How to prove continuity of $g(y)=\max_{x\in K} f(x,y)$ when $K$ is compact?

Question

I have a function $f$ that is continuous over the set of variables, $f: \ K \times M \to \mathbb{R}$, where $K$ is a compact domain of a metric space and $M$ is a metric space. And I want to prove that the function $g(y)=\sup_{x\in K} f(x,y)$ is continuous.

As $K$ is a compact, $f$ is uniformly continuous with respect to its first argument. Also, for the same reason, $g(y)=\sup_{x\in K} f(x,y)=\max_{x\in K} f(x,y)$.

For now, I know that for the case of $f$ is continuous with respect of each variable separately, this doesn't work (see an example here Is supremum over a compact domain of separately continuous function continuous?) and there is a proof for the case when $M$ is also a compact, or $f$ is just uniformly continuous on $M$, too (How prove this $g(x)=\sup{\{f(x,y)|0\le y\le 1\}}$ is continuous on $[0,1]$), which I personally doubt as there is no guatantee that y-s will be close there (in my notations they are x-s).

All my tries fail when I come to the point where I need to say something about argmaxes of even close y-s. For example:

I consider a sequence $y_n \to y_0$ with $n\to\infty$. For each $y_n$ there exists $x_n$ such that $g(y_n)=\max_{x\in K} f(x,y_n)=f(x_n,y_n).$ Then, I try to estimate the difference $$|g(y_{n+k})-g(y_n)|=|f(x_{n+k},y_{n+k})-f(x_n,y_n)|=|f(x_{n+k},y_{n+k})-f(x_{n+k},y_{n})+f(x_{n+k},y_{n})-f(x_n,y_n)|\le|f(x_{n+k},y_{n+k})-f(x_{n+k},y_{n})|+|f(x_{n+k},y_{n})-f(x_n,y_n)|.$$ One can easily see that the first component tends to zero with $k\to\infty$, but I have no idea what to do with the second one...

Any help would be very appreciated!

Answer 1

Consider a sequence $(y_n)_{n\in\mathbb{N}}$ in $M$ that converges to some $y \in M$. We will show that

Claim. $g(y_n) \to g(y)$.

Step 1. We first show that $\liminf_{n\to\infty} g(y_n) \geq g(y)$.

This is an immediate consequence of the general fact that the supremum of a family of lower-semicontinuous functions is again lower-semicontinuous. (In particular, this fact does not require any structure on $K$.) Nevertheless, we write out the proof for self-containedness.

Let $\alpha$ be an arbitrary real satisfying $\alpha < g(y)$. Then there exists $x \in K$ such that $f(x, y) > \alpha$. So by the continuity of $f$, there exists $\delta > 0$ such that $f(x, y') > \alpha$ whenever $d(y', y) < \delta$. This then implies that

$$ g(y_n) \geq f(x, y_n) > \alpha \qquad \text{whenever } d(y_n, y) < \delta, $$

and so, we have $\liminf_{n\to\infty} g(y_n) \geq \alpha$. Since this is true for any $\alpha < g(y)$, the desired assertion follows by letting $\alpha \uparrow g(y)$.

Step 2. Next, we show that $\limsup_{n\to\infty} g(y_n) \leq g(y)$.

Indeed, for each $n$, use the compactness of $K$ to choose $x_n \in K$ so that $f(x_n, y_n) = g(y_n)$. Also,

• find a subsequence $I \subseteq \mathbb{N}$ so that $(g(y_n))_{n\in I} \to \limsup_{n\to\infty} g(y_n)$, and then

• find a further subsequence $J \subseteq I$ such that $(x_n)_{n\in J} \to x $ for some $x \in K$. This is possible by the sequential compactness of $K$.

Then

$$ \limsup_{n\to\infty} g(y_n) = \lim_{J \ni n \to \infty} g(y_n) = \lim_{J \ni n \to \infty} f(x_n, y_n) = f(x, y) \leq g(y). $$

Conclusion. Combining step 1 and 2, the desired claim follows. $\square$

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Discussion. The above proof works whenever $K$ is both compact and sequentially compact. When $K$ is a compact metric space, sequential compactness comes for free. I am not sure if we can come up with a proof that does not rely on sequential compactness.

Answer 2

Here's a proof that's more for fun than anything else (especially since there's already a perfectly good answer here).

Let $X$ be a sober space (every metric space is sober) and let $K$ be compact and sober. Then if $f : X \times K \to \mathbb{R}$ is continuous, the function $g(x) = \max_{k \in K} f(x,k)$ is continuous too.

We use the fact that the extreme value theorem is actually constructively true in the following form:

Let $K$ be a compact, positive, overt locale and $f : K \to \mathbb{R}$ continuous. Then $f$ admits a maximum $\max_K f \in \mathbb{R}$.

This is great because it means this statement is true inside the sheaf topos $\text{Sh}(X)$! Externalizing this statement (and cashing out locales for sober spaces everywhere) gives

Let $\pi : K \to X$ be a proper, open surjection and $f : K \to \mathbb{R}$ continuous. Then $x \mapsto \max_{k \in \pi^{-1} x} f(k)$ is a continuous function.

(This is basically because compact locales in $\text{Sh}(X)$ are proper maps to $X$, positive overt locales in $\text{Sh}(X)$ are open surjections to $X$, and a "real number" in $\text{Sh}(X)$ is a continuous function $X \to \mathbb{R}$)

But now let's interpret the above theorem with

• $\pi : X \times K \to X$, which one easily checks to be a proper open surjection
• $f : X \times K \to \mathbb{R}$, which we're assuming is continuous

Then since $\pi^{-1}(x) = \{x\} \times K$, the theorem tells us that the function $x \mapsto \max_{(x,k)} f(x,k)$ is continuous, as desired.

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As a brief soapbox pitch, this is one reason to care about constructive proofs! Constructively true things can be interpreted inside a sheaf topos, which gives us "continuously parameterized" versions of the theorem basically for free!

A few days ago when I saw this question, I recognized it as a version of the usual extreme value theorem where the continuous function $f$ and the resulting max $g$ are both allowed to depend on a bonus parameter from $X$. So I thought if I could find a constructive version of the extreme value theorem, it should give your theorem as a corollary! (Actually it gives something a bit more general than what you asked for). I'm not an expert in locale theory, and I had to wait a few days to find the theorem I cited here, but you can see how, with experience, these kinds of questions can be answered very quickly!

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I hope this helps ^_^