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\begin{document}
\title[Regularity of the composition operator]
{Regularity of the composition operator
in spaces of H\"older functions.}
\author[R. de la Llave]{R. de la Llave$^{\,1}$\footnotemark}
\author[R. Obaya]{R. Obaya$^{\,2}$\footnotemark}
\addtocounter{footnote}{1}
\footnotetext{Department of Mathematics,
University of Texas at Austin,
Austin TX 78712, USA}
\addtocounter{footnote}{1}
\footnotetext{Departamento de Matem\'{a}tica Aplicada a la Ingenier\'{\i}a,
ETSII, Universidad de Valladolid,
Valladolid, Spain.}
\subjclass{}
\begin{abstract}
We study the regularity of the
composition operator ($(f,g) \mapsto g\circ f) $
in spaces of H\"older differentiable functions.
Depending on the smooth norms used to
topologize $f,g$ and their composition, the
operator has different differentiability properties.
We give complete and sharp results for
the classical H\"older spaces of
functions defined on geometrically well behaved
open sets in Banach spaces. We also provide
examples that show that the regularity conclusions are
sharp and also that if the geometric conditions fail,
even in finite dimensions, many elements of
the theory of functions (smoothing, interpolation,
extensions) can have somewhat unexpected properties.
\end{abstract}
\maketitle
\section{Introduction.} \label{introd}
The goal of this paper is to establish
differentiability properties of composition
in the classical spaces of
H\"{o}lder functions.
The operator that to a pair of
functions $(f,g)$ associates $h = g\circ f$,
(where $\circ$ denotes composition) can
be considered as defined in function spaces
equipped with different norms.
Depending on the Banach spaces where we think $f$, $g$
and $h$ to be ranging, the composition operator
may have different degrees of differentiability.
In this paper we consider the classical spaces of
H\"older
functions defined on open sets in a
Banach space satisfying some geometric
conditions and we give
a complete characterization of the regularity
of composition in all the spaces of this scale;
that is, we provide counterexamples for all the
regularities that we do not establish.
We also give examples that show that if the
geometric properties in our assumptions do not hold,
there are different pathologies that
may arise and the classical H\"older spaces
may fail to have smoothing operators -- that approximate
functions in a controlled way by smoother ones --, satisfy
interpolation inequalities or have consistent ways
of extending them to a larger domain.
We have emphasized the methods based on
elementary estimates, triangle
inequality, mean value theorem and the like.
Therefore, the results presented
here will be valid in a great generality
-- e.g., open sets of Banach spaces
satisfying some geometric properties.
The methods and the results we present also carry out to
manifolds and it is easy to lift them to this case.
In finite dimension, it is sometimes possible
to obtain similar results using
characterizations of H\"{o}lder spaces by
approximation properties by analytic
functions \cite{Ste}, \cite{Kr}, \cite{Ni} but this
does not seem to generalize easily to Banach spaces.
As intermediate steps
we will also prove elementary
versions of interpolation inequalities that work
in infinite dimensional
domains with good geometric properties.
There are several motivations for
the study we undertake in this paper.
It seemed to us that often the papers in the literature
developed the results they needed from scratch,
and hence there was great deal of repetition
while, at the same time, there
was not a systematic survey; indeed, the
precise boundaries were not well known
and several important phenomena were
not even appreciated at all.
Let us mention just a few applications, which we have been
drawn to at some time or another.
The list
does not intend to be exhaustive or to
reflect any other criterion than the history
of the authors of this paper
and we would welcome any information about new uses.
A first motivation is the study of diffeomorphism
groups when the group operation is composition.
These groups, besides being the natural symmetry groups
of differentiable topology, have been found
to have an increasing number of
applications. A classical one is
a formulation of hydrodynamics and elasticity
as geometry in the group of diffeomorphisms
(see \cite{EM}, \cite{Ar} and references there
to previous work).
Other applications to mathematics include the
study of classifying spaces of
foliations and, more modernly, to string theory.
An application that has been important for us
is the theorem
that the
commutator subgroups of many groups of diffeomorphisms
are simple \cite{Epstein}, and that many diffeomorphism
groups are their own commutator \cite{Ma}.
One of our original motivations was the study of
factorizations and decomposition theorems in
spheres \cite{LO1}, which in turn
was motivated by inverse problems in optics
and in geodesic fields \cite{LS}.
Unfortunately, as it
will become apparent in this paper, the composition operation is
not very regular, so the standard methods of Lie theory,
even though they are
a useful strategic guide, often have to be
supplemented by hard analysis methods
(notably implicit functions theorems of
Nash-Moser type) that involve
scales of spaces; hence, one needs to
use the properties of composition in scales of spaces.
With that in mind, we have also included some material
on the existence of smoothing operators
that are an indispensable tool. As it turns
out, those are closely related to the interpolation
inequalities that we need to use often.
Another motivation is dynamical systems. In this field,
we often have to consider repeated compositions. Objects
that remain invariant under the dynamics -- and hence
provide useful landmarks -- are often found solving
functional equations that involve composition.
Having available sharp regularity results
for the composition allows us to obtain in a clean
way smoothness results on invariant manifolds.
Indeed, it seems that this was one of
the original motivations of \cite{Ir}.
Similarly, the proof of \cite{Mo2} and the
appendix by J. Mather
to \cite{Sm} show that problems
such as structural stability can be formulated as
implicit function theorems whose unknowns are
homeomorphisms. Some recent results such as
the differentiability of the homeomorphisms
appearing in structural
stability with respect to the maps involved
\cite{LMM} or the differentiability of
entropy \cite{KKPW} can be
rewritten as functional equations
involving the composition operator.
Once one obtains differentiability
of the conjugacy, it is possible to
study the differentiability with respect to
parameters of SRB measures, a problem
considered recently by D. Ruelle \cite{Ru}.
With the results developed here, this seems to lead to some
improvements in the regularities
stated in those papers.
We also mention the recent work
of \cite{NT} on triviality
cocycles over an Anosov system taking
values in groups of diffeomorphism of the $n$-torus $\T^n$.
As it was discovered in the mentioned paper, one
can use their results to prove rigidity of
some non-hyperbolic actions.
By using systematically the point of view of
diffeomorphisms groups, it is possible to improve
the regularity and generalize the results to other manifolds
\cite{Ll}.
Let us also point out that there are other natural spaces
that, presumably, deserve study similar to that presented
here. For example, spaces
of diffeomorphisms with a Sobolev topology. Some beginnings
of this are done in the literature. For
some applications of groups of diffeomorphisms
with Sobolev topologies and some
of the regularity properties of composition, we refer to \cite{EM}.
Similar remarks apply to the $\Lambda_r$ spaces (which agree with
the $C^r$ spaces considered here for $r \notin \N$).
At the end, we also compare with other theories of functions,
such as Whitney theory, which allows us
to define functions on closed sets.
Of course, for functions defined on closed sets with possibly
empty interior, composition is delicate to define.
We call attention to the paper \cite{Vi}, which considers
two-dimensional hydrodynamics with initial data
and borderline smoothness. The main technical tool
there are very accurate estimates for composition in
Besov spaces.
Let us finish this introduction by paying homage to previous
surveys in the field that, even if restricted
to integer regularities, have been influential and
useful \cite{Ir}, \cite{Eels} and also to the program
sketched in \cite{Pa}. A recent survey, including
other useful spaces in partial differential
equations is \cite{AZ}.
\section{Basic definitions and notation.}\label{dfn}
Let $E, F$ be Banach spaces. We will denote
the Banach space of bounded
linear functions from $E$ to $F$ by $L(E,F)$ and, in general,
$L_n(E,F)=L(E,L_{n-1}(E,F))$.
For each $x \in E$ and $r>0$ we represent by $B_r^E(x)=\{ y\in E |
\n{y-x}0$.
\end{prop}
\begin{pf}
First, we will study both cases for $n=0$ and then
we will establish the claim using induction on $n$.
If $r= \alpha=0$ then $\beta=\gamma(1- \mu)$ and we have
\[
\begin{array}{rcl} H_\beta(f) &=&
\sup_{x\neq y} \n{f(x)-f(y)}^\mu
\Frac{\n{f(x)-f(y)}^{1-\mu}}
{\n{x-y}^{\gamma(1-\mu)} } \\
&\leq& 2^\mu \n{f}_{C^0}^{\mu }H_\gamma(f)^{1-\mu }.
\end{array}
\]
Hence $C^t(U,F) \subset C^s(U,F)$ and
$ \n{f}_{C^s} \leq 2\n{f}_{C^r}^{\mu }\n{f}_{C^t}^{1-\mu }$
for every $f \in
C^t(U,F).$
Besides, if $r=\alpha>0$ then
\[
\begin{array}{rcl}
H_\beta(f) &=&
\sup_{x\neq y} \Frac{ \n{f(x)-f(y)}^\mu}
{\n{x-y}^{\alpha \mu} }
\Frac{\n{f(x)-f(y)}^{1-\mu}}
{\n{x-y}^{\gamma(1-\mu)} } \\[.2 cm]
&\leq& H_{\alpha}(f)^{\mu }H_\gamma(f)^{1-\mu},
\end{array}
\]
which provides
\[
\n{f}_{C^s} \leq \n{f}_{C^r}
^{\mu } \n{f}_{C^t}^{1-\mu }
\]
for $n = 0$.
