# Exercise 3.23 from Real and Complex Analysis of Walter Rudin [duplicate]

I need help in the exercise cited in the title.

Suposse $\mu$ is a positive measure on $X$, $\mu(X)<\infty$, $f \in L^\infty(\mu)$, $\|f\|_\infty >0$, and $$\alpha_n = \int_X |f|^n d\mu, \forall n \in \mathbb{N}$$

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

I tried this.

As we know that $\lim_{n \rightarrow \infty} (\int |f|^n)^{1/n} d\mu = \|f\|_\infty$ then for every $m \in \mathbb{N}$ existis $N_m \in \mathbb{N}$ such that for all $n \geq N, n \in \mathbb{N}$ we have the following assert

$$|(\int |f|^n)^{1/n} d\mu - \|f\|_\infty| < \frac{1}{m}$$

Then

$$\|f\|_\infty - \frac{1}{m} < (\int |f|^n)^{1/n} d\mu < \|f\|_\infty + \frac{1}{m}$$ $$(\|f\|_\infty - \frac{1}{m})^n < \int |f|^n d\mu < (\|f\|_\infty + \frac{1}{m})^n$$

Similar way as $n+1>N_m$ we have

$$(\|f\|_\infty - \frac{1}{m})^{n+1} < \int |f|^{n+1} d\mu < (\|f\|_\infty + \frac{1}{m})^{n+1}$$

Then for sufficiently large $m$ such that all terms are positve and taking care that $\|f\|_\infty>0$

$$\frac{(\|f\|_\infty - \frac{1}{m})^{n+1}}{(\|f\|_\infty + \frac{1}{m})^n} < \frac{\int |f|^{n+1} d\mu}{\int |f|^n d\mu} < \frac{(\|f\|_\infty + \frac{1}{m})^{n+1}}{(\|f\|_\infty - \frac{1}{m})^n}$$

Now, making $m \rightarrow \infty$ and as a consecuence making $n \rightarrow \infty$

$$\lim_{n \rightarrow \infty} \frac{\int |f|^{n+1} d\mu}{\int |f|^n d\mu} = \|f\|_\infty$$

• Did you tried something? – Tomás Nov 20 '13 at 17:29
• – derivative Nov 20 '13 at 19:40