214 CHAPTER 8. POSITIVE LINEAR FUNCTIONALS

17. ↑From Problem 1 above and f ∈ L1 (Rp), there exists h ∈Cc (Rp) , continuous andspt(h) a compact set, such that

∫| f −h|dmn < ε . Now consider h ∗ψk. Show that

this function is in C∞c(spt(h)+B

(0, 2

k

)). The notation means you start with the com-

pact set spt(h) and fatten it up by adding the set B(0, 1

k

). It means x+y such that

x ∈ spt(h) and y ∈ B(0, 1

k

). Show the following. For all k large enough,∫

| f −h∗ψk|dmn < ε

so one can approximate with a function which is infinitely differentiable and alsohas compact support. Also show that h ∗ψk converges uniformly to h. If h is afunction in Ck

(Rk)

in addition to being continuous with compact support, show thatfor each |α| ≤ k,Dα (h∗ψk)→ Dα h uniformly. Hint: If you do this for a singlepartial derivative, you will see how it works in general.

18. ↑Let f ∈ L1 (R). Show that limk→∞

∫f (x)sin(kx)dm = 0 Hint: Use the result of the

above problem to obtain g ∈ C∞c (R) , continuous and zero off a compact set, such

that∫| f −g|dm < ε. Then show that

limk→∞

∫g(x)sin(kx)dm(x) = 0.

You can do this by integration by parts. Then consider this.∣∣∣∣∫ f (x)sin(kx)dm∣∣∣∣ =

∣∣∣∣∫ f (x)sin(kx)dm−∫

g(x)sin(kx)dm∣∣∣∣

+

∣∣∣∣∫ g(x)sin(kx)dm∣∣∣∣

≤∫| f −g|dm+

∣∣∣∣∫ g(x)sin(kx)dm∣∣∣∣

This is the celebrated Riemann Lebesgue lemma which is the basis for all theoremsabout pointwise convergence of Fourier series.

19. As another application, here is a very important result. Suppose f ∈ L1 (Rp) andfor every ψ ∈ C∞

c (Rp) ,∫

f ψdmn = 0. Show that then it follows that f (x) = 0 fora.e.x. That is, there is a set of measure zero such that off this set f equals 0. Hint:What you can do is to let E be a measurable which is bounded and let Kk ⊆ E ⊆ Vkwhere mn (Vk \Kk)< 2−k. Here Kk is compact and Vk is open. By an earlier exercise,Problem 12 on Page 154, there exists a function φ k which is continuous, has values in[0,1] equals 1 on Kk and spt(φ k)⊆V. To get this last part, show there exists Wk opensuch that Wk ⊆Vk and Wk contains Kk. Then you use the problem to get spt(φ k)⊆Wk.Now you form ηk = φ k ∗ψ l where {ψ l} is a mollifier. Show that for l large enough,ηk has values in [0,1] ,spt(ηk)⊆Vk and ηk ∈C∞

c (Vk). Now explain why ηk→XEoff a set of measure zero. Then∣∣∣∣∫ f XEdmn

∣∣∣∣ =

∣∣∣∣∫ f (XE −ηk)dmn

∣∣∣∣+ ∣∣∣∣∫ f ηkdmn

∣∣∣∣=

∣∣∣∣∫ f (XE −ηk)dmn

∣∣∣∣