# $\frac{\int|f|^{n+1}}{\int|f|^n} \rightarrow ||f||_{\infty}$ [closed]

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.

• Wrong title. "I could show αn's converge to a limit" Really? You should include in the question how you did this, then. – Did Jan 19 '19 at 20:45
• @Did Yes. It's easy because $\alpha_{n+1}<M \alpha_n$ and these are positive numbers. – mathvc_ Jan 19 '19 at 20:50
• And you think this proves $(a_n)$ converges? Well, there are easy counterexamples to that. – Did Jan 19 '19 at 21:11
• Title even wronger now... The assertion that $a_n\to b_n$ is meaningless (unless $b_n$ is a constant sequence). – Did Jan 19 '19 at 21:12
• math.stackexchange.com/q/92147 – 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 $$\lim_{n}\frac{\alpha_{n+1}}{\alpha_{n}}=M$$.
• To avoid triviality, we assume that $\mu(X)>0$. – Danny Pak-Keung Chan Jan 19 '19 at 23:10