Let $f$ be a convex function from $\Bbb R^n$ to $\Bbb R$ and $x\in \mathbb R^n$
Suppose $f$ has derivatives in every direction in $x$.
Prove that $f$ is differentiable at $x$.

I already proved that a convex function is continuous and locally-Lipschitz, but I don't see how this can help here.

I know the statement is not true without the assumption that $f$ is convex, see for example this question: $f$ not differentiable at $(0,0)$ but all directional derivatives exist

Since the function is convex, I guess something can be done using subdifferentials.

  • $\begingroup$ $f(x) = \| x \|$ is convex everywhere, but not differentiable at $x =0$. $\endgroup$ – Martin R Aug 23 '19 at 13:19
  • $\begingroup$ @TonyK I know, but it is a part of statement, not of solution. $\endgroup$ – dead slug Aug 23 '19 at 13:20
  • $\begingroup$ @MartinR: You are not the first to interpret the question that way! I have edited it to remove the ambiguity. $\endgroup$ – TonyK Aug 23 '19 at 13:21
  • $\begingroup$ @TonyK: Changing “It is known” to “Suppose it is known” changes the meaning of the question significantly. I wouldn't do that without conformation from OP. $\endgroup$ – Martin R Aug 23 '19 at 13:21
  • 1
    $\begingroup$ Now all OK. Don't edit anymore. $\endgroup$ – dead slug Aug 23 '19 at 13:23

Theorem 25.1 in Rockafellar's Convex Analysis states that if the subdifferential of $f$ at $x$ is a singleton, then $f$ is differentiable at $x$.

Let $u,v\in \partial f(x)$. Note that for any direction $y$, if $\alpha >0$ $$\langle y,u \rangle = \frac 1{\alpha} \langle y,\alpha u \rangle \leq \frac{f(x+\alpha y)-f(x)}{\alpha}$$ and if $\alpha <0$ $$\langle y,u \rangle = \frac 1{\alpha} \langle y,\alpha u \rangle \geq \frac{f(x+\alpha y)-f(x)}{\alpha}$$ Letting $\alpha \to 0$ yields $\langle y,u \rangle = f'(x,y)$.

Replacing $u$ with $v$, $\langle y,v \rangle = f'(x,y) = \langle y,u \rangle$. Hence $\forall y\in \mathbb R^n, \langle y,u-v\rangle =0$ and $u-v=0$. $\partial f(x)$ is thus a singleton.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.