Bourbaki

Exercise 25, § 1

Integration, VII, p. 85

Bourbaki marks this one as harder than the rest.

a) Let ff be a positive numerical function defined on R\mathbf{R}, integrable (for the Lebesgue measure μ\mu on R\mathbf{R}), bounded, and with compact support. Let γ=sup⁡t∈Rf(t)\gamma = \sup_{t \in \mathbf{R}} f(t). For every w∈Rw \in \mathbf{R}, denote by Uf(w)U_f(w) the set of t∈Rt \in \mathbf{R} such that f(t)≥wf(t) \geq w; set νf(w)=μ∗(Uf(w))\nu_f(w) = \mu^*(U_f(w)). Show that for every α>1\alpha > 1,

∫−∞+∞fα(t) dt=∫0γνf(w)αwα−1 dw. \int_{-\infty}^{+\infty} f^\alpha(t) \, dt = \int_0^\gamma \nu_f(w) \alpha w^{\alpha-1} \, dw .

b) Let gg be a second numerical function satisfying the same conditions as ff, and set δ=sup⁡t∈Rg(t)\delta = \sup_{t \in \mathbf{R}} g(t). Let hh be the function defined on R2\mathbf{R}^2 by h(u,v)=f(u)+g(v)h(u, v) = f(u) + g(v) if f(u)g(v)≠0f(u)g(v) \neq 0, and h(u,v)=0h(u, v) = 0 otherwise; finally, set k(t)=sup⁡u+v=th(u,v)k(t) = \sup_{u+v=t} h(u, v), so that kk is positive, integrable, bounded, and has compact support. Show that for every α>1\alpha > 1,

∫−∞+∞kα(t) dt≥(γ+δ)α(1γα∫−∞+∞fα(t) dt+1δα∫−∞+∞gα(t) dt). \int_{-\infty}^{+\infty} k^\alpha(t) \, dt \geq (\gamma + \delta)^\alpha \left( \frac{1}{\gamma^\alpha} \int_{-\infty}^{+\infty} f^\alpha(t) \, dt + \frac{1}{\delta^\alpha} \int_{-\infty}^{+\infty} g^\alpha(t) \, dt \right) .

(Observe that for 0<w<10 < w < 1, one has Uk(γw+δw)⊃Uf(γw)+Ug(δw)U_k(\gamma w + \delta w) \supset U_f(\gamma w) + U_g(\delta w), and use a) and Exer. 23 d).)

c) Let μn\mu_n be Lebesgue measure on Rn\mathbf{R}^n, A and B two integrable subsets of Rn\mathbf{R}^n. Show that (Brunn–Minkowski inequality)

((μn)∗(A+B))1/n≥(μn(A))1/n+(μn(B))1/n. ((\mu_n)_*(A+B))^{1/n} \geq (\mu_n(A))^{1/n} + (\mu_n(B))^{1/n} .

(Reduce to the case that A and B are compact. Then argue by induction on nn using Exer. 23 d), the Lebesgue–Fubini theorem, the inequality proved in b), as well as Hölder’s inequality.)

Solution

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Transcribed from the printed volume. Not translated.