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

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

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 Didier.

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In the very first displayed inequality, are you assuming that $\mu(X) = 1$? – Jesse Madnick Dec 17 '11 at 1:07
Oh yes. Please see my edited answer. – Paul Dec 17 '11 at 1:15
@Jonas, excellent point: consider $a_{2n}=a_{2n+1}=4^n$. – Did Dec 17 '11 at 9:54
@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... – Did Dec 17 '11 at 12:00
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. – Did Dec 19 '11 at 13:22

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, $$a_{n+1}\geqslant u\int_A f^n=ua_n-u\int_{X\setminus A}f^n\geqslant ua_n-\mu(X\setminus A)u^n,$$ 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 $$\frac{a_{n+1}}{a_n}\geqslant u-\frac{\mu(X\setminus A)}{\mu(B)}\left(\frac{u}v\right)^n.$$ 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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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?) – sdcvvc Feb 6 '12 at 9:44
@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). – Did Feb 6 '12 at 10:19
Understood, thank you. – sdcvvc Feb 6 '12 at 10:22
@sdcvvc: You are welcome. – Did Feb 6 '12 at 10:27