TITLE: Is there an example of a convex differentiable function which is not continuously differentiable?
QUESTION [5 upvotes]: It is easy to see that any convex differentiable function of one variable is also continuous differentiable. But the proof is based on monotony of the derivative and doesn't work in multiple dimensions. So the question is, has any convex differentiable function of many variables to be also continuously differentiable? Or is there some counterexample?
Thanks in advance.

REPLY [2 votes]: I think the derivative will be continuous.
Here is an intuitive argument. Consider a closed convex body ( the epigraph of the convex function). Assume that the function is differentiable at every point. This means that the boundary of this convex body has a unique supporting hyperplane at each boundary point. We would like to show that the dependence of this hyperplane ( the directions of its normal) is continuous on the point $P$ on the boundary.
It is easier to think in the case of a compact convex body $C$ Surround it completely with a sphere $S$. The map from $S$ to the boundary
$\partial C$ of $C$ given by $x \mapsto \phi(x)$, the closest point in $C$ to $x$ is a contraction, so continuous, and surjective. The vector $\phi(x)-x$ is perpendicular to a supporting hyperplane at $\phi(x)$.
Now assume that every point on the boundary of $C$ has a unique supporting hyperplane. Then the map $x\mapsto \phi(x)$ is bijective from $S$ to $\partial C$, and so a homeomorphism, since $S$ is compact.  We conclude that the inverse map $\phi(x) \mapsto x$ is also continuous. Now note that $\frac{\phi(x) - x}{\|\phi(x) - x\|}$ is the unit normal vector to the supporting hyperplane at $\phi(x)$.