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The statement I need to prove is following. Let $S$ be a closed subspace of Lebesgue space $L^1[0,1].$ Assume that for every $f\in S$ there exists a number $p(f)>1$ such that $f\in L^{p(f)}[0,1].$ Then there exists a number $p>1$ such that $S\subset L^p[0,1].$

As far as I understand, this problem is related to Baire theorem. Hence, I write

$S=\bigcup^{\infty}_{n=1}( S\cap L^{1+1/n} [0,1])$

and conclude that for some $n$ closure of $S\cap L^{1+1/n} [0,1]$ has nonempty interior. But how can I proceed further to show that it lies in some $L^{1+1/m}$?

Davide Giraudo
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Georgii Riabov
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1 Answers1

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Apply Baire category theorem to the closed sets $$F_n:=\{f\in S,\lVert f\rVert_{L^{1+n^{-1}}}\leqslant n\}$$ (these one are closed by Fatou's lemma).

Davide Giraudo
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  • Hello, what can we do once we used Baire Category theorem to find that there exists an integer $n_0$ such that ${n_0}$ is not a nowhere dense set ? Can we just say that a subspace of a normed vector space with non-empty interior has to be the space itself so $F{n_0} = S$ and $S \subset L^{1+\frac{1}{n_0}}([0,1])$ ? – yrual Jan 13 '23 at 14:00
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    @yrual See here for the details: https://math.stackexchange.com/questions/167475/inclusion-of-lp-spaces – Davide Giraudo Jan 13 '23 at 14:15