.
			\end{equation*}
			To prove the reverse inequality, it is enough to swap the roles of $x$ and $y$.

			(i)	First of all, without loss of generality, we can assume that $m_{\varphi} \neq \ell_{\infty}$ (or, which is the same thing,
			that the function $\varphi$ is unbounded), because otherwise the sequence $\{\varphi(n)\}_{n=1}^{\infty}$ is equivalent to a constant
			sequence and, almost trivially, we have the following identifications
			\begin{equation*}
				M(m_{\varphi},Y) = M(\ell_{\infty},Y) = Y = Y\left( \frac{1}{\varphi} \right) = \left[ Y\left( \frac{1}{\varphi} \right) \right]^{(*)}.
			\end{equation*}

			Continuing the proof, we will now consider two cases.

			$1^{\text{st}}$. Assume that $\left\{ \frac{1}{\varphi(n)} \right\}_{n=1}^{\infty} \in Y$. Then it is clear that
			$\ell_{\infty} \hookrightarrow Y\left( \frac{1}{\varphi} \right)$. Moreover, since for all $n \in \mathbb{N}$ we have
			\begin{equation*}
				 \frac{\varphi_Y(n)}{\varphi(n)}x_n^* = \varphi_Y(n) \left[ \frac{x_n^*}{\varphi(n)} \right]^*
					\leqslant \frac{\varphi_Y(n)}{n} \sum_{k=1}^{n} \frac{x_k^*}{\varphi(k)},
			\end{equation*}
			where the first equality follows from a simple fact that the product of two non-increasing functions is non-increasing,
			and the second inequality is due to \cite[Proposition~3.2, p.~52]{BS88}, so in view of the embedding $Y \hookrightarrow m_{\varphi_Y}$
			(see \cite[Proposition~5.1, p.~70]{BS88}) we infer that
			\begin{equation*}
				\ell_{\infty} \equiv (\ell_{\infty})^{(*)} \hookrightarrow \left[ Y\left( \frac{1}{\varphi} \right) \right]^{(*)}
					\hookrightarrow \left[ m_{\varphi_Y}\left( \frac{1}{\varphi} \right) \right]^{(*)}
					\hookrightarrow \left[ \ell_{\infty}\left( \frac{\varphi_Y}{\varphi} \right) \right]^{(*)}.
			\end{equation*}
			Now, because $\sup_{n \in \mathbb{N}} \frac{\varphi_Y(n)}{\varphi(n)} < \infty$ (otherwise, the sequence $\left\{ \frac{1}{\varphi(n)} \right\}_{n=1}^{\infty}$
			does not belongs to $\left[ \ell_{\infty}(\varphi_Y) \right]^{(*)}$, which is impossible!), so
			\begin{equation*}
				\left[ \ell_{\infty}\left( \frac{\varphi_Y}{\varphi} \right) \right]^{(*)} \hookrightarrow [\ell_{\infty}]^{(*)} \equiv \ell_{\infty}.
			\end{equation*}
			This gives us that
			\begin{equation} \label{EQ: Y(w)^s = l infty}
				\left[ Y\left( \frac{1}{\varphi} \right) \right]^{(*)} = \ell_{\infty}.
			\end{equation}

			On the other hand, if $\left\{ \frac{1}{\varphi(n)} \right\}_{n=1}^{\infty} \in Y$, then $\ell_{\infty}(\varphi) \hookrightarrow Y$ and
			\begin{equation*}
				m_{\varphi} = \left[ \ell_{\infty}(\varphi) \right]^{(*)} \hookrightarrow Y^{(*)} \equiv Y,
			\end{equation*}
			where the first equality follows from our assumption that $p_{\varphi}^{\infty} > 1$. This imply, however, that $\ell_{\infty} \hookrightarrow M(m_{\varphi},Y)$
			and, because the space $M(m_{\varphi},Y)$ is rearrangement invariant (see the argument given at the beginning of this proof), so also
			$M(m_{\varphi},Y) \hookrightarrow \ell_{\infty}$. In conclusion
			\begin{equation*}
				M(m_{\varphi},Y) = \ell_{\infty}
			\end{equation*}
			and, not forgetting about (\ref{EQ: Y(w)^s = l infty}), we finally get the equality
			\begin{equation*}
				M(m_{\varphi},Y) = \ell_{\infty} = \left[ Y\left( \frac{1}{\varphi} \right) \right]^{(*)}.
			\end{equation*}

