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i'm trying to prove the next problem, and I wanted to know if my answer is correct.

Problem: Let $(\Omega,\mathcal{F},\mu)$ be a $\sigma$-finite measurable space. If $f\in L^p$ for all $p\in [1,\infty)$, show that $$\|f\|_{\infty} \leq \liminf_{p\to \infty}\|f\|_p.$$

Solution: Suppose $0<\mu (\Omega)\leq \infty$. Let $0\leq M\leq\|f\|_{\infty}$, and $A=\{x\in \omega:|f(x)|>M\}$, then $\mu (A)>0$ and $\liminf_{p\to \infty} \mu(A)^{1/p}=1$. Then, $$\liminf_{p\to \infty} \|f\|_p \geq M \liminf_{p\to \infty} \mu(A)^{1/p}=M,$$ for all $M\in [0,\|f\|_{\infty})$. Hence, $\|f\|_{\infty}\leq \liminf_{p\to \infty}\|f\|_p$.

czzzzzzz
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1 Answers1

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You need to define $A$ differently. The current definition yields $\mu(A)=0$. You should let $\epsilon >0$ and then define $A = \{x \in \Omega \mid |f(x)| > M- \epsilon\}$. Then you have, by definition of the essential supremum, $\mu(A)>0$. Now it could happen that $\mu(A)=\infty$, so $\liminf\limits_{p \to \infty} \mu(A)^{1/p}=\infty$... However by sigma finiteness there must be a set $B$ with $\infty > \mu(B)>0$ and $\mu(A \cap B) >0$. Now substitute $A$ by $A \cap B$ and the rest of your proof works.

h3fr43nd
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