text:
\section{Main concepts, terminology and notation}\label{secnot}
In this section, we compile terminology and notation
concerning differential calculus over topological
fields, together with basic facts.
Most of these facts are easy to take
on faith, and we recommend to skip
the proofs on a first reading.
If desired, the proofs can be looked up in
Appendix~\ref{appbore}.\\[2.5mm]
All topological fields occurring in this article
are assumed Hausdorff and non-discrete;
all topological vector spaces are assumed Hausdorff.
Given a field~${\mathbb K}$, as usual
we write ${\mathbb K}^\times:={\mathbb K}\setminus \{0\}$
for its group of invertible elements.
A~valued field is a field~${\mathbb K}$,
equipped with an absolute value
$|.|\colon {\mathbb K}\to [0,\infty[$
which defines a non-discrete topology on~${\mathbb K}$.
If $|.|$ satisfies the ultrametric
inequality, we call $({\mathbb K},|.|)$
an \emph{ultrametric field}.
Totally disconnected, locally compact
topological fields will be referred to as \emph{local
fields}. It is well known
that each locally compact field
admits an absolute value defining its
topology. We fix such an absolute value,
and thus consider~${\mathbb K}$ as a valued field.
On ${\mathbb R}$ and ${\mathbb C}$, we shall always use the
usual absolute value.
We write ${\mathbb N}=\{1,2,\ldots\}$
and ${\mathbb N}_0:={\mathbb N}\cup\{0\}$.
\subsection*{{\normalsize{\boldmath$C^k$}-maps in the sense of
Bertram, Gl\"{o}ckner and Neeb}}
We
recall the approach to $C^k$-maps
between open subsets of topological vector spaces
over a topological field
developed in~\cite{BGN}
(and its extension to maps on
non-open domains from~\cite{IM2}).
More information concerning
this approach can be found in the
survey~\cite{SUR}.
Cf.\ \cite{Ber}
for applications of the corresponding
differential calculus over topological rings
in differential geometry.
We are mostly interested in mappings on open domains,
but some results will hold more generally.\\[3mm]
Let $E$ and $F$ be topological vector
spaces over a topological field~${\mathbb K}$
and $f\colon  U\to F$ be a map,
defined on a subset $U\subseteq E$
with dense interior.
Then the directional difference quotient
\[
f^{]1[}(x,y,t)\; :=\;
\frac{f(x+ty)-f(x)}{t}
\]
makes sense for all $(x,y,t)$ in the subset
\[
U^{]1[}\; :=\; \{(x,y,t)\in U\times E\times {\mathbb K}^\times\colon x+ty\in U\}
\]
of $E\times E\times {\mathbb K}$.
To define directional derivatives,
we need to allow also
the value $t=0$.
Hence, we consider
\[
U^{[1]}\;:=\; \{(x,y,t)\in U\times E\times {\mathbb K}
\colon x+ty\in U\}\,.
\]
Then $U^{[1]}=U^{]1[}\cup (U\times E\times \{0\})$, as a disjoint union.
If $U$ is open, then $U^{[1]}$ is an open
subset of the topological ${\mathbb K}$-vector space
$E^{[1]}=E\times E\times {\mathbb K}$.
In the general case, $U^{[1]}\subseteq E^{[1]}$
has dense interior.
Recursively, we define
$U^{[k]}:=(U^{[1]})^{[k-1]}$
and $U^{]k[}:=(U^{]1[})^{]k-1[}$
for $2\leq k\in {\mathbb N}$.
Then $U^{]k[}$ is dense in
$U^{[k]}$ (see \cite[Remark~1.6]{IM2}).
\begin{defn}\label{def1}
The map $f\colon U\to F$ is called $C^1_{BGN}$
if $f$ is continuous (i.e., $C^0$, or $C^0_{BGN}$),
and there exists a continuous map
$f^{[1]}\colon U^{[1]}\to F$
which extends
$f^{]1[}\colon U^{]1[}\to F$.
Given $k\in {\mathbb N}$ with $k\geq 2$,
we say that $f$ is
$C^k_{BGN}$ if $f$ is $C^1_{BGN}$ and $f^{[1]}\colon U^{[1]}\to F$
is $C^{k-1}_{BGN}$.
We define
$f^{[k]}:=(f^{[1]})^{[k-1]}\colon
U^{[k]}\to F$ in this case.
The map~$f$ is $C^\infty_{BGN}$
if it is $C^k_{BGN}$ for all $k\in {\mathbb N}_0$.
\end{defn}
Since $U^{]1[}$ is dense in~$U^{[1]}$,
$f^{[1]}$ is unique if
it exists (and likewise each $f^{[k]}$).
\begin{numba}\label{exalin}
For example, every continuous linear
map
$\lambda\colon  E\to F$ is $C^\infty_{BGN}$
with $\lambda^{[1]}(x,y,t)=\lambda(y)$
for all $(x,y,t)\in E\times E\times {\mathbb K}$
(thus also $\lambda^{[1]}$ is continuous linear).
Also each continuous multilinear map
is $C^\infty_{BGN}$ (see \cite{BGN}).
\end{numba}
\begin{numba}\label{chainr}
(Chain Rule).
If $E$, $F$ and $H$ are topological
${\mathbb K}$-vector spaces, $U\subseteq E$ and $V\subseteq F$ are
subsets with dense interior, and $f\colon  U\to V\subseteq F$,
$g\colon  V\to H$ are $C^k_{BGN}$-maps,
then also the composition
$g\circ f\colon  U\to H$,
$x\mapsto g(f(x))$ is~$C^k_{BGN}$.
If $k\geq 1$, we have
$(\widehat{T}f)(x,y,t):=(f(x),f^{[1]}(x,y,t),t)\in V^{[1]}$
for all $(x,y,t)\in U^{[1]}$, and
\begin{equation}\label{formchain}
(g\circ f)^{[1]}(x,y,t)=g^{[1]}(f(x),f^{[1]}(x,y,t),t)\, .
\end{equation}
Thus $(g\circ f)^{[1]}=g^{[1]}\circ \widehat{T}f$
with $\widehat{T}f\colon U^{[1]}\to V^{[1]}$
(see \cite[Proposition~3.1 and Proposition~4.5]{BGN},
also \cite[\S\,1]{IM2}).
\end{numba}
We recall from \cite[Lemma~4.9]{BGN}
and \cite[\S\,1]{IM2} that
being $C^k$ is a local property.
\begin{la}\label{Crlocal}
Let $E$ and $F$ be topological ${\mathbb K}$-vector spaces,
and $f\colon  U\to F$ be a map, defined
on a subset $U\subseteq E$
with dense interior.
Let $k\in {\mathbb N}_0\cup\{\infty\}$.
If there is an open cover $(U_i)_{i\in I}$
of~$U$ such that
$f|_{U_i}\colon  U_i\to F$ is
$C^k_{BGN}$ for each $i\in I$,
then $f$ is~$C^k_{BGN}$.\nopagebreak\hspace*{\fill}$\Box$
\end{la}
\subsection*{{\normalsize{\boldmath $C^k$}-maps in the sense
of Schikhof and De Smedt}}
In this section, we give a definition
of $C^k$-maps of several variables
based on continuous extensions
to certain partial difference quotient
maps,
which generalizes special cases
considered by Schikhof \cite[\S\,84]{Sch}
and De Smedt \cite{DSm}.
Our notation differs from the one used
in~\cite{DSm} and~\cite{Sch},
because we find it more convenient
to use multi-indices in
higher dimensions.
\begin{numba}
Until Remark~\ref{onlynois},
let ${\mathbb K}$ be a topological field,
$d\in {\mathbb N}$, $U\subseteq {\mathbb K}^d$
be an open subset
(where ${\mathbb K}^d$ is equipped with the product
topology), and $F$ be a topological
${\mathbb K}$-vector space.
As usual, for $i\in \{1,\ldots, d\}$
we set $e_i:=(0,\ldots,0,1,0,\ldots, 0)\in {\mathbb K}^d$,
with $i$-th entry~$1$.
\end{numba}
\begin{numba}\label{notconv}
As usual, given a ``multi-index'' $\alpha=(\alpha_1,\ldots,\alpha_d)
\in {\mathbb N}_0^d$,
we write $|\alpha|:=\sum_{i=1}^d \alpha_i$.
The definition of a $C^k$-map $f\colon U\to F$ in the
sense of Schikhof and De Smedt
will involve a certain continuous extension
$f^{<\alpha>}$ of a partial difference
quotient map $f^{>\alpha<}$ corresponding to
each multi-index $\alpha\in {\mathbb N}_0^d$
such that $|\alpha|\leq k$.
It is convenient to define
the domains $U^{<\alpha>}$ and $U^{>\alpha<}$
of these mappings first.
They will be subsets
of ${\mathbb K}^{d+|\alpha|}$.
It is useful to write elements
$x\in {\mathbb K}^{d+|\alpha|}$ in the form
$x=(x^{(1)}, x^{(2)},\ldots, x^{(d)})$,
where $x^{(i)}\in {\mathbb K}^{1+\alpha_i}$
for $i\in \{1,\ldots, d\}$.
We write $x^{(i)}=
(x^{(i)}_0,x^{(i)}_1,\ldots, x^{(i)}_{\alpha_i})$
with $x^{(i)}_j\in {\mathbb K}$
for $j\in \{0,\ldots,\alpha_i\}$.
\end{numba}
\begin{numba}
Given $\alpha\in {\mathbb N}_0^d$, we now define
$U^{<\alpha>}$ as the set of all
$x\in {\mathbb K}^{d+|\alpha|}$
such that, for all
$i_1\in \{0,1,\ldots,\alpha_1\},\ldots,
i_d \in \{0, 1,\ldots,\alpha_d\}$,
we have
\[
(x^{(1)}_{i_1},\ldots,
x^{(d)}_{i_d})\in U\, .
\]
We let $U^{>\alpha<}$ be the set of all
$x\in U^{<\alpha>}$ such that,
for all $i\in \{1,\ldots, d\}$
and $0\leq j<k\leq \alpha_i$,
we have $x^{(i)}_j\not=x^{(i)}_k$.
It is easy to see that~$U^{<\alpha>}$ and~$U^{>\alpha<}$
are open in ${\mathbb K}^{d+|\alpha|}$
and~$U^{>\alpha<}$
is dense in~$U^{<\alpha>}$.
\end{numba}
\begin{example}\label{mimpex}
If $U=U_1\times\cdots\times U_d$
with open sets $U_i\subseteq {\mathbb K}$,
then simply
\begin{equation}\label{spcas}
U^{<\alpha>}\; =\; U_1^{1+\alpha_1}\times
U_2^{1+\alpha_2}\times\cdots
\times U_d^{1+\alpha_d}\,.
\end{equation}
Only this case (in fact only special
cases thereof) was considered
in~\cite{DSm}~and~\cite{Sch}.
\end{example}
\begin{rem}
A simple induction on $|\alpha|$ shows that
the sets $U^{<\alpha>}$ can be defined
alternatively by recursion on $|\alpha|$,
as follows: Set $U^{<0>}:=U$.
Given $\alpha\in {\mathbb N}_0^d$ such that
$|\alpha|\geq 1$, pick $\beta\in {\mathbb N}_0^d$
such that $\alpha=\beta+e_i$
for some $i\in \{1,\ldots, d\}$.
Then
$U^{<\alpha>}$
is the set of all elements $x\in {\mathbb K}^{d+|\alpha|}$
such that\linebreak
$(x^{(1)},\ldots, x^{(i-1)},
x^{(i)}_0,x^{(i)}_1,\ldots,x^{(i)}_{\alpha_i-1},
x^{(i+1)},\ldots, x^{(d)})\in U^{<\beta>}$
holds as well as\linebreak
$(x^{(1)},\ldots, x^{(i-1)},
x^{(i)}_{\alpha_i},x^{(i)}_1,\ldots,x^{(i)}_{\alpha_i-1},
x^{(i+1)},\ldots, x^{(d)})\in U^{<\beta>}$.
\end{rem}
We now define certain mappings
$f^{>\alpha<}\colon U^{>\alpha<}\to F$ and show afterwards
that they can be interpreted
as partial difference
quotient maps.
\begin{defn}
We set $f^{>0<}:=f$.
Given a multi-index $\alpha\in {\mathbb N}_0^d$
such that $|\alpha|\geq 1$,
we define $f^{>\alpha<}(x)$ as the sum
\begin{equation}\label{badbd}
\sum_{j_1=0}^{\alpha_1}\cdots
\sum_{j_d=0}^{\alpha_d}
\left(\prod_{k_1\not=j_1} \frac{1}{x^{(1)}_{j_1}-x^{(1)}_{k_1}}
\cdot\ldots\cdot
\prod_{k_d\not=j_d}\frac{1}{x^{(d)}_{j_d}-x^{(d)}_{k_d}}\right)
f(x^{(1)}_{j_1},\ldots, x^{(d)}_{j_d})
\end{equation}
for $x\in U^{>\alpha<}$,
using the notational conventions
from {\bf\ref{notconv}}.
The products are taken over all
$k_\ell\in \{0,\ldots, \alpha_\ell\}$
such that $k_\ell\not=j_\ell$,
for $\ell\in \{1,\ldots, d\}$.
\end{defn}
The map $f^{>\alpha<}$ has
important symmetry properties.
\begin{la}\label{presymm}
Assume that $\alpha\in {\mathbb N}_0^d$,
$i\in \{1,\ldots, d\}$
and $\pi$ is a permutation of\linebreak
$\{0,1,\ldots, \alpha_i\}$.
Then
$(x^{(1)},\ldots, x^{(i-1)},
x^{(i)}_{\pi(0)},\ldots, x^{(i)}_{\pi(\alpha_i)},
x^{(i+1)},\ldots, x^{(d)})\in U^{>\alpha<}$
for each $x\in U^{>\alpha<}$,
and
\begin{equation}\label{preeqsmm}
f^{>\alpha<}(x^{(1)},\ldots, x^{(i-1)},
x^{(i)}_{\pi(0)},\ldots, x^{(i)}_{\pi(\alpha_i)},
x^{(i+1)},\ldots, x^{(d)})\;=\; f^{>\alpha<}(x)\, .
\end{equation}
\end{la}
The following lemma
shows that $f^{>\alpha<}$ can indeed be
interpreted as a partial difference quotient map.
\begin{la}\label{interpr}
For each $i\in \{1,\ldots, d\}$ and $x\in U^{>e_i<}$,
the element $f^{>e_i<}(x)$ is given by
\[
\frac{f(x^{(1)},\ldots, x^{(i-1)},x^{(i)}_0, x^{(i+1)},\ldots, x^{(d)})
-
f(x^{(1)},\ldots, x^{(i-1)},x^{(i)}_1,
x^{(i+1)},\ldots, x^{(d)})}{x^{(i)}_0-x^{(i)}_1}.
\]
If $\alpha\in {\mathbb N}_0^d$ such that $|\alpha|\geq 2$,
let
$\beta\in {\mathbb N}_0^d$ be a multi-index such that
$\alpha=\beta+e_i$
for some
$i\in \{1,\ldots,d\}$. Then $f^{>\alpha<}(x)$ is given by
\begin{eqnarray}
\hspace*{-5mm}\lefteqn{\frac{1}{x^{(i)}_0-x^{(i)}_{\alpha_i}}\cdot
\Big(f^{>\beta<}(x^{(1)},\ldots, x^{(i-1)},
x^{(i)}_0, x^{(i)}_1, \ldots, x^{(i)}_{\alpha_i-1},
x^{(i+1)},\ldots, x^{(d)})}\quad\quad\quad \notag\\
& & \;\;\; -
f^{>\beta<}(x^{(1)},\ldots, x^{(i-1)},
x^{(i)}_{\alpha_i}, x^{(i)}_1, \ldots, x^{(i)}_{\alpha_i-1},
x^{(i+1)},\ldots, x^{(d)})\Big)
\label{schlmm}
\end{eqnarray}
for all $x\in U^{>\alpha<}$.
\end{la}
\begin{defn}\label{defSDS}
We say that $f$ is $C^0_{SDS}$ if it is continuous,
and define $f^{<0>}:=f$ in this case.
Recursively, given an integer $k\geq 1$
we say that
$f$ is $C^k_{SDS}$
if $f$ is $C^{k-1}_{SDS}$
and, for each multi-index
$\alpha\in {\mathbb N}_0^d$ such that $|\alpha|=k$,
there exists
a continuous map
$f^{<\alpha>}\colon U^{<\alpha>}\to F$
such that $f^{<\alpha>}|_{U^{>\alpha<}}=f^{>\alpha<}$.
As usual, $f$ is called $C^\infty_{SDS}$
if $f$ is $C^k_{SDS}$ for each $k\in {\mathbb N}_0$.
