Family of bounded functions which are pointwise bounded is bounded.

Question

I am working on a problem from Simmons' Introduction to Topology and Modern Analysis.

Let $X$ be a compact metric space and $F$ a closed subspace of $\mathcal{C}(X, \mathbb{R})$ or $\mathcal{C}(X, \mathbb{C})$. Show that $F$ is compact if it is equicontinuous and $F_x = \{f(x): f \in F\}$ is a bounded set of numbers for each point $x\in X$.

Now, if we show that $F$ is bounded with respect to sup norm, it will follow from Ascoli's theorem that $F$ is compact. Hence, it suffices to show that $\sup_{x\in X} |f(x)|$ exists for each $f \in F$. Here is the progress I have made so far:

We know that each of the $F_x$ is bounded. Thus $M_x := \sup_{f\in F} |f(x)| < \infty$.

Further, the image of $X$ under each $f \in F$ is bounded (since, $X$ is compact, $f(X)$ is a compact subspace of $\mathbb{R}$ and is thus bounded). So, $K_f := \sup_{x\in X} |f(x)| < \infty$.

If we show that $\sup_{x \in X} M_x$ or $\sup_{f\in F}K_f$ exists, then, it will follow that $F$ is bounded.

Hence, all that remains to prove is that $\{M_x:x\in X\}$ or $\{K_f:f\in F\}$ is bounded.

So, we can restate the problem as follows:

$F$ is a family of continuous, bounded functions on a compact set $X$. Such that:

1. At each point $x \in X$, $$M_x := \sup_{f\in F} |f(x)| < \infty$$
2. For each function $f \in F$, $$K_f := \sup_{x\in X} |f(x)| < \infty$$

We need to show that either, $$\sup_{x \in X} M_x = \sup_{x\in X}\sup_{f\in F} |f(x)| < \infty$$ or $$\sup_{f \in F} K_f = \sup_{f \in F}\sup_{x\in X} |f(x)| < \infty$$

I am stuck here. I am aware that continuous functions on a compact set attain their extreme points, but I am not sure if that fact will be helpful here.

Any suggestions/hints are welcome.

Answer 1

The pointwise bounded to globally bounded is a straightforward application of compactness of $X$ and equicontinuity of $F$:

We have that $F$ is equicontinuous, so:

$$\forall \varepsilon >0: \exists \delta >0: x_0 : \forall x, x' \in X: \forall f \in F: d(x,x') < \delta \to |f(x)-f(x')| < \varepsilon$$

Now also suppose that $F$ is pointwise bounded: $M_x = \sup \{|f(x)| : x \in F\} < \infty$ for all $x \in X$.

Now find $\delta >0$ for $\varepsilon=1$ in the definition of equicontinuity. We can find finitely many $x_1,\ldots,x_m$ such that the balls $B(x_i, \delta)$ are a cover of $X$ (or apply total boundedness of $X$ with this $\delta$). So $M := \max(M_{x_1}, \ldots,M_{x_m}) < \infty$

Now, if $x, x'$ are any two points of $X$, we have $x_{i_1}$ with $x \in B(x_{i_1}, \delta)$ and $x_{i_2}$ with $x' \in B(x_{i_2}, \delta)$. Then for any $f \in F$ we have

$$|f(x)-f(x')| \le |f(x)-f(x_{i_1})| + |f(x_{i_1})- f(x_{i_2})| + |f(x_{i_2}) - f(x')| \le 1 + |f(x_{i_1})| + |f(x_{i_2})| + 1 \le 2M+2$$

where the $1$'s come from the equicontinuity.

But this easily implies that $\sup \{|f(x)|: x \in X, f \in F\} < \infty$ as well.