Lemma 21.2.334526AA883B2

Proof. Fix a Euclidean inner product $y\cdot w$ on $W$, with norm $|y|$.

(1) Choose $y_ k\in C$ with $|z-y_ k|\to \inf _{w\in C}|z-w|$. The sequence $(y_ k)$ is bounded, so a subsequence converges to a point $y$, which lies in $C$ because $C$ is closed. Thus $y$ minimizes the distance to $z$ on $C$, and $y\neq z$. For $w\in C$ and $0\leq t\leq 1$ the point $y+t(w-y)$ lies in $C$, so the function $t\mapsto |z-y-t(w-y)|^2$ on $[0,1]$ is minimal at $t=0$, and its derivative there, $-2(z-y)\cdot (w-y)$, is nonnegative. Hence $\ell (w)=(z-y)\cdot w$ satisfies $\ell (w)\leq \ell (y)=\ell (z)-|z-y|^2<\ell (z)$ for every $w\in C$.

(2) Since $z$ is a boundary point of $C$, there are points $z_ k\notin C$ with $z_ k\to z$. By (1) there are unit vectors $a_ k$ with $a_ k\cdot w<a_ k\cdot z_ k$ for all $w\in C$. A subsequence of $(a_ k)$ converges to a unit vector $a$, and in the limit $a\cdot w\leq a\cdot z$ for all $w\in C$. As $z\in C$, the functional $\ell (w)=a\cdot w$ has the required property.

(3) Let $x=\sum _{i=1}^ q\lambda _ ix_ i$ with $x_ i\in S$, $\lambda _ i>0$ and $\sum _ i\lambda _ i=1$. If the points $x_ i$ are affinely dependent, the vectors $(x_ i,1)\in W\times \mathbb R$ are linearly dependent, say $\sum _ i\mu _ i(x_ i,1)=0$ with not all $\mu _ i$ zero. Since $\sum _ i\mu _ i=0$, some $\mu _ i$ is positive. Replacing each $\lambda _ i$ by $\lambda _ i-s\mu _ i$, where $s=\min \{ \lambda _ i/\mu _ i:\mu _ i>0\} $, preserves $\sum _ i\lambda _ i$ and $\sum _ i\lambda _ ix_ i$, keeps all coefficients nonnegative and makes at least one of them zero. We discard the points with zero coefficient and repeat. Eventually the remaining points are affinely independent, and an affinely independent subset of $W$ has at most $n+1$ elements.

(4) Let $A$ be the affine span of $C$, of dimension $k$, and write $B_ A(x,r)$ for the open ball of $A$ with center $x$ and radius $r$. If $k=0$, then $C$ is a point, equal to its relative interior and to its closure, and its relative boundary is empty; so assume $k\geq 1$. A maximal affinely independent subset of $C$ spans an affine subspace containing $C$, hence equal to $A$, so it consists of $k+1$ points $x_0',\ldots ,x_ k'$. Every point $y\in A$ has unique barycentric coordinates $\lambda _ i(y)$, which are affine functions of $y$ with $\sum _ i\lambda _ i(y)=1$ and $y=\sum _ i\lambda _ i(y)x_ i'$. At the barycenter $c$ of the simplex with vertices $x_ i'$ all of them equal $1/(k+1)$, so they are positive on a ball $B_ A(c,r)$; every point of this ball is then a convex combination of the $x_ i'$ and lies in $C$. Hence $c\in \operatorname {ri}C$.

Now let $x_0\in \operatorname {ri}C$, choose $r>0$ with $B_ A(x_0,r)\subset C$, and let $y\in \overline C$ and $0\leq t<1$. If $y\in C$, convexity gives $B_ A((1-t)x_0+ty,(1-t)r)=(1-t)B_ A(x_0,r)+ty\subset C$. In general choose $y_ k\in C$ with $y_ k\to y$; then $B_ A((1-t)x_0+ty_ k,(1-t)r)\subset C$, and for large $k$ this ball contains $B_ A((1-t)x_0+ty,(1-t)r/2)$. Thus $(1-t)x_0+ty\in \operatorname {ri}C$. Letting $t\uparrow 1$, we see that every point of $\overline C$ is a limit of points of $\operatorname {ri}C$, so $\overline{\operatorname {ri}C}=\overline C$. Next, $\operatorname {ri}C\subset \operatorname {ri}\overline C$, since both are interiors in the same affine space $A$. Conversely, if $x\in \operatorname {ri}\overline C$, then $y=x+\varepsilon (x-c)$ lies in $\overline C$ for small $\varepsilon >0$, and $x=(1-t)c+ty$ with $t=1/(1+\varepsilon )<1$; hence $x\in \operatorname {ri}C$.

For the measure assertion, translate so that $c=0$ and identify $A$ with $\mathbb R^ k$ by a linear isomorphism, which changes Lebesgue measure on $A$ only by a constant factor. Let $Z=\overline C\setminus \operatorname {ri}C$ be the relative boundary; it is closed in $A$, because $\operatorname {ri}C$ is open in $A$, hence Borel, and $0\notin Z$. We claim that the dilates $sZ$, $s>0$, are pairwise disjoint. Indeed, suppose that $sz=s'z'$ with $z,z'\in Z$ and $0<s<s'$. Then $z'=(s/s')z$ is of the form $(1-t)\cdot 0+tz$ with $z\in \overline C$ and $0<t<1$, so $z'\in \operatorname {ri}C$ by what we have just shown, which is impossible. Now fix $R>0$ and put $Z_ R=Z\cap B(0,R)$. The sets $sZ_ R$ with $s$ rational in $[\frac12,1]$ are pairwise disjoint subsets of $B(0,R)$, and each has measure $s^ k|Z_ R|\geq 2^{-k}|Z_ R|$. Since infinitely many of them fit into a set of finite measure, $|Z_ R|=0$. Letting $R\to \infty $, we obtain $|Z|=0$. $\square$