1
$\begingroup$

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?

$\endgroup$
2
  • 1
    $\begingroup$ 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. $\endgroup$
    – Junekey Jeon
    Jan 6, 2021 at 9:38
  • $\begingroup$ @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. $\endgroup$
    – Zim
    Jan 6, 2021 at 15:53

0

Reset to default

You must log in to answer this question.