Question: Suppose $f_n : [0,1] \rightarrow \mathbb{R}$ is an equi-continuous sequence of functions and suppose $\lvert f_n(0) \lvert \leq 1$ for all $n$. Then, $\{f_n\}$ is uniformly bounded.

Work so far: Given $\varepsilon>0$, there exists a $\delta > 0$ such that for all $\lvert x \lvert <\delta$, we have: $$\lvert f_n(x) \lvert = \lvert f_n(x) - f_n(0) + f_n(0) \lvert \leq \lvert f_n(x) - f_n(0) \lvert + \lvert f_n(0) \lvert \leq \varepsilon + 1$$

However, this seems like it could potentially bound $\lvert f_n(x) \lvert$ for only a small portion of $[0,1]$, i.e. $x \in (-\delta,\delta)$.

Is there some way to extend this to a cover of open intervals on which $\lvert f_n(x) \lvert$ is bounded, so that we can take advantage of the compactness of $[0,1]$?

At this point I am a bit lost. Any direction would be much appreciated.

Edit: The family of functions is uniformly equi-continuous on $[0,1]$.


You'll want to show that an equicontinuous family of functions on a compact space is uniformly equicontinuous, i.e. that for all $\varepsilon > 0$, there is some $\delta$ such that for all $n$ and $|x-y| \leq \delta$, $|f_n(x)-f_n(y)| < \varepsilon$, that is $\delta$ is independent of $x$. The proof should be similar to proving uniform continuity of continuous functions on compact spaces.

This way, you can keep the $\delta$ the same throughout and chain it to your initial bound. That is, given $\varepsilon > 0$, choose a corresponding $\delta > 0$ and let $n = \lceil \frac{1}{\delta} \rceil$. Then for $x \in [0,1]$, $$|f(x)| \leq \sum_{k=1}^{n}|f(\tfrac{kx}{n})-f(\tfrac{(k-1)x}{n})| + |f(0)| \leq n\varepsilon + 1$$ and so the $f_n$ are uniformly bounded.

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  • $\begingroup$ I am learning Analysis form baby Rudin, where he doesn't talk about (to the best of my knowledge) equi-contrinuity at a point. He only discusses uniform equi-continuity. I will edit above. $\endgroup$ – Greyson Dec 13 '15 at 20:31


A point $x \in [0,1]$ can be reached by a chain of points starting at $0$ $$0 \le \delta \le 2\delta \le \dots \le n\delta \le x <(n+1) \delta$$ Moreover for all $x \in [0,1]$, you can bound $n$ with $\frac{1}{\delta}+1$.

Note: I would suggest that rather than saying "given $\epsilon > 0$", you pickup one $\epsilon$, like $1$, $\pi$... Or whatever you want as here you only need one $\epsilon$ to work with.

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  • $\begingroup$ This makes perfect sense, and I agree that fixing $\varepsilon$ simplifies things. $\endgroup$ – Greyson Dec 13 '15 at 20:30

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