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\begin{document}
\title[]{Extension of smooth functions in infinite dimensions, I: unions of
convex sets}
\author[]{C. J. Atkin}
\address{School of Mathematical and Computing Sciences, Victoria university
of Wellington, P.O. Box 600, Wellington, New Zealand}
%\curraddr{}
\email{Chris.Atkin@vuw.ac.nz}
%\dedicatory{}
\date{March 1999}
\thanks{}
% \translator{}
%\keywords{}
\subjclass{46T20}
\begin{abstract}
Let $f$ be a smooth function defined on a finite union $U$ of open convex
sets in a locally convex Lindel\"of space $E$. If, for every $\,x\in U\,$,
the restriction of $f$ to a suitable neighbourhood of $x$ admits a smooth
extension to the whole of $E$, then the restriction of $f$ to a union of
convex sets that is strictly smaller than $U$ also admits a smooth extension
to the whole of $E$.
\end{abstract}
\maketitle

\section{General remarks on extension problems.}\label{section1}

\label{para1}The guiding idea of this paper, which is intended as the first
of a series, is to see what can be done to avoid the common and highly
restrictive assumption that the model of a smooth manifold admits C$^\infty$
(or C$^p$, for some $\,p\geq 1\,$) partitions of unity. In this first paper I
shall be exclusively concerned with the models themselves, and with some
theorems which at first sight may seem too special to be of much consequence.
In later papers I shall show that they are, in fact, the foundation of a
substantial theory.

There has been much work done on the question of the existence or
non-existence of partitions of unity of various degrees of differentiability
(in the sense of Fr\a'echet) in particular Banach spaces. Bonic and Frampton
\cite{bonicframpton} showed that there are restrictions on their existence in
many classical spaces, and pointed out some remarkable consequences of this
fact. Toru$\acute{\text{n}}$czyk \cite{torunczyk} showed that C$^\infty$
partitions of unity exist in Hilbert spaces of any dimension. A copious
reference for the detailed and subtle theory of differentiable functions in
Banach spaces, as it has developed since then, is the book of Deville,
Godefroy, and Zizler \cite{devillegodefroyzizl}. (A perhaps more recent and
startling result is given by Deville \cite{deville}, who shows that, if a
Banach space $E$ does not contain a copy of $\,c_0\,$, then the question of
C$^\infty$ partitions of unity for $E$ is, as it were, determined at the
level of polynomials). In all this work, however, some basic questions about
Banach manifolds have remained unsolved.

The ideal theory of Banach manifolds, if such a thing may be imagined, and if
it were to generalize finite-dimensional constructions satisfactorily, would
be valid for all, or at least for all separable, Banach spaces as models; it
might, for instance, assert, irrespective of the existence of smooth
partitions of unity, that every separable C$^\infty$ Banach manifold is
C$^\infty$-diffeomorphic to a manifold of a certain standard kind, such as a
closed submanifold of some multiple of the model or an open subset of the
model. Individual peculiarities of the model should not unduly influence the
basic theory, although they must inevitably have consequences elsewhere. The
crucial difficulty in the way of such a theory is the absence, in general, of
a suitably rich collection of smooth functions with which to perform familiar
constructions. C$^\infty$ partitions of unity, when they exist, enable one to
imitate many finite-dimensional techniques, but their rarity is the reason
why, for instance, Elworthy in \cite{elworthy} employs considerable effort to
weaken the differentiability class he requires. As far as I know, even the
question he raises in the introduction to \cite{elworthy}, whether every
smooth Banach manifold (in fact he mentions only the models $C$ and
$\,\ell^1\,$, which one would expect to be the most refractory) admits a
non-constant smooth function, has never been answered. In due course,
although not in this paper, I shall give a positive answer to this question,
at least for separable manifolds, and show that much more is true. \ \ 

To establish results of such generality, hard analysis is unlikely to be
appropriate. Indeed, the arguments I employ might be described as
``combinatorical'' --- they prove that a positive solution to an extension
problem in the small leads, by a complicated procedure of piecing-together,
to a positive solution in the large. It will turn out in the end that such a
technique is quite adequate for many purposes, and it has the advantage of
requiring only very weak assumptions. Specifically, the model $E$ must be at
least a Lindel\"of locally convex space (that is, one of which any open cover
has a countable subcover --- see \cite{engelking}, 3.8, pp. 247 et seqq.
Recall that any Lindel\"of space is paracompact). Any separable metrizable
locally convex space is Lindel\"of, and so is the dual of any normed space in
the weak* topology, since it is $\sigma$-compact. I shall also assume, for
verbal and conceptual convenience, that $E$ is Hausdorff. The arguments could
easily be modified to the non-Hausdorff case, which, in the natural cases
where one is dealing with continuous functions, is a trivial corollary
anyway. The notion of `smoothness' need not, however, be made precise, the
only demands being that the class of `smooth' functions should satisfy
certain conditions which obviously ought to hold, and do indeed hold, for
most reasonable theories of differentiability, and for many other classes of
functions and their generalizations as well.

In this first paper, I shall consider an extension problem for finite unions
of convex sets in the model itself. If the model is paracompact and admits
partitions of unity of the appropriate kind, which can be used to piece
together functions in the usual way, the results presented here are quite
trivial and their limitation to convex sets is unnecessary. In general,
however, even these apparently weak results have surprisingly strong
consequences. 

It seems preferable for clarity's sake not to present the construction at the
outset in the greatest generality possible; it can be adapted in many ways,
some of which will be studied later. I have, however, included some
potentially useful \ modifications of the basic idea that require relatively
little extra argument. The subset ``$X$'', which appears in
(\ref{extensioncomplex}), (\ref{extensionfromstars}), (\ref{bigthm}) and
elsewhere, is introduced with a view to classes of functions, such as Besov
spaces, that may be defined on ``thin'' subsets of $\,\mathbb{R}^n$ (see, for
instance, Triebel \cite{triebel1}, \cite{triebel2}), and (\ref{glimp}),
(\ref{homeatlast}) are motivated by the example of abstract Wiener manifolds,
for which the coordinate transformations preserve a dense subspace of the
dual (\cite{eellselworthy}, \cite{elworthy2}). The countability restrictions
on the model already appear several times as an essential hypothesis. 

The simplest extension problem is perhaps the following.

\begin{problem}
Let $\;E,F\,\label{para2}\;$ be normed spaces, with $E$ separable, and let
$\;f:U\longrightarrow F\;$ be a C$^\infty$ map defined on the neighbourhood
$U$ of $0$ in $E$. Does there exist a C$^\infty$ map $\;g:E$ $\longrightarrow
F\;$ which agrees with $f$ on a neighbourhood $V$ of $0$ in $U$? 
\end{problem}

In this formulation ``C$^\infty$'' is understood in the Fr\a'echet sense. A
similar question might be asked about other classes of functions, such as
C$^\infty$ maps with compact derivatives at all points, or maps which are
bounded or have uniformly bounded derivatives. For the above case, one has
for instance the simple result \ 

\begin{theorem}
\label{number}Let \label{para3}$E$ be a normed space in which there is a
C$^\infty$ mapping $\;h:E\longrightarrow E\;$ that is the identity on some
neighbourhood $W$ of the origin, and such that $\,h(E)\,$ is bounded. Let $F$
be any normed space. Then, if $U$ is a neighbourhood of $0$ in $E$ and
$\;f:U\longrightarrow F\;$ is C$^\infty$, there are a neighbourhood $V$ of
$0$ in $U$ and a C$^\infty$ function $\;g:E\longrightarrow F\;$ such that
$\;g\mid V=f\mid V\,$.
\end{theorem}

\begin{proof}
Suppose that $\,h\mid B_E(0;\epsilon)\,$ is the identity, where
$\,B_E(0;\epsilon)\,$ of course denotes the ball of radius $\epsilon$ about
the origin in $E$, and that $\;h(E)\subseteq B_E(0;r)\,$. (Thus
$\,r>\epsilon\,$, of course). Choose $\,\kappa>0\,$ so that $\;B_E(0;\kappa
r)\subseteq U\,$, and define
\begin{equation*}
(\forall x\in E)\quad g(x):=f(\kappa h(\kappa^{-1}x))\,.
\end{equation*}
Then take $\,V:=B_E(0;\kappa\epsilon)\,$ and the result follows.
\end{proof}

\begin{remark}
The proof\label{para4} of (\ref{para3}) shows also that, in this case, the
extension operator $\;f\mapsto g:=\mathcal{E}f\;$ is linear (with respect to
pointwise operations), and that the image of $\,\mathcal{E}f\,$ is included
in the image of $f$. If, therefore, $f$ is bounded with respect to the norm
in $F$, so is $\,\mathcal{E}f\,$, and its bound is no greater than that of
$f$. If the mapping $h$ has other properties --- for instance if its $n$th
derivative is bounded on $E$ for all $\,n\leq N\,$ --- $\mathcal{E}$ may have
associated properties, such as being bounded with respect to the C$^N$ norm
on the space of functions on $U$ for which all derivatives up to the $N$th
are bounded on $U$.

The existence of a function such as $h$ in many ``classical'' Banach spaces
is easily established; the most immediate example is $C$, in which $h$ may
simply be composition with a bounded C$^\infty$ function
$\;\mathbb{R}\longrightarrow\mathbb{R}\;$ which is the identity on a
neighbourhood of $0$. It is not much harder to construct a C$^\infty$
function $h$ in the other ``extreme'' space, $\,\ell^1\,$. On the other hand,
there are classes of functions other than C$^\infty$, most obviously
C$^\omega$, for which the question analogous to (\ref{para2}) has, in
general, a negative answer. 

If one assumes in (\ref{para2}) that $\,$cl$_E(V)\,$ is included in $U$, $f$
must behave well on the boundary of $V$; there can be no ``local''
obstruction to the extension, such as a failure of local boundedness in $E$.
It seems, therefore, quite conceivable, although I think it unlikely, that
the answer to (\ref{para2}) is positive in any normed space. It is perhaps
more plausible, especially in non-separable spaces, that the class of all
functions C$^\infty$ in the sense of Fr\a'echet may be either too large or
too small to have satisfactory extension properties for all normed spaces $E$
and $F$, but that some other class contains all the mappings of practical
interest and has better properties. Generally speaking, the questions of the
form (\ref{para2}) for various choices of function class are all different.

I shall therefore develop a theory which, in essence, defines the question
away; my functions will have a suitable extension property by definition. It
will eventually appear that the results are much more satisfactory than one
might expect.
\end{remark}

\begin{definition}
\label{locally:defined}Let $\mathcal{F}$ \label{para5}be a class of functions
defined on the open sets of the topological space $X$. One may say that
$\mathcal{F}$ is a class \emph{on }$X$.\emph{ }$\mathcal{F}$ is said to be
\emph{locally defined }if\\
\hspace*{1cm}\emph{(i})\quad if $\;f:U\longrightarrow Y\;$ is a function
defined on the open set $U$ such that, for any $\,x\in U$, there is a
neighbourhood $V$ of $x$ in $U$ for which $\;f\mid V\in\mathcal{F}\,$, then
necessarily $\,f\in\mathcal{F}\,$; and\\
\hspace*{1cm}\emph{(ii})\quad if $\;U,V\;$ are open in $X$, $\,V\subseteq
U\,$, and $\;f:U\longrightarrow Y\;$ is in $\mathcal{F}$, then $\;f\mid
V\in\mathcal{F}\,$.
\end{definition}


\begin{definition}
A fu\label{f:extensible}n\label{para6}ction $\;f:U\longrightarrow Y\;$ in the
class $\mathcal{F}$ on $X$ is $\mathcal{F}$\emph{-extensible}, or extensible
in $\mathcal{F}$, if, for any $\,x\in U\,$, there are an open neighbourhood
$V$ of $x$ in $U$ and a function $\;g:X\longrightarrow Y\;$ in $\mathcal{F}$
such that $\;g\mid V=f\mid V\,$. (If every function in $\mathcal{F}$ is
$\mathcal{F}$-extensible, $\mathcal{F}$ itself may be described as
extensible).
\end{definition}


\begin{definition}
A \label{para7}fun\label{finitely:determined}ction $f$, defined on an open
subset $U$ of the locally convex Hausdorff space $E$, is \emph{finitely
determined }if there are a continuous linear surjection
$\;P:E\longrightarrow\mathbb{R}^k\;$ (for some natural number $k$) and a
function $\phi$ on the open subset $\,P(U)\,$ of $\;\mathbb{R}^k\;$ such that
$\;f=\phi\circ P\,$. In effect, a finitely determined function on $U$ is one
which may be expressed as a function of finitely many continuous linear
functionals. $f$ will be described as C$^p$ if and only if $\phi$ is C$^p$.
(This is consistent with any reasonable \emph{a priori} definition.)
\end{definition}

The finitely determined functions on $U$ with values in a vector space $F$
clearly form a vector space; if they take values in an algebra, they form an
algebra. (It is also possible to restrict the linear surjections considered
in the definition, and thus have a somewhat larger class of admissible
functions. See (\ref{parc6})).

\begin{definition}
Le\label{admissible}t $X$ be \label{para8}a subset of the locally convex
space $E$, and let $Y$ be a convex set in a vector space $F$. The class
$\mathcal{F}$ on $X$ is $Y$-\emph{admissible }if it contains at least one
mapping $\;X\longrightarrow Y\,$, and if, whenever $\;f,g:X\longrightarrow
Y\;$ are mappings in $\mathcal{F}$ and $\;\psi:W\longrightarrow[0,1]\;$ is a
finitely determined C$^\infty$ function on an open subset $W$ of $E$ which
includes $X$, then $\;h:X\longrightarrow Y\;$ is also in $\mathcal{F}$, where
$h$ is defined by 
\begin{equation*}
(\forall x\in X)\quad h(x)=\psi(x)f(x)+(1-\psi(x))g(x)\,.
\end{equation*}
\end{definition}

\begin{lemma}
Let\label{ext:on:segments} th\label{para9}e space $X$ be a subset of a
metrizable locally convex Hausdorff space $E$, and let $\mathcal{F}$ be a
$Y$-admissible class of mappings from the open sets of $X$ (in the subspace
topology) to the convex subset $Y$ of the vector space $F$. Suppose that $A$
is an open subset of $E$, $B$ is a closed subset of $E$, and $\;B\subseteq
A\,$. Then, for any $\mathcal{F}$-extensible $\;f:A\cap X\longrightarrow Y\;$
in $\mathcal{F}$, and any closed straight-line segment $\;J:=[a,b]\;$ in $A$,
there exists a convex open neighbourhood $W$ of $0$ in $E$ such that
$\;J+W\subseteq A\;$ and there is a mapping $\;g:X\longrightarrow Y\;$ in
$\mathcal{F}$ for which 
\begin{equation*}
g\mid(J+W)\cap B\cap X=f\mid(J+W)\cap B\cap X\,.
\end{equation*}
\end{lemma}

\begin{proof}
\ For each $\,x\in A\cap X\,$, there is a convex open neighbourhood $\,U_x\,$
of $0$ in $E$ such that $\;x+U_x\subseteq A\,$, and there is a function
$\;h_x:X\longrightarrow Y\;$ in $\mathcal{F}$ such that 
\begin{equation*}
h_x\mid(x+U_x)\cap X=f\mid(x+U_x)\cap X\,.
\end{equation*}
Define $\;A_1:=\bigcup_{x\in B\cap X}(x+\frac{1}{2}U_x)\,$, an open subset of
$A$ including$\,\,B\cap X\,$. 

