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I've been wrestling with the following proof off and on for a number of days, and I'm in need of a nudge in the right direction.

Let $(E,\mathcal{M},\mu)$ be a measure space with $0 < \mu(E) < \infty$. Consider $f \in L^\infty(E)$ with $\|f\|_\infty > 0$; show that

$$ \lim_{n\to\infty} \|f||_n = \lim_{n\to\infty} \frac{\|f\|_{n+1}^{n+1}}{\|f\|_n^n} = \|f\|_\infty $$

Now I'm familiar with the result and proof that under these circumstances $\lim_{n\to\infty} \|f||_n = \|f\|_\infty$, so I've been focusing on somehow showing the two limits to be the same.

bosmacs
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1 Answers1

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From Holder's inequality, $$\|f\|_n^n=\int_E|f|^n\leq\left(\int_E|f|^{n\frac{n+1}{n}}\right)^\frac{n}{n+1}\left(\int_E1\right)^\frac{1}{n+1}=\|f\|_{n+1}^n\mu(E)^\frac{1}{n+1}\Rightarrow \frac{\|f\|_{n+1}^{n+1}}{\|f\|_n^n}\geq \|f\|_{n+1}\mu(E)^{-\frac{1}{n+1}},$$ and also $$\|f\|_{n+1}^{n+1}=\int_E|f|^{n+1}\leq\|f\|_{\infty}\int_E|f|^n=\|f\|_{\infty}\|f\|_n^n\Rightarrow \frac{\|f\|_{n+1}^{n+1}}{\|f\|_n^n}\leq \|f\|_{\infty}.$$

Now, let $n\to\infty$.

detnvvp
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