Let $g(a)=\max_{x\in \mathcal{C}}f(a,x)$ where $f(a,x)$ is a differentiable and continuous function of $x$ and $a$, and $\mathcal{C}$ is a bounded convex region. Is $g(a)$ also continuous?
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1$g(a)$ isn't always defined; there isn't always a max for $f$ as you vary $x$. For instance, $f(a, x) = ax$, which gives $g(0) = 0$, and $g(a)$ undefined for any other $a$. – Arthur Aug 15 '17 at 07:50
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You should require that $f$ is bounded and not the region. E.g. $(0,1)$ is bounded and convex, but still $f(x)=1/x$ does not have a maximum on it. – Hyperplane Aug 15 '17 at 08:00
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If you want to fix the problem I mentioned above that way, then you need $\mathcal C$ to be a compact region. I would rather go with what @Hyperplane suggested and say that $f$ is bounded above with respect to $x$ (and change $\max$ to $\sup$, since even though it's bounded above, $-e^x$ doesn't have a max). – Arthur Aug 15 '17 at 08:01
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@Hyperplane You means that if $f$ is bounded at the region, then $g(a)$ is continuous? Right? – Dave Aug 15 '17 at 08:07
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I mean that at least then we have a $g$ in the first place. Once we do, we can analyse continuity. – Arthur Aug 15 '17 at 08:10
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No, the function would not necessarily be continuous. Take, for instance, $$ f(a, x) = \frac{1}{1+e^{-a^2x}} $$ Here, $f(0, x) = \frac12$, so $g(x) = \frac12$, but for any $a\neq 0$, you can make $f(a, x)$ arbitrarily close to $1$ by making $x$ large. Thus we have $$ g(a) = \cases{\frac12& if $a = 0$\\1 & otherwise} $$ which is not continuous.
Note: this assumes the condition that $f$ is bounded, but $x\in \Bbb R$ (and $g(a) = \sup_x f(a, x)$ instead of $\max$). If the domain of $x$ is restricted to an open interval $(i, j)$, then $f$ may be adapted to that interval by instead using $f(a, 1/(j-x))$. If the domain of $x$ is half-open, then $f$ can still be adapted.
However, if the domain of $x$ is a closed, bounded interval, then I am pretty certain that $g$ is continuous, although I haven't got a proof.
Arthur
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