Bourbaki

Exercise 1, § 3

Functions of a Real Variable, II, 86

Let I be an arbitrary interval in R\mathbf{R}, let A be a set, and g a finite real function defined on I×AI \times A such that for every α∈A\alpha \in A the map t↦g(t,α)t \mapsto g(t, \alpha) is decreasing and ≥0\geq 0 on I; suppose further that there is a number M independent of α\alpha such that g(t,α)≤Mg(t, \alpha) \leq M on I×AI \times A.

a) Show that if ff is a regulated function on I such that the integral ∫If(t) dt\int_I f(t) \, dt converges, then the integral ∫If(t)g(t,α) dt\int_I f(t) g(t, \alpha) \, dt is uniformly convergent for α∈A\alpha \in A (use the second mean value theorem; cf. II, p. 82, exerc. 16).

b) Suppose that I=[a,+∞[I = [a, +\infty[ and also that g(t,α)g(t, \alpha) tends uniformly to 0 (for α∈A\alpha \in A) when tt tends to +∞+\infty. Show that if ff is a regulated function on I and if there is a number k>0k > 0 such that ∥∫Jf(t) dt∥≤k\| \int_J f(t) \, dt \| \leq k for every compact interval J contained in I, then the integral ∫If(t)g(t,α) dt\int_I f(t) g(t, \alpha) \, dt is uniformly convergent for α∈A\alpha \in A (same method).

c) Suppose that I=[a,+∞[I = [a, +\infty[ with a>0a > 0, A=[0,+∞[A = [0, +\infty[, and g(t,α)=φ(αt)g(t, \alpha) = \varphi(\alpha t), where φ\varphi is a convex decreasing function and ≥0\geq 0 on A, tending to 0 as xx tends to +∞+\infty and such that φ(0)=1\varphi(0) = 1. Suppose further that f=h′′f = h'', where hh is a twice differentiable function on [a,+∞[[a, +\infty[, and, together with h′h', is bounded on this interval. The integral ∫a+∞f(t)φ(αt) dt\int_a^{+\infty} f(t) \varphi(\alpha t) \, dt is then convergent for α>0\alpha > 0: show that when α\alpha tends to 0, it tends to −h′(a)-h'(a) (integrate by parts and use the second mean value theorem).

d) Take φ(x)=e−cx\varphi(x) = e^{-cx}, where c>0c > 0, take a=1a = 1, and for ff take the complex function t↦etlog⁡t/tt \mapsto e^{t \log t}/t, which is the derivative of a bounded function on I; show that for a suitable choice of cc the integral ∫1+∞euαtetlog⁡tt dt\int_1^{+\infty} e^{u \alpha t} \frac{e^{t \log t}}{t} \, dt, which is absolutely convergent for α>0\alpha > 0, does not tend to any limit when α\alpha tends to 0 (after integrating by parts, make the change of variables αt=u\alpha t = u; use Laplace transform theory to see that ∫0+∞e−cu+tlog⁡u du\int_0^{+\infty} e^{-cu + t \log u} \, du does not vanish for certain c>0c > 0).

Solution

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