Let $\left\{f_{n}\right\}$ be a sequence of equicontinuous functions where $f_n: [0,1] \rightarrow \mathbf{R}$. If $\{f_n(0)\}$ is bounded, why is $\left\{f_{n}\right\}$ uniformly bounded?

6The compactness of $[0,1]$ plays an important role. – Daniel Fischer Oct 31 '16 at 16:28

1@DanielFischer Compactness is not essential, but boundedness is. The result would also hold on $(1,1).$ – zhw. Oct 31 '16 at 18:26

@zhw. Only if you interpret equicontinuity as uniform equicontinuity. If you have the topological definition of equicontinuity(1), compactness is essential. (1) Let $\mathscr{F}$ be a family of functions $X\to Y$, where $X$ is a topological space, and $Y$ a uniform space (let $Y$ be metric if you're more comfortable with that). Then for $x\in X$, the family $\mathscr{F}$ is equicontinuous at $x$, if for every entourage $V$ of $Y$ there is a neighbourhood $U$ of $x$ such that for all $f\in \mathscr{F}$ and all $z\in U$ we have $(f(x), f(z)) \in V$. – Daniel Fischer Oct 31 '16 at 18:32

1The family is called equicontinuous if it is equicontinuous at every $x\in X$. – Daniel Fischer Oct 31 '16 at 18:32

I took equicontinuous to mean "equicontinuous on $[0,1]$" and for the latter the standard definition in metric spaces as in Rudin PMA – zhw. Oct 31 '16 at 18:47

@zhw: Even with the uniform definition, not just boundedness, but total boundedness is essential. It holds on $(1,1)$ with the uniform definition because it is totally bounded. Total boundedness is equivalent to having compact completion (assuming the axiom of choice), and this result isn't true for all bounded domains (ones that aren't totally bounded), so in a way compactness is more essential than boundedness. – Jonas Meyer Feb 09 '17 at 05:25
4 Answers
Let's do it for an equicontinuous family $\mathcal F$ of functions on $[0,1]$ such that
$$\sup_{f\in \mathcal F}f(0) =C < \infty.$$
Choose $m \in \mathbb N$ such that $yx\le 1/m$ implies $f(y)f(x) \le 1$ for all $f\in \mathcal F.$ Then for any $f\in \mathcal F,$
$$f(k/m) = [f(k/m)  f((k1)/m) ]+ [f((k1)/m)  f((k2)/m)]\,+$$ $$ \cdots + [f(1/m) f(0)] + f(0).$$
for $k= 1,\dots , m.$ Take absolute values to see this implies $f(k/m) \le k + C \le m+C.$ Now any $x\in [0,1]$ is within $1/m$ of one of the $k/m$ points. It follows that $f(x) \le 1 +m + C,$ for all $x\in [0,1],$ for all $f\in \mathcal F.$
 102,095
 6
 47
 106
Since $[0,1]$ is compact, $f_n$ is uniform continuous. So given $\epsilon>0$, there is a $\delta'>0$ such that for any $x,y\in [0,1], \:xy<\delta'$ $$ f_n(x)f_n(y)<\epsilon\tag1 $$ Since $f_n$ is equicontinuous, $(1)$ holds for all $n$.
Take $\delta=\delta'/2$. For any open cover on $\bigcup_{x\in [0,1]}(x\delta, x+\delta)$ on $[0,1]$, there is a finite cover $\bigcup_{i\in \{1, \cdots, l\}}(x_i\delta, x_i+\delta)$ covers $[0,1]$ because it is compact. Since $f_n(0)<M'$ and assume $0\in (x_1\delta, x_1+\delta)$, by $(1)$ for any $x\in (x_1\delta, x_1+\delta)$ and any $n$ $$ f_n(x)<f_n(0)+\epsilon<M'+\epsilon $$ Similarly for any $x\in (x_2\delta, x_2+\delta)$ and any $n$ $$ f_n(x)<f_n(y)+\epsilon<M'+2\epsilon $$ where $y\in (x_1\delta, x_1+\delta)$. Repeat this process and we have, for any $x\in (x_l\delta, x_l+\delta)$ and any $n$ $$ f_n(x)<f_n(y)+\epsilon<M'+l\epsilon $$ where $y\in (x_{l1}\delta, x_{l1}+\delta)$.
Take $M=M'+l\epsilon$. Then $f_n(x)<M$ on $[0,1]$ for any $n$. So $f_n$ is uniform bounded.
 15,810
 2
 23
 46
Let me try to prove this using real induction. You can find some basic description of this proof technique together with some references in this answer. I have tried to give some informal description of real induction here.
Real induction basically says that if we have some subset $S\subseteq[0,1]$, to show that $S=[0,1]$ it suffices to verify there conditions:
(RI1) $0\in S$.
(RI2) If $0\le x<1$, then $x\in S$ $\implies$ $[x,y]\subseteq S$ for some $y > x$.
(RI3) If $0 < x \le 1$ and $[0,x)\subset S$, then $x \in S$.
We will show that (RI1), (RI2), (RI3) is true for the set $S=\{x\in[0,1]; \text{ the sequence }f_n\text{ is uniformly bounded on }[0,x]\}$, i.e., $$S=\{x\in[0,1]; (\exists M\in\mathbb R)(\forall t\in[0,x])(\forall n) f_n(t)\le M\}.$$
(RI1) is exactly the assumption that the given sequence is bounded in $0$.
(RI2) Let us assume that $x\in S$, which means that there exists $M$ such that $$ (\forall t\in[0,x])(\forall n) f_n(t)\le M. $$ In particular, we have also $f_n(x)\le M$.
Let us choose some $\varepsilon<0$. Now from equicontinuity we get that there exists $\delta>0$ such that $$tx<\delta \implies f_n(t)f_n(x)<\varepsilon$$ for every $n$. So for every $t\in [x,x+\delta/2]$ and any $n$ we have $$f_n(t) \le f_n(t)f_n(x) + f_n(x) < M+\varepsilon.$$
Let $y=x+\delta/2$. We see that all $f_n$'s are bounded by $M+\varepsilon$ on both intervals $[0,x]$ and $[x,y]$, so it is uniformly bounded on the whole interval $[0,y]$. This shows that $[x,y]\subset S$.
(RI3) Let $x$ be such that $[0,x)\subset S$. Let $\varepsilon>0$. We will again use that we have $\delta>0$ such that $$yx<\delta \implies f_n(t)f_n(x)<\varepsilon$$ for every $n$. Now we choose any $y<x$ such that $y>0$ and $y>x\delta$.
Since $y\in S$, there exists $M$ such that $$ (\forall t\in[0,y]) f_n(t)\le M $$ for every $n$. I.e., on the interval $[0,y]$ the sequence is uniformly bounded by $M$; it remains to show what happens on $[y,x]$.
However, for $t\in [y,x]$ we have $$f_n(t) \le f_n(t)f_n(y) + f_n(y) < M+\varepsilon.$$ Again, we get that $(f_n)$ is uniformly bounded (by $M+\varepsilon$) on the whole interval $[0,x]$ and that $x\in S$.
You may also notice that we have never used the fact that we work with a countable family of functions. So the same argument works for arbitrary family of equicontinuous functions.
 50,316
 18
 169
 342