# When is a convex function continuous on its domain?

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 $$\Vert \nabla f(x_n)\Vert \to \infty$$)

• Maybe in the last condition you mean $x\in\partial C$ ? Jun 10, 2021 at 4:24
• Keep in mind simple examples in dim 1, e.g. $f(x):=-\sqrt{1-x^2}$ on $[-1,1)$, and $f(1):=17$ Jun 10, 2021 at 4:30
• edited, thanks. Jun 10, 2021 at 4:53

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 $$g(x) = -\log( 1 - \| x - (0,1)\|^2 ).$$