Let $X$ be a compact Polish space and $Y$ be a separable real Banach space. Assume $U \subseteq X \times Y$ is open, bounded in $Y$-norm, and s.t. for any $x \in X$, $\{y \in Y \mid (x,y) \in U\}$ is convex and nonempty. Does it follow that there is a continuous function $f: X \rightarrow Y$ s.t. the graph of $f$ is contained in $U$?
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Yes, and this follows directly from a selection theorem in
Michael, Ernest. “Continuous Selections. I.” Annals of Mathematics, vol. 63, no. 2, 1956, pp. 361–382. Second Series, www.jstor.org/stable/1969615.
Theorem 3.1''' on page 368 shows that a lower hemicontinuous nonempty-valued correspondence from a perfectly normal space to separable Banach space such that all values are convex and, moreover, finite-dimensional, closed, or have an interior point, then the correspondence admits a global section. The statement in the paper contains a typo, as is clear from the statement right after it (the space $\mathcal{K}(Y)$ has to be replaced by $\mathcal{D}(Y)$).
answered Jan 22, 2017 at 10:18
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$\begingroup$ Thank you Michael! This theorem seems practically sufficient for my purposes, although I don't entirely follow how it answers the question as stated? In Michael's selection theorem, the mulitmap has to have closed values. So, I can take the closure of my open multimap and it will still be lower hemicontinuous (I think?). But, on the face of it, the resulting selection may take values on the boundary (outside of $U$)? $\endgroup$– VanessaCommented Jan 22, 2017 at 10:43
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$\begingroup$ @squark You are right. I do think a version in Michael's original paper adresses the case though. Let me take a look. $\endgroup$ Commented Jan 22, 2017 at 10:47
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$\begingroup$ I'm slightly confused. It looks like Theorem 3.1''' in the original paper is exactly what I need. The confusing part is that the comment under the theorem says "it would be nice if $D(Y)$ could be replaced by $K(Y)$ in (c) above, but Example 6.3 shows this is impossible." Now, as it's written, we actually have $K(Y)$ in (c) rather than $D(Y)$. I assume this is an error, and this error doesn't spoil anything since in section 5 it's explained that any convex set with an interior point is in $D(Y)$. Only I'm still confused because Example 6.3 uses open sets but a non-separable Banach space. $\endgroup$– VanessaCommented Jan 22, 2017 at 11:44
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$\begingroup$ Oh, I haven't noticed you edited the answer. This still doesn't solve my confusion about Example 6.3, but I guess the answer is legit. Thanks again! $\endgroup$– VanessaCommented Jan 22, 2017 at 12:09
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1$\begingroup$ @Squark A theorem that also does the trick with a more accessible proof is Theorem 3.1. in Yannelis, N. C., & Prabhakar, N. D. (1983). Existence of maximal elements and equilibria in linear topological spaces. Journal of Mathematical Economics, 12(3), 233-245.. $\endgroup$ Commented Jan 22, 2017 at 13:38