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I have a question that I need help with getting started (possibly I would be back for more help).

I have a measure space $(X,A,\mu)$ that is finite, and $f \in L^{\infty}(\mu)$. Also, defined is $a_n = \int_X\,|f|^n\,d\mu$. I need to show that the limit is: $$\lim_{n\to\infty}\,\frac{a_{n+1}}{a_n} = \|f\|_{\infty} .$$

I am stuck on getting started, anybody have any suggestions?

thanks much

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    $\begingroup$ nate: Did you realize the proof in the accepted answer is wrong? Or maybe you do not read the comments... $\endgroup$
    – Did
    Feb 6, 2012 at 6:07
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    $\begingroup$ nate: I see you have been visiting the site regularly since I posted the comment above, this can only mean you do not care. Good to know. $\endgroup$
    – Did
    Feb 25, 2012 at 20:52
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    $\begingroup$ @DidierPiau, I never shut off my Mac, nor close Safari - just sleep the computer. So, when I wake it up it "reactivates" all my webpages, which is why you see that I am here a lot. With 20 some pages having about 10 tabs open, no, I do not check all the tabs. So your assumption that I have been actively visiting this site is wrong. Furthermore, I do care - I am even trying to find time to write up an answer myself. Please don't assume I have as much free time as you think I should have. Thank you for all your help though - it is appreciated. $\endgroup$
    – nate
    Feb 27, 2012 at 1:51
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    $\begingroup$ Still no problem with the accepted answer being wrong? $\endgroup$
    – Did
    Nov 3, 2012 at 14:38
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    $\begingroup$ @nate: several people have commented, and Paul has verified, that his answer is flawed. In the interest of the site, and future readers, it would be best to accept a correct answer. $\endgroup$
    – robjohn
    Feb 24, 2014 at 17:13

5 Answers 5

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The result holds as soon as $\|f\|_\infty$ is positive and finite.

To prove this, assume without loss of generality that $f\geqslant0$ almost everywhere and $\|f\|_\infty=1$. Then $0\leqslant f^{n+1}\leqslant f^n$ almost everywhere hence $0\leqslant a_{n+1}\leqslant a_n$. Since $a_{n}\ne0$, this yields $$ \limsup\limits_{n\to\infty}\ a_{n+1}/a_n\leqslant1. $$ In the other direction, note that for every positive $u\lt v\lt1$, $A=[f\geqslant u]$ and $B=[f\geqslant v]$ both have positive measure, and that, for every $n\geqslant0$, $$ a_n\geqslant\int_Bf^n\geqslant v^n\mu(B). $$ Hence, $$ a_{n+1}\geqslant u\int_A f^n=ua_n-u\int_{X\setminus A}f^n\geqslant ua_n-\mu(X\setminus A)u^{n+1}, $$ where the first inequality comes from the fact that $f^{n+1}\geqslant uf^n$ on $A$ and $f^{n+1}\geqslant 0$ everywhere, and the second inequality comes from the fact that $f^n\lt u^n$ on $X\setminus A$.

Together, these two lower bounds on $a_n$ and $a_{n+1}$ yield $$ \frac{a_{n+1}}{a_n}\geqslant u-\frac{u \cdot \mu(X\setminus A)}{\mu(B)}\left(\frac{u}v\right)^n. $$ Since $u\lt v$ and $\mu(X\setminus A)$ is finite, $\liminf\limits_{n\to\infty}\ a_{n+1}/a_n\geqslant u$. This holds for every $u\lt1$, hence $$ \lim\limits_{n\to\infty}\ a_{n+1}/a_n=1. $$

