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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?

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  • $\begingroup$ Are you assuming that $f$ is convex? I don't think $f(x) = x^2 \sin(x)$ has a nonempty subdifferential anywhere. $\endgroup$ – David Kraemer Sep 16 at 0:18
  • $\begingroup$ @DavidKraemer Yes, every nice assumptions. lower semicontinuous, proper, convex $\endgroup$ – Sanjay Gupta Sep 16 at 0:19
  • $\begingroup$ Note that convexity implies lower semicontinuity (in fact, convex functions are locally Lipschitz). $\endgroup$ – Chris Sep 16 at 0:46
  • $\begingroup$ @Chris The convex indicator function of a bounded open set is an example of a convex function that is not lower continuous. $\endgroup$ – littleO Sep 16 at 0:51
  • $\begingroup$ 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. $\endgroup$ – Chris Sep 16 at 2:07
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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.

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  • $\begingroup$ 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. $\endgroup$ – Sanjay Gupta Sep 16 at 1:23

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