			$2^{\text{nd}}$. Next, assume that $\left\{ \frac{1}{\varphi(n)} \right\}_{n=1}^{\infty} \notin Y$. Our plan is to show that
			\begin{equation} \tag{$\star$} \label{EQ: idea KLM14}
				\left[ Y\left( \frac{1}{\varphi} \right) \right]^{(*)} \overset{2}{\hookrightarrow}
					M(m_{\varphi},Y) \overset{1}{\hookrightarrow} \left[ Y\left( \frac{1}{\varphi} \right) \right]^{(*)}.
			\end{equation}
			To do that, we will need the following preparatory result whose proof we borrow from \cite[p.~653]{KLM14}.

			\vspace{5pt} \noindent {\bf Claim~$\triangle$.}
			{\it Let $Y$ be a rearrangement invariant sequence space. Let, further, $\varphi$ be a positive, increasing and concave function on $(0,\infty)$
				with $1 < p_{\varphi}^{\infty}$. Then for any $x \in M(m_{\varphi},Y)$ we have the following equality}
			\begin{equation*} \label{EQ: equality}
				\norm{x}_{M(m_{\varphi},Y)} = \sup\limits_{\varpi} \norm{\frac{x}{\varphi(\varpi)}}_Y,
			\end{equation*}
			{\it where the supremum is taken over all measure-preserving transformations $\varpi$.}
			\begin{proof}[Proof of Claim~$\triangle$]
			Take $x \in M(m_{\varphi},Y)$. Let $\varpi$ be a measure preserving transformation. Since $m_{\varphi} = [\ell_{\infty}(\varphi)]^{(*)}$
			and $\left( \frac{1}{\varphi(\varpi)} \right)^{*} = \frac{1}{\varphi} \in \ell_{\infty}(\varphi)$, so $\frac{1}{\varphi(\varpi)} \in m_{\varphi}$.
			In consequence, $\frac{x}{\varphi(\varpi)} \in Y$ and $\norm{\frac{x}{\varphi(\varpi)}}_Y \leqslant \norm{x}_{M(m_{\varphi},Y)}$.
			Taking the supremum over all measure-preserving transformations, we immediately see that
			\begin{equation*}
				\sup\limits_{\varpi} \norm{\frac{x}{\varphi(\varpi)}}_Y \leqslant \norm{x}_{M(m_{\varphi},Y)}.
			\end{equation*}

			To prove the reverse inequality, suppose that for any measure-preserving transformation $\varpi$ the function $\frac{x}{\varphi(\varpi)}$
			belongs to $Y$. Take $z \in m_{\varphi}$ with $\norm{z}_{m_{\varphi}} \leqslant 1$. As we explained above, we can always assume that
			$m_{\varphi} \neq \ell_{\infty}$. Therefore, $m_{\varphi} \hookrightarrow c_0$ and, consequently, $z^*(\infty) = 0$. Moreover, Ryff's
			theorem guarantee the existence of a measure-preserving transformation, say $\varpi_0$, with $\abs{z} = z^*(\varpi_0)$. Putting these
			facts together, we get
			\begin{equation*}
				\abs{xz} = \abs{x}z^*(\varpi_0) \leqslant \frac{\abs{x}}{\varphi(\varpi_0)} \in Y.
			\end{equation*}
			But this means that $x \in M(m_{\varphi},Y)$ and
			\begin{equation*}
				\norm{x}_{M(m_{\varphi},Y)} \leqslant \sup\limits_{\varpi} \norm{\frac{x}{\varphi(\varpi)}}_Y.
			\end{equation*}
			The proof is finished.
			\end{proof}

