text:
\section{Introduction}

A Banach space $X$ is said to have the \emph{Dunford-Pettis property} if, for any Banach space $Y$ every
weakly compact operator $T:X\to Y$ is completely continous.
Further, $X$ is said to have the \emph{reciprocal Dunford-Pettis property} if, for any Banach space $Y$ every  completely continous
operator $T:X\to Y$ is weakly compact.

Let us recall that $T$ is \emph{weakly compact} if the image by $T$ of the unit ball of $X$ is relatively weakly compact in $Y$. Further, $T$ is \emph{completely continous} if it maps weakly convergent sequences to norm convergent ones, or, equivalently, if it maps weakly Cauchy sequence to norm Cauchy (hence norm convergent) ones.

In general, these two classes of operators are incomparable.
For example, the identity on $\ell_2$ is weakly compact (due to reflexivity of $\ell_2$) but not completely continuous. On the other hand, the identity on $\ell_1$ is completely continuous
(by the Schur property) and not weakly compact.

It is obvious that reflexive spaces have the reciprocal Dunford-Pettis property (as any operator with reflexive domain is weakly compact) and that the Schur property implies the Dunford-Pettis property (as any operator defined on a space with the Schur property is completely continuous). Moreover, the space $L^1(\mu)$ has the Dunford-Pettis property for any non-negative $\sigma$-additive measure $\mu$ by (see  \cite[Theorem 1]{gro} or \cite[p. 61--62]{JoLi}). The space $C_0(\Omega)$, where $\Omega$ is a locally compact space, has both the Dunford-Pettis property
and the reciprocal Dunford-Pettis property (see \cite[p. 153, Theorem 4]{gro}).

In the present paper we investigate quantitative versions of the reciprocal Dun\-ford-Pettis property. It is a kind of a continuation of a recent paper \cite{qdpp} where quantification of the Dunford-Pettis property is studied.
It is also related to many  results on quantitative versions of certain theorems and properties. In particular,
quantitative versions of Krein's theorem were studied in \cite{f-krein,Gr-krein,GHM,CMR}, quantitative versions of Eberlein-\v{S}mulyan and Gantmacher theorems were investigated in \cite{AC-meas}, a quantitative version of James' compactness theorem was proved in \cite{CKS,Gr-james}, quantification of weak sequential continuity and of the Schur property was addressed in \cite{wesecom,qschur}.

The main idea behind quantitative versions is an attempt to replace the respective implication by an inequality. So, in case of the reciprocal Dunford-Pettis property we will try to replace the implication
$$T\mbox{ is completely continuous}\Rightarrow T\mbox{ is weakly compact} $$
by an inequality of the form
\begin{multline*} \mbox{measure of weak non-compactness of }T \\ \leq C\cdot \mbox{measure of non-complete-continuity of }T.\end{multline*}

We will use the same quantities as in \cite{qdpp} and in addition some equivalent ones.

In \cite{qdpp} it is proved, in particular, that both $L^1(\mu)$ spaces and $C_0(\Omega)$ spaces enjoy the strongest possible version of quantitative Dunford-Pettis property. In the present paper, we show that $C_0(\Omega)$ spaces have also a quantitative version of the reciprocal Dunford-Pettis property.

\section{Preliminaries}

In this section we define the quantities used in the present paper
and recall some known relationships between them. Most of the quantities we investigate are taken from \cite{qdpp} but we will
need a few more.

We will need to measure how far a given operator is from being
weakly compact, completely continuous or Mackey compact.

Our results are true both for real and complex spaces. In the real case sometimes better constants are obtained. By $\ef$ we will denote $\er$ or $\ce$, depending on whether we consider real or complex spaces.

\subsection{Measuring non-compactness of sets}
In this subsection we define measures on non-compactness, weak non-compactness and Mackey non-compactness of sets. We start by recalling the Hausdorff measure of non-compactness in metric spaces
and one of its equivalents.

Let $(X,\rho)$ be a metric space. If $A,B\subset X$ are two
nonempty sets, their non-symmetrized Hausdorff distance is defined
by
$$\widehat{\operatorname{d}}(A,B)=\sup\{\operatorname{d}(x,B):x\in A\}.$$
The Hausdorff measure of non-compactness of a nonempty set $A\subset
X$ is defined by
$$\chi(A)=\inf\left\{\widehat{\operatorname{d}}(A,F):F\subset X\mbox{ finite}\right\}.$$
Then $\chi(A)=0$ if and only if $A$ is totally bounded. In case $(X,\rho)$ is complete this is equivalent to relative compactness of $A$. We will need the following ``absolute'' equivalent:
$$\chi_0(A)=\inf\left\{\widehat{\operatorname{d}}(A,F):F\subset A\mbox{ finite}\right\}.$$
The quantity $\chi_0(A)$ depends only on the metric stucture of $A$ itself, not on the space $X$ where it is embedded. It is easy to check that
\begin{equation}
\chi(A)\le\chi_0(A)\le 2\chi(A)
\label{eq:chi-chi0}
\end{equation}
for any nonempty set $A\subset X$.

If $X$ is a Banach space and $A\subset X$ a nonempty bounded set,
we define the following two measures of weak noncompactness of $A$:
$$\begin{aligned}
\omega(A)&=\inf\left\{\widehat{\operatorname{d}}(A,K): K\subset X\mbox{ weakly compact}\right\},\\
\wk{A}&=\widehat{\operatorname{d}}\left(\wscl{A},X\right).
\end{aligned}$$
The quantity $\omega(A)$ is the de Blasi measure of weak noncompactness introduced in \cite{deblasi} and later investigated
for example in \cite{tylli-cambridge,AC-meas,qdpp}. The quantity $\wk{A}$ was used (with various notations) for example in \cite{f-krein,f-subswcg,Gr-krein,GHM,Gr-james,AC-meas,AC-jmaa,CMR,CKS}. These two quantities are not equivalent (see \cite{tylli-cambridge,AC-meas,qdpp}), while there
are several other natural quantities equivalent to the second one (see the papers quoted above). In general we have the following inequalities
\begin{equation}
\label{eq:kompaktnost} \wk{A}\le\omega(A)\le \chi(A)
\end{equation}
for any nonempty bounded subset $A\subset X$. These inequalities are easy,
the first one is proved for example in \cite{AC-meas}. The second one is obvious as finite sets are weakly compact.

Although the quantities $\omega(A)$ and $\wk{A}$ are not equivalent in general,
in some spaces they are equal. In particular, by \cite[Proposition 10.2 and Theorem 7.5]{qdpp} we have
\begin{equation}\begin{aligned}
\label{eq:c0L1}X=c_0(\Gamma)\mbox{ or }X=&L^1(\mu)\mbox{ for a non-negative $\sigma$-additive measure }\mu \\ &\Rightarrow \wk{A}=\omega(A)\mbox{ whenever }A\subset X\mbox{ is bounded.}
\end{aligned}
\end{equation}

We continue by measuring Mackey non-compactness. Let $X$ be still a Banach space.
Suppose that $A\subset X^*$ is a nonempty bounded set. Let us recall that the Mackey topology on $X^*$ is the topology of uniform convergence on weakly compact subsets of $X$. Moreover, the Mackey topology is complete, hence relatively compact subsets
coincide with totally bounded ones. So, $A$ is relatively Mackey compact if and only
if
$$A|_L=\{x^*|_L:x^*\in A\}$$
is totally bounded in $\ell_\infty(L)$ for each weakly compact set $L\subset X$.
This inspires the following definition:
$$\chi_m(A)=\sup\{\chi_0(A|_L): L\subset B_X\mbox{ weakly compact}\}.$$
This quantity measures Mackey non-compactness in the sense that $\chi_m(A)=0$ if and only if $A$ is relatively Mackey compact.

\subsection{Measuring of non-compactness of operators}
Let $T:X\to Y$ be a bounded operator. Since $T$ is compact (weakly compact)
if and only if $TB_X$ is relatively compact (relatively weakly compact) in $Y$,
it is natural to measure (weak) noncompactness of $T$ by a quantity applied
to $TB_X$. To simplify the notation we set
$$\chi(T)=\chi(T B_X),\chi_0(T)=\chi_0(T B_X),\omega(T)=\omega(T B_X),
\wk[Y]{T}=\wk[Y]{TB_X}.$$
Similarly, if $Y=Z^*$ for a Banach space, we set
$$\chi_m(T)=\chi_m(TB_X).$$

There are quantitative versions of Schauder's and Gantmacher's theorems
on compactness and weak compactness of the dual operators. More precisely,
if $T:X\to Y$ is a bounded linear operator, we have
\begin{gather}\label{eq:q-Schauder}
\frac12\chi(T^*)\le\chi(T)\le 2\chi(T^*),\qquad \frac12\chi_0(T^*)\le\chi_0(T)\le 2\chi_0(T^*),\\
\label{eq:Gant}\frac12\wk[X^*]{T^*}\le\wk[Y]{T}\le2\wk[X^*]{T^*},\\
\label{eq:noneq}\mbox{the quantities $\omega(T)$ and $\omega(T^*)$ are incomparable in general.}
\end{gather}

The first part of \eqref{eq:q-Schauder} follows from \cite{GM},
the second part follows for example from Lemma~\ref{lm-TA} below applied to the identity operator on $X$ and $A=B_X$. The assertion  \eqref{eq:Gant} follows from \cite[Theorem 3.1]{AC-meas}. The last assertion is proved in \cite[Theorem 4]{tylli-cambridge}.

\subsection{Measuring non-complete-continuity}
In this subsection we introduce two quantities which measure how far
an operator is from being completely continuous. The first one is that used in \cite{qdpp} and it is based on the definition of complete continuity given in the introduction.  Let us start by defining a quantity measuring how far
a given sequence is from being norm-Cauchy.

Let $(x_k)$ be a bounded sequence in a Banach space. Following
\cite{qschur,qdpp} we set
$$\ca{x_k}=\inf_{n\in\en}\sup\{\|x_k-x_l\|: k,l\ge n\}.$$
It is clear that $\ca{x_k}=0$ if and only if $(x_k)$ is norm-Cauchy.

Further, let $T:X\to Y$ be a bounded operator between Banach spaces.
Following \cite{qdpp} we set
$$\operatorname{cc}(T)=\sup\{\ca{Tx_k}:(x_k)\mbox{ is a weakly Cauchy sequence in }B_X\}.$$
It is clear that $T$ is completely continuous if and only if $\operatorname{cc}(T)=0$.

