The intrinsic core and minimal faces of convex sets in general vector spaces

The part (ii) is straightforward: since $0$ must belong to every linear subspace, if $0\notin K$, then no linear subspace is contained in $K$. Likewise, if $0\in K$, then $\{0\}$ is a trivial linear subspace contained in $K$, and the lineality space can only be larger than that.

The necessary part of ((i)) is evident: assuming $x\in\operatorname{lin}K\subseteq K$$x\in\operatorname{linspace}K\subseteq K$, and recalling that $\operatorname{lin}K$ $\operatorname{linspace}K$ is a linear subspace, we must have $-x\in\operatorname{lin}K$$-x\in\operatorname{linspace}K$.

To show the sufficiency, that is, if $x,-x\in K$, then $x\in\operatorname{lin}K$$x\in\operatorname{linspace}K$, first observe that it follows from convexity that $0\in[-x,x]\subset K$, and hence by (i) $\operatorname{lin}K\neq\emptyset$$\operatorname{linspace}K\neq\emptyset$. Assume that nevertheless $x\notin\operatorname{lin}K$. Let $x\notin\operatorname{linspace}K$. Let

$$L=\operatorname{span}({\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\sout{\operatorname{lin}}}{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\uwave{\operatorname{linspace}}}K\cup\{x\}).$$

For any $w\in L$ we have $w=v+u$, with $v\in\operatorname{lin}K\subseteq K$ $v\in\operatorname{linspace}K\subseteq K$ and $u=\lambda x$ for some $\lambda\in\mathbb{R}$. Now $\lambda x\in K$, irrespective of the sign of $\lambda$, since $K$ is a cone and both $x$ and $-x$ are in $K$. Now $u,v\in K$, hence $w=u+v\in K$ (as $u+v=2(\frac{1}{2}u+\frac{1}{2}v)\in K$ by convexity and homogeneity), and we have demonstrated that $L$ is a linear subspace contained in $K$ that is strictly larger than $\operatorname{lin}K$$\operatorname{linspace}K$, which is impossible. ∎

Let $a,b,u,v\in X$, and assume that $a\in(u,v)$ and $c\in(a,b)$ (see Fig. 2).

There are $\alpha,\beta\in(0,1)$ such that

$$a=\alpha u+(1-\alpha)v,\quad c=\beta a+(1-\beta)b.$$

(1)

We have from (1)

	$\displaystyle c$	$\displaystyle=\beta(\alpha u+(1-\alpha)v)+(1-\beta)b$
		$\displaystyle=\alpha\beta u+(1-\alpha\beta)\left(\frac{\beta(1-\alpha)}{(1-\alpha\beta)}v+\frac{1-\beta}{(1-\alpha\beta)}b\right).$		(2)

Let

$$w=\frac{\beta(1-\alpha)}{1-\alpha\beta}v+\frac{1-\beta}{1-\alpha\beta}b.$$

It is evident that $w\in(b,v)$, while from (2.1) we have $c\in(u,w)$. ∎

We first show that

$$C\cap({\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\sout{\operatorname{lin}}}{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\uwave{\operatorname{linspace}}}\operatorname{cone}(C-x)+x)\subseteq F_{\min}(x,C).$$

Let $y\in C\cap(\operatorname{lin}\operatorname{cone}(C-x)+x)$$y\in C\cap(\operatorname{linspace}\operatorname{cone}(C-x)+x)$. Then $y=x+u$ for some $u\in\operatorname{lin}\operatorname{cone}(C-x)$$u\in\operatorname{linspace}\operatorname{cone}(C-x)$, and by Proposition 2.3 (i) we have $-u\in\operatorname{lin}\operatorname{cone}(C-x)\subseteq\operatorname{cone}(C-x)$$-u\in\operatorname{linspace}\operatorname{cone}(C-x)\subseteq\operatorname{cone}(C-x)$. This means that there exists some $t>0$ such that $x-tu\in C$. We have

$$\frac{t}{1+t}(x+u)+\frac{1}{1+t}(x-tu)=x,$$

hence $x\in(x+u,x-tu)$ and for every face $F$ of $C$ containing $x$ we must have $y=x+u\in F$. We conclude that

$$y\in\bigcap_{\begin{subarray}{c}F\unlhd C\\ x\in F\end{subarray}}F=F_{\min}(x,C).$$