			Now we are ready to show (\ref{EQ: idea KLM14}). Let's first focus on the embedding
			$M(m_{\varphi},Y) \hookrightarrow \left[ Y\left( \frac{1}{\varphi} \right) \right]^{(*)}$.
			Take $x \in M(m_{\varphi},Y)$. In view of Claim~$\triangle$, it is enough to prove that
			\begin{equation*}
				\norm{\frac{x^*}{\varphi}}_{Y} = \norm{\frac{x^*(\varpi)}{\varphi(\varpi)}}_Y \leqslant \sup\limits_{\varpi} \norm{\frac{x}{\varphi(\varpi)}}_Y.
			\end{equation*}
			However, if it is true that there is a measure-preserving transformation, say $\varpi_0$, with $\abs{x} = x^*(\varpi_0)$,
			then the above inequality is basically obvious. This, however, will be an immediate consequence of Ryff's theorem once
			we know that $x^*(\infty) = 0$. Let us show the latter. To do this recall that it follows from our assumptions that
			\begin{equation*}
				\ell_{\infty}(\varphi) \not\hookrightarrow Y.
			\end{equation*}
			Then it is clear that
			\begin{equation*}
				m_{\varphi} = \left[ \ell_{\infty}(\varphi) \right]^{(*)} \not\hookrightarrow Y^{(*)} = Y.
			\end{equation*}
			But this means that $\ell_{\infty} \not \hookrightarrow M(m_{\varphi}, Y)$. Indeed, otherwise, due to the fact that the space
			$M(m_{\varphi}, Y)$ is rearrangement invariant, we would have that $M(m_{\varphi}, Y) = \ell_{\infty}$. However, this in turn
			would mean that $m_{\varphi} \hookrightarrow Y$. Clearly, this is impossible, so using once again the fact that the space
			$M(m_{\varphi}, Y)$ is rearrangement invariant, we get
			\begin{equation*}
				M(m_{\varphi}, Y) \hookrightarrow c_0.
			\end{equation*}
			This is exactly what we needed.

			To complete the proof, we only need to show the second embedding. Take
			$x = \{x_n\}_{n=1}^{\infty} \in \left[ Y\left( \frac{1}{\varphi} \right) \right]^{(*)}$. Let $\varpi$ be a measure-preserving transformation.
			Applying \cite[Property~$10^{\circ}$, p.~67]{KPS82} and remembering that the function $t \mapsto \frac{1}{\varphi(t)}$ is non-increasing, we get
			\begin{equation*}
				\left( \frac{x}{\varphi(\varpi)} \right)^* \leqslant D_2(x^*)D_2\left( \left( \frac{1}{\varphi(\varpi)} \right)^* \right)
					= D_2(x^*)D_2\left( \frac{1}{\varphi} \right)
					= D_2\left( \frac{x^*}{\varphi} \right).
			\end{equation*}
			Moreover, since the space $Y$ is rearrangement invariant, so
			\begin{equation*}
				\norm{\frac{x}{\varphi(\varpi)}}_Y = \norm{\left( \frac{x}{\varphi(\varpi)} \right)^*}_Y \leqslant
					\norm{D_2 \colon Y \rightarrow Y} \norm{\frac{x^*}{\varphi}}_Y \leqslant 2 \norm{x}_{\left[ Y\left( \frac{1}{\varphi} \right) \right]^{(*)}},
			\end{equation*}
			where the last inequality is essentially due to the Calder{\' o}n--Mityagin interpolation theorem (see \cite[p.~130]{LT79}; cf. \cite[pp.~96--98]{KPS82}).
			Taking the supremum over all measure-preserving mappings ends the proof of (\ref{EQ: idea KLM14}).

			(ii) Remembering about our assumptions we have that
			\begin{align*}
				M\left( X, \lambda_{\varphi} \right) & = M\left( (\lambda_{\varphi})^{\times}, X^{\times} \right) \quad \text{(using \cite[Property~(vii), p.~879]{KLM12})} \\
					& = M\left( m_{t/\varphi(t)}, X^{\times} \right) \quad \text{(using \cite[Theorem~5.2, p.~112]{KPS82})} \\
					& = \left[ X^{\times}\left( \frac{\varphi(t)}{t} \right) \right]^{(*)} \quad \text{(by (i) proved above)}.
			\end{align*}
			That's all.
		\end{proof}

		Next, we need a lemma in the spirit of Kolwicz's results from \cite{Kol16} (precisely, see Theorem~3.9 ibidem).

		\begin{lemma} \label{LEMMA: a'la Kolwicz OC}
			{\it Let $X$ be a rearrangement invariant sequence space with the Fatou property. Moreover, let $w$ be a positive and non-increasing
				function on $(0,\infty)$ such that $w \notin X$. Assume that $[X(w)]^{(*)}$ is a Banach space up to an equivalent norm. Then}
			\begin{equation*}
				\left[ X(w)^{(*)} \right]_o = \left[ X_o(w) \right]^{(*)}.
			\end{equation*}
		\end{lemma}
		\begin{proof}
			First, in order not to overload the notation, we will omit some parentheses here and there. This should not lead to misunderstandings.
			Secondly, it is more or less obvious that $X(w)_o = X_o(w)$, so we only need to show that
			\begin{equation*}
				\left[X(w)_o\right]^{(*)} = \left[X(w)^{(*)}\right]_o.
			\end{equation*}
			We will justify both inclusions separately.