Using the inductive hypothesis we find, for $n\ge 1$,
\[
\n{df}_{C^{s-1}} \leq M_\alpha \n{df}_{C ^{r-1}}^{\mu}
\n{df}_{C^{t-1}}^{1-\mu} \leq
M_\alpha \n{f}_{C ^r}^{\mu} \n{f}_{C^t}^{1-\mu}
\]
and, therefore, the desired conclusion.
\end{pf}
In contrast with
Proposition \ref{int}, interpolation inequalities for different
$n$ are more complicated and
only hold under additional hypotheses on the domain.
\begin{exam}\label{noint}
Let $U=\{(x,y) \in \R^2\,|\;x>0, y>1, xy<1\}$. We take $\varphi$ as a
$C^{\infty}$ cut-off function with
$\n{\varphi'}_{C^0} \leq 2$ which takes
the value $1$ on $[-1/3,1/3]$ and
zero outside of $[-1,1]$. We define on $U$ the functions
\[
f_n(x,y)=\log(1+x)\, \varphi(y-n)\,,\quad n \geq 1 \,.
\]
It is immediate to check that
\begin{align*}
\Frac{ \partial^j f_n}{\partial y^j} (x,y)
& = \log{(1+x)} \, \varphi^{j)} (y-n)\quad \mbox{\rm for every}
\; j\geq 0,\\[.2cm]
\Frac{ \partial^{i+j} f_n}{\partial x^{i} \partial y^j } (x,y)
& = \Frac{(-1)^{i+1}(i-1)!}{(1+x)^i} \, \varphi^{j)} (y-n)\quad
\mbox{\rm for every} \; i \geq 1,j \geq 0.
\end{align*}
Observe that the support of $ \varphi (y-n) $ is centered around
$ y=n $, which means that $ x \approx 1/n $.
In consequence $ \n{f_n}_{C^2} $ remains bounded and $ \n{f_n}_{C^0}$
converges to zero.
Since $ \partial f_n/ \partial x$ evaluated at
$ (x,y)=(1/2n,n) $ does not converge to zero,
we conclude that there cannot be an inequality of the form
\[
\n{f_n}_{C^1} \leq M \n{f_n}^{1/2}_{C^0} \n{f_n} ^{1/2} _{C^2}.
\]
More generally, notice that there is not interpolation inequalities
for the range of values $0 \leq r < 1^- \leq s \leq t$.
\end{exam}
\begin{prop}\label{ico1}
Let $n \in \N$ and $0\leq \alpha < \beta \leq 1^-$.
If $E$, $F$ are finite dimensional Banach spaces and
$U\subset E$ is an open bounded set, then the embedding
$i: C^{n+\beta} (U,F) \rightarrow C^{n+\alpha}(U,F)$ is compact.
\end{prop}
\begin{pf}
In the finite dimensional case, here considered,
the successive derivatives of each
function of $C^{n+\beta}(U,F)$ admit a continuous extension to
${\overline U}$. Thus, the bounded sets of $C^{n+\beta}(U,F)$
are compact in $C^{n+ \alpha}(U,F)$.
In effect, for $\alpha =0$ this is a consequence of Ascoli-Arzel\`a
theorem. For $0 < \alpha < \beta \leq 1^-$ we use the
previous result for $\alpha =0$ and
the known interpolation inequality (\ref{itp}).
\end{pf}
We also quote Proposition A3 from \cite{Lan}. It generalizes
the elementary fact that the pointwise limit of
uniformly H\"older functions is also H\"older.
\begin{prop}\label{Lanford}
Let $n \in \N$ and $0 < \alpha \leq 1^-$. Take a sequence
$(f_k)_{k \in \N}$ of functions of
$C^{n+\alpha}(U,F)$ with $\n{f_k}_{C^{n+\alpha}} \leq M$ and
assume that for each $x \in U$, $f_k(x)$ converges weakly
to $f(x)$. Then,
\begin{itemize}
\item[i)] $f$ lies in $C^{n+\alpha}(U,F)$ and
$\n{f}_{C^{n+\alpha}} \leq M$,
\item[ii)] $d^jf_k(x)$ converges weakly to $d^j f(x)$ for
every $x \in U$ and $1 \le j \le n$.
\end{itemize}
\end{prop}
We refer to the paper \cite{La} for the proof. The theorem there
is stated only for the exponent $\alpha = 1^-$, but the proof goes through
without any modification for the statement here.
We call attention to the fact that, in contrast with the usual
Ascoli-Arzel\`a theorem, the space is not necessarily
separable but, on the contrary, the weak convergence of $f_k(x)$
is assumed for every $x \in U$.
We also note that Proposition \ref{Lanford}
can be rephrased as stating that the unit ball in
$C^{n + \alpha}$ is sequentially
closed when we give the functions the topology of
the pointwise weak limit.
\subsection{Compensated domains.} \label{comdom}
Notice that in Proposition \ref{int} we have not claimed that
if $ r = n + \alpha < s = m + \beta $ then $ C^s(U,F)\subset C^r(U,F)$.
The fact that this is not true in general can
be seen in the following example.
\begin{exam}\label{cas}
{\it Let $U$ be the domain in Example \ref{domain}
and $d_U$ as there denote the minimum length of
arcs in $U$ joining two points. Consider the function
\[
f(x,y) = \varphi( d_U(\, (x,y),\, (1/2, 1/2)))\,,
\]
where $\varphi$ is a $C^\infty$ function of the real
line that equals the identity in $[10^{-2}, \sqrt{2}/2 + 10^{-3}]$,
is constant in $[ \sqrt{2}/2 + 10^{-2}, \infty)$ and on
$( -\infty, 10^{-3}]$ and, moreover, satisfies $| \varphi'(x)| \le 1$.
Then, $f$ is continuously differentiable in $U$
and $ \n{df}_{C^0} \leq 1 $.
Nevertheless, there are no constants $ k > 0, \alpha > 0 $ such that
\[
|f(x,y) - f(x',y') | \leq k |(x,y) - (x',y') | ^{\alpha}\,.
\]
Hence, in the domain $U$, $ C^1(U) \not \subseteq C^{ \alpha} (U) $
for any $ 0 < \alpha \leq 1^- $.}
\end{exam}
\begin{pf}
Let us describe somewhat informally why the function $f$
is differentiable.
The gradient of $d_U( (x,y), (1/2,1/2))$ (the differential is
just the inner product with the gradient) is, roughly, the unit vector
in the shortest line that joins the two points going through the point.
More precisely, the expression for $d_U( (x,y),(1/2,1/2) )$ is a
piecewise analytic function that can be expressed as
either a) the Euclidean distance of $(x,y)$ to $(1/2,1/2)$,
b) the distance
of $(x,y)$ to $(0,0)$ plus $\sqrt{2}/2$ (which is the distance of
$(0,0)$ to $(1/2,1/2)$),
or c) the distance
of $(x,y)$ to $(1,1)$ plus $\sqrt{2}/2$ (the distance of
$(1,0)$ to $(1/2,1/2)$).
Note that even if the straight lines that go through
$(1,0)$ or $(0,0)$ are not contained in the domain, they
can be approximated by paths with arbitrarily closed length
in the domain.
The gradients will be, the unit vectors
pointed respectively to $(1/2, 1/2)$ in the case of a),
to $(0,0)$ in the case of b), and to $(1,0)$
in the case of c).
The function $d_U$ will be continuously differentiable
in the points where these unit vectors
depend continuously on the point. This includes all the points
in $U$ except $(1/2,1/2)$ and the boundary
between the regions a), b), c). As it turns out, the
boundaries between a) and b) and a) and c) are not
a problem: the boundaries are
precisely those points in the
segment joining $(1/2, 1/2)$ to
$(0,0)$ and to $(1,0)$. Nevertheless, the gradient
of $d_U( (x,y), (1/2, 1/2))$ is indeed
discontinuous at $(1/2,1/2)$ and
at the boundary between b) and c), which is the
segment $\big\{ $(1/2, y)$ \ \big|\ y \le 0\big\}$.
The chain rule establishes that, except for these
points, the function $f$ is continuously differentiable.
On the other hand, our choice of the function $\varphi$
forces $f$ to be constant on a neighborhood of these points so
that the derivative is $0$.
\end{pf}
In many respects this example is a \lq\lq pathology".
Its origins are clear
if one tries to use the mean value theorem.
\begin{equation}\label{vme}
f(x) - f(y)
= \int ^1_0 df( \gamma(s)) \dot{ \gamma} (s) ds
\end{equation}
where $ \gamma $ is a differentiable path joining
$ x $ and $y$.
This equality gives $ \n{f(x) - f(y)}
\leq \n{f}_{C^1} d_U(x,y) $.
In the domain $U$ of the example
this could be much larger than $\n{x- y}$.
