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Let $(X, \mathcal F, \mu)$ be a finite measure space and let $f\in L^\infty(X, \mu)$. Define $\alpha_n=\int_X |f|^n\ d\mu$. Then $$\lim_{n\to \infty}\frac{\alpha_{n+1}}{\alpha_n}=\|f\|_\infty$$

I am a bit lost here. Can somebody please help.

Thanks.

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    $\begingroup$ HINT I think this may help. Note that $$ \frac{\alpha_{n+1}}{\alpha_n}=\frac{\|f\|_{n+1}^{n+1}}{\|f\|_{n}^{n}} = \left( \frac{\|f\|_{n+1}}{\|f\|_{n}} \right)^n \|f\|_{n+1} $$ Now, use the fact that when the space has finite measure, $\lim_{p\to \infty }\| f\|_p=\|f\|_\infty$, which has already been proven here math.stackexchange.com/questions/242779/limit-of-lp-norm $\endgroup$ – Alonso Delfín Apr 21 '15 at 13:43
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Since $f\in L^\infty(X,\mu)$ and $\mu(X) < \infty$, then $f\in L^n(X,\mu)$ for all $n \ge 1$. Since

$$\alpha_{n+1} \le \|f\|_\infty \int_X |f|^n\, d\mu = \|f\|_\infty \alpha_n,$$

we have $$\varlimsup_{n\to \infty} \frac{\alpha_{n+1}}{\alpha_n} \le \|f\|_\infty.$$

On the other hand, since $\alpha_n^{1/n} = \|f\|_n \to \|f\|_\infty$ as $n\to \infty$,

$$\varliminf_{n\to \infty} \frac{\alpha_{n+1}}{\alpha_n} \ge \varliminf_{n\to \infty} \alpha_n^{1/n} = \|f\|_\infty.$$

Therefore $$\lim_{n\to \infty} \frac{\alpha_{n+1}}{\alpha_n} = \|f\|_\infty.$$

Note. We can get $\varliminf_{n\to \infty} \frac{\alpha_{n+1}}{\alpha_n} \ge \|f\|_\infty$ alternatively by applying Holder's inequality with conjugate exponents $n+1$ and $(n+1)/n$. Doing so, we obtain $\alpha_n \le \mu(X)^{1/(n+1)}\alpha_{n+1}^{n/(n+1)}$. Thus

$$\varliminf_{n\to \infty} \frac{\alpha_{n+1}}{\alpha_n} \ge \varliminf_{n\to \infty} \mu(X)^{-1/(n+1)} \alpha_{n+1}^{1/(n+1)} = \lim_{n\to \infty} \|f\|_{n+1} = \|f\|_\infty.$$

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  • $\begingroup$ That's great. Thanks! $\endgroup$ – caffeinemachine Apr 21 '15 at 14:11

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