We will need one more equivalent quantity. This is inspired by an equivalent description of completely continuous operators. The operator $T$ is completely
continuous if and only if $T(L)$ is norm-compact for any weakly compact set $L\subset X$. Such operators are sometimes called Dunford-Pettis, so we will use the following
notation.
$$\operatorname{cc}_{DP}(T)=\sup\{\chi_0(TL):L\subset B_X\mbox{ weakly compact} \}.$$

The two quantities are equivalent. More precisely, we have
\begin{equation}
\label{eq:ccdp} \operatorname{cc}_{DP}(T)\le\operatorname{cc}(T)\le 2\operatorname{cc}_{DP}(T)
\end{equation}

Let us provide a proof. Suppose that $\operatorname{cc}_{DP}(T)>c>0$. Fix a weakly compact
set $L\subset B_X$ with $\chi_0(TL)>c$. It is easy to construct by induction
a sequence $(y_k)$ in $TL$ with $\|y_k-y_l\|>c$ for any $1\le l<k$.
Let $x_k\in L$ be such that $Tx_k=y_k$. By weak compactness of $L$ we can without
loss of generality suppose that $(x_k)$ is weakly convergent and hence weakly Cauchy.
Since $\ca{Tx_k}\ge c$, we get $\operatorname{cc}(T)\ge c$. This completes the proof of the first inequality.

To show the second one, suppose $\operatorname{cc}(T)>c>0$. Let $(x_k)$ be a weakly Cauchy sequence in $B_X$ with
$\ca{Tx_k}>c$. We can find two sequences $(m_k)$ and $(n_k)$ of natural numbers
such that for each $k\in\en$ we have $m_k<n_k<m_{k+1}$ and $\|Tx_{n_k}-Tx_{m_k}\|>c$. Set $y_k=\frac12(x_{n_k}-x_{m_k})$.
Then $(y_k)$ is a weakly null sequence in $B_X$ and hence $L=\{y_k:k\in\en\}\cup\{0\}$ is a weakly compact subset of $B_X$.

We claim that $\chi_0(TL)\ge\frac c2$. Suppose not. Then there is a finite set
$F\subset TL$ and
with $\widehat{\operatorname{d}}(TL,F)<\frac c2$. Since $F$ is finite, there is $h\in F$ and  a subsequence $(y_{k_l})$ such that $\|T y_{k_l}-h\|<\frac c2$ for each $l\in\en$.
Since $(T y_{k_l}-h)$ weakly converges to $-h$, we get $\|h\|\le \frac c2$, so
$h=0$. (Any other element of $TL$ has norm strictly greater than $\frac c2$.)

So,  $\|T y_{k_l}\|<\frac c2$ for each $l\in\en$. But this is a contradiction with the choice of the sequence $(y_k)$.

Therefore $\chi_0(TL)\ge\frac c2$ and so $\operatorname{cc}_{DP}(T)\ge\frac c2$.
This completes the proof of the second inequality.

\section{Main results}

Our first main result is the following theorem which says that
spaces $C_0(\Omega)$ enjoy a quantitative version of the reciprocal Dunford-Pettis property. The formulation combines this result with
a result of \cite{qdpp} on the quantitative Dunford-Pettis property. We thus get that for operators defined on a $C_0(\Omega)$ space
the weak compactness and complete continuity are quantitatively equivalent.

\begin{thm}\label{main-t} Let $X=C_0(\Omega)$, where $\Omega$ is a Hausdorff locally compact space. Let $Y$ be any Banach space and $T:X\to Y$ be a bounded linear operator. Then we have
$$\frac1{4\pi}\wk[Y]{T}\le\operatorname{cc}(T)\le 4\wk[Y]{T}.$$
More precisely, the following inequalities hold:
\begin{equation}
\label{eq:main1}
\begin{aligned}
\frac1{4\pi}\wk[Y]{T}\le\frac1{2\pi}&\wk[X^*]{T^*}= \frac1{2\pi}\omega(T^*)
\le\operatorname{cc}_{DP}(T)\\&\le \operatorname{cc}(T)\le 2\omega(T^*) = 2\wk[X^*]{T^*}\le 4\wk[Y]{T}.
\end{aligned}
\end{equation}
In case of real-valued functions the constant $\pi$ in the above inequalities can be everywhere replaced by the constant $2$.
\end{thm}

This theorem says, in particular, that the quantities $\operatorname{cc}(T)$, $\wk[Y]{T}$, $\wk[X^*]{T^*}$ and $\omega(T^*)$ are equivalent
for any bounded linear operator $T:C_0(\Omega)\to Y$.

The first inequality, as well as the last one follows from \eqref{eq:Gant}. The two equalities follow from \eqref{eq:c0L1}
as $C_0(\Omega)^*$ is of the form $L^1(\mu)$. The inequality $\operatorname{cc}(T)\le 2\omega(T^*)$ follows from \cite[Theorem 5.2]{qdpp} as
$C_0(\Omega)$ has the Dunford-Pettis property. The inequality
$\operatorname{cc}_{DP}(T)\le\operatorname{cc}(T)$ follows from \eqref{eq:ccdp}.

Finally, the main new result is the inequality
$\frac1{2\pi}\omega(T^*)\le\operatorname{cc}_{DP}(T)$ which follows from
Theorems~\ref{cc-Mc} and~\ref{t:M=w} below.

\medskip

It is natural to ask whether also the quantity $\omega(T)$ is equivalent
to the remaining ones. The answer is positive in case $\Omega$ is scattered. Indeed, then $X^*$ is isometric to $\ell_1(\Omega)$ and hence we can use \cite[Theorem 8.2]{qdpp}.

In general the answer is negative as witnessed by the following example.

\begin{example}\label{main-ex} There is a separable Banach space $Y$ such that for any uncountable separable metrizable locally compact space $\Omega$ there is
a sequence $(T_n)$ of bounded operators $T_n:C_0(\Omega)\to Y$ such that
$$\lim_{n\to\infty}\frac{\wk[Y]{T_n}}{\omega(T_n)}
=\lim_{n\to\infty}\frac{\operatorname{cc}(T_n)}{\omega(T_n)}=0.$$
\end{example}

This example is proved in Section~\ref{sec-exa} below.

\section{Complete continuity and Mackey compactness}

In this section we prove a quantitative version of a particular case
of \cite[Lemma 2]{gro}. The quoted lemma implies that an operator
$T:X\to Y$ is completely continuous if and only if its adjoint $T^*$
is Mackey-compact (i.e., $T^*(B_{Y^*})$ is relatively Mackey compact in $X^*$). A quantitative version is the following theorem.

\begin{thm}\label{cc-Mc}
Let $X$ and $Y$ be Banach spaces and $T:X\to Y$ an operator.
Then
$$\frac12\chi_m(T^*)\le\operatorname{cc}_{DP}(T)\le 2\chi_m(T^*)$$
\end{thm}

The proof of this theorem is done by a refinement of the arguments
in \cite{gro}. The first tool used in \cite{gro} is the  Arzel\`a-Ascoli theorem. We will use its quantitative version for a special case of $1$-Lipschitz functions on a metric space. It is contained in the following lemma.

\begin{lemma}\label{lm-ascoli} Let $(M,\rho)$ be a metric space and $A\subset \ell_\infty(M)$ be a bounded set formed by $1$-Lipschitz functions.
Then $\chi_0(A)\le 2\chi_0(M)$.
\end{lemma}

\begin{proof}
Fix an arbitrary $c>\chi_0(M)$ and $\varepsilon>0$. Then there is a finite set $F\subset M$ such that $\widehat{\operatorname{d}}(M,F)<c$. Let us define the mapping
$\Phi:A\to\ef^F$ by $\Phi(f)=f|_F$ for $f\in A$. Let us equip $\ef^F$ with the $\ell_\infty$ norm. Then $\Phi(A)$ is
a bounded subset of $\ef^F$, so it is also totally bounded (as $F$
is finite). It follows that there is a finite set $B\subset A$ such that $\widehat{\operatorname{d}}(\Phi(A),\Phi(B))<\varepsilon$.

We will show that $\widehat{\operatorname{d}}(A,B)\le 2c+ \varepsilon$. To this end take an arbitrary $f\in A$. By the choice of $B$ there is some $g\in B$ with $\|\Phi(f)-\Phi(g)\|< \varepsilon$. Fix an arbitrary $x\in M$.
We can find $x_0\in F$ with $\rho(x_0,x)<c$. Then
$$\begin{aligned}
|f(x)-g(x)|&\le |f(x)-f(x_0)|+|f(x_0)-g(x_0)|+|g(x_0)-g(x)|
\\&\le \rho(x,x_0)+ \|\Phi(f)-\Phi(g)\| + \rho(x_0,x)
<2c+ \varepsilon.
\end{aligned}$$

Hence $\|f-g\|<2c+ \varepsilon$, so $\operatorname{d}(f,B)<2c+ \varepsilon$. Since $f\in A$ is arbitrary, we get
$\widehat{\operatorname{d}}(A,B)\le2c+ \varepsilon$, in particular $\chi_0(A)\le 2c+ \varepsilon$. Finally, since $c>\chi_0(M)$ and $\varepsilon>0$ are arbitrary, we get $\chi_0(A)\le 2\chi_0(M)$.
\end{proof}

The next lemma is a quantitative version of a part of \cite[Lemma 3]{gro}. It is formulated in a very abstract setting.

\begin{lemma}\label{twosets}
Let $A$ be a nonempty set, $B$ a nonempty bounded subset of $\ell_\infty(A)$. Let $\varphi:A\to\ell_\infty(B)$ be
defined by
$$\varphi(a)(b)=b(a),\qquad a\in A, b\in B.$$
Then
$$\frac12\chi_0(B)\le\chi_0(\varphi(A))\le 2\chi_0(B).$$
\end{lemma}

\begin{proof} It is clear that $\varphi(A)$ is a bounded subset of
$\ell_\infty(B)$. Moreover, it is formed by $1$-Lipschitz functions.
Indeed, let $a\in A$ be arbitrary. Fix $b_1,b_2\in B$. Then
$$|\varphi(a)(b_1)-\varphi(a)(b_2)|=|b_1(a)-b_2(a)|\le\|b_1-b_2\|.$$
So, by Lemma~\ref{lm-ascoli} we have
$$\chi_0(\varphi(A))\le 2\chi_0(B).$$
To show the second inequality, let us define a canonical
embedding $\psi:B\to\ell_\infty(\varphi(A))$ by
$$\psi(b)(\varphi(a))=b(a),\qquad b\in B, a\in A.$$
This is a well-defined mapping. Indeed, if $a_1,a_2\in A$ are such that $\varphi(a_1)=\varphi(a_2)$, then for any $b\in B$
we have
$$b(a_1)=\varphi(a_1)(b)=\varphi(a_2)(b)=b(a_2).$$
Moreover, $\psi$ is an isometry of $B$ onto $\psi(B)$. Indeed, let $b_1,b_2\in B$ be arbitrary. Then
$$\begin{aligned}\|\psi(b_1)-\psi(b_2)\|& = \sup \{|\psi(b_1)(x)-\psi(b_2)(x)|: x\in \varphi(A)\}
\\&= \sup \{|\psi(b_1)(\varphi(a))-\psi(b_2)(\varphi(a))|: a\in A\}
\\&= \sup \{|b_1(a)-b_2(a)|: a\in A\}  = \|b_1-b_2\|.\end{aligned}$$
It follows that $\psi(B)$ is isometric to $B$. Moreover, $\psi(B)$ is a bounded subset of $\ell_\infty(\varphi(A))$ made from $1$-Lipschitz functions (the argument is the same as the one used above in the proof of the first inequality). Hence, using Lemma~\ref{lm-ascoli} we get
$$\chi_0(B)=\chi_0(\psi(B))\le 2\chi_0(\varphi(A)).$$
This completes the proof.
\end{proof}