It remains to show the converse, that

$$F_{\min}(x,C)\subseteq C\cap({\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\sout{\operatorname{lin}}}{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\uwave{\operatorname{linspace}}}\operatorname{cone}(C-x)+x).$$

Since $F_{\min}(x,C)\subseteq C$, it is then sufficient to show that

$$F_{\min}(x,C)=F_{\min}(x,C)\cap({\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\sout{\operatorname{lin}}}{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\uwave{\operatorname{linspace}}}\operatorname{cone}(C-x)+x).$$

(3)

If we prove that

$$F=F_{\min}(x,C)\cap({\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\sout{\operatorname{lin}}}{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\uwave{\operatorname{linspace}}}\operatorname{cone}(C-x)+x)$$

(4)

is a face of $C$, then from $x\in F$, we must have $F\subseteq F_{\min}(x,C)$, hence $F=F_{\min}(x,C)$, and so (3) holds.

Let $y\in F$ and $u,v\in C$ and $\alpha\in(0,1)$ be such that

$$y=(1-\alpha)u+\alpha v.$$

(5)

Our goal is to show that $u\in F$.

Let $p=y-x$. Since $y\in\operatorname{lin}\operatorname{cone}(C-x)+x$$y\in\operatorname{linspace}\operatorname{cone}(C-x)+x$, we have $p\in\operatorname{lin}\operatorname{cone}(C-x)$$p\in\operatorname{linspace}\operatorname{cone}(C-x)$, and by Proposition 2.3 (i) we have $-p\in\operatorname{lin}\operatorname{cone}(C-x)\subseteq\operatorname{cone}(C-x)$$-p\in\operatorname{linspace}\operatorname{cone}(C-x)\subseteq\operatorname{cone}(C-x)$. There exists $t>0$ such that $z:=x-tp\in C$. Applying Proposition 2.5 to the points $y,z,u,v$ and $x$ (cf. Figs. 2 and 3),

we conclude that there exists some $w\in(v,z)$ such that $x\in(u,w)$, which yields $[u,w]\subseteq F_{\min}(x,C)$, hence, $u-x,-(u-x)\in\operatorname{cone}(C-x)$, so $u-x\in\operatorname{lin}\operatorname{cone}(C-x)$$u-x\in\operatorname{linspace}\operatorname{cone}(C-x)$. We conclude that $u\in\operatorname{lin}\operatorname{cone}(C-x)+x$$u\in\operatorname{linspace}\operatorname{cone}(C-x)+x$. Hence, from (4) we conclude that $u\in F$.

∎

It is evident that the set on the right-hand side is a subset of any face that contains $x$, hence, it must be contained in the minimal face of $x$ in $C$ as well. It remains to show that the minimal face of $x$ in $C$ is a subset of the right-hand side. Let $y\in F_{\min}(x,C)$. Then by Proposition 2.4 we have

$$y\in{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\sout{\operatorname{lin}}}{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\uwave{\operatorname{linspace}}}\operatorname{cone}(C-x)+x,$$

equivalently $y-x\in\operatorname{lin}\operatorname{cone}(C-x)$$y-x\in\operatorname{linspace}\operatorname{cone}(C-x)$. This means that $x-y\in\operatorname{lin}\operatorname{cone}(C-x)\subseteq\operatorname{cone}(C-x)$$x-y\in\operatorname{linspace}\operatorname{cone}(C-x)\subseteq\operatorname{cone}(C-x)$, and so there exists some $t>0$ such that $t(x-y)\in C-x$, equivalently $x+t(x-y)\in C$. It remains to note that

$$x=\frac{t}{1+t}y+\frac{1}{1+t}(x+t(x-y)),$$

hence $x\in(y,x+t(x-y))$, and so $y$ belongs to the right-hand side of (6). ∎

First observe that $E$ is a convex subset of $C$: every point $x\in E$ belongs to some face of $C$, and hence belongs to $C$; furthermore, for any $u,v\in E$ there exist $F_{u},F_{v}\in\mathcal{F}$ such that $u\in F_{u}$ and $v\in F_{v}$. Since $\mathcal{F}$ is a chain, without loss of generality we can assume that $F_{u}\subset F_{v}$, and hence $u,v\in F_{v}$. Therefore, $[u,v]\subseteq F_{v}\subseteq C$, and so $C$ is convex.

