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Suppose $\phi({\bf x},{\bf z})$ is a continuous function of two arguments,

$\phi: {\mathbb R}^N \times Z \rightarrow {\mathbb R}$,

where $Z \subset {\mathbb R}^m$ is a (non-empty) compact convex set. Further assume that $\phi({\bf x},{\bf z})$ is convex in ${\bf x}$ for every ${\bf z} \in Z$.

Is the function $f({\bf x}) = \max_{{\bf z} \in Z} \phi({\bf x},{\bf z})$ continuous? For sure it is convex, but what about continuity?

I think one can appeal to the so-called Danskin's theorem to answer this, but I am not sure.

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    $\begingroup$ For purposes of continuity of $f$, convexity is irrelevant, and Danskin's theorem is irrelevant. As $\bf x$ varies, the compact subsets $\phi({\bf x},Z)$ vary continuously in the Hausdorff metric on compact subsets of $\bf R$, and so their maxima vary continuously. But, this is really a question for math.stackexchange.com, not MO. $\endgroup$
    – Lee Mosher
    Commented Apr 27, 2012 at 16:55
  • $\begingroup$ I believe you are right, thanks. Do you know if the continuity is maintained when in addition one imposes that ${\bf x} \in {\bf X}$ where $\bf X$ is (non-empty) compact and convex? $\endgroup$
    – Stewart
    Commented Apr 27, 2012 at 16:58
  • $\begingroup$ Yes. All that is relevant to make this argument work is that $\phi : X \cross Z \to \R$ is continuous, $X$ is a topological space, and $Z$ is a compact topological space. Consult any introductory book on topology. $\endgroup$
    – Lee Mosher
    Commented Apr 27, 2012 at 17:09
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    $\begingroup$ Thanks! I understand I might have underrated the importance of this website with my ignorance - my apologies. $\endgroup$
    – Stewart
    Commented Apr 27, 2012 at 17:21

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