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Let $f: X \times Y \rightarrow \mathbb{R}$ be a continuous map. Show that if $Y$ is compact then the function $g: X \rightarrow \mathbb{R}$ defined by $g(x) = \inf \{f(x,y): y \in Y\}$ is also continuous.

No clue here. Can you please help?

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This is a standard exercise. Maybe if you tell us what have you tried (and you surely have tried something!) we could provide hints. Otherwise, I cannot think of much more than «use the definition of continuity». – Mariano Suárez-Alvarez Feb 3 '11 at 23:02
(By the way, you can type a prettier $\inf$ by saying \inf in $\LaTeX$.) – Mariano Suárez-Alvarez Feb 3 '11 at 23:04
@user6495: At least, start by showing that $g$ is well-defined, that is, that $g(x)\in\mathbb{R}$ for all $x\in X$ (you'll need compactness of $Y$ for this). – Arturo Magidin Feb 3 '11 at 23:08
@user6496: Well, it's not quite right, since $f$ is defined on $X\times Y$, so $f(y)$ does not make sense. But what you mean, I think, is that by compactness (in $\mathbb{R}$), $f_x(Y)$ is bounded and closed, so its infimum is a minimum, so there is a $y_x$ such that $\inf(f_x(Y)) = f_x(y_x)$. – Arturo Magidin Feb 3 '11 at 23:29
It might be useful to show that $g^{-1}(a,\infty)$ and $g^{-1}(-\infty,a)$ are open for all $a\in\mathbb{R}$. – Joe Johnson 126 Feb 3 '11 at 23:39
up vote 1 down vote accepted

Following the suggestion of Joe Johnson (I was going using intervals $(a,b)$, but this is simpler) (do you see why Joe's suggestion is enough? The sets $(a,\infty)$ and $(-\infty,a)$ form a subbasis for the topology of $\mathbb{R}$, and for a function to be continuous, it suffices to show that the inverse image of every set in a given subbasis is open).

Let $a\in\mathbb{R}$. What is $g^{-1}(-\infty,a)$? It consists of all $x\in X$ such that $g(x)\lt a$. If $g(x)\lt a$, there exists $y_x$ such that $f(x,y_x)\lt a$. Since $f$ is continuous, $f^{-1}(-\infty,a)$ is open, and $(x,y_x)\in f^{-1}(-\infty,a)$, so there exist open sets $U_x$ and $V_y$ of $X$ and $Y$, respectively, such that $U_x\times V_y\subseteq f^{-1}(-\infty,a)$.

Now, suppose $x'\in U_x$. Then for all $y\in V_y$ (in particular, for $y_x$) you have $f(x',y_x)\in (-\infty,a)$. What does that tell you about $g(x')$? What does that tell you about $g^{-1}(-\infty,a)$?

Now try to do something along those lines with $g^{-1}(a,\infty)$.

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I think you want that to read "all $x\in X$ such that $g(x)<a$". – Joe Johnson 126 Feb 3 '11 at 23:56
@Joe: It was wrong, but I think I fixed it; you may want to refresh the page...? – Arturo Magidin Feb 3 '11 at 23:57
thanks! got it. – user6495 Feb 4 '11 at 0:33
Your proof does not use the compactness of $Y$. That is because you need it in the (more difficult part) concerning $g^{-1}(a,\infty)$ that you ommited. – Meneldur Oct 30 '12 at 10:48

Perhaps, we can suppose that $X$ and $Y$ are metric spaces (in particular subset of real line $\mathbb R$) and that $f(x,y)$ is real-valued function in $x ,y$. For a simple model we can take $X=\mathbb R$ and $Y=[a,b]$ closed interval in $\mathbb R$.

Set $I(x) = \inf_{y\in Y} f(x,y)$. Fix a point $x_0$ in $X$ .

Let us prove that $I(x)$ is continuous at $x_0$.

Consider a sequence $x_n$ in $X$ such that $x_n$ converges to $x_0$.

Let us prove that $I(x_n)$ converges to $I(x_0)$.

Since $Y$ is compact there is a sequence $(y_n)$ and $y_0$ in $Y$ such that $I(x_n) = f(x_n,y_n)$ and $a_0:= I(x_0)= f(x_0,y_0)$. Since $Y$ is compact there is a subsequence of $y_n$ which converges to $y^0$ in $Y$. By abusing notation we can suppose that $(x_n,y_n)$ converges to $(x_0, y^0)$. By continuity of $f$ in $x ,y$ we conclude that $f(x_n,y_n)$ converges to $f(x_0, y^0):=b_0$. By definition of $I(x_n)$, $f(x_n,y_n) \leq f(x_n,y_0)$ and therefore since $(x_n,y_n)$ converges to $(x_0, y^0)$ and $(x_n,y_0)$ converges to $(x_0, y_0)$, by continuity of $f$, we find

$b_0=f(x_0,y^0)\leq a_0$. On the other hand, by definition $a_0\leq b_0$. Hence $a_0=b_0$.

This shows that the set of cluster points (or accumulation points) of sequence $I(x_n)$ is a single point $a_0$ and therefore $I(x_n)$ converges to $a_0=I(x_0)$.

Therefore $I$ is continuous in $x_0$ .

If necessary, we can give more details.

You can also see Mathematical Analyis I/II. by Vladimir A. Zorich.

Can I get a text book reference for the proof that $g(x) = \inf_{y\in Y} f(x,y)$ is continuous in $x$ given compactness of $Y$ and continuity of $f$ in $x ,y$? - ResearchGate. Available from: [accessed Jul 29, 2015].

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