\end{defn}
Since $U^{>\alpha<}$ is dense in
$U^{<\alpha>}$, the continuous extension
$f^{<\alpha>}$ of~$f^{>\alpha<}$
is unique whenever it exists.
We readily deduce from Lemma~\ref{presymm}:
\begin{la}\label{partsymm}
Let $f$ be a $C^k$-map
for some $k\in {\mathbb N}$, $\alpha\in {\mathbb N}_0^d$
with $|\alpha|=k$,\linebreak
$i\in \{1,\ldots, d\}$
and $\pi$ be a permutation of
$\{0,1,\ldots, \alpha_i\}$.
Then
\begin{equation}\label{nweq}
(x^{(1)},\ldots, x^{(i-1)},
x^{(i)}_{\pi(0)},\ldots, x^{(i)}_{\pi(\alpha_i)},
x^{(i+1)},\ldots, x^{(d)})\; \in \; U^{<\alpha>}
\end{equation}
for each $x\in U^{<\alpha>}$, and
\begin{equation}\label{eqsmm}
f^{<\alpha>}(x^{(1)},\ldots, x^{(i-1)},
x^{(i)}_{\pi(0)},\ldots, x^{(i)}_{\pi(\alpha_i)},
x^{(i+1)},\ldots, x^{(d)})\;=\; f^{<\alpha>}(x)\, .
\end{equation}
\end{la}
The following variant of Lemma~\ref{interpr}
is available for~$f^{<\alpha>}$.
\begin{la}\label{extmore}
Let $f$ be a $C^k_{SDS}$-map
for an integer $k\geq 2$,
$\alpha\in {\mathbb N}_0^d$ such that $|\alpha|=k$,
and $\beta\in {\mathbb N}_0^d$ such that
$\alpha=\beta+e_i$ for some
$i\in \{1,\ldots, d\}$.
Then $f^{<\alpha>}(x)$ is given by
\begin{eqnarray}
\hspace*{-5mm}
\lefteqn{\frac{1}{x^{(i)}_0-x^{(i)}_{\alpha_i}}
\cdot \Big(f^{<\beta>}(x^{(1)},\ldots, x^{(i-1)},
x^{(i)}_0, x^{(i)}_1, \ldots, x^{(i)}_{\alpha_i-1},
x^{(i+1)},\ldots, x^{(d)})}\qquad\quad \notag\\
& & \;\;\; -
f^{<\beta>}(x^{(1)},\ldots, x^{(i-1)},
x^{(i)}_{\alpha_i}, x^{(i)}_1, \ldots, x^{(i)}_{\alpha_i-1},
x^{(i+1)},\ldots, x^{(d)})\Big)\label{schlmm2}
\end{eqnarray}
for all $x\in U^{<\alpha>}$
such that $x^{(i)}_0\not=x^{(i)}_{\alpha_i}$.
\end{la}
\begin{rem}\label{onlynois}
If $U\subseteq {\mathbb K}^d$ is a subset
(possibly with empty interior)
of the form $U=U_1\times\cdots\times U_d$,
where $U_i\subseteq {\mathbb K}$ is a non-empty subset
without isolated points for $i\in \{1,\ldots, d\}$,
then Definition~\ref{defSDS} can be used
just as well to define $C^k_{SDS}$-maps
$f\colon U\to F$.
\end{rem}
If $d=1$, we also write
$f^{<j>}$ in place of
$f^{<j e_1>}$, as in~\cite[\S\,6]{BGN}.
\subsection*{{\normalsize Seminorms and gauges}}
Gauges on topological vector spaces over
valued fields were introduced in~\cite{IM2} as a substitute
for continuous seminorms when dealing with a
general topological vector space,
the topology of which need not come from a family
of continuous seminorms
(cf.\ \cite[\S\,6.3]{Jar} for the real case).
We only recall some essentials here; see~\cite{IM2}
for further information.
\begin{defn}
Let $E$ be a topological
vector space over a valued field
$({\mathbb K},|.|)$. A \emph{gauge on~$E$}
is a map $q \colon E\to [0,\infty[$
(also written $\|.\|_q := q$)
which satisfies
$q (t x) = |t|q(x)$ for all $t\in {\mathbb K}$
and $x\in E$,
and such that $B_r^q(0)$
is a $0$-neighbourhood
for each $r>0$,
where
$B_r^q(x):=\{y\in E\colon \|y-x\|_q<r\}$
for all $x\in E$ and $r>0$.
We also define
$\overline{B}_r^q(x):=\{y\in E\colon \|y-x\|_q\leq r\}$.
If $(E,\|.\|)$ is a normed
space, we relax notation and write
$B^E_r(x):=B^{\|.\|}_r(x)$.
\end{defn}
In~\cite{IM2},
only upper semicontinuous gauges~$q$
were considered, i.e., it was required that
$B^q_r(0)$ is an open $0$-neighbourhood,
for each $r>0$.
\begin{rem}\label{minkow}
Typical examples
of gauges are Minkowski functionals~$\mu_U$
of balanced, open $0$-neighbourhoods~$U$
in a topological vector space~$E$
over a valued field~${\mathbb K}$;
these are upper semicontinuous
(see \cite[Remark\,1.21]{IM2}).
Here $U\subseteq E$ is called
\emph{balanced} if $tU\subseteq U$
for all $t\in {\mathbb K}$ such that $|t|\leq 1$.
The Minkowski functional
is
$\mu_U\colon E\to [0,\infty[$,
$x\mapsto \inf\{|t|\colon \mbox{$t\in {\mathbb K}^\times$
with $x\in tU$}\}$.
\end{rem}
\begin{rem}\label{remfk}
Note that gauges
need not satisfy the triangle inequality.
But we still have a certain substitute:
Given a gauge $q\colon E\to [0,\infty[$,
there always exists a gauge
$p\colon E\to [0,\infty[$
such that
\begin{equation}\label{faketri}
\|x+y\|_q\;\leq\; \|x\|_p+\|y\|_p
\quad\mbox{for all $x,y\in E$}
\end{equation}
(cf.\ \cite[Lemma~1.29]{IM2}). We shall refer to
(\ref{faketri}) as the \emph{fake
triangle inequality}.
\end{rem}
As in the case of continuous seminorms, it
frequently suffices to consider
a sufficiently large set of gauges:
\begin{defn}\label{funsys}
A set $\Gamma$ of gauges
on a topological ${\mathbb K}$-vector space~$E$
is called a \emph{fundamental system
of gauges} if each $0$-neighbourhood in~$E$
contains some finite intersection
of balls of the form $B_r^q(0)$,
with $q\in \Gamma$ and $r>0$.
\end{defn}
Cf.\ \cite[Lemma 1.24]{IM2}
for the following lemma, which is useful
to determine fundamental systems
of gauges.
\begin{la}\label{findfund}
Let $p,q\colon E\to [0,\infty[$
be gauges on a topological vector space~$E$ over
a valued field ${\mathbb K}$.
If there exist $r,s>0$ such that
$B^q_s(0)\subseteq B^p_r(0)$,
then
\[
p \; \leq rs^{-1}|a|^{-1}q\, ,
\]
for each $a\in {\mathbb K}^\times$ such that $|a|<1$.
In particular, $p\leq C q$ for some $C>0$.\,\nopagebreak\hspace*{\fill}$\Box$
\end{la}
\begin{rem}\label{examfun}
Combining Remark~\ref{minkow}
and Lemma~\ref{findfund}, it is easy to see
that upper semicontinuous gauges form a
fundamental system of gauges,
for each topological vector space over
a valued field (cf.\ also \cite[Remark~1.21]{IM2}).
In the real case, continuous
gauges form a fundamental system
(cf.\ \cite[\S\,6.4]{Jar}).
\end{rem}
\begin{examples}
Given $r\in \;]0,1]$,
a gauge $q\colon E\to [0,\infty[$
is called an $r$-seminorm if
$q(x+y)^r\leq q(x)^r+q(y)^r$
for all $x,y\in E$.
If, furthermore,
$q(x)=0$ if and only if $x=0$,
then~$q$ is called an $r$-norm
(cf.\ \cite[\S\,6.3]{Jar}
for the real case).
For examples of $r$-normed spaces over~${\mathbb R}$
and more general non-locally convex
real topological vector spaces,
the reader is referred to \cite[\S\,6.10]{Jar}
and~\cite{Kal}.
For ${\mathbb K}$ a valued field,
the simplest
examples are the spaces
$\ell^p({\mathbb K})$
of all $x=(x_n)_{n\in{\mathbb N}}
\in {\mathbb K}^{\mathbb N}$
such that $\|x\|_p:=\sqrt[p]{\sum_{n=1}^\infty|x_n|^p}<\infty$,
for $p\in \;]0,1[$.
Then $\|.\|_p$
is a $p$-norm
on $\ell^p({\mathbb K})$ defining a Hausdorff
vector topology on this space
(and thus $\{\|.\|_p\}$ is a fundamental
system of gauges).
\end{examples}
\subsection*{{\normalsize Bounded sets and bounded maps}}
Let $E$ be a topological
vector space over a topological
field~${\mathbb K}$.
Recall that a subset $B\subseteq E$
is called \emph{bounded}
if, for each $0$-neighbourhood
$U\subseteq E$, there exists a $0$-neighbourhood
$V\subseteq {\mathbb K}$ such that $VB\subseteq U$.
If ${\mathbb K}$ is a valued field,
we can test boundedness using gauges.
\begin{la}\label{bdviagau}
Let $E$ be a topological vector space
over a valued field~${\mathbb K}$.
Then a subset $B\subseteq E$ is bounded
if and only if the set $q(B)\subseteq {\mathbb R}$ is bounded,
for each gauge $q$ on~$E$.
\end{la}
It suffices to show $\,\sup \|B\|_q<\infty$
for $q$ in a fundamental system
of gauges. In Section~\ref{secmetr},
$C^k$-maps with bounded difference
quotients will play a vital~role.
\begin{defn}
Let $E$ be a topological vector space
over a topological field~${\mathbb K}$.
\begin{itemize}
\item[(a)]
If~$X$ is a topological space,
then $BC(X,E)$ denotes the set
of all continuous maps $\gamma\colon X\to E$
whose image $\gamma(X)$ is bounded in~$E$.
Then $BC(X,E)$ is a vector subspace
of $E^X$.
\item[(b)]
If $k\in {\mathbb N}_0\cup\{\infty\}$ and
$U\subseteq {\mathbb K}$ is a subset without isolated points,
we let
$BC^k(U,E)$ be the space of all $C^k_{SDS}$-maps
$\gamma\colon U\to E$ such that
$\gamma^{<j>}\in BC(U^{j+1},E)$
for all $j\in {\mathbb N}_0$ such that $j\leq k$.
\end{itemize}
\end{defn}
We mention that $BC^k(U,E)$ can be made a topological
vector space~\cite[Definition~1.2]{GaLa},
but we shall not use this topology.
If~${\mathbb K}$ is a valued field,
$j\in {\mathbb N}_0$ with $j\leq k$ and
$q$ a gauge on~$E$, we define
\begin{equation}\label{nogau}
\|\gamma^{<j>}\|_{q,\infty}\; :=\;
\sup\{\|\gamma^{<j>}(x)\|_q\colon x\in U^{j+1}\}\quad
\mbox{for $\gamma\in BC^k(U,E)$.}
\end{equation}
\subsection*{{\normalsize H\"{o}lder continuity}}
Using gauges, we now define a version of H\"{o}lder
continuous maps and record their basic properties.
\begin{defn}\label{deflipcts}
Let $E$ and $F$
be topological vector spaces over a valued field~${\mathbb K}$,
and $U\subseteq E$ be a subset.
A map $f\colon U\to F$
is called \emph{H\"{o}lder continuous
of exponent $\sigma\in\;]0,\infty[$}
(or $C^{0,\sigma}$)
if, for every $x_0\in U$ and gauge $q$
on $F$, there exists a gauge~$p$ on~$E$
and a neighbourhood $V\subseteq U$ of~$x_0$
such that
\begin{equation}\label{hoelcon}
\|f(y)-f(x)\|_q \; \leq \; \big(\|y-x\|_p\big)^\sigma
\quad
\mbox{for all $\, x,y\in V$.}
\end{equation}
$C^{0,1}$-maps are also called
\emph{Lipschitz continuous.}
\end{defn}
We remind the reader that
H\"{o}lder exponents $\sigma>1$
are meaningful in non-archimedian analysis
(see \cite[Exercise~26.B]{Sch} for an
instructive example).
We are mostly interested in H\"{o}lder
exponents $\sigma\in \;]0,1]$,
but some of the results are valid
just as well for $\sigma>1$.
\begin{la}\label{baseHglob}
Let $E$, $F$ and $H$ be topological
vector spaces over a valued field~${\mathbb K}$,
$U\subseteq E$ and $V\subseteq F$ be subsets,
$f\colon U\to V\subseteq F$
and $g\colon V\to H$ be maps,
and $\sigma,\tau > 0 $.
Then the following holds:
\begin{itemize}
\item[\rm (a)]
If $f$ is $C^{0,\sigma}$,
then $f$ is continuous.
\item[\rm (b)]
If $f$ is $C^{0,\sigma}$ and $\sigma\geq \tau$,
then $f$ is also $C^{0,\tau}$.
\item[\rm (c)]
If $f$ is $C^{0,\sigma}$ and $g$ is $C^{0,\tau}$,
then $g\circ f$ is $C^{0,\sigma\cdot\tau}$.
\item[\rm (d)]
If $U$ has dense interior
and $f$ is $C^1_{BGN}$, then $f$ is Lipschitz continuous.
\end{itemize}
\end{la}
In connection with Part\,(d) of the preceding lemma,
note that $\id\colon {\mathbb K}\to{\mathbb K}$, $x\mapsto x$
is $C^\infty_{BGN}$ but not $C^{0,\sigma}$ for any $\sigma>1$.\\[2.5mm]
If $f$ is not H\"{o}lder continuous,
then pairs of points with pathological
behaviour
can always be chosen in a given dense set.
This will become essential later.
\begin{la}\label{densehoel}
Let $E$ and $F$ be topological
vector spaces over a valued field~${\mathbb K}$,
$f\colon U\to F$ be a continuous mapping
on a subset
$U\subseteq E$,
$D\subseteq U$ be a dense subset,
and $\sigma > 0$.
If $f$ is not $C^{0,\sigma}$,
then there exists $x_0\in U$
and a gauge~$q$ on~$F$ such that,
for each neighbourhood $V\subseteq U$
of~$x_0$ and gauge~$p$ on~$E$,
there exist $x,y\in V\cap D$ such that
$\|f(x)-f(y)\|_q>(\|x-y\|_p)^\sigma$.
\end{la}
\subsection*{{\normalsize Two approaches to
H\"{o}lder differentiable maps}}
We define $k$ times H\"{o}lder differentiable maps
and record some properties.
\begin{defn}
Let ${\mathbb K}$ be a valued field,
$E$ and $F$ be topological ${\mathbb K}$-vector spaces,
$U\subseteq E$ be a subset
with dense interior,
and $f\colon U\to F$ be a mapping.
Let $k\in {\mathbb N}_0\cup\{\infty\}$
and $\sigma>0$.
We say that $f$ is $C^{\,k,\sigma}_{BGN}$
if $f$ is $C^k_{BGN}$
and $f^{[j]}\colon U^{[j]}\to F$
is $C^{0,\sigma}$
for all $j\in {\mathbb N}_0$
such that $j\leq k$
(where $f^{[0]}:=f$).
\end{defn}
\begin{rem}\label{autolip}
Note that, if $\sigma\in \;]0,1]$,
then $f^{[j]}$ is $C^1_{BGN}$ for $j<k$
and hence automatically~$C^{0,\sigma}$,
by Lemma~\ref{baseHglob}~(b) and~(d).
In particular, if $\sigma\in \;]0,1]$,
then $f$ is $C^{\infty,\sigma}_{BGN}$ if and only if
$f$ is $C^\infty_{BGN}$.
And if $k$ is finite, then
