Let $(X, B, μ)$ be a finite measure space. Let $f \in L^\infty (μ)$. Let $α_n = \int_X |f|^n$. Show that:

$$\lim_{n\rightarrow\infty} \frac{\alpha_{n+1}}{\alpha_{n}}=||f||_{\infty}$$

I could show $\alpha_n$'s converge to a limit. I am not sure how to use the finite measure property here.

  • $\begingroup$ Wrong title. "I could show αn's converge to a limit" Really? You should include in the question how you did this, then. $\endgroup$ – Did Jan 19 '19 at 20:45
  • $\begingroup$ @Did Yes. It's easy because $\alpha_{n+1}<M \alpha_n$ and these are positive numbers. $\endgroup$ – mathvc_ Jan 19 '19 at 20:50
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    $\begingroup$ And you think this proves $(a_n)$ converges? Well, there are easy counterexamples to that. $\endgroup$ – Did Jan 19 '19 at 21:11
  • $\begingroup$ Title even wronger now... The assertion that $a_n\to b_n$ is meaningless (unless $b_n$ is a constant sequence). $\endgroup$ – Did Jan 19 '19 at 21:12
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    $\begingroup$ math.stackexchange.com/q/92147 $\endgroup$ – Did Jan 19 '19 at 21:25

Claim 1: For any $f\in L^{\infty}(\mu$), $\lim_{n\rightarrow\infty}||f||_{n}=||f||_{\infty}$.

Proof: To ease our discussion, we assume that $f\geq0$. Otherwise, replace $f$ with $|f|$.

Clearly, the above obviously holds if $||f||_{\infty}=0$ because in which case $f=0$ $\mu$-a.e.. Consider the case that $M:=||f||_{\infty}>0$. Let $\alpha\in(0,M)$ be arbitrary and define $A_{\alpha}=\{x\in X\mid f(x)>\alpha\}$. Since $\alpha$ is not an essential upper bound of $f$, $\mu(A_{\alpha})>0$. Note that \begin{eqnarray*} \int f^{n} & \geq & \int_{A}f^{n}\\ & \geq & \mu(A_{\alpha})\alpha^{n}. \end{eqnarray*} Therefore $||f||_{n}\geq\alpha\mu(A_{\alpha})^{\frac{1}{n}}$ and hence $\liminf_{n}||f||_{n}\geq\alpha$. As $\alpha\in(0,M)$ is arbitrary, we have $\liminf_{n}||f||_{n}\geq M$. On the other hand, $\int f^{n}\leq M^{n}\mu(X)$, so $||f||_{n}\leq M\mu(X)^{\frac{1}{n}}.$ Therefore, $\limsup_{n}||f||_{n}\leq M$. It follows that $\lim_{n}||f||_{n}$ exists and $\lim_{n}||f||_{n}=||f||_{\infty}.$


Now, we go back to the original question. Clearly, without loss of generality, we may assume that $f\geq0$ (to ease our discussion, otherwise, replace $f$ with $|f|$). Denote $M:=||f||_{\infty}$. We assume that $M>0$. Otherwise $\alpha_{n}=0$ for all $n$ and $\frac{\alpha_{n+1}}{\alpha_{n}}$ is undefined. Note that $\alpha_{n+1}=\int f^{n+1}\leq\int f^{n}M=M\alpha_{n}$, so $\frac{\alpha_{n+1}}{\alpha_{n}}\leq M$. Put $p=\frac{n+1}{n}$, $q=n+1$, then $p,q\in(1,\infty)$ with $\frac{1}{p}+\frac{1}{q}=1$. By Holder inequality, we have \begin{eqnarray*} \alpha_{n} & = & \int f^{n}\cdot1\\ & \leq & ||f^{n}||_{p}||1||_{q}\\ & = & \left(\alpha_{n+1}\right)^{\frac{n}{n+1}}\left(\mu(X)\right)^{\frac{1}{n+1}}. \end{eqnarray*} Simplifying it yields $\frac{\alpha_{n+1}}{\alpha_{n}}\geq||f||_{n}\left[\frac{1}{\mu(X)}\right]^{\frac{1}{n}}$ and hence, $$ ||f||_{n}\left[\frac{1}{\mu(X)}\right]^{\frac{1}{n}}\leq\frac{\alpha_{n+1}}{\alpha_{n}}\leq M. $$ Note that $||f||_{n}\rightarrow M$ by Claim 1 and $\left[\frac{1}{\mu(X)}\right]^{\frac{1}{n}}\rightarrow1$. We conclude that $\lim_{n}\frac{\alpha_{n+1}}{\alpha_{n}}$ exists and $\lim_{n}\frac{\alpha_{n+1}}{\alpha_{n}}=M$.

  • $\begingroup$ To avoid triviality, we assume that $\mu(X)>0$. $\endgroup$ – Danny Pak-Keung Chan Jan 19 '19 at 23:10

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