Math 2.0

Real functions Theorem

Prékopa–Leindler inequality

Builds on Lebesgue integralMeasurable function

Statement

Let n≥1n \ge 1 and 0<t<10 < t < 1. Let f,g,h ⁣:ℝn→[0,∞)f, g, h \colon \R^n \to [0, \infty) be Lebesgue measurable functions such that h((1−t)x+ty)≥f(x)1−tg(y)tfor all x,y∈ℝn.h((1-t) x + t y) \ge f(x)^{1-t} g(y)^{t} \quad \text{for all } x, y \in \R^n. Then ∫ℝnh dx≥(∫ℝnf dx)1−t(∫ℝng dx)t.\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=1n = 1 uses a transport map between ff and gg after normalisation, and the general case follows by induction on nn with Fubini's theorem. Example: let A,BA, B be nonempty compact sets and f=1Af = 1_A, g=1Bg = 1_B, h=1(1−t)A+tBh = 1_{(1-t)A + tB}. The hypothesis holds, and the conclusion gives λ((1−t)A+tB)≥λ(A)1−tλ(B)t\lambda((1-t) A + t B) \ge \lambda(A)^{1-t} \lambda(B)^{t}. This multiplicative form implies the Brunn–Minkowski theorem by scaling AA and BB. It is a reversed form of Hölder's inequality: Hölder bounds ∫f1−tgt\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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  1. Lebesgue integral concept
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Further reading

  1. Web https://en.wikipedia.org/wiki/Prékopa–Leindler_inequality
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Kindstatement, theorem
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MSC26D15, 52A40
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AboutLebesgue integral
UsesLebesgue integral, Measurable function
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Changed2 November 2026

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