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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.

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  • $\begingroup$ $f(x) = \| x \|$ is convex everywhere, but not differentiable at $x =0$. $\endgroup$ – Martin R Aug 23 at 13:19
  • $\begingroup$ @TonyK I know, but it is a part of statement, not of solution. $\endgroup$ – dead slug Aug 23 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 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 at 13:21
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    $\begingroup$ Now all OK. Don't edit anymore. $\endgroup$ – dead slug Aug 23 at 13:23
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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.

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