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Given a convex, lower semicontinuous, and proper function $f:\mathbb{R}^n\to \mathbb{R}$ which is differentiable on its domain, is it true that its gradient $\nabla f$ is continuous on the interior of the domain of $f$? Here I am taking $\text{dom}f = \{x\in\mathbb{R}^n: f(x)<\infty\}$. What I came up with was that for such a function $f$, it must be true that $f$ is locally Lipschitz continuous on its domain and then by Rademacher's theorem it is locally differentiable a.e.. This doesn't get what I want, however. Anyone have a proof or counter example?

Edit: this is corollary 9.20 in Rockafellar and Wets, as it turns out.

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  • $\begingroup$ Locally Lipschitz doesn't help. Consider $f(x) = |x|$ on $\mathbb R^1$. $\endgroup$ – Stephen Montgomery-Smith Aug 16 '20 at 20:36
  • $\begingroup$ I can prove it for $n = 1$ using the fact that $\nabla f = f'$ is increasing, and the mean value theorem. But this doesn't help for $n > 1$. $\endgroup$ – Stephen Montgomery-Smith Aug 16 '20 at 20:37
  • $\begingroup$ @StephenMontgomery-Smith that's not differentiable to begin with, not applicable. $\endgroup$ – TSF Aug 16 '20 at 21:05
  • $\begingroup$ I wasn't presenting it as a counterexample. I was presenting it as an example that your argument was the wrong direction to go. $\endgroup$ – Stephen Montgomery-Smith Aug 16 '20 at 21:06
  • $\begingroup$ Looks like this is a duplicate: math.stackexchange.com/questions/3591521/… although no-one answered it there. $\endgroup$ – Stephen Montgomery-Smith Sep 9 '20 at 2:54
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Without loss of generality, it is sufficient to prove $\nabla f$ is continuous at $x = 0$ when $\nabla f(0) = 0$. Suppose $x_n \to 0$ is such that $|\nabla f(x_n)| > a > 0$. Given $\epsilon>0$ such that $B(0,2\epsilon) \subset \text{dom}(f)$, pick $n$ so that $x_n \in B(0,\epsilon)$ and $f(x_n) - f(0) > -\epsilon^2$. We know there exists $y \in B(x_n,\epsilon)$, $y \ne x_n$, such that $$ f(y) \ge f(x_n) + a |x_n - y| $$ (that is, choose $y$ in the direction of $\nabla f(x_n)$ close to $x_n$). For $t \in \mathbb R$, let $z_t = t(y-x_n) + x_n$. By convexity, see that for $t \ge 1$ $$ \tfrac1t f(z_t) + (1-\tfrac{1}t) f(z_0) \ge f(z_1) ,$$ that is $$ f(z_t) \ge f(x_n) + a t |x_n-y| .$$ Choose $t = \epsilon / |x_n - y|$. Note that $|z_t| < 2 \epsilon$. Then $$ f(z_t) - f(0) = f(z_t) - f(x_n) + f(x_n) - f(0) \ge a \epsilon - \epsilon^2 . $$ This contradicts that $\nabla f(0) = 0$.

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  • $\begingroup$ Thanks for your help. $\endgroup$ – TSF Aug 16 '20 at 21:22
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I am updating this post with the follow-up question: If $f$ is a convex function defined on some convex set $E\subseteq \mathbb R^n$ and if it is differentiable on $E$, is it true that its gradient must be continuous on $E$ (and not only in the interior) ?

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  • $\begingroup$ I believe this is very much related, math.stackexchange.com/questions/3524930/… $\endgroup$ – TSF Nov 13 '20 at 12:51
  • $\begingroup$ @TSF: thank you, but I believe that the convexity assumption is essential. Indeed, if a convex function is differentiable on $E$, then it is automatically $C^1$ in the interior of $E$, and I am trying to determine whether it is actually $C^1$ on the whole $E$. $\endgroup$ – TrivialPursuit Nov 13 '20 at 13:32

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