Subdifferential of convex function.

Question

Let $X$ be a Banach space.

For continuous convex function $f:X \longrightarrow \mathbb{R}$. The linear functional $\phi\in X^*$ is called subgradient of function $f$ at a point $x\in X$ if $$f(y)-f(x) \geq Re\ \phi (y-x) \ \ \forall\ y\in X. $$ The set $\partial f(x)$ of all subgradients of $f$ at $x$ is called the subdifferential of $f$ at $x$.

Now we can easily prove that $\partial f(x)$ is a closed and convex subset of $X^*.$

But how can we prove that $\partial f(x)$ is a bounded subset of $X^*?$

Answer 1

This is a little bit more difficult. However, since the continuity of $f$ is given, it is not too hard.

The main idea is:

• Prove that $f$ is bounded from above and from below on balls $B_R(x)$ via continuity.
• Via convexity, it follows that $f$ is Lipschitz on $B_r(x)$ for $r \in (0,R)$.
• Conclude.