Let $x\in E$, and let $y,z\in C$ be such that $x\in(y,z)$. Since $E$ is a union of faces of $F$, there must be a face $F_{x}\in\mathcal{F}$ such that $x\in F_{x}$. Since $F_{x}$ is a face of $C$, we have

$$[y,z]\subseteq F_{x}\subseteq E,$$

and so $E$ is a face of $C$. ∎

It is evident from the definition of the affine hull that if $E\subseteq F$, then $\operatorname{aff}E\subseteq\operatorname{aff}F$. It is hence sufficient to show that if $E\subsetneq F$ then there exists some $u\in\operatorname{aff}F\setminus\operatorname{aff}E$. Assume the contrary, then in particular $F\subseteq\operatorname{aff}E$. Since $E\subsetneq F$, there exists $x\in F\setminus E$, which can be represented as

$$x=\sum_{i\in I}\alpha_{i}u_{i},\qquad\sum_{i\in I}\alpha_{i}=1,\quad u_{i}\in{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\sout{C}}{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\uwave{E}}\quad\forall i\in I,\;|I|<\infty.$$

By convexity

$$y=\sum_{i\in I}\frac{1}{|I|}u_{i}\in E.$$

Now there must exist a $t>0$ such that

$$\frac{1}{|I|}\geq t\left(\alpha_{i}-\frac{1}{|I|}\right)\quad\forall i\in I.$$

Let

$$\lambda_{i}=\frac{1}{|I|}-t\left(\alpha_{i}-\frac{1}{|I|}\right),\quad i\in I.$$

Since $\lambda_{i}>0$ for all $i\in I$ and

$$\sum_{i\in I}\lambda_{i}=1,$$

we have

$$z=y+t(y-x)=\sum_{i\in I}\lambda_{i}u_{i}\in E\subset C,$$

and since $y\in E\subseteq C$ and $y\in(x,z)\subset C$, by the definition of a face we must have $x\in E$, which contradicts our assumption. ∎

Let $x,y\in\operatorname{icr}C$ and suppose that $z\in(x,y)$. Then for any $u\in C$ there exist $v,w\in C$ such that $x\in(u,v)$ and $y\in(u,w)$. Then we have for some $s,t,r\in(0,1)$

$$x=u+s(v-u),\quad y=u+t(w-u),\quad z=x+r(y-x).$$

Substituting the expressions for $x$ and $y$ into the one for $z$, we obtain

$$z=u+(s(1-r)+tr)(q-u),$$

where

$$q=\frac{s(1-r)}{s(1-r)+tr}v+\frac{t}{s(1-r)+tr}w\in C$$

by convexity (see Fig. 5).

Hence, $z\in(u,q)$ with $q\in C$, and by the arbitrariness of the choice of $u$ we have $z\in\operatorname{icr}C$. ∎

Consider some $x\in\operatorname{icr}C$ and $y\in C$. Fix an arbitrary $q\in(x,y)$. For any $u\in C$ there exists $v\in C$ such that $x\in(u,v)$ by the definition of the intrinsic core. Applying Proposition 2.5 to $x,y,u,v$ and $q$, we deduce that there exists $w\in(v,y)$ such that $q\in(u,w)$. Since $v,y\in C$, by convexity $w\in C$. Then $q\in\operatorname{icr}C$ by the definition of the intrinsic core. We have hence shown that $[x,y)\subseteq\operatorname{icr}C$. Now since $x\in\operatorname{icr}C$ and $y\in C$, there must be some $z\in C$ such that $x\in(z,y)$. Then by the earlier proven statement we must have $[x,z)\subset\operatorname{icr}C$. Hence $x\in(z,y)\subseteq\operatorname{icr}C$. ∎

Observe that $\operatorname{icr}\operatorname{icr}C\subseteq\operatorname{icr}C$, so we only need to prove the converse. Now take any $x\in\operatorname{icr}C$. Then for any $y\in\operatorname{icr}C$ we also have $y\in C$. Since $x\in\operatorname{icr}C$ and $y\in C$, by Proposition 3.4 there must be $z\in\operatorname{icr}C$ such that $x\in(z,y)\subseteq\operatorname{icr}C$. This shows that $x\in\operatorname{icr}\operatorname{icr}C$. ∎