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  • $\begingroup$ You've got lower bounds on $a_n$ and $a_{n+1}$, why do they imply lower bound on their quotient? (shouldn't you derive upper bound for the denominator?) $\endgroup$
    – sdcvvc
    Feb 6, 2012 at 9:44
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    $\begingroup$ @sdcvvc: No. First step: $a_{n+1}\geqslant ua_n-\mu(X\setminus A)u^n$ hence $a_{n+1}/a_n\geqslant u-\mu(X\setminus A)(u^n/a_n)$ (1). Second step: plug $a_n\geqslant\mu(B)v^n$ into (1). $\endgroup$
    – Did
    Feb 6, 2012 at 10:19
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    $\begingroup$ Isn't $\|f\|_{\infty}$ positive and finite by definition of $\|\cdot\|_{\infty}$? $\endgroup$
    – roo
    Mar 20, 2014 at 20:49
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    $\begingroup$ I wasn't trying to be interesting, I just asked because I was just stuck on that line at the time while trying to skim your argument. But I'm glad to know that manners have been deprecated. After reading the comments in the childish spat above, its no wonder the OP stopped responding to you. $\endgroup$
    – roo
    Mar 21, 2014 at 22:14
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    $\begingroup$ @Bonnaduck Because if $\|f\|_\infty\ne1$, we can solve the question for $g=f/\|f\|_\infty$ since $\|g\|_\infty=1$, then deduce the result for $f$ by homogeneity. $\endgroup$
    – Did
    Apr 22, 2017 at 21:32
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One of my students found this proof, that uses a simpler estimate than Did's, at the cost of using the equality $\|f\|_\infty=\lim_{p\to\infty}\|f\|_p$, which holds in the case we are considering.

To see the hard inequality, we use Hölder (with $p=(n+1)/n$, $q=n+1$) to obtain $$ \|f\|_n^n=\int_X|f|^n\leq\left(\int_X|f|^{n+1} \right)^{n/(n+1)}\,\mu(X)^{1/(n+1)}=\|f\|_{n+1}^n\,\mu(X)^{1/(n+1)}. $$

Then $$ \frac{\alpha_{n+1}}{\alpha_n}=\frac{\|f\|_{n+1}^{n+1}}{\|f\|_n^n}\geq\frac{\|f\|_{n+1}^{n+1}}{\|f\|_{n+1}^n\,\mu(X)^{1/(n+1)}}=\|f\|_{n+1}\,\mu(X)^{-1/(n+1)}. $$ Then, as the right-hand-side converges to $\|f\|_\infty$, $$ \liminf_{n\to\infty}\frac{\alpha_{n+1}}{\alpha_n}\geq\|f\|_\infty. $$

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  • $\begingroup$ To assume the equality $\|f\|_\infty=\lim\limits_{p\to\infty}\|f\|_p$ is nearly equivalent to assuming the conclusion, no? $\endgroup$
    – Did
    Apr 21, 2014 at 9:21
  • $\begingroup$ Using that equality means that this answer bridges the gap left by Paul's answer, so it is a nice supplement. $\endgroup$ Jan 28, 2015 at 2:26
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Disclaimer. This is not an answer, but rather something "too long for a comment" that provides what is missing to make Paul's proof complete.

Paul has shown that $a_n^{1/n}\to \Vert f\Vert_\infty$. As observed above by Jonas, it is enough to show that the sequence $c_n:=\frac{a_{n+1}}{a_n}$ is convergent, because then its limit $L$ will be the same as that of $a_n^{1/n}$.

First note that $a_{n+1}=\int_X \vert f\vert^{n+1}\, d\mu\leq \Vert f\Vert_\infty\,\int_X \vert f\vert^n \, d\mu=\Vert f\Vert_\infty\, a_n$, so that $c_n$ is bounded above by $\Vert f\Vert_\infty$. Hence, it is enough to show that the sequence $(c_n)$ is nondecreasing. In other words, one has to show that $$a_{n+1}^2\leq a_n\, a_{n+2}\, . $$ But this follows from Cauchy-Schwarz's inequality: $$a_{n+1}=\int_X\vert f\vert^{n+1}\, d\mu =\int_X \vert f\vert^{\frac{n}2}\,\vert f\vert^{\frac{n+2}2}\, d\mu\leq \left(\int_X\vert f\vert^n d\mu\right)^{1/2}\left(\int_X\vert f\vert^{n+2} d\mu\right)^{1/2}=a_n^{1/2}a_{n+2}^{1/2}\, . $$

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(Note added 2017.04.23 by @Did.) The note below by the OP describes incorrectly the trouble with this answer. To be brief, not much can be saved from this post and the lack of desire of this OP and of the asker (both still present on the site) to correct the situation is flabbergasting. For more details please see the comments thread.