Clearly, if we want to use arguments
based on the mean value formula (\ref{vme}),
the alternatives are either to consider always the path distance
or to restrict ourselves to
spaces in which the distance
among points is commensurate to the path
distance. In this paper we have chosen the second
alternative. Hence, we isolate the property:
\begin{defi}\label{compensated}
We say that an open set $ U $ is {\em compensated\/}
if there exists a constant $ \kappa_U $ such that
$ d_U(x,y) \leq \kappa_U \n{x-y}$ for every $ x, y \in U $.
\end{defi}
This definition only makes sense for connected set, however
this is not a serious shortcoming. If $U = U_1 \cup \cdots U_n$ and
$U_i \cap U_j = \emptyset $ for $ i \ne j$,
then $C^r(U) = C^r(U_1) \oplus \cdots \oplus C^r(U_n)$.
Similarly for infinite unions of sets that can be decomposed
in infinite disjoint unions but the pairs
are at a distance bounded from below.
This is general enough for the applications we have in mind.
If the domain $U$ is compensated it follows from (\ref{vme})
that $C^{n+1^-}(U,F)$ is a closed subset of $C^{n+1}(U,F)$
and $\n{f}_{C^{n+1}} \leq \kappa_U \n{f}_{C^{n+1^-}}$ for every
$f \in C^{n+1}(U,F)$. More precisely, we can state
\begin{prop}\label{ico2}
Let $U$ be a compensated open set
and take numbers $0 \leq r \leq s$.
Then $C^s(U,F) \subset C^r(U,F)$ and,
for every $f \in C^s(U,F)$, one has
\[
\n{f}_{C^r} \leq 2\kappa_U \n{f}_{C^s}\,.
\]
\end{prop}
\begin{pf}
Write $r=n + \alpha$, $s=m+\beta$ where $n,m \in \N$
and $0 \leq \alpha,\beta \leq1^-$.
Let us fix a function $f \in C^s(U,F)$.
If $n=m$ and $\alpha< \beta$ we have verified in
Proposition \ref{int} that
$C^s(U,F) \subset C^r(U,F)$ and
\[
\n{f}_{C^r} \le 2 \n{f}_{C^n}^{\alpha/\beta}
\n{f}_{C^s}^{1-\alpha/\beta}\leq 2\n{f}_{C^s}\,.
\]
If $n 0$ we consider the complex strip
\[
U_\tau=\{x \in \C^m \;\big|\; |\Im x_i|\leq \tau,\; 1\leq i \leq m \}
\]
and the Banach space $E_\tau$ defined as the set of real holomorphic
functions $f$ on $U_\tau$ with period $1$ in each variable endowed
with the norm $\n{f}_\tau=\sup_{x \in U_\tau}|f(x)|$.
We introduce the concept of smoothing operators which is an abstract
version of the usual controlled approximation of H\"older
functions by smoother ones. The results stated below can be found
in \cite{Mo} and \cite{Ze}.
\begin{theo}\label{wsmo}
There exists an analytic smoothing in the family $(E_\tau)_{\tau \geq 0}$
with respect to $\{\Lambda_r(\T^m)\}_{r\geq 0}$
(and $\{C^r(\T^m,\R)\}_{r\geq 0}$), i.e. a collection of linear continuous
operators $S_\tau \in L(C(\T^m,\R),E_1)$ together with constants
$k_r$, $00$ and $\tau>0$ one has
\begin{itemize}
\item[i)] $\lim_{\tau \rightarrow \infty} \n{(S_\tau-1)f}_{C^0}=0
$ if $ f\in C(\T^m,\R)$,
\item[ii)] $\n{S_\tau f}_{\tau^{-1}} \leq k_r\n{f}_{\Lambda_r}$
if $ f\in \Lambda_r(\T^m)$,
\item[iii)] $\n{(S_\sigma-S_\tau)f}_{\sigma^{-1}}
\leq \tau^{-r} k_r\n{f}_{\Lambda_r}$
if $ f\in \Lambda_r(\T^m), \;\sigma \geq \tau>0$.
\end{itemize}
\end{theo}
In addition, it is interesting to remark that the above quantitative
estimates in the approximation by analytic functions characterize the
$\Lambda_r$-spaces. Observe that relations (ii) and (iii) can be
stated for functions $f \in C^r(\T^m,\R)$ with their usual norms.
The argument of the proof of Theorem \ref{wsmo}, based on
realizing $S\tau$ as a convolution with a smooth kernel that approximates
the identity, also works on $\R^m$.
\begin{theo}\label{csmo}
There exists a $C^\infty$-smoothing in the family
$\{C^r(\T^m,\R)\}_{r\geq 0}$ i.e. a collection $(S_\tau)_{\tau>0}$ of
linear mappings $S_\tau:C(\T^m,\R) \rightarrow C^\infty(\T^m,\R)$,
together with constants $k_{r,s}$, $0\leq r,s < \infty$ satisfying
\begin{itemize}
\item[i)] $ \lim_{\tau\to\infty} \n{(S_\tau -1)f}_{C^0}=0$
if $f \in C(\T^m,\R)$,
\item[ii)] $ \n{S_\tau f}_{C^s} \le \tau^{(s-r)}k_{r,s} \n{f}_{C^r}$
if $f \in C^r(\T^m,\R)$, $ 0 \le r \le s $,
\item[iii)] $ \n{(S_\tau -1)f}_{C^r} \le \tau^{-(s-r)}k_{r,s}
\n{f}_{C^s}$ if $f \in C^s(\T^m,\R)$, $0 \le r \le s$.
\end{itemize}
\end{theo}
More generally, one can find in \cite{Ze}
that smoothing operators
exist for the scale of spaces $\{C^{n+r}(K,\R)\}_{r \geq 0}$,
where $n \in \N$ and $K$ is a compact manifold.
The compactness of $K$ becomes essential for the proof,
e.g. $C^r(\R,\R)$ is not dense in $C(\R,\R)$ for any $r>0$.
The connection with interpolation comes from the following
statement taken from \cite{Ze}.
\begin{prop}\label{Zehnder}
Let us assume that the scale of Banach spaces
$\{E_r\}_{r \geq 0}$
admits a family $S_\tau $ of smoothing operators satisfying
(i)-(iii) of Theorem \ref{csmo}.
If $0 \le r \le t, \; \mu \in (0,1)$ and $x \in E_t$
then, defining $s = \mu r + (1 - \mu)t$, we have
\[
\n{x}_s \le M_{r,s,t}
\n{x}_r^\mu\, \cdot \,
\n{x}_t^{(1-\mu)}\,.
\]
\end{prop}
\begin{pf}
For all $\tau > 0$, we have
\[
\n{x}_s \le
\n{S_\tau x}_s \,+ \,
\n{S_\tau x - x}_s \le
\tau^{(s- r)} k_{r,s}\n{x}_r
+ \tau^{-(t -s)} k_{s,t}\n{x}_s\,.
\]
Evaluating at $\tau=(\n{x}_t/\n{x}_r)^{1/(t-r)}$
we obtain the desired conclusion.
\end{pf}
In consequence the existence of $C^\infty$-smoothing operators
on spaces of differentiable functions is affected
by the geometry of the compact set of definition.
We continue by recalling the definition of H\"older spaces
on closed sets that is used in \cite{Ste} formalization of
Whitney's extension theorem in terms of function spaces.
\begin{defi} \label{extension}
Let $X \subset \R^m$ be a closed set and take $n \in \N$,
$0 < \alpha \leq 1^-$. We say that $f \in C^{n+\alpha}(X,\R)$
if there exist functions $f^{(j)}$ for $j \in \Z^m$, $0 \leq |j|
\leq n$, defined on $X$ such that, if
\[
f^{(j)}(x)= \sum_{|j+l| \leq k} \frac{f^{(j+l)}(y)}{l!}(x-y)^l+
R_j(x,y),
\]
then
\begin{equation}\label{Whol}
|f^{(j)}(x)| \leq N \quad \mbox{\rm and}
\quad |R_j(x,y)| \leq N |x-y|^{n-|j|+ \alpha}
\end{equation}
for every $0 \leq |j| \leq n$ and $x, y \in X$.
Again the norm $\n{f}_{C^{n+\alpha}}$ is taken to be the smallest $N$
satisfying the inequalities (\ref{Whol}).
\end{defi}
If the domain $U$ is badly behaved, it is not true in general
that the extension $\overline {f}$
on $\overline{U}$ of a function $f \in C^r(U,\R)$ lies in the space
$C^r(\overline{U},\R)$ above defined.
An extension operator $E_{X,Y,r}$
for closed sets $X\subset Y$ is
a bounded linear operator from $C^r(X,\R)$ to $C^r(Y,\R)$ such that
$E_{X,Y,r} x $ and $x$ agree on $X$.
We say that an operator $E_{X,Y,r}$ is
is {\em consistent\/} in a set of indices $P$
if, for $s,\,r\in P$ with $s > r $,
$ E_{X,Y,r} $ maps $C^s(X,\R)$ into $C^s(Y,\R)$ and
is bounded.
We note that if there is a coherent extension,
$X$ inherits the interpolation
properties of $Y$ in every subset of $P$.
