Prékopa–Leindler inequality

Builds on Lebesgue integral Measurable function

Statement

Let $n \ge 1$ and $0 < t < 1$ . Let $f, g, h \colon \R^n \to [0, \infty)$ be Lebesgue measurable functions such that $h((1-t) x + t y) \ge f(x)^{1-t} g(y)^{t} \quad \text{for all } x, y \in \R^n.$ Then $\int_{\R^n} h \, dx \ge \left(\int_{\R^n} f \, dx\right)^{1-t} \left(\int_{\R^n} g \, dx\right)^{t}.$

Remarks

The integrals are Lebesgue integrals and may be infinite. The proof for $n = 1$ uses a transport map between $f$ and $g$ after normalisation, and the general case follows by induction on $n$ with Fubini's theorem. Example: let $A, B$ be nonempty compact sets and $f = 1_A$ , $g = 1_B$ , $h = 1_{(1-t)A + tB}$ . The hypothesis holds, and the conclusion gives $\lambda((1-t) A + t B) \ge \lambda(A)^{1-t} \lambda(B)^{t}$ . This multiplicative form implies the Brunn–Minkowski theorem by scaling $A$ and $B$ . It is a reversed form of Hölder's inequality: Hölder bounds $\int f^{1-t} g^t$ from above, while here a function above the "sup-convolution" is bounded from below.

Going deeper

For a student

A main consequence is that log-concave functions and measures are stable: marginals and convolutions of log-concave densities are log-concave. This makes the inequality a basic tool in convex geometry and in the concentration of measure, for example for Gaussian measures.

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Tag	`J9CJX9F42Y`
URI	`m2://J9CJX9F42Y`
Kind	statement, theorem
Writing	W0 stub
MSC	26D15, 52A40
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About	Lebesgue integral
Uses	Lebesgue integral, Measurable function
Same as	Wikidata Q7253815 Wikipedia Prékopa–Leindler inequality