When is a convex function continuous on its domain?

Question

Consider a lower-semicontinuous convex function $f\colon \mathbb{R}^n \to \mathbb{R}$ with domain $C = \{x \in \mathbb{R}^d: f(x) < \infty\}$. I am interested in understanding under what conditions $f$ is continuous over $C$.

It is well known that this is true whenever $C$ is simplicial, but not otherwise (see the discussion of Theorem 10.2 in Rockafellar's convex analysis).

What if $C$ is not simplicial but $f$ very well behaved?

Is the following known: Is $f$ continuous on $C$ if $C$ is bounded and $f$ is lsc, strictly convex and essentially smooth? (essentially smooth means that $f$ is differentiable in the interior of $C$ and for every sequence $(x_n)$ in the interior of $C$, if $x_n$ converges to a point $x$ to the boundary of $C$ then $\Vert \nabla f(x_n)\Vert \to \infty$)

Answer 1

I don't think that this is true. Let us take $$ C := \{ x \in \mathbb R^2 \mid x_1^2 \le x_2 \le 1\}$$ and $$ f(x) = \frac{x_1^2}{x_2} $$ for $x \in C \setminus \{0\}$, $f(0,0) = 0$. This function is convex, lsc but discontinuous in $(0,0)$. However, it is not strictly convex and not essentially smooth. I think that a function with these additional properties can be achieved by considering $f + g$, where $g$ is strictly convex, continuous and essentially smooth.

For example, we can replace $C$ by the (closed) ball with radius $1$ around $(0,1)$ and choose $$ g(x) = -\log( 1 - \| x - (0,1)\|^2 ). $$