Proposition 21.2.472D7B299EAB8

Proof. We first show that $F$ is bounded, hence Lipschitz, near each point of $\operatorname {ri}(\operatorname {dom}F)$, and find a subgradient there by separation; then we prove (1)–(4). Let $A$ be the affine span of $\operatorname {dom}F$, of dimension $k$, and let $x\in \operatorname {ri}(\operatorname {dom}F)$. Choose affine coordinates on $A$ centered at $x$, with the associated Euclidean norm $|\cdot |$; since all norms on $A$ are equivalent, local Lipschitz continuity does not depend on this choice. For small $\delta >0$ the cube $Q$ of points of $A$ whose coordinates are at most $2\delta $ in absolute value lies in $\operatorname {dom}F$. Every point of $Q$ is a convex combination of the $2^ k$ vertices of $Q$; this follows by induction on $k$, since writing the last coordinate of a point as a convex combination of $\pm 2\delta $ exhibits the point as a convex combination of two points of the opposite faces of $Q$, which are cubes of dimension $k-1$. By the finite form of Jensen's inequality, $F\leq C$ on $Q$, where $C$ is the largest value of $F$ at a vertex. If $y\in Q$, then also $2x-y\in Q$, and $F(x)\leq \frac12(F(y)+F(2x-y))$ gives $F(y)\geq 2F(x)-C$. Thus $F$ is bounded on $Q$, and by (21.2.2) it is Lipschitz on the cube of half-side $\delta $ centered at $x$.

Next we construct a subgradient at $x$. Since $\operatorname {dom}F\subset A$, the epigraph of $F$ lies in $A\times \mathbb R$; let $E$ be its closure, a closed convex subset of $A\times \mathbb R$. Because $F$ is continuous at $x$ as a function on $A$, the points $(x,F(x)-\varepsilon )$, $\varepsilon >0$, do not belong to $E$, so $(x,F(x))$ is a boundary point of $E$ in $A\times \mathbb R$. By lemma 21.2.3(2), applied in $A\times \mathbb R$ with origin $(x,F(x))$, there are a linear functional $a$ on the linear subspace $A_0=\{ y-y':y,y'\in A\} $ parallel to $A$ and a real number $c$, not both zero, such that

Letting $s\to +\infty $ shows that $c\geq 0$. If $c=0$, then $a\neq 0$ and $\langle a,y-x\rangle \geq 0$ for all $y\in \operatorname {dom}F$; but $\operatorname {dom}F$ contains a ball of $A$ about $x$, and this ball contains points $y$ with $\langle a,y-x\rangle <0$. Hence $c>0$, and taking $s=F(y)$ we obtain $F(y)\geq F(x)+\langle p_ A,y-x\rangle $ for $y\in \operatorname {dom}F$, where $p_ A=-a/c$. Extend $p_ A$ to a linear functional $p\in M_{\mathbb R}$. The inequality $F(y)\geq F(x)+\langle p,y-x\rangle $ holds for $y\in \operatorname {dom}F$, because then $y-x\in A_0$, and also for $y\notin \operatorname {dom}F$, where $F(y)=+\infty $. Thus $p\in \partial F(x)$.

From now on let $F$ be a finite convex function on the open convex set $\Omega $, so that $A=N_{\mathbb R}$ and $\operatorname {ri}(\operatorname {dom}F)=\Omega $. Fix a basis $e_1,\ldots ,e_ n$ of $N_{\mathbb R}$, let $K\subset \Omega $ be compact, and choose $r>0$ such that the set $K_ r=\{ y+\sum _ is_ ie_ i:y\in K,\ |s_ i|\leq r\} $ lies in $\Omega $. If $x\in K$ and $p\in \partial F(x)$, the subgradient inequality at $x$, evaluated at the points $x\pm re_ i$, gives $\pm r\langle p,e_ i\rangle \leq F(x\pm re_ i)-F(x)$, hence $|\langle p,e_ i\rangle |\leq 2r^{-1}\max _{K_ r}|F|$, where $F$ is bounded on the compact set $K_ r$ because it is continuous. This is the uniform bound in (1). If $x_ k\to x$ in $\Omega $, $p_ k\in \partial F(x_ k)$ and $p_ k\to p$, then letting $k\to \infty $ in $F(y)\geq F(x_ k)+\langle p_ k,y-x_ k\rangle $ and using that $F$ is continuous at $x$, we get $p\in \partial F(x)$. Thus the graph is closed, and each $\partial F(x)$ is closed and bounded, hence compact.

