$||f||_p\to ||f||_\infty$ under general assumptions

Let $(X,\mu)$ be a measure space, and let $f:X\to [0,\infty)$ be measurable.

Thm 1: If $\|f\|_\infty = \infty,$ then $\lim_{p\to \infty}\|f\|_p = \infty.$

Thm 2: If $\|f\|_\infty < \infty$ and $\|f\|_{p_0} < \infty$ for some $p_0 \in (0,\infty),$ then $\lim_{p\to \infty}\|f\|_p = \|f\|_\infty.$

Proof of Thm 1: Note that we have $\mu(\{f>C\}) >0$ for all $C>0$ because $\|f\|_\infty = \infty.$ Suppose $\mu(\{f>C\})=\infty$ for some $C>0.$ Then because

$$\tag 1 \int_X f^p\, d\mu \ge \int_{\{f>C\}} f^p\, d\mu \ge C^p\mu(\{f>C\})$$

for all $p\in (0,\infty),$ all integrals on the left of $(1)$ are $\infty.$ Thus all quantities of interest are stuck at the value $\infty,$ and we're done.

In the case $\mu(\{f>C\})<\infty$ for all $C>0,$ we see by $(1)$ that $\|f\|_p \ge C(\mu(\{f>C\}))^{1/p}.$ It then follows that $\liminf_{p\to \infty} \|f\|_p \ge C.$ Since $C$is arbitrarily large, $\liminf \|f\|_p =\infty,$ and we're done.

Proof of Thm 2: For any $p>p_0,$ we have

$$\int_X f^p\, d\mu = \int_X f^{p-p_0}\cdot f^{p_0}\, d\mu \le \|f\|_\infty^{p-p_0}\int_X f^{p_0}\, d\mu.$$

Now take $p$th roots to get

$$\|f\|_p \le (\|f\|_\infty)^{1-p_0/p}(\int_X f^{p_0}\, d\mu)^{1/p}.$$

This implies $ \limsup \|f\|_p \le \|f\|_\infty.$

On the other hand, if $0<\epsilon<\|f\|_\infty,$ we have $\mu(\{f>\|f\|_\infty-\epsilon\}) >0.$ Thus

$$\int_X f^p\, d\mu \ge \int_{\{f>\|f\|_\infty-\epsilon\}}f^p\, d\mu \ge (\|f\|_\infty-\epsilon\})^p\mu(\{f>\|f\|_\infty-\epsilon\}).$$

Take $p$th roots and let $p\to \infty$ to get $\liminf \|f\|_p \ge \|f\|_\infty-\epsilon.$ Since $\epsilon$ is arbitrarily small, we have $ \liminf \|f\|_p \ge \|f\|_\infty.$

We've shown $\|f\|_\infty \le \liminf \|f\|_p \le \limsup \|f\|_p \le \|f\|_\infty,$ so we're done.

Assume $\|f\|_\infty\lt\infty$.

For $\lambda\gt\|f\|_\infty$, $|\{x\in\Omega:|f(x)|\gt\lambda\}|=0$. Thus, $$ \begin{align} \|f\|_p^p &=\int_\Omega|f(x)|^p\,\mathrm{d}x\\ &=p\int_0^\infty|\{x\in\Omega:|f(x)|\gt\lambda\}|\,\lambda^{p-1}\,\mathrm{d}\lambda\\ &\le p\|f\|_\infty^{p-1}\int_0^\infty|\{x\in\Omega:|f(x)|\gt\lambda\}|\,\mathrm{d}\lambda\\ &=p\|f\|_\infty^{p-1}\|f\|_1\tag{1} \end{align} $$ Therefore, $$ \begin{align} \limsup_{p\to\infty}\|f\|_p &\le\lim_{p\to\infty}\left(p\|f\|_1\right)^{1/p}\|f\|_\infty^{(p-1)/p}\\ &=\|f\|_\infty\tag{2} \end{align} $$ If $\|f\|_\infty=\infty$, then $(2)$ is trivially true. Thus, $(2)$ is true whether $\|f\|_\infty\lt\infty$ or not. Therefore, we can remove the assumption that $\|f\|_\infty\lt\infty$.

For $\lambda\lt\|f\|_\infty$, $|\{x\in\Omega:|f(x)|\gt\lambda\}|\gt0$. Furthermore, for $\lambda\gt0$, $$ \|f\|_p^p\ge|\{x\in\Omega:|f(x)|\gt\lambda\}|\,\lambda^p\tag{3} $$ Therefore, for all $\lambda\lt\|f\|_\infty$, $$ \begin{align} \liminf_{p\to\infty}\|f\|_p &\ge\lim_{p\to\infty}|\{x\in\Omega:|f(x)|\gt\lambda\}|^{1/p}\lambda\\ &=\lambda\tag{4} \end{align} $$ Since $(4)$ is true for all $\lambda\lt\|f\|_\infty$, we have $$ \liminf_{p\to\infty}\|f\|_p\ge\|f\|_\infty\tag{5} $$ Finally, $(2)$ and $(5)$ imply $$ \lim_{p\to\infty}\|f\|_p=\|f\|_\infty\tag{6} $$

Here is another proof to this problem:

Assume $0<\|f\|_\infty<\infty$ and $f\in L_r$ or some $r>0$. Then $|f|/\|f\|_\infty<1$ a.s.. For $p>r$

$$\frac{|f|^p}{\|f\|^p_\infty}\leq \frac{|f|^r}{\|f\|^r_\infty}\in L_1$$

hence $p\in E:=\{s: \|f\|_s<\infty\}$. Integrating on both side leads to

$$\frac{\|f\|_p}{\|f\|_\infty}\leq\Big(\frac{\|f\|_r}{\|f\|_\infty}\Big)^{r/p}\xrightarrow{p\rightarrow\infty}1$$ That is $$\limsup_p\|f\|_p\leq \|f\|_\infty$$

By the Markov-Chebyshev inequality, for any $0<\alpha<\|f\|_\infty$

$$0<\alpha\big(\mu(|f|>\alpha)\big)^{1/p}\leq\|f\|_p$$

Hence $\alpha\leq\liminf_p\|f\|_p$ and so, $\|f\|_\infty\leq\liminf_p\|f\|_p$.

If $\|f\|_p=\infty$ and $f\in L_r$ for some $r>0$ then $0<\mu(|f|<n)\leq\frac{1}{n^r}\|f\|_r<\infty$ and so

$$0<n\big(\mu(|f|>n)\big)^{1/p}\leq\|f\|_p\quad\text{for}\quad p\geq r$$

This implies $n\leq\liminf_p\|f\|p$ for any $n\in\mathbb{N}$.