			To show that
			\begin{equation} \label{OCgwiazdkazawieranie 1}
				\left[X(w)_o\right]^{(*)}\subset\left[X(w)^{(*)}\right]_o,
			\end{equation}
			take $x\in\left[X(w)_o\right]^{(*)}$. This means that $x^*\in X(w)_o.$  Since the space $X(w)^{(*)}$ is rearrangement invariant,
			so in order to prove the inclusion \eqref{OCgwiazdkazawieranie 1} it is enough to show that $x^*\in \left[X(w)^{(*)}\right]_o$
			(see \cite[Theorem~3.6]{KKT22}). From the order continuity of $x$ in $X(w)$ it follows that
			\begin{equation*}
				\norm{x^*\chi_{\{n,n+1,\dots\}}}_{X(w)} \xrightarrow{n \rightarrow \infty} 0.
			\end{equation*}
			Consequently, for a fixed $\varepsilon > 0$ there is $N\in\mathbb{N}$ with
			\begin{equation*}
				\norm{x^*\chi_{\{n,n+1,\dots\}}}_{X(w)} < \varepsilon \text{ for } n \geqslant N.
			\end{equation*}
			Next, because we assumed that $w \notin X$, so $x^*(\infty) = 0$ (otherwise, $\ell_\infty \hookrightarrow X(w)$ and this imply that $w\in X$,
			which is impossible!). Therefore, we can find $M \in \mathbb{N}$ such that
			\begin{equation*}
				x^*_{M}\leqslant \frac{\varepsilon}{\norm{\chi_{\{1,\dots,N\}}}_{X(w)}}.
			\end{equation*}
			For $n \geqslant \max\{M,N\}$ we have
			\begin{align*}
				\norm{x^*\chi_{\{n,n+1,\dots\}}}_{X(w)^{(*)}} & \leqslant \norm{\left( x^*\chi_{\{M,\dots\}} \right)^*w}_X\\
					& \leqslant\norm{x^*_{M}\chi_{\{1,\dots,N\}}w}_X+\norm{x^*\chi_{\{N+1,\dots\}}}_{X(w)} < 2\varepsilon.
			\end{align*}
			This proves that $x^*\in \left[X(w)^{(*)}\right]_o$ and completes the  proof of \eqref{OCgwiazdkazawieranie 1}.

