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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 '12 at 6:07
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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 '12 at 1:51
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    $\begingroup$ Still no problem with the accepted answer being wrong? $\endgroup$ – Did Nov 3 '12 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 '14 at 17:13
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    $\begingroup$ @nate Still not decided to clean the mess? $\endgroup$ – Did Mar 20 '14 at 20:52
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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 '11 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 '11 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 '11 at 13:22
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    $\begingroup$ @Paul: The problem is that even though $\lim\limits_{n\to\infty}a_n^{1/n}=L$, this does not guarantee the existence of $\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}$ $\endgroup$ – robjohn Jan 9 '13 at 15:49
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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. $\endgroup$ – Jonas Meyer Aug 19 '13 at 18:32
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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 '12 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 '12 at 10:19
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    $\begingroup$ Isn't $\|f\|_{\infty}$ positive and finite by definition of $\|\cdot\|_{\infty}$? $\endgroup$ – roo Mar 20 '14 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 '14 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 '17 at 21:32
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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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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 '14 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. $\endgroup$ – Jonas Meyer Jan 28 '15 at 2:26

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