If $(X,\mu)$ is a measure space, $0<p<q<\infty$ (or maybe we can even allow $0\leq p<q\leq \infty$?) and $\mu$ is bounded above, say $\mu(E)\leq M$ for all measurable $E\subseteq X,$ then we get
$$\lVert f\rVert_p\leq M^{\frac{1}{p}-\frac{1}{q}}\lVert f\rVert_q.$$
This may be proved via Jensen's inequality, such as is done by Daniel Fischer here, or by Hölder's inequality as is done by Zhanxiong here.
If the measure is bounded below, $\mu(E)\geq m$ for all measurable $E\subseteq X,$ we have instead
$$m^{\frac{1}{p}-\frac{1}{q}}\lVert f\rVert_q\leq \lVert f\rVert_p.$$
The only proof I've found of this is a little messier, realizing that you can scale the components of a simple function to be less than one, and then raise the components to the power $p/q$ to get a smaller value, and use that and the bound on the measure to relate the norms of the simple functions.
Versions of this proof can be seen around, for example the finite dim proof by Laurent Duval here, or versions with counting measure by 1015 here, by AD here, theorem 0.3 of lecture 17 of these MIT lecture notes by Jeff Viaclovsky, or the proof of the general version by uforoboa here, which I attempt to repeat here.
But there is such a nice duality between the results. Do they not admit dual proofs? Let's write the theorems this way:
$$m\leq \mu \leq M$$
implies
$$m^{\frac{1}{p}-\frac{1}{q}} \leq \frac{\lVert f\rVert_p}{\lVert f\rVert_q}\leq M^{\frac{1}{p}-\frac{1}{q}}$$ with the left- (resp. right-) hand side of the inequality below holding when the left- (resp. right-) hand side of the inequality above hold.
I thought to use the concave version of Jensen's inequality with power $q/p$, but it yields the same bounded above inequality. Is there is a notion of a dual $\sigma$-algebra on $X$ exchanging $\varnothing$ and $X$, for which bounded below measures become bounded above?
