# A concave function as supremum of upper semi continuous is upper semi continuous

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.

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.

• Why $f$ is continuous?
• Is it true that even $f$ was ups not continuous, then the last line works?
• You need usc of $f$ for this argument to work.