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The question says it all. I'd be most interested in hearing the answer for an arbitrary topological vector space, though functions on $\mathbb R ^n$ are also of interest.

I know that if a strictly convex function has a minimum, then it's unique. But I'm not sure whether the minimum is necessarily attained without also assuming the function is continuous (or at least lower semicontinuous).

aduh
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    Take $f(x)=e^{-x}$ on $[0,1]$ and modify it by $f(1)=1$. – A.Γ. Aug 09 '20 at 07:00
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    In $\mathbb{R}^n$ this only fails due to boundary effects, see Proof of “every convex function is continuous”. If a minimizing sequence converges to an interior point then the minimum will be attained at that point (convex functions are continuous in the interior of their effective domains). But in Banach spaces you need to additionally assume that the function is bounded from above. – Conifold Aug 09 '20 at 07:32
  • @Digitallis I specifically asked about arbitrary (i.e. possibly infinite dimensional) topological vector spaces, and Conifold helpfully addressed that in his/her comment. The question as stated in the title says nothing about the space being finite dimensional. – aduh Aug 09 '20 at 11:03
  • Ha yes ! I see that know. Perhaps I should've of read your question more carefully.That's quite ironic :/ Apologies. – Digitallis Aug 09 '20 at 11:14
  • I suppose the following need clarification: 1) What is the codomain of $f$? 2) by "strictly convex", do you mean $f$ is strictly convex on the vector space or only on the compact set? – Zim Aug 10 '20 at 19:57
  • @Zim I had in mind extended-real-valued functions, which I should have said of course. I will edit now. – aduh Aug 10 '20 at 20:18
  • @grndl Ok thanks, and for item 2)? – Zim Aug 12 '20 at 03:37
  • @zim I mean that f is strictly convex on whatever its domain happens to be. In general, we can assume that the domain is a convex set, no more. – aduh Aug 12 '20 at 05:09
  • @grndl You mean that $f$ takes the value of $+\infty$ off of its domain, right? – Zim Aug 12 '20 at 14:26
  • @Zim I'm not sure. I'd like to understand what happens in both cases rather than assume away the case where $f$ assumes the value $+\infty$. – aduh Aug 12 '20 at 21:49
  • @grndl In every convex analysis text I've seen: For a vector space $\mathcal{X}$, definition of the domain of the function $f\colon\mathcal{X}\to\mathbb{R}\cup{+\infty}$ is $\textrm{dom}f={x\in\mathcal{X},|,f(x)\in\mathbb{R}}$, i.e. $f$ is usually not assumed to be "undefined" outside of its domain; it just takes the value of $+\infty$ off of its domain. – Zim Aug 13 '20 at 16:59
  • If you want to generalize to the case when $f$ is not always real-valued, then just add the assumption of lower-semicontinuity and you're done. That addresses @A.Γ. 's counterexample as well, as it is not lower semicontinuous. – Zim Aug 13 '20 at 17:00

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