---
abstract: |
    Working in infinite dimensional linear spaces, we deal with support for closed sets without interior. We generalize the Convexity Theorem for closed sets without interior. Finally we study the infinite dimensional version of Jordan hypersurfaces. Our whole work never assumes smoothness and is based exclusively on non-differential Convex Analysis tools and, in particular, on theory of convex cones. A crucial mathematical tool for our results is obtained solving the decomposition problem for non-closed non pointed cones.

    AMS Classification: 52A07; 53D99; 53B10
author:
- |
    Paolo d’Alessandro\
    Former Professor at the Dept of Math Third Univ. of Rome\
    pdalex45@gmail.com
title: 'Support, Convexity Conditions and Convex Hypersurfaces in Infinite Dimension'
---

Introduction
============

If we consider a convex set $C$ in a real linear topological space $E$ and assume that it has interior, then, by a well known topological separation principle, each point in $\mathcal{B}(C)$ is a support point for $C$.

It is also well known that, conversely, if each point of the boundary of a body $C$ has support, then $C$ is convex. These two results form what we call the Convexity Theorem for Bodies (i.e. sets with interior).

The hypothesis of the Theorem is considered of global nature in the literature. And a main stream of literature is that of deducing the same conclusion from purely local conditions. Usually smoothness is assumed, the local hypothesis is provided by curvature and, naturally, the machinery of differential geometry takes over. Since the present paper does not follow this stream of research we just cite [@JONKER1972] as an example.

Here we move with a different orientation. First and foremost we never assume smoothness of surfaces and stay clear of manifold theory. In this non- smooth setting we lean instead only on non-differential Convex Analysis techniques and find our main toolset in the theory of convex cones. This approach will allow us to give support results for closed sets without interior, generalize the Convexity Theorem to closed sets without interior, and, in a Hilbert space setting, develop a non-linear range space theory of convex hypersurfaces. Moreover, in the final Section, we introduce and investigate the concept of convexification of a hypersurface.

We believe that the basic concept of support should not be considered global, simply because it is non-differential. Indeed it is a local non-smooth condition, a fact which we try to substantiate within our theory.

The conical techniques we use are geared on polarity, and, in our non-smooth setting, the concept of tangent cone is the simpler possible. We have no use here for its differential version, namely the famous Bouligant cone.

Interestingly, within the present vision of geometry of convex sets, linear and non-linear range space theory present some superpositions. Moreover, working on these topics a connection between topology and convexity, typical of Convex Analysis, does persists.

Some Mathematical Premises
==========================

In this Section we make precise some technicalities of Vector Topology, that make possible starting a formal treatment.

When we consider a linear topological space (lts) or locally convex space (lcs), we will assume for simplicity that the space be real. It is instead for substantial reasons, appearing in a short while in connection to boundedness, that we will soon restrict ourselves to Hausdorff lts.

A terminological warning. As in [@KNLTS] we call sphere what in geometry is called ball, and what in geometry is called sphere is for us the boundary of a ball.

Also in this paper all cones are intended to be convex. That is, $C$ is a cone in a lts $E$ if $C+C\subset C$ and $\alpha C\subset C$  for any real $\alpha \geq 0$.

We recall some now some well known concepts, that are important to us relating to Hausdorff lts.

\[NSCCONDITIONHAUSS\] Let $E$ be a lts, then $$\{0\}^{-}=\cap \{U:\text{ }U\text{ is a neighborhood of }0\}$$Moreover, $\{0\}^{-}$ is always a linear subspace. The space $E$ is Hausdorff if and only if:$$\{0\}^{-}=\{0\}$$

(A explicit proof is given for example in [@DALEXVECTORTOP1977] at the end of p.29) Thus if a space is not Hausdorff then $\{0\}^{-}$ is a non-trivial linear subspace so that $\{0\}^{-}\neq \{0\}$) .

Incidentally, what one could (and should) do when the space $E$ is not Hausdorff is to consider the topological quotient space $E/\{0\}^{-}$, which is a Hausdorff lts and is in fact the right Hausdorff representation of $E$. Even more incidentally, this is the correct formalization of the a.e. concept in functions spaces. An in-depth analysis might be advisable though, because this is not an issue to be dismissed as lightly as usual. One for example wants to verify that the linear subspace $\{0\}^{-}$stays fixed across the various topologies considered for a given space. But even more serious consideration deserve the facts stated by Theorem 8 in above reference.

In the present context this facts are relevant because of the following consequence of the equation $\{0\}^{-}=\{0\}$, characterizing Hausdorff spaces, which regards boundedness.

Notice in this respect that boundedness is translation independent. Also we call *ray* (emanating from the origin)* * the conical extension (or hull) of a non-zero vector.

\[BOUNDEDNESSCOROL\] Consider a Hausdorff lts $E$. Then a necessary condition for a set $C$ in $E$ to be bounded is that it does not contain any ray or any translated ray.

Without restrictions of generality we may assume that $0\in C$. Suppose that $C$ is bounded but contains a ray. Then because $C$ is absorbed by every neighborhood of the origin, it follows that all neighborhoods must contain the same ray. But then the intersection of all such neighborhoods contains such a ray and this contradicts the hypothesis that the space be Hausdorff, because the equation $\{0\}^{-}=\{0\}$ cannot hold.

Naturally, we will make in what follows use of many elementary computations on convexity, see Theorem 13.1 in [@KNLTS]. We recall a few statement, that will be of use here. Let $E$ be a lts and $A$ a subset of $E$, then:$$A\text{ convex }\Rightarrow A^{i}\text{, }A^{-}\text{ convex}$$$$A\text{ convex body}\Rightarrow tA^{i}+(1-t)A^{-}\subset A^{i}\text{, }0<t\leq 1$$$$A\text{ convex body}\Rightarrow A^{i}=A^{\Diamond }\text{;}A^{i-}=A^{-}\text{;}A^{-i}=A^{i}$$where $A^{\Diamond }$ denotes the radial kernel of $A$. And for any set $B$:$$\mathcal{C}^{-}(B)=\mathcal{C}(B)^{-}$$where is the convex extension (or hull) of $B$ and $\mathcal{C}^{-}(B)$ is the closed convex extension (or hull) of $B$.

We will make of course intensive use of Separation Principles. We recall two statements from [@KNLTS], so that we can bear in mind the exact hypotheses of each major separation result.

Suppose $E$ is a lts and $A$ and $B$ (non-void) convex sets.  Suppose $A$ has interior and that $B\cap A^{i}=\phi $. Then there exists a linear continuous functional that separates $A$ and $B$.

Notice that $E$ is not supposed to be Hausdorff in this particular result.

