Given that $f \in L^{p_0} \cap L^{\infty}$ show $f \in L^p$ for all $p_0 \leq p \leq \infty$

Question

Title says it all. I feel like Holder's inequality may be useful here but I'm struggling on where to start. Not looking for a solution, just some tips to jump start from.

Answer 1

The general fact is, if $f\in L^p\cap L^q$ with $1\leq p<q\leq \infty$, then $f\in L^r$ for any $r\in (p,q)$.

This is entirely elementary: write $f=f_1+f_2$, where $f_1$ is the restriction of $f$ to $S=\{x\mid \lvert f(x)\rvert\geq 1\}$, then estimate $L^r$ norm of $f$ in terms of the norms of $f_1$ and $f_2$, and possibly the measure of $S$ if $q=\infty$ as in the case you're interested in.

(In fact, this is true even if some of $p,q,r$ are in $(0,1)$; you don't need to use anything about $\lVert\cdot\rVert_p,\lVert\cdot\rVert_q,\lVert\cdot\rVert_r$ but the raw definitions. In particular, you don't need Hölder's inequality or the fact that they're norms.)

Answer 2

Let $p\in(p_0,\infty)$. Then $$ \int_X\lvert f(x)\rvert^p\,dx=\int_X\lvert f(x)\rvert^{p-p_0}\lvert f(x)\rvert^{p_0} \,dx \le \int_X\| f\|^{p-p_0}_\infty\lvert f(x)\rvert^{p_0} \,dx = \| f\|^{p-p_0}_\infty\int_X\lvert f(x)\rvert^{p_0} \,dx. $$ Thus $$ \left(\int_X\lvert f(x)\rvert^p\,dx\right)^{1/p}\le\| f\|^{1-p_0/p}_\infty\left(\int_X\lvert f(x)\rvert^{p_0} \,dx\right)^{1/p}=\| f\|^{1-p_0/p}_\infty \|f\|_{p_0}^{p_0/p}. $$