In effect, with $r < t$, $\mu \in (0,1)$,
$s = \mu r + (1- \mu) t $,
denoting the extension operator just by $E$,
we have
\begin{equation*}
\begin{split}
\n{f}_{C^s(X,\R)}&\le
M_{r,s,t} \n{E f}_{C^r(Y,\R)}^\mu
\n{E f}_{C^t(Y,\R)}^{(1 -\mu)}\\[.2cm]
&\le \n{E}_{ C^r(X,\R) ,C^r(Y,\R)}^\mu
\n{E}_{ C^t(X,\R),C^t(Y,\R)}^{(1 -\mu)}
M_{r,s,t} \n{f}_{C^r(X,\R)}^\mu
\n{f}_{C^t(X,\R)}^{(1 -\mu)}\,.
\end{split}
\end{equation*}
Consider the domain $U$ defined in Example \ref{noint},
it is well-known that
the celebrated Whitney extension theorem
(see e.g. \cite{Ste} VI, 2) provides
extension operators from
$C^r({\overline U,\R})$ to $C^r(\R^2,\R)$ for every $r \in \R-\Z$.
We call attention to the fact that,
according to the above remarks, it
produces consistent extensions only
for the sets of regularities with the same
integer part.
This situation is in contrast with regularities measured
in other scales of spaces, e.g.
Sobolev spaces $ L^p _k $ where it is possible
to find extension operators that work for all the
scale of spaces simultaneously (see \cite{Ste} VI 3).
\section {The range of compositions.}\label{range}
In this section, we will study
the regularity of the composition
of two functions expressed in the scale of
spaces that we have introduced before.
The following results, which estimate composition with
linear and bilinear functions, will be used in the
study of the successive
derivatives of the composition.
\begin{prop} \label{lin}
Let $ l \in L(F,G) $ and $t \geq 0$.
Then $ l_{*} : C^t(U,F) \rightarrow
C^t(U,G) $ defined by
$ l_{*}(f) = l \circ f $ is a linear continuous map.
Moreover, $ \n{l_{*}}=\n{l}$.
\end{prop}
\begin{pf}
Take $t=k+\gamma$ with $k \in \N$
and $0\le\gamma\le 1^-$. When $k=0$ it is
easy to verify that $l\circ f \in C^t(U,G)$ and
\[
\n{l \circ f}_{C^t} \leq \n{l}\,\n{f}_{C^t}\,.
\]
We introduce $l_1: L(E,F) \rightarrow L(E,G),
\;h \rightarrow l\circ h$. For
$k \geq 1$ we find that $d(l\circ f)= (l_1)_{*}(df)$.
Thus the proof
can be carried on immediately by induction.
\end{pf}
\begin{prop} \label{bil}
Let $ b \in L_2 (F,G;H) $ and $t \geq 0$.
Then $ b_{*} : C^t(U,F) \times C^t (U,G) \rightarrow C^t (U,H) $ defined
by $ b_{*} (f,g)(x) = b(f(x), g(x)) $ is a bilinear continuous map.
Moreover, $ \n{b_{*}}\leq 2 \n{b}$.
\end{prop}
\begin{pf}
Take $t=k+\gamma$ with $k \in \N$ and
$0\le \gamma\le 1^-$. When $k=0$ one has
\[
\n{b_{*}(f,g)}_{C^t} \leq 2\n{b}\,\n{f}_{C^t} \n{g}_{C^t}\,.
\]
We introduce $b_1: L(E,F)\times G \rightarrow L(E,H),
\;(h_1,z) \rightarrow [x \rightarrow b(h_1(x),z)]$ and
$b_2:F \times L(E,G) \rightarrow L(E,H),\;(y,h_2) \rightarrow
[x \rightarrow b(y,h_2(x))]$. For $k\geq1$ we obtain
$d(b_{*}(f,g))=(b_1)_{*}(df,g)+(b_2)_{*}(f,dg)$,
which allows us to complete the proof.
\end{pf}
We will use the above propositions in the special case where
$b:L(F,G) \times L(G,H) \rightarrow L(F,H),
\;(h_1,h_2) \rightarrow h_2\circ h_1$
is defined by composition of linear maps, which provides
\[
\begin{array}{ccc}
b_{*} :C^t(U,L(F,G))\times C^t(U,L(G,H))
& \longrightarrow&C^t(U,L(F,H)) \\[.1cm]
(f,g)& \longrightarrow& [x \rightarrow g(x)\circ f(x)]\,.
\end{array}
\]
If $F=K$ then $ L(F,G) \equiv G$, $ L(F,H) \equiv H$ and
$b$ becomes the evaluation map. In consequence,
\[
\begin{array}{ccc}
b_{*} : C^t(U,G)\times C^t(U,L(G,H))
& \longrightarrow& C^t(U,H) \\[.1cm]
(f,g)& \longrightarrow& [x \rightarrow g(x)(f(x))]\,.
\end{array}
\]
To simplify the notation in the above cases, we will denote
the map $b_{*}(f,g)$ by $ g \cdot f$.
The following result that characterizes the regularity of compositions
will be the main result of this section.
\begin{theo} \label{com}
Let $E,\,F,\,G $ be Banach spaces and
$ U \subset E, \; V \subset F $ be compensated
subsets.
Let $ r = n+ \alpha , s =m + \beta$ with
$m, n \in \N $ and $0\le \alpha,\beta\le 1^-.$
If $ f \in C^r (U,F)$, $f(U) \subset V $ and $g \in C^s (V,G) $,
then $ g \circ f $ belongs to $ C^t (U,G) $
where we take $t$ as follows:
\begin{itemize}
\item[i)] If $ n=m = 0,\;t = rs $. In that case, we have\vspace{.1cm}
\begin{itemize}
\item[i.1)] $ \n{g \circ f}_{C^{rs}} \leq \n{g}_{C^s}
\n{f}^s_{C^r} + \n{g}_{C^0} $.
\end{itemize}\vspace{.2cm}
\item[\rm ii)] If $ \max (r,s) \geq 1,\;t = \min (r,s) $.
In that case, we have\vspace{.1cm}
\begin{itemize}
\item[ii.1)] if $ r \geq 1,\:0 < s \leq 1^-, $
\[
\n{g \circ f}_{C^s} \leq \n{g}_{C^s}
\n{f}^s _{C^1}\kappa_U ^s + \n{g}_{C^0} \,,
\]
\item[ii.2)] if $ 0< r \leq 1^-,\: s \geq 1, $
\[
\n{g \circ f} _{C^r} \leq \kappa_V\n{g}_{C^{1}}
\n{f}_{ C^r} + \n{g}_{C^0}\,,
\]
\item[ii.3)] if $ r \geq 1,\: s\geq 1,\;
\exists \, M_t \geq 1 $ such that
\[
\n{g \circ f}_{C^t} \leq M_{t}\n{g}_{C^t}
(1+ \n{f}_{C^t}^t)\,.
\]
\end{itemize}
\end{itemize}
\end{theo}
\begin{pf}
If $n = m = 0, $ we have
\[ \n{g \circ f(x) - g \circ f(y)} \leq
\n{g}_{C^{ \beta}} \n{f(x)- f(y)}^{ \beta} \leq
\n{g}_{C^{ \beta}} \n{f}_{C^{ \alpha}}^\beta
\n{x-y}^{ \alpha \beta}\,.
\]
This establishes i) and i.1). In this case no
geometrical conditions on the domain are required.
The proof of ii.1) and ii.2) can be obtained in a
similar way.
To verify the last case, we assume that
$s = r = t = k + \gamma $ with $ k \in \N, \; k \geq 1 $ and
$ 0\le \gamma \le 1^- $. We proceed by induction on $ k $.
For $ k=1$, $d(g \circ f) = d g \circ f\cdot df $. Applying
Proposition \ref{bil} and the above assertion ii.1) we obtain
\begin{equation*}
\begin{split}
\n{d(g \circ f)}_{C^{\gamma}}
&\le 2(\n{dg}_{C^{\gamma}}
\n{f}^{\gamma} _{C^1}\kappa_U^\gamma+
\n{dg} _{C^0}) \n{df}_{C^\gamma}\\[.1cm]
& \le 2\n{g}_{C^{1+\gamma}}(1+\kappa_U^\gamma
\n{f}_{C^{1+ \gamma}}^\gamma)
\n{f}_{C^{1+\gamma}}\,.
\end{split}
\end{equation*}
Since there is a constant $M \geq 1$ such that
$M (1 + |x|^{1+ \gamma}) \geq 2(1 +\kappa_U^\gamma |x|^\gamma ) |x|\: $
for every $ x \in \R $, we obtain
\[
\n{g \circ f}_{C^{1+\gamma}}
\leq M \n{g}_{C^{1+ \gamma}}
(1 + \n{f}^{1 + \gamma}_{C^{1 + \gamma}})\,.
\]
Let us assume that the above inequality holds until the integer
$ k-1 $ and consider $ t= k + \gamma $.
