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I'm aware of Classifying the compact subsets of $L^p$ and have seen similar posts, but I haven't seen an example of a compact set on an $L^p$ space. Trying to think of a possible simple example, and as it is used as a counterexample of compacity in the space of continuous functions, I asked myself if the set $S=\lbrace f:[0,1]\rightarrow [0,1] : ||f||_p<\infty \rbrace$ is compact on $L^p([0,1])$.

Applying the Frechet-Kolmogorov theorem feels too general to apply and makes me think that it may not be compact, but can't come up with a counterexample apart from the ones used for the set of continuous functions on $[0,1]$. Do you have any hint?

Deeponjit Bose
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Iván
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    The sequence $f_n(x)={1\over 2}[1+\cos(2\pi nt)]$ does not admit a convergent subsequence. By the way, if you assume that $f$ is bounded on $[0,1]$ the assumption $|f|_p<\infty $ becomes redundant. – Ryszard Szwarc Apr 01 '23 at 05:47
  • @RyszardSzwarc, great answer, thank you! Do you know any compact set in the $L^p([a,b])$ space? It seems that they are tricky to find. I've looked at several papers on criteria and sufficient conditions, but none of those had a single example. I know that the set $S=\lbrace f\in C^1([0,1])\ |\ ||f||\infty+ ||f'||\infty\leq M \rbrace, \ M\in \mathbb{R}$, is compact on $C([0,1])$, so I'm now invested in proving that this could be compact on $L^p([0,1])$, do you know any reference? – Iván Apr 05 '23 at 03:02
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    Thanks for nice words. The set $S$ in you comment is compact as it is compact in $C[0,1]$ and the uniform convergence implies $L^p$-convergence. See this for criteria of compactness in $L^p.$ – Ryszard Szwarc Apr 05 '23 at 05:15

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Let $$f_n(x)={1\over 2}[1+\cos(n\pi x)]$$ Then $f_n$ satisfy the requirements although the sequence does not admit an accumulation point. It suffices to show that $g_n(x)=\cos(n\pi x)$ do not accumulate. We have $$\int\limits_0^1\cos(\pi nx)\cos(k\pi x)\,dx=0,\ n>k$$ Any accumulation point $g$ would satisfy $$\int\limits_0^1g(x)\cos(k\pi x)\,dx =0,\ k\ge 0$$ Thus $g=0.$ However $$\int\limits_0^1|\cos(n\pi x)|^p\,dx ={1\over n\pi}\int\limits_0^{n\pi}|\cos x|^p\,dx={1\over \pi}\int\limits_0^{\pi}|\cos x|^p\,dx$$ therefore $g$ cannot be equal $0,$ a contradiction.

Ryszard Szwarc
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