Suppose that $\,y\in E\,$. If $\,y\notin B\,$, there is a neighbourhood
$\,V_y\,$ of $0$ in $E$ such that $\;(y+V_y)\cap B=\emptyset\,$. Take $\;$
$g_y:X\longrightarrow Y\;$ in this case to be any mapping in $\mathcal{F}$
(there is one, by Definition \ref{admissible}). If, on the other hand,
$\,y\in A_1\,$, then there exists some $\,x\in B\cap X\,$ such that $\;y\in
x+\frac{1}{2}U_x\,$. Choose such an $x$, and let $\,V_y:=\frac{1}{2}U_x\,$;
hence $\;y+V_y\subseteq x+U_x\,$. By the definition of $\,U_x\,$, there is a
function $\;h_x:X\longrightarrow Y\;$ which agrees with $f$ on $\,(x+U_x)\cap
X\,$. Let $\;g_y:X\longrightarrow Y\;$ be $\,h_x\,$. Since $\,E\setminus B\,$
and $\,A_1\,$ constitute an open cover of $E$, this defines, for every
$\,y\in E\,$, a neighbourhood $\,V_y\,$ of $0$ in $E$ and a function
$\;g_y:X\longrightarrow Y\;$ such that $\;g_y\mid(y+V_y)\cap B\cap
X=f\mid(y+V_y)\cap B\cap X\,$.

There will be no loss of generality in supposing that $J$ is a rectilinear
segment of the form $\,[-c,c]\,$, since this may be arranged by translating
$\,\frac{1}{2}(a+b)\,$ to $0$ and also translating the class $\mathcal{F}$ in
the obvious way, if necessary. By the compactness of $J$, there is an open
symmetric convex neighbourhood $W$ of the origin in $E$ such that, for any
$\,z\in J\,$, there is a $\,y\in J\,$ for which $\;z+3W\subseteq y+V_y\,$.
There are two cases.

If $\,W\cap\mathbb{R}c\,$ is unbounded, $\,J\subseteq W\,$, and
$\;J+W\subseteq 2W\subseteq y+V_y\;$ for some $\,y\in J\,$. The conclusion of
the Lemma therefore follows on taking $\;g:=g_y\,$.

If $\,W\cap\mathbb{R}c\,$ is bounded, let $\,\kappa c\,$ be a frontier point
of $W$, and take $\phi$ to be a support functional for $W$ at $\,\kappa
c\label{q}\,$, such that $\;1=\phi(\kappa c)>\phi(v)\,$ for all $\,v\in W\,$.
(This is where the local convexity of $E$ is needed). Then
\begin{equation}\label{wzorlength}
\phi(W)=\phi(W\cap\mathbb{R}c)=(-1,1)\,.
\end{equation}

Take $\;x_1,x_2,\ldots,x_n\in J\;$ so that $\;\{x_i+W:1\leq i\leq n\}\;$
covers $J$. For each $i$, there is$,$ by the choice of $W,$ a function
$\;g_i:X\longrightarrow Y\;$ in $\mathcal{F}$ which agrees with $f$ on
$\;(x_i+3W)\cap B\cap X\,$. Construct nonnegative C$^\infty$ \ functions
$\,\psi_i\,$ on $\mathbb{R}$ so that
\begin{equation}\label{wzorsupp}
\quad\left\{
\begin{aligned}
\text{for each
}i\text{,}\qquad\text{supp}\,\psi_i&\subseteq\phi((x_i+W)\cap\mathbb{R}c)\\
&=(\phi(x_i)-1,\phi(x_i)+1)\,\text{, }\\
\text{and}\qquad\sum_{i=1}^{\,n}\psi_i(t)&=1\quad\text{for each
}\;t\in[-1,1]\,\text{,}
\end{aligned}
\right.
\end{equation}
and define $\;g:X\longrightarrow Y\;$ \ by
$\;g(x):=\sum_{i=1}^{\,n}\psi_i(\phi(x))g_i(x)\,$. As $\mathcal{F}$ is
admissible, $\;g\in\mathcal{F}\,$, by definition \ref{admissible}.

Suppose $\;x\in(J+W)\cap B\cap X\;$ and $\;\psi_i(\phi(x))>0\,$. Then there
is some $\,y\in J\,$ such that $\,x\in y+W\,$, and therefore
$\;|\phi(x)-\phi(y)|<1\,$. Since $\;\phi(x)\in(\phi(x_i)-1,\phi(x_i)+1)\;$ by
(\ref{wzorsupp}), it follows that $\;|\phi(y)-\phi(x_i)|<2\;$ and (by
(\ref{wzorlength}), as $\,y,x_i\in J\,$) $\;y\in x_i+2W\,$, whence $\;x\in
x_i+3W\,$. Consequently $\;g_i(x)=f(x)\,$. This holds for each $i$ for which
$\;\psi_i(\phi(x))>0\,$, \ and therefore, as required, 
\begin{equation*}
g(x)=\sum_{i=1}^{\,n}\psi_i(\phi(x))f(x)=f(x)\,.
\end{equation*}
\end{proof}

This proof is a prototype of the more complicated arguments which follow. It
admits many variants, some of which will be mentioned later.

\ 

\section{Collections of convex sets.}\label{section2}

\begin{definition}
\ Let $\,\label{parb1}A,B\,$ be subsets of an abelian topological group $G$.
I shall say that $A$ is strongly included in $B$, or $B$ strongly includes
$A$, as subsets of $G$, and write $\,A\prec B\,$, if there exists a
neighbourhood $U$ of the identity in $G$ such that $\;A+U\subseteq B\,$.
\end{definition}

The definition holds automatically if $\,A=\emptyset\,$, when the sum is also
empty. It is an infinite-dimensional substitute for the relation which is
familiar in finite dimensions, that $A$ should be a subset of a compact
subset of $B$. Of course I shall take $G$ to be a locally convex Hausdorff
space.

\begin{lemma}
\ Le\label{parb2}t $A$ be a convex set in the locally convex Hausdorff space
$E$, and let $\;Q\prec A\,$. Then there exists a closed convex set $R$ such
that $\;$\emph{cl}$_E(Q)\prec\,$\emph{int}$_E(R)\,$, $\;R\prec A\,$. In
particular, the closed convex envelope of $Q$ is strongly included in $A$.
\end{lemma}

\begin{proof}
\ Let $U$ be a convex open neighbourhood \ $U$ of $0$ such that
$\;Q+U\subseteq A\,$, and define \\
\begin{equation*}
R_0:=\{x\in E:x+\textstyle\frac{1}{3}U\subseteq A\}\,.\text{ }
\end{equation*}
Then $\,R_0\;$ is convex, as is $\;R:=\,$cl$_E(R_0)\,$, which is also
strongly included in $A$, as $\;R+\frac{1}{4}U\subseteq A\,$, for instance.
Furthermore, $\;Q+\frac{2}{3}U\subseteq R_0\,$, from which trivially 
\begin{equation*}
\text{cl}_E(Q)\subseteq
Q+\textstyle\frac{1}{3}U\,\text{,}\quad\text{cl}_E(Q)+\frac{1}{3}U\subseteq
R_0\,\text{,}
\end{equation*}
so that
$\;$cl$_E(Q)\prec\,$cl$_E(Q)+\frac{1}{3}U\subseteq\,$int$_E(R_0)\subseteq\,%
$int$_E(R)\,$.
\end{proof}

\begin{proposition}
\ Le\label{parb3}t $U$ be a convex open set in the normed space $E$, and let
$C$ be a nonnull convex subset of $U$ which is closed in $E$. Then there
exists a convex set $V$ open in $E$ such that $\;C\subseteq V\,$,
$\;$\emph{cl}$_E(V)\subseteq U\,$.
\end{proposition}

\begin{proof}
\ By translation, one may assume that $\;0\in C\,$. If $\;U=E\,$, there is
nothing to prove; so suppose that $\;U\neq E\,$. For each $\;x\in U\,$,
define 
\begin{equation}\label{radincl}
r_U(x):=\sup\{\delta\in\mathbb{R}:B_E(x;\delta)\subseteq U\}\,\text{,}
\end{equation}
which is finite. Then $\;B_E(x;r_U(x))\subseteq U\,$,
$\;|r_U(x)-r_U(y)|\leq\left\|x-y\right\|\,$, and $\;r_U(\alpha x+\beta
y)\geq\alpha r_U(x)+\beta r_U(y)\;$ for $\;x,y\in U\;$ and
$\;\alpha,\beta\in[0,1]\;$ such that $\;\alpha+\beta=1\,$. Now define 
\begin{equation*}
V:=\bigcup_{x\in C}B_E(x;\textstyle\frac{1}{3}r_U(x))\,.
\end{equation*}
It is apparent that $V$ is open in $E$.

Suppose $\;y_1,y_2\in V\;$ and $\;\alpha,\beta\geq 0\;$ and
$\;\alpha+\beta=1\,$. Then there are points $\;x_1,x_2\in C\;$ such that
$\;\left\|y_i-x_i\right\|<\frac{1}{3}r_U(x_i)\;$ for $\;i=1,2\,$. But,
therefore, 
\begin{equation*}
\begin{aligned}[b]
\left\|(\alpha y_1+\beta y_2)-(\alpha x_1+\beta
x_2)\right\|&\leq\alpha\left\|y_1-x_1\right\|+\beta\left\|y_2-x_2\right\|\\
<\textstyle\frac{1}{3}\alpha r_U(x_1)+\textstyle\frac{1}{3}\beta
r_U(x_2)&\leq\textstyle\frac{1}{3}r_U(\alpha x_1+\beta x_2)\,\text{,}
\end{aligned}
\end{equation*}
which, since $\;\alpha x_1+\beta x_2\in C\,$, shows that $V$ is convex. 

Its closure in $E$ is included in $U$. For suppose $\;y_n\rightarrow y\,$,
where each $\;y_n\in V\,$. For each $n$, there is $\;x_n\in C\;$ such that
$\;y_n\in B_E(x_n;\frac{1}{3}r_U(x_n))\,$. If
$\;\delta:=\frac{1}{2}\limsup_{n\rightarrow\infty}r_U(x_n)>0\,$, one may take
a subsequence and renumber to ensure that $\;r_U(x_n)\geq\delta\;$ for all
$n$. If $n$ is so large that $\;\left\|y_n-y\right\|<\frac{1}{3}\delta\,$,
then 
\begin{equation*}
\textstyle\left\|y-x_n\right\|\leq\left\|y-y_n\right\|+\left\|y_n-x_n\right%
\|<\frac{1}{3}\delta+\frac{1}{3}r_U(x_n)\leq\frac{2}{3}r_U(x_n)\,.
\end{equation*}
In view of the definition of $\,r_U\,$, this proves that $\;y\in U\;$ in this
case. So $\;$cl$_E(V)\subseteq U\;$ as required.

If, on the other hand, $\;r_U(x_n)\rightarrow 0\,$, then
$\;\left\|x_n-y_n\right\|\rightarrow 0\,$, and so $\;x_n\rightarrow y\;$ too.
As $C$ is closed, this proves that $\;y\in C\subseteq U\;$ in this case.
\end{proof}

\begin{definition}
\ Le\label{defbdedstrincl}t $E$ be a normed space, and let $\,C,U\,$ be sets
in $E$ such that $\;C\subseteq U\,$. $C$ is said to be \emph{boundedly
strongly included }in $U$ (which I write $\;C\preceq U\,)$ if, for every
bounded subset $Q$ of $E$, $\;C\cap Q\prec U\,$. (This concept is only of
importance in \S 5). \ 
\end{definition}

\begin{remark}
\ The f\label{rkonstrincl}unction $\,r_U\,$ defined at (\ref{radincl}) can be
used to express $\prec$ and $\preceq$: $\;C\prec U\;$ is equivalent to
``$\inf_{x\in C}r_U(x)>0\,$'', and, consequently, $\;C\preceq U\;$ is
equivalent to ``$\inf_{x\in C\cap Q}r_U(x)>0\;$ for all bounded sets $Q\,$''. 

It appears that (\ref{parb3}) cannot be directly generalized to locally
convex spaces. The difficulty arises when $C$ has empty interior; otherwise,
there is the simple argument (\ref{intermed}).
\end{remark}

\begin{lemma}
\ Le\label{bdedstrincl}t $E$ be a normed space, and let $C$ and $U$ be convex
sets such that $\;C\preceq U\,$. Then there exists a convex set $V$ open in
$E$ such that $\;C\preceq V\,$, $\;$\emph{cl}$_E(V)\preceq U\,$.
\end{lemma}

\begin{proof}
\ Follow the proof of (\ref{parb3}). That $\;C\preceq V\;$ is clear, since,
for any bounded $Q$, there exists $\;\epsilon>0\;$ such that
$\;r_U(x)\geq\epsilon\;$ for all $\;x\in C\cap Q\,$, and therefore 
\begin{equation*}
\textstyle V\supseteq(C\cap Q)+B_E(0;\frac{1}{3}\epsilon)\,.
\end{equation*}

