Show $\lim \limits_{n \to \infty} \frac{a_{n+1}}{a_n} = \|f\|_{\infty}$ for $f \in L^{\infty}$

Question

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

Answer 1

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 $$ \lim\limits_{n\to\infty}\ a_{n+1}/a_n=1. $$

Answer 2

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

Answer 3

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: $$a_{n+1}=\int_X\vert f\vert^{n+1}\, d\mu =\int_X \vert f\vert^{\frac{n}2}\,\vert f\vert^{\frac{n+2}2}\, d\mu\leq \left(\int_X\vert f\vert^n d\mu\right)^{1/2}\left(\int_X\vert f\vert^{n+2} d\mu\right)^{1/2}=a_n^{1/2}a_{n+2}^{1/2}\, . $$

Answer 4

(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.

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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.

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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 $$\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}=\lim_{n\rightarrow\infty}(a_n)^{\frac{1}{n}}.$$

Answer 5

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.