			Now we can turn our attention to the second inclusion, that is, we will show that
			\begin{equation}
				\left[X(w)^{(*)}\right]_o\subset\left[X(w)_o\right]^{(*)}. \label{OCgwiazdkazawieranie 2}
			\end{equation}
			Take $x\in\left[X(w)^{(*)}\right]_o$. Once again, since the space $X(w)^{(*)}$ is rearrangement invariant, so in view of \cite[Theorem~3.6]{KKT22}
			the fact that $x\in\left[X(w)^{(*)}\right]_o$ is equivalent to $x^*\in\left[X(w)^{(*)}\right]_o$. In order to prove \eqref{OCgwiazdkazawieranie 2}
			we need to show that $x^*\in X(w)_o$, that is, that for any sequence $\left\{ x^{(n)} \right\}_{n=1}^{\infty} \subset X(w)$ with
			\begin{equation*}
				0 \leqslant x^{(n)} \leqslant x^* \text{ and } x^{(n)} \downarrow 0,
			\end{equation*}
			we have $\norm{x^{(n)}}_{X(w)} \rightarrow 0$ as $n \rightarrow \infty$. Take such a sequence $\left\{ x^{(n)} \right\}_{n=1}^{\infty}$.
			Since $x^*(\infty) = 0$ (note that $x^*$ is a point of order continuity in a rearrangement invariant space $X(w)^{(*)}$ and $[X(w)^{(*)}]_o \subset [\ell_{\infty}]_o = c_0$),
			so $\left( x^{(n)} \right)^* \downarrow 0$ and, trivially, $0 \leqslant \left( x^{(n)} \right)^* \leqslant x^*$. This means that
			$\norm{\left( x^{(n)} \right)^*}_{X(w)}\to 0$ (recall, that it follows from our assumption that $x^*\in\left[X(w)^{(*)}\right]_o$).
			In consequence, we get
			\begin{align*}
				\norm{x^{(n)}}_{X(w)} & = \norm{x^{(n)} w}_{X} = \norm{\left( x^{(n)} w \right)^*}_{X} \quad \text{(because $X$ is rearrangement invariant)} \\
					& \leqslant \norm{D_2 \left[ \left( x^{(n)} \right)^* \right] D_2 \left[ w^* \right]}_{X} \quad \text{(using \cite[Property~$10^{\circ}$, p.~67]{KPS82})} \\
					& = \norm{D_2 \left[ \left( x^{(n)} \right)^* w^* \right]}_{X} \\
					& \leqslant \norm{D_2 \colon X \rightarrow X} \norm{\left( x^{(n)} \right)^* w^*}_{X} \quad \text{(because $D_2$ is bounded on $X$)} \\
					& = \norm{D_2 \colon X \rightarrow X} \norm{\left( x^{(n)} \right)^* w}_{X} \quad \text{(since the function $w$ is non-increasing)} \\
					& = \norm{D_2 \colon X \rightarrow X} \norm{x^{(n)}}_{X(w)^{(*)}} \xrightarrow{n \rightarrow \infty} 0.
			\end{align*}
			The proof is finished.
		\end{proof}

		What seems quite surprising is that our proof is (almost) completely elementary, that is, it does not use the concept of an \enquote{$OC^*$-point}
		or the machinery developed in \cite{Kol16}. Led us also add that the first author learned from Kolwicz another, slightly more general,
		proof which directly uses \cite{Kol16}. But let's put it in the drawer for now.

		\begin{remark}[When is the symmetrization of a weighted rearrangement invariant sequence space a Banach space?]
			As we tried to explain earlier, the symmetrization of a K{\" o}the space is in general only a quasi-Banach space (see \cite[Lemma~1.4]{KR09}; cf. \cite[Corollary~1]{KLM19}).
			However, there is a rather simple criterion to determine whether the space $[E(1/\phi)]^{(*)}$, where $E$ is a rearrangement invariant sequence space
			and $\phi$ is a non-decreasing and concave function on $(0,\infty)$, is a Banach space up to an equivalent norm. Indeed, suppose that
			\begin{equation*}
				\frac{1}{\beta_E} < 1 - \frac{1}{p_{\phi}^{\infty}},
			\end{equation*}
			where $\beta_E$ is the upper Boyd index of $E$ and $p_{\phi}^{\infty}$ is the lower index of $\phi$. Then the discrete Hardy operator $\mathscr{H}$
			is bounded on the space $E(1/\phi)$ (see \cite[Theorem~1]{Mal83}; cf. note on Pavlov's result on p.~127 in \cite{KMP07}). In consequence, the functional
			\begin{equation*}
				\vertiii{x} \coloneqq \norm{\frac{x^{**}}{\phi}}_E
					\text{ for } x = \{x_n\}_{n=1}^{\infty} \in \left[ E \left( \frac{1}{\phi} \right) \right]^{(*)},
			\end{equation*}
			where the symbol $x^{**}$ denote the {\it maximal function} of $x^*$, that is, $x^{**} \coloneqq \left\{ \frac{1}{n} \sum_{k=1}^{n} x^*_k \right\}_{n=1}^{\infty}$
			(see \cite[Definition~3.1, p.~52]{BS88}), defines on the space $[E(1/\phi)]^{(*)}$ an equivalent norm. In fact, in view of \cite[p.~54]{BS88} the triangle inequality
			is clear. Moreover, for any $x \in [E(1/\phi)]^{(*)}$ we have the following inequalities
			\begin{equation*}
				\norm{\frac{x^*}{\phi}}_E \leqslant \norm{\frac{x^{**}}{\phi}}_{E} = \vertiii{x}
					\leqslant \norm{\mathscr{H} \colon E\left( \frac{1}{\phi} \right) \rightarrow E\left( \frac{1}{\phi} \right)} \norm{\frac{x^*}{\phi}}_E,
			\end{equation*}
			which means that
			\begin{equation*}
				\norm{x}_{[E(1/\phi)]^{(*)}} \approx \vertiii{x}.
			\end{equation*}
		\end{remark}

