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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 \begin{equation} \alpha_n = \int_X |f|^n d\mu, \forall n \in \mathbb{N} \end{equation}

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

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 $

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

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