Proposition 21.2.6EAC7AAEC7279

Proof. By proposition 21.2.4, $F$ has a subgradient $p$ at some point $x_1\in \operatorname {ri}(\operatorname {dom}F)$, so $F$ lies above the affine function $\ell (y)=F(x_1)+\langle p,y-x_1\rangle $. The closure $E$ of $\operatorname {epi}F$ is convex, and it lies above the graph of $\ell $ because $\ell $ is continuous. Hence, for each $x$, the set $\{ r:(x,r)\in E\} $ is closed and bounded below by $\ell (x)$, and with each of its points $r$ it contains $[r,+\infty )$, because $\operatorname {epi}F$ has this property. So this set is either empty or a closed half-line $[\widetilde F(x),+\infty )$ with $\widetilde F(x)\geq \ell (x)>-\infty $. Setting $\widetilde F(x)=+\infty $ in the first case, we see that $E$ is the epigraph of a convex function $\widetilde F$. It is lower semicontinuous because its epigraph is closed, and $\widetilde F\leq F$, so $\operatorname {dom}\widetilde F\supset \operatorname {dom}F\neq \varnothing $. If $G\leq F$ is lower semicontinuous, then $\operatorname {epi}G$ is closed and contains $\operatorname {epi}F$, hence also $E$, and so $G\leq \widetilde F$. Thus $\widetilde F=\overline F$, which is therefore closed, proper and convex.

Let $x\in \operatorname {ri}(\operatorname {dom}F)$. If $(y_ k,r_ k)\in \operatorname {epi}F$ converge to $(x,r)$, then $y_ k$ lies in the affine span of $\operatorname {dom}F$, on which $F$ is continuous at $x$; hence $r\geq \lim _ kF(y_ k)=F(x)$. This shows $\overline F(x)\geq F(x)$, and so $\overline F(x)=F(x)$.

Now let $F$ be closed, fix $x_0\in \operatorname {ri}(\operatorname {dom}F)$, and let $x\in \overline{\operatorname {dom}F}$. By lemma 21.2.3(4), the points $(1-t)x_0+tx$ with $0\leq t<1$ lie in $\operatorname {ri}(\operatorname {dom}F)$, so the convex function $\phi (t)=F((1-t)x_0+tx)$ is finite on $[0,1)$. Convexity gives $\phi (t)\leq (1-t)\phi (0)+t\phi (1)$, hence $\varlimsup _{t\uparrow 1}\phi (t)\leq \phi (1)$, while lower semicontinuity gives $\varliminf _{t\uparrow 1}\phi (t)\geq \phi (1)$. Therefore

also when $F(x)=+\infty $. Since $F=+\infty $ outside $\overline{\operatorname {dom}F}$, which is the closure of $\operatorname {ri}(\operatorname {dom}F)$, the function $F$ is determined by its restriction to $\operatorname {ri}(\operatorname {dom}F)$.

Finally, let $(F_ j)$ be a sequence of finite convex functions on $\Omega $ converging to a finite limit at every point of a dense set $S\subset \Omega $. We show that near each point of $\Omega $ the functions $F_ j$ are uniformly bounded, hence uniformly Lipschitz, so that convergence on $S$ propagates to uniform convergence. Fix a basis of $N_{\mathbb R}$, write $Q(y,\rho )$ for the closed cube of half-side $\rho $ centered at $y$ in the corresponding coordinates, and let $x\in \Omega $ and $\rho >0$ with $Q(x,2\rho )\subset \Omega $. For each vertex $y$ of $Q(x,\rho )$ choose a point of $S$ whose coordinates differ from those of $y$ by less than $\rho /2$, and let $S_0\subset S\cap Q(x,2\rho )$ be the finite set of these points. Its convex hull $H$ is compact as the image of a simplex under a linear map, and it contains $Q(x,\rho /2)$. Indeed, let $\ell $ be a nonzero linear functional, $\ell (z)=\sum _ ia_ iz_ i$ in the chosen coordinates, and put $\| a\| =\sum _ i|a_ i|$. At a vertex $y^*$ of $Q(x,\rho )$ where $\ell $ is maximal, $\ell (y^*)=\ell (x)+\rho \| a\| $, and the point $s\in S_0$ chosen for $y^*$ satisfies $\ell (s)>\ell (y^*)-\frac\rho 2\| a\| =\max _{Q(x,\rho /2)}\ell $. Hence no point of $Q(x,\rho /2)$ can be separated from $H$ as in lemma 21.2.3(1).

The sequences $(F_ j(s))_ j$, $s\in S_0$, converge, so they are bounded above by some $C$, and the finite form of Jensen's inequality gives $F_ j\leq C$ on $H$ for all $j$. Choose $s_0\in S\cap Q(x,\rho /8)$. Then $Q(s_0,3\rho /8)\subset Q(x,\rho /2)$, and for $y\in Q(s_0,3\rho /8)$ the reflected point $2s_0-y$ lies in the same cube, so that $F_ j(y)\geq 2F_ j(s_0)-C\geq 2\inf _ kF_ k(s_0)-C$. Hence $|F_ j|\leq B$ on $Q(x,\rho /4)\subset Q(s_0,3\rho /8)$ for all $j$, with a constant $B$. By (21.2.2), applied with $\delta =\rho /8$ and $\omega \leq 2B$, the $F_ j$ are Lipschitz on $Q(x,\rho /8)$ with the common constant $L=16B/\rho $, for the Euclidean norm $|\cdot |$ of the chosen coordinates.

Given $\varepsilon >0$, choose a finite set $S_\varepsilon \subset S\cap Q(x,\rho /8)$ such that every point of $Q(x,\rho /8)$ lies within distance $\varepsilon /L$ of $S_\varepsilon $. To this end, cover the compact cube $Q(x,\rho /8)$ by finitely many balls of radius $\varepsilon /(2L)$ centered in the cube, and pick a point of the dense set $S$ in the intersection of each ball with the interior of the cube, a nonempty open set. For all sufficiently large $j,k$ we have $|F_ j-F_ k|<\varepsilon $ on $S_\varepsilon $, hence $|F_ j-F_ k|<3\varepsilon $ on $Q(x,\rho /8)$. Thus $(F_ j)$ is uniformly Cauchy on $Q(x,\rho /8)$. Finitely many such cubes cover any given compact subset of $\Omega $, so $(F_ j)$ converges locally uniformly to a finite continuous function, which is convex because the convexity inequalities pass to pointwise limits. $\square$