Then by induction we have that
$ d(g \circ f) = dg \circ f\cdot df \in C^{t-1} (U) $ and
\begin{equation*}
\begin{split}
\n{d(g \circ f)}_{C^{t-1}}
&\le 2 \n{d g \circ f}_{C^{t-1}}
\n{df} _{C^{t-1}} \\[0.1 cm]
&\le 2M_{t-1} \n{dg} _{C^{t-1}} ( 1 + \n{f}^{t-1}_{C^{t-1}})
\n{df} _{C^{t-1}}\\[0.1 cm]
&\le 2M_{t-1} \n{dg} _{C^{t-1}} ( 1 + (2\kappa_U\n{f}_{C^{t}})^{t-1})
\n{df} _{C^{t-1}}\, , \\
\end{split}
\end{equation*}
according to Proposition \ref{ico2}. The
same inequality used for the case $k=1$ provides
\begin{equation*}
\begin{split}
\n{g \circ f} _{C^t}
&\le \sup ( \n{g \circ f}_{C^0},\n{d(g \circ f)}_{C^{t-1}})\\[.1cm]
&\le M_t \n{g}_{C^t} ( 1 + \n{f}_{C^t} ^t )\,,
\end{split}
\end{equation*}
which completes the proof of Theorem \ref{com}.
Actually assertion (ii.3) only needs that the domain $U$
is compensated.
\end{pf}
The following examples show that the hypotheses in Theorem
\ref{com} are sharp.
\begin{exam}
Let $ \alpha > 0 $ and $ f_{ \alpha } : [-1,1] \rightarrow \R, \;
x \rightarrow |x|^{ \alpha} $. Then
\begin{itemize}
\item[i)] if $ 0 < \alpha, \beta \leq 1^-$,
then $ f_{ \alpha} \circ f_{ \beta}
\in C ^{ \gamma} ([-1,1],\R) $ if and only if
$ \gamma \leq \alpha\beta $,\vspace{.1cm}
\item[ii)] if $ \alpha/2 \not \in \N $,
then $ f_ {\alpha} \circ Id \in C^{\gamma}
([-1,1],\R) $ if and only if $ \gamma \leq (\alpha -1)+1^-$.
\end{itemize}
\end{exam}
\section{Interpolation inequalities for different numbers of
derivatives.} \label{interpoldiff}
In this section, we develop interpolation inequalities which
work for indices that straddle different integer parts. As we
have shown before, these cannot hold
for general domains.
The proof of interpolation inequalities for
spaces involving different numbers of
derivatives was started in \cite{Ha} and \cite{La}
for functions in the line.
The method of proof is rather elementary
and hence it is the basis for
the extensions to general Banach spaces.
The idea of the proof is that if we restricted the function
to a curve, we could use the classical
argument for interpolation of the restriction.
To conclude interpolations for the whole set, it suffices
to find through every point in the set
a rather well behaved family of
curves whose tangent vectors at said point allow us to generate
the complete vector space.
This is the main idea for the definition of regular subset.
The Example \ref{noint}
shows that indeed such a hypothesis on the domain is
needed.
Let $E$ be a Banach space. If
$D \subset E$, the absolutely convex cover of $D$ is defined by
$\Gamma(D)=\{\sum_{i=1}^n \rho_i x_i | \sum_{i=1}^n|\rho_i| \leq 1,
x_i \in D\}$. The closure of $\Gamma(D)$, that we represent by
${\overline \Gamma(D)}$,
is the closed absolutely
convex cover of $D$. Note that this set can be also calculated
in the weak topology.
\begin{defi}\label{reg}
We say that an open subset $U \subset E$ is {\em regular} if there
exist constants $\kappa_U^*>0$ and $0<\delta<1$
for which the following conditions hold:
\begin{itemize}
\item[i)] for each $x \in U$ there exists a bounded set of
directions $D_x \subset B_{\kappa_U^*}$ with $B_1 \subset
\overline{\Gamma(D_x)}$,
\item[ii)] for each $x \in U$ and $h \in D_x$ there is a
map $\psi \in C^{1+\delta}([0,1],U)$ such that $\psi(0)=x$,
$\psi'(0)=h$ and $\n{\psi}_{C^{1+ \delta}} < \kappa_U^*$.
\end{itemize}
\end{defi}
We next show the abundance of this kind of sets.
Let us consider $0 <\delta<1$ and
$f \in C^{1+\delta}_{\text loc}(\R^m,\R)$. Assume that
$K=\{ x\in \R^m \;|\;f(x)=0 \}$ is compact and grad$(f(x)) \neq 0$
for every $x \in K$. Then the open set $U=\{ x \in \R^m\,|\,f(x)<0 \} $
is regular. The following result also holds
\begin{prop}\label{conr}
Every convex and bounded open $U \subset E$ is compensated and regular.
\end{prop}
\begin{pf}
Since every convex set is compensated we next refer to the
regularity of $U$.
Let us fix a constant $\kappa_0>0$ with $U \subset B_{\kappa_0}$.
We take $x_0 \in U$ and $0 <\rho <1$ satisfying that $B_\rho(x_0)
\subset U$.
We set $D_x=\{(2/\rho)(y-x)|\, y \in B_{\rho}(x_0) \}$ and
for each $y \in B_{\rho}(x_0)$, we define
\[
\begin{array}{rccc}
\psi_{x,y}:&[0,1]& \rightarrow &U \\
&t& \rightarrow & x +(\frac{2}{\rho}t-(\frac{2}{\rho}-1)t^2)(y-x).
\end{array}
\]
Then $\n{\psi_{x,y}}_{C^{1+\delta}} \leq 4\kappa_0/\rho=\kappa^*_U$
for every $0<\delta<1$. Besides
\[B_1 \subset \{\frac{1}{\rho}(y-x)-\frac{1}{\rho}(x_0-x) \big| \,
y \in B_{\rho}(x_0) \} \subset \Gamma(D_x)\]
which shows that the domain $U$ is regular.
\end{pf}
\begin{exam} We consider the Banach space of summable sequences
\[ l^1=\{ {\bold x}=(x_n)_{n=1}^\infty \in \R^{\N} \big|
\sum_{n=1}^\infty |x_n|<\infty \} \]
with the usual norm $\n{{\bold x}}_1=\sum_{n=1}^\infty|x_n|$. Then
\[U=\{(x_n)_{n=1}^\infty \in l^1 \big|
\sum_{n=1}^\infty \frac{|x_n|}{n!}<1 \} \]
defines a convex open subset of $l^1$ which is not regular. In consequence,
the boundedness of the domain is an essential
hypothesis in Proposition \ref{conr}.
\end{exam}
We establish the existence of interpolation inequalities in compensated
regular domains.
\begin{theo}\label{itt}
Assume that the domain $U$ is compensated and regular.
Let $r,s,t$ be positive numbers
with $0\leq r~~0$ and fix $f \in C^t(U;F)$.
Under our conditions, the map $d^{n-1}f:U\rightarrow L_{n-1}(E,F)$ is
differentiable and given $\epsilon >0$ we can find
$x \in U$ and $h \in D_{x}$
such that $\n{d^nf}_{C^0}-\epsilon \leq
\n{d^nf(x)h}$. In addition, there exists $0<\delta \leq \gamma$
and $\psi \in C^{1+\delta}([0,1],U)$ with $\psi(0)=x$, $\psi'(0)=h$ and
$\n{\psi}_{C^{1+\delta}}\leq \kappa_U^*$.
We define $z:\R_+\rightarrow L_{n-1}(E,F)$,
$t \rightarrow d^{n-1}f(\psi(1-e^{-t}))$ and
take the real number
$\nu= \delta/(1+\delta- \alpha)$. It follows
from Theorem \ref{com} and relation \ref{cur}
that $z\in C^{1+\delta}([0,1],F)$ and
\[
\n{d^nf}_{C^0}-\epsilon \le \n{z'}_{C^0} \le
2 H_{\alpha}(z)^{\nu}
H_{\delta}(z')^{1-\nu} \le M'
\n{f}_{C^r}^{\nu} \n{f}_{C^{n+\delta}}^{1-\nu}.
\]
This provides
\begin{equation*}
\begin{split}
\n{f}_{C^{n}}
&\le M'
\n{f}_{C^{n-1+\alpha}}^\nu
\n{f}_{C^{n+\delta}}^{1-\nu}\\[.1cm]
&\le M \n{f}_{C^{n-1+\alpha}}^\nu
\n{f}_{C^n}^{(1-\nu)(1-\delta/\gamma)}
\n{f}_{C^{n+\gamma}}^{(1-\nu)\delta/\gamma} \,,
\end{split}
\end{equation*}
according to the assertion (i).
The interpolation inequality (\ref{int2}) is obtained by eliminating
$\n{f}_{C^n}$ in the above expressions.
\item[iii)] The case $0 \leq r ~~~~0$ such that
for every $f \in C^t(U,F)$ and
$g \in C^t(V,G)$ with $f(U) \subset V$ one has
\[
\n{g \circ f}_{C^t} \leq M_t(1+\n{f}_{C^1}^{t-1})
(\n{g}_{C^1}\n{f}_{C^t} +
\n{g}_{C^t} \n{f}_{C^1})
+\n{g}_{C^0} \,.
\]
\end{prop}
\begin{pf}
Write $t=n+\alpha$, where $n\ge 1$ and $0\leq \alpha \leq 1^-$.