Now take some fixed $\;y\notin U\;$ (the case $\;U=E\;$ being trivial), and
set $\;L:=\frac{1}{2}\left\|y\right\|\,$. For any $\;x\in U\,$, $\;r_U(x)\leq
2L+\left\|x\right\|\,$. If $\;K>0\;$ and $\;z\in\,$cl$_E(V)\cap B_E(0;2K)\,$,
there is a sequence $\,(z_n)\,$ in $\,V\cap B_E(0;2K)\,$ such that
$\;z_n\rightarrow z\,$. By the definition of $V$, each $\;z_n\in
x_n+B_E(0;\frac{1}{3}r_U(x_n))\;$ for some $\;x_n\in C\,$, and 
\begin{equation*}
2K>\left\|z_n\right\|>\left\|x_n\right\|-\textstyle\frac{1}{3}r_U(x_n)\geq%
\left\|x_n\right\|-\frac{1}{3}(2L+\left\|x_n\right\|)\,\text{,}
\end{equation*}
so that $\;x_n\in C\cap B_E(0;3K+L)\,$. As $\;C\preceq U\,$, there exists a
$\;\delta>0\;$ (depending only on $K$) such that $\;r_U(x_n)>\delta\;$ for
all $n$, and then, for large enough $n$, 
\begin{equation*}
\begin{aligned}[t]
\textstyle z+B_E(0;\frac{1}{3}\delta)&\subseteq
B_E(z_n;\textstyle\frac{2}{3}\delta)\subseteq
B_E(x_n;\textstyle\frac{1}{3}r_U(x_n)+\textstyle\frac{2}{3}\delta)\\
&\subseteq B_E(x_n;r_U(x_n))\subseteq U\,.
\end{aligned}
\end{equation*}
As this holds for all $\;z\in\,$cl$_E(V)\cap B_E(0;2K)\,$,
$\;$cl$_E(V)\preceq U\,$.
\end{proof}

\begin{lemma}
\label{intermed} Let $E$ be a locally convex space. Suppose $C$ and $U$ are
convex sets such that $\;$\emph{int}$_E(C)\neq\emptyset\,$, $C$ is closed,
$U$ is open, and $\;C\subseteq U\,$. Then there is a closed convex set $V$
such that $\;C\subseteq\,$\emph{int}$_EV\;$ and $\;V\subseteq U\,$.
\end{lemma}

\begin{proof}
\ Let the Minkowski functionals of $\,C,U\,$ be $\,p_C,p_U\,$ respectively.
Take $\;V:=\{x\in E:\frac{1}{2}(p_C(x)+p_U(x))\leq 1\label{hhh}\}\,$.
\end{proof}

\begin{lemma}
\ Le\label{parb4}t $E$ be a normed space; suppose that $\;B\prec A\,$, where
$B$ is bounded, and $A$ is convex and open, in $E$. Then there exists a
bounded convex open set $\,A'\,$ such that $\;B\prec A'\subseteq A\,$.
\end{lemma}

\begin{proof}
\ Let $U$ be a bounded convex neighbourhood of $0$ in $E$ such that
$\;B+U\subseteq A\,$. Then take $\;A'\;$ to be the convex cover of $\;B+U\,$.
\end{proof}

\begin{definition}
\ S\label{complex}uppos\label{parb5}e
$\;\mathcal{A}:=\{A_1,A_2,\ldots,A_u\}\;$ is a finite indexed class
consisting of subsets of a set $X$. (The indices need not in principle be
natural numbers, or such a sequence $\,\{1,2,3,\ldots u\}\,$). By an analogy
which will soon become apparent, I shall call it a \emph{complex }in $X$. The
members of a complex need not be distinct or non-empty. When $X$ is a locally
convex space $E$, a second complex in $E$,
$\;\mathcal{B}:=\{B_1,B_2,\ldots,B_v\}\,$, is described as \emph{strongly
included }in $\mathcal{A}$, which I write as
$\,\mathcal{B}\prec\mathcal{A}\,$, if it is indexed by exactly the same
indices as $\mathcal{A}$ and, for each index $i$, $\;B_i\prec A_i\;$ as
subsets of $E$. In such case it is clear, since the index set is finite, that
there exists some neighbourhood $V$ of the origin in $E$ such that
$\;B_i+V\subseteq A_i\;$ for all indices $i$ simultaneously.

When $X$ is a normed space $E$, I say that the complex $\mathcal{B}$ is
\emph{boundedly strongly included }in the complex $\mathcal{A}$,
$\;\mathcal{B}\preceq\mathcal{A}\,$, if $\mathcal{A}$ and $\mathcal{B}$ have
the same index sets and, for each index $i$, $\;B_i\preceq A_i\,$; that is,
for each bounded set $Q$ of $E$, there is a neighbourhood $V$ of $0$ in $E$
such that, for each index $i$, $\;(B_i\cap Q)+V\subseteq A_i\,$. 
\end{definition}


\begin{definition}
\ Let \label{simplex}$\;\mathcal{A}:=\{A_1,A_2,\ldots,A_v\}\;$ be a complex
in the set $X$. For any $\;\Delta\subseteq\{1,2,3$,$\ldots,v\}\,$, let
$\;A_\Delta:=\bigcap_{u\in\Delta}A_u\,$. Then I call $\Delta$ a
\emph{simplex} of $\mathcal{A}$ whenever $\;A_\Delta\neq\emptyset\,$. (In
particular, this definition makes $\emptyset$ a simplex). The subsets
$\,A_u\,$, for $\;u\in\Delta\,$, are called the \emph{vertices }of the
simplex $\Delta$. The \emph{star }of a simplex $\Delta$, or of any subset
$\Delta$ of the index set, is $\;A^\Delta:=\bigcup_{u\in\Delta}A_u\;$. The
\emph{dimension }of a simplex of $\mathcal{A}$ is its cardinality diminished
by $1$; the dimension of the complex $\mathcal{A}$ is the largest dimension
of a simplex $\Delta$ of $\mathcal{A}$; a simplex is \emph{maximal }if it is
not a proper subset of any other simplex. The \emph{order }of $\mathcal{A}$
is the number of distinct simplices whose dimension is the dimension of
$\mathcal{A}$. (These names are obviously suggested by an imprecise analogy
with the nerves of coverings). Finally, the \emph{carrier }of $\mathcal{A}$
is the set $\;A:=\bigcup_{u=1}^{\,v}A_u=A^{\{1,2,\ldots,v\}}\,$. Analogous
notations are to apply to other complexes; thus the complex denoted
$\mathcal{B}$ gives $\,B_\Delta$,$\,B^\Delta$,$\,B\,$.
\end{definition}

\begin{remark}
\ I\label{parb7}f $\mathcal{A}$, $\mathcal{B}$ are complexes in a locally
convex space $E$, and $\;\mathcal{B}\prec\mathcal{A}\,$, then
$\;B^\Delta\prec A^\Delta\;$ and $\;B_\Delta\prec A_\Delta\;$ for each subset
$\Delta$ of the index set. Indeed, there is a neighbourhood $V$ of the origin
in $E$ such that $\;B^\Delta+V\subseteq A^\Delta\;$ and
$\;B_\Delta+V\subseteq A_\Delta\;$ for all $\Delta$.
\end{remark}

The main result of this paper will be Theorem\label{qinsert}
\label{hole}\ref{bigthm}, which says approximately that, if a complex
$\mathcal{B}$ is strongly included in a complex $\mathcal{A}$ of convex open
sets in a locally convex Hausdorff space $E$, and if $\mathcal{F}$ is an
admissible class, then, for any $\mathcal{F}$-extensible function $f$ defined
on the carrier of $\mathcal{A}$, there is a function in $\mathcal{F}$ defined
on $E$ and agreeing with $f$ on the carrier of $\mathcal{B}$. The difficulty
with the proof is that the complexes may have a complicated combinatorical
structure, that is, the convex sets may intersect in a way that makes local
extensions interfere with each other. There are therefore two steps; in the
first I consider what happens if the simplest case, that of stars, is already
known.

\ \ 

\section{Extension (stars of simplices excepted). }\label{section3}

\begin{proposition}
\ L\label{extensioncomplex}et $E$ be a locally convex space$,$ and $F$ a
vector space, with $X$ a subset of $E$ and $Y$ a convex subset of $F$. Let
$\mathcal{F}$ be a $Y$-admissible class (see \emph{(\ref{admissible})}) of
functions defined on the open subsets of $X$ with values in $Y$. Suppose that
$\;\mathcal{A}:=\{A_u:\,1\leq u\leq v\}\;$ is a complex consisting of open
convex sets of $E$, and $\mathcal{B}$ is a complex of subsets of $E$, such
that $\;\mathcal{B}\prec\mathcal{A}\,$. Then, for any map $\;f:B\cap
X\longrightarrow Y\,$, and any family of maps $\;f_\Delta:X\longrightarrow
Y\;$ in $\mathcal{F}$, one for each simplex $\Delta$ of $\mathcal{A}$, such
that, for each $\Delta$, $\;f_\Delta\mid B^\Delta\cap X=f\mid B^\Delta\cap
X\,$, there exists a map $\;h:X\longrightarrow Y\;$ in $\mathcal{F}$ such
that $\;h\mid B\cap X=f\mid B\cap X$.
\end{proposition}

\begin{remark}
\ Ever\label{parc2}y simplex of $\mathcal{B}$ is a simplex of $\mathcal{A}$,
but the converse is not true; that is, there may be many simplices $\Delta$
of $\mathcal{A}$ ($\,A_\Delta\neq\emptyset\,$) for which
$\;B_\Delta=\emptyset\,$. It is also quite possible that $\;B^\Delta\cap
X=\emptyset\,$, in which case the condition on $\,f_\Delta\,$ would be
vacuously satisfied. The reason for introducing $X$ is \ (see
\S\ref{section1}) that there are some interesting classes of functions which
can only be reasonably defined on ``thin'' sets. See (\ref{remarksonproof})
for further comments.
\end{remark}

\begin{proof}
\ The argument is inductive, and will be presented in steps.\\
\ 

\emph{(a})\quad\quad If the dimension of $\;\mathcal{A}\;$ is $\;-1\,$, all
its members and all the members of $\mathcal{B}$ are null; and, whenever
$\,B=\emptyset\,$, the Proposition holds trivially because, by
(\ref{admissible}), there is \emph{some }map $\;h:X\longrightarrow Y\;$ in
$\mathcal{F}$. \\
\ 

\emph{(b})\quad\quad If the dimension of $\;\mathcal{A}\;$ is greater than
$\,-1\,$, there are some 0-simplices in $\mathcal{A}$, that is, indices $u$
for which $\;A_u\neq\emptyset\,$. Clearly any other indices may simply be
ignored, so I shall assume henceforth that $\,A_u\,$ is non-null for each
$u$. Of course, some of the $\,B_u\,$ may be null.\\
\ 

\emph{(c})\quad\quad If the dimension of $\mathcal{A}$ is $\,v-1\,$, every
0-simplex is a vertex of a unique simplex $\Delta$ of maximal dimension$;$
the order of $\mathcal{A}$ must be $1$, and $\;A=A^\Delta\,$. The proposition
is trivially true, for it suffices to take $\;h:=f_\Delta\,.\,$In particular,
this applies if there is a single vertex ($\,v=1,\,d=0,\,o=1\,$).\\
\ 

\emph{(d})\quad\quad Suppose now that the dimension of $\mathcal{A}$ is
$\,d\geq 0\,$ and the order of $\mathcal{A}$ is $o$. The inductive hypothesis
will be that the Proposition has already been proved for any $\mathcal{A}$
and $\mathcal{B}\;$ satisfying the hypotheses, provided that $\mathcal{A}$
has dimension less than $d$, or, when $\,o\geq 2\,$, dimension equal to $d$
and order less than $o$. The cases $\,d=-1\,$ and $\,d=0,\,o=1\,$ have been
proved at \emph{(a)} and \emph{(c)}.\\
\ 

\emph{(e})\quad\quad\emph{(c)} also applies when $\;d+1=v\,$. So now, suppose
$\;d+1<v\,$. Renumbering, if necessary, I may suppose that 
\begin{equation*}
Q:=A_1\cap A_2\cap\ldots\cap A_{d+1}\neq\emptyset\,
\end{equation*}
is a simplex of the maximum dimension $d$. For each $u$ with $\;d+1<u\leq
v\,$, $\;Q\cap A_u=\emptyset\;$ (or there would be a simplex of dimension
$\,d+1\,$). As $\;Q\;$ and $\;A_u\;$ are disjoint open nonempty convex sets,
one has for each such $u$ a continuous nonzero linear functional
$\,\lambda_u\,$ on $E$, and a scalar $\,\rho_u\,$,$\,$such that$\,$
\begin{equation}\label{lambdas}
\lambda_u(Q)\subseteq\{\tau\in\,\mathbb{R}:\tau<\rho_u\}\,\text{,}\quad%
\lambda_u(A_u)\subseteq\{\tau\in\mathbb{R}:\tau>\rho_u\}\,.
\end{equation}
The functionals $\;\lambda_{d+2},\lambda_{d+3},\ldots,\lambda_v\in E'\;$ need
not be linearly independent. Choose from them a maximal linearly independent
subset, which, by renumbering, may be denoted as
$\,\{\lambda_{v-k+1},\ldots,\lambda_v\}\,$. Define 
\begin{equation}\label{projectionp}
P:E\longrightarrow\mathbb{R}^k:x\longrightarrow(\lambda_{v-k+1}(x),\ldots,%
\lambda_v(x))\,.
\end{equation}
Then $P$ is a continuous epimorphism. Each of the functionals
$\,\lambda_u,\lambda_{u+1},\ldots,\lambda_v\,$ for $\;d+1<u\leq v\;$ is of
the form $\,\mu_u\circ P\,$ for some suitable nonzero linear functional
$\,\mu_u\,$ on $\,\mathbb{R}^k\,$. Hence, from (\ref{lambdas}), 
\begin{equation}\label{separation}
\left\{
\begin{aligned}
P(Q)&\subseteq\{\xi\in\mathbb{R}^k:\,(\forall u)\;d+1<u\leq
v\;\Rightarrow\text{ }\mu_u(\xi)<\rho_u\}\,\text{,}\\
P(A_u)&\subseteq\{\xi\in\mathbb{R}^k:\mu_u(\xi)>\rho_u\}\quad\text{for
}\;d+1<u\leq v\,.
\end{aligned}
\right.
\end{equation}

Now $\,B_u\prec A_u\,$ for each $u$; there exists a convex neighbourhood $V$
of $0$ in $E$ such that, for $\,1\leq u\leq v\,$, $\;B_u+3V\subseteq A_u\,$.
Define for each $u$ 
\begin{equation}\label{primedinclusion}
\left\{
\begin{aligned}
A'_u&:=\{x\in E:\,x+V\prec A_u\}\,\text{,}\quad\text{and}\\
Q'&:=A'_1\cap A'_2\cap\ldots\cap A'_{d+1}\,\text{;}\quad\text{thus}\\
&\;\;B_u+V\subseteq A'_u\,\text{,}\quad A'_u+V\subseteq A_u\,\text{,}
\end{aligned}
\right.
\end{equation}
so that $\;B_u\prec A'_u\prec A_u\,$, and $\,A'_u\,$ is an open convex set.
Also $\;Q'+V\subseteq Q\,$. It is of course possible that $\,Q'=\emptyset\,$.
Since the functionals $\,\lambda_u=\mu_u\circ P\,$ are nonzero, there exists
$\,\epsilon>0\,$ such that, for every $u$,
$\;\mu_u(P(V))\supseteq(-2\epsilon,2\epsilon)\,$. \ Consequently,
(\ref{separation}) gives 
\begin{equation}\label{primedseparation}
\left\{
\begin{aligned}
P(Q')&\subseteq\{\xi\in\mathbb{R}^k:\,(\forall u)\;d+1<u\leq
v\;\Rightarrow\;\mu_u(\xi)<\rho_u-2\epsilon\}\,\text{,}\\
P(A'_u)&\subseteq\{\xi\in\mathbb{R}^k:\mu_u(\xi)>\rho_u+2\epsilon\}\quad%
\text{for }\;d+1<u\leq v\,.
\end{aligned}
\right.
\end{equation}

For each subset $T$ of $\,\{d+2,d+3,\ldots,v\}\,$, set $\;$ 
\begin{equation}\label{uandc}
\left\{
\begin{aligned}
T^c&:=\{d+2,d+3,\ldots,v\}\setminus T\quad\text{and}\\
U(T)&:=\{\xi\in\mathbb{R}^k:(\forall r\in
T)(\mu_r(\xi)>\rho_r-2\epsilon)\;\;\\
&\qquad\qquad\qquad\qquad\qquad\text{\&}\;\;(\forall r\in
T^c)(\mu_r(\xi)<\rho_r+2\epsilon)\}\,\text{,}\\
C(T)&:=\{\xi\in\mathbb{R}^k:(\forall r\in
T)(\mu_r(\xi)\geq\rho_r-\epsilon)\;\;\\
&\qquad\qquad\qquad\qquad\qquad\text{\&}\;\;(\forall r\in
T^c)(\mu_r(\xi)\leq\rho_r+\epsilon)\}\,.
\end{aligned}
\right.
\end{equation}
The sets $\,C(T)\,$ are closed in $\,\mathbb{R}^k\,$; $\,U(T)\,$ is open and
convex, $\;C(T)\prec U(T)\,$. Define, for each $T$ and for $\;1\leq u\leq
v\,$, 
\begin{equation}\label{cutdownab}
A_u(T):=A'_u\cap P^{-1}(U(T))\,\text{,}\qquad B_u(T):=B_u\cap P^{-1}(C(T))\,.
\end{equation}
Thus the sets $\;A_u(T)\;$ are open and convex in $E$.

