4
$\begingroup$

Why is an equicontinuous and pointwise bounded sequence of real-valued functions on a compact metric space uniformly bounded? I couldn't get my head around this. Can anyone please explain?

$\endgroup$
5
$\begingroup$

Heuristically, if you have a family of equicontinuous functions, then you could essentially treat them as a "single continuous function" because for any $\epsilon>0$ you could always find a single $\delta>0$ that will work for your entire family. Moreover, we know a continuous function on a compact metric space attains its maximum, i.e. bounded. Hence together you can see why the family should be uniformly bounded.

Edit:

Here's the proof.

Let $(X, d)$ denote the compact metric space and $\mathcal{F}$ is our equicontinuous family of functions.

Fix $\epsilon>0$. Then for every fixed $x \in X$, by the equicontinuity property of $\mathcal{F}$, we can find a $\delta_x>0$ (i.e. $\delta$ depending on $x$) such that \begin{align} |f_n(x)-f_n(y)|<\epsilon \ \ \text{ whenever } \ \ |x-y|<\delta_x. \end{align} In particular, it follows \begin{align} |f_n(y)| < \epsilon +|f_n(x)| \leq \epsilon +M_x \ \ \text{ whenever } \ \ |x-y|<\delta_x. \end{align} Note that we have used pointwise boundedness of $\mathcal{F}$. Next, observe $X \subset \bigcup_{x \in X} B(x, \delta_x)$ is an open cover of our compact metric space. Hence by compactness there exists a finite subcover say $\bigcup^N_{i=1} B(x_i, \delta_i)$. Let $M = \sup_{1\leq i \leq N} M_{x_i}<\infty$, then we see that \begin{align} |f_n(y)| <\epsilon+M \end{align} for every $y \in X$ and $n\in \mathbb{N}$.

$\endgroup$
1
  • $\begingroup$ Yep, I could see that conceptually, but couldnt come up with a delta-epsilon proof $\endgroup$ – user1559897 Sep 21 '16 at 0:29
3
$\begingroup$

Here's a formal proof, which may be found in many textbooks (Rudin chapter 7, for instance).

Let $\{f_n\}$ be the sequence of functions and let $K$ be their compact domain. By equicontinuity, for any $\epsilon>0$ there exists $\delta>0$ such that for all $f_n$ $|f_n(x)-f_n(y)|<\epsilon$ for $d(x,y)<\delta$. By the compactness of $K$, we can find finitely many points $p_1,\ldots,p_r$ such that $K$ is covered by $\delta$-balls centered at the $p_i$. Now for each $p_i$ there exists $M_i$ such that $|f_n(p_i)|<M_i$ for all $n$ so that letting $M$ be the max of the $M_i$ we find, by using the equicontinuity condition, that $|f_n(x)|<M+\epsilon$ for all $n$ and $x\in K$. By letting $\epsilon$ go to zero the claim follows.

$\endgroup$
1
$\begingroup$

In fact, we can generalize this slightly - we do not need $I$ to be countable.

Let $X$ be a compact space. Let $\{f_i \colon X\to\mathbb R; i\in I\}$ be an equicontinuous system of functions, $I$ being an arbitrary set. Moreover, for each $x\in X$ there exists $M_x$ such that $|f_i(x)|<M_x$ for each $i\in I$; i.e., the functions are pointwise bounded. Then the system of functions is uniformly bounded, i.e., there exists $M$ such that $|f_i(x)|<M$ for each $i\in I$ and for each $x\in X$.

Proof. Let us denote $g(x)=\sup\limits_{i\in I} |f_i(x)|$. (The fact that the given functions are pointwise bounded means that $g$ is a real valued function.)

  • We will show that the sets $A_r=\{x\in X; g(x)<r\}$, $r\in\mathbb R$, are open.
  • Then we can use the fact that this is an open cover of a compact space to get uniform bound.

Each $A_r$ is open. If we already know that supremum of a family of equicontinuous function is continuous, then we get that $A_r=g^{-1}(-\infty,r)$ is open.

But it is also easy to prove this directly. (Here I basically repeat the relevant part from the proof linked above.) Assume that $x\in A_r$, i.e., $$g(x)=\sup_{i\in I} |f_i(x)|<r.$$ Let us denote $\varepsilon=(r-g(x))/2$. Equicontinuity implies that there is an open neighborhood $U$ of $x$ such that $$|f_i(y)-f_i(x)|<\varepsilon$$ whenever $y\in U$ and $i\in I$. From this we get $$|f_i(y)| \le |f_i(x)| + |f_i(y)-f_i(x)| < g(x) + \varepsilon < r$$ for every $i\in I$. So we see that that there is an open set $U$ such that $x\in U\subseteq A_r$. Since this is true for every $x\in A_r$, the set $A_r$ is open.

Open cover and uniform bound. Pointwise boundednes means that $x\in A_r$ for $r=M_x$. So $\{A_r; r\in\mathbb R\}$ is an open cover of $X$. Since $X$ is compact, there is an finite subcover $A_{r_1},\dots,A_{r_n}$. If we put $M=\max\{r_1,\dots,r_n\}$, then we have $$g(x)=\sup\limits_{i\in I} |f_i(x)| \le M,$$ which means that the set $\{f_i; i\in I\}$ is uniformly bounded. $\square$

This proof is similar to the proof posted in other answer. However, it still seemed useful to me to stress the point that this is in fact related to the fact that the supremum is continuous. (In fact, we only use upper semicontinuity.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.