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The problem this:

Assume that $f$ is a complex measurable function on $X$, $\mu$ is a positive measure on $X$, $\|f\|_\infty > 0$ and $\|f\|_r<\infty$ for some $r<\infty$. Prove that $$\|f\|_p\to\|f\|_\infty ~~~\mbox{as}~~~p\to\infty.$$

This is the Exercise 4(e) from Real and Complex Analysis (Rudion), Chapter 3 ($L^p$-Spaces). In the internet, I've found some proofs. One is this:

From internet

I couldn't understand why one can conclude $\lim_{p\to\infty}\|f\|_p=\infty?$ How can I fix this problem in the proof?

EDIT: Okay, what Frank Lu did justify $\lim_{p\to\infty}\|f\|_p=\infty$. Now I have other question. Since $\|f\|_p\leq \|f\|^{\frac{r}{p}}_r$, one can conclude that $\limsup \|f\|_p\leq 1.$ How can I conclude the equality?

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    This is a terribly written proof: $x \in x$, $\lim_{n \to \infty}...$ where no $n$ occurs. – 2017-02-04
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    @TheGeekGreek is is true. thought you could thing about the n as a p ... rsrs – 2017-02-04
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    I am just figuring out what is going on...since if you take the $p$-th root on the inequality, the limit tends not to $\infty$. – 2017-02-04
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    @TheGeekGreek yeap. You can consider $\mu\{|f|>2^p\}>0 \forall p$ but we cannot conclude that $\mu\{|f|>2^p\}^{-1/p}$ has a positive low bound. – 2017-02-04

1 Answers 1

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Here's the idea: Let $t>0$ be any real number, set $$E_t=\{x\in X: |f(x)|\geq t\}.$$ If you assume that $\|f\|_\infty=+\infty$, then $\mu(E_t)>0$. Then it follows from Chebyshev's inequality that $$t^p\mu(E_t)\leq\|f\|_p^p\quad\Leftrightarrow\quad t\mu(E_t)^{\frac{1}{p}}\leq\|f\|_p.$$ Now if $\mu(E_t)=\infty$ then we can see that $\|f\|_p=\infty$ for every $p$ thus $\lim_{p\to\infty}\|f\|_p=+\infty$. If $\mu(E_t)<\infty$, then one can see that $$t\leq\liminf_{p\to\infty}\|f\|_p.$$ Observe that $E_s\subseteq E_t$ whenever $s\geq t$ so in this case $0<\mu(E_s)\leq\mu(E_t)<\infty$, thus by the same argument we can prove that $$s\leq\liminf_{p\to\infty}\|f\|_p,\quad\forall s\geq t.$$ Therefore $\liminf_{p\to\infty}\|f\|_p=\infty$. Note that here we take $\liminf$ because we don't know whether the limit exist or not.

For the second question, what we want to prove is $\liminf_{p\to\infty}\|f\|_p\geq\|f\|_\infty$. For any $0

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    What is the purpose of the "s" part? – 2017-02-04
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    To show that the liminf is greater than every positive number. – 2017-02-04
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    Thought $t$ already is any positive number... I edited my question and add another one. Can you help me? – 2017-02-04
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    I think in order for the result to hold, the measure space should be $\sigma-$finite. Otherwise it seems not possible to prove the reversed inequality. – 2017-02-04
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    yes, it is possible (I have a proof right here). I'll show to you tomorrow! Thanks for the help. – 2017-02-04
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    You're right, we don't actually need the $\sigma-$finite assumption. I'll add the proof to my answer. – 2017-02-05
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    In the second part, what if $\mu (E_t) =0$? – 2017-02-05
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    That's impossible as $\|f\|_\infty=1$ is assumed. – 2017-02-05
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    Yeah, you toke t<1. One last thing. I think you have to take $t\to 1^-$ instead of $t\to 1^+$ – 2017-02-05
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    You are right，that is a typo. – 2017-02-05