Bourbaki

Exercise 14, § 6

Integration, IV, p. 137

Bourbaki marks this one as harder than the rest.

Let pp and qq be any two finite real numbers >0> 0 and let f(x1,x2,…,xn)f(x_1, x_2, \ldots, x_n) be a continuous numerical function defined on Rn\mathbf{R}^n.

a) Let μ\mu be Lebesgue measure on X=[0,1]X = [0, 1]. In order that, for every system of nn functions gk∈Lpg_k \in \mathscr{L}^p, the function f(g1,g2,…,gn)f(g_1, g_2, \ldots, g_n) belong to Lq\mathscr{L}^q, it is necessary and sufficient that there exist a number a>0a > 0 such that

∣f(x1,…,xn)∣q⩽a(1+∣x1∣+…+∣xn∣)p. |f(x_1, \ldots, x_n)|^q \leqslant a (1 + |x_1| + \ldots + |x_n|)^p.

(To see that the condition is necessary, argue by contradiction, supposing that for every integer m>0m > 0, there exists a point (x1m,…,xnm)(x_{1m}, \ldots, x_{nm}) of Rn\mathbf{R}^n such that

∣f(x1m,…,xnm)∣q≥m(1+∣x1m∣+…+∣xnm∣)p. |f(x_{1m}, \ldots, x_{nm})|^q \geq m(1 + |x_{1m}| + \ldots + |x_{nm}|)^p .

Show that there would then exist in X a sequence (Am)(A_m) of pairwise disjoint intervals such that, on setting gk(t)=xkmg_k(t) = x_{km} for every t∈Amt \in A_m and gk(t)=0g_k(t) = 0 for every point t that does not belong to any of the AmA_m, each of the functions gkg_k would belong to Lp\mathscr{L}^p but f(g1,…,gn)f(g_1, \ldots, g_n) would not belong to Lq\mathscr{L}^q.)

b) Let μ\mu be Lebesgue measure on R\mathbf{R}. In order that, for every system of n functions gk∈Lpg_k \in \mathscr{L}^p, the function f(g1,…,gn)f(g_1, \ldots, g_n) belong to Lq\mathscr{L}^q, it is necessary and sufficient that there exist a number b>0b > 0 such that

∣f(x1,…,xn)∣q≤b(∣x1∣+∣x2∣+…+∣xn∣)p |f(x_1, \ldots, x_n)|^q \leq b(|x_1| + |x_2| + \ldots + |x_n|)^p

(same method).

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.