There is a convex neighbourhood $W$ of $0$ in $E$ such that $\,W\subseteq
V\,$ and, for each $u$, $\;\lambda_u(W)\subseteq(-\epsilon,\epsilon)\,$. If
$\;x\in B_u(T)\,$, then (see (\ref{primedinclusion})) $\;x+V\subseteq
A'_u\,$. Also, for all $\;r>d+1\,$, $\;P(x)\in C(T)\;$ and 
\begin{equation*}
\lambda_r(x+W)\subseteq(\lambda_r(x)-\epsilon,\lambda_r(x)+\epsilon)\,\text{;}
\end{equation*}
for $\,r\in T\,$, therefore, (\ref{uandc}) shows $\;(\mu_r\circ
P)(x+W)\subseteq(\rho_r-2\epsilon,\infty)\,$, whilst, on the other hand, for
$\,r\in T^c\,$, $\;(\mu_r\circ P)(x+W)\subseteq(-\infty,\rho_r+2\epsilon)\,$.
So $\;P(B_u(T)+W)\subseteq U(T)\,$; hence, $\;B_u(T)+W\subseteq A_u(T)\,$,
and 
\begin{equation*}
\mathcal{B}(T):=\{B_1(T),\ldots,B_v(T)\}\prec\mathcal{A}(T):=\{A_1(T),\ldots,%
A_v(T)\}\,.
\end{equation*}
Each set of $\,\mathcal{A}(T)\,$ or of $\,\mathcal{B}(T)\,$ is a subset of
the corresponding set of $\mathcal{A}$ or of $\mathcal{B}$, and each simplex
of $\;\mathcal{A}(T)\;$ is a simplex of $\mathcal{A}$. Consequently, the
hypotheses of the Proposition are satisfied for the pair
$\;\mathcal{A}(T),\mathcal{B}(T)\,$, if one takes instead of $f$ its
restriction to $\,B(T)\cap X\,$ and the same $\,f_\Delta\,$ as before for
every simplex $\Delta$ of $\;\mathcal{A}(T)\,$.\\
\ 

\emph{(f})\quad\quad From (\ref{cutdownab}) and (\ref{primedseparation}), one
sees that, for $\;d+1<u\leq v\,$, 
\begin{equation*}
\lambda_u(A_u(\emptyset))\subseteq\lambda_u(A'_u)\subseteq(\rho_u+2\epsilon,%
\infty)\,.
\end{equation*}
But (\ref{uandc}) and (\ref{cutdownab}) ensure that
$\;\lambda_u(A_u(\emptyset))\subseteq\mu_u(U(\emptyset))\subseteq(-\infty,%
\rho_u+2\epsilon)\;$ for $\;d+1<u\leq v\,$. So $\;A_u(\emptyset)=\emptyset\;$
for $\;d+1<u\leq v\,$, and the only 0-simplices (non-null members) of
$\;\mathcal{A}(\emptyset)\;$ must be among
$\;A_1(\emptyset),A_2(\emptyset),\ldots,A_{d+1}(\emptyset)\,$. Therefore
either the dimension of $\,\mathcal{A}(\emptyset)\,$ is $d$, and its order is
1, or its dimension is less than $d$. The first case was discussed at
\emph{(c)}. In the second case, the inductive hypothesis \emph{(d)} applies
to the pair $\,\mathcal{A}(\emptyset),\mathcal{B}(\emptyset)\,$. In both
cases, there is a function $\;h_\emptyset:X\longrightarrow Y\;$ such that
$\;h_\emptyset\mid B(\emptyset)\cap X=f\mid B(\emptyset)\cap X\,$.\\
\ 

\emph{(g})\quad\quad If $\;T\neq\emptyset\,$, let $\;r\in T\:$. Then, for all
$u$, $\;1\leq u\leq v\,$, (\ref{uandc}) and (\ref{cutdownab}) entail
\begin{equation}\label{wzor10}
\lambda_r(A_u(T))\subseteq\mu_r(U(T))\subseteq(\rho_r-2\epsilon,\infty)\,.
\end{equation}
On the other hand, (\ref{cutdownab}) and (\ref{primedseparation}) show that,
for $\;d+1<r\leq v\,$, \ 
\begin{equation}\label{wzor11}
\lambda_r\left(\bigcap_{u=1}^{d+1}A_u(T)\right)\subseteq\lambda_r\left(%
\bigcap_{u=1}^{d+1}A_u'\right)=\lambda_r(Q')\subseteq(-\infty,\rho_r-2%
\epsilon)\,.
\end{equation}
By (\ref{wzor10}) and (\ref{wzor11}), necessarily $\;A_1(T)\cap
A_2(T)\cap\ldots\cap A_{d+1}(T)=\emptyset\,$. Now every simplex of
$\;\mathcal{A}(T)\;$ is a simplex of $\mathcal{A}$, and at least one
$d$-simplex of $\mathcal{A}$ is not a simplex of $\;\mathcal{A}(T)\,$; hence
either the dimension of $\;\mathcal{A}(T)\;$ is less than $d$ or its order is
less than $o$. In either case, the inductive hypothesis \emph{(d)} applies to
the pair $\,\mathcal{A}(T),\mathcal{B}(T)\,$.\\
\ 

\emph{(h})\quad\quad For all choices of $T$, then, there is a function
$\;h_T:X\longrightarrow Y\;$ such that $\;h_T\mid B(T)\cap X=f\mid B(T)\cap
X\,$. Note from (\ref{cutdownab}) that$\,$ $\;$
\begin{equation*}
B(T)=B\cap P^{-1}(C(T))\,.
\end{equation*}

From (\ref{uandc}), any $\;\xi\in\mathbb{R}^k\;$ lies in the interior of
$\;C(T_\xi)\;$, if $\,T_\xi\,$ denotes the set of indices
$\,r\in\{d+2,d+3,\ldots,v\}\,$ for which $\,\mu_r(\xi)\geq\rho_r\,$. So the
interiors of the sets $\,C(T)\,$ cover $\,\mathbb{R}^k\,$. There is a
subordinate C$^\infty$ partition of unity $\,\{\psi_T\}\,$ on
$\,\mathbb{R}^k\,$, where $\;C(T)\;$ includes the support of $\,\psi_T\,$ for
each $T$. Certainly $\;\psi_T\circ P\;$ is finitely determined (see
(\ref{finitely:determined})). Thus $\;h:X\longrightarrow Y\,$, defined by 
\begin{equation}\label{partofun}
h(x):=\sum_T\;\psi_T(P(x))h_T(x)\,\text{,}
\end{equation}
where the sum extends over all $\,2^{v-d-1}\,$ choices of
$\;T\subseteq\{d+2,d+3,\ldots,v\}\,$, is in $\mathcal{F}$ by
(\ref{admissible}), and it takes values in $Y$ because $Y$ is convex. Suppose
$\;x\in B\cap X\,$. For any $T$ such that $\;\psi_T(P(x))\neq 0\,$, then
$\;P(x)\in C(T)\;$ and, from (\ref{cutdownab}), $\,x\in B(T)\cap X\,$. Ergo,
$\;h_T(x)=f(x)\;$ by the construction. So $\;h\mid B\cap X=f\mid B\cap X\;$
as required. This completes the inductive step, in the only case
($\,d+1<v\,$) for which it was non-trivial. The proof is therefore finished.
\end{proof}

\begin{remark}
\ In t\label{remarksonproof}\label{parc3}he argument above, there is no need
to assume any countability hypothesis for $E$, or to demand that the class
$\mathcal{F}$ is locally defined, because the construction of $h$ terminates
after finitely many steps. This is no longer true in \S 4. Also, notice that
the only extensions of $f$ that are needed are explicitly demanded in the
conditions of the theorem, and, consequently, that $f$ is automatically
$\mathcal{F}$-extensible. 

As is evident in the proof, the r\^ole of the complex $\mathcal{A}$ in the
statement of the theorem is merely to furnish a convenient class of simplices
to which the conditions that have to be imposed may apply; it does not appear
in the conclusion. It would be more natural to begin with the complex
consisting of the convex envelopes $\,\Gamma(B)\,$ of the elements $B$ of
$\mathcal{B}$, but, no matter how small \ a neighbourhood $V$ of $0$ were
taken, the complex $\;\{\Gamma(B)+V:B\in\mathcal{B}\}\;$ might have simplices
additional to those of $\mathcal{B}$.

The underlying idea is that linear functionals and smooth combinations of
them give a large enough class of C$^\infty$ functions for some purposes.
Specifically, the Hahn-Banach theorem enables one to use them to separate
combinations of open convex sets. It might\label{question1} theoretically be
possible to argue similarly on the basis of other separation theorems, which
might be a way to improve the results and even to extend them to
non-separable spaces; but I do not know of any such theorems. For instance,
can two disjoint closed polynomially-convex sets always be separated by a
smooth function? As things are, the requirements that $\mathcal{A}$ be finite
and consist of convex sets force the later discussion to be restricted to
spaces with some countability restriction; if, for instance, $\mathcal{A}$
could be locally finite but uncountable, more might be proved.

There are several possible addenda to the Proposition, two of which may be
given here. It is convenient to establish the notation $\,\mathcal{F}(M;N)\,$
to mean the set of functions $\,M\longrightarrow N\,$ that belong to
$\mathcal{F}$. Any algebraic operations in $\,\mathcal{F}(M;N)\,$ are
understood to be defined pointwise when they are defined. The proof of
(\ref{extensioncomplex}) yields, without serious alteration:
\end{remark}

\begin{corollary}
\ Let $\,\label{complaff}\mathcal{A},\mathcal{B}\,$ and
$\,\mathcal{F},E,F,X\,\,$be as in \emph{(\ref{extensioncomplex})}. Suppose
that $\mathcal{E}$ is a convex set of functions $\;A\cap X\longrightarrow
F\;$ and that, for each simplex $\Delta$ of $\mathcal{A}$, there is an
affine-linear mapping
$\;\mathcal{L}_\Delta:\mathcal{E}\longrightarrow\mathcal{F}(X;F)\;$ such that
$\;\mathcal{L}_\Delta(f)\mid B^\Delta\cap X=f\mid B^\Delta\cap X\;$ for each
$\,f\in\mathcal{E}\,$. Then there is an affine-linear map
$\;\mathcal{L}:\mathcal{E}\longrightarrow\mathcal{F}(X;F)\;$ such that, for
each $\;$f$\in\mathcal{E}\,$, $\;\mathcal{L}(f)\mid B\cap X=f\mid B\cap X\,$.
Furthermore, the image of $\,\mathcal{L}(f)\,$ is included in the convex
envelope of all the images of the mappings
$\,\mathcal{L}_\Delta(f)\,$.\hfill{\qedsymbol}
\end{corollary}

\begin{remark}
\ The \label{parc5}Corollary arises from the observation that all the
constructions in the proof of (\ref{extensioncomplex}) --- the linear
functionals $\,\lambda_u\,$, the pairs $\,\mathcal{A}(T),\mathcal{B}(T)\,$,
and the partition of unity $\,\psi_T\,$ --- are made to accord only with the
geometry of $\mathcal{A}$ and $\mathcal{B}$, and do not depend on the
functions $\,f,f_\Delta\,$. 
\end{remark}

\begin{definition}
\ Suppos\label{parc6}e that $G$ is a subspace of the topological dual
$\,E'\,$ of the locally convex Hausdorff space $E$. A linear surjection
$\;P:E\longrightarrow\mathbb{R}^k\;$ will be called a $G$-\emph{surjection}
if its composition with each of the coordinate functions of
$\,\mathbb{R}^k\,$ is a linear functional belonging to $G$. (Its composition
with any linear functional on $\,\mathbb{R}^k\,$ then also belongs to $G$). A
function $\;f:U\longrightarrow\mathbb{R}\;$ defined on an open set $U$ of $E$
may be described as $G$-\emph{finitely determined} if there exist a
$G$-surjection $\;P:E\longrightarrow\mathbb{R}^k\;$ and a function
$\;\phi:P(U)\longrightarrow\mathbb{R}\;$ such that $\,f=\phi\circ P\,$. If
$X$ is a subset of $E$ and $Y$ is a convex set in the vector space $F$, a
class $\mathcal{F}$ of functions $\;X\longrightarrow Y\;$ is
$\,(G,Y)$-admissible if, whenever $\;f,g:X\longrightarrow Y\;$ are mappings
in $\mathcal{F}$ and $\;\psi:W\longrightarrow[0,1]\;$ is a $G$-finitely
determined C$^\infty$ function on an open subset $W$ of $E$ which includes
$X$, then $\;h:X\longrightarrow Y\;$ is also in $\mathcal{F}$, where
\begin{equation*}
(\forall x\in X)\quad h(x)=\psi(x)f(x)+(1-\psi(x))g(x)\,.
\end{equation*}
This is of course a less demanding condition, in principle, than the
$Y$-admissibility of (\ref{admissible}).
\end{definition}

\begin{lemma}
\label{limp} Suppose in \emph{(\ref{extensioncomplex})} or in
\emph{(\ref{complaff})} that $E$ is a normed space and all the convex sets
belonging to the complex $\mathcal{A}$ are bounded. Let $G$ be a norm-dense
subspace of the dual space $\,E'\,$. Then the previous conclusions, in
\emph{(\ref{extensioncomplex})} or in \emph{(\ref{complaff}), }hold on the
weaker condition that $\mathcal{F}$ is $\,(G,Y)$-admissible instead of
$Y$-admissible.
\end{lemma}

