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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.

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    $\begingroup$ 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. $\endgroup$
    – Xiang Li
    Commented Feb 13, 2016 at 7:53

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One can prove even more general result which coincides with your particular case.

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.

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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 $y\in\partial \varphi(x)$.

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