The next lemma is a quantitative version of \cite[Lemma 2]{gro}
applied to a single set rather than to a family of sets.

\begin{lemma}\label{lm-TA} Let $T:X\to Y$ be an operator between Banach spaces.
Let $A\subset X$ be a bounded set.  Let $\psi: X^*\to\ell_\infty(A)$ denote the restriction mapping.
Then
$$\frac12\chi_0(TA)\le \chi_0(\psi(T^*B_{Y^*}))\le 2\chi_0(TA).$$
\end{lemma}

\begin{proof}
Let us denote by $\kappa$ the canonical embedding of $Y$ into $\ell_\infty(B_{Y^*})$. Let us define an embedding $\varphi:B_{Y^*}\to \ell_\infty(\kappa(TA))$ by
$$\varphi(y^*)(\kappa(y))=y^*(y),\qquad y\in TA, y^*\in B_{Y^*}.$$
Using Lemma~\ref{twosets} and the fact that $\kappa$ is an isometry,
we obtain
$$\frac12\chi_0(TA)\le \chi_0(\varphi(B_{Y^*}))\le 2\chi_0(TA).$$
Further, $\psi(T^*B_{Y^*})$ is isometric to $\varphi(B_{Y^*})$.
Indeed, the mapping
$\alpha:\varphi(B_{Y^*})\to\psi(T^*B_{Y^*})$
defined by
$$\alpha(\varphi(y^*))=\psi(T^*y^*),\qquad y^*\in B_{Y^*},$$
is an onto isometry. Let $y_1^*,y_2^*\in B_{Y^*}$ be arbitrary.
Then
$$\begin{aligned}\|\psi(T^*y_1^*)-\psi(T^*y_2^*)\|
&=\sup\{|(T^*y_1^*)(a)-(T^*y_2^*)(a)|:a\in A\}
\\&=\sup\{|y_1^*(Ta)-y_2^*(Ta)|:a\in A\}
=\|\varphi(y_1^*)-\varphi(y_2^*)\|.\end{aligned}$$
It follows that $\alpha$ is a well-defined isometry. Moreover,
it is clear that it is surjective.
Hence we get $\chi_0(\psi(T^*B_{Y^*}))=\chi_0(\varphi(B_{Y^*}))$.
So,
$$\frac12\chi_0(TA)\le \chi_0(\psi(T^*B_{Y^*}))\le 2\chi_0(TA).$$
and the proof is completed.
\end{proof}

Now we are ready to proof the theorem.

\begin{proof}[Proof of Theorem~\ref{cc-Mc}]
The inequalities follow from Lemma~\ref{lm-TA} by taking supremum over all weakly compact sets $A\subset B_X$.
\end{proof}

\section{Weak compactness and Mackey compactness in spaces of measures}

Within this section $\Omega$ will denote a locally compact space
and $\M(\Omega)$ will be the space of all finite Radon measures on $\Omega$ equipped with the total variation norm and considered as the dual space to $C_0(\Omega)$. We will consider simultaneuously the real version (i.e., $C_0(\Omega)$ are real-valued functions and $\M(\Omega)$ are signed measures) and complex version (i.e., $C_0(\Omega)$ are complex functions and $\M(\Omega)$ are complex measures) of these spaces.

We will prove a quantitative version of a result of \cite{gro} saying that in $\M(\Omega)$ weakly compact sets coincide with Mackey compact ones. In \cite{gro} this result is hidden in Corollary to Theorem 2 on page 149 and in the first two lines on page 150.
The promised quantitative version is the following theorem.

\begin{thm}\label{t:M=w}
Let $A\subset \M(\Omega)$ be a bounded set.
Then
$$\frac12\chi_m(A)\le \omega(A)\le \pi\chi_m(A).$$
In case of real measures, the constant $\pi$ can be replaced by $2$.
\end{thm}

The first step to the proof is a quantiative version of a modification of \cite[Theorem 2]{gro}. In the quoted theorem several equivalent conditions for weak compactness of a subset of $\M(\Omega)$ are summarized. We will prove  quantitative versions of some of them and of some others. They are contained in
the following proposition.

Let us comment this result a bit.
The second quantity is inspired by condition (2) of \cite[Theorem 2]{gro}. The first inequality follows directly from \cite{gro}.
The third quantity is inspired by condition (3) of the quoted theorem. The second inequality is easy and is done by copying the
respective proof from \cite{gro}.

It is also easy to quantify the implication (3)$\Rightarrow$(4)
from \cite{gro}, but we were not able to quantify the last implication saying that the condition (4) implies weak compactness.
Instead, we used the fourth quantity. The proof of the third inequality required a new idea. The last inequality is proved using
technics from \cite{qdpp}.

\begin{prop}\label{q-dual}
Let $A$ be a bounded subset of $\M(\Omega)$. Then
$$\begin{array}{c}
\omega(A)\\
\rotatebox{90}{$\,\leq\ $}
\\
\sup\left\{ \limsup\limits_{k\to\infty} \sup\limits_{\mu\in A} \left|\int f_k\,\mbox{\rm d}\mu\right| : (f_k)\mbox{ is a weakly null sequence in }B_{C_0(\Omega)}\right\}
\\ \rotatebox{90}{$\,\leq\ $} \\
\sup\left\{ \limsup\limits_{k\to\infty} \sup\limits_{\mu\in A} |\mu(U_k)| :\begin{aligned}(U_k)&\mbox{ is a sequence of}\\ &\mbox{ pairwise disjoint}\mbox{ open subsets of }\Omega\end{aligned}\right\}\\
\rotatebox{90}{$\,\leq\ $}
\\
\sup\left\{ \limsup\limits_{k\to\infty} \sup\limits_{\mu\in A} |\mu(F_k)| : \begin{aligned}(F_k)&\mbox{ is a sequence of}\\ &\mbox{ pairwise disjoint}\mbox{ compact subsets of }\Omega\end{aligned}\right\}
\\ \rotatebox{90}{$\,\leq\ $} \\
\frac1\pi\omega(A).
\end{array}$$
In the real case the constant $\frac1\pi$ can be replaced by $\frac12$.
\end{prop}

\begin{proof} The first two inequalites easily follow from \cite{gro}.
Indeed, let $(f_k)$ be any weakly null sequence in $B_{C_0(\Omega)}$. Fix an arbitrary $c>\omega(A)$. Then there
is a weakly compact set $H\subset \M(\Omega)$ such that $A\subset H+c B$, where $B$ denotes the unit ball of $\M(\Omega)$.  By \cite[Theorem 2]{gro}
$$\lim \int f_k\,\mbox{\rm d}\mu = 0\qquad \mbox{uniformly for }\mu\in H.$$
Thus
$$\begin{aligned}
\limsup\limits_{k\to\infty} \sup\limits_{\mu\in A} \left|\int f_k\,\mbox{\rm d}\mu\right|
&\le\limsup\limits_{k\to\infty} \sup\limits_{\mu\in H+cB} \left|\int f_k\,\mbox{\rm d}\mu\right|
\\&\le
\limsup\limits_{k\to\infty}\left( \sup\limits_{\mu\in H} \left|\int f_k\,\mbox{\rm d}\mu\right|+c
\sup\limits_{\mu\in B} \left|\int f_k\,\mbox{\rm d}\mu\right|\right)
\\&\le \limsup\limits_{k\to\infty}\left( \sup\limits_{\mu\in H} \left|\int f_k\,\mbox{\rm d}\mu\right|+c\right)= c.\end{aligned}$$
Since $c>\omega(A)$ is arbitrary, we get
$$\limsup\limits_{k\to\infty} \sup\limits_{\mu\in A} \left|\int f_k\,\mbox{\rm d}\mu\right|\le\omega(A),$$
so the first inequality follows.

Let us show the second inequality. If the third quantity is zero, the inequality is obvious. So, suppose that the quantity is strictly positive and fix an arbitrary smaller positive constant $c$.
Then there is a sequence $(U_k)$ of pairwise disjoint open sets in $\Omega$ and a sequence $(\mu_k)$ in $A$ such that $|\mu_k(U_k)|>c$ for each $k\in\en$. For each $k\in\en$ we can find a continuous function $f_k:\Omega\to[0,1]$ supported by a compact subset of $U_k$ such that $\left|\int f_k\,\mbox{\rm d}\mu_k\right|>c$. Since $(f_k)$ is a bounded sequence in $C_0(\Omega)$ which pointwise converges to zero, it is weakly null. This completes the proof of the second inequality.

Let us proceed with the third inequality. Obviously, it is enough
to prove the following lemma.

\begin{lemma}\label{compact-open}
Let $(\mu_k)$ be a bounded sequence in $\M(\Omega)$, let $(F_k)$ be a sequence of pairwise disjoint compact subsets of $\Omega$. Let $c>0$ be such that $|\mu_k(F_k)|>c$ for each $k\in\en$. Then for any $\varepsilon>0$
there is a subsequence $(\mu_{k_n})$ and a sequence $(U_n)$ of  pairwise disjoint open subsets such that $|\mu_{k_n}(U_n)|>c- \varepsilon$ for each $n\in\en$:
\end{lemma}

\begin{proof} Let $(\mu_k)$, $(F_k)$ and $c$ satisfy the assumptions. Let $\varepsilon>0$ be arbitrary. Set
$$\gamma=\sup\{\|\mu_k\|:k\in\en\}$$
and find $N\in\en$ with $\frac1N<\frac{\varepsilon}{\gamma}$.
We will construct by induction for each $n\in\en$
a natural number $k_n$, sets of natural numbers $M_n$
and an open set $U_n\subset \Omega$ such that the following
conditions are satisfied for each $n\in\en$:

\begin{itemize}
	\item $k_n\in M_n$,
	\item $M_{n+1}\subset M_n\setminus\{k_n\}$, $M_{n+1}$ is infinite,
	\item $|\mu_k|(F_k\cap \overline{U_n})<\frac{\varepsilon}{2^n}$ for each $k\in M_{n+1}$,
	\item $\overline{U_n}\cap\overline{U_j}=\emptyset$ for $j<n$,
	\item $|\mu_{k_n}(U_n)|>c- \varepsilon$.
\end{itemize}

We start by setting $M_1=\en$.