\begin{proof}
\ Suppose that all the sets of $\mathcal{A}$ lie in $\,B_E(0;L)\,$. Define
the functionals $\,\lambda_u\,$ as at (\ref{lambdas}), and the sets
$\,A_u',Q'\,$ as at (\ref{primedinclusion}). Then there exists an
$\,\epsilon>0\,$ such that, for each $u$ such that $\;d+1<u\leq v\,$,
$\;\sup\lambda_u(Q')+6\epsilon<\inf\lambda_u(A_u')\,$. If $\,\lambda_u'\in
E'\,$ and $\;\left\|\lambda_u-\lambda_u'\right\|<\epsilon/L\,$, then
$\;\sup\lambda_u'(Q')<\sup\lambda_u(Q')+\epsilon\;$ and
$\;\inf\lambda_u'(Q')>\sup\lambda_u'(Q')-\epsilon\;$ for each $u$,
$\;d+1<u\leq v\,$. Hence, one may choose elements $\,\lambda_u'\,$ of $G$
such that $\;\sup\lambda_u'(Q')+4\epsilon<\inf\lambda_u'(A_u')\,$. Select a
maximal linearly independent subset (for infinite-dimensional $E$, the
$\,\lambda_u'\,$ might already have been chosen to be linearly independent),
and proceed as before; (\ref{primedseparation}) becomes a consequence of the
new construction.
\end{proof}

\begin{remark}
\ Wha\label{remonsep}t was needed at (\ref{extensioncomplex})\emph{(e)} was a
condition on two disjoint convex sets $\,Q,R\,$ that would ensure the
existence of $\;\lambda\in E'\;$ such that the intervals
$\,\lambda(Q),\lambda(R)\,$ of $\mathbb{R}$ are ``separated'' --- in effect,
have disjoint closures. The condition used was that $\;(Q+V)\cap
R=\emptyset\;$ for some neighbourhood of $0$ in $E$, which one might call
\emph{convex separation}; then (\ref{limp}) says that, if $Q$ and $R$ are
non-null separated convex sets and bounded in the normed space $E$, the set
of possible $\lambda$ is norm-open in $\,E'\,$. If $Q$ is unbounded, there is
often only one possible $\lambda$, whilst, if $Q$ and $R$ are disjoint
non-null, convex, bounded and closed, but not separated, the set of
separating linear functionals is not norm-open. 
\end{remark}

\ \ \ \ 

\section{Extension from stars of simplices.}\label{section4}

The question of existence of the extension $h$ is reduced by
(\ref{extensioncomplex}) to the existence of the functions $\,f_\Delta\,$,
the ``extensions from stars'' of $f$. From a logical point of view, it is
slightly curious that I now use (\ref{extensioncomplex}) itself to construct
extensions such as $\,f_\Delta\,$. The argument is made possible by
(\ref{ext:on:segments}), which furnishes extensions in special cases. The
construction is by an \emph{infinite} induction; it assumes $\mathcal{F}$ is
locally defined, and $E$ is to be Lindel\"of (\cite{engelking}, 3.8), so
that, at each stage, only finitely many sets, all convex, need be considered.
``Countable'' arguments of a similar kind can be carried out for some
purposes in spaces that are merely paracompact, by virtue of the theorem of
Stone that open coverings have $\sigma$-discrete refinements
(\cite{engelking}, 5.1.12), but such refinements must be obtained by
modifying the geometrical shape of the original sets of the cover. One cannot
assume here that the sets of the refinement would still be convex. Moreover,
at each stage one may also have to deal with infinitely many `predecessors',
namely the individual sets of the preceding discrete classes. 

For remarks about the details of the proof below, see (\ref{pardrem}).

\begin{remark}
\ Let \label{strongstar}$\,\mathcal{U},\mathcal{V}\,$ be open coverings of a
topological space $S$. $\mathcal{V}$ is a \emph{strong star-refinement} of
$\mathcal{U}$ if, whenever $\;U\in\mathcal{V}\,$, there exists some member
$T$ of $\mathcal{U}$ such that, whenever $\;V,W\in\mathcal{V}\;$ and $\;U\cap
V\neq\emptyset\neq V\cap W\,$, necessarily $\;U\cup V\cup W\subseteq T\,$.

This is not quite the same as a ``star-refinement'' as usually defined (see
\cite{engelking}, 5.1.11, p. 376, and the discussion afterwards), but it
remains true that any open covering of a paracompact space has a strong
star-refinement; indeed, it suffices to take a star-refinement of a
star-refinement.
\end{remark}

\begin{proposition}
\ Let $E\label{clag}$ be\label{extensionfromstars} a Hausdorff locally convex
Lindel\"of space$,$ and $F$ a vector space, with $X$ a subset of $E$ and $Y$
a convex subset of $F$. Let $\mathcal{F}$ be a $Y$-admissible (see
\emph{(\ref{admissible})}) and locally defined (see\emph{
(\ref{locally:defined})}) class of functions on the open subsets of $X$ in
the subspace topology, with values in $Y$. Suppose that
$\;\mathcal{A}:=\{A_u:\,1\leq u\leq v\}\;$ is a complex consisting of open
convex sets of $E$, and $\mathcal{B}$ is a complex of subsets of $E$, such
that $\;\mathcal{B}\prec\mathcal{A}\;$ and
$\;\bigcap_{u=1}^vA_u\neq\emptyset\,$. Write $\;A:=\bigcup_{u=1}^vA_u\;$ and
$\;B:=\bigcup_{u=1}^vB_u\;$ (the carriers of $\mathcal{A}$ and
$\mathcal{B}$). If a function$\;f:A\cap$X$\longrightarrow Y\;$ is
$\mathcal{F}$-extensible (see \emph{(\ref{para6})}), then there exists a
function $\;g:$X$\longrightarrow Y\;$ in $\mathcal{F}$ such that $\;g\mid
B\cap$X$=f\mid B\cap$X$\,$.
\end{proposition}

\begin{proof}
\ \emph{(a)} \quad Since $\mathcal{F}$ itself may if necessary be translated,
there is no loss of generality in supposing $\;0\in\bigcap_{u=1}^vA_u\,$.
Thus $\;N:=\bigcap_{u=1}^vA_u\;$ is a convex open neighbourhood of $0$ in
$E\,$. 

As $\;\mathcal{B}\prec\mathcal{A}\,$, there is a convex neighbourhood $V$ of
$0$ in $E$ such that $\;B_u+V\subseteq A_u\;$ for each $u$, and,
consequently, \ $\;$
\begin{equation*}
(B_u\cup\{0\})+(V\cap N)\subseteq
A_u\,\text{,}\quad\text{co}(B_u\cup\{0\})+(V\cap N))\subseteq A_u\,.
\end{equation*}
\ Set 
\begin{equation*}
B_u'':=\text{co}(B_u\cup\{0\})+\textstyle\frac{1}{2}(V\cap N)
\end{equation*}
($\,$``co''$\,$ denotes the convex cover); this is a convex open
neighbourhood of $0$ in $E$. Let $\;B_u':=\:$cl$_E(B_u'')\subseteq
B_u''+\frac{1}{6}(V\cap N)\,$, $\;B_u^\ast:=B_u'+\frac{1}{6}(V\cap N)\,$;
thus 
\begin{equation*}
B_u^\ast+\textstyle\frac{1}{6}(V\cap N))\subseteq A_u\cap(2B_u'')\,.
\end{equation*}
By taking $\,A_u\cap(2B_u'')\,$ instead of $\,A_u\,$, $\,B_u'\,$ instead of
$\,B_u\,$, and $V$ instead of $\;\frac{1}{6}(V\cap N)\,$, I may therefore
assume that, for each $u$, \ 
\begin{equation}\label{assumptions}
\left\{%
\begin{tabular}{l}
$\,B_u\,$ is a closed convex neighbourhood of $0$;\\
$\,B_u^\ast\,$ is another, such that $\;B_u+V\subseteq B_u^\ast\,$,
$\;B_u^\ast+V\subseteq A_u\,$;\\
and $\;A_u\subseteq 2($int$_E(B_u))\,$.
\end{tabular}%
\right.
\end{equation}
In particular, each $\,A_u\,$ and $\,B_u\,$, and the unions
$\;A:=\bigcup_{u=1}^vA_u\;$ and $\;B:=\bigcup_{u=1}^vB_u\,$, are all
star-shaped about $0$ (a fact constantly employed below), and
$\;\frac{1}{2}A\subseteq\,$int$_E(B)\,$. \\
\ 

\emph{(b)}\quad\quad There are two classes of points in $E$. The first class,
which I shall call $\mathfrak{X}$, consists of those points $x$ (including
$0$) such that, for all $\;t\in(0,\infty)\,$, $\;tx\in A\,$. The second,
$\mathfrak{Y}$, consists of those $\;y\in E\setminus\{0\}\;$ for which
$\;ty\notin A\;$ for some $\;t\in(0,\infty)\,$. $\,$ 

Since $A$ is a finite union of the sets $\,A_u\,$,
$\;\mathfrak{X}=\bigcup_{u=1}^v\mathfrak{X}_u\,$, where $\,\mathfrak{X}_u\,$
consists of the elements $\;x\in\mathfrak{X}\;$ such that
$\;(0,\infty)x\subseteq A_u\,$. (This is not a disjoint union). If
$\;y\notin\mathfrak{X}_u\,$, there is some $\;t>0\;$ such that $\;ty\notin
A_u\,$; then $\;(t+1)y\notin$cl$_E(A_u)\,$, and $\;(t+1)z\notin$cl$_E(A_u)\;$
for all $z$ in a suitably small neighbourhood of $y$. Thus
$\,\mathfrak{X}_u\,$ is closed in $E$. (It is a convex cone, since $\,A_u\,$
is convex, and, by (\ref{assumptions}),$\,$one also has
$\;(0,\infty)x\subseteq B_u\;$ for any $\;x\in\mathfrak{X}_u\,$). Hence
$\mathfrak{Y}$, the complement of the union of the $\,\mathfrak{X}_u\,$, is
open in $E$. \ \\
\ 

\emph{(c)}\quad\quad If $\;y\in\mathfrak{Y}\;$ and $\;1\leq u\leq v\,$, there
exists $\;\lambda_u(y)>0\;$ such that $\,\lambda_u(y)y\,$ is in the frontier
of $\,B_u^\ast\,$ in $E$; namely, $\;\lambda_u(y)=(p_u(y))^{-1}\,$, where
$\,p_u\,$ is the Minkowski functional of $\,$int$_E(B_u^\ast)\,$, which takes
positive values on $\mathfrak{Y}$ by definition. The function $\,\lambda_u\,$
is continuous on $\mathfrak{Y}$, since $\,p_u\,$ is. 

If $\,\partial B^\ast\,$ denotes the frontier of $\,B^\ast\,$ in $E$, then
$\;\partial B^\ast\subseteq\mathfrak{Y}\,$. In particular,
$\;\mathfrak{Y}=(0,\infty)\partial B^\ast\,$; to any $\;y\in\mathfrak{Y}\,$
there corresponds $\,$a unique point $\;\Pi(y):=\lambda(y)y\;$ of $\,\partial
B^\ast\,$, where $\;\lambda(y)\in(0,\infty)\,$. It is clear that
$\;\lambda(y)=\max_{1\leq u\leq v}\lambda_u(y)\,$, so that $\lambda$ is a
continuous positive-valued function on $\mathfrak{Y}$. Hence

\begin{equation}\label{homeo}
\left\{
\begin{aligned}
&\text{the mapping }\;(0,\infty)\times\partial
B^\ast\longrightarrow\mathfrak{Y}:(\alpha,x)\mapsto\alpha x\;\\
&\text{is}\,\text{a homeomorphism when}\,\,\partial B^\ast\,\text{ is given
the subspace}\\
&\text{topology; its inverse is }\;y\mapsto(\lambda(y)^{-1},\lambda(y)y)\,.
\end{aligned}
\right.
\end{equation}
\\
\ 

\emph{(d)}\quad\quad For each $\;y\in\partial B^\ast\,$, the segment
$\,[\frac{1}{2}y,y]\,$ lies in $\,A\cap\mathfrak{Y}\,$. There exists some
$\;\epsilon_y\in(0,\frac{1}{6})\;$ such that
$\;[(\frac{1}{2}-\epsilon_y)y,(1+\epsilon_y)y]\subseteq A\,$. Apply
(\ref{ext:on:segments}): there are a convex open neighbourhood $\,W_y\,$ of
$0$ in $E$ and a function $\;k_y:X\longrightarrow Y\;$ in $\mathcal{F}$ such
that 
\begin{equation}\label{inclext}
\left\{
\begin{aligned}
&\qquad\quad[\textstyle(\frac{1}{2}-\epsilon_y)y,(1+\epsilon_y)y]+2W_y%
\subseteq A\qquad\text{and}\\
&k_y\mid([\textstyle(\frac{1}{2}-\epsilon_y)y,(1+\epsilon_y)y]+W_y)\cap
B^\ast\cap X\\
\textstyle&\qquad\qquad=f\mid([(\textstyle\frac{1}{2}-\epsilon_y)y,(1+%
\epsilon_y)y]+W_y)\cap B^\ast\cap X\,.
\end{aligned}
\right.\;
\end{equation}
I may also suppose that $\;(y+2W_y)\cap B=\emptyset\,$, which necessarily
implies that 
\begin{equation}\label{down}
([1,\infty)(y+2W_y))\cap B=\emptyset\,.
\end{equation}
By (\ref{homeo}), and by compactness in the product space, there exists a
\emph{relatively }open set $\,W_y^\ast\,$ in $\,\partial B^\ast\,$ which
satisfies 
\begin{equation}\label{incl}
\left\{
\begin{aligned}
W_y^\ast&\subseteq y+W_y\qquad\text{and}\\
\textstyle(\frac{1}{2}-\epsilon_y,1+\epsilon_y)W_y^\ast&\subseteq[(\textstyle%
\frac{1}{2}-\epsilon_y)y,(1+\epsilon_y)y]+W_y\,
\end{aligned}
\right.
\end{equation}
(which is open in $\mathfrak{Y}$). The sets $\,W_y\,$ and $\,W_y^\ast\,$ are
to be chosen for each $\;y\in\partial B^\ast\,$.

Now $\,\partial B^\ast\,$ is closed in $E$, which is Lindel\"of; thus
$\,\partial B^\ast\,$ is Lindel\"of, and so paracompact. The \emph{relatively
}open covering $\,\{W_y^\ast:y\in\partial B^\ast\}\,$ of $\,\partial
B^\ast\,$ has a \emph{relatively} open strong star-refinement
$\,\{W_\gamma'':\gamma\in\Gamma\}\,$ (see (\ref{strongstar}) above). 

