# Why is this sequence of equicontinuous functions uniformly bounded?

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?

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

Since $[0,1]$ is compact, $f_n$ is uniform continuous. So given $\epsilon>0$, there is a $\delta'>0$ such that for any $x,y\in [0,1], \:|x-y|<\delta'$ $$|f_n(x)-f_n(y)|<\epsilon\tag1$$ 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 $y\in (x_{l-1}-\delta, x_{l-1}+\delta)$.

Take $M=M'+l\epsilon$. Then $|f_n(x)|<M$ on $[0,1]$ for any $n$. So $f_n$ is uniform bounded.

Hint:

$$|f_{n}(x)|=|f_{n}(x) +f_{n}(0) -f_{n}(0)|\leq|f_{n}(x) -f_{n}(0)|+|f_{n}(0)|$$

now use the Equi-continuity and $$\{f_{n}(0)\}$$ bounded

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 $|y-x|\le 1/m$ implies $|f(y)-f(x)| \le 1$ for all $f\in \mathcal F.$ Then for any $f\in \mathcal F,$

$$f(k/m) = [f(k/m) - f((k-1)/m) ]+ [f((k-1)/m) - f((k-2)/m)]\,+$$ $$\cdots + [f(1/m) -f(0)] + f(0).$$

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 $f\in \mathcal F.$

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 $$|t-x|<\delta \implies |f_n(t)-f_n(x)|<\varepsilon$$ 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 $$|y-x|<\delta \implies |f_n(t)-f_n(x)|<\varepsilon$$ 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.