Second Theorem . This time we speak of locally convex spaces (again not necessarily Hausdorff):

Suppose $E$ is a lcs and $A$ and $B$ (non-void) disjoint convex sets, with $A $ compact and $B$ closed.  Then there exists $f\in E^{\ast }$ that strongly separates $A$ and $B$.

The Basic Convexity Theorem
===========================

We consider a convex body $C$ (that is, a convex set with interior) in a real Hausdorff lts $E$, and prove the following Theorem, which gives a little additional information with respect to the Convexity Theorem cited at the beginning. In this respect it is useful to bear in mind that whereas any point of $\mathcal{B}(C)$ is a support point of $C$ in view of the first Separation Theorem, it is instead obvious that no point of $C^{i}$ can be a support point, because each such point has a neighborhood entirely contained in $C^{i}$.

\[BODYCONVTH\] Consider a closed body $C$ in a lts. Then $C$ is convex if and only if each of its boundary points is a support point for $C$, and, if this is the case, then $C$ is the intersection of any family of closed semispaces obtained choosing a single support closed semispace for each point of $\mathcal{B}(C)$ itself.

The only if is immediate consequence of the first cited separation Theorem. Then we prove that if each of the boundary points of $C\ $is a support point for $C$ itself, then $C$ is the intersection specified in the statement, and consequently also a closed convex body. Without restriction of generality and since translation is a homeomorphism that leaves invariant all convexity properties, we can assume that $0\in C^{i}$. Consider the intersection $\Psi 
$ specified in the statement. Then $\Psi $ is a closed convex set containing $C$ and thus a closed convex body too. Suppose there is a point $y$ in $\Psi 
$ such that $y\notin C$ or $y\in \overline{C}$. Since $\overline{C}$  is an open set, there is a neighborhood $U$ of $y$ contained in $\overline{C}$ itself. Also there is a neighborhood $W$ of the origin with $W\subset C^{i}$  Consider the ray $\rho $ generated by $y.$ By radiality of neighborhoods there is a segment $[0:z]\subset W$ with $z\neq 0$ and $[0:z]\subset \rho $. We take $v=\beta y$ with $\beta =\sup \{\gamma :\gamma y\in C^{i}\}$. This sup exists because $y\notin C^{i}$ and because $C^{i}$ is convex and so no point $\gamma y$ with $\gamma >1$ can be in $C^{i}$. Also $\beta >0$ in view of the existence of the vector $z$. Moreover, $v$ cannot be in $C^{i}$ because otherwise it would have a neighborhood contained in $C^{i}$ and since neighborhood of $v$ are radial at $v$ we would contradict that $\beta $ is the given sup. But then it must be $v\in \mathcal{B}(C)$, since there is a net (in $\rho \cap C^{i}$)  that converges to $v$ by construction, and therefore by the first Separation Theorem it is a support point for $C$. At this point if $v=y$ we are done, because we have found a contradiction. If not it must be $\beta <1$. Consider the separating continuous linear functional $f$   corresponding to $v$. We have $f(0)=0$, and to fix the ideas and without restriction of generality, $f(v)=\beta f(y)>0$. But then, because $f(y)>f(v)$, the supporting functional leaves the origin and $y$ on opposite open semispaces. This would imply  that $y\notin \Psi $, which is again a contradiction. Thus the proof is completed.

Of course all the more $C$ is the intersection of all the closed supporting semispaces of $C$.

Notice that this Convexity Theorem is an instance of the loose connection with topology often met in Convex Analysis. In fact if a closed set has no interior we may well try to strenghten the topology of the ambient space to the effect of forcing the interior to appear. Another example of this loose connection, but going the other way around, is the celebrated Krein -Milman Theorem, where we may weaken the topology in an effort to induce compactness of a convex set, to the effect of making the KM Theorem applicable.

A Refinement of The Convexity Theorem, Cones and Support for Void Interior Sets
===============================================================================

We Start making more explicit from the preceding proof an important property of convex bodies.

\[RAYSSINGLETONINTERSECTION\] Suppose $C$ is a closed convex body in a lts and that, without restriction of generality, $0\in C^{i}$. Then$$C=\cup \{l\cap C\text{, }l\text{ is a ray}\}$$and, if a ray $l$ meets $\mathcal{B}(C)$, the intersection is a singleton. Therefore if $\{z\}$ is such an intersection $[0:z)\subset C^{i}$.

The first part is obvious, since if $y\in C$ then $[0:y]\subset C$.  As to the second statement suppose that $z,w\in \rho \cap \mathcal{B}(C)$ and assume for example that $z$ is closer to the origin than $w.$ Then, since $C$ has support at $z$, the supporting hyperplane will leave the origin and $w$ on opposite open semispaces. This is a contradiction and so the second part is proved too. Finally the last part is a direct consequence of elementary computations on convex sets.

With reference to the above Theorem, we prove, regarding boundedness and unboundedness the following

\[RAYSINTERSECTION\] Suppose $C$ is a closed convex body in a Hausdorff lts and that, without restriction of generality, $0\in C^{i}$. If there exists a ray, that does not meet $\mathcal{B}(C)$, then $C$ is unbounded. Thus if $C$ is bounded all the rays meet $\mathcal{B}(C)$ (in a unique point).

Consider a ray $\rho $ which does not meet $\mathcal{B}(C)$ and suppose that there exists a vector $z\in \rho \cap \overline{C}$. Because $\overline{C}$ is open $\rho \cap \overline{C}$ is open in $\rho $. Consider $$\beta =\inf \{\alpha :\alpha z\in \rho \cap \overline{C}\}$$Notice that $\beta >0$ since $C$ has interior. It cannot be $w=\beta z\in
\rho \cap \overline{C}$, because there would be a neighborhood of $w$ contained in $\overline{C}$ and, by radiality of neighborhoods in a lts, we would contradict the definition of $\beta $. Therefore $w\in C$ and, more precisely, $w\in \mathcal{B}(C)$ since each of its neighborhoods meets $\overline{C}$. This contradicts that the ray does not meet $\mathcal{B}(C)$, and therefore such a vector $z$ cannot exist. Thus, if a ray $\rho $ does not meet $\mathcal{B}(C)$, then $\rho \subset C$. But then by Corollary [BOUNDEDNESSCOROL]{}, $C$ is unbounded. This completes the proof.