Suppose that $n\in\en$ and that we have already constructed $M_j$ for $j\le n$ and $k_j$, $U_j$ for $j<n$. Since $M_n$ is infinite,
we can fix a subset $H\subset M_n$ of cardinality $2^n N$. We can find  open sets $(V_h)_{h\in H}$ with pairwise disjoint closures such that $F_h\subset V_h$ for each $h\in H$.

For any $k\in M_n\setminus H$ there is some $h(k)\in H$ such that
$|\mu_k|\left(F_k\cap \overline{V_{h(k)}}\right)<\frac{\varepsilon}{2^n}$.
Fix $k_n\in H$ such that
$$M_{n+1}=\{k\in M_n\setminus H: h(k)=k_{n} \}$$
is infinite.

By the induction hypothesis we have $|\mu_{k_n}|(F_{k_n}\cap\overline{U_j})<\frac{\varepsilon}{2^j}$ for any $j<n$. It follows that $|\mu_{k_n}(F_{k_n}\setminus\bigcup_{j<n} \overline{U_j})|>c- \varepsilon$. It follows that there is a compact set $L\subset F_{k_n}\setminus\bigcup_{j<n} \overline{U_j}$ with $|\mu_{k_n}(L)|>c- \varepsilon$.
Finally, we can find an open set $U_n$ such that
\begin{itemize}
	\item $L\subset U_n\subset V_{k_n}$,
	\item $\overline{U_n}\cap\overline{U_j}=\emptyset$ for $j<n$,
	\item $|\mu_{k_n}(U_n)|>c- \varepsilon$.
\end{itemize}
This completes the induction step and the lemma is proved.
\end{proof}

Now we come back to the proof of Proposition~\ref{q-dual}.
It remains to prove the last inequality. If $\omega(A)=0$, it is trivial. Suppose that $\omega(A)>0$ and fix an arbitray $c\in(0,\omega(A))$. We will proceed in three steps.

In the first step we reduce the problem to a statement on
$L^1$ spaces on a finite measure space. Since $\M(\Omega)$ is an $L^1$ space on a $\sigma$-additive non-negative measure (infinite, of course), by \cite[Theorem 7.5]{qdpp} there is a sequence $(\mu_k)$ in $A$ such that $\operatorname{d}(\clu{\M(\Omega)}{\mu_k})>c$.
Set $\mu=\sum_{k=1}^\infty 2^{-k}|\mu_k|$. Then $\mu$ is a finite measure and $L^1(\mu)$ is canonically isometrically embedded into $\M(\Omega)$ onto a subspace containing the sequence $(\mu_k)$.
Set $\tilde A=\{\mu_k:k\in \en\}$.
By \cite[Proposition 7.1]{qdpp}, the quantity $\omega(\tilde A)$ is the same in $\M(\Omega)$ as in the subspace identified with $L^1(\mu)$. In particular $\omega(\tilde A)>c$ in $L^1(\mu)$.

The second step will be the following lemma.

\begin{lemma} Let $\mu$ be a finite $\sigma$-additive nonnegative measure and $\tilde A\subset L^1(\mu)$ be a bounded subset satisfying $\omega(\tilde A)>c>0$. Then for any $\varepsilon>0$ there is a sequence $(f_k)$ in $\tilde A$ and a sequence $(H_k)$ of pairwise disjoint measurable sets satisfying $\int_{H_k}|f_k|>c- \varepsilon$ for each $k\in\en$.
\end{lemma}

\begin{proof}
We will use the construction from the proof of \cite[Proposition 7.1]{qdpp}. Let $B$ denote the unit ball of $L^\infty(\mu)$. Then $B$ is a weakly compact subset of $L^1(\mu)$ and hence $\widehat{\operatorname{d}}(\tilde A,\alpha B)\ge\omega(\tilde A)>c$ for each $\alpha>0$.
Set $\gamma=\sup\{\|f\|:f\in\tilde A\}$.
We will construct by induction positive numbers $\alpha_k$ and functions $f_k\in\tilde A$ such that
\begin{itemize}
	\item $\operatorname{d}(f_k,\alpha_k B)>c$,
	\item $\alpha_{k+1}>\alpha_k$,
	\item $\int_E |f_j|\,\mbox{\rm d}\mu<\frac{\varepsilon}{2^k}$ whenever $j\le k$ and $\mu(E)\le\frac\gamma{\alpha_{k+1}}$.
\end{itemize}

Set $\alpha_1=1$. Having $\alpha_k$, we can find $f_k\in\tilde A$ satisfying the first condition. Further, by absolute continuity we can find $\alpha_{k+1}>\alpha_k$ such that the third condition is satisfied. This completes the construction.

Set $E_k=\{t:|f_k(t)|>\alpha_k\}$. Then
$$\int_{E_k} |f_k|\,\mbox{\rm d}\mu=\operatorname{d}(f_k,\alpha_k B)>c.$$
Further,
$\mu(E_k)\le\frac{\|f_k\|}{\alpha_k}\le\frac{\gamma}{\alpha_k}$.
It follows that for any $j<k$ we have
$$\int_{E_k}|f_j|\,\mbox{\rm d}\mu<\frac{\varepsilon}{2^{k-1}}.$$

Finally, set $H_k=E_k\setminus\bigcup_{n>k}E_n$. Then $(H_k)$ is a sequence of pairwise disjoint measurable sets and
$$\int_{H_k} |f_k|\,\mbox{\rm d}\mu
\ge \int_{E_k} |f_k|\,\mbox{\rm d}\mu - \sum_{n>k}\int_{E_n} |f_k|\,\mbox{\rm d}\mu
>c- \varepsilon.$$
This completes the proof.
\end{proof}

Now we return to the proof of Proposition~\ref{q-dual}.
Using the first step and the above lemma we obtained, given $\varepsilon>0$, a sequence $(\mu_k)$ in $A$ (a subsequence of the sequence chosen in the first step) and a sequence $(H_k)$ of pairwise disjoint sets which are measurable for each $\mu_k$ such that $|\mu|(H_k)>c-\varepsilon$.

As the third step we will use the following lemma.

\begin{lemma} Let $\mu$ be a finite complex measure. Let $H$ be a measurable set with $|\mu|(H)>d>0$. Then there is a measurable subset $\tilde H\subset H$ with $|\mu(\tilde H)|>\frac d\pi$.

If $\mu$ is real-valued, $\tilde H$ can be found to satisfy  $|\mu(\tilde H)|>\frac d2$.
\end{lemma}

\begin{proof} The real-valued case is easy using the Hahn decomposition $\mu=\mu^+-\mu^-$.

Let us prove the general case. Since $|\mu|(H)>c$, there are
pairwise disjoint measurable sets $D_1,\dots,D_p\subset H$ such that
$\sum_{j=1}^p |\mu(D_j)|>c$. By \cite[Lemma 6.3]{rudin} there is a subset $J\subset\{1,\dots,p\}$ such that $\left|\sum_{j\in J} \mu(D_j)\right|>\frac c\pi$. It is enough to take $\tilde H=\bigcup_{j\in J}D_j$.
\end{proof}

Now we are ready to finish the proof. Find $\mu_k$-measurable sets $\tilde H_k\subset H_k$ with $|\mu_k(\tilde H_k)|>\frac {c-\varepsilon}\pi$
and then a compact subset $F_k\subset \tilde H_k$ with
$|\mu_k(F_k)|>\frac {c-\varepsilon}\pi$.

In case of real measures we can obtain $|\mu_k(F_k)|>\frac {c-\varepsilon}2$.
This completes the proof of the last inequality.
\end{proof}

The last lemma of this section is a quantitative version
of \cite[p. 134, Corollary to Lemma 3]{gro}.

\begin{lemma}\label{mac-seq} Let $X$ be a Banach space and $A\subset X^*$ a bounded set. Then
\begin{multline*}\frac18\chi_m(A)\le \sup\left\{\limsup_{k\to\infty}\sup_{x^*\in A}
|x^*(x_k)| : (x_k)\mbox{ is a weakly null sequence in }B_X\right\}
\\ \le\chi_m(A)\end{multline*}
\end{lemma}

\begin{proof}
Let $(x_k)$ be a weakly null sequence in $B_X$. Set
$L=\{0\}\cup\{x_k:k\in\en\}$. Then $L$ is a weakly compact
subset of $B_X$, so $\chi_0(A|_L)\le\chi_m(A)$. Let $\varepsilon>0$
be arbitrary. Then there is a finite set $F\subset A$ such
that $\widehat{\operatorname{d}}(A|_L,F|_L)<\chi_m(A)+\varepsilon$. For each $k\in\en$ we have
$$\sup_{x^*\in A} |x^*(x_k)|\le \sup_{x^*\in F} |x^*(x_k)| + \chi_m(A)+\varepsilon.$$
Since
$$\lim_{k\to\infty}\sup_{x^*\in F} |x^*(x_k)|=0,$$
we get
$$\limsup_{k\to\infty}\sup_{x^*\in A} |x^*(x_k)|\le \chi_m(A)+\varepsilon.$$
Since $\varepsilon>0$ is arbitrary, this completes the proof of the
second inequality.

Let us prove the first one. If $\chi_m(A)=0$, the inequality is obvious. So, suppose that $\chi_m(A)>c>0$. Fix a weakly compact
set $L\subset B_X$ such that $\chi_0(A|_L)>c$.
Let $\varphi:X\to\ell_\infty(A)$ be the canonical mapping
defined by $$\varphi(x)(x^*)=x^*(x), \qquad x\in X, x^*\in A.$$
Further, let us define $\varphi_0:L\to \ell_\infty(A|_L)$ by
$$\varphi_0(x)(x^*|_L)= x^*(x) \qquad x\in L, x^*\in A.$$
It is clear that $\varphi_0$ is well defined and that $\varphi_0(L)$ is isometric to $\varphi(L)$.
By Lemma~\ref{twosets} we have
$\chi_0(\varphi_0(L))>\frac c2$, hence also $\chi_0(\varphi(L))>\frac c2$. Therefore,
we can construct by induction a sequence $(x_k)$ in $L$ such that
$$\|\varphi(x_k)-\varphi(x_l)\|>\frac c2,\qquad 1\le l<k.$$
Since $L$ is weakly compact, we can suppose without loss of generality that the sequence $(x_k)$ weakly converges to some $x\in L$. Then
$$\|\varphi(x_k)-\varphi(x)\|>\frac c4$$
for all $k\in\en$ with at most one exception. So, suppose without loss of generality that it holds for each $k\in\en$. Hence, if we set
$y_k=\frac12(x_k-x)$, then $(y_k)$ is a weakly null sequence in $B_X$ and
$$\sup_{x^*\in A}|x^*(y_k)|=\|\varphi(y_k)\|>\frac c8.$$
This completes the proof of the first inequality.
\end{proof}

Finally, we are ready to prove the theorem.

\begin{proof}[Proof of Theorem~\ref{t:M=w}.]
Let $A\subset \M(\Omega)$ be a bounded set.
It follows from Proposition~\ref{q-dual}
and Lemma~\ref{mac-seq} that
$$\frac18\chi_m(A)\le\omega(A)\le\pi\chi_m(A).$$
The second inequality is the announced one, the first one still
needs to be improved. So, suppose that $\omega(A)<c$. Fix $H\subset \M(\Omega)$ weakly compact such that $\widehat{\operatorname{d}}(A,H)<c$. Then $H$ is a bounded subset of $\M(\Omega)$ satisfying $\omega(H)=0$, hence also
$\chi_m(H)=0$ (we apply the above inequality to $H$).