only $f^{[k]}$ requires attention.
\end{rem}
\begin{rem}
$k$ times Lipschitz differentiable mappings
($C^{k,1}$-maps)
form a particularly nice class
of maps (see~\cite{IM2}).
Notably, \cite[Theorem~5.2]{IM2} provides
an implicit function theorem
for $C^{k,1}$-maps
from arbitrary topological vector spaces
to Banach spaces, for each valued field
and $k\in {\mathbb N}\cup\{\infty\}$.
\end{rem}
We need some basic information on mappings
into direct products.
\begin{la}\label{prodhoel}
Let $E$ be a topological ${\mathbb K}$-vector space
over a valued field~${\mathbb K}$, $(F_i)_{i\in I}$
be a family of topological ${\mathbb K}$-vector spaces
and $f
\colon U\to F$ be a map into
$F:=\prod_{i\in I}F_i$,
defined on a non-empty subset $U\subseteq E$
with dense interior. Let $k\in {\mathbb N}_0$ and $\sigma>0$;
for $i\in I$, let $\pr_i\colon F\to F_i$ be the projection.
Then $f$ is $C^{k,\sigma}_{BGN}$ if and only if
each of its components
$f_i:=\pr_i\circ f \colon U\to F_i$ is $C^{k,\sigma}_{BGN}$.
\end{la}
The Chain Rule is available in the following form.
\begin{la}\label{chainLip}
Let ${\mathbb K}$ be a valued field,
$E$, $F$ and $H$ be topological ${\mathbb K}$-vector spaces,
$U\subseteq E$ and $V\subseteq F$ be subsets with dense interior,
$\sigma\in \;]0,1]$, $\tau >0$,
$f\colon U\to V$ be $C^{\,k,\sigma}_{BGN}$,
and $g\colon V\to H$ be
$C^{\,k,\tau}_{BGN}$.
Then $g\circ f\colon U\to H$
is $C^{\,k,\sigma\cdot\tau}_{BGN}$.
\end{la}
The following variant even holds if $\sigma>1$:
\begin{la}\label{chainlin}
Let ${\mathbb K}$ be a valued field,
$E$, $F$ and $H$ be topological ${\mathbb K}$-vector spaces,
$\lambda\colon F\to H$ be continuous linear,
$U\subseteq E$ be a subset with dense interior,
$\sigma> 0$
and $f\colon U\to F$ be a
$C^{\,k,\sigma}_{BGN}$-map, where $k\in {\mathbb N}_0$.
Then $\lambda \circ f\colon U\to H$
is $C^{\,k,\sigma}_{BGN}$,
and $(\lambda\circ f)^{[k]}=\lambda
\circ f^{[k]}$.
\end{la}
\begin{la}\label{hoelloc}
Let $E$ and $F$ be topological vector spaces over a
valued field~${\mathbb K}$,
and $f\colon  U\to F$ be a map, defined
on a subset $U\subseteq E$
with dense interior.
Let $k\in {\mathbb N}_0\cup\{\infty\}$
and $\sigma\in \;]0,1]$.
If there is an open cover $(U_i)_{i\in I}$
of~$U$ such that
$f|_{U_i}\colon  U_i\to F$ is
$C^{k,\sigma}_{BGN}$ for each $i\in I$,
then $f$ is~$C^{k,\sigma}_{BGN}$.
\end{la}
H\"{o}lder differentiable maps
can also be defined using the approach of Schikhof and De Smedt.
\begin{defn}
Let ${\mathbb K}$ be a valued field, $d\in {\mathbb N}_0$,
$F$ be a topological ${\mathbb K}$-vector space
and $f\colon U\to F$ be a mapping,
where $U\subseteq {\mathbb K}^d$ is open or
$U=U_1\times \cdots \times U_d$
for certain sets $U_1,\ldots, U_d \subseteq {\mathbb K}$ without
isolated points.
Let $k\in {\mathbb N}_0\cup\{\infty\}$
and $\sigma > 0 $.
We say that $f$ is $C^{\,k,\sigma}_{SDS}$
if $f$ is $C^k_{SDS}$
and $f^{<\alpha>}\colon U^{<\alpha>}\to F$
is $C^{0,\sigma}$
for all $\alpha \in {\mathbb N}_0^d$
such that $|\alpha|\leq k$
$($where $f^{<0>}:=f)$.
\end{defn}
\section{{\boldmath$C^k_{BGN}$}-maps and
{\boldmath$C^k_{SDS}$}-maps coincide}\label{seccoinc}
In this section, we show that the approach of
Bertram, Gl\"{o}ckner and Neeb
and the approach of Schikhof and De Smedt
give rise to the same classes of~$C^k$-maps
and $C^{k,\sigma}$-maps
on open domains (and more generally).
Throughout the section,
${\mathbb K}$ is a topological field,
$d\in {\mathbb N}$ and
$f\colon U\to F$ a map
to a topological ${\mathbb K}$-vector space~$F$,
where $U\subseteq {\mathbb K}^d$ is open or of the
form $U=U_1\times\cdots\times U_d$
for certain sets $U_1,\ldots, U_d\subseteq {\mathbb K}$
with dense interior.
\begin{thm}\label{mainthm}
The following holds
for each $k\in {\mathbb N}_0\cup\{\infty\}$:
\begin{itemize}
\item[\rm (a)]
$f$ is
$C^k_{SDS}$
if and only if $f$ is $C^k_{BGN}$.
\item[\rm(b)]
If ${\mathbb K}$ is a valued field
and $\sigma\in \;]0,1]$,
then $f$ is
$C^{\,k,\sigma}_{SDS}$
if and only if $f$ is $C^{\,k,\sigma}_{BGN}$.
\end{itemize}
\end{thm}
Various lemmas are useful for the proof
of Theorem~\ref{mainthm}.
\begin{la}\label{enindSDS}
Let $k\in {\mathbb N}_0$.
\begin{itemize}
\item[\rm(a)]
If $f\colon {\mathbb K}^d\supseteq U\to F$
is $C^1_{SDS}$ and $f^{<e_i>}$
is $C^k_{SDS}$ for each $i\in \{1,\ldots, d\}$,
then $f$ is $C^{k+1}_{SDS}$.
\item[\rm (b)]
Let ${\mathbb K}$ be a valued field
and $\sigma > 0$.
If $f\colon {\mathbb K}^d\supseteq U\to F$
is $C^{\,1,\sigma}_{SDS}$ and $f^{<e_i>}$
is $C^{\,k,\sigma}_{SDS}$ for each $i\in \{1,\ldots, d\}$,
then $f$ is $C^{\,k+1,\sigma}_{SDS}$.
\end{itemize}
\end{la}
\begin{proof}
Given $\alpha\in {\mathbb N}_0^d$ such that $1\leq |\alpha|\leq k+1$,
there is $i\in \{1,\ldots, d \}$
such that $\alpha_i>0$. Then $\beta:=\alpha-e_i\in
{\mathbb N}_0^d$ and $|\beta|=|\alpha|-1\leq k$.
Write $\beta=(\beta_1,\ldots, \beta_d)$
and set $\beta':=(\beta_1,\ldots,\beta_i,0,\beta_{i+1},\ldots,
\beta_d)\in {\mathbb N}_0^{d+1}$.
For
$x\in U^{>\alpha<}$,
using the notational conventions from~{\bf\ref{notconv}}, we have
\[
f^{>\alpha<}(x)=
(f^{>e_i<})^{>\beta'\!<}
(x^{(1)};\ldots; x^{(i-1)};
x^{(i)}_0,x^{(i)}_2,\ldots, x^{(i)}_{\alpha_i};
x^{(i)}_1;
x^{(i+1)};\ldots; x^{(d)}),
\]
as is clear from the definitions.
Thus
\[
f^{<\alpha>}(x):=
(f^{<e_i>})^{<\beta'\!>}
(x^{(1)};\ldots; x^{(i-1)};
x^{(i)}_0,x^{(i)}_2,\ldots, x^{(i)}_{\alpha_i};
x^{(i)}_1;
x^{(i+1)};\ldots; x^{(d)})
\]
for $x\in U^{<\alpha>}$
defines a continuous (resp., $C^{\,0,\sigma}$-)
extension
$f^{<\alpha>}\colon U^{<\alpha>}\to F$
of $f^{>\alpha<}$,
whenever $|\alpha|\leq k+1$.
Therefore, $f$ is $C^{k+1}_{SDS}$
(resp., $C^{\,k+1,\sigma}_{SDS}$).
\end{proof}
The next lemma establishes one implication
in Theorem~\ref{mainthm} (a) and (b).
Note that $\sigma$ need not be $\leq 1$
here.
\begin{la}\label{BGNSDS}
If $f$ is $C^k_{BGN}$
for some $k\in {\mathbb N}_0$
$($resp., $C^{\,k,\sigma}_{BGN}$
if ${\mathbb K}$ is a valued field
and $\sigma>0 )$,
then $f$ is $C^k_{SDS}$
$($resp., $C^{\,k,\sigma}_{SDS})$.
\end{la}
\begin{proof}
The proof is by induction
on $k\in {\mathbb N}_0$. The case $k=0$ is trivial.
If $k\geq 1$,
let $i\in \{1,\ldots, n\}$.
For each $x\in U^{>e_i<}$,
we then have
\[
f^{>e_i<}(x)\;=\; f^{[1]}(x^{(1)},\ldots,x^{(i-1)},
x^{(i)}_1,x^{(i+1)},\ldots, x^{(d)}; e_i; \, x^{(i)}_0-x^{(i)}_1)\,.
\]
Thus
\begin{equation}\label{RHint}
f^{<e_i>}(x)\;:=\; f^{[1]}(x^{(1)},\ldots,x^{(i-1)},
x^{(i)}_1,x^{(i+1)},\ldots, x^{(d)}; e_i; \, x^{(i)}_0-x^{(i)}_1)
\end{equation}
for $x\in U^{<e_i>}$
defines a continuous extension
of $f^{>e_i<}$
and hence $f$ is $C^1_{SDS}$,
with $f^{<e_i>}$ as just described.
Note that the right hand side of (\ref{RHint})
expresses $f^{<e_i>}$
as a composition of the $C^{k-1}_{BGN}$-map
(resp., $C^{\,k-1,\sigma}_{BGN}$-map)
$f^{[1]}$ and the restriction
of a map ${\mathbb K}^{d+1}\to {\mathbb K}^d\times{\mathbb K}^d\times{\mathbb K}$
which is
continuous affine-linear and hence $C^\infty_{BGN}$.
By the Chain Rule (resp., Lemma~\ref{chainLip}),
$f^{<e_i>}$ is $C^{k-1}_{BGN}$
(resp., $C^{\,k-1,\sigma}_{BGN}$)
and hence $C^{k-1}_{SDS}$
(resp., $C^{\,k-1,\sigma}_{SDS}$),
by induction.
Hence $f$ is $C^{k+1}_{SDS}$ (resp.,
$C^{\,k+1,\sigma}_{SDS}$), by Lemma~\ref{enindSDS}.
\end{proof}
\begin{la}\label{preSDS}
If $f$ is $C^k_{SDS}$ for some $k\in {\mathbb N}_0$
$($resp., if ${\mathbb K}$ is a valued field
and $f$ is $C^{\,k,\sigma}_{SDS}$
for some $\sigma>0 )$,
then $f^{<\alpha>}$ is $C^{k-|\alpha|}_{SDS}$
$($resp., $C^{k-|\alpha|,\sigma}_{SDS})$
for each $\alpha\in {\mathbb N}_0^d$ such that
$|\alpha|\leq k$.
\end{la}
\begin{rem}\label{remeazy}
The proof of Lemma~\ref{preSDS}
will provide the following formula
for $(f^{<\alpha>})^{<\beta>}$,
if $\alpha\in {\mathbb N}_0^d$ such that $|\alpha|\leq k$
and $\beta\in {\mathbb N}_0^{d+|\alpha|}$
such that $|\beta|\leq k-|\alpha|$:
\begin{equation}\label{simpfml}
(f^{<\alpha>})^{<\beta>}\;=\;f^{<\alpha+\bar{\beta}>}\,,
\end{equation}
where $\bar{\beta}\in {\mathbb N}_0^d$ is defined by
$\bar{\beta}_j:=\sum_{i=s_j}^{s_{j+1}-1}\beta_i$
for $j\in \{1,\ldots, d\}$,
with $s_j:=j+\sum_{i=1}^{j-1}\alpha_i$
and $s_{d+1}:=d+|\alpha|+1$.
\end{rem}
\textbf{Proof of Lemma~\ref{preSDS}.}
Given $\alpha\in {\mathbb N}_0^d$,
we first show by induction on\linebreak
$\ell \in \{0,\ldots,k-|\alpha|\}$ that
\begin{equation}\label{simpfml2}
(f^{>\alpha<})^{>\beta<}\;=\;f^{>\alpha+\bar{\beta}<}\,,
\end{equation}
for each $\beta\in {\mathbb N}_0^{d+|\alpha|}$
such that $|\beta|=\ell$,
where~$\bar{\beta}$ (and $s_1,\ldots, s_{d+1}$)
are as in Remark~\ref{remeazy}.
The case $\ell=0$ being trivial,
let us assume now that (\ref{simpfml2})
holds for some $\beta$ with $|\beta|<k-|\alpha|$.
For each $i \in \{1,\ldots, d+|\alpha|\}$,
we have to show that~(\ref{simpfml2})
holds with $\beta$ replaced by $\gamma:=\beta+e_i$
and $\bar{\beta}$ replaced by the corresponding~$\bar{\gamma}$.
There is a unique
$j \in \{1,\ldots, d\}$
such that $s_j\leq i <s_{j+1}$.
Then $\bar{\gamma}=\bar{\beta}+e_j$.
Given $x\in (U^{>\alpha<})^{>\gamma<}
=(U^{>\alpha<})^{>\beta+e_i<}\subseteq ({\mathbb K}^{d+|\alpha|})^{>\beta+e_i<}$,
we abbreviate
$y:=(x^{(1)},\ldots, x^{(s_j-1)})$
and $z:=(x^{(s_j+1)},\ldots, x^{(d+|\alpha|)})$.
Define $t:=
x_0^{(i)}-x^{(i)}_{\beta_i+1}$.
Then $t \cdot
(f^{>\alpha<})^{>\gamma<}(x)$
is given by
\begin{eqnarray*}
\!\!\! & &\!\!\!
(f^{>\alpha<})^{>\beta<}(
x^{(1)};\ldots; x^{(i-1)};
x^{(i)}_0, x^{(i)}_1,\ldots, x^{(i)}_{\beta_i};
x^{(i+1)};\ldots; x^{(d+|\alpha|)})\\
\!\!\! &&\!\!\! -\,
(f^{>\alpha<})^{>\beta<}(x^{(1)};\ldots; x^{(i-1)};
x^{(i)}_{\beta_i+1},x^{(i)}_1,\ldots, x^{(i)}_{\beta_i};
x^{(i+1)};\ldots; x^{(d+|\alpha|)})\\
\!\!\!&=&
\!\!\! f^{>\alpha+\bar{\beta}<}(x^{(1)};\ldots; x^{(i-1)};
x^{(i)}_0,x^{(i)}_1,\ldots, x^{(i)}_{\beta_i};
x^{(i+1)};\ldots; x^{(d+|\alpha|)})\\
\!\!\! &&\!\!\! -\,
f^{>\alpha+\bar{\beta}<} (x^{(1)};\ldots; x^{(i-1)};
x^{(i)}_{\beta_i+1},x^{(i)}_1,\ldots, x^{(i)}_{\beta_i};
x^{(i+1)};\ldots; x^{(d+|\alpha|)})\\
\!\!\!&=& \!\!\!
f^{>\alpha+\bar{\beta}<}(y; x^{(s_j)};\ldots; x^{(i-1)};
x^{(i)}_0,x^{(i)}_1,\ldots, x^{(i)}_{\beta_i};
x^{(i+1)},\ldots, x^{(s_{j+1}-1)}; z)\\
\!\!\! &&\!\!\! -\,
f^{>\alpha+\bar{\beta}<}(y; x^{(s_j)};\ldots; x^{(i-1)};
x^{(i)}_{\beta_i+1},x^{(i)}_1,\ldots, x^{(i)}_{\beta_i};
x^{(i+1)};\ldots;  x^{(s_{j+1}-1)}; z)\\
\!\!\!&=&\!\!\!
f^{>\alpha+\bar{\beta}<}(y;
x^{(i)}_0,x^{(i)}_1,\ldots, x^{(i)}_{\beta_i};
x^{(s_j)};\ldots;x^{(i-1)};
x^{(i+1)};\ldots;x^{(s_{j+1}-1)}; z)\\
\!\!\! &&\!\!\! -\,
f^{>\alpha+\bar{\beta}<} (y;
x^{(i)}_{\beta_i+1},x^{(i)}_1,\ldots, x^{(i)}_{\beta_i};
x^{(s_j)};\ldots;x^{(i-1)};
x^{(i+1)};\ldots;x^{(s_{j+1}-1)}; z)\\
\!\!\!&=&\!\!\!\!
t f^{>\alpha+\bar{\gamma}<}(y;
x^{(i)}_0,x^{(i)}_1,\ldots, x^{(i)}_{\beta_i};
x^{(s_j)};\ldots;x^{(i-1)};
x^{(i+1)};\ldots;x^{(s_{j+1}-1)}; x^{(i)}_{\beta_i+1}; z)\\
\!\!\!&=&\!\!\!
t f^{>\alpha+\bar{\gamma}<}(y;x^{(s_j)};\ldots;
x^{(s_{j+1}-1)};z)\;=\;
t f^{>\alpha+\bar{\gamma}<}(x)\,,
\end{eqnarray*}
using~(\ref{simpfml2}) for the first equality
and the symmetry properties
of $f^{>\alpha+\bar{\beta}<}$
and $f^{>\alpha+\bar{\gamma}<}$
(as in Lemma~\ref{presymm})
for the third and penultimate
equality.
This completes the inductive proof of (\ref{simpfml2}).\\[2.5mm]
As a consequence of (\ref{simpfml2}),
the continuous (resp., $C^{0,\sigma}$-) map
$f^{<\alpha+\bar{\beta}>}$
extends the map $(f^{>\alpha<})^{>\beta<}$.
It hence also extends $(f^{<\alpha>})^{>\beta<}$,
for each $\beta\in {\mathbb N}_0^{d+|\alpha|}$
with $|\beta|\leq k-|\alpha|$.
Hence $f^{<\alpha>}$ is $C^{k-|\alpha|}_{SDS}$
(resp., $C^{k-|\alpha|,\sigma}_{SDS}$)
and (\ref{simpfml}) holds.\vspace{2.5mm}\nopagebreak\hspace*{\fill}$\Box$

