If $f \in L^{\infty}$ and $\exists r < \infty$ so that $\|f\|_r < \infty$, show $\lim_{p \rightarrow \infty} \|f\|_p = \|f\|_{\infty}$

Question

Question: This is the last part of a 5 part question I am working on. Let $(X,\mu)$ be a possibly infinite measure space. Assume $\exists r < \infty$ with $\|f\|_r < \infty$ and that $\|f \|_{\infty} < \infty$. Show that $\lim_{p \rightarrow \infty} \|f_p\| = \|f\|_{\infty}$.

This is from Real and Complex by Rudin, chapter 3 exercise 14.

Progress: I have shown that $\|f\|_{\infty} \le \lim_{p \rightarrow \infty} \|f\|_p$ as follows,

Fix $\epsilon > 0$. Let $E = \{x : |f(x)| > \|f\|_{\infty} - \epsilon \}$. Then observe $$ \|f\|_p \ge \left( \int_{E} |f|^p d\mu \right)^{1/p} > \left( \int_{E}(\| f \|_{\infty} - \epsilon)^{p} d\mu \right)^{1/p} = \left( \|f\|_{\infty} - \epsilon \right) \mu(E)^{1/p}, $$ thus, $\lim_{p \rightarrow \infty} \|f\|_p \ge \|f\|_{\infty} - \epsilon$ since $\mu(E) < \infty$.

I attempted something similar for the other direction, but could not say the measure of a set was finite like (I think) I need for this argument to work. Here is what I tried:

Since $\|f\|_r < \infty, \exists R$ so that $|x| > R \implies f(x) < \frac{1}{2}$. Let $A = \{ x : |x| \le R \}$ and $B = \{x : |x| > R \}$. Then, $$\|f\|_{p} \le \left( \int_{A} |f|^p d \mu + \int_{B} \frac{1}{2^p} d\mu\right)^{1/p} = \left( \int_{A} |f|^p d\mu + \frac{1}{2^p} \mu( B ) \right)^{1/p}.$$

If $\mu(B) < \infty$ this can easily show the desired result. Moreover, if I could show that there is a family of sets $\{B_p\}$ that act similarly so that $\mu(B_p)$ grows slower than $e^p$, then I can also complete the proof.

Thoughts?

Answer 1

For the first part you have to use $\liminf$, as you don't still know that the limit exists.

For the second part, you are thinking as if $X$ was $\mathbb R^n$, which it might not be. One way of attacking the problem along your line of thought would be to assume $\|f\|_\infty=1$ (i.e., work with $f/\|f\|_\infty$). Then, for $p>r$, $$ \left(\int_X|f|^pd\mu\right)^{1/p}\leq\left(\int_X|f|^rd\mu\right)^{1/p}=\|f\|_r^{r/p} $$ Then $$ \limsup_{p\to\infty}\|f\|_p\leq1. $$ Now you can scale back with $\|f\|_\infty$ to get $$ \limsup_{p\to\infty}\|f\|_p\leq\|f\|_\infty. $$

Another way of doing this second part is to use Hölder's inequality: $$ \|f\|_p^p=\int_X|f|^pd\mu=\int_X|f|^r|f|^{p-r}d\mu\leq \|f\|_\infty^{p-r}\,\|f\|_r^r. $$ So $$ \limsup_{p\to\infty}\|f\|_p\leq\limsup_{p\to\infty}\|f\|_\infty^{(p-r)/p}\,\|f\|_r^{r/p}=\|f\|_\infty. $$