Given $L\subset B_{C_0(\Omega)}$ weakly compact and $\varepsilon>0$, there is a finite set $F\subset H$ such
that $\widehat{\operatorname{d}}(H|_L,F|_L)<\varepsilon$. Then clearly
$\widehat{\operatorname{d}}(A|_L,F|_L)<c+ \varepsilon$, so $\chi(A|_L)\le c+ \varepsilon$.
By \eqref{eq:chi-chi0} we get
$\chi_0(A|_L)\le 2(c+ \varepsilon)$.

It follows that $\chi_m(A)\le 2c$, so $\chi_m(A)\le 2\omega(A)$.
This completes the proof of the first inequality.
\end{proof}

\section{Proof of Example~\ref{main-ex}}\label{sec-exa}

Let $\Delta=\{-1,1\}^\en$ be the Cantor space.
Let $\Omega$ be an uncountable separable metrizable locally compact space.
Denote by $K$ its one-point compactification. Then $K$ is an uncountable metrizable compact space, therefore $C(K)$ is isomorphic to $C(\Delta)$ by Milyutin's theorem \cite{mil}
(see, e.g., \cite[Theorem 2.1]{rosenthal}). So, $C_0(\Omega)$, which is a hyperplane in $C(K)$, is isomorphic to
a hyperplane in $C(\Delta)$. Since hyperplanes in $C(\Delta)$ are isomorphic to $C(\Delta)$, we can conclude that $C_0(\Omega)$ is isomorphic to $C(\Delta)$.

Fix an onto isomorphism $Q:C_0(\Omega)\to C(\Delta)$. Then
$$\frac1{\|Q^{-1}\|} B_{C(\Delta)}\subset Q(B_{C_0(\Omega)})\subset\|Q\|
 B_{C(\Delta)},$$
 and hence for any bounded opearator $T:C(\Delta)
\to Y$ we have
$$
\begin{aligned}
\operatorname{cc}(TQ)&\le \|Q\|\operatorname{cc}(T),\\
\wk[Y]{TQ}&=\wk[Y]{TQ(B_{C_0(\Omega)})}\le\wk[Y]{T(\|Q\|B_{C(\Delta)})}
=\|Q\|\wk[Y]{T},\\
\omega(TQ)&=\omega(TQ(B_{C_0(\Omega)}))\ge \omega\left(T\left(\frac1{\|Q^{-1}\|}B_{C(\Delta)}\right)\right)=\frac1{\|Q^{-1}\|}\omega(T).
\end{aligned}$$

It follows that it is enough to restrict ourselves to the case $\Omega=\Delta$.

Let $\mu$ denotes the product probability measure on $\Delta$. I.e.,
$\mu$ is the countable power of the uniform probability measure
$\frac12(\delta_1+\delta_{-1})$ on the two-point set $\{-1,1\}$.
Let us define an equivalent norm $\|\cdot\|_n$ on $C(\Delta)$ by
$$\|f\|_n = \frac1n\|f\|+\int_\Delta |f| \,\mbox{\rm d}\mu,\qquad f\in C(\Delta).$$
Set $Y_n=(C(\Delta),\|\cdot\|_n)$ and let
$$Y =\left(\;\bigoplus_{n\in\en} Y_n\right)_{c_0}$$
be the $c_0$-sum of the spaces $Y_n$.

Let $Q_n: C(\Delta)\to Y_n$ be the identity mapping, $I_n:Y_n\to Y$ the canonical inclusion made by completing by zeros and $P_n:Y\to Y_n$ be the canonical projection. Let us define $T_n= I_nQ_n$. Then $T_n$ is an operator from $C(\Delta)$ to $Y$.
The proof will be completed if we show that
$$\frac1{4\pi}\wk[Y]{T_n}\le\operatorname{cc}(T_n)\le \frac2n,\qquad \omega(T_n)\ge \frac12.$$
The first inequality follows from Theorem~\ref{main-t}. Let us show the second one. Let $(f_k)$ be a weakly Cauchy sequence in $B_{C(\Delta)}$.
Then the sequence $(f_k)$ pointwise converges to a (not necessarily continuous) function $f$. Since the sequence is uniformly bounded,
the Lebesgue dominated convergence theorem shows that
$$\lim_{k\to\infty}\int_\Delta |f_k-f|\,\mbox{\rm d}\mu=0.$$
In particular, $(f_k)$ is Cauchy in the norm of $L^1(\mu)$. Thus,
given $\varepsilon>0$ there is $k_0\in\en$ such that whenever
$k,l\ge k_0$ we have $\|f_k-f_l\|_{L^1(\mu)}<\varepsilon$, hence
$$\|f_k-f_l\|_n = \frac1n \|f_k-f_l\| + \int_\Delta |f_k-f_l|\,\mbox{\rm d}\mu
<\frac2n+ \varepsilon.$$
It follows that for each $n\in\en$ we have
$$\ca{T_n f_k}=\ca{Q_n f_k}\le \frac2n+\varepsilon.$$
Since $\varepsilon>0$ is arbitrary and $(f_k)$ is an arbitrary weakly Cauchy sequence in $B_{C(\Delta)}$, we obtain $\operatorname{cc}(T_n)\le \frac2n$.

We finish by proving the third inequality. We will prove it by contradiction. Suppose that $\omega(T_n)<c<\frac12$. Let us fix a weakly compact set $L_0\subset Y$ with $\widehat{\operatorname{d}}(T_n(B_{C(\Delta)}),L_0)<c$. Since
$T_n(B_{C(\Delta)})\subset I_n(Y_n)$, we have
$$\widehat{\operatorname{d}}(Q_n(B_{C(\Delta)}),P_n(L_0))\le\widehat{\operatorname{d}}(T_n(B_{C(\Delta)}),L_0)<c.$$
Set $L=P_n(L_0)$. Then $L$ is a weakly compact subset of $Y_n$.

For any $k\in\en$ let $\pi_k:\Delta\to \{-1,1\}$ be the projection on the $k$-th coordinate. It is a continuous function from $B_{C(\Delta)}$.
So, there is $y_k\in L$ such that $\|y_k-Q_n(\pi_k)\|_n<c$.
Since $L$ is weakly compact, there is a subsequence $(y_{k_j})$ weakly
converging to some $y\in L$. Set $f_{k_j}=Q_n^{-1}(y_{k_j})$.
 Since $Q_n$ is an isomorphism, the sequence
$(f_{k_j})$ is weakly convergent in $C(\Delta)$. So it is uniformly bounded and pointwise convergent, hence by the Lebesgue dominated theorem it
is Cauchy in the $L^1$ norm. Let $0<\varepsilon<1-2c$. Fix $j_0$ such that
for $i,j\ge j_0$ we have
$$\int_\Delta |f_{k_i}-f_{k_j}|\,\mbox{\rm d}\mu <\varepsilon.$$
Fix $i>j\ge j_0$. Then
$$\begin{aligned}1&=\int_\Delta |\pi_{k_i}-\pi_{k_j}|\,\mbox{\rm d}\mu
\\&\le \int_\Delta |\pi_{k_i}-f_{k_i}|\,\mbox{\rm d}\mu
+\int_\Delta |f_{k_i}-f_{k_j}|\,\mbox{\rm d}\mu
+\int_\Delta |f_{k_j}-\pi_{k_j}|\,\mbox{\rm d}\mu
\\&<\|Q_n(\pi_{k_i})-y_{k_i}\|+ \varepsilon+\|y_{k_j}-Q_n(\pi_{k_j})\|
<2c+ \varepsilon,\end{aligned}$$
which is a contradiction completing the proof.

\section{Final remarks and open questions}

The first natural question is the following one:

\begin{question}
Are the constants in the inequalities in Theorem~\ref{main-t} optimal?
\end{question}

Another natural problem concerns other spaces with the reciprocal Dunford-Pettis property.

\begin{question} Is there a Banach space which enjoys the reciprocal Dunford-Pettis property but not a quantitative version?
\end{question}

By a quantitative version we mean  the existence of a constant $C$ such that the inequality
$$\wk[Y]{T}\le C\cdot\operatorname{cc}(T)$$
holds for any operator $T:X\to Y$.

Let us remark that for the Dunford-Pettis property there is a quantitative version which is automatically satisfied, see \cite[Theorem 5.2]{qdpp}. We do not know whether a similar thing
holds for the reciprocal Dunford-Pettis property. Our proofs strongly used the structure of $C_0(\Omega)$ spaces.

Let us explain what seems to be a difference between these two properties.

It follows from \cite[Proposition 1]{gro} that
\begin{equation}
\label{eq:DPP}
\begin{aligned}X&\mbox{ has the Dunford-Pettis property} \\&\qquad\qquad\Leftrightarrow
\mbox{any weakly compact subset of $X^*$ is Mackey compact.}
\end{aligned}
\end{equation}
Futher, \cite[Proposition 8]{gro} implies that
\begin{equation}
\label{eq:RDPP}
\begin{aligned}X&\mbox{ has the reciprocal Dunford-Pettis property} \\&\qquad\qquad\Leftrightarrow
\mbox{any Mackey compact subset of $X^*$ is weakly compact.}
\end{aligned}
\end{equation}

Hence, suppose that $X$ has the Dunford-Pettis property. Then any bounded set $A\subset X^*$ satisfies $\chi_m(A)\le 2\omega(A)$,
see the final part of the proof of Theorem~\ref{t:M=w}. And this yields an automatic quantitative version of the Dunford-Pettis property.

We are not able to proceed similarly for the reciprocal Dunford-Pettis property. If $X$ has the reciprocal Dunford-Pettis
property and $A\subset X^*$ is bounded, we do not know how to control $\omega(A)$ by $\chi_m(A)$. We know that any Mackey compact
is weakly compact, thus, if we define
$$\omega_m(A)=\inf\{\widehat{\operatorname{d}}(A,H): H\subset X^*\mbox{ Mackey compact}\},$$
we obtain $\omega(A)\le\omega_m(A)$. But it is not clear, whether
$\omega_m(A)$ can be controlled by $\chi_m(A)$. (Conversely, $\chi_m(A)\le 2\omega_m(A)$ by the final part of the proof of Theorem~\ref{t:M=w}.) This inspires the following question:

\begin{question} Is the quantity $\omega_m$ defined above equivalent
to $\chi_m$?
\end{question}

For $X=C_0(\Omega)$ it is the case by Theorem~\ref{t:M=w}. But the proof essentially used the structure of $X$. We do not know the answer for general Banach spaces.