Note added: I make a mistake here. First I thought that $\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}=L$ would imply $\lim_{n\rightarrow\infty}(a_n)^{\frac{1}{n}}=L$. But this is not correct. Please refer to the proof by @Did.


First it's easy to see that $$(a_n)^{\frac{1}{n}}=\Big(\int_X|f|^nd\mu\Big)^{\frac{1}{n}}\leq\|f\|_\infty\mu(X)^\frac{1}{n},$$ which implies that $$\lim_{n\rightarrow\infty}(a_n)^{\frac{1}{n}}\leq\|f\|_\infty,$$ where we have used the fact that $\mu(X)$ is finite. On the other hand, by definition of $\|f\|_\infty$, for all $\epsilon>0$, there exists a measurable set $E$ in $X$ such that $\mu(E)>0$ and $f\geq \|f\|_\infty-\epsilon$ on $E$. Hence, we have $$(a_n)^{\frac{1}{n}}=\Big(\int_X|f|^nd\mu\Big)^{\frac{1}{n}}\geq\Big(\int_E|f|^nd\mu\Big)^{\frac{1}{n}}=(\|f\|_\infty-\epsilon)\mu(E)^{\frac{1}{n}}.$$ As $n\rightarrow\infty$, we have $$\lim_{n\rightarrow\infty}(a_n)^{\frac{1}{n}}\geq(\|f\|_\infty-\epsilon).$$ Combining the above inequalities, we have $$\|f\|_\infty\geq\lim_{n\rightarrow\infty}(a_n)^{\frac{1}{n}}\geq(\|f\|_\infty-\epsilon).$$ Since $\epsilon>0$ is arbitrary, we have $$\lim_{n\rightarrow\infty}(a_n)^{\frac{1}{n}}=\|f\|_\infty.$$

Now the result follows easily from the fact that $$\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}=\lim_{n\rightarrow\infty}(a_n)^{\frac{1}{n}}.$$

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    $\begingroup$ @Jonas, excellent point: consider $a_{2n}=a_{2n+1}=4^n$. $\endgroup$
    – Did
    Dec 17, 2011 at 9:54
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    $\begingroup$ @Jonas, Indeed. This also confirms the empirical fact that trivially or easily or other similar expressions appearing in a proof should operate as a signal to the reader to become even more careful... $\endgroup$
    – Did
    Dec 17, 2011 at 12:00
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    $\begingroup$ Paul: $\lim a_{n+1}/a_n=L$ DOES imply that $\lim(a_n)^{1/n}=L$. It is the other implication which is wrong in general. $\endgroup$
    – Did
    Dec 19, 2011 at 13:22
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    $\begingroup$ Paul: I find a little annoying that, more than one month later, the fact that your proof is wrong is not even mentioned in your post. Sure, one can always read the fine print, aka the comments, nevertheless... $\endgroup$
    – Did
    Feb 6, 2012 at 6:06
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    $\begingroup$ The note added at the end of the answer incorrectly describes the mistake; see Did's comment from Dec 19 2011 at 13:22. $\endgroup$ Aug 19, 2013 at 18:32
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There is an alternative proof, also relying on the fact $\|f\|_p \rightarrow \|f\|_\infty$ as $p\rightarrow \infty$ ($\star$).

The inequality $\frac{a_{n+1}}{a_n} \le \|f\|_\infty$ being evident, the other one follows then from the following:

It holds true that

\begin{equation} \liminf \frac{a_{n+1}}{a_n} \ge \liminf \sqrt[n]{a_n} \end{equation} for any strictly positive sequence $(a_n) \in \mathbb R^\mathbb N$. The same equality holds for $\limsup$, but with reversed inequality sign. This fact is the essential reason why the root test implies the ratio test for numerical series.

And yes, $(\star)$ and the fact to be proven here are linked, as pointed out by Did above, and the proofs are similar. But, in my opinion, very much as in elementary analysis, though their proofs are similar, the ''root'' result implies the ''ratio'' result, but not inversely. That implication is exactly what I wanted to point out here.

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