Let $p\in [0,\infty)$, $(\Omega,\mathcal{F},\mathbb{P})$ a probability space and $E$ a closed subspace of $L^p(\Omega)$. We suppose that $E\subset L^{\infty}(\Omega)$. How do I prove that $\exists M>0$ such that $||f||_{\infty}\le M||f||_2\ \forall f\in E$ ? I managed to prove using the closed graph theorem that $\exists K>0$ such that $||f||_{\infty}\le K ||f||_p\ \forall f\in E$. If $p\le 2$ then I can take $M=K$. But what if $p>2$ ?
Thank you in advance for your help.
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Tengen
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1If $p>2$ I think that the claim does not hold. – Giuseppe Negro Nov 17 '19 at 19:09
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@GiuseppeNegro I thought that too, but I couldn't come up with an example where the claim doesn't hold for $p>2$. – Tengen Nov 17 '19 at 19:16
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I am sorry, it seems that my thought was wrong. – Giuseppe Negro Nov 17 '19 at 19:24
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@GiuseppeNegro No problem mate. – Tengen Nov 17 '19 at 19:33
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For $p > 2$ simply integrate the inequality $\displaystyle |f|^p \le \lVert f \rVert_{\infty}^{p-2}|f|^2$ to write $$ \lVert f \rVert_{p}^p \le \lVert f \rVert_{\infty}^{p-2}\lVert f \rVert_{2}^2 \le K^{p-2}\lVert f \rVert_{p}^{p-2}\lVert f \rVert_{2}^2$$ which now implies $\displaystyle \lVert f \rVert_{p} \le K^{(p-2)/2}\lVert f \rVert_{2}$ and hence $\displaystyle \lVert f \rVert_{\infty} \le K^{p/2}\lVert f \rVert_{2}$.
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