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If you love the flavor of ripe red raspberries, you'll love this delicious homemade ice cream topping. Raspberry sauce is especially delicious on a dish of rich vanilla ice cream on a hot summer day.
To make homemade raspberry sauce you'll need ten ounces of ripe raspberries, one half cup of raspberry jam, one tablespoon of cold water, and one and one half teaspoons of cornstarch. If you don't have access to fresh raspberries, a ten-ounce package of frozen raspberries may be used.
Combine the jam and raspberries in a saucepan, and heat the ingredients until they come to a boil. In a small cup, mix the water and cornstarch, and after the mixture is smooth, pour it into the pan of raspberries and jam. Bring the mixture to a boil once again, and stir it constantly so it doesn't burn. Allow it to cool sufficiently before serving it over a bowl of ice cream. This sweet and tangy homemade topping is sure to become one of your favorites.


This is an ancient Scandinavian Viking silver finger ring, dating to the 9th - 10th century A.D. The ring is formed from two rods of silver twisted together, the terminals hammered and welded (to enable the ring to be adjustable). This example is a good piece of Viking age jewelry, similar examples recorded SUSS-588B26, KENT-4D4DB5and SUR-C33FB2.
The form of the ring is a miniature version of Viking-period gold and silver neck- and arm-rings of around the 10th-11th centuries, like those from Skaill, Orkney, and Wipholm, Germany. As noted by the Finds Liaison Officer, Stephanie Smith, a close parallel from near Wold Newton, North-East Lincolnshire has been recently reported as potential Treasure (Treasure ref. 2013 T285; PAS ref. DUR-A1E866) and a slightly more elaborate version from Boynton near Bridlington, East Yorkshire, incorporating a beaded wire, was reported as Treasure in 2007 (Portable Antiquities and Treasure Annual Report 2007, no. 216). This type of ring appears to have continued in use into the 12th century, or perhaps slightly later (F. Lindahl, 2003, Symboler i guld og sølv. Nationalmuseets fingerringe 1000-1700-årene, Copenhagen, nos. 8, 10, and 13; J. Graham-Campbell, 2011, The Cuerdale Hoard and related Viking-Age silver and gold from Britain and Ireland in the British Museum, London, British Museum Press, 107-108).
DATE: c. 9th - 10th century A.D.


Music is one of the most enriching activities that a young person can engage in and it is an integral component of Avila College's curriculum and extra-curricular activities.
The Music Department at Avila College is equipped with outstanding facilities. The purpose-built complex comprises two music technology classrooms equipped with computers and keyboards, five practice rooms and a spacious Orchestra Room for rehearsing which also allows tuition on orchestral percussion.
Avila's private instrumental/singing program operates during school hours and is an extension of the classroom based music learning experience offered to all Avila College students from year 7 to 12.
the opportunity to be part of the many instrumental ensembles and choirs at Avila College.


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What's The 'Right' Debate On Net Neutrality?
How is net neutrality being debated in India?
InnovationOperationsTechnology
Keshav Murugesh
3Jun
803,723 mails in TRAI's inbox… it sure is a deluge of 'opinion' flooding the government's corridors. The recent debate on 'Net-neutrality' has many layers, but what has taken the nation by storm is the top-most layer on being given the RIGHT to access information without being punished by forces wanting to ride the principles of a 'market economy'. The mandate is clear from the 0.005 per cent of the population that uses smartphones in India.
I am all for freedom to choose where and how I want to consume content. Especially when the age of Internet freedom has created so many different products and solutions that are lifelines for most of us today. Be it the videos on YouTube or the chats on What'sApp. It's difficult to even fathom a telecom company getting a hold of trying to 'unintentionally' or 'unknowingly' leading us to consume content as strategized by its marketing and strategy team.
Not just me, but an entire industry, the business process management or BPO industry, looks at any disruption to this freedom with a really close lens. After all, we are poised to grow to $50 billion in size from the $26 billion we are at today, riding on the promise of innovation, domain specialisation and the ability to build the right technology platforms for our global clientele. All of this on the back of access to the right data and the right channels in any part of the world… and the last thing we would want as an industry is to be led by any telecom company as per its 'innovative constructs'.
However, peeling through the layers of this debate, I find market economics levelling out the playing field in many ways. Let us look at some extremely basic examples in real life. For instance, we still have a VIP darshan in holy places, despite this one place presumably being 'equal to all'. We have come to terms with cable operators who provide free access to certain channels on their networks while either not providing or asking for a fee on others. We still find the 'kirana' or 'mom and pop' shops being in happy co-existence with the big superstores.
The thought I am putting forth is that 'preferential constructs' or 'discrimination in the form of differentiation' exist in various areas even today.
Looking closely at the government's Digital India vision, we know that increased investments would be required in infrastructure. Broadband penetration in the country has been at an abysmal low with media reports claiming that out of the 175 million broadband connections set as a target for 2017, 85.74 million have been achieved till December 2014. The wireless broadband speed in India is 1.7 megabytes per second or MBPS as against 10.5 MBPS in the US and 9.9 MBPS in the UK.
The moot point is that we as a nation need a very strong prop up in terms of our digital infrastructure and connect, which will require large telecom players to come forth and put their money in. Why would they do it if they were not allowed to survive and grow profitably as a business?
I am not being a capitalist, and would not recommend that we look at it as a nation in a capitalist sense. I am merely suggesting that without capitalism, the next phase of transformational growth cannot happen. But we are protected as a democracy, as a country that understands the divide between the rich and the poor. We are a country whose government can set right arm-twisting tactics with firm policies and governance. I believe that it is the fear of the unknown or rather limited understanding of this debate that is creating an unprecedented uproar.
All the Net activism has clearly showcased that no policy can be thrust upon the citizens of this country. But the activism alludes to restrictive policies that telecom companies can create, completely hampering the rate of innovation in the country and thereby threatening the interests of start-ups, which I find hard to digest at this point.
I belong to an industry that co-exists with the government in many ways. For instance, the government looks to the BPM industry to set up offices in tier-2 and tier-3 cities, thereby generating large-scale employment and pulling up the standard of living in such places. Large investments are made by corporate houses, while continuing to seek government incentives in various forms to do so. Capitalism co-exists with socialism.
I am still wondering what the debate is. Is it about 'neutrality' in gaining access to information? In being unrestricted about learning or gaining information? Then I am all for it. Any restriction of any kind has always hindered overall growth and history stands testimony to it.
But I am quite certain that that is not the debate here. It is about 'how to do business' wherein private players are involved from a business stand-point and the government is involved through its agencies to ensure that citizens are not being affected or discriminated against in any manner, and their fundamental rights are not violated in any sense by large business houses. It is about creating policies and frameworks that enthuse players to invest in telecom infrastructure while allowing the citizens the freedom to view what they want from wherever they want.
The Big Sleep: Big Data Helps Scientists Tackle Lack of Quality Shut Eye
How Artificial Intelligence Can Benefit Email Marketing
IBM's blockchain service goes multicloud
Apple set to release three new iPhones
Uber selects Melbourne for flying taxi tests
IT outages left seven million Britons unable to use cards last year
Google launches AI language library TensorFlow.Text
The impact of 5G and IoT on blockchain technology
UK found to be Europe's tech unicorn leader
Apple to buy Drive.ai and add self-driving experts to its roster
Weekend update: The death of iTunes, cloud investment from Microsoft, Google and Oracle...
Uber takes to the skies with new helicopter service
Retail is reaping the rewards of predictive analysis
Leading Innovation into the Mainstream

redpajama_set_name: RedPajamaCommonCrawl


Olive drab canvas teardrop type backpack with brown leather accents & brown leather bottom. Top carry handle is made of brown leather. Padded and adjustable shoulder straps. Large main compartment with zipper closure, interior cell phone pocket and cushioned section is perfect for laptop. Front zippered pouch with exterior zippered pocket. Small top zippered pouch. Dimensions: 5.5" x 13" x 17". For work, school or play, this backpack can be used to carry all your books, gear and equipment. Olive drab canvas teardrop pack w/leather accents.


Thanksgiving is nearly here, and ROBLOX is bringing you a cornucopia of awesome events, sales, and cool prizes to earn. This month's ROBLOX event, BLOXgiving, takes three of the most popular games on ROBLOX and adds a Thanksgiving twist!
The three games, The Mad Murderer, ROBLOX High School, and Super Bomb Survival, have all added a Thanksgiving theme, and have special badges for you to earn sweet prizes! Each game has two unique prizes to earn. The first is earned by completing a specific task in-game; the second is earned by finding a special Thanksgiving-themed item that randomly spawns in-game. Completing each objective earns you an exclusive hat and badge. Get all six prizes and you'll earn a pilgrim BLOXikin!
All the details and prizes are listed on the official BLOXgiving page, so check it out and get to playing!
Hey, Pilgrim, can you pass me one of those traditional Thanksgiving pineapples?
In addition to BLOXgiving, we've got two awesome sale events coming up. This Friday (November 21st) marks our second annual Bright Friday event. What's Bright Friday? It's a totally rad event we made up, where the brightest colored items go on sale. So, if neon is your jam, you're in luck!
Then, the following Friday, (November 28th) is the beginning of our annual Black Friday sale. And we're letting you decide what goes on sale. Take this survey to tell us which items you want to see in the sale, and we'll pick items based on what the community as a whole wants the most. The Black Friday sale runs the entire weekend, items will go on sale at different times, and some items may only be available for an hour. We'll have more information about the sale as next weekend draws near, but be sure to check the ROBLOX catalog regularly that weekend! In the meantime, be sure to take the Black Friday survey by the end of November 23rd to make sure your vote counts.


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Ustawa z dnia 27 marca 2003 roku o planowaniu i zagospodarowaniu przestrzennym – polska ustawa, uchwalona przez Sejm RP, która reguluje zasady kształtowania polityki przestrzennej przez jednostki samorządu terytorialnego i organy administracji rządowej, oraz zakres i sposoby postępowania w sprawach przeznaczania terenów na określone cele oraz ustalania zasad ich zagospodarowania i zabudowy.

Przebieg procesu legislacyjnego
Projekt ustawy wpłynął do Sejmu 20 czerwca 2002 roku. 13 lutego 2003 roku, po trzecim czytaniu projektu ustawy, Sejm przyjął ustawę. Głosowało 380 posłów. Za głosowało 249, przeciw 129, a 2 wstrzymało się od głosu. 10 marca 2003 roku Senat RP przyjął uchwałę o wniesieniu poprawek do ustawy i skierowaniu jej ponownie do prac Sejmu. 27 marca 2003 roku Sejm przyjął część poprawek Senatu. 15 kwietnia 2003 roku, Prezydent Rzeczypospolitej Polskiej podpisał ustawę.