\noindent
{\bf Proof of Theorem~\ref{mainthm},
completed.}
It remains to show that if $f$ is $C^k_{SDS}$
(resp., $C^{k,\sigma}_{SDS}$ with $\sigma\in \;]0,1]$),
then $f$ is $C^k_{BGN}$ (resp., $C^{k,\sigma}_{BGN}$).
We assume first that
$U=U_1\times\cdots\times U_d$
for certain subsets $U_i\subseteq {\mathbb K}$.
The proof is by induction
on $k\in {\mathbb N}_0$. The case $k=0$ being trivial,
assume now that $f$ is $C^k_{SDS}$ (resp.,
$C^{\,k,\sigma}_{SDS}$) for some $k\geq 1$.
For each $(x,y,t)\in U^{]1[}$,
we have
\begin{eqnarray}
\hspace*{-1mm}\lefteqn{f^{]1[}(x,y,t)}\quad\notag\\
& \!\! = \!\! & \frac{f(x+ty)-f(x)}{t}
\, =\,  \sum_{j=1}^d
\frac{f(x+t\sum_{i=1}^jy_ie_i)-f(x+t\sum_{i=1}^{j-1}y_ie_i)}{t}\notag\\
& \!\! = \!\! &  \sum_{j=1}^d
y_jf^{<e_j>}(x_1\! +\! ty_1;\ldots;x_{j-1}\! +\! ty_{j-1};
x_j,x_j\! + \! ty_j;x_{j+1},\ldots, x_d),\;  \label{rhs}
\end{eqnarray}
because
$a_j:=\frac{f(x+t\sum_{i=1}^jy_ie_i)-f(x+t\sum_{i=1}^{j-1}y_ie_i)}{t}
=y_j\frac{f(x+t\sum_{i=1}^jy_ie_i)-f(x+t\sum_{i=1}^{j-1}y_ie_i)}{y_jt}$
coincides with
$b_j\! :=\! y_jf^{<e_j>}(x_1+ty_1;\ldots;x_{j-1}+ty_{j-1};
x_j,x_j+ty_j;x_{j+1},\ldots, x_d)$
if $y_j\not=0$,
while both~$a_j$ and~$b_j$
vanish if $y_j=0$.
Since the right hand side of~(\ref{rhs})
defines a continuous (resp., $C^{0,\sigma}$-) map on all of $U^{[1]}$,
we see that $f$ is $C^1_{BGN}$ (resp., $C^{\,1,\sigma}_{BGN}$) with
\begin{eqnarray}
\lefteqn{f^{[1]}(x,y,t)}\quad \notag\\
\!\!\!\!\! &\! \!\! \!\! =\!\! &
\sum_{j=1}^d y_jf^{<e_j>}(x_1 + ty_1;\ldots;
x_{j-1} +  ty_{j-1}; x_j, x_j + ty_j;x_{j+1},\ldots, x_d)\;\;\; \label{fviaphi}
\end{eqnarray}
for all $(x,y,t)\in U^{[1]}$.
Here $f^{<e_j>}$
is a $C^{k-1}_{SDS}$-map (resp.,
a $C^{\,k-1,\sigma}_{SDS}$-map)
on $U^{<e_j>}=U_1\times\cdots\times U_{j-1}\times
U_j\times U_j\times U_{j+1}\times\cdots\times U_d$
by Lemma~\ref{preSDS}
and hence a $C^{k-1}_{BGN}$-map (resp.,
a $C^{\,k-1,\sigma}_{BGN}$-map),
by induction.
Formula (\ref{fviaphi})
now shows that $f^{[1]}$ is
built up from $f^{<e_j>}$
and various smooth maps,
whence~$f^{[1]}$ is $C^{k-1}_{BGN}$ (resp.,
$C^{\,k-1,\sigma}_{BGN}$), by the Chain Rule
(resp., Lemma~\ref{chainLip}).
Hence~$f$ is $C^k_{BGN}$ (resp.,
$C^{\,k,\sigma}_{BGN}$). This finishes
the proof if $U=U_1\times\cdots\times U_d$.\\[2.5mm]
If $U$ is open but not necessarily
of the form $U_1\times\cdots\times U_d$,
then every point $x\in U$
has an open neighbourhood~$V$
of the form $V=V_1\times\cdots\times V_d$
for certain open subsets $V_1,\ldots, V_d\subseteq {\mathbb K}$.
It is clear that $f|_V$ is $C^k_{SDS}$ (resp.,
$C^{\,k,\sigma}_{SDS}$) if so is~$f$.
Thus $f|_V$ is
$C^k_{BGN}$ (resp.,
$C^{\,k,\sigma}_{BGN}$)
by the case
already settled,
and thus $f$ is
$C^k_{BGN}$ (resp.,
$C^{\,k,\sigma}_{BGN}$)
as these properties can be checked locally
(see Lemma~\ref{Crlocal}, resp., Lemma~\ref{hoelloc}).\,\vspace{2.5mm}\nopagebreak\hspace*{\fill}$\Box$

