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Suppose that $||f||_p \le K$ for all $1 \le p <\infty$ for some $K>0$. How to show that the essential supremum exists and bounded by $K$ that i s$||f||_\infty \le K$?

I know how to prove that if $f \in L^\infty$ then \begin{align} lim_{p \to \infty} ||f||_p=||f||_\infty \end{align} but this already assume that $f \in L^\infty$ in this question we have to show that $f$ has an essential supremum. To be more precise I don't think I can use a technique when I define \begin{align} A_\epsilon =\{ x | \ |f(x)|>||f||_{\infty}-\epsilon \} \end{align}

I feel like here we have to use some converges theorem. Thanks for any help

Boby
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  • Any condition on the measure space? –  Sep 24 '15 at 21:32
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    Assume it's finite – Boby Sep 24 '15 at 21:33
  • possible duplicate of Limit of $L^p$ norm –  Sep 24 '15 at 21:35
  • Thanks. But I am asking something different. I am not assuming that $f$ has an essential supremum before hand we have to show it. – Boby Sep 24 '15 at 21:39
  • If $|f|_\infty =\infty$, then for all $n$ there is $A_n$ with $\mu(A_n)>0$ so that $|f|\ge n$ on $A_n$. Then $|f|_n \ge (\mu(A_n) n^n)^\frac{1}{n} \to \infty$. Thus that contradicts your assumption (But I do agree that this question is not a direct duplicate, but your problem can be solved using that anyway). –  Sep 24 '15 at 21:46
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    @JohnMa What if $\mu(A_n)$ goes to zero faster than $n^{-n}$? – Ian Sep 24 '15 at 21:47
  • Um.... Good point @Ian –  Sep 24 '15 at 21:47

2 Answers2

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Assume $\| f \|_\infty=\infty$. Let $M>0$. Define $A_M=\{ |f| \geq M \}$. Then $\mu(A_M)>0$. Take $p$ so large that $\mu(A_M)^{1/p} \geq 1/2$, then $\| f \|_p \geq (\mu(A_M) M^p)^{1/p} \geq M/2$. Since $M$ was arbitrary, $f$ is not uniformly bounded in $L^p$, and your result follows by contraposition.

This is essentially the argument suggested by John Ma in the comments, but decoupling $M$ and $p$.

Ian
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  • Ok. I get it thank you. – Boby Sep 24 '15 at 22:08
  • The result still it holds if $K=K(p)$ and $K(p) \to C$ as $p\to +infty$? In $\mathbb{R}^n$ it holds as well? – Rodolfo Ferreira de Oliveira Jul 13 '23 at 07:48
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    @RodolfoFerreiradeOliveira If $K \to C$ as $p \to \infty$ then $K$ is uniformly bounded anyway. The only way that would fail would be if $K$ were sometimes infinite, but now you just restrict attention to $p$'s larger than all those where $K$ is infinite. And yes, this holds in whatever measure space. It also holds in the sense of the $p$ norms in $\mathbb{R}^n$ since these can be viewed as $L^p$ norms under the counting measure in a finite set. – Ian Jul 13 '23 at 12:10
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I am going to assume that $f\geq 0$ (otherwise replace $f$ with $|f|$). Consider $$g = \min(f,K + 1)\in L^{\infty}.$$ Then $\|g\|_p\leq \|f\|_p\leq K$ hence $$\|g\|_\infty = \lim_{p \rightarrow \infty} \|g\|_p \leq K$$ so $$ \min(f,K + 1) \leq K \ \ \ a.e. \implies f\leq K a.e.$$

themaker
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