Subdifferential is closed, convex and bounded

Question

If $f: \mathbb{R}^n \to \mathbb{R} \cup \{+\infty\}$ is convex, how can I show that $\partial f(x_0)$ (sub differential) is closed and convex, and also bounded (bounded when f over the entire domain)

Answer 1

$\partial f(x) = \{ h | f(y) - f(x) \ge \langle h, y-x \rangle \ \forall $y$\}$.

It is straightforward to see that $D_y = \{ h | f(y) - f(x) \ge \langle h, y-x \rangle \}$ is a closed half space, hence convex. Since $\partial f(x) = \cap_y D_y$, we see that $\partial f(x)$ is closed and convex.

If $f$ is finite on some open set $U$ containing $x$, then there is some open $V \subset U$ containing $x$ and some $L$ such that $f$ is Lipschitz of rank $L$ on $V$. In particular, we have $f(y)-f(x) \le L \|x-y\|$.

Let $B = \{ h | \langle h, d \rangle \le L \|d\| \ \forall d \}$ and note that $B = \overline{B}(0,L)$. It is straightforward to check that $\partial f(x) \subset B$, hence bounded.