Theorem \[RAYSSINGLETONINTERSECTION\] has a Corollary which, in a way, generalizes the finite dimensional fact asserting that if $C$ is compact $\mathcal{C}^{-}(C)=\mathcal{C}(C)$. This peculiar fact for finite dimension can be proved leaning on the celebrated Caratheodory’s Theorem, which is finite dimensional only as well.

\[PLAINCEXTWITHOUTCOMPACT\] Suppose $C$ is a closed bounded convex body in a Hausdorff lts and that, without restriction of generality, $0\in C^{i}$. Let $y(\rho )$ be the unique point where each ray $\rho $ meets $\mathcal{B}(C)$. Then$$C=\cup \{[0:y(\rho )]\}$$$$\mathcal{B}(C)=\{y(\rho ):\rho \text{ is a ray}\}$$$$C=\cup \{[x:y]:x,y\in \mathcal{B}(C)\}$$and$$\mathcal{C}^{-}(\mathcal{B}(C))=\mathcal{C}(\mathcal{B}(C))=C$$

The first three statements are either obvious consequences of the preceding Theorem or of immediate proof. As to the last statement, to simplify notations, let $\Psi =\cup \{[x:y]:x,y\in \mathcal{B}(C)\}$ and notice that:$$C\supset \mathcal{C}^{-}(\mathcal{B}(C))\supset \mathcal{C}(\mathcal{B}(C))\supset \Psi =C$$and so we are done.

This result allow us to rephrase the Convexity Theorem for bounded bodies in an interesting way.

\[BODYCONVEXTH2\] Consider a closed bounded body $C$ in a Hausdorff lts. Then $C$ is convex if and only each point of $\mathcal{B}(C)$supports $\mathcal{B}(C)$ itself, and if this is the case, then $C$ is the intersection of any family of semispaces obtained choosing a single closed semispace supporting $\mathcal{B}(C)$ for each point of $\mathcal{B}(C)$.

the proof follows from the fact that $C=\mathcal{C}(\mathcal{B}(C))$, and thus a closed supporting semispace contains $C$ if and only if it contains $\mathcal{B}(C)$.

Again a fortiori $C$ is the intersection of all semispaces that support $\mathcal{B}(C)$.

That a set (not necessarily convex and possibly without interior) has support at a point in its boundary is equivalent to the fact that the normal cone at such point is not trivial (meaning that such a cone is $\neq \{0\}$). The normal cone is nothing but the polar of the tangent cone. We recall a few basic concepts and properties first about cones and then about polarity.

Let $V$ be a cone in a lts $E$. The lineality space of $V$ is the linear subspace $lin(V)$ :$$lin(V)=V\cap (-V)$$Clearly $lin(V)$ is the maximal subspace contained in $V$. We say that $V$ is pointed if $lin(V)=\{0\}$.

Beware that Phelps ([@Phelps2001]) calls proper cone what is for us a pointed cone. Here instead a proper cone is a cone whose closure is a proper subset of the ambient lts.

A cone in a lts $E$ is proper if it is not dense in the whole space. Thus a closed cone is proper if it is not the whole space, or, equivalently, is a proper subset of $E$.

We next cite from [@DALEXCUBO2011]  a first major result regarding support for void interior convex sets, illustrating the mathematical facts and reasoning from which it originates. These recalls will be instrumental in the sequel.

\[properclosedcone\] A cone in a lcs $E$ is proper (equivalently its closure is a proper subset) if and only if it is contained in a closed half-space.

Sufficiency is obvious. As to necessity, let $C$ be a closed proper cone. Then there is a singleton $\{y\}$ disjoint from $C$. Singletons are convex and compact and therefore the second Separation Theorem cited at the beginning applies. Thus there exists a continuous linear functional $f$ such that:$$f(x)<f(y)\text{, }\forall x\in C$$and since $C$ is a cone this implies:$$f(x)\leq 0\text{, }\forall x\in C$$Thus the condition is also necessary and we are done.

\[CLOSUREPOINTEDCONE\]The closure of a pointed cone in a linear topological space is a proper cone.

Suppose that it is not true, that is there is a pointed cone $C$ in a linear topological space $E$, such that $C^{-}=E$. Consider a finite dimensional subspace $F$, with its (unique) relative topology, which intersect $C$ in a non trivial, necessarily pointed, cone. Actually we can take instead of $F$, its subspace $\mathcal{L}(F\cap C)$, without restriction of generality. For simplicity we leave the symbol $F$ unchanged. Next notice that, as is well known, because $F$ is the finite dimensional, the pointed convex cone $\Upsilon =F\cap C$  has interior. Thus it can be separated by a continuous linear functional from the origin and therefore it is contained in a closed semi-space. It follows that the closure of $\Upsilon $ in $F$ is contained in a closed half-space and therefore is a proper cone. But by Theorem 1.16 in [@KELLEY55], such closure is $C^{-}\cap F$. By the initial assumption $C^{-}\cap F=E\cap F=F$. This is a contradiction and therefore the proof is finished.

Let $C$  be a closed subset of a lts $E$ and $x\in \mathcal{B}(C)$. Then the Tangent cone $\mathcal{T}_{C}(x)$ to $C$ at $x$ is the cone:$$\mathfrak{T}_{C}(x)=\mathcal{C}o(C-x)$$

We now turn to lcs spaces, because we start using duality for linear topological spaces.

Let $C$ be a subset of a lcs $E$. Then the polar of $C$ is the (always closed) convex cone:$$C^{p}=\{f:f\in E^{\ast },f(x)\leq 0,\forall x\in C\}$$

Let $C$  be a closed subset of a lcs $E$ and $x\in \mathcal{B}(C)$. Then the Normal cone $\mathfrak{N}_{C}(x)$ to $C$ at $x$ is the cone:$$\mathfrak{N}_{C}(x)=\mathfrak{T}_{C}(x)^{p}$$

Also we will use the other important notion of Normal Fan. Let $sp(C)$ the set of all points at which a closed set $C$ has support.

Let $C$  be a closed subset of a lcs $E$. Then the Normal Fan $NF(C)$ is the set:$$NF(C)=\cup \{\mathfrak{N}_{C}(x):x\in sp(C)\}$$

The following Theorem has an immediate proof and nevertheless in conjunction with the above Lemma \[properclosedcone\] and Theorem [CLOSUREPOINTEDCONE]{} it allows to state a major result concerning support for sets with void interior.

\[EXTREMEEQUIVPOINTED\]Consider a closed set $C$ in a lcs. Then a point $y\in \mathcal{B(}C)$ is extreme if and only if the cone $\mathcal{T}(y)$ is pointed.

Straightforward consequence of the definition of extreme point.

