Supporting direction at a boundary point of a finite full-dimensional polytope

Enumerate $X$ as $x_1,\dots ,x_m$ and, for $\lambda \in {\mathbb{R}}^m$, write ${\lambda }_j:=\lambda (j-1)$ for $1\le j\le m$. Let $\mathrm{\Delta }=\{\lambda \in [0,1{]}^m:{\sum }_{j=1}^m{\lambda }_j=1\}$. The set $\mathrm{\Delta }$ is nonempty (take ${\lambda }_1=1$ and all other coordinates zero) and bounded. Its coordinate maps and sum map are continuous, since each coordinate inequality $\mathrm{\mid }{\lambda }_j-{\lambda }_j^{\mathrm{\prime }}\mathrm{\mid }\le \mathrm{\parallel }\lambda -{\lambda }^{\mathrm{\prime }}{\mathrm{\parallel }}_2$ follows from the norm formula [F9], while $\mathrm{\mid }{\sum }_j({\lambda }_j-{\lambda }_j^{\mathrm{\prime }})\mathrm{\mid }\le m\mathrm{\parallel }\lambda -{\lambda }^{\mathrm{\prime }}{\mathrm{\parallel }}_2$ follows from the real triangle inequality; hence $\mathrm{\Delta }$, an intersection of their closed inverse images, is closed. By [F1], $\mathrm{\Delta }$ is compact. For $\lambda ,{\lambda }^{\mathrm{\prime }}\in \mathrm{\Delta }$, the triangle inequality and coordinate bound give ${\parallel {\sum }_j{\lambda }_{j}T(x_j)-{\sum }_j{\lambda }_j^{\mathrm{\prime }}T(x_j)\parallel }_2\le {\sum }_j\mathrm{\mid }{\lambda }_j-{\lambda }_j^{\mathrm{\prime }}\mathrm{\mid }\mathrm{\parallel }T(x_j){\mathrm{\parallel }}_2\le \mathrm{\parallel }\lambda -{\lambda }^{\mathrm{\prime }}{\mathrm{\parallel }}_2{\sum }_j\mathrm{\parallel }T(x_j){\mathrm{\parallel }}_2,$ so the map $\lambda \mapsto {\sum }_j{\lambda }_{j}T(x_j)$ is continuous. Thus [F2] makes its image $K$ compact; [F3] makes $K$ closed.

For each natural $r\le k$, define $P_r:=\{f:r\to \mathbb{Q}\}$. The set $P_0=\{\mathrm{\varnothing }\}$ is finite. Restriction and the last coordinate give a bijection $P_{r+1}\to P_r\times \mathbb{Q}$, namely $f\mapsto (f{\mathrm{\mid }}_r,f(r))$; its inverse extends $g:r\to \mathbb{Q}$ by the specified last value. By induction on $r$, the binary-product result [F6] makes every $P_r$ at most countable. Since $P_k$ contains the zero function, [F6] supplies one surjection $q:\mathbb{N}\to P_k={\mathbb{Q}}^k$. Embed its values coordinatewise in ${\mathbb{R}}^k$ using [F7] and write them as $q_j$. These vectors are dense: for $p\in {\mathbb{R}}^k$ and $\epsilon >0$, each interval $(p(i)-\epsilon \mathrm{/}(2k),p(i)+\epsilon \mathrm{/}(2k))$, $i<k$, contains a rational by [F7]. Select these finitely many coordinates by induction on $i<k$, obtaining $a\in P_k$. Then $\mathrm{\parallel }a-p{\mathrm{\parallel }}_2^2<k(\epsilon \mathrm{/}(2k){)}^2<{\epsilon }^2$ by [F9], so some $q_j=a$ lies in $B(p,\epsilon )$. Only finite induction and a single enumeration are used; no countable family of selections is made.

Fix an integer $n\ge 1$. Noninteriority of $\mu$ gives a point $p\mathrm{\notin }K$ with $\mathrm{\parallel }p-\mu {\mathrm{\parallel }}_2<1\mathrm{/}(2n)$. Since $K$ is closed, its complement is open, so some ball $B(p,\rho )$ of radius $\rho >0$ lies outside $K$. Set $r=\frac{1}{2}\min \{\rho ,1\mathrm{/}n-\mathrm{\parallel }p-\mu {\mathrm{\parallel }}_2\}>0$. Step 1.2 gives a rational vector $q_j\in B(p,r)$; then $q_j\mathrm{\notin }K$ and $\mathrm{\parallel }{q}_j-\mu {\mathrm{\parallel }}_2<1\mathrm{/}n$. Thus $J_n=\{j\in \mathbb{N}:q_j\mathrm{\notin }K,\mathrm{\parallel }{q}_j-\mu {\mathrm{\parallel }}_2<1\mathrm{/}n\}$ is nonempty. This argument proves nonemptiness separately for each $n$ without choosing a sequence of $p,\rho ,r$. Let $j_n$ be the least member of $J_n$ and set $y_n=q_{j_n}$. This least-index construction uses no choice and $y_n\to \mu$.