\noindent
In the following, we shall often simply
refer to $C^k_{BGN}$-maps as~$C^k$-maps,
and to $C^{k,\sigma}_{BGN}$-maps as~$C^{k,\sigma}$-maps.
\section{Comparison with Ludkovsky's concepts}\label{seclud}
In this section, we give a definition of $C^k$-maps
following an idea of Ludkovsky,
and show that such maps need not be
$C^k_{BGN}$ (nor $C^k_{SDS}$)
in the case of ground fields
of positive characteristic.
\begin{numba}
To define $C^k$-maps in Ludkovsky's
sense, we find it useful to introduce the
following notations for $U$ an open subset
of a topological vector space~$E$
over a topological field~${\mathbb K}$:
We define $\Phi_1(U):=U^{]1[}$ and
$\overline{\Phi}_1(U):=U^{[1]}$. Given an integer
$k\geq 2$, we let
$\overline{\Phi}_k(U)$ be the set of all
$(x,\xi_1,\ldots, \xi_k,t_1,\ldots, t_k)\in
U\times E^k\times {\mathbb K}^k$
such that
$(x,\xi_1,\ldots, \xi_{k-1},t_1,\ldots, t_{k-1})\in \overline{\Phi}_{k-1}(U)$
holds as well as
$(x+t_k\xi_k,\xi_1,\ldots, \xi_{k-1},t_1,\ldots, t_{k-1})\in
\overline{\Phi}_{k-1}(U)$.
Finally, we let $\Phi_k(U)$
be the set of all
$(x,\xi_1,\ldots, \xi_k,t_1,\ldots, t_k)\in \overline{\Phi}_k(U)$
such that $t_k\not=0$.
\end{numba}
\begin{defn}\label{deflud}
Let $E$ and $F$ be topological
vector spaces over a topological
field ${\mathbb K}$, and $f \colon U\to F$ be a map on an open
subset $U\subseteq E$.
We say that $f$ is $C^1_{Lud}$
if $f$ is $C^1_{BGN}$,
i.e., if the directional difference quotient map
$\Phi_1 (f):=f^{]1[}$
admits a continuous extension
$\overline{\Phi}_1 (f):=f^{[1]}$ to
$\overline{\Phi}_1(U)=U^{[1]}$.
Recursively, having declared
when $f$ is a $C^{k-1}_{Lud}$-map
and defined a map $\overline{\Phi}_{k-1}(f)\colon \overline{\Phi}_{k-1}(U)\to F$
in this case,
we say that $f$ is $C^k_{Lud}$
if~$f$ is $C^{k-1}_{Lud}$
and if the map $\Phi_k(f)\colon \Phi_k(U)\to F$
taking $(x,\xi_1,\ldots, \xi_k,t_1,\ldots, t_k)$ to
\[
\frac{
\overline{\Phi}_{k-1}(x+t_k\xi_k,\xi_1,\ldots, \xi_{k-1},t_1,\ldots, t_{k-1})
-
\overline{\Phi}_{k-1}(x,\xi_1,\ldots, \xi_{k-1},t_1,\ldots, t_{k-1})}{t_k}
\]
admits a continuous extension
$\overline{\Phi}_k(f)\colon \overline{\Phi}_k(U)\to F$.
We say that $f$ is $C^\infty_{Lud}$
if $f$ is $C^k_{Lud}$ for each $k\in {\mathbb N}$.
\end{defn}
\begin{rem}
Ludkovsky defined $C^k$-maps only
for certain ultrametric fields
of characteristic~$0$ and $E$, $F$
locally convex, but of course
the preceding definition is meaningful
in the stated generality.
Furthermore, he only required the existence
of $\overline{\Phi}_1(f)$ locally
on a neighbourhood of $(x,0,0)$ for
each given point $x\in U$
(and similarly for $\overline{\Phi}_k(f)$).
We find it more convenient to define
$\overline{\Phi}_k(f)$
globally on $\overline{\Phi}_k(U)$
(which is equivalent to the local existence).
\end{rem}
\begin{rem}
If $f$ is $C^k_{Lud}$,
we define
\[
d^{(j)}f(x,\xi_1,\ldots,\xi_j)\; :=\;
\overline{\Phi}_j(f)(x,\xi_1,\ldots,\xi_j,0,\ldots, 0)
\]
for $x\in U$, $j\in {\mathbb N}$ with $j\leq k$,
and $\xi_1,\ldots, \xi_j\in E$.
We also write $df(x,\xi):=d^{(1)}f(x,\xi)$.
Then $d^{(j)}f\colon U\times E^j\to F$
is continuous, being a partial map of $\overline{\Phi}_j(f)$
(i.e., a map obtained from $\overline{\Phi}_j(f)$
by fixing some of its arguments).
Furthermore,
$f^{(j)}(x):= d^{(j)}f(x,{\scriptscriptstyle \bullet})\colon E^j\to F$
is a symmetric $j$-linear map,
by a reasoning similar to that
used to prove \cite[Lemma~4.8]{BGN}.\footnote{In
the example discussed in Theorem~\ref{countludk} below,
the $j$-linearity will be obvious.}
We remark that Ludkovsky made the $j$-linearity
of the maps $f^{(j)}(x)$ part of his
definition of a $C^k$-map;
by the preceding, this requirement is
redundant and can be omitted.
\end{rem}
\begin{rem}
We mention that Definition~\ref{deflud}
captures the basic idea of\linebreak
Ludkovsky's
approach, but differs slightly from his
actual definition which imposes
additional boundedness conditions.
In the example discussed in Theorem~\ref{countludk} below,
the domain~${\mathbb O}$ will be an open and compact
set, whence these additional
conditions will be satisfied
automatically.
\end{rem}
\begin{rem}
It is clear that each $C^k_{BGN}$-map
is also $C^k_{Lud}$; a suitable
partial map of $f^{[j]}$ serves
as the continuous extension
$\overline{\Phi}_j(f)$ of $\Phi_j(f)$,
for each $j\in {\mathbb N}$ such that $j\leq k$.
To make this more precise,
let us write
$E^{[j]}=E\times H_j\times {\mathbb K}$,
where $H_j$ collects all factors
in the middle.
Explicitly, we have
$H_1:=E$,
$H_j:= H_{j-1}\times {\mathbb K}\times E^{[j-1]}$
if $j\geq 2$.
Let $0_{H_j}$ be the zero element
in~$H_j$.
Then a simple induction on $k\in {\mathbb N}$ shows
that, if $f$ is $C^k_{BGN}$,
then $f$ is $C^k_{Lud}$, with
\[
\overline{\Phi}_k(x,\xi_1,\ldots,\xi_k,t_1,\ldots,t_k)\, :=\,
f^{[k]}(x,\xi_1,t_1;\xi_2,0_{H_1},t_2;\ldots;
\xi_k,0_{H_{k-1}},t_k)
\]
for $(x,\xi_1,\ldots,\xi_k,t_1,\ldots, \xi_k)\in \overline{\Phi}_k(U)$
providing the continuous extension
of $\Phi_k(f)$ to a map on $\overline{\Phi}_k(U)$.
\end{rem}
Let ${\mathbb K}$ be a local field of positive
characteristic now.
Thus, up to isomorphism,
${\mathbb K}={\mathbb F}_q(\!(X)\!)$ is a field of formal
Laurent series over a finite field ${\mathbb F}_q$
with $q=p^\ell$ elements
for some $\ell\in {\mathbb N}$.
We let ${\mathbb O}:={\mathbb F}_q[\![X]\!]$
be the ring of formal power series,
which is an open, compact subring
of~${\mathbb K}$. Its elements
are of the form $x=\sum_{k=0}^\infty a_k X^k$,
where $a_k\in {\mathbb F}_q$.
Recall that if $x\not=0$, then
its absolute value is given by $|x|=q^{-k}$,
where $k\in {\mathbb N}_0$ is chosen minimal
such that $a_k\not=0$.
We define a mapping
\begin{equation}\label{defbd}
f\colon {\mathbb O}\to {\mathbb K}\, ,
\quad \sum_{k=0}^\infty a_kX^k
\mapsto \sum_{k=0}^\infty a_kX^{[\frac{3}{2}k]}\, ,
\end{equation}
where $[r]$ denotes the Gau\ss{} bracket
(integer part) of a real number $r\geq 0$.
\begin{thm}\label{countludk}
The map $f\colon {\mathbb O}\to {\mathbb K}$
defined in {\rm(\ref{defbd})}
is $C^\infty_{Lud}$,
but not $C^2_{BGN}$.
\end{thm}
\begin{proof}
It is useful to note first that $f$ is a homomorphism
of additive groups, i.e., $f(x+y)=f(x)+f(y)$
for all $x,y\in {\mathbb O}$.
For $x,y\in {\mathbb O}$, we have
\[
|x-y|^{\frac{3}{2}}\; \leq \;
|f(x-y)|\;\leq\;
q |x-y|^{\frac{3}{2}}\, ,
\]
where $|f(x-y)|=|f(x)-f(y)|$.
As a consequence, $f$ is $C^1_{BGN}$
with derivative $f'(x)=0$ for all $x\in {\mathbb O}$
(see \cite[Lemma 2.1]{NOA};
cf.\ \cite[Theorem 29.12]{Sch}).
Furthermore, $f$ is not $C^2_{BGN}$
because it does not admit a second order
Taylor expansion (see \cite[Lemma 2.2]{NOA}).
We now show that $f$ is $C^\infty_{Lud}$.
First, we note that $f$ is $C^1_{Lud}$
because it is $C^1_{BGN}$,
with
\begin{equation}\label{simsh}
\overline{\Phi}_1(f)(x,\xi,0)\, =\, df(x,\xi)\, =\, f'(x)\xi \, =\, 0
\quad \mbox{for all $x\in {\mathbb O}$ and $\xi\in {\mathbb K}$.}
\end{equation}
Using that $f$ is a homomorphism,
for all $x\in {\mathbb O}$, $\xi\in {\mathbb K}$
and $t\in {\mathbb K}^\times$ such that $x+t\xi \in {\mathbb O}$,
we obtain
\begin{equation}\label{essl}
\overline{\Phi}_1(f)(x,\xi,t)
= \frac{f(x+t\xi)-f(x)}{t}
= \frac{f(x)+f(t\xi)-f(x)}{t}
= \frac{f(t\xi)}{t} .
\end{equation}
Since the right hand side of (\ref{essl})
is independent of~$x$, we obtain
for all $x\in {\mathbb O}$, $\xi_1,\xi_2\in {\mathbb K}$
and $t_1,t_2\in {\mathbb K}^\times$
such that $(x,\xi_1,\xi_2,t_1,t_2)\in \Phi_2({\mathbb O})$:
\begin{equation}\label{essl2}
\Phi_2(f)(x,\xi_1,\xi_2,t_1,t_2)=
\frac{\overline{\Phi}_1(f)(x+t_2\xi_2,\xi_1,t_1)-\overline{\Phi}_1(f)(x,\xi_1,t_1)}{t_2}
= 0 \, .
\end{equation}
By (\ref{simsh}), we also have
$\Phi_2(f)(x,\xi_1,\xi_2,t_1,t_2)=0$
for all $x\in {\mathbb O}$, $\xi_1,\xi_2\in {\mathbb K}$,
$t_1=0$ and $t_2\in {\mathbb K}^\times$
such that $(x,\xi_1,\xi_2,0,t_2)\in \Phi_2({\mathbb O})$.
Thus
\[
\overline{\Phi}_2(f)\colon \overline{\Phi}_2({\mathbb O})\to F\, ,
\quad (x,\xi_1,\xi_2,t_1,t_2)\mapsto 0
\]
is a continuous map which extends
$\Phi_2(f)$, and thus $f$ is
$C^2_{Lud}$ with $\overline{\Phi}_2(f)=0$.
It now readily follows by induction
that $f$ is $C^k_{Lud}$ for each
$k\geq 2$, with $\overline{\Phi}_k(f)=0$.
\end{proof}
\begin{rem}
It would be interesting to clarify whether
$C^k_{Lud}$-maps between locally convex
spaces over an ultrametric field~${\mathbb K}$ of characteristic~$0$
(as originally defined by Ludkovsky)
coincide with $C^k_{BGN}$-maps (for $k\geq 2$),
notably for ${\mathbb K}={\mathbb Q}_p$.
It is also unknown whether
$C^k_{Lud}$-maps into real
non-locally convex spaces
are $C^k_{BGN}$.
The author conjectures that neither
is the case, but has not found counterexamples
so far.
\end{rem}
\section{H\"{o}lder differentiable maps on
metrizable spaces}\label{secmetr}
This section is devoted
to the proof of the following
characterization of
$C^{\ell,\sigma}_{BGN}$-maps
on open subsets of metrizable spaces.
\begin{thm}\label{dfc}
Let $({\mathbb K},|.|)$ be ${\mathbb R}$ or
an ultrametric field.
Let $E$ and $F$ be topological ${\mathbb K}$-vector spaces
and $f\colon U\to F$ be a map,
defined on an open subset $U\subseteq E$.
Let $\ell\in {\mathbb N}_0$ and $\sigma\in \; ]0,1]$.
If $E$ is metrizable,
then $f$ is $C^{\,\ell,\sigma}$
if and only if $f\circ \gamma\colon {\mathbb K}^{\ell+1}\to F$
is $C^{\,\ell,\sigma}$, for each
$C^\infty$-map
$\gamma \colon {\mathbb K}^{\ell+1}\to U$.
\end{thm}
The proof of Theorem~\ref{dfc}
heavily relies on tools
developed in~\cite{GaLa},
which are variants of standard methods
of differential calculus in
real locally convex spaces (cf.\ \cite{KaM}).
To describe these tools,
we need the auxiliary notion
of a ``calibration''
on a topological vector space~$E$
over a valued field~${\mathbb K}$.
\begin{defn}\label{defcalib}
A sequence $(q_n)_{n\in {\mathbb N}_0}$
of gauges on~$E$ is called a \emph{calibration} if
\begin{equation}\label{ptycalib}
(\forall n\in{\mathbb N}_0)(\forall x,y\in E)\quad
q_n(x+y)\;\leq\; q_{n+1}(x)+q_{n+1}(y)\,.
\end{equation}
If $q$ is a gauge on~$E$, then there always
exists a calibration $(q_n)_{n\in {\mathbb N}_0}$
such that $q_0=q$ (cf.\ Remark~\ref{remfk});
we then say that $q$ extends to $(q_n)_{n\in {\mathbb N}_0}$.
\end{defn}
\begin{rem}\label{increas}\label{excalib}
If $(q_n)_{n\in {\mathbb N}_0}$ is a calibration,
then $q_n\leq q_{n+1}$ for each $n\in {\mathbb N}_0$
because $q_n(x)=q_n(x+0)\leq q_{n+1}(x)+q_{n+1}(0)=q_{n+1}(x)$
for each $x\in E$.
Also note that if $q\colon E\to [0,\infty[$ is a continuous
seminorm, then $(q)_{n\in {\mathbb N}_0}$
is a calibration.
If $(q_n)_{n\in {\mathbb N}_0}$
is any calibration extending the seminorm~$q$,
then $q_n\geq q$ for each $n$,
by the preceding remark.
Thus $(q)_{n\in {\mathbb N}_0}$ is the smallest
calibration extending~$q$.
\end{rem}
The following two lemmas are
the main results of~\cite{GaLa}.
They are variants of \cite[Lemma~12.2]{KaM}.
In the first lemma, ${\mathbb O}:=\overline{B}^{\mathbb K}_1(0)$.
\begin{la}[Ultrametric General Curve Lemma]\label{thmgencu}
Let $E$ be a topological vector space
over an ultrametric field~${\mathbb K}$,
$\rho\in {\mathbb K}^\times$ with $|\rho|<1$
and $(\gamma_n)_{n\in {\mathbb N}}$ be a family
of smooth maps $\gamma_n\in BC^\infty(\rho^n{\mathbb O}, E)$
which become small sufficiently fast
in the sense that,
for each gauge $q$ on~$E$,
there exists a calibration $(q_n)_{n\in{\mathbb N}_0}$
extending $q$ such that
\begin{equation}\label{condgen}
(\forall a>0)\, (\forall k, m\in {\mathbb N}_0)\quad
\lim_{n\to\infty}\,
a^n\|\gamma_n^{<k>}\|_{q_{n+m}, \infty}\;=\;  0\, .
\end{equation}
Then there exists a smooth curve
$\gamma\colon {\mathbb K}\to E$ with
$\im(\gamma)=\{0\}\cup\bigcup_{n\in {\mathbb N}}\im(\gamma_n)$,
such that
$\gamma(\rho^{n-1}+t)=
\gamma_n(t)$
for all $n\in {\mathbb N}$ and $t\in \rho^n{\mathbb O}$.\,\nopagebreak\hspace*{\fill}$\Box$
\end{la}
\begin{rem}\label{symetr}
Let $E$ in Lemma~\ref{thmgencu} be metrizable
and suppose that there exists
a calibration $(p_n)_{n\in {\mathbb N}_0}$
such that $\{p_n\colon n\in {\mathbb N}_0\}$
is a fundamental system of gauges,
and $C>0$ such that
\begin{equation}\label{shalluse}
(\forall k\in {\mathbb N}_0)\, (\forall n\geq k)\quad
\|\gamma_n^{<k>}\|_{p_{2n},\infty}\;\leq\; C n^{-n}\,.
\end{equation}
Then the hypothesis (\ref{condgen})
of Lemma~\ref{thmgencu} is satisfied:
Given~$q$, we can extend it to a suitable calibration
via $q_n:=rp_{n+n_0}$ for $n \in {\mathbb N}$,
with $r>0$ and $n_0\in {\mathbb N}_0$ sufficiently large.
\end{rem}
\begin{la}[Real Case of General Curve Lemma]\label{thmrealcu}
Let $E$ be a real to\-pological
vector space and $(s_n)_{n\in {\mathbb N}}$
as well as $(r_n)_{n\in {\mathbb N}}$
be sequences of positive reals such that
$\sum_{n=1}^\infty s_n<\infty$ and $r_n\geq
s_n+\frac{2}{n^2}$ for each $n\in {\mathbb N}$. Furthermore,
let $(\gamma_n)_{n\in {\mathbb N}}$ be a sequence of smooth
maps $\gamma_n\colon [{-r_n}, r_n] \to E$
which become small sufficiently
fast in the sense that, for each gauge~$q$ on~$E$,
there exists a calibration $(q_n)_{n\in {\mathbb N}_0}$
extending $q$ such that
\begin{equation}\label{dfco}
(\forall k,\ell, m\in {\mathbb N}_0)\qquad
\lim_{n\to\infty} \, n^\ell\|\gamma_n^{<k>}\|_{q_{n+m},\infty}\,=\,0\,.
\end{equation}
Then there exists a curve $\gamma\in BC^\infty({\mathbb R}, E)$
with $\,\im(\gamma)\subseteq [0,1]\cdot\bigcup_{n\in {\mathbb N}}\im(\gamma_n)$
and a convergent sequence $(t_n)_{n\in {\mathbb N}}$
of real numbers such that
$\gamma(t_n+t) = \gamma_n(t)$
for all $ n\in {\mathbb N}$ and $ t\in [-s_n,s_n]$.\nopagebreak\hspace*{\fill}$\Box$
\end{la}
Again, (\ref{shalluse}) enables
to manufacture calibrations
satisfying~(\ref{dfco}).\\[2.5mm]
In our applications, the maps $\gamma_n$
are restrictions
of affine-linear maps to balls.
The following simple lemma will help us to
verify the hypotheses of the General
Curve Lemmas in this case.
\begin{la}\label{deg1}
Consider the map $\gamma\colon \overline{B}_r^{\mathbb K}(0)\to E$,
$x\mapsto xa+b$,
where
$E$ is a topological vector space
over a valued field~${\mathbb K}$,
$a,b\in E$ and $r>0$.
Let $q_1$ and $q_2$ be gauges
on~$E$ such that $q_1(x+y)\leq q_2(x)+q_2(y)$
for all $x,y\in E$.
Then
$\|\gamma\|_{q_1,\infty}\leq r\|a\|_{q_2}+\|b\|_{q_2}$,
$\|\gamma^{<1>}\|_{q_1,\infty}=\|a\|_{q_1}$
and $\|\gamma^{<k>}\|_{q_1,\infty}=0$
for $k\geq 2$.
\end{la}
\begin{proof}
Since $\|\gamma(x)\|_{q_1}=\|xa+b\|_{q_1}\leq
|x|\cdot\|a\|_{q_2}+\|b\|_{q_2}\leq
r\cdot\|a\|_{q_2}+\|b\|_{q_2}$ for each $x\in \overline{B}_r^{\mathbb K}(0)$,
the first inequality holds.
The remaining assertions follow
from the observations that $\gamma^{<1>}(x,y)=a$
for all $x,y\in \overline{B}_r^{\mathbb K}(0)$
and $\gamma^{<k>}=0$
for all $k\geq 2$.
\end{proof}
Another simple observation will be used.
\begin{la}\label{ctoff}
Let $({\mathbb K},|.|)$ be
either ${\mathbb R}$ or
an ultrametric field.
Let $E$ and $F$ be topological ${\mathbb K}$-vector spaces
and $f\colon U\to F$ be a map,
defined on an open subset $U\subseteq E$.
Let $\ell, d \in {\mathbb N}_0$ and $\sigma\in \;]0,1]$.
If $f\circ \gamma\colon {\mathbb K}^d\to F$
is $C^{\,\ell,\sigma}$, for each
$C^\infty$-map
$\gamma \colon {\mathbb K}^d \to U$,
then also
$f\circ \gamma\colon  V \to F$
is $C^{\,\ell,\sigma}$, for each
$C^\infty$-map
$\gamma \colon V\to U$
defined on an open subset $V\subseteq {\mathbb K}^d$.
\end{la}
\begin{proof}
Given $x_0\in V$, there exists a smooth map
$\kappa\colon {\mathbb K}^d\to V$
such that $\kappa|_W=\id_W$
for some open neighbourhood $W\subseteq V$
of~$x_0$.
In fact, if~${\mathbb K}$ is ultrametric,
we can choose an open, closed neighbourhood
$W\subseteq V$ of~$x_0$ and define $\kappa(x):=x$
if $x\in W$, $\kappa(x):=x_0$ if $x\in V\setminus W$.
In the real case, we can manufacture~$\kappa$
by standard arguments, using a cut-off
function. Then $\eta:=\gamma\circ \kappa\colon {\mathbb K}^d\to U$
is smooth and hence
$f\circ \eta$
is $C^{\ell,\sigma}$.
Then $(f\circ\gamma)|_W=