Given $\;z\in\partial B^\ast\,$, take $\;\gamma\in\Gamma\;$ such that $\;z\in
W_\gamma''\,$. By construction, $\,W_\gamma''\,$ is included in some
$\,W_y^\ast\,$. By (\ref{homeo}),
$\;(\frac{1}{2}-\epsilon_y,1+\epsilon_y)W_\gamma''\;$ is an open
neighbourhood of $z$ in $\mathfrak{Y}$. So there is an open convex
neighbourhood $\,N_z\,$ of $0$ in $E$ such that $\;3N_z\subseteq B\;$ and 
\begin{equation}\label{hangdog}
\textstyle[\frac{1}{2},1](z+3N_z)\subseteq(\frac{1}{2}-\epsilon_y,1+%
\epsilon_y)W_\gamma''\;
\end{equation}
(notice that $\,[\frac{1}{2},1](z+3N_z)\,$ is open in $E$). Next, set 
\begin{equation}\label{defds}
\textstyle D(z):=[\frac{1}{2},\infty)(z+N_z)\text{ ,}\quad
D^\ast(z):=[\frac{1}{2},\infty)(z+3N_z)\,.
\end{equation}
Notice that $\;\,$cl$_E(D(z))\subseteq D(z)+N_z\subseteq D^\ast(z)\,$.

Finally, as $\,\partial B^\ast\,$ is Lindel\"of, its open cover
$\,\{(z+N_z)\cap\partial B^\ast:z\in\partial B^\ast\}\,$ has a countable
subcover $\,\{z_n+N_{z_n}:n=1,2,3\ldots\}\,$. Then
$\,\{D(z_n):n\in\mathbb{N}\}\,$ is an open cover of
$\;[\frac{1}{2},\infty)\partial B^\ast\,$ by convex sets, and 
\begin{equation}\label{strongincl}
D_n:=D(z_n)\prec D_n^\ast:=D^\ast(z_n)\qquad\text{for each }\;n\geq 1\,.
\end{equation}

If $\;i,j,k\in\mathbb{N}\;$ and $\;D_i^\ast\cap D_j^\ast\neq\emptyset\neq
D_j^\ast\cap D_k^\ast\,$, the construction above gives corresponding
$\;\gamma_i,\gamma_j,\gamma_k\in\Gamma\;$ such that $\;W_{\gamma_i}''\cap
W_{\gamma_j}''\neq\emptyset\neq W_{\gamma_j}''\cap W_{\gamma_k}''\,$; the
radial projection on $\,\partial B^\ast\,$ of a common point of the $D^\ast$s
is a common point of the $W''$s. Then, by construction, there is some
$\;y(i)\in\partial B^\ast\;$ with $\;W_{\gamma_i}''\cup W_{\gamma_j}''\cup
W_{\gamma_k}''\subseteq W_{y(i)}^\ast\,$. \ Let 
\begin{equation}\label{zdef}
\begin{aligned}[t]
&Z_i:=(\textstyle\frac{1}{2}-\epsilon_{y(i)},\infty)(y(i)+W_{y(i)})\,\text{;}
\end{aligned}
\end{equation}
\ then, by (\ref{incl}) and (\ref{hangdog}), $\;$
\begin{equation}\label{diinclzi}
D_i^\ast\cup D_j^\ast\cup D_k^\ast\subseteq Z_i\,.
\end{equation}
\ 

\emph{(e)}\quad\quad For each $\;n\geq 1\;$ and each $u$, $\;1\leq u\leq
v\,$, $\,[1,\infty)(y(n)+2W_{y(n)})\,$ and $\,B_u\,$ are open, convex, and
disjoint by (\ref{down}). Thus there are a continuous linear functional
$\,\lambda_{nu}\,$ and a positive number $\,\epsilon_{nu}\,$ such that 
\begin{equation*}
\begin{aligned}[t]
&\lambda_{nu}(x)>1\qquad\quad\text{when }\;x\in
y(n)+W_{y(n)}\,\text{,}\quad\text{and}\\
&\lambda_{nu}(x)<1-\epsilon_{nu}\quad\text{when }\;x\in B_u\,.
\end{aligned}
\end{equation*}
If $\;x\in Z_n\;$ and $\;\lambda_{nu}(x)\leq 1\,$, then
$\;x\in(\frac{1}{2}-\epsilon_{y(n)},1)(y(n)+W_{y(n)})\subseteq A\;$ (see
(\ref{zdef}) and (\ref{inclext})).\\
\ 

\emph{(f)}\quad\quad Consider now, for $\;n=0,1,2,\ldots$, the three
complexes 
\begin{equation*}
\mathcal{D}(n):=\{D_1,\ldots,D_n\}\,\text{,}\quad\mathcal{D}^\ast(n):=\{D_1^%
\ast,\ldots,D_n^\ast\}\,\text{,}\quad\mathcal{Z}(n):=\{Z_1,\ldots,Z_n\}\,.
\end{equation*}
$\,$For $\;n=0\,$, these complexes are all empty. 

Suppose that, for each simplex $\Delta$ of $\;\mathcal{D}^\ast(n)\,$, a
function $\;g_n(\Delta):X\longrightarrow Y\;$ in $\mathcal{F}$ is
constructed, in such a way that the following two conditions are satisfied.
If $\;\Delta=\emptyset\,$, and in particular when $\;n=0\,$,
$\,g_n(\Delta)\,$ may be arbitrary (by
(\label{k}\ref{para8}),$\;\mathcal{F}\neq\emptyset\,$). $\,$

\hspace*{1cm}(1)\quad For any simplices $\;\Delta_1,\Delta_2\;$ of
$\;\mathcal{D}^\ast(n)\,$, and any $\;i\in\Delta_1\cap\Delta_2\,$,
\begin{equation*}
g_n(\Delta_1)\mid D_i\cap X=g_n(\Delta_2)\mid D_i\cap X\;.
\end{equation*}
As a consequence, the mappings $\;g_n(\Delta)\;$ for the various simplices
$\Delta$ of $\;\mathcal{D}^\ast(n)\;$ fit together to define a mapping
$\;g_n:D(n)\cap X\longrightarrow Y\;$ in $\mathcal{F}$, although they may not
agree elsewhere. Recall from (\ref{simplex}) that
$\,D(n):=\bigcup_{i=1}^nD_i\,$. 

\hspace*{1cm}(2)\quad For $\;1\leq i\leq n\,$,
\begin{equation}\label{linfun}
\left\{
\begin{gathered}
g_n\mid D_i\cap R_n\cap X=f\mid D_i\cap R_n\cap X\,\text{,}\quad\text{where}\\
R_n:=\bigcap_{j=1}^n\{y\in E:(\exists k)(1\leq k\leq v\qquad\qquad\qquad\\
\qquad\qquad\qquad\qquad\;\;\text{\&}\;\;\lambda_{jk}(y)\leq
1-\textstyle\frac{n}{n+1}\epsilon_{jk})\}\,.
\end{gathered}
\right.
\end{equation}
\ $\,R_n\,$ is closed in $E$, and a moment's thought shows (with \emph{(e)})
that 
\begin{equation}\label{humbug}
R_{n+1}\subseteq\,\text{int}_E(R_n)\,\text{,}\quad Z_n\cap R_n\subseteq
B^\ast\text{,}\quad B\subseteq R_n\,.
\end{equation}

The aim now, given the inductive assumptions \emph{(f)}(1) and (2), is to
construct $\;g_{n+1}(\Delta)\;$ for each simplex $\Delta$ of
$\;\mathcal{D}^\ast(n+1)\,$. The induction may start when $\;n=0\,$; then
$\;\mathcal{D}^\ast(0)=\emptyset\;$ and $\;D^\ast(0)=\emptyset\,$, and the
argument below still works (rather trivially). \\
\ 

\emph{(g)}\quad\quad Suppose first that $\Delta$ is a maximal simplex of
$\,\mathcal{D}^\ast(n+1)\,$ such that $\;D^\ast(n+1)^\Delta\cap
D^\ast_{n+1}=\emptyset\,$. Then $\Delta$ must be a (maximal) simplex of
$\;\mathcal{D}^\ast(n)\,$. Thus it makes sense to let
$\;g_{n+1}(\Delta):=g_n(\Delta)\,$.\\
\ 

\emph{(h)}\quad\quad For convenience of notation, temporarily rearrange the
indices $\,\{1,2,\ldots,n+1\}\,$, so that all the vertices of the simplices
$\Delta$ of $\,\mathcal{D}^\ast(n+1)\,$ for which $\;D^\ast(n+1)^\Delta\cap
D^\ast_{n+1}\neq\emptyset\;$ are listed as $\;\{p+1,p+2,\ldots,n+1\}\,$, with
$\;q+1,q+2,\ldots,n+1\;$ all the indices $j$ for which $\;D^\ast_j\cap
D^\ast_{n+1}\neq\emptyset\,$. (Here $p$ and $q$ of course depend on $n$, and
$\;q\geq p\,$; furthermore, if $\;q=n\,$, then necessarily $\;p=q\,$). Thus,
whenever $\;p<i\leq n+1\,$, there are a maximal simplex $\;\Delta\ni i\;$ in
$\;\mathcal{D}^\ast(n+1)\,$, and a vertex $\;j\in\Delta$,$\;$ such that
$\;q<j\leq n+1\;$ and $\;D^\ast_i\cap D^\ast_j\neq\emptyset\neq D_j^\ast\cap
D_{n+1}^\ast\,$. By the construction in \emph{(d)}, then, 
\begin{equation}\label{starinc}
D_{n+1}^\ast\subseteq Z_i\quad\text{and}\quad D^\ast_i\subseteq
Z_{n+1}\quad\text{for }\;p\leq i\leq n+1\,.
\end{equation}
(However, remember that the labels $i$ and $\,n+1\,$ are only temporary).\\
\ 

\emph{(i)}\quad\quad Introduce the temporary notations
$\;\mathcal{D}_{n,p}^\ast:=\{D^\ast_{p+1},\ldots,D^\ast_n\}\;$ and
$\;\mathcal{D}_{n,p}:=\{D_{p+1},\ldots,D_n\}\,$. These complexes may be
empty, if $\;p=n\,$. Certainly
$\;\mathcal{D}_{n,p}\prec\mathcal{D}_{n,p}^\ast\,$, as was pointed out at
(\ref{strongincl}). Any simplex $\Delta$ of $\mathcal{D}_{n,p}^\ast$ is a
simplex of $\;\mathcal{D}^\ast(n)\,$, so $\;g_n(\Delta):X\longrightarrow Y\;$
is defined. The conditions of (\ref{extensioncomplex}) are satisfied if I
choose ``$\mathcal{A}$'' to be $\,\mathcal{D}_{n,p}^\ast\,$ and
``$\mathcal{B}\,$'' to be $\;\mathcal{D}_{n,p}\,$, and let the function
$\;g_n:D_{n,p}\cap X\longrightarrow Y\;$ (whose existence was assured by the
inductive hypothesis \emph{(f)}(1)$;\,$here $\;D_{n,p}:=D_{p+1}\cup
D_{p+2}\cup\ldots\cup D_n\,$) take the place of the ``$f\,$'' of
(\ref{extensioncomplex}). From (\ref{extensioncomplex}), then, there exists a
function, which I call $\;h_n:X\longrightarrow Y\,$, in $\mathcal{F}$, such
that 
\begin{equation}\label{handf}
h_n\mid D_{n,p}\cap X=g_n\mid D_{n,p}\cap X\,.
\end{equation}
\ However, I have no control over the behaviour of $\,h_n\,$ near the origin
(see (\ref{defds})).

In addition, $\,Z_{n+1}\,$ was defined (see (\ref{inclext}$)$, (\ref{zdef}),
and (\ref{humbug})) so that there is a function $\;k_{n+1}:X\longrightarrow
Y\;$ in $\mathcal{F}$ (previously called $\,k_{y(n+1)}\,$)\label{hotspot} for
which 
\begin{equation}\label{mumble}
k_{n+1}\mid Z_{n+1}\cap B^\ast\cap X=f\mid Z_{n+1}\cap B^\ast\cap X\text{ .}
\end{equation}
\\
\ 

\emph{(j)}\quad\quad Let
$\;\Lambda_n:E\longrightarrow(\mathbb{R}^v)^n:x\longrightarrow(%
\lambda_{j1}(x),\lambda_{j2}(x),\ldots,\lambda_{jv}(x))_{j=1}^{\,n}\,$.
Represent $\,(\mathbb{R}^v)^n\,$ as the space of $\,n\times v\,$ matrices
$\,M_{nv}\,$, denoting by $\;e_{ji}:M_{nv}\longrightarrow\mathbb{R}\;$ the
map which selects the $\;(j,i)$-th. entry. 

