Convexity of a minimum function

Question

I was reading a proof of $9g-9$ theorem which states that $9g-9$ length parameters are sufficient the parametrize the Teichmuller space of a closed surface of genus $g$. The proof uses the following fact.

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Theorem: Let $f:\mathbb{R}^m\times \mathbb{R}^n\rightarrow \mathbb{R}$ be a strictly conves function. If the function $F:\mathbb{R^m}\rightarrow \mathbb{R}$ is defined by $$F(x) = \min \left\{ f(x,y) ; y \in \mathbb{R}^n \right\}$$ is well defined, i.e., if the minimum always exists then $F$ is always strictly convex.

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Can someone please give me any proof or at least idea of the proof of this fact.

P.S: I am reading the book "A primer on mapping class group." And I don't want a reference.

Answer 1

This is a standard result in convex analysis. See for example, $\S$3.2.5 of Convex Optimization by Boyd and Vandenberghe (just slightly modify their proof to conclude strictness).

Answer 2

An idea of the proof. For the convexity of $F$: a function is convex iff its epigraph is convex; the epigraph of $F$ is the projection of the epigraph of $f$; the projection of a convex set is convex. Note that this part also work with $\inf$ more generally that $\min$ in the definition.

For the strict convexity: assume that $F$ is convex but not strictly convex. Then, up to adding a linear form to $F$, the $F$ has more than a minimum point. Any minimum point of $F$ is the projection of a minimum point of $f$, and since $F$ has more than a minimum point, so does $f$, and $f$ is not strictly convex.

Finally, note that a strictly convex and bounded below function $f$ does not produce in general a strictly convex $F(x):=\inf_y f(x,y)$, like the example of $f(x,y):=\exp(x^2/2+y)$ shows, for $f$ is strictly convex and $F$ is identically $0$.