Prékopa–Leindler inequality

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 → [0, +∞) be non-negative real-valued measurable functions defined on n-dimensional Euclidean space Rⁿ. Suppose that these functions satisfy

${\displaystyle h\left((1-\lambda )x+\lambda y\right)\geq f(x)^{1-\lambda }g(y)^{\lambda }}$

1

for all x and y in Rⁿ. Then

${\displaystyle \|h\|_{1}:=\int _{\mathbb {R} ^{n}}h(x)\,\mathrm {d} x\geq \left(\int _{\mathbb {R} ^{n}}f(x)\,\mathrm {d} x\right)^{1-\lambda }\left(\int _{\mathbb {R} ^{n}}g(x)\,\mathrm {d} x\right)^{\lambda }=:\|f\|_{1}^{1-\lambda }\|g\|_{1}^{\lambda }.}$

Essential form of the inequality

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 inequality

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 statistics

Log-concave distributions

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

${\displaystyle M((1-\lambda )y_{1}+\lambda y_{2})\geq M(y_{1})^{1-\lambda }M(y_{2})^{\lambda },}$

which is the definition of log-concavity for M.

To see how this implies the preservation of log-convexity by independent sums, suppose that X and Y are independent random variables with log-concave distribution. Since the product of two log-concave functions is log-concave, the joint distribution of (X,Y) is also log-concave. Log-concavity is preserved by affine changes of coordinates, so the distribution of (X + Y, X − Y) is log-concave as well. Since the distribution of X+Y is a marginal over the joint distribution of (X + Y, X − Y), we conclude that X + Y has a log-concave distribution.

Applications to concentration of measure

The Prékopa–Leindler inequality can be used to prove results about concentration of measure.

Theorem^{[citation needed]} Let ${\textstyle A\subseteq \mathbb {R} ^{n}}$, and set ${\textstyle A_{\epsilon }=\{x:d(x,A)<\epsilon \}}$. Let ${\textstyle \gamma (x)}$ denote the standard Gaussian pdf, and ${\textstyle \mu }$ its associated measure. Then ${\textstyle \mu (A_{\epsilon })\geq 1-{\frac {e^{-\epsilon ^{2}/4}}{\mu (A)}}}$.

Proof of concentration of measure

The proof of this theorem goes by way of the following lemma: Lemma In the notation of the theorem, ${\textstyle \int _{\mathbb {R} ^{n}}\exp(d(x,A)^{2}/4)d\mu \leq 1/\mu (A)}$. This lemma can be proven from Prékopa–Leindler by taking ${\textstyle h(x)=\gamma (x),f(x)=e^{\frac {d(x,A)^{2}}{4}}\gamma (x),g(x)=1_{A}(x)\gamma (x)}$ and ${\textstyle \lambda =1/2}$. To verify the hypothesis of the inequality, ${\textstyle h({\frac {x+y}{2}})\geq {\sqrt {f(x)g(y)}}}$, note that we only need to consider ${\textstyle y\in A}$, in which case ${\textstyle d(x,A)\leq ||x-y||}$. This allows us to calculate: ${\displaystyle (2\pi )^{n}f(x)g(x)=\exp({\frac {d(x,A)}{4}}-||x||^{2}/2-||y||^{2}/2)\leq \exp({\frac {||x-y||^{2}}{4}}-||x||^{2}/2-||y||^{2}/2)=\exp(-||{\frac {x+y}{2}}||^{2})=(2\pi )^{n}h({\frac {x+y}{2}})^{2}.}$ Since ${\textstyle \int h(x)dx=1}$, the PL-inequality immediately gives the lemma. To conclude the concentration inequality from the lemma, note that on ${\textstyle \mathbb {R} ^{n}\setminus A_{\epsilon }}$, ${\textstyle d(x,A)>\epsilon }$, so we have ${\textstyle \int _{\mathbb {R} ^{n}}\exp(d(x,A)^{2}/4)d\mu \geq (1-\mu (A_{\epsilon }))\exp(\epsilon ^{2}/4)}$. Applying the lemma and rearranging proves the result.