real analysis - $||f||_pto ||f||_infty$ under general assumptions

I am trying to show that for nonnegative $f$ on $\mathbb{R}$ if $||f||_1<\infty$, we have
$$\lim_{p\uparrow \infty}||f||_p=||f||_\infty$$
I have tried to fiddle around with
$$\left(\int_\mathbb{R}|f|^p|f|^{1-p} \right)^{1/p}<\infty$$

but haven't found a nice way to get an estimate on $\int_E|f|^p$ from this. Honestly I am pretty uncomfortable with non finite measure spaces.

Any hints would be awesome! Any references for problems similar to this would also be awesome.

edit: this is not a duplicate of the linked question as 1) my measure space is not finite, 2) my function is in $L^1$ and not necessarily in $L^\infty$

Answer

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

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

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.