Under the assumptions of the proposition, suppose that $x\in\operatorname{icr}S$. Then by Proposition 3.4 for any $y\in S$ there exists $z\in S$ such that $x\in(y,z)$. Since $S\subset C$, Corollary 2.6 then yields $S\subseteq F_{\min}(x,C)$. Hence the minimal face of $S$ must be a subset of $F_{\min}(x,C)$, however there can be no smaller face containing $x$ (and hence $S$), therefore, $F_{\min}(x,C)=F_{\min}(S,C)$.

Now for any $x\in\operatorname{icr}S$ we have $x\in\operatorname{icr}F_{\min}(x,C)=F_{\min}(S,C)$, hence, $\operatorname{icr}S\subseteq\operatorname{icr}F_{\min}(S,C)$. ∎

Since $C$ is a convex set, the set $\operatorname{cone}(C-x)$ is a linear subspace if and only if $\operatorname{cone}(C-x)=-\operatorname{cone}(C-x)$ (additivity holds naturally), equivalently for every $y=t(u-x)$, where $t>0$ and $u\in C$ there exists $s>0$ and $v\in C$ such that $-y=s(v-x)$. This is in turn equivalent to having, for each $u\in C$, the existence of $v\in C$ and $r=s/t>0$ such that $u-x=-r(v-x)$, which is equivalent to

$$x=\frac{1}{1+r}u+\frac{r}{1+r}v\in(u,v).$$

∎

Let $x\in C$ and suppose (i) holds. Take any $y\in\operatorname{aff}C$. Then

$$y=\sum_{i=1}^{m}\lambda_{i}u_{i}+\lambda_{0}x,\quad\sum_{i\in I}\lambda_{i}+\lambda_{0}=1,\quad u_{i}\in C\setminus\{x\}\;\;\forall\,i\in\{1,\dots,m\}.$$

(8)

If $m=0$, then $y=x$ and (ii) holds with $z=x=y$. Otherwise for any $i\in\{1,\dots,m\}$ there exists $v_{i}\in C$ such that $x\in(u_{i},v_{i})$; explicitly, there is some $t_{i}\in(0,1)$ such that

$$x=t_{i}u_{i}+(1-t_{i})v_{i}.$$

(9)

Now let

$\displaystyle\alpha_{i}:=\begin{cases}\displaystyle\sum_{\begin{subarray}{c}\lambda_{i}<0,\\ i\geq 1\end{subarray}}\frac{\lambda_{i}}{t_{i}}+\lambda_{0},&i=0\\ \lambda_{i},&\lambda_{i}\geq 0,i\geq 1,\\ \frac{(t_{i}-1)\lambda_{i}}{t_{i}},&\lambda_{i}<0,i\geq 1,\\ \end{cases}\qquad w_{i}=\begin{cases}u_{i},&\lambda_{i}\geq 0,\\ v_{i},&\lambda_{i}<0.\end{cases}$

It is easy to see using (8) and (9) that the affine representation of $y$ can be rewritten in this new notation as

$$y=\sum_{i=1}^{m}\alpha_{i}w_{i}+\alpha_{0}x.$$

(10)

Here $\sum_{i=1}^{m}\alpha_{i}+\alpha_{0}=1$, $\alpha_{i}\geq 0$ and $w_{i}\in C$ for $i\in\{1,\dots,m\}$ Now if $\alpha_{0}\geq 0$, then $y\in C$, hence $[x,y]\subseteq C$ and for any $z\in(x,y)$ we have $[x,z]\subseteq C$. Otherwise (if $\alpha_{0}<0$) let

$$z:=\frac{1}{1-\alpha_{0}}y+\frac{-\alpha_{0}}{1-\alpha_{0}}x.$$

Observe that $z\in(x,y)$ and also substituting the expression for $y$ from (10) we have

$\displaystyle z$

$\displaystyle=\sum_{i=1}^{m}\frac{\alpha_{i}}{1-\alpha_{0}}w_{i}.$

It is evident that $z\in C$, and therefore $[x,z]\subseteq C$. We have shown that (i) yields (ii). It remains to show the converse.