Theorem 21.3.2011FA81B0000

Proof. The conjugate is the supremum of the continuous affine functions $p\mapsto \langle p,x\rangle -F(x)$, $x\in \operatorname {dom}F$, so it is convex and lower semicontinuous. Since $\operatorname {dom}F\neq \varnothing $, we have $F^*(p)\geq \langle p,x_0\rangle -F(x_0)>-\infty $ for any $x_0\in \operatorname {dom}F$. By proposition 21.2.4, $F$ has a subgradient $p_0$ at some point $x_1\in \operatorname {ri}(\operatorname {dom}F)$, and then $F^*(p_0)=\langle p_0,x_1\rangle -F(x_1)<+\infty $. Thus $F^*$ is closed and proper. The definition of $F^*$ gives $F(x)\geq \langle p,x\rangle -F^*(p)$ for all $p$, that is, $F^{**}\leq F$.

We next show that $F^{**}=F$ when $F$ is closed. An affine function $\langle p,\cdot \rangle -c$ lies below $F$ if and only if $c\geq F^*(p)$. Hence $F^{**}$ is the supremum of all affine minorants of $F$, and it suffices to find, for every $x$ and every real $r<F(x)$, an affine minorant of $F$ whose value at $x$ exceeds $r$. The point $(x,r)$ does not belong to the closed convex set $\operatorname {epi}F$, so by lemma 21.2.3(1), applied in $N_{\mathbb R}\times \mathbb R$, there are $a\in M_{\mathbb R}$, $c\in \mathbb R$ and $\delta >0$ with

Letting $s\to +\infty $ shows that $c\geq 0$. If $c>0$, taking $s=F(y)$ gives the affine minorant $y\mapsto r+(\delta -\langle a,y-x\rangle )/c$, whose value at $x$ is $r+\delta /c>r$. If $c=0$, then $\langle a,y\rangle \geq \langle a,x\rangle +\delta $ for all $y\in \operatorname {dom}F$. Let $L_0$ be an affine minorant of $F$, which exists by the first paragraph. For $k>0$ the affine functions

are again minorants of $F$, since the added term is at most $-k\delta /2$ on $\operatorname {dom}F$, and their values at $x$ tend to $+\infty $ as $k\to \infty $. In both cases the required minorant exists, so $F^{**}=F$. For a general proper convex $F$, a continuous affine function lies below $F$ if and only if it lies below the largest lower semicontinuous minorant $\overline F$. Therefore $F^*=(\overline F)^*$, and since $\overline F$ is closed and proper by proposition 21.2.6, $F^{**}=(\overline F)^{**}=\overline F$.

Let $F$ be closed. By definition, $p\in \partial F(x)$ means that $x\in \operatorname {dom}F$ and $\langle p,y\rangle -F(y)\leq \langle p,x\rangle -F(x)$ for all $y$, that is, the supremum defining $F^*(p)$ is attained at $x$; equivalently, $F(x)+F^*(p)=\langle p,x\rangle $. (If $x\notin \operatorname {dom}F$, both conditions fail, since then $F(x)+F^*(p)=+\infty $.) Applying this to the closed proper function $F^*$ and using $F^{**}=F$, we find that $x\in \partial F^*(p)$ if and only if $F^*(p)+F(x)=\langle p,x\rangle $. This proves the equivalences. If $p\in \operatorname {ri}(\operatorname {dom}F^*)$, then $F^*$ has a subgradient $x$ at $p$ by proposition 21.2.4, and $p\in \partial F(x)$ by the equivalences; this is the first inclusion. If $p\in \partial F(x)$, then $F^*(p)=\langle p,x\rangle -F(x)<+\infty $, which gives the second. The closures coincide because $\operatorname {ri}(\operatorname {dom}F^*)$ is dense in $\operatorname {dom}F^*$ by lemma 21.2.3(4). Finally, $\chi _ P^*=h_ P$ by the definitions, and $\chi _ P$ is a closed proper convex function, so $h_ P^*=\chi _ P^{**}=\chi _ P$. $\square$