		\begin{proof}[Proof of Proposition~\ref{Prop: computing multipliers from m and to lambda}]
		(i) Under the assumption that $\left\{ \frac{\varphi(n)}{n} \right\}_{n=1}^{\infty} \notin X^{\times}$,
		we have the following sequence of equivalences
		\begin{align*}
			M_{\lambda} \colon X \rightarrow \lambda_{\varphi} \text{ is compact }
			& \Longleftrightarrow \lambda \in M(X,\lambda_{\varphi})_o \quad \text{(using Theorem~\ref{THM: Essential norm Kothe sequence})} \\
			& \Longleftrightarrow \lambda \in \left[ \left[ X^{\times}\left( \frac{\varphi(t)}{t} \right) \right]^{(*)} \right]_o \quad \text{(by Lemma~\ref{LEMMA: a'la KLM12/14}(ii))} \\
			& \Longleftrightarrow \lambda \in \left[ \left( X^{\times} \right)_o \left( \frac{\varphi(t)}{t} \right) \right]^{(*)} \quad \text{(by Lemma~\ref{LEMMA: a'la Kolwicz OC})} \\
			& \Longleftrightarrow \left\{ \frac{\varphi(n)}{n}\lambda_n^* \right\}_{n=1}^{\infty} \in \left( X^{\times} \right)_o \quad \text{(because $X$ is reflexive)} \\
			& \Longleftrightarrow \left\{ \frac{\varphi(n)}{n}\lambda_n^* \right\}_{n=1}^{\infty} \in X^{\times}.
		\end{align*}
		Consequently, it follows from Theorem~\ref{THM: Essential norm Kothe sequence} that
		\begin{equation*}
			M_{\lambda} \colon X \rightarrow \lambda_{\varphi} \text{ is compact if, and only if, } \left\{ \frac{\varphi(n)}{n}\lambda_n^* \right\}_{n=1}^{\infty} \in X^{\times}.
		\end{equation*}

		Next, assume that $\left\{ \frac{\varphi(n)}{n} \right\}_{n=1}^{\infty} \in X^{\times}$. Then the first part of the proof of Lemma~\ref{LEMMA: a'la KLM12/14}
		together with duality results teaches us that
		\begin{equation*}
			M(X, \lambda_{\varphi}) = M\left( (\lambda_{\varphi})^{\times},X^{\times} \right) = M(m_{t/\varphi(t)},X^{\times}) = \ell_{\infty}.
		\end{equation*}
		Therefore, applying Theorem~\ref{THM: Essential norm Kothe sequence} one more time, we see immediately that
		\begin{equation*}
			M_{\lambda} \colon X \rightarrow \lambda_{\varphi} \text{ is compact if, and only if, } \lambda \in (\ell_{\infty})_o = c_0.
		\end{equation*}

		(ii) Similarly, assuming that $\left\{ \frac{1}{\varphi(n)} \right\}_{n=1}^{\infty} \notin Y$, we have the following equivalences
		\begin{align*}
			M_{\lambda} \colon m_{\varphi} \rightarrow Y \text{ is compact }
			& \Longleftrightarrow \lambda \in M(m_{\varphi},Y)_o \quad \text{(using Theorem~\ref{THM: Essential norm Kothe sequence})} \\
			& \Longleftrightarrow \lambda \in \left[ \left[ Y\left( \frac{1}{\varphi} \right) \right]^{(*)} \right]_o \quad \text{(by Lemma~\ref{LEMMA: a'la KLM12/14}(i))} \\
			& \Longleftrightarrow \lambda \in \left[ Y_o\left( \frac{1}{\varphi} \right) \right]^{(*)} \quad \text{(by Lemma~\ref{LEMMA: a'la Kolwicz OC})} \\
			& \Longleftrightarrow \left\{ \frac{1}{\varphi(n)}\lambda_n^* \right\}_{n=1}^{\infty} \in Y_o \quad \text{(because $Y \in (OC)$)} \\
			& \Longleftrightarrow \left\{ \frac{1}{\varphi(n)}\lambda_n^* \right\}_{n=1}^{\infty} \in Y.
		\end{align*}

