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Let $f:\mathbb{R}\times[0,1]\to\mathbb{R}$ continuous and $c:\mathbb{R}\to[0,1]$ continuous. Consider $$F:\mathbb{R}\to\mathbb{R},\ \ F(x)=\max_{t\in[0,c(x)]}f(x,t)$$ Is $F$ continuous? I believe it is true, but I've difficulties to prove it. I managed to prove that fixing the parameter in one of the two places then the obtained function is continuous, i.e. $$x\mapsto\max_{t\in[0,c(x_0)]}f(x,t) \qquad\text{and}\qquad x\mapsto\max_{t\in[0,c(x)]}f(x_0,t) \qquad\text{are continuous.}$$ But now it seems not trivial to conclude by the triangle inequality...

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Note that the set $C:=\{(x,t)\mid x\in\Bbb R,\ t\in[0,c(x)]\}$ is closed and contains the graph of $c$, the set $G(c) = \{(x,c(x))\mid x\in\Bbb R\}$, which is closed in $\Bbb R \times [0,1]$

Show that the preimage of an open subbase set $(-\infty,a)$ under $F$ is open: This is the set $\Bbb R-\{x\mid \exists t\le c(x),f(x,t)\ge a\}$. The complement can be expressed as $\pi_{\Bbb R}(f^{-1}([a,\infty)\cap C)$. Since $[a,\infty)$ is closed, its preimage under $f$ is closed. But if we now apply the projection $\pi_{\Bbb R}$, we get a closed set again. This is because the projection $X\times Y\to X$ is closed if $Y$ is compact, so it is in the case $\Bbb R\times[0,1]\to\Bbb R$.

Still, there is a problem if we want to apply the same argument to an open subbase set $(a,\infty)$ since the restriction of an open map, like the projection, to a closed subset isn't necessarily open. In this case, however, $\pi_{\Bbb R}|_C$ is indeed open. This is obvious on $C\setminus G(c)$. And on $G(c)$, we have $\pi_{\Bbb R}((U×V)\cap G(c))=c^{-1}(V)\cap U$.

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  • $\begingroup$ Thank you, I like this topologial proof. I tried to generalize it when the domain and the function depend on two independent variable x and y: it should still hold. But I'd like to ask you how to prove the fact that the projection is closed when the other factor is compact. $\endgroup$
    – qwertyuio
    Jun 21 '13 at 14:12
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    $\begingroup$ Maybe there's a way not to use this result about projections: as continuity is a local property, w.l.o.g. substitute $\mathbb{R}$ in the statement with a compact interval $[a,b]$, then just use the fact that a continous function from a compact set is closed. Is this reasoning correct? $\endgroup$
    – qwertyuio
    Jun 21 '13 at 14:21
  • $\begingroup$ Yes, it's correct, but you still have to find a way to deduce the global closedness from the local closedness. Here it is easy, you can use the local finiteness of the family of intervals into which you were going to subdivide the real line, I guess. You could also use the following result: A proper map into a locally compact Hausdorff space is closed. A proper map is a continuous map such that the preimage of each compact set is compact. $\endgroup$ Jun 21 '13 at 14:47
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    $\begingroup$ @qwertyuio: Take a closed set $C$ in $X\times Y$. Let $x$ be a point outside of $\pi(C)$, the projection in $X$. The compact set $\{x\}\times Y$ is contained in the open complement of $C$. For each $y\in Y$ there is an open box $U_y\times V_y$ containing $(x,y)$ and not intersecting $C$. By compactness finitely many of these boxes cover $\{x\}\times Y$. Then the intersection of the corresponding $U_y$'s is an open neighborhood of $x$. $\endgroup$ Jun 21 '13 at 14:54
  • $\begingroup$ You may also be interested in the Tube Lemma, of which this proof uses a special case. The special case mentioned in the article is actually equivalent to the closedness of the projection. $\endgroup$ Jun 21 '13 at 14:57
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Hint: If $c(x)=c_0 \in (0,1)$ then it may be that $f$ has a value exceeding $f(c_0)$ which occurs at some $x<c$. In this case a small movement of $x$ will only move $c(x)$ near $c_0$ so the max will stay the same.

On the other hand it may be that $f$ has its maximum value on the interval $[0,c_0]$ of $f(c_0)$ i.e. it is maximal at $c(x)=c(x_0).$ In this case as $x$ moves near $x_0$ it may either cause $c(x)$ to go back down below $c(x_0)$, or move $c_(x)$ above $c_0$, but either way the continuity of $f$ should keep the max from jumping.

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