The Existence of a Maximum Point for a Continuous Function on a Metric Space

Question

Assume that $(X,d)$ is a metric space and that $f \colon X \to [0, \infty)$ is a continuous function. Assume that for each $\epsilon > 0$, there is a compact set $K_{\epsilon} \subseteq X$ such that $f(x) < \epsilon$ when $x \notin K_{\epsilon}$. Our task is to show that $f$ has a maximum point.

My process for solving this problem is the following. We first need a wise choice for $\epsilon$, let that be $f(z)$ for some $z \in X$. We now know that $f(x) < f(z)$ for all $x \notin K_{f(z)}$. Now, $z$ is clearly a candidate for the global maximum, but we turn our attention to the compact subset $K_{f(z)}$. The latter itself has a maximum point, say $c$, for which $f(c) \geq f(x)$ for all $x \in K_{f(z)}$. Here is the part where I cannot proceed. My guess is that it is either the case that $z \in K_{f(z)}$ in which case we have a global maximum at $c$. But this does not have to be the case; I mean, the only thing that we know so far is (roughly speaking) that $z$ is the maximum for $K_{f(z)} \setminus X$ and $c$ for $K_{f(z)}$, it is also not obvious to me why $f(c) \geq f(z)$. How do I draw a conclusion here? Can I just say that either $z$ or $c$ is the maximum?

Answer 1

I claim that if $f(c) \geq f(z)$, then $c$ is a maximum point. Otherwise, if $f(c) \leq f(z)$, then $z$ is a maximum point. In either case, a maximum point exists.

Indeed, if $f(c) \geq f(z)$, when $x \notin K_{f(z)}$, we have $f(x) < f(z) \leq f(c)$. When $x \in K_{f(z)}$, by definition of $c$ we have $f(x) \leq f(c)$. Hence, for all $x$, $f(x) \leq f(c)$.

If $f(c) \leq f(z)$, when $x \notin K_{f(z)}$, we have $f(x) < f(z)$. When $x \in K_{f(z)}$, by definition of $c$ we have $f(x) \leq f(c) \leq f(z)$. Again, we have that, for all $x$, $f(x) \leq f(z)$.

In either case, we have shown the existence of a maximum point, so we're done.

Answer 2

If $f\equiv 0$ there is nothing to prove. Since $f$ is bounded on $K_1$ and less than $1$ in the complement it follows that $M=\sup f$ is finite.

There is a sequence $(x_n)$ such that $f(x_n) \to M$. Note that $M>0$. Taking $\epsilon <M$ we see that $f(x_n) >\epsilon$ for sufficiently large $n$ so $x_n \in K_\epsilon$ for such $n$. There is,therfore, a subsequence converging to some point $x$. Can you finish the proof by showing that $f(x)=M$?

Answer 3

We first need a wise choice for $\epsilon$, let that be $f(z)$ for some $z \in X$.

This sentence is a bit clumsy, and like all clumsy sentences in a mathematical reasoning, it hides away several subtleties. It would be better to be more explicit.

Importantly, you need to make a disjunction of cases. Either $\forall x \in X, f(x) = 0$, or $\exists x \in X, f(x) > 0$. In the first case, $K_{\varepsilon}$ is going to be the empty set no matter what we choose for $\varepsilon > 0$, but happily, if $f$ is constant equal to 0, then the maximum of $f$ is 0 and we're done. For the remainder of the proof you can assume we're on the second case, and so you are now allowed to write things like "Let $z \in X$ such that $f(z) > 0$." and then "Let $\varepsilon = f(z)$."

We now know that $f(x) < f(z)$ for all $x \notin K_{f(z)}$. Now, $z$ is clearly a candidate for the global maximum, but we turn our attention to the compact subset $K_{f(z)}$. The latter itself has a maximum point, say $c$, for which $f(c) \geq f(x)$ for all $x \in K_{f(z)}$.

So far so good. Note that this is where you're using the assumption that $f$ is continuous, so it would be good to cite it. "$f$ is continuous and $K_{f(z)}$ is compact, therefore $f$ has a maximum on $K_{f(z)}$. Let $c \in K_{f(z)}$ such that $f(c)$ is the maximum of $f$ on $K_{f(z)}$."

Here is the part where I cannot proceed. My guess is that it is either the case that $z \in K_{f(z)}$ in which case we have a global maximum at $c$.

Yes, $z \in K_{f(z)}$ by definition of $K_{f(z)}$, and we have a global maximum at $c$.

But this does not have to be the case; I mean, the only thing that we know so far is (roughly speaking) that $z$ is the maximum for $K_{f(z)} \setminus X$ and $c$ for $K_{f(z)}$

No? $K_{f(z)}$ is a subset of $X$, so $K_{f(z)} \setminus X$ is the emptyset. What we know, and I think that might be what you were trying to write, is that $f(z)$ is an upper-bound of $f$ on $X \setminus K_{f(z)}$.

it is also not obvious to me why $f(c) \geq f(z)$.

Ah, yes. Well that's the crucial point. Without this inequality it's impossible to conclude. Happily this inequality is easy to get. This inequality is given to us because both $c$ and $z$ are in $K_{f(z)}$. How do we know that $z \in K_{f(z)}$? This is given to us by the definition of $K_{f(z)}$. By definition of $K_{f(z)}$, we know that $\forall x \in X \setminus K_{f(z)}, f(x) < f(z)$. In particular, $\forall x \in X \setminus K_{f(z)}, x \neq z$. In other words, $z \notin X \setminus K_{f(z)}$. In other words, $z \in K_{f(z)}$.

Now you have all the facts:

• $\forall x \in K_{f(z)}, f(x) \leq f(c)$;
• $z \in K_{f(z)}$;
• $\forall x \in X \setminus K_{f(z)}, f(x) < f(z)$.

Since $K_{f(z)} \cup (X \setminus K_{f(z)}) = X$, you should be able to conclude: $$\forall x \in X, f(x) \leq f(c).$$

Answer 4

The Extreme Value Theorem says that every continuous function on a compact set achieves a maximum. So once you show the existence of $x$ such that $f(x)>0$ and $x\in K_{f(x)}$, then $K_{f(x)}$ has a maximum, it's at least $f(x)$, so it's greater than all values outside the set, so it's a global maximum.