(f\circ \eta)|_W$ is~$C^{\ell,\sigma}$.
Hence $f\circ\gamma$ is locally $C^{\ell,\sigma}$
and thus $C^{\ell,\sigma}$, by Lemma~\ref{hoelloc}.
\end{proof}
{\bf Proof of Theorem~\ref{dfc}.}
If $f$ is $C^{\ell,\sigma}$, then $f\circ \gamma$ is $C^{\ell,\sigma}$
for each $C^\infty$-map $\gamma \colon {\mathbb K}^{\ell+1} \to U$,
by Lemma~\ref{chainLip}
and Remark~\ref{autolip}.
To prove the converse direction,
we first assume that ${\mathbb K}$ is an ultrametric field.
We start with the case $\ell=0$.
If~$f$ is not $C^{0,\sigma}$,
then the condition formulated
in Definition~\ref{deflipcts}
is violated by some $x_0\in U$.
Hence, there exists a gauge~$q$ on~$F$ such that,
for each neighbourhood $V\subseteq U$
of~$x_0$ and gauge~$p$ on~$E$,
there are $x,y\in V$ such that
$\|f(x)-f(y)\|_q>(\|x-y\|_p)^\sigma$.
After a translation, we may assume
that $x_0=0$.
Pick a gauge $q_0$ on~$E$
such that $B_1^{q_0}(0)\subseteq U$,
and extend it to a calibration
$(q_n)_{n\in {\mathbb N}_0}$
on~$E$
such that $\{q_n\colon n\in {\mathbb N}_0\}$
is a fundamental system of gauges.
Also, pick $\rho\in {\mathbb K}^\times$
such that $|\rho|<1$.
After replacing $q_1, q_2,\ldots$
by large multiples if necessary, we
may assume that
\begin{equation}\label{preprp}
q_0 \;\leq\;
\big({\textstyle
\frac{1}{2}+\frac{1}{|\rho|}}
\big)^{-1}  q_1\, .
\end{equation}
Applying the above property of~$q$ for a given $n\in {\mathbb N}$
to $V:=\overline{B}^{q_{2n+3}}_{\frac{1}{2}n^{-n}|\rho|^n}(0)$
and $p:=n^{\frac{1}{\sigma}}n^n q_{2n+2}$,
we find $x_n,y_n\in E$ such that
\begin{equation}\label{firstprp}
\|x_n\|_{q_{2n+3}},\|y_n\|_{q_{2n+3}}
\;\leq\; \frac{1}{2}n^{-n}|\rho|^n
\end{equation}
and
\begin{equation}\label{secprp}
\|f(x_n)-f(y_n)\|_q\; >\;
n\cdot n^{\sigma n}\big(\|x_n-y_n\|_{q_{2n+2}}\big)^\sigma\,.
\end{equation}
\emph{Case}~1. If $\|x_n-y_n\|_{q_{2n+2}}\not=0$,
let $k_n$ be the unique integer such that
\begin{equation}\label{stasta}
|\rho|^{k_n}\; \leq \; n^n\|x_n-y_n\|_{q_{2n+2}}\; < \; |\rho|^{k_n-1}\, .
\end{equation}
Since $n^n\|x_n-y_n\|_{q_{2n+2}}
\leq n^n(\|x_n\|_{q_{2n+3}}+\|y_n\|_{q_{2n+3}})\leq
|\rho|^n$ by (\ref{firstprp}),
we have $k_n\geq n$.\\[3mm]
\emph{Case}~2. If $\|x_n-y_n\|_{q_{2n+2}}=0$,
we choose the integer $k_n\geq n$ so large that
$\|f(x_n)-f(y_n)\|_q\geq n (|\rho|^{k_n})^\sigma$.\\[3mm]
In either case, we define
\[
\gamma_n\colon \overline{B}^{{\mathbb K}}_{|\rho|^n}(0)\to E\,,\quad
\gamma_n(t)\, :=\, x_n+\frac{t}{\rho^{k_n}}(y_n-x_n)\,.
\]
By Lemma~\ref{deg1}, we then have
$\gamma_n^{<k>}=0$ for $k\geq 2$,
furthermore
\[
\|\gamma_n^{<1>}\|_{q_{2n},\infty}
\;=\;
\frac{\|x_n-y_n\|_{q_{2n}}}{|\rho|^{k_n}}
\;\leq\;
\frac{\|x_n-y_n\|_{q_{2n+2}}}{|\rho|^{k_n}}
\;<\; \frac{n^{-n}}{|\rho|}
\]
by definition of $k_n$,
and finally
\[
\|\gamma_n\|_{q_{2n},\infty}\leq
|\rho^n|\frac{\|x_n-y_n\|_{q_{2n+1}}}{|\rho|^{k_n}}
+\|x_n\|_{q_{2n+1}}
<
{\textstyle \frac{n^{-n}}{|\rho|}+\frac{1}{2}n^{-n}}|\rho|^n
< {\textstyle \big(\frac{1}{2}+\frac{1}{|\rho|}\big)}n^{-n}\, ,
\]
entailing that $\|\gamma_n \|_{q_0,\infty}<1$
(see (\ref{preprp})) and thus
$\,\im \gamma_n\subseteq B^{q_0}_1(0)\subseteq U$.
In view of the preceding,
(\ref{shalluse}) in Remark~\ref{symetr}
is satisfied by the calibration $(q_n)_{n\in {\mathbb N}_0}$
with $C=\frac{1}{2}+\frac{1}{|\rho|}$.
Therefore the General Curve Lemma (Lemma~\ref{thmgencu})
provides a smooth map $\gamma\colon {\mathbb K}\to E$
with $\gamma({\mathbb K})\subseteq U$ such that
$\gamma(t)=\gamma_n(t-\rho^{n-1})$
for each $n\in {\mathbb N}$ and $t\in \overline{B}_{|\rho|^n}^{\mathbb K}(\rho^{n-1})$.
In particular,
$\gamma(\rho^{n-1})=\gamma_n(0)=x_n$
and $\gamma(\rho^{n-1}+\rho^{k_n})=\gamma_n(\rho^{k_n})
=y_n$ for each $n\in {\mathbb N}$.
Hence
\begin{eqnarray}
\hspace*{-2cm}
\lefteqn{\|f(\gamma(\rho^n))-f(\gamma(\rho^{n-1} +\rho^{k_n})) \|_q}
\qquad \notag \\[.5mm]
&=& \|f(x_n)-f(y_n)\|_q\;>\;
n\cdot n^{\sigma n} (\|x_n-y_n\|_{q_{2n+2}})^\sigma \label{sta}\\
& = &
n\cdot n^{\sigma k}\left(
\frac{\|x_n-y_n\|_{q_{2n+2}}}{|\rho^{k_n}|} \right)^\sigma
|\rho^{k_n}|^\sigma
\;\geq\; n |\rho^{k_n}|^\sigma\label{oneref}
\end{eqnarray}
in Case~1, using (\ref{secprp})
in (\ref{sta}) and then (\ref{stasta})
for the final inequality.
In Case~2, we have
\begin{equation}\label{stp}
\|f(\gamma(\rho^n))-f(\gamma(\rho^{n-1}+\rho^{k_n}))\|_q
\,=\, \|f(x_n)-f(y_n)\|_q\, \geq \, n\cdot (|\rho|^{k_n})^\sigma
\end{equation}
by choice of~$k_n$.
Thus (\ref{stp})
holds for each~$n$,
and hence $f\circ\gamma$ is not $C^{0,\sigma}$.
In fact, if $f\circ  \gamma$ were $C^{0,\sigma}$,
there would be a $0$-neighbourhood $J\subseteq {\mathbb K}$
and a gauge $g$ on ${\mathbb K}$
such that
$\|f(\gamma(t))-f(\gamma(s))\|_q\leq
(\|t-s\|_g)^\sigma$ for all $s,t\in J$.
As a consequence of Lemma \ref{findfund},
there is $C>0$ such that
$g\leq C|.|$. Hence
$\|f(\gamma(t))-f(\gamma(s))\|_q\leq
C^\sigma |t-s|^\sigma$ for all $s,t\in J$,
which contradicts (\ref{stp}).\\[3mm]
The general case: Let $\ell$
be a positive integer now.
If $f\circ\gamma$ is $C^{\ell,\sigma}$
for each smooth map
$\gamma\colon {\mathbb K}^{\ell+1}\to U$,
then
$f\circ\gamma$ is $C^\ell$
in particular and hence $f$ is $C^\ell$,
by \cite[Theorem~12.4]{BGN}.
To prove that $f$ is
$C^{\ell,\sigma}$,
it only remains to show
that $f^{[\ell]}$ is
$C^{0,\sigma}$.
We assume
that $f^{[\ell]}$ is
not $C^{0,\sigma}$
and derive a contradiction.
Since $U^{]\ell[}$
is dense in the domain
$U^{[\ell]}$ of the continuous map
$f^{[\ell]}$,
Lemma~\ref{densehoel} shows that
there exists $x_0\in U^{[\ell]}$ and
a gauge~$q$ on~$F$ such that,
for each neighbourhood $V\subseteq U^{[\ell]}$
of~$x_0$ and gauge~$p$ on~$E^{[\ell]}$,
there are $x,y\in V\cap U^{]\ell[}$ such that
$\|f^{[\ell]}(x)-f^{[\ell]}(y)\|_q>(\|x-y\|_p)^\sigma$.
We now pick
$x_n,y_n \in U^{]\ell[}$
as above in the case $\ell=0$,
applied to $f^{[\ell]}$
instead of~$f$,
and obtain a smooth curve $\gamma\colon
{\mathbb K}\to U^{[\ell]}$ such that $\gamma(\rho^{n-1})=x_n$
and $\gamma(\rho^{n-1}+\rho^n)=y_n$.
Applying \cite[Lemma~12.3]{BGN}
with $m:=1$, $V:={\mathbb K}$,
$D:=\{\rho^{n-1}\colon n\in {\mathbb N}\}
\cup \{\rho^{n-1}+\rho^n \colon n\in {\mathbb N}\}$
and $X_0:=\{0\}$,
we obtain a smooth map
$\Gamma\colon W\to U$,
defined on an open subset
$W\subseteq {\mathbb K}^{\ell+1}$,
an open neighbourhood $Y$ of~$0$
in~${\mathbb K}$,
and a smooth map
$g\colon Y\to W^{[\ell]}$
such that
\begin{equation}\label{givctr}
(\forall t\in D\cap Y)\quad
f^{[\ell]}(\gamma(t))\;=\;
(f\circ\Gamma)^{[\ell]}(g(t))\,.
\end{equation}
There is $N\in {\mathbb N}$ such that
$\rho^{n-1}\in Y$
and $\rho^{n-1}+\rho^n \in Y$
for each integer $n\geq N$.
The hypothesis implies
that $f\circ \Gamma$
is $C^{\ell ,\sigma}$ (see Lemma~\ref{ctoff}).
As a consequence,
$(f\circ\Gamma)^{[\ell]}$
is $C^{0,\sigma}$ and hence also
$(f\circ\Gamma)^{[\ell]}\circ g$
is $C^{0,\sigma}$.
However, by construction of~$\gamma$
and (\ref{givctr}), for each $n\geq N$ we have
\begin{eqnarray*}
\lefteqn{\|(f\circ\Gamma)^{[\ell]}(g(\rho^{n-1}))-
(f\circ\Gamma)^{[\ell]}(g(\rho^{n-1}+\rho^n))\|_q}\quad\\
&=&
\|f^{[\ell]}(\gamma(\rho^{n-1}))
- f^{[\ell]}(\gamma(\rho^{n-1}+\rho^n))\|_q\; =\;
\|f^{[\ell]}(x_n)
- f^{[\ell]}(y_n)\|_q\\
& \geq & n |\rho^n|^\sigma\,,
\end{eqnarray*}
arguing as in (\ref{oneref}) to pass to the last
line.
Hence $(f\circ\Gamma)^{[\ell]}\circ g$
is not $C^{0,\sigma}$, contradicting
the preceding. This closes
the proof in the ultrametric case.\\[3mm]
Now assume that ${\mathbb K}={\mathbb R}$,
and pick $r\in \;]0,1[$.
If~$f$ is not $C^{0,\sigma}$,
then there exists $x_0\in U$ and a gauge~$q$ on~$F$ such that,
for each neighbourhood $V\subseteq U$
of~$x_0$ and gauge~$p$ on~$E$,
there are $x,y\in V$ such that
$\|f(x)-f(y)\|_q>(\|x-y\|_p)^\sigma$.
After a translation, we may assume
that $x_0=0$.
Take a gauge~$q_0$ on~$E$
such that $\overline{B}_1^{q_0}(0)\subseteq U$,
and extend it to a calibration
$(q_n)_{n\in {\mathbb N}_0}$
on~$E$
such that $\{q_n\colon n\in {\mathbb N}_0\}$
is a fundamental system of gauges.
We may assume that
$q_0 \leq \frac{2}{7} q_1$.
Applying the above property of~$q$ for a given $n\in {\mathbb N}$
to $V:=\overline{B}^{q_{2n+3}}_{\frac{1}{2}n^{-n}r^n}(0)$
and $p:=n^{\frac{1}{\sigma}}n^n q_{2n+2}$,
we find $x_n,y_n\in E$ such that
\[
\|x_n\|_{q_{2n+3}},\|y_n\|_{q_{2n+3}}
\;\leq\; \frac{1}{2}n^{-n} r^n
\]
and
$\|f(x_n)-f(y_n)\|_q >
n\cdot n^{\sigma n}\big(\|x_n-y_n\|_{q_{2n+2}}\big)^\sigma$.\\[2.5mm]
\emph{Case}~1: If $\|x_n-y_n\|_{q_{2n+2}}\not=0$,
define $s_n:=
n^n\|x_n-y_n\|_{q_{2n+2}}\leq r^n$.
\emph{Case}~2:
If $\|x_n-y_n\|_{q_{2n+2}}=0$,
choose
$s_n\in \; ]0, r^n]$ such that
$\|f(x_n)-f(y_n)\|_q\geq n \cdot (s_n)^\sigma$.
In either case, we define
$r_n:= s_n+\frac{2}{n^2}$ and
\[
\gamma_n\colon [{-r_n}, r_n] \to E\,,\quad
\gamma_n(t)\, :=\, x_n+\frac{t}{s_n}(y_n-x_n)\,.
\]
By Lemma~\ref{deg1}, we then have
$\gamma_n^{<k>}=0$ for $k\geq 2$,
furthermore
$\|\gamma_n^{<1>}\|_{q_{2n},\infty}
=\frac{\|y_n-x_n\|_{q_{2n}}}{s_n}
\leq n^{-n}$
by definition of $s_n$,
and finally
$\|\gamma_n\|_{q_{2n},\infty}<
\frac{7}{2} n^{-n}$ because
\[
\|\gamma_n(x)\|_{q_{2n}} \leq
\|x_n\|_{q_{2n+1}}+r_n\frac{\|y_n-x_n\|_{q_{2n+1}}}{s_n}
\leq
{\textstyle \frac{1}{2}}n^{-n}r^n+\left(r^n+{\textstyle \frac{2}{n^2}}\right)
n^{-n}\leq
{\textstyle \frac{7}{2}n^{-n}} ,
\]
entailing that $\|\gamma_n\|_{q_0,\infty}\leq 1$
and thus
$\,\im \gamma_n\subseteq \overline{B}^{q_0}_1(0)\subseteq U$.
In view of the preceding,
(\ref{shalluse}) in Remark~\ref{symetr}
is satisfied with $C=\frac{7}{2}$.
Therefore the General Curve Lemma (Lemma~\ref{thmrealcu})
provides a smooth map $\gamma\colon {\mathbb R}\to E$
with $\gamma({\mathbb R})\subseteq [0,1]\overline{B}^{q_0}_1(0)
=\overline{B}^{q_0}_1(0)\subseteq U$,
and a convergent sequence $(t_n)_{n\in {\mathbb N}}$
of reals such that
$\gamma(t_n+t)=\gamma_n(t)$
for each $n\in {\mathbb N}$ and $t\in {\mathbb R}$ such that
$|t|\leq s_n$.
In particular,
$\gamma(t_n)=\gamma_n(0)=x_n$
and $\gamma(t_n+s_n)=\gamma_n(s_n)
=y_n$ for each $n\in {\mathbb N}$.
Hence
\begin{eqnarray*}
\hspace*{-2cm}
\lefteqn{\|f(\gamma(t_n))-f(\gamma(t_n+s_n)) \|_q}
\qquad \\[.5mm]
&=& \|f(x_n)-f(y_n)\|_q\;>\;
n\cdot n^{\sigma n} (\|x_n-y_n\|_{q_{2n+2}})^\sigma\\
& = &
n\cdot n^{\sigma k}\left(
\frac{\|x_n-y_n\|_{q_{2n+2}}}{s_n} \right)^\sigma
(s_n)^\sigma
\;=\; n \cdot (s_n)^\sigma
\end{eqnarray*}
in Case~1.
In Case~2, we have
\begin{equation}\label{stpR}
\|f(\gamma(t_n))-f(\gamma(t_n+s_n))\|_q
\, =\, \|f(x_n)-f(y_n)\|_q\, \geq \, n \cdot (s_n)^\sigma
\end{equation}
by choice of~$s_n$.
Thus (\ref{stpR})
holds for each~$n$,
and hence $f\circ\gamma$ is not $C^{0,\sigma}$.\\[3mm]
The general case: If~$\ell$
is a positive integer and
$f$ is not $C^{\ell,\sigma}$
although $f\circ\gamma$ is $C^{\ell,\sigma}$
for each smooth map
$\gamma\colon {\mathbb K}^{\ell+1}\to U$,
we reach a contradiction
along the lines of the ultrametric case.
First, applying the case $\ell=0$
to~$f^{[\ell]}$ instead of~$f$,
we find a gauge
$q$ on~$F$ and $x_n, y_n \in U^{]\ell[}$,
positive reals~$s_n$ such that
$\sum_{n=1}^\infty s_n<\infty$,
a smooth curve $\gamma\colon {\mathbb R}\to U^{[\ell]}$
and a convergent sequence $(t_n)_{n \in {\mathbb N}}$
of reals such that
$\gamma(t_n)=x_n$, $\gamma(t_n+s_n)=y_n$
and $\|f^{[\ell]}(x_n)-f^{[\ell]}(y_n)\|_q\geq n (s_n)^\sigma$
for each $n\in {\mathbb N}$.
Let $t_\infty :=\lim_{n\to\infty} t_n$.
Applying \cite[Lemma~12.3]{BGN}
with $m:=1$, $V:={\mathbb K}$,
$D:=\{t_n \colon n\in {\mathbb N}\}
\cup \{t_n + s_n \colon n\in {\mathbb N}\}$
and $X_0:=\{t_\infty\}$,
we obtain a smooth map
$\Gamma\colon W\to U$,
defined on an open subset
$W\subseteq {\mathbb R}^{\ell+1}$,
an open neighbourhood $Y$ of~$t_\infty$
in~${\mathbb R}$,
and a smooth map
$g\colon Y\to W^{[\ell]}$
such that~(\ref{givctr}) holds.
There is $N\in {\mathbb N}$
such that $t_n\in Y$ and $t_n+s_n\in Y$
for all $n\geq N$.
The hypothesis implies
that $f\circ \Gamma$
is $C^{\ell ,\sigma}$ (Lemma~\ref{ctoff}).
As a consequence,
$(f\circ\Gamma)^{[\ell]}$
is $C^{0,\sigma}$ and hence also
$(f\circ\Gamma)^{[\ell]}\circ g$
is $C^{0,\sigma}$.
However, by construction of~$\gamma$
and (\ref{givctr}), we have
$\|(f\circ\Gamma)^{[\ell]}(g(t_n))-
(f\circ\Gamma)^{[\ell]}(g(t_n+s_n))\|_q
=
\|f^{[\ell]}(\gamma(t_n))
- f^{[\ell]}(\gamma(t_n +s_n))\|_q=
\|f^{[\ell]}(x_n)
- f^{[\ell]}(y_n)\|_q
\geq n (s_n)^\sigma$
for each $n\geq N$,
whence
$(f\circ\Gamma)^{[\ell]}\circ g$
is not $C^{0,\sigma}$, which is absurd.\nopagebreak\hspace*{\fill}$\Box$
\section{Weakly H\"{o}lder differentiable maps}\label{secweak}
If~${\mathbb K}$ is a topological
field and $E$ a topological ${\mathbb K}$-vector space,
we let $E'$ be the space of all continuous
linear functionals $\lambda\colon E \to{\mathbb K}$.
\begin{defn}
Let $E$ and $F$ be topological
vector spaces over a valued field~${\mathbb K}$
and $f\colon U\to F$ be
a map on a subset $U\subseteq E$.
Let $\sigma>0 $.
We say that $f$ is \emph{weakly $C^{0,\sigma}$}
if $\lambda\circ f\colon U\to{\mathbb K}$
is $C^{0,\sigma}$ for each $\lambda\in F'$.
If $U$ has dense interior
and $k\in {\mathbb N} \cup\{\infty\}$,
we say that
$f$ is \emph{weakly $C^{k,\sigma}$}
if $\lambda\circ f\colon U\to{\mathbb K}$
is $C^{k,\sigma}$ for each $\lambda\in F'$.
\end{defn}
\begin{rem}\label{simplwea1}
Note that
each $C^{k,\sigma}$-map
is weakly $C^{k,\sigma}$
(cf.\ Lemma~\ref{chainlin}).
\end{rem}
\begin{rem}\label{simplwea2}
Let $f\colon U\to F$
be a weakly $C^{k,\sigma}$-map
on a subset $U\subseteq E$
with dense interior and
$g\colon V\to U$ be a $C^{k,1}$-map
on a subset $V$ with dense
interior of a topological
${\mathbb K}$-vector space~$H$ (e.g.,
a $C^{k+1}$-map).
Then $\lambda\circ (f\circ g)=(\lambda\circ f)\circ g$
is $C^{k,\sigma}$ for each $\lambda\in F'$
(by Lemma~\ref{chainLip})
and thus $f\circ g$ is weakly~$C^{k,\sigma}$.
\end{rem}
Recall that a topological vector space
over an ultrametric field
is called\linebreak
\emph{locally convex}
if its vector topology can be defined
by a family of ultrametric seminorms
(cf.\ \cite{Mon} for further
information).
\begin{la}\label{cpinnormed}
Let $(E,\|.\|)$ be a normed space over a locally compact
field~${\mathbb K}$,
$F$ be a locally convex space over~${\mathbb K}$
and $f\colon K\to F$ be a
map on a compact set $K\subseteq E$.
Let $\sigma > 0 $.
Then the following
conditions are equivalent:
\begin{itemize}
\item[\rm(a)]
$f$ is a $C^{0,\sigma}$-map.
\item[\rm(b)]
For each continuous seminorm
$q$ on~$F$,
there is $C\in [0,\infty[$
such that
\begin{equation}\label{ordinhoel}
\|f(y)-f(x)\|_q\;\leq\; C\, (\|y-x\|)^\sigma \quad
\mbox{for all $x,y\in K$.}
\end{equation}
\item[\rm(c)]
$f$ is weakly $C^{0,\sigma}$.
\end{itemize}
\end{la}
\begin{proof}
(a)$\Rightarrow$(b):
Let $q$ be as in~(b).
If $f$ is $C^{0,\sigma}$,
then for each $z\in K$
there exists an open neighbourhood
$U_z\subseteq K$ of~$z$ and a gauge
$p_z$ on~$E$ such that
\[
\|f(y)-f(x)\|_q\;\leq\; (\|y-x\|_{p_z})^\sigma\quad
\mbox{for all $x,y\in U_z$.}
\]
Let $V_z$ be an open
neighbourhood of~$z$ in~$K$ with compact closure
$\overline{V_z}\subseteq U_z$.
There exists a finite subset
$\Phi\subseteq K$
such that $K=\bigcup_{z\in \Phi}V_z$.
For each $z\in \Phi$, there exists
$r_z>0$ such that $p_z\leq r_z\|.\|$
(cf.\ Lemma~\ref{findfund}).
Let $r:=\max\{r_z\colon z\in \Phi\}$.
The sets $\overline{V_z}$ and $K\setminus U_z$
being compact and disjoint, we can define
\[
s\; :=\; \sup\,\{\|y-x\|^{-\sigma} \colon
\mbox{$z\in \Phi$, $x\in \overline{V_z}$, $y\in K\setminus U_z$}\}\;
\in \; [0,\infty[\,.
\]
Then (\ref{ordinhoel}) holds
with $C :=\max\{r^\sigma , 2s\max \|f(K)\|_q\}$.
In fact, given \mbox{$x,y\in K$,}
there exists $z\in \Phi$
such that $x\in V_z $.
If $y\in U_z$, then
$\|f(y)-f(x)\|_q\leq$\linebreak
$(\|y-x\|_{p_z})^\sigma
\leq (r_z)^\sigma\|y-x\|^\sigma
\leq C \|y-x\|^\sigma$.
If $y\not\in U_x$, then
$\|f(y)-f(x)\|_q\leq \frac{\|f(y)\|_q+\|f(x)\|_q}{\|y-x\|^\sigma}
\|y-x\|^\sigma\leq 2s\max\|f(K)\|_q \|y-x\|^\sigma\leq
C\|y-x\|^\sigma$ as well.\vspace{1.3mm}

