Given $f$ a convex function, $x_0 \in int(dom(f))$, then the subdifferential $\partial f(x_0)$ has only one point iff $f$ is differentiable in $x_0$.

Asked Nov 23 '23 at 17:18

Active Nov 23 '23 at 18:12

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I am currently working on convex optimization and I have approached the mentioned statement. The right-to-left implication is straightforward but I am finding myself in trouble trying to prove the other way around. When I say differentiable, I mean in a general sense not is Gâteaux differentiable.Can anyone help with it?

pstd: I have seen the post How to show $\partial f(x) =\{\nabla f(x) \}$ for a convex differentiable function? that works on this topic. I am not so sure that the left-to-right implication is right, because to take $\phi(x)'$ is assuming that $f$ is differentiable (and so has gradient), but it is pretty much what we are trying to proof.

Any suggestion will be appreciated! (edited)

edited Nov 23 '23 at 18:12

asked Nov 23 '23 at 17:18

Cristian

see https://math.stackexchange.com/questions/4478569/assume-a-is-open-convex-and-f-convex-continuous-then-f-is-g%C3%A2teaux-differe – daw Nov 23 '23 at 17:39

Given $f$ a convex function, $x_0 \in int(dom(f))$, then the subdifferential $\partial f(x_0)$ has only one point iff $f$ is differentiable in $x_0$.

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