Let’s gather for locally convex spaces some important consequences of the analysis developed in this Section. The fact that the closure of a pointed cone is a proper cone (Theorem \[CLOSUREPOINTEDCONE\]) and Lemma [properclosedcone]{} (valid for locally convex spaces) imply that the polar cone of a pointed cone is always non-trivial. At this point we can apply Theorem \[EXTREMEEQUIVPOINTED\] to conclude that for a closed set (without interior) $\ $if a point in its boundary is extreme, then the tangent cone at that point is pointed and hence its polar, or normal cone at the point in question, is non-trivial. We state this result for sets without interior, since if there were a non-void interior, support would be insured anyway. We summarize the above discussion in the following

\[EXTREMEEQUALSUPPORT\]Consider a closed set without interior in a lcs. Then it has support at all points of its boundary that are extreme.

This is a first fundamental result on support for convex closed sets with void interior. It has been applied by the author to the Theory of Maximum Principle as outlined in the following

This result (Theorem \[EXTREMEEQUALSUPPORT\]) was stated in [DALEXCUBO2011]{}, where it was used in a Hilbert Space setting to demonstrate that the Maximum Principle holds for all candidate targets. We briefly sketch the meaning of this assertion. Consider a linear PDE in a Hilbert space setting and suppose that a target $\zeta $ is reachable from the origin in an interval $[0,T]$. Then that a minimum norm control $u_{o}$ (steering the origin to $\zeta $ in the time interval $[0,T]$) exists is an immediate consequence of the fact that the forcing operator $L_{T}$ is continuous and of the Projection Theorem. Let $\rho $ be the norm of $u_{o}$. The Maximum Principle holds if $\zeta $ is a support point for $L_{T}(S_{\rho })$ (where $S_{\rho }$ is the closed sphere of radius $\rho $ in the $L_{2}$ space of input functions), which is a closed convex set without interior in general. Now it is shown, in the cited paper, that $\zeta $ is always a support point for $L_{T}(S_{\rho })$ because $\zeta $ is always extreme, and hence Theorem \[EXTREMEEQUALSUPPORT\] applies. The fact that $\zeta $ is extreme is proved exploiting a property that all Hilbert space have, namely strict convexity. For details proofs and much more the interested reader is referred to the cited paper

 Support for sets with void interior in Hilbert Spaces
======================================================

We saw that the closure of a pointed cone is proper. What if the cone is not pointed? We will investigate this issue in the real Hilbert Space environment (in the sequel all Hilbert spaces are assumed to be real without further notice), and then, after developing the appropriate mathematical tools, we will introduce a completely general necessary and sufficient condition of support for closed set with void interior.

Naturally in a Hilbert space environment we can take advantage of the Riesz Theorem and so we can consider Normal Cones as subsets of the Hilbert space itself.

Let $C$  be a closed subset of a lcs $E$. Then the Applied Normal Fan $ANF(C)$ is the set:$$ANF(C)=\cup \{x+\mathfrak{N}_{C}(x):x\in sp(C)\}$$

To begin with we recall some relevant material from [@DALEXAJMAA2].

We start from an elementary Lemma.

\[sumorthosets\] Suppose that $F$  and $G$ are closed subspaces of Hilbert space $H$, that $\ F$ $\bot $ $G$, and that, for two non-void subsets $C$ and $D$ of $H$, $C\subset F$ and $D\subset G$. Then $C+D$ is closed if and only if both $C$ and $D$ are closed. Moreover:$$y=P_{F}y+P_{G}y\in C+D\Longleftrightarrow P_{F}y\in C\text{ and }P_{G}y\in D$$and if $C$ and/ or $D$ are not closed$$(C+D)^{-}=C^{-}+D^{-}$$

We can assume without restriction of generality that $G=F^{\bot }$ because, if this were not the case we can take in lieu of $H$ the Hilbert space $H_{1}=F+G$, which is in fact a closed subspace of $H$. Suppose that $C$ and $D$ be closed and consider a sequence $\{z_{i}\}$ in $C+D$ with $\{z_{i}\}\rightarrow z$. We can write in a unique way:$$z_{i}=P_{F}z_{i}+P_{G}z_{i}$$with $P_{F}z_{i}\in C$ and $P_{G}z_{i}\in D$. By continuity of projection and the assumption that $C$ and $D$ be closed , $\{P_{F}z_{i}\}\rightarrow
P_{F}z\in C$ and $\{P_{G}z_{i}\}\rightarrow P_{G}z\in D$, but since $P_{F}z+P_{G}z=z$ it follows $z\in C+D$ and we are done. Conversely suppose that, for example $D$ is not closed so that there exists a sequence $\{d_{i}\}$ in $D$, that converges to $d\notin D$, but, of course, $d\in G$. Take a vector $c\in C$. The sequence $\{c+d_{i}\}$ converges to $c+d$.  But $c=P_{F}(c+d)$ and $d=P_{G}(c+d)$, and therefore, by uniqueness of the decomposition, it is not possible to express $c+d$ as a sum of a vector in $C $ plus a vector in $D$. It follows that $c+d\notin C+D$. The second statement follows immediately from uniqueness of decomposition of a vector $y $ in the sum $P_{F}y+P_{G}y$, and since the proof of the last statement is immediate, we are done.

Next it is in order to recall from the same source the fundamental theorem on decomposition of non-pointed cones. We give a slightly expand the statement and include the proof for subsequent reference.

\[conedecompgeneral\]Consider a convex cone $C$ in a Hilbert space $H$ and assume that its lineality space be closed. Then $$(lin(C)^{\bot }\cap C)=P_{lin(C)^{\bot }}C$$where the cone $lin(C)^{\bot }\cap C$ is pointed. Consequently, if $C$ is closed the cone $P_{lin(C)^{\bot }}C$ is closed too. Moreover, the cone $C$ can be expressed as:$$C=lin(C)+(lin(C)^{\bot }\cap C)=lin(C)+P_{lin(C)^{\bot }}C$$Finally if a cone $C$  has the form $C=F+V$, where $F$ is a closed subspace and $V$ a pointed cone contained in $F^{\bot }$, then $F$ is its lineality space and the given expression coincides with the above decomposition of the cone.