		On the other hand, if we assume that $\left\{ \frac{1}{\varphi(n)} \right\}_{n=1}^{\infty} \in Y$, then the following equality
		\begin{equation*}
			M(m_{\varphi},Y) = \ell_{\infty}
		\end{equation*}
		is a straightforward consequence of the first part of the proof of Lemma~\ref{LEMMA: a'la KLM12/14}. Thus,
		\begin{equation*}
			M_{\lambda} \colon m_{\varphi} \rightarrow Y \text{ is compact if, and only if, } \lambda \in (\ell_{\infty})_o = c_0.
		\end{equation*}

		(iii) After noticing that the space $\lambda_{\varphi}$ is order continuous, this part can be proved in the same way as (i)
		(see \cite[Lemma~5.1, p.~110]{KPS82}; cf. also Lemma~\ref{LEMMA: a'la Kolwicz OC}).

		(iv) Assume that $\limsup_{n \rightarrow \infty} \frac{\psi(n)}{\varphi(n)} = \infty$. In view of the embedding
		\begin{equation*}
			\lambda_{\psi} \hookrightarrow m_{\psi}
		\end{equation*}
		(see \cite[Proposition~5.9, p.~70]{BS88}), it is clear that
		\begin{equation*}
			\text{ if } \limsup_{n \rightarrow \infty} \frac{\psi(n)}{\varphi(n)} = \infty \text{, then }
				\left\{ \frac{1}{\varphi(n)} \right\}_{n=1}^{\infty} \notin m_{\psi} \text{ and, in consequence, }
				\left\{ \frac{1}{\varphi(n)} \right\}_{n=1}^{\infty} \notin \lambda_{\psi}.
		\end{equation*}
		We have
		\begin{align*}
			M(m_{\varphi}, \lambda_{\psi}) & = \left[ \lambda_{\psi}\left( \frac{1}{\varphi} \right) \right]^{(*)} \quad \text{(by Lemma~\ref{LEMMA: a'la KLM12/14}(i))}\\
				& = \left[ \left[ \ell_1(\Delta \psi) \right]^{(*)} \left( \frac{1}{\varphi} \right) \right]^{(*)}
					\quad \text{(because $\lambda_{\phi} \equiv [\ell_1(\Delta \phi)]^{(*)}$, $\Delta \phi(n) \coloneqq \phi(n+1) - \phi(n)$)} \\
				& = \left[ \ell_1(\Delta \psi) \left( \frac{1}{\varphi} \right) \right]^{(*)} \quad \text{(in view of the equality $\left( \frac{x^*}{\varphi} \right)^* = \frac{x^*}{\varphi}$)} \\
				& = \left[ \ell_1 \left( \frac{\Delta \psi}{\varphi} \right) \right]^{(*)},
		\end{align*}
		where $\Delta \psi(t) \coloneqq \psi(n+1) - \psi(n)$ for $n \in \mathbb{N}$. Since the sequence $\left\{ \frac{\psi(n+1)-\psi(n)}{\varphi(n)} \right\}_{n=1}^{\infty}$
		is non-increasing and $\ell_1$ is separable, so applying Lemma~\ref{LEMMA: a'la Kolwicz OC} we get that
		\begin{equation*}
			\left[ \left[ \ell_1 \left( \frac{\Delta \psi}{\varphi} \right) \right]^{(*)} \right]_o
				= \left[ (\ell_1)_o \left( \frac{\Delta \psi}{\varphi} \right) \right]^{(*)}
				= \left[ \ell_1 \left( \frac{\Delta \psi}{\varphi} \right) \right]^{(*)}.
		\end{equation*}
		Let us make it clear that the use of Lemma~\ref{LEMMA: a'la Kolwicz OC} was completely justified in this situation, because
		$\left\{ \frac{\Delta \psi(n)}{\varphi(n)} \right\}_{n=1}^{\infty} \notin \ell_1$. In fact, if this were not the case, then
		$\left\{ \frac{1}{\varphi(n)} \right\}_{n=1}^{\infty} \in \lambda_{\psi}$ which, due to our assumptions, is impossible.
		All this means is that the sequence $\lambda = \{\lambda_n\}_{n=1}^{\infty}$ belongs to $M(m_{\varphi}, \lambda_{\psi})_o$ if,
		and only if, the sequence $\left\{ \frac{\psi(n+1)-\psi(n)}{\varphi(n)} \lambda_n^* \right\}_{n=1}^{\infty}$ belongs to $\ell_1$.
		Since the space $\lambda_{\psi}$ is order continuous, so the rest follows as before from Theorem~\ref{THM: Essential norm Kothe sequence}.

