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Let $E$ be a measurable set with $m(E)< \infty$.Then $L^{\infty}\subset L^p(E)$ for each $p$ with $1\le p\lt \infty$.Furthermore,if $f\in L^{\infty}$,then $$\|f\|_{ \infty}=\lim_{p\to \infty}\|f\|_{ p}$$

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Proof:Let $f\in L^{\infty}(E)$ and $\mu=\|f\|_{\infty}$

$\implies \vert f(x)\vert^p \le \mu^p$ a.e. on $E$

$\implies \int_E \vert f(x)\vert^p \ dx \le \mu^p.m(E) \lt \infty$ a.e. on $E$

$\implies f\in L^p(E)$

$\implies L^{\infty}(E)\subset \ L^p(E)$.

How to show if $f\in L^{\infty}$,then $\|f\|_{ \infty}=\lim_{p\to \infty}\|f\|_{ p}$?

I've tried hard to prove this. My professor suggested me to use the concept of limit superior,but i'm failing to make that idea practical.I know it is easy as it is consequence of above result but i'm lacking somewhere.So it will be heplful to give the detailed explanation of proof.

Thank you!

This is Davide Giraudo's answer for Limit of $L^p$ norm

Fix $\delta>0$ and let $S_\delta:=\{x,|f(x)|\geqslant \lVert f\rVert_\infty-\delta\}$ for $\delta<\lVert f\rVert_\infty$. We have $$\lVert f\rVert_p\geqslant \left(\int_{S_\delta}(\lVert f\rVert_\infty-\delta)^pd\mu\right)^{1/p}=(\lVert f\rVert_\infty-\delta)\mu(S_\delta)^{1/p},$$ since $\mu(S_\delta)$ is finite and positive. This gives $$\liminf_{p\to +\infty}\lVert f\rVert_p\geqslant\lVert f\rVert_\infty.$$ As $|f(x)|\leqslant\lVert f\rVert_\infty$ for almost every $x$, we have for $p>q$, $$ \lVert f\rVert_p\leqslant\left(\int_X|f(x)|^{p-q}|f(x)|^qd\mu\right)^{1/p}\leqslant \lVert f\rVert_\infty^{\frac{p-q}p}\lVert f\rVert_q^{q/p},$$ giving the reverse inequality.

My question is how $\left(\int_X|f(x)|^{p-q}|f(x)|^qd\mu\right)^{1/p}\leqslant \lVert f\rVert_\infty^{\frac{p-q}p}\lVert f\rVert_q^{q/p}$ happened?

Picaso
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  • This has already been answered here. – Miguel Apr 28 '17 at 15:14
  • @Miguel:Will you please explain how does liminf enters in the scenario-$\liminf_{p\to +\infty}\lVert f\rVert_p\geqslant\lVert f\rVert_\infty.$? – Picaso Apr 28 '17 at 16:50
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    Sure: You know $a_n \geq b_n \rightarrow b$ but you don't know yet whether $a_n$ converges. What you do know is that $\lim \inf a_n \geq b$, since the $\lim \inf$ is always defined. If you prove the other inequality for the $\lim \sup$ then you know that the limit of $a_n$ exists and is equal to $b$. – Miguel Apr 28 '17 at 16:58
  • @Miguel:Why lim inf $a_n$ is defined only why not lim sup $a_n$? – Picaso Apr 28 '17 at 18:13
  • @Miguel:How $\left(\int_X|f(x)|^{p-q}|f(x)|^qd\mu\right)^{1/p}\leqslant \lVert f\rVert_\infty^{\frac{p-q}p}\lVert f\rVert_q^{q/p}$ happened? – Picaso Apr 28 '17 at 18:46
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    (Please don't spam the conversation with repeated posts) lim sup is of course also defined for every sequence (check your notes for analysis 1!). As I said, you also have to prove a bound for it later. As to your other question: $f$ is bounded by its supremum norm. – Miguel Apr 28 '17 at 21:30
  • @Miguel:Apology for the repeated posts,I'm new on this site and do not know how to use it properly. – Picaso Apr 28 '17 at 21:50
  • @Miguel:I got it.I thought that this is the application of some standard inequality . – Picaso Apr 28 '17 at 21:54
  • Great! I'm glad it's clear now. :) – Miguel Apr 28 '17 at 22:10
  • https://en.wikipedia.org/wiki/H%C3%B6lder%27s_inequality#Generalization_of_H.C3.B6lder.27s_inequality - look at the end of this section. Then use boundedness. – Martin Apr 30 '17 at 07:23
  • Miguel:How does $\delta $ vanished in $\liminf_{p\to +\infty}\lVert f\rVert_p\geqslant\lVert f\rVert_\infty.$ – Styles Apr 30 '17 at 18:45

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