First we prove that $$C=lin(C)+(lin(C)^{\bot }\cap C)$$That the rhs is contained in the lhs is obvious. Consider any vector $x\in C$ and for brevity let $\Gamma $ $=lin(C)$. Decompose $x$ as follows:$$x=x_{\Gamma }+x_{\Gamma ^{\bot }}  \label{ortdec}$$where $x_{\Gamma }\in \Gamma $ and $x_{\Gamma ^{\bot }}\in \Gamma ^{\bot }$. Next $x_{\Gamma ^{\bot }}=x-x_{\Gamma }$ as sum of two vectors in $C$ is in $C$ and hence in $\Gamma ^{\bot }\cap C$. Thus we have proved that the $lhs$ is contained in the $rhs$. Next we show that the cone $lin(C)^{\bot }\cap C$ is pointed.  Suppose that both a vector $x\neq 0$ and its opposite $-x$ belong to $lin(C)^{\bot }\cap C$ and decompose $x$ as above (\[ortdec\]). Because $lin(C)^{\bot }\cap C\subset lin(C)^{\bot }$, $x_{\Gamma }=0$, so that $x=x_{\Gamma ^{\bot }}\neq 0$. Do the same for $-x$, to conclude that $x_{\Gamma ^{\bot }}$  and $-x_{\Gamma ^{\bot }}$ are in $C$ (but obviously not in $lin(C)$). Because this is a contradiction, $lin(C)^{\bot }\cap C$ is pointed. Finally we prove that:$$lin(C)^{\bot }\cap C=P_{lin(C)^{\bot }}C$$In fact,$$P_{\Gamma ^{\bot }}(\Gamma ^{\bot }\cap C)=\Gamma ^{\bot }\cap C\subset
P_{\Gamma ^{\bot }}(C)$$On the other hand if $z\in P_{\Gamma ^{\bot }}(C)$, for some $w\in C$, $z=P_{\Gamma ^{\bot }}w=w-P_{\Gamma }w$  so that $z\in C$. Hence $z\in
\Gamma ^{\bot }\cap C$ and  so $P_{\Gamma ^{\bot }}(C)\subset \Gamma ^{\bot
}\cap C$. Finally, let $x\in C=F+V$ as defined in the last statement. Write $x=x_{F}+x_{F^{\bot }}$. According to Lemma \[sumorthosets\] $x_{F}\in F$ and $x_{F^{\bot }}\in V$. Now $-x=-x_{F}-x_{F^{\bot }}$ and, by the same Lemma, $-x\in C$ if and only if $-x_{F}\in F$ and $-x_{F^{\bot }}\in V$. But $V$ is assumed to be pointed and so $-x_{F^{\bot }}=0$, which shows that $F=lin(F+V)$. On the other hand the pointed cone of the decomposition is $P_{F^{\bot }}(F+V)=V$ and thus the proof is complete.

In general, and in particular for tangent cones, we cannot make any closedness assumption. Thus we need to develop a mathematical tool, which allow us to deal with the case non closed cones. We will tackle this problem momentarily, right after taking a short break to dwell on polarity.

Polarity can be viewed as the counterpart of orthogonality in the context of cone theory, as already seen from the following basic computations for polars. We do not pursue this parallel in depth here, limiting ourselves to what is needed here (for more details see [@DALEXAJMAA2]). Many of the following formulas are for generic sets, however we will be interested mostly in the special cases of polars of cones.

\[ELCOMPPOLARS\]The following formulas regarding polars hold where $C$ and $D$ are arbitrary sets $$(-C)^{p}=-C^{p}$$$$(C^{p})^{p}=C^{-}$$$$C^{-p}=C^{p}$$$$C\subset D\Rightarrow D^{p}\subset C^{p}$$$$(C+D)^{p}=C^{p}\cap D^{p}$$Moreover, if $F$ is a linear subspace (closed or non closed it doesn’t matter) then$$F^{p}=F^{\bot }$$

The polar cone of a convex cone is the normal cone at the origin to the given convex cone. Also, the polar cone of a closed convex cone is the set of all points in the space, whose projection on the cone, coincides with the origin.

We still have to handle the case where the lineality space of a cone is not closed, and here comes our solution for this issue.

In the next Theorem we exclude the case that the cone $C$ is contained in $lin(C)$, for in this case since the reverse inclusion always holds we have $C=linC$ that is, the cone is a linear subspace. Such case is trivial since we have only to ask, for our purposes, that the cone is not dense.

\[CLOSUREOFLINEALITY\]Consider a non pointed non closed cone $C$ and let $\Gamma $ be its lineality space, which is assumed to be not dense. We also assume that $C$ is not contained in $\Gamma $ and that $\Gamma $ is not closed. Consider the set$$\Psi =((C\backslash \Gamma ^{-})\cup \{0\})\cap \Gamma ^{\bot }\subset C$$Then $\Psi $ is a non-void pointed cone, and if $$\Delta \mathfrak{=}\Gamma ^{-}+\Psi$$then$$lin\Delta \mathfrak{=}\Gamma ^{-}$$(so that and Theorem \[conedecompgeneral\] applies to $\Delta $) and: $$C\subset \Delta \mathfrak{\subset }C^{-}$$ Moreover, $C$ and $\Delta $ have the same polar cone.

We first show that $\Psi $ is a cone. Referring to non-zero vectors to avoid trivialities, first of all it is obvious that $x\in \Psi $ implies $\alpha
x\in \Psi $ $\forall \alpha \geq 0$.  If $x,y\in \Psi $ then $z=x+y\in
\Gamma ^{\bot }$ and $z\in C$, and these two imply that $z\in \Psi $. Thus $\Psi $ is a cone. At this point we also know that $\Delta $ is a cone, because it is the sum of two cones. Next if there where two non-zero opposite vectors in $\Psi $ they would also be in $\ C\backslash \Gamma $ (since $C\backslash \Gamma ^{-}\subset C\backslash \Gamma $) contradicting that $\Gamma $ is the lineality space of $C$. Therefore $\Psi $ is pointed. Now we look at the intersection:$$(\Gamma ^{-}+\Psi )\cap (-\Gamma ^{-}-\Psi )=(\Gamma ^{-}+\Psi )\cap (\Gamma
^{-}-\Psi )$$with the aid of Lemma \[sumorthosets\]. Clearly, for a vector with a component in $\Psi $, to be in the intersection, it is required that we find two opposite vectors in $\Psi $, which is impossible because $\Psi $ is pointed. Thus a non-zero vector in the intersection can only be in $\Gamma
^{-}$ and so we have proved that $lin\Delta \mathfrak{=}\Gamma ^{-}$. As the first inclusion in the statement (namely $C\subset \Delta $), suppose $x\neq
0$ is in $C$.  Either $x\in \Gamma ^{-}$ and thus $x\in \Delta $ or $x\in
C\backslash \Gamma ^{-}$. In this latter case we write $x=x_{\Gamma
^{-}}+x_{\Gamma ^{\bot }}$. Now arguing as in the proof of Theorem [conedecompgeneral]{}, $P_{\Gamma ^{\bot }}(C)\subset \Gamma ^{\bot }\cap C$ (by the way this also shows that $\Psi $ is non-void) and so $x_{\Gamma
^{\bot }}\in \Gamma ^{\bot }\cap C$. Thus it must be $x_{\Gamma ^{\bot }}\in
\Psi $. So we have proved that in any case $x\in \Delta $ and so we have proved that $C\subset \Delta $. Next, applying Lemma \[sumorthosets\] to $\Delta $, we get:$$\Delta ^{-}=\Gamma ^{-}+\Psi ^{-}$$But $\Gamma \subset C\Rightarrow \Gamma ^{-}\subset C^{-}$ and, similarly, $\Psi \subset C\Rightarrow \Psi ^{-}\subset C^{-}$. Thus $\Delta ^{-}$ is the sum of two cones, both contained in $C^{-}$, and therefore we got $\Delta
^{-}\subset C^{-}$ and, a fortiori, $\Delta \mathfrak{\subset }C^{-}$. This completes the proof of the statement about inclusions. Now applying the elementary computations on polars we have:$$C^{-p}\subset \Delta ^{p}\subset C^{p}$$and since, again by the elementary computations, $C^{-p}=C^{p}$ it follows:$$C^{p}=\Delta ^{p}$$Thus the proof is finished.

