Timeline for Is a convex, lower semicontinuous function that is bounded from below, actually continuous?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Jun 10 at 18:25 | answer | added | user529441 | timeline score: 1 | |
| Apr 27, 2022 at 9:25 | vote | accept | iolo | ||
| Apr 1, 2022 at 10:58 | comment | added | gerw | I would like to add that some assumption on $X$ is necessary. Indeed, if we consider an infinite-dimensional Hilbert space $H$ equipped with its weak topology, then $f = \|\cdot\|$ is convex, lower semicontinuous but not continuous. | |
| Apr 1, 2022 at 10:34 | comment | added | Christian Clason | (Shameless plug: Section 3.3 in arxiv.org/abs/2001.00216) | |
| Apr 1, 2022 at 9:16 | comment | added | Christian Clason | Yes. This is a standard result in convex analysis that can be found in most textbooks. Note that lower semicontinuity (which you assume explicitly) is an essential requirement in infinite dimensions; you also have to assume that you are in the interior of the effective domain (which you assume implicitly since your $f$ is real-valued and cannot attain the value $+\infty$). | |
| Apr 1, 2022 at 9:04 | history | edited | YCor | CC BY-SA 4.0 |
fixed typo
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| Apr 1, 2022 at 7:19 | answer | added | iolo | timeline score: 3 | |
| Mar 31, 2022 at 16:05 | comment | added | iolo | Indeed, see e.g this question. Any counterexample should have to be in the infinite-dimensional setting. | |
| Mar 31, 2022 at 15:58 | comment | added | usul | Hm, I think a convex function defined on all of $\mathbb{R}^2$ is already continuous... | |
| Mar 31, 2022 at 15:56 | comment | added | iolo | I tried to find some and the closest I got was seminorms that are measurable with respect to some Gaussian measure. But these seem overly abstract and are usually not lower semicontinuous. Your example is not a contradiction, since it is not defined on all of $\mathbb{R}^2$. | |
| Mar 31, 2022 at 15:42 | comment | added | usul | Did you look at examples of lower semicontinuous convex functions that are not continuous? [This answer]math.stackexchange.com/a/2487999/9759) seems to give one that is bounded below: $f(x,y) = x^2/y$ defined on the set $\{(x,y) : y \geq x^2\}$ where we define $f(0,0) = 0$. | |
| Mar 31, 2022 at 14:43 | history | asked | iolo | CC BY-SA 4.0 |