(b)$\Rightarrow$(a): Given a gauge $g$ on~$F$,
by local convexity there exists a continuous
seminorm~$q$ (which can be chosen ultrametric if~${\mathbb K}$
is a local field) such that
$g \leq q$ (cf.\ Lemma \ref{findfund}).
Let~$C$ be as
in~(b). Then $p:=C^{\frac{1}{\sigma}}\|.\|$
is a gauge on~$E$ such that
$\|f(y)-f(x)\|_g\leq \|f(y)-f(x)\|_q\leq (\|x-y\|_p)^\sigma$
for all $x,y\in K$. Thus (a) holds.\vspace{1.3mm}

(a)$\Rightarrow$(c): See Remark~\ref{simplwea1}.\vspace{1.3mm}

(c)$\Rightarrow$(b): Pick $\rho\in {\mathbb K}^\times$
with $|\rho|<1$ and
define $\beta(r):=\rho^k$
for $r\in \;]0,\infty[$,
where $k\in {\mathbb Z}$ is the unique
integer such that $|\rho|^{k+1}< r^\sigma\leq
|\rho|^k$.
Then
\begin{equation}\label{estbet}
|\rho|\cdot |\beta(r)|\; < \; r^\sigma\;\leq\;
|\beta(r)|\quad\mbox{for all $r\in \;]0,\infty[$.}
\end{equation}
If $f$ is weakly $C^{0,\sigma}$, define
\[
B\;:=\;
\left\{\frac{f(y)-f(x)}{\beta(\|y-x\|)} \colon
\mbox{$x,y\in K$ such that $x\not=y$.}\right\}
\]
We claim that $B$ is bounded.
If this is true, then
$M:=\sup \|B\|_q<\infty$ for each
continuous seminorm (or gauge) $q$ on~$F$
(see Lemma~\ref{bdviagau})
and hence
\begin{eqnarray*}
\|f(y)-f(x)\|_q
& = &
\frac{\|f(y)-f(x)\|_q}{|\beta(\|y-x\|)|} \, |\beta(\|y-x\|)|
\; \leq\; M\,|\beta(\|y-x\|)|\\
& \leq &M |\rho|^{-1} \|y-x\|^\sigma
\end{eqnarray*}
for all $x,y\in K$ such that $x\not=y$,
using (\ref{estbet})
for the final inequality.
Hence (\ref{ordinhoel}) holds
with $C:=M |\rho|^{-1}$.\\[3mm]
It remains to show that~$B$ is bounded,
or equivalently, that $\lambda(B)\subseteq {\mathbb K}$
is bounded for each $\lambda\in F'$
(see \cite[Theorem~3.18]{Rud} for the real
case (from which the complex case
follows) and
\cite[Theorem~4.21]{Thi}
for the case where~${\mathbb K}$ is a local
field). However, for each
$\lambda\in F'$,
the map $\lambda\circ f\colon K\to{\mathbb K}$ is
$C^{0,\sigma}$
and hence, by (a)$\Rightarrow$(b)
already established, there exists $C\in [0,\infty[$
such that
\[
|\lambda(f(y))-\lambda(f(x))|\;\leq\;
C\,\|y-x\|^\sigma\quad\mbox{for all $x,y\in K$.}
\]
But then $\sup|\lambda(B)|\leq C$
(whence $\lambda(B)$ is bounded),
since
\begin{equation}\label{mayb}
\left| \lambda\Big(\frac{f(y)-f(x)}{\beta(\|y-x\|)}\Big)\right|
=
\frac{|\lambda(f(y))-\lambda(f(x))|}{|\beta(\|y-x\|)|}
\leq
\frac{|\lambda(f(y))-\lambda(f(x))|}{\|y-x\|^\sigma} \leq
C
\end{equation}
for all $x,y\in K$ such that $x\not=y$,
using (\ref{estbet}) to obtain the first inequality.
\end{proof}
\begin{rem}\label{onlyult}
If ${\mathbb K}$ is a local field
in the situation
of Lemma~\ref{cpinnormed},
it suffices to consider
\emph{ultrametric} continuous seminorms in~(b)
(as the proof shows).
\end{rem}
Recall that a topological vector space
$E$ over a topological field~${\mathbb K}$
is called \emph{sequentially complete}
if every Cauchy sequence in $E$ is convergent.
We say that $E$ is \emph{Mackey complete}
if every Mackey-Cauchy sequence in~$E$
is convergent.
Here, a sequence $(x_n)_{n\in {\mathbb N}}$ in~$E$
is called a \emph{Mackey-Cauchy sequence}
if there exists a bounded subset $B\subseteq E$
and elements $\mu_{n,m}\in {\mathbb K}$
such that $x_n-x_m\in \mu_{n,m}B$ for all $n,m\in {\mathbb N}$
and $\mu_{n,m}\to 0$ in~${\mathbb K}$ as both
$n,m\to\infty$.\\[2.5mm]
Note that every Mackey-Cauchy sequence also is
a Cauchy sequence; hence every sequentially complete
topological ${\mathbb K}$-vector space is
Mackey complete.
In the real locally convex case,
Mackey completeness is
a (particularly weak) standard
completeness property,
which is of great
usefulness for infinite-dimensional
calculus (see \cite[notably \S2]{KaM}
for an in-depth discussion).
\begin{thm}\label{holvsweak}
Let ${\mathbb K}\not={\mathbb C}$ be a locally compact field,
$E$ and~$F$ be topological
${\mathbb K}$-vector spaces, $f\colon U\to F$
be a map on an open set $U\subseteq E$,
$k\in {\mathbb N}_0\cup\{\infty\}$
and $\sigma\in \;]0,1]$.
If $E$ is metrizable
and $F$ is both Mackey complete
and locally convex,
then $f$ is $C^{k,\sigma}$
if and only if $f$ is weakly $C^{k,\sigma}$.
\end{thm}
\begin{proof}
The other implication being trivial,
we only need to show that if $f$ is weakly
$C^{k,\sigma}$, then $f$ is $C^{k,\sigma}$.
As a consequence of Theorem~\ref{dfc},
$f$ will be $C^{k,\sigma}$
if we can show that $g:=f\circ\gamma\colon {\mathbb K}^\ell\to F$
is $C^{k,\sigma}$ for each $\ell\in {\mathbb N}$
and each smooth map $\gamma \colon {\mathbb K}^\ell \to U$.
Note that $g$ is weakly $C^{k,\sigma}$
since so is~$f$ (see Remark~\ref{simplwea2}).
Hence, after replacing~$f$ with~$g$,
we may assume that $U=E={\mathbb K}^\ell$ for some
$\ell\in {\mathbb N}$.
We may assume that $k\in {\mathbb N}_0$;
the proof is by induction on~$k$.\\[2.5mm]
If $k=0$ and $f\colon E={\mathbb K}^\ell\to F$ is weakly $C^{0,\sigma}$,
let $x\in E$ and
$K \subseteq E$ be a compact neighbourhood
of~$x$. Then $f|_K$ is $C^{0,\sigma}$
by Lemma~\ref{cpinnormed}.
Hence $f$ is $C^{0,\sigma}$ locally
and hence $f$ is $C^{0,\sigma}$, by
Lemma~\ref{hoelloc}.\\[2.5mm]
\emph{Induction step}.
If $k\geq 1$ and $f\colon  E={\mathbb K}^\ell\to F$
is weakly $C^{k,\sigma}$,
given $x,y\in E$
choose a sequence $(t_n)_{n\in {\mathbb N}}$ of pairwise distinct elements
in $\overline{B}^{\mathbb K}_1(0) \setminus \{0\}$
such that $t_n\to 0$. Set
\[
B\; :=\; \Big\{
\frac{f^{]1[}(x,y,t_m)
-f^{]1[}(x,y,t_n)}{\beta(|t_m-t_n|)} \colon n,m\in {\mathbb N}\Big \}\, ,
\]
where $\beta\colon \;]0,\infty[\, \to {\mathbb K}^\times$
is as in the proof of Lemma~\ref{cpinnormed}
(as well as $\rho$ used to define~$\beta$).
Then $\lambda(B)\subseteq {\mathbb K}$
is bounded for each $\lambda\in F'$
and hence
$B$ is bounded
(by \cite[Theorem~3.18]{Rud}, resp.,
\cite[Theorem~4.21]{Thi}).
In fact, since $\lambda\circ f$ is
$C^{1,\sigma}$, it follows
that $(\lambda\circ f)^{[1]}$
is
$C^{0,\sigma}$.
Applying now Lemma~\ref{cpinnormed}
to the restriction of
$(\lambda\circ f)^{[1]}$
to the compact set $\{x\}\times\{y\}\times \overline{B}^{\mathbb K}_1(0)$,
we find $C\in [0,\infty[$
such that
\[
|(\lambda\circ f)^{[1]}(x,y, t)
- (\lambda\circ f)^{[1]}(x,y, s)|\;\leq\;
C|t-s|^\sigma\quad
\mbox{for all $s,t\in \overline{B}^{\mathbb K}_1(0)$.}
\]
Repeating the calculation
in~(\ref{mayb}),
we find that $\sup\, |\lambda(B)|\leq C$.
Hence $B$ is indeed bounded.\\[2.5mm]
Since $f^{]1[}(x,y,t_m)
-f^{]1[}(x,y,t_n) \in \beta(|t_m-t_n|) B$,
where $B$ is bounded and
$\beta(|t_m-t_n|)\to 0$ as both $n,m\to\infty$,
we deduce that $(f^{]1[}(x,y,t_n))_{n\in {\mathbb N}}$
is a Mackey-Cauchy sequence in~$F$ and thus convergent;
we let $g(x,y,0)$ be its limit.
Then $\lambda(g(x,y,0))=\lim_{n\to\infty}(\lambda\circ f)^{]1[}(x,y,t_n)=
(\lambda\circ f)^{[1]}(x,y,0)$
for each $\lambda$. Furthermore, trivially
$\lambda(g(x,y,t))=(\lambda\circ f)^{[1]}(x,y,t)$
for $g(x,y,t):=f^{]1[}(x,y,t)=t^{-1}(f(x+ty)-f(x))$
if $(x,y,t)\in E\times E\times {\mathbb K}^\times$.
Thus $\lambda\circ g=(\lambda\circ f)^{[1]}$
is $C^{k-1,\sigma}$ for each $\lambda$, whence
$g$ is $C^{k-1,\sigma}$,
by induction. Hence~$f$ is~$C^{1,\sigma}$,
with $f^{[1]}=g$ a $C^{k-1,\sigma}$-map.
As a consequence,
$f$ is $C^{k,\sigma}$.
\end{proof}

redpajama_set_name: RedPajamaArXiv


Newsletter Title Ideas newsletter title ideas newsletter titles examples studiojpilates. newsletter title ideas 74 attention grabbing blog titles that actually work inc download. newsletter title ideas newsletter title ideas newsletter titles examples sample newsletter templates. Newsletter Title Ideas newsletter title ideas 75 catchy creative newsletter names brandongaille ideas. newsletter title ideas newsletter titles ideas studiojpilates. newsletter title ideas 35 newsletter names titles and design ideas stocklayouts blog printable. Newsletter Title Ideas newsletter title ideas how to choose the right newsletter title i activetrail free.
Newsletter Title Ideas newsletter title ideas 74 attention grabbing blog titles that actually work inc download. newsletter title ideas newsletter title ideas newsletter titles examples sample newsletter templates. newsletter title ideas 75 catchy creative newsletter names brandongaille ideas. newsletter title ideas newsletter titles ideas studiojpilates. newsletter title ideas 35 newsletter names titles and design ideas stocklayouts blog printable.