Sometimes, when a cone is contained in a subspace, it is convenient to consider the polar of a cone within the subspace itself, regarded as the ambient space. When we do so we put a subscript, indicating the subspace, under the symbol of the polar.

We now establish another important result on polar cones.

\[GENERALPOLARLINCLOSED\] Suppose that a cone $C$ has the form $F+\Psi $, where $F$ is a closed proper subspace  and $\Psi $ is a pointed cone in $F^{\bot }$. Then$$C^{p}=\Psi _{F^{\bot }}^{p}$$Therefore the polar of the cone is contained in $F^{\bot }$ and in view of Theorem \[CLOSUREPOINTEDCONE\] cannot be trivial.

By direct computation. First of all (by elementary computations on polars) $C^{p}=F^{\bot }\cap \Psi ^{p}$, which implies $C^{p}\subset F^{\bot }$. Thus if $x\in C^{p}$ then $x=x_{F^{\bot }}$. Therefore for $y\in C$, the inequality $(x,y)\leq 0$ implies $x\in \Psi _{F^{\bot }}^{p}$, as we wanted to prove.

Applying the last two Theorems, we reach the conclusion contained in the statement of the following Theorem, which obviously requires no proof:

Consider a non pointed non closed cone $C$ and let $\Gamma $ be its lineality space, which is assumed to be not dense. We also assume that $C$ is not contained in $\Gamma $ and that $\Gamma $ is not closed. Then by Theorem \[CLOSUREOFLINEALITY\] it is true that:$$C^{p}=\Delta ^{p}$$where$$\Delta \mathfrak{=}\Gamma ^{-}+\Psi$$and $\Psi $ is the pointed cone contained in $\Gamma ^{\bot }$ defined in the statement of \[CLOSUREOFLINEALITY\]. Thus $$C^{p}=\Psi _{F^{\bot }}^{p}$$in view of Theorem \[GENERALPOLARLINCLOSED\]

The mathematical tools that we have been developing so far, put together, allow us to state the following fundamental Theorem on support for closed sets with void interior in Hilbert spaces.

Consider a closed set $C$ with void interior in a Hilbert space $H$. Then a point $y\in \mathcal{B}(C)$ has support if and only if the tangent cone $\mathfrak{T}_{C}(y)$ is either pointed (that is, $y$ is an extreme point) or is not pointed but its lineality space is not dense.

The proof is contained in the theory developed so far and does not require any additional effort.

Generalizing the Convexity Theorem to Sets without Interior
===========================================================

We recalled at the beginning the Convexity Theorem asserting that a closed body is convex if and only if each point of its boundary has support.

Here, sticking to the Hilbert space environment we remove the condition that the set has interior. Throughout the rest of the paper the ambient space is always understood to be a real infinite dimensional Hilbert space (with the only exception of our very last Theorem).

We start with a Lemma.

Consider a closed set $C$ without interior (so that $C=\mathcal{B}(C)$) and suppose $x\in C$ has support. Then for all the points of $x+\mathfrak{N}_{C}(x)$ the minimum distance problem from both $C$ and $\mathcal{C}^{-}(C)$ has the unique solution given by $x$.

By definition of normal cone, for any the point $z$ in $x+\mathfrak{N}_{C}(x) $ the minimum distance problem of $z$ from $x+\mathfrak{T}_{C}(x)^{-} $ has a unique solution given by $x$. Let $\delta =\Vert
z-x\Vert $. Now, since $C\subset \mathfrak{T}_{C}(x)^{-}$, $\Vert w-x\Vert
>\delta $, $\forall w\in C$. Therefore $x$, which is in $C$, is also the unique minimum distance solution of $z$ from $C$. Since $\mathcal{C}^{-}(C)\subset x+\mathfrak{T}_{C}(x)^{-}$ the same argument shows that $x$, which is also in $\mathcal{C}^{-}(C)$, is the unique minimum distance solution of $z$ from $\mathcal{C}^{-}(C).$This completes the proof .

Consider a closed set $C$ without interior (so that $C=\mathcal{B}(C)$) and suppose $x,y\in C$ (with $x\neq y$) have support. Then $x+\mathfrak{N}_{C}(x)\cap y+\mathfrak{N}_{C}(y)=\phi $.

In fact if the two translated normal cones had a non-void intersection, then for points in the intersection the minimum distance from $C$ problem would have a non unique solution. But this in view of the preceding Lemma is a contradiction and therefore our statement is proved.

The last Theorem and Corollary show that the support condition can arguably be viewed as local. There may be a closed set that at a single point, or in a whole region, behaves locally like a convex set, although it is not convex at all.

The two above results imply the following:

Consider a closed set $C$ without interior and suppose the set $sp(C)$ of all support points of $C$ is such that the set $ANF(sp(C))$ covers $\overline{C}$. Then the projection Theorem hold good for $C$.

This line of arguing makes it possible to state the following generalization of the Convexity Theorem for closed sets without interior.

\[VOIDINTERIORCONVEXITYTHEOREM\]Consider a closed set $C$ without interior. Then $C$ is convex if and only if the set $sp(C)$ of all support points of $C$ is such that the set $ANF(sp(C))$ covers $\overline{C}$. And if this is the case $C$ is equal to the intersection of the family of closed supporting semispaces of $C$.

The only if is an immediate consequence of the Projection Theorem. Conversely let $\Psi $ be the intersection of semispaces defined in the statement of the Theorem, which is obviously a closed convex set containing $C$. Suppose there exists $y\in \Psi $ such that $y\notin C$. Then by the preceding Theorem we can project $y$ on $C$. The unique projection $z\in C$ has support because $y-z$ is in the normal cone to $C$ at $z$ and so $(y-z,w=z)\leq 0$, $\forall w\in C$. But then the functional $(y-z,.)$ leaves $y$ in an open semispace opposite to the closed one containing $C$, thereby contradicting that $y\in \Psi $. Thus $\Psi =C$ and we are done.

