This question comes up after going over Arzela-Ascoli theorems.

For a set of continuous functions $\mathbb F$ from $\mathbb R$ to $\mathbb R$ that is equicontinuous. How do I show that if sup{$|f(0)|:f \in F$} $< \infty$ , then $\mathbb F$ is pointwise bounded?

I know I need to show that since sup{$|f(0)|:f \in F$} = $M_0 < \infty$ then for each $x \in \mathbb R$ sup{$|f(x)|:f \in F$} = $M_x < \infty$.

What I have gotten so far is that I should fix $\epsilon$ and use the fact that the domain is $\mathbb R$.

I think I just need some help using equicontinuity and sup together.


As an alternative to mookid's answer (which looks good), consider this one, which uses compactness (actually, mookid's answer reminds me of the proof of the compactness of $[0, 1]$, so maybe I'm just doing this at a less fundamental level).

Anyway, consider any $x \in \mathbb{R}$, and consider the set $[0, x]$ (we suppose WLOG that $x > 0$; if $x < 0$, we just examine $[-x, 0]$ and the rest of the proof is the same).

For $\varepsilon = 1$, for each $y \in [0, x]$, choose $\delta_y > 0$ so that $z \in (y - \delta_y, y + \delta_y) \Rightarrow |f(z)-f(y)| < 1$ for each $f \in F$; $\delta_y$'s existence is guaranteed by the equicontinuity of $F$.

Define $A_y = (y - \delta_y, y + \delta_y)$. Then $\{A_y\}_{y \in [0,x]}$ is an open cover of $[0, x]$, a compact set, and thus has a finite subcover, $\{A_{y_i}\}_{i=1}^n$.

Now, we know $M_0 < \infty$. If we order the $y_i$'s so that $y_1 < y_2 < \ldots < y_n$, then by the definition of equicontinuity of $F$, we know that $y \in A_{y_1} \Rightarrow |f(y)| \leq M_0 + 1$ for all $f \in F$, and more generally, $y \in A_{y_i} \Rightarrow |f(y)| \leq M_0 + i$ for all $f \in F$ (this might require a more formal proof--perhaps one using induction--but it shouldn't be too difficult). So, since $x \in A_{y_i}$ for some $i \in \{1, \ldots, n\}$, we know that $|f(x)| \leq M_0 + n$ for all $f \in F$, and therefore that $M_x \leq M_0 + n < \infty$.

I think explicitly using compactness makes the intuition a little more clear: because there is a bound for every function in $F$ at $0$, and because there is a bound on how much every function can change in a neighborhood of each point of $\mathbb{R}$, it can never be the case that some subset of $F$ increases locally at such a rate that it "breaks away" from the rest of the group toward infinity.

  • $\begingroup$ This helps, thanks. $\endgroup$ – Koate Mar 27 '14 at 21:04

Assume that the set $ Z=\{x: M_x<\infty \} $is non empty.

Then let $u = \inf Z$.

As the familly is equicontinuous, consider $\delta$ corresponding to the definition of equicontinuity with $\epsilon=1$.

Consider $u-\frac\delta2<x<u$, $y<u+\frac\delta 2$ such as $y\in Z$.

For $A= M_x + 1$, you then have for a certain $f\in F$: $$ |f(y)|>A $$ but you also have $$ |f(y)| \le |f(x)| + |f(y)-f(x)|\le M_x + 1 = A$$ This is impossible.


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