Proof of that the graph of subdifferential is closed when $f$ is closed proper convex function on $\mathbb{R}^n$.

Question

In the Theorem 24.4 of Convex Analysis written by R. Tyrrell Rockafellar, if $f$ is a closed proper convex function on $\mathbb{R}^n$, then the graph of $\partial f$ is a closed subset of $\mathbb{R}^n \times \mathbb{R}^n$.

In the proof of the theorem, the author takes "lim inf" and uses the fact that $f$ and $f^*$ are closed.

I don't fully understand the proof because I'm not familiar with detailed convex analysis. Can anybody explain it in detail for me?

For example, why do we have to take "lim inf" instead of "lim"? Further, how is the closedness of $f$ and $f^*$ used there?

Answer 1

I kept considering this problem, and I think that I found the detailed proof. Here's the ingradients:

1. For any sequence $(a_i)$ and $(b_i)$, if $a_i \geq b_i, \forall i \Rightarrow \liminf_{i \rightarrow \infty} a_i \geq \liminf_{i \rightarrow \infty} b_i$.
2. Let sequences $(x_i)$ and $(x_i^*)$ such that $x_i \rightarrow x$ and $x_i^* \rightarrow x^*$. Then, $\liminf_{i \rightarrow \infty} \left< x_i, x_{i}^{*} \right> = \left <x, x^{*} \right>$ cuz $<\cdot, \cdot>$ is continuous function.
3. "$f$ is closed" means that all sublevel sets are closed. Note that "all sublevel sets are closed" $\Rightarrow$ "$f$ is lower semi-continuous (l.s.c.)" (according to wikipedia. I didn't actually know this).
4. A function $f$ is l.s.c. if $\liminf_{i \rightarrow} f(x_i) \geq f(x)$ where $x_i \rightarrow x$.

Then, the proof of the theorem that I asked is very natural :)

Answer 2

Since $f$ is a closed proper function we know that $f^*$ is a closed proper function.

Then both $f,f^*$ are lsc. and so we have $\lim_i f(x_i) \ge f(x)$, $\lim_i f^*(x_i^*) \ge f^*(x^*)$.

Applying Theorem 23.5 we get $\langle x, x^* \rangle = \lim_i\langle x_i, x_i^* \rangle = \liminf_i\langle x_i, x_i^* \rangle = \liminf_i f(x_i)+f^*(x_i^*) \ge f(x)+f^*(x^*)$ and applying Theorem 23.5 again we get $x^* \in \partial f(x)$.