The Bounded Case in Hilbert Spaces and Hypersurfaces
====================================================

For the case where $C$ is bounded body in a Hilbert space, we will study convex non-smooth hypersurfaces. We will prove, in our infinite dimensional and non-smooth setting, that there exists a homeomorphism of the boundary of any convex bounded body onto the boundary of the sphere . Thus the boundary of any convex bounded body is what we will define to be a convex hypersurface.

Our final Theorem in the Section will state a sort of converse of this result, showing that if we consider a convex bounded hypersurface, this is exactly the boundary of both its convex and its closed convex extension (assumed to have interior), and that these two sets are the same.

We note that we will make no use of the weak topology and compactness, since weak topology seems to be of no help in the present matter. Also notice that in our approach we have a *global* description of a hypersurface as the range of a non-linear function (the homeomorphism) defined on a set that we can fix to be the boundary of the unit sphere. In this sense we talk of a (non-linear) range space theory. We will comment at the end on the fact that a linear range space theory can also be used for the same purposes.

We define now the closed hypersurface, where closed stands for “without boundary”. The word closed will be tacitly understood in the sequel to avoid confusion with topological closedness (and actually an hypersurface is assumed to be closed).  On the other hand we will not deal here with hypersurfaces with boundary.

In finite dimension the definition can assume that there is a one to one continuous mapping on the surface of a sphere onto the hypersurface. Then the hypersurface is actually homeomorphic to the surface of the sphere for free, since there is a Theorem in [@KELLEY55], stating that a continuous one to one function on a compact set to a Hausdorff topological space has a continuous inverse. Thus in the present infinite dimensional case it seems natural to assume a homeomorphism from scratch.

Also we will deal with the case where convex sets have interior and this seems intuitively founded just in view of such homeomorphism to the boundary of a convex body like the sphere.

A (Jordan) hypersurface $\Sigma $ in a Hilbert space is a set which is homeomorphic (in the strong topology sense) to the boundary of a closed sphere (and therefore it is a closed set). A hypersurface is convex if it has support at each of its points.

Our first aim is to show that if we consider a bounded convex body $C$ then its boundary $\mathcal{B}(C)$ is a bounded convex hypersurface.

\[BOUNDEDCONVEXBODYVSHYPERSURF\] Let $C$ be a bounded convex body. Then its boundary $\mathcal{\ }$of $C$ is a bounded convex hypersurface.

The proof is constructive in the sense that we will exhibit an appropriate homeomorphism. First of all, as usual, we may assume that $0\in C^{i}$ and so we may consider a closed sphere $S_{r}$ around the origin of radius say $r>0$, entirely contained in $C^{i}$. Each ray $\rho $ emanating from the origin meets obviously the boundary of $S_{r}$ (which we denote by $\Phi $)  in a unique point, but also meets $\mathcal{B}(C)$ (which we denote by $\Omega )$ in another unique point thanks to Theorem \[RAYSINTERSECTION\]. Now define by $\varphi $ the map that associates each point $y$ of $\Phi $ with the unique point $\varphi (y)$ which is the intersection of the ray generated by $y$ with $\Omega $. The map $\varphi $ is obviously onto, since each point of $\Omega $ generates a ray on its own. It is also one to one because distinct rays meet only at the origin, and so two different values of $\varphi $ in $\Omega $ can only be generated by distinct vectors in $\Phi $. Next we show that the map is continuous. In fact consider any point $z\in \Omega $. We can define a base of the neighborhood system of $z$, intersecting closed spheres around $z$ with $\Omega $.  Consider one of those spheres $\Sigma _{\varepsilon }$ with radius $\varepsilon >0$. Notice that $\mathcal{C}o(\Sigma _{\varepsilon })\supset \Sigma _{\varepsilon }$. Now if we take the cone generated by a closed sphere around $r\frac{z}{\Vert
z\Vert }$ of radius $\frac{\varepsilon }{2}\frac{r}{\Vert z\Vert }$ this cone will contain a sphere of radius $\varepsilon /2$ around $z.$This proves continuity of $\varphi $. On the other hand $\varphi ^{-1}(z)=\frac{z}{\Vert
z\Vert }$, and since $\Vert z\Vert $ is bounded from above and from below on $\Omega $, $\varphi ^{-1}$ is continuous too. This completes the proof.

Next we tackle the problem of stating a result going in the reverse direction, that is, from the convex hypersurface to the appropriate convex body.

\[HYPERSURFASBOUNDCONVBODY\] Suppose $\Omega $ is a bounded convex hypersurface and assume that the (necessarily bounded) set $\mathcal{C}^{-}(\Omega )$ has interior. Then $$\Omega =\mathcal{B}(\mathcal{C}^{-}(\Omega ))$$Moreover, (in view of Corollary \[PLAINCEXTWITHOUTCOMPACT\])$$\mathcal{C}^{-}(\Omega )=\mathcal{C}(\Omega )$$and so also $$\Omega =\mathcal{B}(\mathcal{C}(\Omega ))$$

