# Is it true that the subdifferential is empty if and only if $f$ is infinite?

I am interested in conditions when the subdifferential is empty.

Claim:

If $$f$$ is lower semicontinuous, convex, proper, then $$\partial f(x) = \emptyset$$ if and only if $$f(x) = +\infty$$

Is this true?

I know that $$f(x) = +\infty \implies \partial f(x) = \emptyset$$.

But is the reverse true?

$$\partial f(x) = \emptyset \implies f(x) = +\infty$$?

If not, is there a general condition that says when $$\partial f(x)$$ is empty?

• Are you assuming that $f$ is convex? I don't think $f(x) = x^2 \sin(x)$ has a nonempty subdifferential anywhere. – David Kraemer Sep 16 at 0:18
• @DavidKraemer Yes, every nice assumptions. lower semicontinuous, proper, convex – Sanjay Gupta Sep 16 at 0:19
• Note that convexity implies lower semicontinuity (in fact, convex functions are locally Lipschitz). – Chris Sep 16 at 0:46
• @Chris The convex indicator function of a bounded open set is an example of a convex function that is not lower continuous. – littleO Sep 16 at 0:51
• Sorry, I meant real-valued, but I guess that the author of the question is specifically wondering about when the function takes the value infinity. Oops. – Chris Sep 16 at 2:07

The function $$f(x) = \begin{cases} -\sqrt{x} &\quad \text{if } x \geq 0, \\ \infty & \quad \text{otherwise} \end{cases}$$ is closed and convex, and $$\partial f(0) = \emptyset$$ despite the fact that $$f(0)$$ is finite.
• This is really strange. I thought that intuitively, a function only becomes non-subdifferentiable if the function goes to $\infty$ at some point at the boundary. Maybe I am stuck thinking of the x^2 over open domain and $+\infty$ else where. But now looking at the graph of -sqrt(x), it seems that the condition is the slope of the line becomes vertical. – Sanjay Gupta Sep 16 at 1:23