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I am studying $L^p$ spaces and I would like a proof why the unit sphere in $L^p([0,1])$ is not compact. I know that unit sphere is not compact in infinite dimensional spaces, but I think there is an elementary proof, tailored to this example. I would appreciate your help.

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  • $\begingroup$ I would suggest looking for a counter-example via sequential compactness $\endgroup$ – Thoth Nov 22 '15 at 4:49
  • $\begingroup$ Yes, that is the point, I just could not make one, I thought it was more complicated, but as Hans and Prahland proved, it can be made easy. $\endgroup$ – Iniciador Nov 22 '15 at 5:19
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One way to do this is to produce a sequence $(u_n)_n$ of functions in the unit ball of $L^p(0,1)$ such that $\|u_i - u_j \|_{L^p} \ge c$ for some $c > 0$, all $i \ne j$. Then no subsequence can converge. You can get such a sequence by defining $$ u_j(x) = \begin{cases} 2^{k/p}\quad (\frac{\ell}{2^k} \le x \le \frac{\ell+1}{2^k}\\ 0 \quad \text{otherwise} \end{cases} $$ if $j = 2^k + \ell$ and $0 \le \ell < 2^k$, with $k = 0, \, 1, \, 2, \dots$..

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Consider $$ f_n := n\chi_{[0,1/n]} \in L^p[0,1] $$ Then if $n<m$, $$ \|f_n - f_m\|_p^p = \int_0^{1/m} (m-n)^pdt + \int_{1/m}^{1/n} n^pdt \geq \frac{(m-n)^p}{m} $$ Hence if $p\geq 1$, this sequence cannot have a convergent subsequence.

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  • $\begingroup$ Ahh, just realized that this sequence is norm bounded only if $p=1$. Else you probably need @HansEngler's example $\endgroup$ – Prahlad Vaidyanathan Nov 22 '15 at 5:21
  • $\begingroup$ mmm, can't it be picked in a similar way so it is bounded? $\endgroup$ – Iniciador Nov 22 '15 at 6:03
  • $\begingroup$ How about $n^{1/p} \, \chi_{[0,1/n]}$? $\endgroup$ – gerw Nov 22 '15 at 11:42
  • $\begingroup$ @gerw: That would work! Except the estimate would be $(m-n)$ instead of $(m-n)^p$. $\endgroup$ – Prahlad Vaidyanathan Nov 23 '15 at 2:34

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