		(v) Assume that $\limsup_{n \rightarrow \infty} \frac{\varphi(n)}{\psi(n)} = \infty$. Using the known equality between the fundamental
		functions of the space $X$ and the fundamental function of its K{\" o}the dual $X^{\times}$ (see \cite[Theorem~5.2, p.~66]{BS88}), that is,
		\begin{equation*}
			\varphi_X(n)\varphi_{X^{\times}}(n) = n \text{ for } n \in \mathbb{N},
		\end{equation*}
		and arguing as at the beginning of the proof of (iv), it is straightforward to see that
		\begin{equation*}
			\text{if } \limsup_{n \rightarrow \infty} \frac{\varphi(n)}{\psi(n)}
				= \infty \text{, then } \left\{ \frac{\varphi(n)}{n} \right\}_{n=1}^{\infty} \notin (\lambda_{\psi})^{\times} = m_{t/\psi(t)}.
		\end{equation*}
		In consequence, we have
		\begin{align*}
			M(\lambda_{\psi}, \lambda_{\varphi}) & = \left[ \left( \lambda_{\psi} \right)^{\times}\left( \frac{\varphi(t)}{t} \right) \right]^{(*)}
					\quad \text{(by Lemma~\ref{LEMMA: a'la KLM12/14}(ii))} \\
				& = \left[ m_{t/\psi(t)} \left( \frac{\varphi(t)}{t} \right) \right]^{(*)} \quad \text{(using \cite[Theorem~5.2, p.~112]{KPS82})}\\
				& = \left[ \left[ \ell_{\infty}\left( \frac{t}{\psi(t)} \right) \right]^{(*)} \left( \frac{\varphi(t)}{t} \right) \right]^{(*)}
					\quad \text{(because $m_{\phi} = [\ell_{\infty}(\phi)]^{(*)}$ as long as $p_{\phi} > 1$)}.
		\end{align*}
		Next, keeping in mind that the function $t \mapsto \frac{\varphi(t)}{t}$ is non-increasing and using the obvious fact that the product of
		two non-increasing functions is still non-increasing, we see that
		\begin{equation*}
			\left[ \left[ \ell_{\infty}\left( \frac{t}{\psi(t)} \right) \right]^{(*)} \left( \frac{\varphi(t)}{t} \right) \right]^{(*)}
				= \left[ \ell_{\infty}\left( \frac{t}{\psi(t)} \right) \left( \frac{\varphi(t)}{t} \right) \right]^{(*)}
				= \left[ \ell_{\infty}\left( \frac{\varphi}{\psi} \right) \right]^{(*)}.
		\end{equation*}
		Now, since $\left\{ \frac{\varphi(n)}{\psi(n)} \right\}_{n=1}^{\infty} \notin \ell_{\infty}$ (otherwise,
		$\left\{ \frac{\varphi(n)}{n} \right\}_{n=1}^{\infty} \in m_{t/\psi(t)}$ and this is contrary to our assumptions), so we can use
		Lemma~\ref{LEMMA: a'la Kolwicz OC} and, consequently, we get
		\begin{equation*}
			\left[ \left[ \ell_{\infty}\left( \frac{\varphi}{\psi} \right) \right]^{(*)} \right]_o
				= \left[ (\ell_{\infty})_o \left( \frac{\varphi}{\psi} \right) \right]^{(*)}
				= \left[ c_0 \left( \frac{\varphi}{\psi} \right) \right]^{(*)}.
		\end{equation*}
		Hence it is clear that the sequence $\lambda = \{\lambda_n\}_{n=1}^{\infty}$ belongs to $M(\lambda_{\psi}, \lambda_{\varphi})_o$ if, and only if,
		the sequence $\left\{ \frac{\varphi(n)}{\psi(n)} \lambda_n^* \right\}_{n=1}^{\infty}$ belongs to $c_0$. The use of Theorem~\ref{THM: Essential norm Kothe sequence}
		together with the information that the space $\lambda_{\varphi}$ is order continuous completes the proof.
		\end{proof}

redpajama_set_name: RedPajamaArXiv