We will make use of the two following well-known results:
\begin{itemize}
\item[i)] given $0\leq \gamma \leq1$ there is
$M>0$ satisfying $x^\gamma y^{1-\gamma}\leq M (x+y)$
for every $x,y \geq 0$.
\item[ii)] given $\kappa, C \geq 0$ there is $M>0$ satisfying
$C(1+\kappa^\alpha x^\alpha)x^{n-1}\leq M(1+x^{n+\alpha-1})$
for every $x\geq 0$.
\end{itemize}
By a repeated application of the chain rule we can find constants
$c_{k,j_1,\ldots,j_k}$ such that
\begin{equation}\label{dmgf}
d^m(g\circ f)(x)=\sum_{k=1}^m \;\sum_{j_1+\cdots j_k=m}
c_{k,j_1,\cdots,j_k}d^kg(f(x))\{d^{j_1}f(x),\ldots,d^{j_k}f(x)\}\,,
\end{equation}
with $j_i \geq 1$ and $1\leq m \leq n$. Using the interpolation
inequalities for both $f$ and $g$ with $10$
satisfying the claim.
\end{pf}
\section{Differentiability properties of composition.} \label{prco}
Throughout this section, we will systematically study
differentiability properties of the map
\[
\begin{array}{rccc}
\Comp:\:[{\cal U} \subset&\!\!\!\! C^r(U,F)] \times C^s(V,G) &
\longrightarrow& C^t(U,G) \\
&\!\!(f,g)& \longrightarrow &\Comp(f,g)=g \circ f
\end{array}
\]
under the following hypotheses:
\begin{itemize}
\item[h.1)]the exponents $r,s\geq 0$ and
\[
t \leq \left \{ \begin{array}{ll}
rs \;\;\;& \mbox{if} \; 0\leq s<1,\; 0 \leq r<1 \\
\min(r,s) & \mbox{if} \; 1 \leq \max (r,s)\,,
\end{array}
\right.
\]
\item[h.2)] ${\cal U} \subset C^r(U,F)$ is open
and dist$\,(f(U),V^c)>0$ for every $f \in {\cal U}$,
\item[h.3)] $U \subset E$ and $V \subset F$ are open compensated sets.
\item[h.4)] for every $1>\varepsilon >0$, there is $\delta(\varepsilon)>0$
and an open compensated $V_\varepsilon \subset V$ such that
$\{y \in V| \; \text{dist}\,(y,V^c)>\varepsilon \} \subset
V_\varepsilon \subset \{y \in V| \; \text{dist}\,(y,V^c)> \delta(\varepsilon) \}$.
\end{itemize}
Hypotheses h.1) and h.2) are needed
to obtain a correct definition of composition on a open
neighborhood in the $f$-component. Finally, h.3) and h.4) require
geometrical conditions on the domains which simplify
the behavior of $\Comp$; in particular, they assure that small
translations of differentiable paths joining points of
each domain $V_\varepsilon$ remain in $V$.
Observe that all these conditions
are satisfied in most of the applications.
We will study regularity with respect to each of the factors,
as a previous step to deduce the conditions for the
joint continuity and differentiability of the map.
The following result is a direct consequence of Theorem \ref{com}.
\begin{prop}\label{pa1}
Let $ r,s\geq 0 $ and
$ f \in C^r (U,F) $. Then the map
\[
\begin{array}{ccc}
f^*: C^s(V,G)& \longrightarrow &C^t(U,G) \\
g &\longrightarrow &g \circ f
\end{array}
\]
is linear and continuous.
\end{prop}
Now we consider the variation of the first factor.
\begin{prop} \label{pa2}
Let $ r,s,t \geq 0$ and $g \in C^s(V,G)$. The map
\[
\begin{array}{ccc}
g_* : {\cal U} \subset C^r(U,F)& \longrightarrow& C^t (U,G) \\
f& \longrightarrow& g \circ f
\end{array}
\]
is continuous in the following cases:
\begin{itemize}
\item[i)] for $t=0\:$, when $\: r\geq 0 ,\; s>0,$
\item[ii)] for $\: 0t,\; r\geq t$ and $rs>t,$
\item[iii)] for $\: t=k+ \gamma$ with $k\geq 1$ and
$ 0 \leq \gamma <1^-$, when $ s>t$ and $ r \geq t,$
\item[iv)] for $t=k+1^-$ with $k\geq 0$, when $s>k+1$ and $r\geq t.$
\end{itemize}
Besides, for every $f_1 \in {\cal U}$ there exist
positive real numbers $\delta$, $\rho$ and
$M$, irrespective of $g$, such that if
$f_2 \in C^r(U,F)$ with $\| f_1-f_2 \|_{C^r} <\delta$ one has
that $f_2 \in {\cal U}$ and
\begin{equation}\label{loch}
\n{g \circ f_1 - g \circ f_2}_{C^t} \leq
M \n{g}_ {C^{s}}\n{f_1 - f_2}^{\rho}_{C^r} .
\end{equation}
In particular, $\rho$ does not depend on $f$, i.e.
$\rho=\rho(r,s,t)$, and $g_{*} \in C_{\text{\rm loc}}^\rho({\cal
U},C^t(U,G))$. \end{prop}
\begin{pf}
We make use of different properties of $\Comp$ obtained
in Sections \ref{geom} and \ref{range}. Since we expound in detail
borderline cases, the exponents $0<\rho\leq 1$ obtained
in the proof are not always optimum; sometimes they can be improved
using interpolation inequalities. We fix $f_1 \in {\cal U}$
for all what follows.
\begin{itemize}
\item[i)] It is similar to i) of Theorem \ref{com}.
Thus, if $0< s \leq 1^-$ then
\begin{equation}\label{con1}
\n{g_*(f_1)-g_*(f_2)}_{C^0}
\leq \n{g}_{C^s} \n{f_1-f_2}_{C^0}^s.
\end{equation}
\item[ii.1)]
First we consider $0~~~~0 $, there
exist, according to hypothesis h.2), a real positive number
$\delta(\lambda)$ and an open compensated set $V_\lambda$ such that
$f_1(U) \subset V_\lambda \subset \{y \in V|\; \text{dist}\,(y,V^c)
> \delta \}$. We can also assume that
$\text{dist}\,(f_1(U),V_\lambda^c)>0$.
Take $ x,y \in U $ and represent $ z_1 = f_1(x), w_1 = f_1 (y) $.
For each $\varepsilon>0$ there is a path $ \gamma _1 $ joining
$ z_1 $ and $ w_1 $ in $V_\lambda$ with
\[
\int_0^1\n{\dot{\gamma}_1(t)}\, dt \leq
(\kappa_\lambda+\varepsilon)\n{z_1-w_1}
\quad (\text{where}\;\kappa_\lambda=\kappa_{V_\lambda})\,.
\]
Let $ f_2 \in {\cal U}$ with $ \| f_2-f_1 \| _r < \delta $.
First of all we have
\[ \n{g_*(f_1)-g_*(f_2)}_{C^0} \leq \kappa_V \n{g}_{C^1}
\n{f_1-f_2}_{C^0}.\]
If we denote $ z_2 = f_2(x) $ and $ w_2 = f_2(y) $ then
\[
\gamma _2 (t) = \gamma _1 (t) + (1-t) (z_2-z_1) + t(w_2-w_1)
\]
is a path joining $z_2$ and $w_2$ in $V$. We next introduce
\begin{align*}
I_1 & = \int ^1_0 \n{[dg(\gamma _1 (t)) - dg (\gamma _2(t))]
\dot{ \gamma}_1 (t)}\, dt \\[.1cm]
& \leq \n{g}_{C^{1 + \beta}} (\n{z_1 - z_2} +
\n{w_1 -w_2})^{\beta}( \kappa_\lambda+\varepsilon)
\n{z_1 - w_1}\,,\\[.1cm]
I_2 & = \int^1_0 \n{dg(\gamma _2 (t))
( \dot{ \gamma}_1 (t)- \dot{ \gamma} _2 (t))}\,dt \leq
\n{g}_1\n{z_2 - z_1 + w_1 - w_2}\,.
\end{align*}
We obtain
\begin{eqnarray*}
\lefteqn{\!\!\!
\n{g(f_1(x)) - g(f_2(x)) - g(f_1(y)) + g(f_2(y))}}\\[.1cm]
&&\quad\le \int ^1 _0\n{(dg(\gamma_1(t))\dot{\gamma_1}(t)-
dg(\gamma_2(t))\dot{\gamma_2}(t)}\,dt\le I_1+I_2 \\[.1cm]
&&\quad\le 2\n{g}_{C^{1+ \beta}}\n{f_1 - f_2}^{\beta}_{C^0}
\kappa_\lambda \n{f_1}_{C^r}\n{x-y}^r \\[.1cm]
&&\quad\quad + \n{g}_{C^1} \n{f_1-f_2}_{C^r}\n{x-y}^r.
\end{eqnarray*}
This shows that
\[
\n{g \circ f_1 - g \circ f_2}_{C^r}
\le \n{g}_{C^{1+ \beta}}
\left(2\kappa_\lambda \n{f_1}_{C^r}\n{f_1 - f_2}^{\beta}_{C^0}
+\kappa_V\n{f_1 - f_2}_{C^r}\right)\!.
