Suppose that $f\in L^p\cap L^\infty$ for some finite $p$. Then $f\in L^q$ for all $q>p$ and
$\|f\|_\infty=\lim_{q\to\infty}\|f\|_q$
I've managed to prove that $\int_E |f|^q<\int_E |f|^p<\infty$ with $E=\{x:|f(x)|<1\}$
and that $\int_F|f|^q=0$, with $F=\{x: |f(x)|>\|f\|_\infty\}$.
However I don't know how to proceed in the case where $1\leq f(x) \leq \|f\|_\infty$ and to prove that this convergence.
Assuming we're working on a finite measure space:
Take $t< \|f\|_\infty$. Then let $A = \{|f| >t \}$. Then $m(A) >0$ by assumption. Now $$\|f\|_p^p = \int |f|^p = \int_A |f|^p + \int_{A^c} |f|^p$$ $$ > \int_{A^c} |f|^p \ge t^p\mu(A^c).$$
So $\|f\|_p > t \mu(A^c)^{1/p}$ and sending $p \to \infty$ gives $ \liminf_p\|f\|_p >t$. Since $t< \|f\|_\infty$ was arbitrary we get $\|f\|_\infty \le \liminf_p \|f\|_p$.
For the other direction, we have $$\|f\|_p^p = \int |f|^p \le \int \|f\|_\infty^p = \|f\|_\infty^p \mu(X).$$
So $\|f\|_p \le \|f\|_\infty \mu(X)^{1/p}$ and thus $$\limsup_p \|f\|_p \le \|f\|_\infty.$$
By normalizing $f$ to $f/\|f\|_{\infty}$, we can assume $\|f\|_{\infty}=1$, so $\|f\|_{q}$, so $\|f\|_{q}\leq\|f\|_{p}^{p/q}$, taking $q\rightarrow\infty$, then $\limsup_{q\rightarrow\infty}\|f\|_{q}\leq 1$. For $\epsilon\in(0,1)$, then $S_{\epsilon}=\{|f|>1-\epsilon\}$ has measure strictly greater than zero, then $\|f\|_{q}$, then $(1-\epsilon)^{q}|S_{\epsilon}|\leq\|f\|_{q}^{q}$, so $(1-\epsilon)|S_{\epsilon}|^{1/q}\leq\|f\|_{q}$, taking $q\rightarrow\infty$, so $1-\epsilon\leq\liminf_{q\rightarrow\infty}\|f\|_{q}$, taking limit $\epsilon\downarrow 0$, so $1\leq\liminf_{q\rightarrow\infty}\|f\|_{q}$.
