# Let $\mu$ be finite measure and $\lVert f\rVert_\infty>0$ and $a_n=\int |f|^n d\mu$. Prove $\lim {a_{n+1}\over a_n}=\lVert f\rVert_\infty$

Here $$\lVert f\rVert_\infty=\text{ess sup } |f|=\inf\{c\in\mathbb{R}|\ \mu(|f|\ge c)=0\}$$ i.e. $$\lVert f\rVert_\infty$$ denotes least essential upper bound of $$|f|$$.

Now $$\lVert f\rVert_\infty>0\implies \exists k>0$$ such that $$f\ge k$$ almost everywhere $$\mu$$. So $$|f|^n\ge k^n$$ almost everywhere, this implies $$a_n\ge k^n\mu(\Omega)$$ for all $$n$$ where $$\Omega$$ is the measure space.

Now $$\left|\frac{a_{n+1}}{a_n}-\lVert f\rVert_\infty\right|\\\le\frac{1}{|a_n|}\int |f|^n[|f|-\lVert f\rVert_\infty]\ d\mu\\\le \frac{2\lVert f\rVert_\infty}{a_n}\int |f|^n\ d\mu\\=2\lVert f\rVert _\infty$$

But I need to have $$\left|\frac{a_{n+1}}{a_n}-\lVert f\rVert_\infty\right|<\epsilon$$ for all but finitely many $$n$$.

Can anyone help me in this regard? Thanks for your help in advance.

Hint: By Holder's inequality, $$\int |f|^{n}d\mu \leq (\int |f|^{n+1}d\mu)^{\frac n {n+1}} C^{\frac 1 {n+1}}$$ where $$C=\mu (X)$$. Now use the fact that $$\int |f|^{n+1}d\mu \geq \int_E |f|^{n+1}d\mu$$ where $$E=\{x: |f(x)| >\|f\|_{\infty} -\epsilon\}$$. Can you finish?
• Yes. From here I get $\left[\int |f|^{n+1}\right]^{1/{n+1}}\ge (\lVert f\rVert _\infty -\epsilon)D^{1/{n+1}}$ where $D=\mu(E)$. And using this I get $\frac{a_{n+1}}{a_n}\ge \lVert (f\rVert_\infty-\epsilon)(D/C)^{1/{n+1}}$. Again, $\frac{a_{n+1}}{a_n}\le \lVert f\rVert_\infty$ Commented Mar 10, 2021 at 7:08