\]
Hence there is a constant $M$, irrespective of $g$, such that
\begin{equation}\label{li2}
\n{g \circ f_1 - g \circ f_2}_{C^r} \le M
\n{g}_{C^{1+\beta}} \n{f_1 - f_2}^{\beta}_{C^r}.
\end{equation}
Since the same above constants can be taken in a complete
neighborhood of $f_1$, we can deduce that
$g_*\in C_{\text{\rm loc}}^\beta({\cal U}, C^t(U,G))$.
\item[iii)]
Assume that $ t=r = k + \alpha$ and $ s = k+ \beta$ with
$k \geq 1$ and $0 \leq \alpha <\beta \le 1^- $.
Take $\delta>0$ such that
if $f_2 \in C^r(U,F)$ and $\n{f_1-f_2}_{C^r} < \delta$ then
$f_2 \in {\cal U}$.
We proceed by induction in $k$. For the integer $ k=1 $ we have
\[
\n{g_*(f_1)-g_*(f_2)}_{C^0} \le
\kappa_V \n{g}_{C^1} \n {f_1-f_2}_{C^0}.
\]
Moreover, $ d(g \circ f ) \in C^{\alpha} (U,L(E,G)) $ and
\begin{eqnarray}
\lefteqn{\!\!\!\nonumber
\n{d(g \circ f_1)-d(g \circ f_2 )}_{C^{\alpha}}}\\[.1cm]
&& =\n{d g \circ f_1\cdot d(f_1-f_2)-(dg\circ f_2-dg\circ f_1)
\cdot df_2}_{C^{\alpha}} \label{li3}\\[.1cm]
&& \le 2 \n{dg \circ f_1}_{C^{\alpha}}\n{d (f_1-f_2)}_{C^{\alpha}}
+ 2 \n{dg \circ f_2 - dg \circ f_1}_{C^{\alpha}}
\n{df_2}_{C^{\alpha}}\,.\nonumber
\end{eqnarray}
On the other hand,
\begin{eqnarray*}
\lefteqn{\!\!\!
\n{dg \circ f_2 - dg \circ f_1}_{C^{\alpha}}}\\[.1cm]
&&\quad
\leq 2 \n{dg \circ f_2 - dg \circ f_1}^{(\beta-\alpha)/\beta}_{C^0}
\left(\n{dg \circ f_2}_{C^{\beta}} +
\n{dg \circ f_1}_{C^{\beta}}\right)^{\alpha/\beta}\\[.1cm]
&&\quad
\le 2\n{g}_{C^{1+ \beta}}\left(\n{f_1}^{ \beta}_{C^1} \kappa_U^\beta
+ \n{f_2}^{ \beta}_{C^1}\kappa_U^{\beta}+2\right)^{\alpha/\beta}
\n{f_1 - f_2}^{\beta-\alpha}_{C^0} .
\end{eqnarray*}
By substitution in (\ref{li3}) we find a constant $M_{1+\alpha}$,
irrespective of $g$, such that if $\| f_1- f_2 \|_{C^{1+\alpha}}
< \delta $ then
\[
\n{g \circ f_1 - g \circ f_2}_{C^{1+\alpha}}
\leq M_{1+ \alpha} \n{g}_ {C^{1+ \beta}}
\n{f_1 - f_2}^{\beta-\alpha}_{C^{1+\alpha}}.
\]
Let us assume that an inequality
of the type (\ref{loch}) with $\rho(r,s,t)= \beta- \alpha$
holds for every integer lower or equal to
$k-1$, and analyze the composition for the integer $k$. We obtain
\[
\n{dg\circ f_2-dg\circ f_1}_{C^{t-1}}\le M_{t-1}\n{g}_{C^s}
\n{f_1-f_2}_{C^{t-1}}^{\beta-\alpha}.
\]
From relation (\ref{li3}) and ii.3) of Theorem \ref{com} we deduce that
\begin{equation*}
\begin{split}
\n{d(g\circ f_2)-d(g\circ f_1)}_{C^{t-1}}
& \le 2M'_t\n{g}_{C^t}\left(1+\n{f}_{C^{t-1}}^{t-1}\right)
\n{f_1-f_2}_{C^t}\\[.1cm]
& \quad + M_{t-1}\n{g}_{C^s}
\n{f_1-f_2}_{C^{t-1}}^{\beta-\alpha}
\end{split}
\end{equation*}
which yields
\[\n{g\circ f_2-g \circ f_1}_{C^{t}}
\leq M_{t}\n{g}_{C^s}
\n{f_1-f_2}_{C^{t}}^{\beta-\alpha}\]
for an adequate constant $M_t>0$. This completes the proof
of the assertion for the range of exponents considered above.
\item[iv)]
Assume that $t=r=k+1^-$ and $s=k+1+\beta$ with $k\geq 0$ and
$ 0 <\beta \leq 1^-$.
The continuity of $g_*$ for $k=0$ is a consequence of
the arguments used in ii). There exist constants
$\delta$ and $M$ such that if $\| f_1-f_2 \|_{C^r}
< \delta$ then $f_2 \in {\cal U}$ and
\begin{equation}\label{li5}
\n{g_*(f_1)-g_*(f_2)}_{C^r}
\le M \n{g}_{C^s} \n{f_1-f_2}_{C^r}^\beta .
\end{equation}
The proof of the inequality (\ref{li5}) for $k \geq 1$ can be derived
by an inductive argument similar to the one detailed in iii). In this
case, one has $\rho(r,s,t)=\beta$.
\end{itemize}
Notice that the hypothesis h.4) is an important ingredient in the proof
of the assertions ii.2) and iv). On the contrary, no geometrical
conditions on the domains are required in the simplest situations.
\end{pf}
\begin{defi}\label{ind}
Let $ \Omega \subset \R ^3_+ $ be the set of exponents $ (r,s,t) $
satisfying some of the conditions stated in i)-iv)
of Proposition \ref{pa2}.
We say that $ \Omega $ is the {\em set of the exponents of continuity.}
\end{defi}
\begin{exam}\label{dis}
{\it Let $ (r,s,t) \in \R^3_+ -\Omega$.
There are open balls $U\subset E$,
$V \subset F$ of Banach spaces and $ g \in C^s (V, \R ) $ such that
$ g_* :\:{\cal U}\subset C^r (U, F ) \longrightarrow C^t (U, \R) $
is discontinuous.}
\end{exam}
\begin{pf}
We first introduce several examples concerning non integer exponents.
If $ 0< s\leq 1^-, 0m$.
Then $ |g \circ f_n - g \circ f_m
| (x) = 2 |x|^{rs} $ when $ x \in [-2^{(-n-1)/r}, 2^{(-n-1)/r}] $,
so that $ \n{g\circ f_n - g \circ f_m}_{C^{rs}} \geq 2 $.
Hence, it is impossible that $ g \circ f_n $ converges in $ C^{rs} $.
If we consider in ii) or iii) the option $ s=t $ and $r\geq \max(1,t)$,
the continuity also fails.
This fact is well known for integer exponents and
we will sharpen it below. If $ s= t=k + \gamma $
with $ k \in \N $ and $ 0 < \gamma \leq 1^- $, it suffices to take
$ g = x^k |x| ^{ \gamma} $ and $ f_n = x+1/n$.
Clearly, $ \n{f_n-Id}_{C^r} \rightarrow 0 $ for any $r$.
If $k=0$ we get
\[
\left|(g \circ f_n - g \circ Id)(\frac{-1}{n})
- (g \circ f_n - g \circ Id) (0) \right|
= 2 \frac{1}{n^{ \gamma}} \,,
\]
hence $ \n{g \circ f_n - g \circ Id}_{C^\gamma} \geq 2$.
Besides, for $k \geq 1$
\[
( g \circ f_n)^{(k)} (x) =\left|x+\frac{1}{n}\right|^{ \gamma}
(k + \gamma)\cdots ( \gamma + 1) \,;
\]
and
\[
\left|(g \circ f_n - g \circ Id)^{(k)}(\frac{-1}{n})
- (g \circ f_n - g \circ Id)^{(k)} (0) \right|
= 2 \frac{1}{n^{ \gamma}} (k + \gamma) \cdots ( \gamma + 1)\,,
\]
so that $ \n{g \circ f_n - g \circ Id}_{C^t} >2\,k!\,$. In
consequence, there is not convergence in the $C^t$-topology.
We detail the following examples in infinite-dimensional Banach spaces.
For $p\geq 1$ we introduce the usual space of summable sequences,
\[
l^p=\left\{{\bold x}=(x_n)_{n=1}^{\infty} \in \R^{\N}
\,\left| \; \sum_{n=1}^{\infty}|x_n|^p<\infty \right. \right\}
\]
with the norm
$\| {\bold x} \|=(\sum_{n=1}^{\infty} |x_n|^p)^{1/p}$,
which converts it into a Banach space.
We denote by $e_n$ the sequence with
all the components $0$ except the $n$-th one which takes the value $1$.