Here are some guideline pointers – decide which you feel most comfortable following. If you're just starting out, you may want to work with just one paint color on the walls throughout your home. You can always build on that later. Better to take it gradually, than to feel that you've overstepped the mark & don't like it, can't live with it, & have to redo it.


How to exposure compensation ?
Feature Update 4 for XF with Credo Back ?
Small Leaf Credo Cases still available?
Does anyone own the Credo50 in this forum?
Credo 60 Liveview is too hazy for focusing—now solved!
usb 3 with latest retina problems.
How to change Credo LCD display to B+W?
credo live view over USB?
Anyway to disable / turn off the touch screen?
Why does Credo 40 back turn on when battery is inserted?
Credo 80 LCD has digital line through center!!


text:
Minivan News – Archive
FIRST FOR INDEPENDENT NEWS IN THE MALDIVES
Archives Select Month July 2015 June 2015 May 2015 April 2015 March 2015 February 2015 January 2015 December 2014 November 2014 October 2014 September 2014 August 2014 July 2014 June 2014 May 2014 April 2014 March 2014 February 2014 January 2014 December 2013 November 2013 October 2013 September 2013 August 2013 July 2013 June 2013 May 2013 April 2013 March 2013 February 2013 January 2013 December 2012 November 2012 October 2012 September 2012 August 2012 July 2012 June 2012 May 2012 April 2012 March 2012 February 2012 January 2012 December 2011 November 2011 October 2011 September 2011 August 2011 July 2011 June 2011 May 2011 April 2011 March 2011 February 2011 January 2011 December 2010 November 2010 October 2010 September 2010 August 2010 July 2010 June 2010 May 2010 April 2010 March 2010 February 2010 January 2010 December 2009 November 2009 October 2009 September 2009 July 2009 June 2009 May 2009 April 2009 January 2009 December 2008 September 2008 April 2008 February 2008 January 2008 October 2007 August 2007 July 2007 June 2007 May 2007 March 2007 January 2007 December 2006 November 2006 September 2006 August 2006 July 2006 May 2006 April 2006 March 2006 January 2006 December 2005 November 2005 October 2005 September 2005 August 2005 July 2005 June 2005 May 2005 April 2005 March 2005 February 2005 January 2005 December 2004 November 2004 November 1999 0
Features & Comment
Maldives Decides 2013 – select a candidate
Candidate 1: Gasim Ibrahim
Candidate 2: Dr Mohamed Waheed
Candidate 3: Abdulla Yameen
Candidate 4: Mohamed Nasheed
The forgotten coup in the Maldives: Conservativehome
It all came to an end on February 7 when Nasheed was forced to resign, effectively at gunpoint, writes political and digital strategist Samuel Coates for Conservativehome.
"Only one side had the luxury of making advance preparations for handling the aftermath, so inevitably the true nature of his resignation was lost in the fog of war. For the critical first 48hrs, history was written by the victor — almost everyone fell for the initial spin that Nasheed had taken it upon himself to resign due to public pressure and a police revolt.
That anyone had an inkling of what was really going on at that time was in large part down to Nasheed's British international press adviser, Paul Roberts — who managed to blow the whistle about the coup to a few journalists whilst hiding in a President's Office toilet cubicle, before fleeing the country. William Hague was the first major figure to condemn the situation and an article for ConHome by MPs John Glen and Karen Lumley was one of the first to tell it how it was. But overall international reaction was muted and mixed — the US and India recognised the new government immediately.
Since then, the world has kept turning and few seem to remember or care what happened in what is a strategically insignificant nation. Shortly afterwards, President Waheed had the pleasure of joining other Commonwealth leaders for dinner with the Queen to mark her Diamond Jubilee. As they politely tucked into brie and avocado terrine, wild sea bass and apple crumble soufflé, one wonders if Her Majesty was aware that Waheed's allies had been publicly trashing her days before.
World famous eco-resort embroiled in local politics
"We call on the police and the army to pledge allegiance to the Vice President": Umar Naseer, Jan 31
Comment: Constructing the truth – CoNI and the coup, part one
Male's day of protests: Islam and women key themes
Transcript: President Waheed's testimony to the Commission of National Inquiry
"We have been off the streets for some time – now we are back": MDP
Posted on August 7, 2012 Author Minivan NewsCategories News in BriefTags conservative, coup, maldives, maldives news, minivan, minivan news, waheed
4 thoughts on "The forgotten coup in the Maldives: Conservativehome"
mody says:
Agree Nasheed was forced to resign from his post . This pressure was extorted by majority of Maldivian due to his continuous threat to Judiciary and his continuos breaching of the constitution.
If he had been decent enough to abide by the law and had threaten the public to be part of his gangs then he could have rule the country for next 10 years.
Sorry the guy was arrogant and was implanted by the some influential people to eradicate the Islam from this country.
InternationalObserver says:
I am truly ashamed of my country (USA) for condoning the coup by recognizing the traitor Waheed's illegitimate "government". Perhaps the Clinton's planned to holiday in the Maldives or something.
The Commonwealth certainly has taken the high road by calling the coup a coup and demanding restoration of the democratically elected President.
Anni the American people are with you and we will try to get our government to make amends by imposing crippling sanctions on the Gayoom/Waheed regime.
jameel says:
How can it be at gun point when Nasheed himself has told publicly that he did not say gun point?
He himself claimed that this gun point was made up by the media.
Was this a coup? Most probably since he was forced to resign. Was it a military coup?Most certainly not.
this was a coup but circumstances around Anni the dictator had forced him to resign.
Maldivian majority wanted him to resigned and i respect for him that but he is doing after resignation is not something which is good for the country.
What ever said and done he is only after the money and power and not for basic rights of people.
Previous Previous post: Government must assess financial, investor impacts of airport renationalisation: Thasmeen
Next Next post: Raajje TV service terminated after intruders break in, cut control room cables

redpajama_set_name: RedPajamaCommonCrawl


Our interactive games and sports are a great way to keep both children and adults entertained and moving! With so many options to choose from, you'll be sure to find something for everyone! These are a great way to keep everyone at your event at a high excitement level! Take your event to the next level by adding these to your corporate event, school & community event, block party, or at your party at home. All pieces and balls are included - we take care of it all!
Your Party Rental & Event Specialists!
Licensed, Insured, Park & PTA Approved Vendor.
Great customer service & safety is our priority!
On-time delivery - you'll feel confident that we will be on-time to provide the fun.
Professionally cleaned, sanitized, & maintained interactive sports and games.
Large selection of high-quality & lead free commercial grade inflatables.
Professional full-service company - relax while we set-up for your special event!
Proudly serving greater Orange County, CA.
Featuring 24/7 Live Online Availability & Reservations. We always sell out - Don't wait, BOOK NOW!
Awesome service! Delivered early picked up late. Very clean bouncer and excellent customer service. I will definitely rent from them again!
Jump n Bounce was great, they were easy to schedule with and arrived promptly to set up for the party. Would definitely recommend!
Great service bounce rental place! Mary was very friendly and communicated with me through out the whole rental process! They allowed me to keep the jumper over night ! Birthday girl was very pleased ! I was please and happy with the service and price! Thank you and yes I will use them again!
Very easy to work with, great selection, and the delivery/pick up driver was extremely courteous and professional. Will definitely use again for kids party.
This must've been the 5th or 6th time I've rented a bounce house and it was the first time I was happy with it! The JumpNBounce website is easy to navigate, booking was easy as well and most importantly the bounce house was so clean! Customer service is excellent as well! I will definitely recommend them to my family and friends!
We rented a cotton candy machine for a graduation party and had a great experience working with Mary. Mary was very accommodating for a last minute rental and even met with us again for a special request. Will definitely rent from them again!


Creative writing has been a long term passion of mine, and I wrote a little while ago about living my life purpose – it's time to stop procrastinating. In it I talk about my dream to become a writer. I started a journal, enlisted an editor, and began taking those necessary steps to start turning my dream into a reality. My book – The Wall – is currently in the capable hands of a wonderful editor that I have found. I am extremely hopeful about what I will learn by having an expert cast their eyes over my work. I realise that I have a long way to go, to get both myself and my book to the finished product, but I am determined to enjoy the journey, and to keep learning on the way.
In all aspects of my working life, I am not a person driven by excessive planning. If I have a goal in mind, whatever that goal is, I always find a way to get there. I simply focus on what I want to achieve, what I am doing now, and what I am doing next. What I don't do in plan every single step that I will need to take, just the immediate and the distant. Is this the right approach? Will it work for writing? I honestly don't know. For the moment I am very focused on getting this book edited and to quality, but I am already starting to understand more about publishing, agents, self publishing, and the things I need for my next step. I do know that a lot of people send synopsis and first chapters and wait for interest before finishing a book. For me this didn't really appeal. I had to finish a whole book just so that I knew that I could. As part of my research I have found some awesome blogs out there, with great information on improving the quality of writing, and on publishing itself. I am also a recent, but now avid, podcast listener. My daily commute to and from work provides the perfect opportunity to read blogs or listen to a podcast on writing, and provides plenty of ideas to use when I get home in the evening. I guess one of the things that sticks in my mind about writers is the stamina you need to see your book through to publication, and how incredibly hard it is to put your heart and soul on the line. The bottom line is, if we want to achieve our goals, and if we want to stop wishing we were a writer, and to begin accepting that we are a writer – we have to take that risk – take the risk of being a fool or a failure.
Perseverance is the key, and having the confidence to follow your dream! 🙂 Great post!
A great post that really resonated with me. If you want to be a writer, then be one 🙂 Follow those dreams!
Well said, ladies. Get in that chair and write!


text:
How Sweet It Is: Alabama Beats Auburn in a Classic Iron Bowl
John Armstrong November 27, 2009 College Football
Alabama vs. Auburn is very often a classic college football game. Today was no exception. Alabama was a heavy favorite, had a perfect season and had to play Auburn on the road. Auburn had two weeks to prepare and prepare they did. They came out firing and running fast, surprising and gutsy plays. Every Auburn fan should be proud of the effort the Tigers put out this afternoon. They out played Alabama until the final 8:30 of the fourth quarter, when the Tide showed the character and determination that has made them a contender for the second year in a row to win 26-21. What can I say about two straight 12-0 regular seasons? It has never been done in Alabama's storied history. I am proud of this team. They could have lost at least three games this year but refused to lose. And to think that many people thought Nick Saban would not do this well until year four and year three is not over yet. The guy is a winner.
The Crimson Tide drove the field, consuming time and making four huge third down conversions, to win with 1:34 to go. And the play call at the end, of a pass to Roy Upchurch, was a Nick Saban classic. He decided against the call of his offensive coordinator for what appeared to be a certain running play and threw a safe and effective pass to go up by five points. A field goal might have won this game but it would have left Auburn with a serious chance to still win by a field goal at the end. Instead, Saban made the right call (with the guts of a winning coach) and proved again why he is truly a great coach. He can recruit, manage a great staff, handle the pressure and seems to always call a great game. (Consider Les Miles poor management of the LSU vs. Old Miss game last week and you can see why real LSU fans still wish they had Nick Saban.) This guy is one of the five greatest coaches in the game.
Now it is on to the SEC Championship game in Atlanta next Saturday against Florida. Like last year the Gators are the favorite but Alabama will come to play. Both teams have not yet played a truly "great" game so who knows what will happen? I would not bet against Tim Tebow but the Tide will play hard and play to the end. I can't wait but then I need the next eight days to recover from the game today and get ready for more of this kind of excitement. I expect this team is spent tonight but they will be ready when the time comes. Roll Tide. "Every Bama man's behind you, hit your stride!"


Amazing home located in master planned community Fall Creek. Home sits just steps away from one of several community parks w/ stocked pond & jogging trail, also just a short walk to the FC Sports Complex where several community events take place. Home features a beautiful stone turret w/ an iron gate leading into a private courtyard. Once you step into this lovely home, you\'ll find a private office & guest bath to the right. To the left you\'ll see the elegant dining room & butler\'s pantry & straight ahead you\'ll see a beautiful curved staircase & 2-story living room. The home boasts an open kitchen w/ custom cabinets, huge eating area for everyday dining & enormous kitchen island. The master retreat is conveniently located on the first floor & features a beautiful bathroom w/ his & hers walk-in closets, an XL shower & jacuzzi tub. Upstairs you\'ll find a great gameroom, outdoor balcony, 3 generous bedrooms, & 2 full baths. Make your appointment today!


text:
Pierre Dress, död december 1646, var en svensk brukspatron av vallonskt ursprung. Han var bror till Andry Dress.

Pierre Dress torde ha kommit till Sverige tillsammans med sin bror, och förvaltade 1624–1627 Åkers styckebruk tillsammans med denne och gick senare i Andry Dress tjänst, bland annat som förvaltare på Kvarnbacka bruk. 1633 köpte han en gård i Linde socken och 1636 gården Vedevåg med tillhörande hytta. 1638 lät han uppföra en hammare där, och blev 1642 ensam ägare av bruket. Pierre Dress var även 1643 en av de första tomtägarna i den 1643 grundade staden Lindesberg.

Källor

Personer inom Sveriges näringsliv under 1600-talet
Födda okänt år
Avlidna 1646
Män

redpajama_set_name: RedPajamaWikipedia


We offer a comprehensive range of quality smash repair solutions for Kew, Hawthorn and Richmond locals to get you back on the road fast, including panel beating, spray painting, car detailing, car body repair and towing. All work is covered by our Golden Lifetime written Guarantee.
Get the results you deserve with Kew Accident Repair Centre. Call 0385 778 377 today.


text:
Mansfield Town badge - Link to home
2019 pre-season fixtures
U21 pre-season fixtures
U18s fixtures/results
Family tickets
Quarry Lane memberships
Youth Setup
The RH Academy
3G pitch
Sandy's Bar & Kitchen
Mansfield Town Blind FC
Stags Supporters' Association
First time guide
Sammy & Sally the Stag!
Conference and banqueting
Enjoy family fun prior to final home game!
There's plenty going on in our family room ahead of Saturday's clash against Stevenage, open to supporters who purchase a family ticket, which admit one adult and one child to the match for £18!
We will have arts and crafts set up for our mini-Stags to allow their imaginations to run wild, including paint, glitter and glue!
We'll have a wide range of board games available for the children to use, along with music for our young fans to dance to.
There will also be face painting, which is FREE of charge for our final home game of the Sky Bet League Two campaign.
Supporters' liaison officer Danielle Hett says that today is a great opportunity for our mini-Stags to let their creativity flow.
"With this being our last home match, we really see this as an opportunity for everyone to come together and cheer the Stags on," she said.
"We're calling today 'free play,' so the only limits are the children's imaginations!"
Our family room is open for those with a valid family ticket for today's match, but children aged 17 and under must be accompanied by a parent or guardian.
The sensory area, built for those prone to experiencing a sensory overload, will also be available for those with a family ticket to use.
Our family room has free fruit for our young Stags, which has been the case for every home matchday.
There will also be a food bank donation point located in the family room on matchday, and all donations from supporters are welcomed and greatly appreciated.
What's going on in our family room this weekend?
There are plenty of activities taking place in our family room ahead of Saturday's match against Cheltenham Town, open to supporters who purchase a family ticket, which admit one adult and one child...
Family fun at a discount price for Forest Green match
There's plenty going in our family room prior to Saturday's clash against Forest Green Rovers - open to those who purchase a family ticket, which admit one adult and one child to the match for just...
Tickets stil on sale for Cardiff replay
We will welcome Championship side Cardiff City to One Call Stadium on Tuesday 16 January (7.45pm kick-off) in our Emirates FA Cup third round replay.
Tickets on sale for home friendlies
Tickets for our four home pre-season friendlies, to be played this month against Championship opposition, are now on sale.
Tickets for Carabao Cup clash with Morecambe on sale
Tickets for our first round Carabao Cup clash against Morecambe at One Call Stadium on Tuesday 13 August (7.45pm kick-off) are now on sale.
Tickets on sale for first game of season against Newport
Tickets for Stags' trip to Newport County on Saturday 03 August (3pm kick-off) are now on sale.
Quarry Lane memberships available for collection
Quarry Lane Membership holders for the 2019-20 season are now able to collect their membership cards.
Tickets for home friendlies on sale
Tickets for our three home pre-season friendlies against Championship opposition are now on sale.


All other things being equal, people buy from those they trust the most. If it's the right fit and they can afford it, all objections come down to trust. Do they trust you? That's the million dollar question. So, how do you gain trust in the modern marketplace? With buyer skepticism at an all-time high, how do you set yourself apart from all the scammers and charlatans out there? How do you connect with people on a real level, without having to fake it till you make it? Your market's BS meter is hyper tuned. This is how you win.