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

Question

Let $(E, \mathbb{E}, \mu)$ be a measure space, and assume that $\mu(E) < \infty$, and suppose that $f \in L^\infty$. Prove that $f \in L^p$ for any $p$, and show that $$\lim_{n\to \infty} \|f\|_n=\lim_{n\to \infty} \frac{\|f\|_{n+1}^{n+1}}{\|f\|_{n}^{n}}=\|f\|_\infty$$

I tried to do it but I stuck in the middle , also Idk how to show that last equality in the limit.

a) since $f \in L^\infty$, we have that $\exists M$ such that $|f|\leq M$ $$\|f\|^p_p = \int_E|f|^pd\mu= \int_\mathbb{R} \chi_E |f|^pd\mu \leq \|\chi_E\|_1\||f|^p\|_\infty\leq M^p \mu(E) < \infty$$

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

$\textit{claim}$: $\lim_{n\to \infty}\frac{\|f\|_{n}^{n+1}}{\|f\|_{n+1}^{n+1}}=1$ $$\lim_{n\to \infty}\frac{\|f\|_{n}^{n+1}}{\|f\|_{n+1}^{n+1}}=\lim_{n\to \infty}\bigg(\frac{\|f\|_{n}}{\|f\|_{n+1}}\bigg)^{n+1}$$

Answer 1

You cannot assume that the limits exist.

First note that $\|f\|_n \leq \|f\|_{\infty} (\mu (E))^{1/n}$.

Let $\epsilon >0$ and consider $F=\{x: |f(x)| >\|f\|_{\infty}-\epsilon\}$. We have $\int |f|^{n} \geq \int_F |f|^{n}\geq (\|f\|_{\infty}-\epsilon)^{n}\mu(F)$. Hence $\|f\|_n \geq (\|f\|_{\infty}-\epsilon) (\mu(F))^{1/n}$. Let $n \to \infty$ and then $\epsilon \to 0$ to complete the proof of the fact that $\|f\|_n \to \|f\|_{\infty}$.

Now it is obvious that $\|f\|_{n+1}^{n+1} \leq \|f\|_{n}^{n} \|f\|_{\infty}$. For completing the proof I will give you a hint and let you handle the rest: $\|f\|_n^{n}=\int |f|^{n} \leq (\int |f|^{n+1})^{n/(n+1)} (\mu (E))^{1/n+1}$ by Holder's Inequality.