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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.

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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.

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