Gradient of a convex function is continuous on the interior of its domain 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.
 A: 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 $\nabla f(0) = 0$.
A: 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) ?