Budowa
Ustawa nie zawiera preambuły i zbudowana jest z 9 rozdziałów:

 Rozdział 1. Przepisy ogólne (art. 1 - 8)
 Rozdział 2. Planowanie przestrzenne w gminie (art. 9 - 37n)
 Rozdział 2a. Planowanie przestrzenne na obszarze metropolitalnym (art. 37o - 37q)
 Rozdział 3. Planowanie przestrzenne w województwie (art. 38 - 45)
 Rozdział 4. Planowanie przestrzenne na szczeblu krajowym (art. 46 - 49)
 Rozdział 4a. Obszary funkcjonalne (art. 49a - 49g)
 Rozdział 5. Lokalizacja inwestycji celu publicznego i ustalanie warunków zabudowy w odniesieniu do innych inwestycji (art. 50 - 67)
 Rozdział 6. Zmiany w przepisach obowiązujących (art. 68 - 82)
 Rozdział 7. Przepisy przejściowe i końcowe (art. 83 - 89).

Nowelizacje
Do 2022 roku włącznie zostało wydanych ponad 60 aktów zmieniających przepisy ustawy o planowaniu i zagospodarowaniu przestrzennym. Większość z nich to nowelizacje ustaw. Ostatnia zmiana weszła w życie w 2022 roku.

Przypisy

Prawo administracyjne
Ustawy III Rzeczypospolitej
2003 w Polsce

redpajama_set_name: RedPajamaWikipedia


I was delighted when Claire & Gary asked me to take some pictures of their family whilst the autumn leaves were still on the ground and the weather dry. Claire chose an awesome location for us to meet, the Duck Pond next to Hartley Wintney cricket ground in Hampshire.
Now if ever you happen to visit Hartley Wintney Duck Pond, you should be aware that there are in fact TWO Duck Ponds! The one near The Waggon & Horses and the one by the Cricket Pitch. I found this out after standing in the cold for ten minutes by the wrong one! Oops!
The Sunday morning we chose to meet was pretty chilly, a bit misty, damp but perfect for taking pictures. Gary & Claire's girls are adorable and I loved photographing them. Despite the cold weather we had a great time and captured some lovely images of them all.
I have made Claire & Gary promise that we will do it again in the spring, when the flowers are out and colour is returning to the trees (oh and its a bit warmer!). I have made them promise to bring their dog too!


text:
Cover Reveal - Teach Me by Ella Sheridan
Teach Me by Ella Sheridan
Southern Nights Book One
A woman determined to heal…
Shy researcher Jess Kingston spent the last eight weeks recovering from her ex-boyfriend's brutal attack. Body healed, she's ready to put her life back together—except her ex isn't ready to let go. She won't cower in a corner while Brit tortures her, but she's powerless to fight back.
A man determined to resist…
Ex-military security specialist Conlan James avoids commitment like the plague. His job, his Harley, and the occasional one-night stand are all he needs, until the day he rescues Jess from a tense situation and realizes he can't get her off his mind. He can teach her to protect herself, but protecting his heart is another matter.
A madman determined to win…
As the deadly game of cat-and-mouse with Brit heats up, so does the hunger between Con and Jess. Safety might be found in numbers, but in bed, all bets are off—and the wrong move could lead to heartbreak. Or death.
Content/Theme(s): Bodyguard, Military, Thriller
Publisher: Ella Sheridan
Purchase links: Amazon ARe Kobo B&N
What the hell are you doing here?
This wasn't the first time in the last five minutes that Conlan had asked himself the same question. Maybe if he had an answer, the revolving door in his brain would stop spinning, but that didn't seem likely. Not anytime soon. Not with the beautiful brunette he'd come to see sitting close enough that, if he let himself look, he could detect the light dusting of freckles across her nose. But he wasn't looking, and he shouldn't be here, so how had he ended up standing in line behind the thirtysomething latte league? It sure as hell wasn't for the coffee.
Legs braced wide, he shifted from one hip to the other, the creak of his motorcycle chaps reminding him he could be enjoying a few extra minutes on the Harley before work instead of spending that precious time here, mooning over a woman. Doe Eyes. The first time he'd seen her all those months ago, he'd thought her eyes reminded him of sweet Georgia pecans and skittish does. The name stuck, as had the memory of her eyes—and a hundred other glimpses he shouldn't have taken.
Another name called, another latte dispensed, another shuffle forward.
He hadn't seen those eyes in eight weeks, and yet still he'd shown up every Monday, like clockwork, hoping for one more glimpse and calling himself an idiot. Wasn't like he planned on asking her out. So why the hell did he torture himself with these weekly forays into enemy territory?
Sex. Or sex appeal, at least.
Another step closer to the counter. The move didn't ease the constriction behind the zipper of his jeans. This was what she did to him, thinking about her. Especially now, after so long apart.
The thought had a snort escaping. Ahead of him, Mr. Suit and Tie startled and glanced over a shoulder, but Conlan ignored the look. He was too busy figuring out when "this" had become enough like a relationship in his head that he would think things like "after so long apart." Doe Eyes might appear prominently in his thoughts from time to time—especially certain times—but he'd never seen her outside of this coffee shop. And he wouldn't. A quick roll in the hay was one thing, but Doe Eyes wasn't the kind of woman who had one-night stands. He could tell that much just by looking at her. She was a relationship kind of woman, and he was a relationship-phobic kind of guy. Which meant he seriously needed to get a grip—and not on the part of him growing even harder at the idea.
Idiot was right.
He should be at work. Southern summer heat brought out the crazies almost as well as full moons did, and JCL Security was feeling the impact, juggling cases like they had eight arms, which they didn't. Too many sleepless nights had been spent at his office, especially with the Bennett case coming up. Just a couple more weeks before Thea Bennett had her bastard of a husband before a judge and hopefully out of her life, but the paper- and prep work to get the high-profile bastard there had been a bitch. He seriously needed to—
"Conlan, hey!"
For a passing moment he was convinced the voice belonged to the woman filling his thoughts. But when the high, candied voice called again, he realized it was coming from the counter. The cashier. Tonya, Tammy? Tracy? He couldn't remember. She was blonde with a deep tan he would've deemed impossible in a landlocked city like Atlanta, the shade a stark contrast to her white smile. Stepping up, he threw her a grin. "Hey."
She batted long lashes, almost hiding the way her glance slid down to the crotch of his jeans, framed in his leather chaps. "Long time, no see."
He winked automatically. "It's a long wait between Mondays."
The girl giggled. "Your usual?"
"That's right. Thanks," he said, passing over a ten-dollar bill.
She made change, certain to caress his hand as she laid the money in his palm. Conlan was more interested in the dark Colombian roast another employee was walking toward them. High-octane all the way. The sight of the near-black brew had him salivating for something other than Doe Eyes for the first time that morning.
He reached the condiment counter just as his phone buzzed in his back pocket. Probably Jack. Retrieving the cell confirmed his suspicion.
Where the hell are you? his partner had texted.
Piss off, Con replied, a grin tugging at his lips. The irony that he'd spent too much time asking himself the very same question didn't escape him. In a half hour he'd be at the office and they could both stop wondering.
With a little back-and-forth he managed to cram the phone back in his tight jeans. He glanced around absently, and his gaze snagged on a pair of amber-brown eyes that suddenly met his.
He froze.
Doe Eyes dropped her chin and shifted over the slightest bit, enough that her friend's position blocked her from view, but not before he caught the blush coloring her creamy cheeks.
His cock banged against his zipper as if begging to be let out. The bite of pain caught his breath in his throat. Jesus, what the hell was he—
Don't! Ask. Again. He knew what the hell he was doing here, and he needed to go; he really did. He needed to stop letting his dick run this show, grab his coffee, and get back to reality.
He was restless, that was all. He was a man who needed action. Needed to be doing something, anything, not sitting behind a desk like he'd been for weeks while prepping Thea's case. Usually he worked off his frustration in a way that involved cool silk sheets and bare skin and satisfaction on both sides, but there'd been no damn time. Just his hand and the additional chafing it provided, which wasn't near as effective—or satisfying. That had to be the reason he couldn't stop thinking about his mystery woman.
Of course. That had to be it.
Popping the lid off his cardboard cup released the rich aroma of ground coffee beans into the air. He lifted his cup and blew across the hot liquid, the sound almost a sigh of relief. He was already reaching for the packets of sugar when black squiggles caught his eye. There. On the part of the paper sleeve now facing him, he could see a name and number were clearly written: Tiffany. A 470 area-code phone number.
So that was her name. Sounded like an eighties pop star. A glance over his shoulder found the cashier leaning across the bar where drinks were picked up, her mounded breasts shelved there, on display. Come back soon, she mouthed, her shoulders doing a little wiggle. On reflex, he threw her a grin, but her seemingly seductive move couldn't pull his glance downward. His dick didn't even twitch. Apparently only one thing could trigger his runaway libido this morning.
He added the sugar, trying to ignore the panic in his gut and his one-track mind. The latter was impossible. He wanted to know Doe Eyes' name, her phone number. Were her breasts as full as they looked beneath that starched white button-down? Was her hair as soft as he swore it would be when he fisted it between his fingers?
He stirred a bit too vigorously, and coffee sloshed over the side of the cup.
Don't look. Don't. He realized he'd closed his eyes. A sigh escaped as he rubbed a thumb and finger against them, but as soon as the lids popped open, he searched for her. Had to see her. Felt his heartbeat pick up knowing she might meet his eyes.
He was so screwed—and smart enough to admit it. He let go, let the conflict and the churning in his gut and the tension cramping his muscles go. And then he looked toward her table.
It was empty.
"Well shit."
He stood for a moment, cursing himself, the coffee, and everything else he could think of. When another customer stepped up behind him and cleared his throat, wanting access to the counter, Con grabbed his cup and headed out the door. On his way, he chucked the coffee in the trash without a single sip.
Other titles by Ella Sheridan:
Teach Me
Book 1 Trust Me
Book 2 Take Me
Secret Just A
More Naughty
Christmas Taste
Find Ella Sheridan at:
www.EllaSheridanAuthor.com
Twitter: @AuthorESheridan
Ella Sheridan Facebook page
Ella Sheridan Goodreads author page
Ella Sheridan Newsletter sign up
Ella Sheridan Amazon author page
More Ella Sheridan on Cover Reveals
Be on the lookout for Ella Sheridan's upcoming releases: Trust Me coming Dec 2014, Southern Nights Anthology coming May 2015, Take Me coming early Summer 2015, and Secrets To Hide book 4 coming late Summer 2015
Categories: Contemporary Romance, cover reveal, Ella Sheridan, excerpt reveal, Military Romance, Oct 2014 releases, Romantic Suspense, Southern Nights series, Teach Me


