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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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    nate: Did you realize the proof in the accepted answer is wrong? Or maybe you do not read the comments...2012-02-06
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    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.2012-02-25
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    @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.2012-02-27
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    nate: Two points. 1. @Jonas mentioned there was a problem with Paul's proof on Dec'17, more than two months ago. Nevertheless you *accepted* this answer on Feb'7. I mentioned right away once again the problem, with no visible reaction from you. Since then, the accepted answer is a wrong one. 2. Why the fact that you would try *to write up an answer yourself* should prevent you to react in any way until 4 hours ago is still a mystery to me, tabs and all notwithstanding.2012-02-27
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    Still no problem with the accepted answer being wrong?2012-11-03
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    @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.2014-02-24
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    @nate Still not decided to clean the mess?2014-03-20
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    @nate I finally got it: you are after the record of the wrong answer accepted during the longest period of time. Silly me, in retrospect this is so obvious...2016-09-02

4 Answers 4

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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 $\|f\|_\infty$, $$ \liminf_{n\to\infty}\frac{\alpha_{n+1}}{\alpha_n}\geq\|f\|_\infty. $$

  • 0
    To assume the equality $\|f\|_\infty=\lim\limits_{p\to\infty}\|f\|_p$ is nearly equivalent to assuming the conclusion, no?2014-04-21
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    Using that equality means that this answer bridges the gap left by Paul's answer, so it is a nice supplement.2015-01-28
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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 $$ \lim\limits_{n\to\infty}\ a_{n+1}/a_n=1. $$

  • 0
    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?)2012-02-06
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    @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).2012-02-06
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    Understood, thank you.2012-02-06
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    @sdcvvc: You are welcome.2012-02-06
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    Isn't $\|f\|_{\infty}$ positive and finite by definition of $\|\cdot\|_{\infty}$?2014-03-20
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    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.2014-03-21
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    @Kyle Isn't this the best motivation there could be NOT answering you?2014-03-21
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    Have a better day.2014-03-21
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    @Did Would you mind clarifying why we may assume $\|f\|_\infty=1$?2017-04-18
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    @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.2017-04-22
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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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    In the very first displayed inequality, are you assuming that $\mu(X) = 1$?2011-12-17
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    Oh yes. Please see my edited answer.2011-12-17
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    I know that in general if $(a_n)$ is a sequence of positive numbers such that $\lim\frac{a_{n+1}}{a_n}$ exists, then $\lim\frac{a_{n+1}}{a_n}=\lim\sqrt[n]{a_n}$. But in general I suspect it is possible for $\lim\sqrt[n]{a_n}$ to exist while $\lim\frac{a_{n+1}}{a_n}$ does not. Am I wrong, or is there something special about the sequence $\|f\|_n^n$ that makes it work regardless?2011-12-17
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    @Jonas, excellent point: consider $a_{2n}=a_{2n+1}=4^n$.2011-12-17
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    @DidierPiau: Nice example, thank you. So this does leave the question of why $\lim \|f\|_{n+1}^{n+1}/\|f\|_n^n$ should exist.2011-12-17
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    @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...2011-12-17
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    @Paul: Will you please explain that last step?2011-12-18
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    @Jonas: Your comment is right. I make a mistake here. First I thought that $\lim\frac{a_{n+1}}{a_n}=L$ would imply $\lim (a_n)^\frac{1}{n}=L$. After looking at your comment, I find that it's the other way around as you have said. Sorry for the mistake.2011-12-19
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    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.2011-12-19
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    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...2012-02-06
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    @Didier: I added the note in my answer. Then people can refer to your proof.2012-02-06
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    @Paul: The problem is that even though $\lim\limits_{n\to\infty}a_n^{1/n}=L$, this does not guarantee the existence of $\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}$2013-01-09
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    The note added at the end of the answer incorrectly describes the mistake; see Did's comment from Dec 19 2011 at 13:22.2013-08-19
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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: $$a_{n+1}=\int_X\vert f\vert^{n+1}\, d\mu =\int_X \vert f\vert^{\frac{n}2}\,\vert f\vert^{\frac{n+2}2}\, d\mu\leq \left(\int_X\vert f\vert^n d\mu\right)^{1/2}\left(\int_X\vert f\vert^{n+2} d\mu\right)^{1/2}=a_n^{1/2}a_{n+2}^{1/2}\, . $$