# Continuity of subdifferential mapping

I'm reading Cedric Villani's book topics in optimal transportation, and I have a problem on page 53:

If $\varphi$ lower semi-continuous, then the subdifferential mapping $\partial\varphi$ is always continuous on the whole $R^n$, in the sense that, if $x_k\to x$ , $\partial\varphi(x_k)\ni y_k \to y$, then $y\in\partial\varphi(x)$

How to prove this? Thanks in advance.

• The correct statement is the subdifferential map of a closed proper convex function is closed, i.e., has a closed graph. This is Theorem 24.4 in Rockafellar's Convex Analysis. Under some conditions, having a closed graph implies upper-hemicontinuity. But, the subdifferential map is generally not lower-hemicontinuous. A simple example is the absolute value function on the real line. The subdifferential map of this function is not lower-hemicontinuous at 0. So, saying the subdifferential map is always continuous is very misleading. Commented Feb 13, 2016 at 7:53

Consider a lsc and convex function $$\varphi:V\to \mathbb{R}$$ (V is a Banach space). If $$(x_n) \subset V$$, $$(\xi_n) \subset V^*$$ with $$\xi_n\in\partial \varphi(x_n)$$, $$x_n \to x$$ and $$\xi_n \to \xi$$ $$\,$$in $$w^*$$-$$V^*$$, then $$\xi \in \partial \varphi (x)$$.
From the definition of convex subdifferential, we have $$\varphi(v)-\varphi(x_n)\geq \langle \xi_n,v-x_n\rangle_{V^*\times V}$$ for all $$v\in V$$. Taking limit infimum and using the fact that $$\varphi$$ is lsc, we easily deduce that $$\varphi(v)-\varphi(x)\geq \langle \xi,v-x\rangle_{V^*\times V}.$$ The right-hand side is true since $$w^*$$-convergance of $$(\xi_n)$$ combined with strong convergane of $$(x_n)$$ implies the duality pairing convergance. If $$V=\mathbb{R}^n$$, then it is relfexive and all three types of topologies coincide.
Let $z\in \mathbb{R}^n$,by assumption, $\varphi(z)\ge \varphi(x_k)+\langle y_k,z-x_k\rangle$, let $k\to \infty$ and use lsc of $\varphi$ at $x$ and joint continuity of inner product, you can get
$$\varphi(z)\ge \varphi(x)+\langle y,z-x\rangle$$
which holds for all $z\in \mathbb{R}^n$. Hence $y\in\partial \varphi(x)$.