NEW### The Player Development Committee in conjunction with Hockey Canada provides technical information, oversight, and monitors the delivery of Long Term Player Development (LTPD). The Player Development Committee is also responsible for the development of new hockey technical components as well as administrative and operational standards for player development.
NEW### The Player Development Delivery Group is responsible to deliver the program components to member associations at the expected level.
Builds Association resources and advises Associations on implementation.
Oversee the implementation of and updates to the Instructional stream material dealing with progressive player skill development.
Delivery group members would be personnel such as the BC Hockey Cross Ice Leads.
This resolution has been endorsed by both the Development and Minor Committees.


text:
Dr. Jim Loehr: Why Leading with Character Matters (Ep #89)
Glenn and Michael Parker: How to be a Positive Influence Leader (Ep #88)
Tomas Chamorro-Premuzic and Amy Edmondson: Vulnerability, Not Bravado (Ep #87)
Chester Elton and Adrian Gostick: Finding Gratitude This Christmas Season (Ep #86)
Michael Canic: How to be Ruthlessly Consistent (Episode #85)
Servant Leadership Panel of Experts (Episode #76)October 2, 2020 - 24
Tomas Chamorro-Premuzic and Amy Edmondson: Vulnerability, Not Bravado (Ep #87)December 18, 2020 - 24
Glenn and Michael Parker: How to be a Positive Influence Leader (Ep #88)January 8, 2021 - 24
Dr. Jim Loehr: Why Leading with Character Matters (Ep #89)January 15, 2021 - 24
Marcel's Story
Love In Action Podcast
Richard Boyatzis: Helping People Change (Episode #24)
By Marcel Schwantes 1 year agoNo Comments
Home / Podcast / Richard Boyatzis: Helping People Change (Episode #24)
This week's guest on the Love In Action podcast is Richard Boyatzis, a renowned expert on emotional intelligence and bestselling author. Richard and his colleagues at Case Western University have recently released a new book entitled Helping People Change, which reveals their findings about a more effective approach to helping people learn and change behaviors.
Listen and Subscribe on Apple Podcasts | Stitcher | Spotify | Google Play | Google Podcasts
The big idea in the book is that we want to help others learn or change, but how we go about it causes them to close down to new ideas, which is the opposite of what we're aiming for. The way to help people open up to change is to inspire them into the psychophysiological state he calls the Positive Emotional Attractor (PEA). Richard advocates the Coaching With Compassion approach. [04:55]
Marcel asks why we tend to try to fix people or coach for compliance. Richard lists both negative and positive reasons why we tend to do this. One negative reason is if we're in a position of authority, we may think we know what other people should do and we try to make them do it. A positive reason is that we may really care for the person. [06:58]
Coaching with compassion is a way of approaching someone that helps to stimulate a psychological state of mood that focuses on more positive than negative. It allows you to ameliorate the ravages of stress and, being more in the empathic network of your brain, makes you open to new ideas and people. This state can be invoked through two main approaches: asking a person about their dreams, which makes them feel hopeful; and building and maintaining more caring relationships with other people which stimulates compassion and gratitude. [12:08]
The PEA state is when a person is open to new ideas, people and emotions. Before any learning or change can happen, you need to be in this state because it helps to create the Tipping Point. You want to go into this state periodically every day to tone down the stresses of life. [16:05]
Richard says that it's the responsibility of every leader or manager to understand the dreams of the people who report to them. Most organizations are sub-optimizing their human capital. If you activate your human capital by understanding and inspiring them, they will make magic for your organization. [18:22]
Marcel asks Richard to unpack an interesting story from his book about diabetic patients. Richard describes how a shared vision of their long-term future with their physicians, increased patients' treatment adherence. This research can be translated into the workplace, he agrees. [25:25]
No organization in the world exists to make money, Richard says. They exist to serve people. Making money is how to measure how effectively you're doing that. [33:15]
Instead of focusing on the task, focus on engaging the person doing the task. [36:15]
You can create a sustainable culture of caring in your organization by helping your people to form coaching groups where they care for and rely on each other. [ 37:18]
Richard shares how our minds and bodies react in a fear-based environment versus a loving one. [38:30]
Marcel asks the question "Why do people still lead through fear?" Richard shares his thoughts on narcissism and its impacts. [43:50]
Richard on LinkedIn
Book on Amazon
Case Western Reserve University faculty page
Melissa Hughes: Employee Engagement Starts in the Brain (Episode #20)
Mark C. Crowley: The Physiological Impacts of Bad Management (Episode #14)
Bob Chapman: The Extraordinary Power of Caring (Episode #5)
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As we challenge current management thinking and practices with a clear goal to raise more awareness and to build skills around human leadership and human-centered work cultures, we would be delighted to accept your sponsorship for future episodes. If interested, contact Marcel.
Richard Boyatzis
Joel Peterson: Building the Bonds of Trust (Episode #25)
Sally Helgesen: How Women Rise (Episode #23)
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Marcel Schwantes
Marcel Schwantes is the founder and chief human officer of Leadership from the Core, a global leadership training and executive coaching boutique with one core purpose: to grow profitable and powerful servant leaders through "Love in Action."
© Copyright 2017. All Rights Reserved, Leadership from the Core LLC.


when he wasn't completely wasted.
if you're unlucky enough to board the C train.
writing. Maybe he had stopped telling jokes.
Scotch, please ... for me too. On the rocks .. with a (cruel) twist. Rest in peace, dear Robin.
Yes. Maybe. And whether it is true or not, we are all taking stock as if he were each of our younger brother. I love this poem that slowly reveals the link, moves out across the world to him, our mirror, and back to the self. This is writing, Marian! Very fine writing.
I feel this, Marian. I wish that I had written this; I wish I had the strength. All of the news coverage has said over and over - he was only 63, so young! He was a year older than my dying mother. Perhaps he had run out of jokes. How do you cope, then?
Thank you for this, Marian.
So much in the media event horizon is more noise -- why then write? Why add to the freight? I found your poem compact and powerful and on the nose asking that question. There's also an unwritten creative maxim, write or die, tell jokes or die -- I mean, what is living without writing? Yet I'm pondering a hiatus too. How can words like this mean anything unless held against their silence? Watched an episode of "The Sopranos" this weekend and thought, o yeah, James Gandolfini's dead too. And Charlie Haden. And Richard Attenborough. And Philip and Lauren and Robin. And a nobody in Monrovia. And my father's partner. And everyone who died in Flanders a hundred years ago and a cat down the street who got nailed by a coyote. Why write? Perhaps for them. Or for what remains, of them, of us. Great poem, Marian.
Great title. My first husband committed suicide, a friend's son hung himself two years ago... yesterday a neighbor found his 20 year old son hanging by a sheet in a tree.... so tragic when one no longer has anything to live for - they think... write because there are words that must be said.
Woah! I read the first sentence, then marvelled at the length of the poem, Marian. Then I read it, and understood the source of this out-pouring.. some things remained bottled up until it is time for them to flow, unsullied. Your viewpoint is one I very much agree with - it gets harder to see the funny side of life.
There are some splendid imagery in this -- the six-foot depression.. what a way to express what many a comedian try to cover with that thin layer of comedy, and especially moving is of course.. did we really care, or is it just a show-off -- like an ice-bucket challenge.. It's hard to be honest on yourself.. -- This resonated a lot with me..
Back for a second read, now the black dog has backed off a bit. Depression. It is its own realm. Blessedly, age, and perhaps wisdom gained, have made my battles with it more easily fought and more rapidly won.
How glad I am to revisit this.
I hope your new job is pleasing you and that you find life astounding with or without writing.
thank you! writing more now, actually. whew.


text:
Business Companies 26 Apr 2019 Fitch Ratings sees risks to OMCs as govt seeks more returns
Business, Companies
Fitch Ratings sees risks to OMCs as govt seeks more returns
Published : Apr 26, 2019, 2:26 pm IST
Updated : Apr 26, 2019, 2:26 pm IST
Struggling to meet budget targets, the government had asked cash-rich PSUs to pay second interim dividend.
"The higher shareholder returns will put more pressure on the financial profiles of the companies, which have large investment plans for the next two years," Fitch said. (Photo: File)
New Delhi: Financial profiles of state-owned oil marketing companies such as Indian Oil Corp (IOC) may be at risk in the near to medium term due to pressure from the government to increase shareholder returns, Fitch Ratings said on Friday.
Struggling to meet budget targets, the government had in the just concluded fiscal (2018-19) asked cash-rich PSUs to pay second interim dividend as well as undertake share buyback. The share buybacks, wherein a company repurchases its shares and extinguishes them, helped the government get money while maintaining its control over them.
"The oil marketing companies (OMCs) IOC, Bharat Petroleum Corporation Ltd (BPCL) and upstream entity Oil India Ltd (OIL) declared high interim dividends of 67.5 per cent to 110 per cent of the face value of their shares, and undertook share buybacks in the financial year ended March 31, 2019 ," Fitch said in a statement.
These, it said, were likely to have been driven by pressure from the Indian government to increase shareholder returns to shore up the weak fiscal position and finance promises made ahead of elections in April and May 2019.
"The higher shareholder returns will put more pressure on the financial profiles of the companies, which have large investment plans for the next two years," Fitch said.
IOC and BPCL are in the process of upgrading and expanding their refineries to meet the deadline for producing ultra-clean Euro-VI grade fuel by April 2020, and improving downstream integration in petrochemicals. OIL plans to augment its domestic production and reserves.
In addition, BPCL and OIL are likely to invest in an upcoming Mozambique LNG project after a final investment decision on the project, which is expected in the second half of 2019, it said.
"OIL has limited headroom in its current standalone credit profile of 'bbb-', which may be revised down if weakening of its financial profile results in net leverage (adjusted net debt/operating EBITDAR) exceeding 2.5x. However, in such an event, OIL will benefit from one notch of support from the state, resulting in the final rating remaining unchanged at 'BBB-'," the rating agency said.
In comparison, IOC and BPCL have more headroom in their 'bb+' standalone credit profiles and are likely to remain resilient even in case of any modest weakening in their financial profiles.
The OMCs additionally face the risk of government intervention in fuel prices during the elections.
"However, Fitch does not foresee any significant impact on OMCs' financial profiles on account of fuel price controls. We expect the OMCs to make up the losses in the subsequent period during the year in such an event," the statement said.
Fitch said it believes OMCs will continue to revise prices daily to reflect the market prices over the medium term.
The ratings on IOC and BPCL are equalised with that of the state, in line with Fitch's Government Related Entities Rating Criteria.
Tags: oil marketing companies, indian oil corp, ioc, bpcl, oil, fitch ratings


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