For (2) and (4) we use the following slope inequalities for a convex function $\phi $ on an interval: if $s_1<s_2<s_3$, then

Indeed, $s_2=\lambda s_1+(1-\lambda )s_3$ with $\lambda =(s_3-s_2)/(s_3-s_1)\in (0,1)$, and the convexity inequality $\phi (s_2)\leq \lambda \phi (s_1)+(1-\lambda )\phi (s_3)$ is equivalent to each of the two inequalities. Now fix $x$ and $v$, and apply (21.2.5) to $\phi (t)=F(x+tv)$, which is convex on the open interval of those $t$ with $x+tv\in \Omega $: with $s_1=0$ it shows that the difference quotient in (2) is nondecreasing in $t>0$, so its limit as $t\downarrow 0$ exists in $[-\infty ,+\infty )$. Each $p\in \partial F(x)$ gives $F(x+tv)-F(x)\geq t\langle p,v\rangle $; hence $F'(x;v)\geq \max _{p\in \partial F(x)}\langle p,v\rangle $, the maximum being attained by compactness. Conversely, choose $p_ t\in \partial F(x+tv)$. The subgradient inequality at $x+tv$, evaluated at $x$, gives $F(x+tv)-F(x)\leq t\langle p_ t,v\rangle $. By (1) the vectors $p_ t$ stay bounded as $t\downarrow 0$, and each of their cluster points lies in $\partial F(x)$. Therefore $F'(x;v)\leq \varlimsup _{t\downarrow 0}\langle p_ t,v\rangle \leq \max _{p\in \partial F(x)}\langle p,v\rangle $.

For (3), suppose that $\partial F(x)=\{ p\} $, and choose $p_ h\in \partial F(x+h)$. By (1), every cluster point of $p_ h$ as $h\to 0$ lies in $\partial F(x)$, so $p_ h\to p$. The subgradient inequalities at $x$ and at $x+h$ give

so $F$ is differentiable at $x$, with $\nabla F(x)=p$. Conversely, if $F$ is differentiable at $x$ and $p\in \partial F(x)$, dividing $F(x+tv)-F(x)\geq t\langle p,v\rangle $ by $t$ and letting $t\downarrow 0$ gives $\langle \nabla F(x),v\rangle \geq \langle p,v\rangle $ for every $v$, hence $p=\nabla F(x)$. As $\partial F(x)\neq \varnothing $, it equals $\{ \nabla F(x)\} $.

For (4), fix $i$, and write $D_ i^+F(x)=F'(x;e_ i)$ and $D_ i^-F(x)=-F'(x;-e_ i)$ for the one-sided partial derivatives; they are finite by (2). Let $\phi $ be the restriction of $F$ to the intersection of $\Omega $ with a line parallel to $e_ i$, parametrized as $\phi (s)=F(y+se_ i)$ on an open interval. By (21.2.5), $\phi $ has one-sided derivatives $\phi '_\pm $ at every point, $\phi '_-(s)\leq \phi '_+(s)$, and $\phi '_+(s)\leq \phi '_-(s')$ whenever $s<s'$, because both are bounded by the slope of the chord from $s$ to $s'$. Hence the open intervals $(\phi '_-(s),\phi '_+(s))$ are pairwise disjoint; each nonempty one contains a rational number, so only countably many of them are nonempty, and $\phi '_-=\phi '_+$ outside a countable set. Next, $D_ i^+F$ is the pointwise limit on $\Omega $ of the Borel functions equal to $k(F(x+e_ i/k)-F(x))$ where $x+e_ i/k\in \Omega $ and to $0$ elsewhere, so it is a Borel function, and similarly for $D_ i^-F$. Thus the set $Z_ i\subset \Omega $ where $D_ i^+F\neq D_ i^-F$ is a Borel set that meets each line parallel to $e_ i$ in a countable set, and Fubini's theorem, in the coordinates of the basis $e_1,\ldots ,e_ n$, shows that $Z_ i$ is a null set. At a point $x\notin Z_1\cup \cdots \cup Z_ n$, (2) gives

for every $i$, so all subgradients at $x$ coincide, and $F$ is differentiable at $x$ by (3). Conversely, if $F$ is differentiable at $x$, then $D_ i^+F(x)=D_ i^-F(x)=\langle \nabla F(x),e_ i\rangle $ for every $i$. Therefore $\Omega _ F=\Omega \setminus (Z_1\cup \cdots \cup Z_ n)$ is a Borel set with null complement, and the coordinates $\langle \nabla F,e_ i\rangle =D_ i^+F$ of $\nabla F$ are Borel functions on $\Omega _ F$. $\square$