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We define a affine(concave), upper semi continuous function and bounded function $f:X \to \mathbb{R}$, where $X \subset \mathbb{R}^{k}$ is compact and convex set. Assume that $T$ is an affine and upper semi continuous function on $X$. Let $S$ be a concave function on a compact and convex set $A$ defined by $S(t)=\sup\{f(x), T(x)=t\}$.

$\textbf{Question}$: Is $S$ upper semi-continuous?

My attempt: Since $S$ is concave, $S$ is continuous in the interior of $X$. On the other hand, the supremum of upper semicontinuous function is not necessarily upper semi-continuous.

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Yes. Your assumptions imply $f$ is continuous and $X$ is compact by assumption. It should not be too hard to prove $S$ is usc using the $\varepsilon$-$\delta$ definition of usc.

Here is a more geometric argument:

One generally answers these questions by going through epigraphs. Function is lsc ($-f$ here) if the epigraph $\mathrm{epi}(-f) = \{(x,y)|y\geq -f(x)\}$ is closed.

So, you need to show that the epigraph of $-S$ is closed. But, $$ \mathrm{epi}(-S) = \{(t,y)|y\geq -S(t)\} = \{(t,y)|y\geq -f(x),\ Tx=t\} $$ i.e. $\mathrm{epi}(-S)$ is the projection of the set $\mathrm{epi}(-f)$, $Tx=t$. So, by compactness of $X$ and continuity of $f$, also $S$ is usc.

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  • $\begingroup$ Why $f$ is continuous? $\endgroup$
    – Adam
    Dec 12 '20 at 16:08
  • $\begingroup$ Could you a bit explain about the projection? $\endgroup$
    – Adam
    Dec 12 '20 at 16:18
  • $\begingroup$ Is it true that even $f$ was ups not continuous, then the last line works? $\endgroup$
    – Adam
    Dec 12 '20 at 16:23
  • $\begingroup$ You need usc of $f$ for this argument to work. $\endgroup$
    – masi
    Dec 12 '20 at 23:14
  • $\begingroup$ The projection thing is this: $\endgroup$
    – masi
    Dec 12 '20 at 23:14

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