How to show that $ \|f\|_\infty\leq m^{-\frac{1}{p}}\|f\|_p $ when every set of positive measure in $X$ has measure at least $m$?

Question

If $f \in L^p$ for some $0 < p < \infty$, and every set of positive measure in $X$ has measure at least $m$, show that $f \in L^q$ for all $p < q \leq \infty$, with $\|f\|_{L^q} \leq m^{\frac{1}{q}-\frac{1}{p}} \|f\|_{L^p}$.

This is a generalization of the statement mentioned in another question:

Understanding the proof of $l^p(A)\subset l^q(A)$ where $0<p<q\leq\infty$ in Folland's Real Analysis

A related question:

When $L^p \subset L^q$ for $p <q$.

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I think this can be seen by first checking the $q=\infty$ case, and then using log-convexity of the $L^p$ norm (which is equivalent to Hölder) to get the remaining cases.

Here is my question:

How shall I prove the $q=\infty$ case: $$ \|f\|_\infty\leq m^{-\frac{1}{p}}\|f\|_p? $$

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[Attempt.] Suppose $f\in L^p$ and define $A_n=\{x:n\leq|f(x)|<n+1\}$. Then $$ \infty>\int_X |f|^p\ d\mu=\sum_{n=0}^\infty\int_{A_n}|f|^p\ d\mu \geq\sum_{n=0}^\infty\mu(A_n) n^p\geq m \sum_{n=0}^\infty n^p $$ Note that the sum above must be a finite sum and thus there exists a nonnegative integer $N$ such that $ \mu(A_N)>0 $ and $\mu(A_n)=0$ for all $n>N$. It follows that $$ \|f\|_p^p\geq m\sum_{n=0}^N n^p $$ For now I only have $\|f\|_\infty\leq N+1$. I don't think I can get $$ \sum_{n=0}^N n^p\geq (N+1)^p $$ which would give the result.

Answer 1

For each positive integer $n$, the set $A_n:=\left\{x\mid \left\lvert f(x)\right\rvert\gt \left\lVert f\right\rVert_\infty-n^{-1} \right\} $ has a positive measure hence by assumption, $\mu\left(A_n\right)\geqslant m$. Now, notice that $$ \left\lVert f\right\rVert_p^p =\int_X \left\lvert f(x)\right\rvert^p\mathrm d\mu(x)\geqslant \int_{A_n} \left\lvert f(x)\right\rvert^p\mathrm d\mu(x) \geqslant\mu\left(A_n\right) \left(\left\lVert f\right\rVert_\infty-n^{-1} \right)^p \geqslant m \left(\left\lVert f\right\rVert_\infty-n^{-1} \right)^p. $$
Since $n$ is arbitrary, we get the wanted result.