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In a proof, I implicitly used the following 'fact':

Let $\mathcal C([a,b])$ be the set of continuous functions $f:[a,b]\to \mathbb R$. Then the set $S=\{f\in C([a,b]):||f||_\infty = 1\}$ is compact.

I tried proving this using the Arzelà–Ascoli theorem, since $\bar S = S$ but I Wasn't able to prove that $S$ is equicontinuous.

Is this true? If so, how can this be proven?

Eduardo Magalhães
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  • It is not true. – copper.hat Jan 02 '23 at 19:12
  • Why? @copper.hat – Eduardo Magalhães Jan 02 '23 at 19:12
  • just use the sequence $f_n(x)=1-nx$ for $x \in [0,\frac{1}{n}]$ (and null otherwise), its converges to 0 pointwise . – Bongo Jan 02 '23 at 19:17
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    Guys, are you sure OP is familiar with the (not very trivial) theorem that the closed unit ball is not compact in infinite dimensional normed spaces? Because if not then it is not a duplicate. Anyway, if you want a direct proof for this space, take for example $[a,b]=[0,1]$ and consider the sequence $f_n(x)=x^n$. It's not difficult to prove that it doesn't have any convergent subsequences. – Mark Jan 02 '23 at 19:30
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    If he knows Arzelà–Ascoli but not Riesz, it will be a good opportunity for him to learn it. – Anne Bauval Jan 02 '23 at 20:04
  • "What does not kill me makes me stronger." – Anne Bauval Jan 02 '23 at 20:10

2 Answers2

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You are in a metric context, so compactness is just sequential compactness. Following Bongo, consider the sequence $_()=\max(0,1−)$ on $[0,1]$. This sequence is contained in $S$. If $S$ were compact, it should contain a convergent subsequence (in the uniform norm). But it clearly does not, since $\|f_n-f_m\|_{\infty}=1-\frac{m}{n}>\frac 1{2}$ for every $n>2m$. Of course you can also invoke Riesz' Theorem, but there is no need.

GReyes
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Another example: $f_n(x)=\sin(\pi n x).$ Then for $n\neq m$ we get $$\|f_n-f_m\|_\infty ^2\\ \ge \int\limits_0^1[\sin(\pi nx)-\sin(\pi m x)]^2\,dx = 1$$

Ryszard Szwarc
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