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Consider a bounded sequence of continuous functions $f_n:\left[0,1\right]\to\mathbb{R}$ such that $\left\Vert f_{n}-f_{m}\right\Vert _{\infty}=\sup_{x\in\left[0,1\right]}\left|f_{n}\left(x\right)-f_{m}\left(x\right)\right|=1$ whenever $n\neq m$. Can such a sequence be equicontinuous?

I want to say no and my reasoning is that if you have such a sequence, then it must act like $\left\{x^n\right\}$ which can be shown to be not equicontinuous. However, I am not sure.

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    Show that such a sequence can not have a uniformly convergent subsequence. (If $\{f_n\}$ were equicontinuous, the Arzela-Ascoli theorem would tell you it would have a uniformly convergent subsequence.)2012-02-20

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Note, as Jose27 points out in the comments, that the boundedness assumption is not needed, as it is implied by the norm condition.

From the norm condition, $\tag{1} \Vert f_n-f_m\Vert_\infty=1,\quad \text{whenever }\ n\ne m, $ it follows that the sequence $\{f_n\}_{n=1}^\infty$ is uniformly bounded over $[0,1]$. Equation (1) also implies that $\{f_n\}_{n=1}^\infty$ cannot be equicontinuous over $[0,1]$:

Suppose $\{f_n\}_{n=1}^\infty$ were equicontinuous over $[0,1]$. Then by the Arzelà-Ascoli Theorem, there is a subsequence $\{f_{n_k}\}_{k=1}^\infty$ of $\{f_n\}_{n=1}^\infty$ that converges uniformly on $[0,1]$. In particular, $\{f_{n_k}\}_{k=1}^\infty$ is uniformly Cauchy. That is, for each $\epsilon>0$, there is a positive integer $N$ so that $ |f_{n_k}(x ) -f_{n_{l}}(x )|<\epsilon, \quad \text{for all }\ k,l\ge N\ \text{ and all }\ x\in[0,1]. $

But, setting $\epsilon={1\over2}$ and fixing a positive integer $N$, we have by equation (1) the existence of some $x_{\scriptscriptstyle N}\in[0,1]$ with $|f_{n_N}(x_{\scriptscriptstyle N}) -f_{n_{N+1}}(x_{\scriptscriptstyle N})|>{1\over2}$. As $N$ was arbitrary, this contradicts the fact that $\{f_{n_k}\}_{k=1}^\infty$ is uniformly Cauchy.

It follows that $\{f_n\}_{n=1}^\infty$ is not equicontinuous over $[0,1]$.

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    The assumption that the sequence is bounded is not necessary, since by the distance condition and the triangle inequality, for any n>1 we have $\| f_n\|_{\infty} \leq 1+\| f_1\|_{\infty}$.2012-02-21