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Is this true that "Any bounded sequence in $L^4[0,1]$ has a convergent sbsequence in $L^2[0,1]$?"

Of course I tried so much for this but I could not get any solution till now.

Any hints. is appreciated.

Thank you

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    $\begingroup$ $f_n(t) = \sin(nt)$. $\endgroup$ Commented Nov 26, 2017 at 18:41

2 Answers 2

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Take the sequence of functions

$$f_n(x)=\sin(n\pi x).$$

$f_n$ weakly converges to $0$ both in $L^2$ and $L^4$, it is a bounded sequence in $L^4$, but it has no strongly convergent subsequences in $L^2$; in fact such subsequences would converge to $0$, but $\|f_n\|_{L^2}=\frac{1}{\sqrt{2}}$ for all $n$. So the statement is false.

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Maybe I may add to Pozz's correct answer that although it has no strongly convergent subsequence (i.e. in the norm sense), there is always a weakly convergent subsequence in $L^2(0,1)$. This is because on compact domains, $L^p \subset L^q$ continuously whenever $\infty \geq p \geq q \geq 1$. So a bounded sequence in $L^4(0,1)$ is also a bounded sequence in $L^2(0,1)$. By Banach-Alaoglu and since $L^2$ is reflexive (it is a Hilbert space), we get a weakly convergent subsequence.

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