For $z,z^{\mathrm{\prime }}\in K$, the triangle inequality gives $\mathrm{\parallel }{y}_n-z{\mathrm{\parallel }}_2\le \mathrm{\parallel }{y}_n-z^{\mathrm{\prime }}{\mathrm{\parallel }}_2+\mathrm{\parallel }z-z^{\mathrm{\prime }}{\mathrm{\parallel }}_2$, and exchanging $z,z^{\mathrm{\prime }}$ gives $\mid \mathrm{\parallel }{y}_n-z{\mathrm{\parallel }}_2-\mathrm{\parallel }{y}_n-z^{\mathrm{\prime }}{\mathrm{\parallel }}_2\mid \le \mathrm{\parallel }z-z^{\mathrm{\prime }}{\mathrm{\parallel }}_2$. Hence $z\mapsto \mathrm{\parallel }{y}_n-z{\mathrm{\parallel }}_2$ is continuous on $K$ in its Euclidean metric. The nonempty compact set $K$ therefore has a minimizer $z_n\in K$ by [F5]. It is unique: if distinct $z,z^{\mathrm{\prime }}\in K$ both minimize, convexity puts $(z+z^{\mathrm{\prime }})\mathrm{/}2$ in $K$ and $\mathrm{\parallel }{y}_n-(z+z^{\mathrm{\prime }})\mathrm{/}2{\mathrm{\parallel }}_2^2=(\mathrm{\parallel }{y}_n-z{\mathrm{\parallel }}_2^2+\mathrm{\parallel }{y}_n-z^{\mathrm{\prime }}{\mathrm{\parallel }}_2^2)\mathrm{/}2-\mathrm{\parallel }z-z^{\mathrm{\prime }}{\mathrm{\parallel }}_2^2\mathrm{/}4$ is strictly smaller. For any $v\in K$ and $0<s\le 1$, convexity gives $z_n+s(v-z_n)\in K$. Minimality then implies $\mathrm{\parallel }{y}_n-z_n{\mathrm{\parallel }}_2^2\le \mathrm{\parallel }{y}_n-z_n-s(v-z_n){\mathrm{\parallel }}_2^2$; expanding the inner products, dividing by $s$, and letting $s\downarrow 0$ gives $(y_n-z_n)\cdot (v-z_n)\le 0$.

Since $\mu \in K$, minimality gives $\mathrm{\parallel }{y}_n-z_n{\mathrm{\parallel }}_2\le \mathrm{\parallel }{y}_n-\mu {\mathrm{\parallel }}_2$, so the triangle inequality gives $\mathrm{\parallel }{z}_n-\mu {\mathrm{\parallel }}_2\le 2\mathrm{\parallel }{y}_n-\mu {\mathrm{\parallel }}_2\to 0$. Put $u_n=(y_n-z_n)\mathrm{/}\mathrm{\parallel }{y}_n-z_n{\mathrm{\parallel }}_2$; the denominator is positive because $y_n\mathrm{\notin }K$, and homogeneity of the norm gives $\mathrm{\parallel }{u}_n{\mathrm{\parallel }}_2=1$. Thus $(u_n)$ is bounded in the Euclidean metric; [F8] gives a subsequence $u_{n_j}$ converging to $u$. The reverse triangle inequality, obtained by applying the triangle inequality in both orders, gives $\mid \mathrm{\parallel }{u}_{n_j}{\mathrm{\parallel }}_2-\mathrm{\parallel }u{\mathrm{\parallel }}_2\mid \le \mathrm{\parallel }{u}_{n_j}-u{\mathrm{\parallel }}_2$, hence $\mathrm{\parallel }u{\mathrm{\parallel }}_2=1$. For every $v\in K$, step 3.1 gives $u_{n_j}\cdot (v-z_{n_j})\le 0$. By [F10], $\mathrm{\mid }{u}_{n_j}\cdot (v-z_{n_j})-u\cdot (v-\mu )\mathrm{\mid }\le \mathrm{\parallel }{u}_{n_j}-u{\mathrm{\parallel }}_2\mathrm{\parallel }v-z_{n_j}{\mathrm{\parallel }}_2+\mathrm{\parallel }u{\mathrm{\parallel }}_2\mathrm{\parallel }\mu -z_{n_j}{\mathrm{\parallel }}_2⟶0,$ since $z_{n_j}\to \mu$ and hence $(v-z_{n_j})$ is bounded. Passing to the limit gives $u\cdot (v-\mu )\le 0$. Taking $v=T(x)\in K$ proves the claimed support inequality.