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This answer gives the "meat" of the proof that $\lim_{p\to\infty}\|f\|_p=\|f\|_\infty$, where $f\in L^p(\mu)$ for all $p\in[1,\infty)$ and for an arbitrary measure space $(X,\mathscr A,\mu)$. But there are several technical issues and unclear points in this proof, as pointed out in the 23 (!) comments, some of which were responded to better than others. Could someone please provide a complete and clear proof?

In particular, I find the following issues with the proof in the linked answer:

  • Why are we using $\liminf$ and $\limsup$ instead of just $\lim$?
  • How does the last line in the above-cited proof "give the reverse inequality"?
  • Does the case $f=0$ need to be treated separately?
  • The case $\|f\|_\infty=\infty$ is ignored in the above-cited proof. How can we include it?
WillG
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  • Because we do not know a priori that the limit exists. The reverse inequality follows from taking the limit $p\rightarrow \infty$. The case $f=0$ is trivial anyway. – Severin Schraven Mar 16 '20 at 18:25
  • @SeverinSchraven But if we "try" taking the (regular) limit $p\to\infty$ on the last expression in the linked proof, don't we immediately get that $\lim_{p\to\infty} |f|p \leqslant |f|\infty$? We don't know a priori that the limit exists, but we know after this manipulation. – WillG Mar 16 '20 at 18:28
  • The unbounded case follows from a similar bound as the first inequality in the linked answer. Namely define $S_a:={x \ : \ \vert f(x) \vert \geq a }$. Then we have $$ \Vert f \Vert_p \geq a \mu(S_a)^\frac{1}{p}$$ Taking the limit superior $p\rightarrow \infty$ yields $$ \limsup_{p\rightarrow \infty} \Vert f \Vert_p \geq a$$ as $a$ was arbitrary we get what you want. – Severin Schraven Mar 16 '20 at 18:34
  • No, why would you know that the limit exists? – Severin Schraven Mar 16 '20 at 18:36
  • Oh I see; I was thinking we just had to show the limit was bounded. – WillG Mar 16 '20 at 18:38
  • I would suggest you to work out the details yourself and post it as an answer. – Severin Schraven Mar 16 '20 at 18:40

1 Answers1

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  • We do not know a priori that $\lim_{p\to +\infty}\lVert f\rVert_p$ exists be we know that the $\liminf$ and $\limsup$ exists.

  • In the inequality $$ \lVert f\rVert_p \leqslant \lVert f\rVert_\infty^{\frac{p-q}p}\lVert f\rVert_q^{q/p}, $$ take the $\limsup_{p\to +\infty}$ to get $$ \limsup_{p\to +\infty}\lVert f\rVert_p \leqslant \limsup_{p\to +\infty}\lVert f\rVert_\infty^{\frac{p-q}p}\lVert f\rVert_q^{q/p}, $$ since $f\in L^q$, $\lVert f\rVert_q^{q/p}\to 1$ and $\lVert f\rVert_\infty^{\frac{p-q}p}\to\lVert f\rVert_\infty$.

  • Purely technically speaking, yes, since the choice $0<\delta<\lVert f\rVert_\infty$ could not be made. However, the case $f=0$ is trivial.

  • Saying that the case $\lVert f\rVert_\infty=+\infty$ is ignored is a big word, my point was to address the question of the post where the function was supposed to be in $L^\infty$. Anyways it is a consequence of the case where $\lVert f\rVert_\infty$ is finite: let $f_k=\lvert f\rvert\mathbf{1}\{\lvert f\rvert\leqslant k\}$. For all $k$ and $p$,

    $$\lVert f\rVert_p\geqslant \lVert f_k\rVert_p $$ hence $$\liminf_{p\to +\infty}\lVert f\rVert_p\geqslant \liminf_{p\to +\infty}\lVert f_k\rVert_p =\lVert f_k\rVert_\infty\geqslant (k-1) c_k,$$ where $c_k=1$ if $\mu\{k -1\leqslant \lvert f\rvert \leqslant k\}$ is positive and $0$ otherwise. Since $\lVert f\rVert_\infty$ is infinite, there are infinitely many $k$ such that $c_k=1$.

Community
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Davide Giraudo
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