# Nontrivial example of a discontinuous real-valued convex function?

Let $$\mathcal{X}$$ be a real Hilbert space, and let $$f\colon\mathcal{X}\to\mathbb{R}$$ be convex.

It's known that if $$\mathcal{X}$$ is finite-dimensional, then $$f$$ is also continuous. However, if $$\mathcal{X}$$ is infinite dimensional, this result is no longer true, since there are discontinuous linear functionals (and linear functionals are convex).

I'd like to get a better intuition on this class of discontinuous, convex, real-valued functions, so I'm looking for more nonlinear examples. A simple affine example is constructed here. However, that example is not incredibly enlightening since it is based off of the linear case. Are there other examples?

• IIRC, an everywhere defined real-valued convex function on a Banach space is continuous if and only if locally bounded. This might help finding a nonlinear example. Jan 6, 2021 at 9:38
• @KaviRamaMurthy I'm not sure I follow, since there are explicit examples of discontinuous linear functionals. For instance, in the space of polynomials over $[0,1]$ under the $\sup$ norm, the linear map $D\colon f\mapsto f'(1)$ is discontinuous, since it maps the bounded sequence $(x^k)_{k\in\mathbb{N}}$ to the unbounded sequence $(k)_{k\in\mathbb{N}}$. Now that I think of it, I can't come up with an explicit example in a Hilbert space. However, it's a much stronger statement to say it's impossible.
– Zim
Jan 6, 2021 at 15:53