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Let $C[0,1]$ be the space of continuous functions, equipped with the $\sup$-norm.

My question is, how one can prove, that this space is not locally compact? Is it possible to show this explicitly, by providing a sequence of continuous functions, which does not contain a convergent subsequence?

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3 Answers 3

up vote 2 down vote accepted

Consider the intervals $E_n=[1/n,1/(n+1)]$, $n\in\mathbb N$. Define continuous functions $f_n$ such that $f_n=0$ outside $E_n$, and $f=1$ on the mid point of $E_n$.

Then $\|f_n\|_\infty=1$ for all $n$, and $\|f_n-f_m\|_\infty=1$ whenever $n\ne m$, so the sequence admits no convergent subsequence.

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Try $f_n(x):=x^n$.

The deeper reason is that $C[0,1]$ is infinite dimensional, so by Riesz theorem its unit ball is not compact.

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It's possible to show something even more amazing: Any locally compact Hausdorff real or complex topological vector space is finite dimensional and every finite dimensional Hausdorff topological vector space has the usual topology. See Terry Tao's blog for proofs.

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