Since the functionals $\,\lambda_{ji}\,$ may not be linearly independent,
$\,\Lambda_n\,$ need not be surjective; call its image $\;T_n\subseteq
M_{nv}\,$, a finite-dimensional real vector space. Let$\,$
\begin{equation*}
\begin{aligned}
K_n&:=\left\{\xi\in T_n:(\forall j)(\exists k)\;1\leq k\leq
v\;\;\text{\&}\;\;e_{jk}(\xi)\leq
1-\textstyle\frac{n+1}{n+2}\epsilon_{jk}\right\}\,\text{, }\\
J_n&:=\left\{\xi\in T_n:(\forall j)(\exists k)\;1\leq k\leq
v\;\;\text{\&}\;\;e_{jk}(\xi)<1-\textstyle\frac{n}{n+1}\epsilon_{jk}\right\}\,%
\text{,}
\end{aligned}
\end{equation*}
so that $\;J_n,K_n\;$ are convex sets in $\,T_n\,$; $\,K_n\,$ is closed,
$\,J_n\,$ is open, $\;K_n\prec J_n\,$. Construct a C$^\infty$ function
$\;\psi_n:T_n\longrightarrow[0,1]\;$ with the properties that
$\;\psi_n(K_n)=\{1\}\;$ and $\;\psi_n(E\setminus J_n)=\{0\}\,$. Consequently,
$\;\psi_n\circ\Lambda_n\;$ takes the value $1$ on
$\;R_{n+1}\subseteq\Lambda_n^{-1}(K_n)\;$ (see (\ref{linfun})) and the value
$0$ on $\;E\setminus($int$_E(R_n))\,$, since
$\;\Lambda_n^{-1}(J_n)\subseteq\,$int$_E(R_n)\,$. Now combine the functions
$\,h_n\,$ and $\,k_n\,$ of \emph{(i)} by defining, for $\;x\in X\,$, 
\begin{equation}\label{htilde}
h'_n(x):=\psi_n(\Lambda_n(x))k_{n+1}(x)+(1-\psi_n(\Lambda_n(x)))h_n(x)\,.
\end{equation}
As $\mathcal{F}$ is admissible, $\,h'_n\,$ also belongs to $\mathcal{F}$.\\
\ 

\emph{(k)}\quad\quad If $\;x\in D_i\cap X\;$ for some $i$ with $\;p<i\leq
n\,$, there are two cases. If $\;x\notin\,$int$_E(R_n)\,$, then
(\ref{htilde}) gives $\;h'_n(x)=h_n(x)=g_n(x)\,$, by (\ref{handf}) and
\emph{(f)}(1). If $\;x\in D_i\cap R_{n+1}\cap X\,$, then $\;g_n(x)=f(x)\;$ by
the inductive hypothesis \emph{(f)}(2) and therefore $\;h_n(x)=f(x)\;$ by
(\ref{handf}). At the same time, from (\ref{diinclzi}) and (\ref{humbug}), 
\begin{equation*}
D_i\cap R_{n+1}\subseteq Z_i\cap R_i\subseteq B^\ast\,\text{,}
\end{equation*}
so that $\;D_i\cap R_{n+1}\cap X\subseteq Z_{n+1}\cap B^\ast\cap X\;$ by
(\ref{starinc}). Consequently, by (\ref{mumble}), $\;k_{n+1}(x)=f(x)\;$ too
--- so $\;h'_n(x)=g_n(x)\;$ once more. Hence 
\begin{equation}\label{htildeandg}
\left\{
\begin{aligned}
h_n'\mid D_i\cap R_{n+1}\cap X&=f\mid D_i\cap R_{n+1}\cap X\quad\text{and}\\
h'_n\mid D_i\cap X&=g_n\mid D_i\cap X\quad\text{for }\;p<i\leq n\,.
\end{aligned}
\right.
\end{equation}

If $\;x\in Z_n\cap R_{n+1}\cap X\,$, $\;h'_n(x)=k_{n+1}(x)\,$, by
(\ref{htilde}). From (\ref{starinc}) and (\ref{mumble}), 
\begin{equation}\label{humble}
h'_n\mid D_{n+1}\cap R_{n+1}\cap X=k_{n+1}\mid D_{n+1}\cap R_{n+1}\cap X\,.
\end{equation}
\ 

\emph{(l)}\quad\quad If $\;D^\ast(n+1)^\Delta\cap
D^\ast_{n+1}\neq\emptyset\,$, where $\Delta$ is a maximal simplex in the
complex $\;\mathcal{D}^\ast(n+1)\,$, then let $\;g_{n+1}(\Delta):=h'_n\,$, as
just constructed in \emph{(k)}. This, with \emph{(g)}, defines
$\;g_{n+1}(\Delta)\;$ for all \emph{maximal} simplices of
$\;\mathcal{D}^\ast(n+1)\;$.\\
\ 

\emph{(m)}\quad\quad To each \emph{non-maximal} simplex $\Delta$ of
$\;\mathcal{D}^\ast(n+1)\,$, associate (once and for all) a maximal simplex
$\;\Delta'\supseteq\Delta\,$, and define
$\;g_{n+1}(\Delta):=g_{n+1}(\Delta')\;$. \\
\ 

\emph{(n)}\quad\quad If $\;\Delta_1,\Delta_2\;$ are simplices of
$\,\mathcal{D}^\ast(n+1)\,$ and $\;i\in\Delta_1\cap\Delta_2\,$, then take the
maximal simplices $\;\Delta_1',\Delta_2'\;$ of $\,\mathcal{D}^\ast(n+1)\,$
that were fixed at \emph{(m)}; certainly $\;i\in\Delta_1'\cap\Delta_2'\,$. If
the stars of both $\;\Delta_1'\;$ and $\;\Delta_2'\;$ meet
$\,D^\ast_{n+1}\,$, as at \emph{(l)}, then $\,g_{n+1}(\Delta_1')\,$ and
$\,g_{n+1}(\Delta_2')\,$ are both $\,h'_n\,$, and automatically $\;$
\begin{equation*}
g_{n+1}(\Delta_1')\mid D_i\cap X=g_{n+1}(\Delta_2')\mid D_i\cap X\,.
\end{equation*}
If $\;i\leq n\,$, (\ref{htildeandg}) shows that both of them agree with
$\;g_n\mid D_i\cap X\,$.

If only the star of $\,\Delta_1'\,$ in $\,\mathcal{D}^\ast(n+1)\,$ meets
$\,D^\ast_{n+1}\,$, then, again, $\;i\leq n\;$ and $\,\Delta_2'\,$ is a
simplex of $\,\mathcal{D}^\ast(n)\,$ and
$\;g_{n+1}(\Delta_2')=g_n(\Delta_2')\;$ by \emph{(g)}; whilst, by
(\ref{htildeandg}), 
\begin{equation*}
g_{n+1}(\Delta_1')\mid D_i\cap X=h'_n\mid D_i\cap X=g_n\mid D_i\cap
X=g_n(\Delta_2')\mid D_i\cap X\;.
\end{equation*}

Lastly, if the star in $\,\mathcal{D}^\ast(n+1)\,$ of neither $\;\Delta_1'\;$
nor $\;\Delta_2'\;$ meets $\;D^\ast_{n+1}\,$, then, by \emph{(g)} and
\emph{(f)}(1), 
\begin{equation*}
\begin{aligned}[t]
g_{n+1}(\Delta_1')\mid D_i\cap X&=g_n(\Delta_1')\mid D_i\cap X\\
&=g_n(\Delta_2')\mid D_i\cap X=g_{n+1}(\Delta_2')\mid D_i\cap X\,.
\end{aligned}
\end{equation*}
Thus the inductive hypothesis \emph{(f)}($1$) still holds when $n$ is
substituted by $\,n+1\,$, and also $\;g_{n+1}\mid D_i\cap X=g_n\mid D_i\cap
X\;$ for $\;1\leq i\leq n\,$, or 
\begin{equation}\label{stability}
g_{n+1}\mid D(n)\cap X=g_n\mid D(n)\cap X\,.
\end{equation}
\ 

\emph{(o)}\quad\quad The hypothesis \emph{(f)}(2) must also be considered. If
$\;1\leq i\leq p\,$, and $\Delta$ is a maximal simplex of
$\,\mathcal{D}^\ast(n+1)\,$ with $\;i\in\Delta\,$, then \emph{(g) }applies
because of the definition of $p$ (see \emph{(h)}). Thus 
\begin{equation*}
\begin{aligned}[t]
g_{n+1}(\Delta)\mid D_i\cap R_{n+1}\cap X&=g_n(\Delta)\mid D_i\cap
R_{n+1}\cap X\\
&=f\mid D_i\cap R_{n+1}\cap X\,\text{,}
\end{aligned}
\end{equation*}
by \emph{(f)}(2) --- recall from (\ref{linfun}) that $\;R_{n+1}\subseteq
R_n\,$. If $\;p<i\leq n+1\,$, choose a maximal simplex
$\;\Delta\subseteq\{p+1,p+2,\ldots,n+1\}\;$ with $\;i\in\Delta\;$ and
$\;D^\ast(n+1)^\Delta\cap D_{n+1}^\ast\neq\emptyset\,$, which is possible,
again by the definition of $p$. Hence, by \emph{(k)} and by
(\ref{htildeandg}), for $\;p<i\leq n\,$ 
\begin{equation*}
\begin{aligned}[t]
g_{n+1}(\Delta)\mid D_i\cap R_{n+1}\cap X&=h'_n\mid D_i\cap R_{n+1}\cap X\\
&=f\mid D_i\cap R_{n+1}\cap X\,\text{,}
\end{aligned}
\end{equation*}
whilst, for $\;i=n+1\,$, (\ref{humble}), (\ref{starinc}),
and$\,$(\ref{mumble}) yield \ 
\begin{equation*}
\begin{aligned}[t]
g_{n+1}(\Delta)\mid D_{n+1}\cap R_{n+1}\cap X&=k_{n+1}\mid D_{n+1}\cap
R_{n+1}\cap X\\
&=f\mid D_i\cap R_{n+1}\cap X\,.
\end{aligned}
\end{equation*}
In either case, the inductive hypothesis \emph{(f)}(2) still holds when
$\;n+1\;$ takes the place of $n$. Now restore the original order of the first
$\,n+1\,$ indices. The inductive construction may be carried out for all $n$,
giving a mapping $\;g_n(\Delta):X\longrightarrow Y\;$ for each $n$ and each
simplex $\Delta$ of $\,\mathcal{D}^\ast(n)\,$. \\
\ 

\emph{(}$p$\emph{)}\quad\quad Let
$\;D(\infty):=\bigcup_{n=1}^{\,\infty}\:D_n\,$. By the definition of the sets
$\,D_n\,$ in \emph{(d)}, $\;D(\infty)\supseteq[\frac{1}{2},\infty)\partial
B^\ast\,$. If $\;x\in E\setminus D(\infty)\,$, then either
$\;x\in(0,\frac{1}{2})\partial B^\ast\;$ or $\;(0,\infty)x\subseteq
B^\ast\,$. Since $\;B^\ast\subseteq A\;$ and $\;\frac{1}{2}A\subseteq B\;$
(see (\ref{assumptions})), both these possibilities give
$\;x\in\,$int$\,B\,$; hence $\;D(\infty)\cup($int$\,B)=E\,$. A mapping \ 
\begin{equation*}
g\text{(}\infty):D(\infty)\cap X\longrightarrow Y\;
\end{equation*}
\ may be defined by setting $\,g(x):=g_n(x)\,$, where $n$ is the first index
such that $\;x\in D(n)\,$. Because of (\ref{stability}), any larger index
could also be taken, so that $\;g(\infty)\mid D_p=g_p\mid D_p\;$ for any $p$;
as $\;g_p\in\mathcal{F}\,$, $\,D_p\,$ is open, and $\mathcal{F}$ is locally
defined, it follows that $\,g(\infty)\,$ is in $\mathcal{F}$.

However, by \emph{(f)}(2) (with (\label{here}\ref{humbug})), $\;g(\infty)\mid
D_n\cap B\cap X=f\mid D_n\cap B\cap X\;$ for all $n$. Thus $\,g(\infty)\,$
agrees with $f$ on $\;D(\infty)\cap B\cap X\,$, and we may define
$\;g:X\longrightarrow Y\;$ by 
\begin{equation*}
g\mid D(\infty)\cap X=g(\infty)\,,\qquad g\mid(\text{int}\,B)\cap
X=f\mid(\text{int}\,B)\cap X\,.
\end{equation*}
This function $g$ is in $\mathcal{F}$, and has the required properties. 
\end{proof}

\begin{remark}
\ The\label{pardrem} details of the proof could be varied in many ways. The
technical difficulty of the argument here is the necessity of obtaining
strong star-refinements by radial projection on $\,\partial B^\ast\,$, which
is awkward because the $\,D_n\,$ have to be convex in $E$ (to allow for the
introduction of linear functionals) as well as refining the cones on the
$\,W_\gamma''\,$. \ 

The introduction of the sets $\;R_n\;$ is needed to ``correct'' the mapping
$\,k_n\,$ given by (\ref{extensioncomplex}), and ensure it agrees with $f$ on
$B$. It appears impossible to arrange that $\;g_n(\Delta)\mid
D^\ast(n)=g_{n+1}(\Delta)\mid D^\ast(n)\;$ for all the simplices $\Delta$ of
$\;\mathcal{D}^\ast(n)\,$; the reason is that the application of
(\ref{extensioncomplex}) preserves the restriction of the function to
$\;D(n)^\Delta\,$, but may change it on the larger set $\;D^\ast(n)^\Delta\,$.

In (\ref{extensioncomplex}), the hypothesis that
$\;\mathcal{B}\prec\mathcal{A}\;$ was important for (\ref{primedseparation}).
In (\ref{extensionfromstars}), which is perhaps only interesting in
conjunction with (\ref{extensioncomplex}), it is natural and convenient, but
not really necessary. The only point at which it is used seems to be the
construction in \emph{(a)} of the `intermediate' closed convex sets
$\,B_u^\ast\,$ such that $\;B_u\subseteq\,$int$_E(B_u^\ast)\;$ and
$\;B_u^\ast\subseteq A_u\,$. If $E$ is a \emph{normed }space, (\ref{parb3})
shows that this construction is possible as soon as $\,B_u\,$ is a closed
convex subset of $\,A_u\,$. I do not know how this result can be extended to
more general locally convex spaces. However, if $\,B_u\,$ has non-empty
interior, (\ref{intermed}) constructs $\,B_u^\ast\,$. 

On the other hand, the proof of (\ref{extensionfromstars}) does not, in fact,
require the convexity of $\,B_u^\ast\,$, but only the facts that it is
star-shaped (about the origin, by convention) and that its frontier meets
each radius in at most one point. The construction of a $\,B_u^\ast\,$
satisfying these weaker conditions can be carried out much more generally.
The interest of such questions is that, given some specific class
$\mathcal{F}$ on $E$, one would like to know for exactly which pairs
$\,(U,C)\,$ of an open set $U$ and a closed subset $C$ in $E$ any function
$f$ defined on $U$ and belonging to $\mathcal{F}$ corresponds to some
function $g$ of class $\mathcal{F}$ on $E$ such that $\;g\mid C=f\,$. The
remarks at (\ref{remarksonproof}) on other notions of convexity are relevant
to this.

A rough summary of (\ref{extensionfromstars}) and of
(\ref{extensioncomplex}\label{c}) would be that a function which can be
extended \emph{locally }to $E$ (is ``extensible'') can also be extended
\emph{globally }from suitable sets. It is global extensibility which is
required in practice.
\end{remark}

\begin{definition}
\ Let $\mathcal{F}$ be an affine space of functions on the topological space
$X$, with values in the convex subset $Y$ of the vector space $F$. Suppose
$U$ is an open subset of $X$, and $\mathcal{E}$ is a convex set of functions
$\;U\longrightarrow F\,$. $\mathcal{E}$ is said to admit local affine
$\mathcal{F}$-extension operators if, for every point $\,x\in U\,$, there is
an open neighbourhood $V$ of $x$ in $U$ and an affine mapping
$\;\mathcal{L}_V:\mathcal{E}\longrightarrow\mathcal{F}(X,Y)\;$ such that, for
each $\;f\in\mathcal{E}\,$, $\;\mathcal{L}_V(f)\mid V=f\mid V\,$.