We know from the preceding Theorem that $\mathcal{B}(\mathcal{C}^{-}(\Omega
))$ is a bounded convex hypersurface and elementary computations in [KNLTS]{} show that $\mathcal{C}^{-}(\Omega )^{i}$ is a convex set. Furthermore, because each point of $\Omega $ is a support point for $\Omega $ itself and hence also for $\mathcal{C}^{-}(\Omega )$, it follows that $\Omega \subset $ $\mathcal{B}(\mathcal{C}^{-}(\Omega ))$.  As usual we assume without restriction of generality that $0\in \mathcal{C}^{-}(\Omega
)^{i}$. In order to simplify notations in what follows, and again without restriction of generality, we assume that $\mathcal{C}^{-}(\Omega )^{i}$ contains a closed sphere of radius $2$ around the origin (if not we multiply everything by a positive scalar factor and nothing changes for the purposes of the present proof). From our previous theory we know that each ray emanating from the origin meets $\mathcal{B}(\mathcal{C}^{-}(\Omega ))$ (and hence possibly $\Omega \subset $ $\mathcal{B}(\mathcal{C}^{-}(\Omega )$) in a unique point. Let $\varphi $ be the homeomorphism of $\mathcal{B}(S_{2})$ onto $\Omega $. What we want to do now is to extend this homeomorphism to an homeomorphism on the whole of $S_{2}$ onto the set $[0:1]\Omega $. This will be done in two pieces: first we extend it to $S_{2}\backslash S_{1}^{i}$ and then we will take the identity map on $\ S_{1}^{i}$. We call $\psi $ the extended map. Bear in mind that$$M\geq \Vert z\Vert \geq m>2\text{, }\forall z\in \Omega$$for some $M>m$. Thus also:$$M\geq \Vert z\Vert \geq m>2\text{, }\forall z\in \mathcal{C}^{-}(\Omega )$$ For $y\in S_{2}\backslash S_{1}^{i}$ to simplify notations we put:$$\gamma (y)=\varphi (2\frac{y}{\Vert y\Vert })$$which is clearly a continuous function. Then on $S_{2}\backslash S_{1}^{i}$, $\psi $ is defined by:$$\psi (y)=(\Vert y\Vert -1)\gamma (y)+(2-\Vert y\Vert )\frac{y}{\Vert y\Vert }$$In this way when $y\in \mathcal{B}(S_{2})$ or $\Vert y\Vert =2$, we have that $\psi (y)=\gamma (y)$. When $y\in \mathcal{B}(S_{1})$ or $\Vert y\Vert
=1$, then $\psi (y)=y$. In between we have a convex combination of these two vectors. In the rest of  $S_{2}$, $\psi $ is defined to be the identity map. Now it is readily seen that $\psi $ is one to one and the inverse function  on $([0:1]\Omega )\backslash S_{1}$is:$$\psi ^{-1}(z)=\frac{z}{\Vert z\Vert }(1+\frac{\Vert z\Vert -1}{\Vert \gamma
(z)\Vert -1})$$It is easily verified by inspection that both $\psi $ and $\psi ^{-1}$ are continuous. Thus $\psi $ is an homeomorphism and it maps $S_{2}$ onto the set $[0:1]\Omega $. Because its range is bounded, the image of $S_{2}^{i}$ , which is an open set containing the origin and whose boundary is $\Omega $, must be bounded. Thus each ray emanating from the origin must meet $\Omega $ and so, in view of Corollary \[PLAINCEXTWITHOUTCOMPACT\] it must be $\Omega =$ $\mathcal{B}(\mathcal{C}^{-}(\Omega ))$. This completes the proof .

There is another sort of range space theory, developed in [@DALEXAJMAA2]) that still studies convex structures, but using the range of a linear (instead of non-linear) continuous functions, that is, of operators. We know that a convex closed set is the intersection of closed half spaces. Now in a separable Hilbert space environment (which may be represented by the space $l_{2}$) it is well known that we can limit the intersection to a countable number of semispaces. Under mild assumption the corresponding countable set of inequalities takes on the form $Lx\leq v$ where $L$ is an operator and $v$ is a vector (called the bound vector). The range space point of view consists at looking at this equation from the side of the of the range $\mathcal{R}(L)$ of $L$, instead of the side of the unknown $x$. Just to give a taste of it, from the range space point of view, this inequality has solution if and only if $v\in \mathcal{R}(L)+P$, where $P$ is the closed pointed cone of all non-negative (componentwise) vectors (which, incidentally has void interior!). Note that $\mathcal{R}(L)+P$ is a cone albeit a non pointed one. With this range space approach an extensive theory of geometry, feasibility, optimization and approximation has been developed. For details, proofs and more the interested reader may look at the paper cited above.

Convexification of Hypersurfaces
================================

Intuitively, what is meant here by convexification of a bounded hypersurface, can be illustrated colloquially and in three dimension by this image: wrap a non-convex bounded hypersurface with a plastic kitchen pellicle and the then pellicle takes the form of a surface, which is the convexification of the original surface. Here is a formal definition

Consider a closed bounded hypersurface $\Phi $ in a a real Hilbert space and assume that $\mathcal{C}^{-}(\Phi )=\mathcal{C}(\Phi )$ (where the equality follows from Corollary \[PLAINCEXTWITHOUTCOMPACT\]) has non-void interior. Then $\Omega =\mathcal{B(C}(\Phi ))$, which by Theorem [BOUNDEDCONVEXBODYVSHYPERSURF]{} is a bounded convex hypersurface, is called the convexification of $\Phi $.

Naturally the convexification of a bounded convex hypersurface satisfying the above condition is the surface itself. So to speak, convexity is the fixed point of the convexification operator.

A straightforward consequence of the work carried out so far is the following:

Let $\Phi $ be a closed bounded hypersurface in a real Hilbert space and let $\Omega $ its convexification. Then $\Phi $ is homeomorphic to $\Omega $.

Lets indicate by the symbol $\sim $ the relation “is homeomorphic to” and by $S$ the closed unit sphere around the origin. Then we know that, by definition, $\Phi \sim \mathcal{B}(S)$. But by Theorem [BOUNDEDCONVEXBODYVSHYPERSURF]{} we also know that $\Omega \sim \mathcal{B}(S)$. Because $\sim $ is an equivalence relation, it follows also $\Omega \sim
\Phi $.

The above (and recurring) hypothesis that $\mathcal{C}(\Phi )$ has interior is not needed in finite dimension. In fact we state the following Theorem. By the way, recall that in finite dimension, from the only fact that $\Phi $ is compact it follows that $\mathcal{C}(\Phi )$ is compact too.

\[MAINTHCLOSEDSURFACE\] Let $\Omega $ be a compact convex hypersurface in $R^{n}$. Then $\mathcal{A}(\Omega )=R^{n}$, where $\mathcal{A}(\Omega )$ is the affine extension of $\Omega $. Thus $\mathcal{C}(\Omega )$ is a convex body.

In finite dimension every convex set has relative interior. Thus we can argue momentarily in $E=\mathcal{L}(\mathcal{C}(\Omega ))$, where $\mathcal{C}(\Omega )$ has interior. In this space, in view of Theorem [HYPERSURFASBOUNDCONVBODY]{} we can say that:$$\Omega =\mathcal{B}(\mathcal{C}^{-}(\Omega ))$$$$\mathcal{C}^{-}(\Omega )=\mathcal{C}(\Omega )$$and $$\Omega =\mathcal{B}(\mathcal{C}(\Omega ))$$Next, as we did before we extend the homeomorphism $\varphi $ of $\mathcal{B}(S)$ onto $\Omega $ to the whole of $S$. Thus $\Pi =\psi (S^{i})$ is an open set and $\Omega $ is its boundary. We intersect each coordinate axis with $\Pi $ to obtain a relatively open subset and hence a segment of the form $[0:z^{i})$ with $z^{i}\neq 0$ and laying on the $ith$ coordinate axis. But then $z^{i}$ is in the boundary of $\ \Pi $, that is, in $\Omega $. Thus $\mathcal{A}(\Omega )=R^{n}$ and so we established our thesis.

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