First, we take $00$ big enough we define
\[
\begin{array}{rccc}
{\bold g}_k:\!\!&B_\sigma \subset l^p& \longrightarrow& \R\\
&(x_j)_{j\in \N}& \longrightarrow & \sum_{j=1}^\infty g_{k,j}(x_j)
\end{array}\]
Given ${\bold x}=(x_j)_{j \in \N} \in B_\sigma$, there exist indices $j_1,
\ldots ,j_{m_p}$ with $m_p \leq (2\sigma)^p$ such that if
${\bold y}\in B_{1/2}({\bold x})$ then $|y_j|<1$ for
$j \in \N-\{j_1,\ldots,j_{m_p}\}$. If we denote
\[
\begin{array}{rccc}
\pi _x:\!\!&l^p &\longrightarrow& \R^{m_p}\\
&(y_j)_{j\in\N}&\longrightarrow&(y_{j_1},\ldots,y_{j_{m_p}})
\end{array}
\]
and
\[
\begin{array}{rccc}
{\bold g}_{k,x}:\!\!&\R^{m_p}&\longrightarrow&\R\\
&(y_1,\ldots,y_{m_p})&\longrightarrow&\sum_{i=1}^{m_p} g_{k,j_i}(y_i) \,,
\end{array}
\]
then ${\bold g}_k({\bold y})={\bold g}_{k,x}\circ \pi_x({\bold y})$
for every ${\bold y} \in B_{1/2}({\bold x})$,
and hence ${\bold g}_k \in C^k(B_\sigma,\R)$.
Let $h$ be a $C^\infty$ cut-off function which takes the value $1$ on
$[-1/3,1/3]$ but zero outside of $(-2/3,2/3)$,
and set $f(x)=(1+|x-1|^r)h(x-1)$. We define
\[
\begin{array}{rccc}
{\bold f}:\!\!&B_{3/2} \subset l^1& \longrightarrow& l^p\\
&(x_j)_{j\in \N}& \longrightarrow &( f(x_j))_{j\in \N}
\end{array}
\]
and
\[
\begin{array}{rccc}
{\bold h}:\!\!&B_{3/2} \subset l^1& \longrightarrow& l^p\\
&(x_j)_{j\in \N}& \longrightarrow & (h(x_j-1))_{j\in \N}\,.
\end{array}
\]
The same previous argument shows that ${\bold h} \in
C^\infty(B_{3/2} \subset l^1,l^p)$ (this map will be used in the
study of the differentiability of ${\bold g}_k$).
On the other hand,
\[
\n{{\bold f}({\bold x})-{\bold f}({\bold y})}=
\left(\sum_{j=1}^{\infty}|f(x_j)-f(y_j)|^p\right)^{1/p} \leq
\n{f}_{C^r} \n{{\bold x-y}}^r
\]
and ${\bold f}_n={\bold f}+e_n/2n$ satisfies this same inequality for every
$n \in \N$. It is obvious that ${\bold f},{\bold f}_n \in C^r(B_{3/2}
\subset l^1,l^p)$ and $\n{{\bold f}_n-{\bold f}}_{C^r}\rightarrow 0$.
However, if $x \in [1,1+(1/2n)^{p})$ for $n$ big enough,
\begin{equation*}
\begin{split}
{\bold g}_1\circ{\bold f}_n(x e_n)-{\bold g}_1\circ{\bold f}(xe_n)
&=g_{1,n}\left(f(x) + \frac{1}{2n}\right)-g_{1,n}(f(x))\\[.1cm]
&= \frac{|x-1|^r}{2}+\frac{1}{8n}\:,
\end{split}
\end{equation*}
which provides
$\|{\bold g}_1\circ{\bold f}_n-{\bold g}_1\circ{\bold f}\|_{C^r}\geq 1/2$.
In this example there is not continuity for $s=1$ and $t=r$. This
completes ii).
Finally, if we consider $p=1$ and $f(x)=xh(x-1)$ in the above
relations, then ${\bold f} \in C^\infty(B_{3/2} \subset l^1,l^p))$ and
$\n{{\bold f}_n- {\bold f}}_{C^r} \rightarrow 0$ for every $r>0$.
Take $x \in [1,1+1/2n)$; therefore
\[
{\bold g}_0\circ{\bold f}_n(x e_n)-{\bold g}_0\circ{\bold f}(xe_n)=
\frac{1}{2}\:,
\]
which shows that the continuity in i) requires the uniform
continuity of the second factor $g$.
Besides, in the same conditions, one has
\begin{equation*}
\begin{split}
d^k({\bold g}_{k+1}\circ{\bold f}_n-{\bold g}_{k+1}\circ{\bold f})(xe_n)
\{e_n,\stackrel{(k)}{\ldots},e_n\}
&=g_{1,n}(f(x) + \frac{1}{2n})-g_{1,n}(f(x))\\[.1cm]
&= \frac{x-1}{2}+\frac{1}{8n}\:,
\end{split}
\end{equation*}
so that $\|{\bold g}_{k+1}\circ{\bold f}_n-{\bold g}_{k+1}
\circ{\bold f}\|_{C^{k+1^-}}\geq 1/2$. In this example, there is not
continuity for $s=k+1,\;t=k+1^-$ and neither for $s=k+1,\;t=k+1$.
\end{pf}
\begin{rema}
If $F$ is a finite dimensional space,
$\overline{f(U)}$ is a compact subset of $V$.
Every continuous function on $V$
is uniformly continuous on each compact neighborhood of
$f(U)$. Notice that under the additional hypothesis
that $g \in C_{u}^s(V,F)$ or
$F$ is finite dimensional, we can supplement the cases of continuity
in Proposition (\ref{pa2}) with the following ones:
\begin{itemize}
\item[v)] $00$.
We consider the open sets
$V_1=\cup_{x \in U}B^F_{\lambda/3}(f_1(x))$,
$V_2=\cup_{x \in U}B^F_{2\lambda/3}(f_1(x))$, then
$\overline{V}_1\subset V_2\subset V$.
There exists a constant $\delta>0$ such that if
$f \in C^r(U,F)$ and $\n{f-f_1}_{C^r} \leq \delta$, then
$f \in {\cal U}$ and $f(U) \subset V_1$. The Banach
space $F$ is superreflexive, this allows us to take
a uniformly continuously differentiable function
$\varphi:F \rightarrow \R$ such that $0\leq\varphi\leq 1$,
$\varphi|_{V_1}= 1$ and
$V_2= \{y \in F\,|\, \varphi(y) \neq 0 \}$ (see \cite{Su}).
Since $V_2$ is bounded,
it is easy to check that $\varphi \in C^1_u(F,G)$.
Besides, for every $g \in
C^1_u(V,G)$ one has that $\varphi \cdot g \in C^1_u(F,G)$ and
$C^1_u(V,G) \rightarrow C^1_u(F,G)$, $g \rightarrow
\varphi \cdot g$ is a bounded linear map. Finally,
$(\varphi \cdot g) \circ f|_U=g \circ f|_U$ for every $f\in
{\cal B}_{\delta}(f_1)$, which proves that Comp is continuous.
Observe that for $0 < t \leq 1^-$, $s >1$, $t \geq 1$,
i.e. $(r,s,t) \in \Omega$, all the maps $g_*$ admit a common
modulus of continuity when $g$ varies in a bounded subset of
$C^s(V,G)$. This result will be used in the study of the
differentiability of Comp.
For the case
$t=k+1^-$, $s=k+1$ and $r \geq k+1^-$ with $k \in \N$,
we follow
an inductive process, similar to the one detailed in iv)
of Proposition \ref{pa2}, to deduce the continuity of Comp.
In each one of steps
we make use of the above arguments,
based in a convenient extension to $F$ of continuously
differentiable functions on $V$.
In addition, for $ t = k+1^-$, $s >k+1$, $r \geq k+1$,
i.e. $(r,s,t) \in \Omega$, we remark again that
all the maps $g_*$ admit a common
modulus of continuity when $g$ varies in a bounded subset of
$C^s(V,G)$.
We now consider $m\geq 1$. The corresponding versions,
under the present conditions, of Proposition \ref{dip}
and Proposition \ref{pad} allow us to conclude that
Comp is $m$ times continuously differentiable.
On the other hand, hypothesis h.4) was not required to check the
differentiability of Comp for the rest of values $(r,s,t)$
with $(r,s-m,t) \in \Omega_u$; thus the previous proof is also
valid for this case.
\end{pf}
We finally remark that, since the considered sets of definition
are open subsets of Banach spaces,
there may be functions which are continuously differentiable
but unbounded or with unbounded derivatives.
In these cases, besides the topology induced
by the supremum norm, there are other natural topologies,
e.g., the Withney topology associated with the uniform convergence
on closed bounded sets. This topology is a projective limit
of the considered in this paper, and the results on continuity can be
immediately lifted. In addition,
Theorems \ref{teo} and \ref{teu} yield the starting point to complete the
study of the differentiability of the composition in the mentioned topology.
\section{Acknowledgments.}\label{acknow}
The work of R.L. has been supported by NSF grants.
Also R.L wants to acknowledge the hospitality
enjoyed at Valladolid during several visits.
The work of R. O. has been partially supported
by Junta de Castilla y Le\'{o}n.
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\end{document}
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