How is the Prékopa Leindler inequality different from Holder's inequality?

Question

The following is an extract from these notes (Pg. 30), and I'm not able to understand the part towards the end. I would appreciate any help!

Prékopa-Leindler Inequality: Let $f$, $g$ and $m$ be non-negative measurable functions on $\mathbb R^n$, $\lambda\in(0,1)$ and for all $x,y\in\mathbb R^n$, \begin{equation} m(\lambda x + (1-\lambda)y) \geq f(x)^{\lambda}g(y)^{1-\lambda}. \end{equation} Then, $$\int_{\mathbb R^n} m \geq \left(\int_{\mathbb R^n}f\right)^\lambda \left(\int_{\mathbb R^n}g\right)^{1-\lambda}$$ It is perhaps helpful to notice that the Prékopa–Leindler inequality looks like Hölder’s inequality, backwards. If $f$ and $g$ were given and we set $$m(z) = f(z)^{1-\lambda} g(z)^\lambda$$ (for each $z$), then Hölder’s inequality says that $$\int_{\mathbb R^n} m \leq \left(\int_{\mathbb R^n}f\right)^{1-\lambda} \left(\int_{\mathbb R^n}g\right)^{\lambda}$$ (Hölder’s inequality is often written with $1/p$ instead of $1 − λ$, $1/q$ instead of $λ$, and $f , g$ replaced by $F^p , G^q$ .) The difference between Prékopa–Leindler and Hölder is that, in the former, the value $m(z)$ may be much larger since it is a supremum over many pairs $(x, y)$ satisfying $z = (1 − λ)x + λy$ rather than just the pair $(z, z)$.

Could someone please explain the part in bold? I don't understand the part where the author talks about the suprema over pairs $(x,y)$ and $(z,z)$ in the two cases. Lastly, what's $F^p, G^q$?

Thank you.

Answer 1

In favour of symmetry, I've switched $\lambda$ and $1-\lambda$ in the Hölder's inequality you have stated.
Hölder's inequality says that given non-negative functions $f$ and $g$, if we define $$m(z)=f(z)^\lambda g(z)^{1-\lambda}$$ for each $z$, $$\int m \leq \left(\int f\right)^\lambda \left(\int g\right)^{1-\lambda}.$$

In the Prékopa-Leindler Inequality on the other hand, each $m(z)$ is bounded below by some parameter depending on all the $f(x)$ and $g(y)$ for points $x$,$y$ satisfying $z=\lambda x + (1-\lambda)y$.
Since the inequality is a lower bound on the integral of $m$, the worst-case scenario is when $m$ attains the minimum possible value at every point, which is the supremum that is mentioned. That is, if we define $$m(z)=\sup_{\substack{x,y\in\mathbb{R}^n \\ z=\lambda x + (1-\lambda)y}} f(x)^\lambda g(y)^{1-\lambda}$$ for each $z$ (even for any general $m$ satisfying the condition, $m(z)$ must be at least the quantity above), $$\int m\geq \left(\int f\right)^\lambda \left(\int g\right)^{1-\lambda}.$$

Comparing the definition of $m$ in the two inequalities, it is seen that we only take the pair $(x,y)=(z,z)$ in Hölder's inequality.
The two inequalities are rather similar in terms of basic structure, but also quite different due to the supremum involved in the Prékopa-Leindler inequality, which causes the direction of the inequality to be reversed.

Lastly, what's $F^p$,$G^q$?

Hölder's inequality usually isn't stated in the format the author (and I) have stated it above. In the Wikipedia article which contains the "usual" statement of the theorem, it is given as

Let $(S, \Sigma, \mu)$ be a measure space and let $p, q \in [1, \infty)$ with $1/p + 1/q = 1$. Then, for all measurable real- or complex-valued functions $F$ and $G$ on $S$, $$\|FG\|_1 ≤ \|F\|_p \|G\|_q. $$

The equivalence of this and the statement described in the notes is seen on writing $f=F^p$, $g=G^q$, and $1/p=\lambda$.