In mathematics, the Prékopa–Leindler inequality is an integral inequality closely related to the reverse Young's inequality, the Brunn–Minkowski inequality and a number of other important and classical inequalities in analysis. The result is named after the Hungarian mathematicians András Prékopa and László Leindler.^[1][2]

Statement of the inequality

Let 0 < λ < 1 and let f, g, h : Rⁿ → edit

Recall that the essential supremum of a measurable function f : Rⁿ → R is defined by

${\displaystyle \mathop {\mathrm {ess\,sup} } _{x\in \mathbb {R} ^{n}}f(x)=\inf \left\{t\in -\infty ,+\infty \mid f(x)\leq t{\text{ for almost all }}x\in \mathbb {R} ^{n}\right\}.}$

This notation allows the following essential form of the Prékopa–Leindler inequality: let 0 < λ < 1 and let f, g ∈ L¹(Rⁿ; 0, +∞)) be non-negative absolutely integrable functions. Let

${\displaystyle s(x)=\mathop {\mathrm {ess\,sup} } _{y\in \mathbb {R} ^{n}}f\left({\frac {x-y}{1-\lambda }}\right)^{1-\lambda }g\left({\frac {y}{\lambda }}\right)^{\lambda }.}$

Then s is measurable and

${\displaystyle \|s\|_{1}\geq \|f\|_{1}^{1-\lambda }\|g\|_{1}^{\lambda }.}$

The essential supremum form was given by Herm Brascamp and Elliott Lieb.^[3] Its use can change the left side of the inequality. For example, a function g that takes the value 1 at exactly one point will not usually yield a zero left side in the "non-essential sup" form but it will always yield a zero left side in the "essential sup" form.

Relationship to the Brunn–Minkowski inequalityedit

It can be shown that the usual Prékopa–Leindler inequality implies the Brunn–Minkowski inequality in the following form: if 0 < λ < 1 and A and B are bounded, measurable subsets of Rⁿ such that the Minkowski sum (1 − λ)A + λB is also measurable, then

${\displaystyle \mu \left((1-\lambda )A+\lambda B\right)\geq \mu (A)^{1-\lambda }\mu (B)^{\lambda },}$

where μ denotes n-dimensional Lebesgue measure. Hence, the Prékopa–Leindler inequality can also be used^[4] to prove the Brunn–Minkowski inequality in its more familiar form: if 0 < λ < 1 and A and B are non-empty, bounded, measurable subsets of Rⁿ such that (1 − λ)A + λB is also measurable, then

${\displaystyle \mu \left((1-\lambda )A+\lambda B\right)^{1/n}\geq (1-\lambda )\mu (A)^{1/n}+\lambda \mu (B)^{1/n}.}$

Applications in probability and statisticsedit

Log-concave distributionsedit

The Prékopa–Leindler inequality is useful in the theory of log-concave distributions, as it can be used to show that log-concavity is preserved by marginalization and independent summation of log-concave distributed random variables. Since, if ${\displaystyle X,Y}$ have pdf ${\displaystyle f,g}$, and ${\displaystyle X,Y}$ are independent, then ${\displaystyle f\star g}$ is the pdf of ${\displaystyle X+Y}$, we also have that the convolution of two log-concave functions is log-concave.

Suppose that H(x,y) is a log-concave distribution for (x,y) ∈ R^m × Rⁿ, so that by definition we have

${\displaystyle H\left((1-\lambda )(x_{1},y_{1})+\lambda (x_{2},y_{2})\right)\geq H(x_{1},y_{1})^{1-\lambda }H(x_{2},y_{2})^{\lambda },}$

(2)

and let M(y) denote the marginal distribution obtained by integrating over x:

${\displaystyle M(y)=\int _{\mathbb {R} ^{m}}H(x,y)\,dx.}$

Let y₁, y₂ ∈ Rⁿ and 0 < λ < 1 be given. Then equation (2) satisfies condition (1) with h(x) = H(x,(1 − λ)y₁ + λy₂), f(x) = H(x,y₁) and g(x) = H(x,y₂), so the Prékopa–Leindler inequality applies. It can be written in terms of M as

${\displaystyle M((1-<!--\; -\; -->\mathrm{\lambda <!--\; \lambda \; -->})y_1+\mathrm{\lambda <!--\; \lambda \; -->}{y}_2)\ge <!--\; \ge \; -->M(y_1{)}^{1-<!--\; -\; -->\mathrm{\lambda <!--\; \lambda \; -->}}M(y_2{)}^{}}$Zdroj:https://en.wikipedia.org?pojem=Prékopa–Leindler_inequality
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