There is a similar definition for ``linear extension operators''.
\end{definition}

\begin{corollary}
\ Let\label{family} \label{simaff}$\,\mathcal{A},\mathcal{B}\,$ and
$\,\mathcal{F},E,F,X\,$ be as in \emph{(\ref{extensionfromstars})}, and let
$Y$ be a convex set in $F$. Let $\mathcal{F}$ be a $Y$-admissible and locally
defined affine space of functions on the open sets of $X$, with images in
$Y$. Suppose that $\mathcal{E}$ is a convex set of functions $\;A\cap
X\longrightarrow Y\;$ in $\mathcal{F}$ which admits local affine
$\mathcal{F}$-extension operators. Then there is an affine-linear map
$\;\mathcal{L}:\mathcal{E}\longrightarrow\mathcal{F}(X;$Y$)\;$ such that
$\;\mathcal{L}(f)\mid B\cap X=f\mid B\cap X\;$ for each
$\,f\in\mathcal{E}\,$. \ 
\end{corollary}

\begin{proof}
\ Every step of the constructions in (\ref{ext:on:segments}) and
(\ref{extensionfromstars}) is by a pointwise convex combination of the values
of extensions either already constructed or given by the hypotheses. Where,
at (\ref{extensionfromstars})\emph{(i)}, appeal was made to
(\ref{extensioncomplex}), it should now be made to (\ref{complaff}).
\end{proof}

There is of course a ``linear'' version, which is an easy corollary. Now
recall (\ref{parc6}).


\begin{lemma}
\ Suppo\label{glimp}se in \emph{(\ref{extensionfromstars})} or in
\emph{(\ref{family})} that $E$ is a normed space and all the convex sets
belonging to the complex $\mathcal{A}$ are bounded. Let $G$ be a norm-dense
subspace of the dual space $\,E'\,$. Then the previous conclusions in
\emph{(\ref{extensionfromstars})} or in \emph{(\ref{family})} hold on the
weaker condition that $\mathcal{F}$ is $\,(G,Y)$-admissible instead of
$Y$-admissible.
\end{lemma}

\begin{proof}
\ The use of (\ref{extensioncomplex}) or of (\ref{complaff}) at
(\ref{extensionfromstars})\emph{(i) }should now be substituted by
(\ref{limp}). The neighbourhoods $\,W_y\,$ of the origin may be norm-balls,
which are bounded, so that, as was shown in (\ref{limp}), the linear
functionals $\,\lambda_{nu}\,$ of (\ref{extensionfromstars})\emph{(e)} may be
chosen from $G$ too.

The full dual space $\,E'\,$ also appeared in the proof of (\ref{para9}),
when $\phi$ was chosen as a support functional for $W$ at $\,\kappa c\,$ to
arrange (\ref{wzorlength}). However, one might instead choose any $\;\phi\in
G\;$ such that $\;\phi(\kappa c)=1\,$, and substitute
$\;W':=W\cap\phi^{-1}(-1,1)\;$ for $W$ throughout the rest of the proof. ($W$
had only to be sufficiently small, convex and symmetric). 
\end{proof}

\ \ 

\section{ The main theorems.}

\begin{theorem}
\ \ Let $E\label{bigthm}$ be a Hausdorff locally convex Lindel\"of space, and
$F$ a vector space, with $X$ a subset of $E$ and $Y$ a convex subset of $F$.
Let $\mathcal{F}$ be a $Y$-admissible and locally defined class of functions
on the open subsets of $X$, with values in $Y$. Suppose that
$\;\mathcal{A}:=\{A_1,A_2,\ldots,A_v\}\;$ is a complex of open convex sets of
$E$ and $\;\mathcal{B}:=\{B_1,B_2,\ldots,B_v\}\prec\mathcal{A}\,$. Let $A$
denote the carrier of $\mathcal{A}$ and $B$ the carrier of $\mathcal{B}$, and
suppose that there are given a mapping $\;f:B\cap$X$\longrightarrow Y\;$ and,
for each simplex $\Delta$ of $\mathcal{A}$, an $\mathcal{F}$-extensible
mapping\label{h} $\;f_\Delta:A^\Delta\cap X\longrightarrow Y\;$ such that
$\;f_\Delta\mid B^\Delta\cap X=f\mid B^\Delta\cap X\,$. Then there exists a
function $\;h:X\longrightarrow Y\;$ in $\mathcal{F}$ \ such that $\;h\mid
B\cap X=f\mid B\cap X\,$.
\end{theorem}

\begin{proof}
\ For each simplex $\Delta$ of $\mathcal{A}$, (\ref{extensionfromstars}) may
be applied to the complexes
\begin{equation*}
\mathcal{A}^\Delta:=\{A_u:u\in\Delta\}\,,\quad\mathcal{B}^\Delta:=\{B_u:u\in%
\Delta\}\,\text{;}
\end{equation*}
so there exists $\;g_\Delta:X\longrightarrow Y\;$ in $\mathcal{F}$ such that
$\;g_\Delta\mid B^\Delta\cap X=f\mid B^\Delta\cap X\,$.
(\ref{extensioncomplex})$\,$may be applied in its turn to give the result.
\end{proof}

The statement above is convenient for later applications, because it does not
require that the mappings $\,f_\Delta\,$ should fit together to define a
mapping of the whole of $\,A\cap X\,$, but it is less striking than the
following.

\begin{corollary}
\ Let $\,\label{coroll}\mathcal{A},\mathcal{B}\,$ and
$\,\mathcal{F},E,F,X,Y,A,B\,$ be as in \emph{(\ref{bigthm})}, and let
$\;k:A\cap X\longrightarrow Y\;$ be $\mathcal{F}$-extensible. Then there
exists a function $\;q:X\longrightarrow Y\;$ in $\mathcal{F}$ such that
$\;q\mid B\cap X=k\mid A\cap X\,$.
\end{corollary}

\begin{proof}
\ Take in (\ref{bigthm}) $\;f_\Delta:=k\mid A^\Delta\cap X\;$ for each
simplex $\Delta$ of $\mathcal{A}$.
\end{proof}

\begin{theorem}
\ \ Let \label{familyend}$\,\mathcal{A},\mathcal{B}\,$ and
$\,\mathcal{F},E,F,X,Y,A,B\,$ be as in \emph{(\ref{bigthm})}. Let
$\mathcal{F}$ be a $Y$-admissible and locally defined affine space of
functions on the open sets of $X$, with images in $Y$. Suppose that
$\mathcal{E}$ is a convex set of functions $\;A\cap X\longrightarrow Y\;$
which admits local affine $\mathcal{F}$-extension operators. Then there is an
affine-linear map
$\;\mathcal{L}:\mathcal{E}\longrightarrow\mathcal{F}(X;$Y$)\;$ such that 
\begin{equation*}
\mathcal{L}(f)\mid B\cap X=f\mid B\cap X\;
\end{equation*}
for each $\,f\in\mathcal{E}\,$. \ 
\end{theorem}

\begin{proof}
\ Collate (\ref{family}) and (\ref{complaff}), exactly as in the proof of
(\ref{bigthm}).
\end{proof}

Of course (\ref{coroll}) is the special case of (\ref{familyend}) when
$\;\mathcal{E}:=\{f\}\,$.

\begin{remark}
\ In n\label{remlast}ormed spaces, the results above may be improved; that
is, one may weaken the conditions on $\mathcal{F}$ to require only
$\,(G,Y)$-admissibility as defined at (\ref{parc6}). This is probably a
significant fact. At (\ref{limp}) and (\ref{glimp}), it depended on the
boundedness of the sets of the complexes, but this restriction can now be
removed. (It is not clear, however, that the further theory benefits
\emph{substantially }from this generalization).

A normed (or metric) space is Lindel\"of if and only if it is separable.
\end{remark}

\begin{theorem}
\ \ Let $E\label{homeatlast}$ be a separable normed space, and $F$ a vector
space, with $X$ a subset of $E$ and $Y$ a convex subset of $F$. Let $G$ be a
norm-dense subspace of the dual space $\,E'\,.\,$Let $\mathcal{F}$ be a
$\,(G,Y)$-admissible and locally defined class of functions on the open
subsets of $X$, with values in $Y$. Suppose that
$\;\mathcal{A}:=\{A_1,A_2,\ldots,A_v\}\;$ and
$\;\mathcal{B}:=\{B_1,B_2,\ldots,B_v\}\;$ are complexes of subsets of $E$,
where $\mathcal{A}$ consists of open convex sets of $E$ and
$\;\mathcal{B}\preceq\mathcal{A}\;$ (see \emph{(\ref{defbdedstrincl})} and
\emph{(\ref{parb5})}), and let $A$ denote the carrier of $\mathcal{A}$ and
$B$ the carrier of $\mathcal{B}$. Let $\mathcal{E}$ be a convex set of
functions $\;A\cap X\longrightarrow Y\;$ in $\mathcal{F}$ which admits local
affine $\mathcal{F}$-extension operators. Then there is an affine-linear map
$\;\mathcal{L}:\mathcal{E}\longrightarrow\mathcal{F}(X;$Y$)\;$ such that
$\;\mathcal{L}(f)\mid B\cap X=f\mid B\cap X\;$ for each $\,f\in\mathcal{E}\,$.
\end{theorem}

\begin{proof}
\ By (\ref{bdedstrincl}), construct a sequence
$\,(\mathcal{B}^n)_{n=1}^\infty\,$ of complexes of open convex sets such that
$\;\mathcal{B}\preceq\mathcal{B}^{n+1}\preceq\mathcal{B}^n\preceq\mathcal{A}%
\;$ for all $n$. Here $\,\mathcal{B}^n\,$ stands for
$\,\{B_1^n,B_2^n,\ldots,B_v^n\}\,$. I shall write $\,B^n\,$ for the carrier
of $\,\mathcal{B}^n\,$, and let $\;\mathcal{B}^0:=\mathcal{A}\,$. Define, for
$\;n=0,1,2,\ldots\,$, 
\begin{equation}\label{defd}
\left\{
\begin{aligned}
\mathcal{C}_n&:=\{B_E(0;n),B_E(0;n+2)\cap B_1^n,B_E(0;n+2)\cap B_2^n,\ldots\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\ldots,B_E(0;n+2)\cap B_v^n\}\,,\\
\mathcal{D}_n&:=\{B_E(0;n-1),B_E(0;n+1)\cap B_1^{n+1},\ldots\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\ldots,B_E(0;n+1)\cap
B_v^{n+1}\}\,\text{,}
\end{aligned}
\right.
\end{equation}
where, of course, $\;B_E(0;\rho)=\emptyset\;$ when $\;\rho\leq 0\,$, and the
index set of the complexes is understood to be $\,\{0,1,2,\ldots,v\}\,$. It
is clear that $\;\mathcal{D}_n\prec\mathcal{C}_n\;$ for all $n$, and that all
the sets in $\,\mathcal{C}_n\,$ are bounded. Let $\,C_n,D_n\,$ denote the
carriers of $\,\mathcal{C}_n\,$ and of $\,\mathcal{D}_n\,$.

There is an inductive procedure, which may be described as follows. Suppose
that $\;f\in\mathcal{E}\subseteq\mathcal{F}(A\cap X,Y)\,$. Since
$\;B_E(0;0)=\emptyset\,$, $\;C_0\cap X\subseteq A\cap X\,,$ and there is a
linear restriction operator $\,i_0\,$, where $\;i_0(f):=f\mid C_0\cap X\,$.
Set $\;\mathcal{E}_0:=i_0(\mathcal{E})\,$. Certainly $\,\mathcal{E}_0\,$
admits local affine $\mathcal{F}$-extension operators, and, moreover,
$\;f_0:=i_0(f)\;$ \ agrees with $f$ on $\;D_0\cap X=B_E(0;1)\cap B^1\cap
X\,$. Use (\ref{glimp}) to construct a function $\;f_1\in\mathcal{F}(X,Y)\;$
which depends linearly on $\,f_0\,$, and therefore on $f$, and agrees with
$f$ on $\,D_0\cap X\,$. 

Inductively, suppose for $\;n\geq 1\;$ that $\;f_n\in\mathcal{F}(X,Y)\;$ is
given and depends linearly on $f$; and that $\,f_n\,$ agrees with $f$ on
$\;B_E(0;n+1)\cap B^{n+1}\cap X\,$. We may then define
$\;g_n\in\mathcal{F}(C_{n+1}\cap X,Y)\;$ by 
\begin{equation}\label{zloz}
\left\{
\begin{aligned}
g_n\mid B_E(0;n+3)\cap B^{n+1}\cap X&=f\mid B_E(0;n+3)\cap B^{n+1}\cap X\,,\\
g_n\mid B_E(0;n+1)\cap X&=f_n\mid B_E(0;n+1)\cap X\,.
\end{aligned}
\right.
\end{equation}
As $\mathcal{F}$ is locally defined, the function thus defined is indeed in
$\mathcal{F}$, and it admits local affine $\mathcal{F}$-extension operators,
as each of the restrictions does. 

Since $\;\mathcal{D}_{n+1}\prec\mathcal{C}_{n+1}\,$, (\ref{glimp}) now
constructs a function $\;f_{n+1}\in\mathcal{F}(X,Y)\;$ that agrees with
$\,g_n\,$ on \ $\,D_{n+1}\cap X\,$. (I recall that this is for $\;n\geq
1\,$). Furthermore, $\,f_{n+1}\,$ depends linearly on $f$. This completes the
inductive construction.

As $\;f_{n+1}\mid D_{n+1}\cap X=g_n\mid D_{n+1}\cap X\,$, (\ref{defd}) and
(\ref{zloz}) show that 
\begin{equation*}
f_{n+1}\mid B_E(0;n)\cap X=f_n\mid B_E(0;n)\cap X\,
\end{equation*}
for all $\,n\geq 1\,$. One may therefore define a function
$\;g:X\longrightarrow Y\;$ by the prescription 
\begin{equation}\label{defofg}
g(x):=f_n(x)\quad\text{when }\;x\in B_E(0;n)\cap X\,.
\end{equation}
Since $\mathcal{F}$ is locally defined, $\;g\in\mathcal{F}(X,Y)\,$, and it
depends linearly on $f$; we may write $\;\mathcal{L}(f):=g\,$. However, look
at the ``other part'' of $\,D_{n+1}\,$ in (\ref{defd}) and (\ref{zloz}): 
\begin{equation*}
\begin{aligned}[t]
f_{n+1}\mid B_E(0;n+2)\cap B^{n+2}\cap X&=g_n\mid B_E(0;n+2)\cap B^{n+2}\cap
X\\
&=f\mid B_E(0;n+2)\cap B^{n+2}\cap X\,\text{,}
\end{aligned}
\end{equation*}
for all $\;n\geq 1\,$. Hence, from definition (\ref{defofg}) (recall
$\;B\subseteq B^{n+2}\;$ for all $n$), 
\begin{equation*}
\begin{aligned}[t]
g\mid B_E(0;n+1)\cap B\cap X&=f_{n+1}\mid B_E(0;n+1)\cap B\cap X\\
&=f\mid B_E(0;n+1)\cap B\cap X\,\text{,}
\end{aligned}
\end{equation*}
which implies that $\;g\mid B\cap X=f\mid B\cap X\,\;$ as required. \ 
\end{proof}


\begin{conclusion}
\ In the sequel, I shall show how the above results may be extended to
manifolds. The methods to be used are not unlike the proof of
(\ref{extensionfromstars}); that is, they depend on an induction of a similar
type. It is worth noting that the proof of (\ref{homeatlast}) already
involves three inductions, none of them altogether trivial.
\end{conclusion}

\ 

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\ 
\end{document}
