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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.

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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?

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  • $\begingroup$ 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, $\frac{a_{n+1}}{a_n}\le \lVert f\rVert_\infty$ $\endgroup$ Commented Mar 10, 2021 at 7:08

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