Lecture Notes, Singular Integrals

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Lecture Notes, Singular Integrals Lecture notes, Singular Integrals Joaquim Bruna, UAB May 18, 2016 Chapter 3 The Hilbert transform In this chapter we will study the Hilbert transform. This is a specially important operator for several reasons: • Because of its relationship with summability for Fourier integrals in Lp- norms. • Because it constitutes a link between real and complex analysis • Because it is a model case for the general theory of singular integral op- erators. A main keyword in the theory of singular integrals and in analysis in general is cancellation. We begin with some easy examples of what this means. 3.1 Some objects that exist due to cancellation One main example of something that exists due to cancellation is the Fourier 2 d 2 transform of functions in L (R ). In fact the whole L -theory of the Fourier transform exists thanks to cancellation properties. Let us review for example 1 d 2 d the well-known Parseval's theorem, stating that for f 2 L (R )\L (R ) one has kf^k2 = kfk2, a result that allows to extend the definition of the Fourier 2 d transform to L (R ). Formally, Z Z Z Z jf^(ξ)j2 dξ = f(x)f(y)( e2πiξ·(y−x)dξ)dxdy: Rd Rd Rd Rd This being equal to R jf(x)j2 dx means formally that Rd Z 2πiξ·x e dξ = δ0(x): Rd Along the same lines, let us look at the Fourier inversion theorem, stating 1 d 1 d that whenever f 2 L (R ); f^ 2 L (R ) one has 1 Z f(x) = f^(ξ)e2πiξ·x dξ: Rd Again, the right hand side, by a formal use of Fubini's theorem becomes Z Z Z Z ( f(y)e−2πiξ·y dy)e2πiξ·x dξ = f(y)( e−2πiξ·(y−x) dy) dξ: Rd Rd Rd Rd If this is to be equal to f(x) we arrive to the same formal conclusion, namely that superposition of all frequencies is zero outside zero. An intuitive way to understand this is by noting that (d = 1) Z R sin(2πRx) e2πiξx dξ = ; R πx 1 is zero for x an integer multiple of R , so when R ! +1 the zeros become more and more dense. 1 d The Fourier transform exists trivially if f 2 L (R ), but not so trivially 2 d 2 if f 2 L (R ). It exists, in the L -sense Z −2πixξ f^(ξ) = lim fn(x)e dx; n R 1 d 2 d 2 d with fn arbitrary in L (R ) \ L (R ) such that fn ! f in L (R ). The most trivial choice is to take fR(x) to be equal to f(x) if jxj < R and zero elsewhere, so that Z R f^(ξ) = lim f(x)e−2πixξ dx; R −R the limit being in the L2-sense. It is a general and very deep result due to L. Carleson that this limits exists a.e. To illustrate all this we will consider the Fourier transform of two partic- 1 ular functions, none of which in L (R), and study the pointwise behaviour of the limit above. For instance, R sin x dx is conditionally convergent, meaning that R x Z R sin x lim dx (3.1) R!1 −R x R sin x exists while j x j dx = +1. The later can be proved by estimating from R π below the contribution to the integral at points jx − ( 2 + kπ)j < δ; k 2 Z. Alternatively, noting that the Fourier transform of 1[−a;a] is Z a sin 2πaξ e−2πxξ dx = ; a πξ 2 we can argue that it cannot be integrable, because if it were, by Fourier in- version theorem, 1[−a;a] would be continuous. We note too that by Parseval's theorem Z Z sin 2πaξ 2 2 ( ) dξ = 1[−a;a] = 2a: R πξ R Regarding the value of the integral, we will see that the limit Z R sin 2πaξ 2πiξx lim e dξ = 1[−a;a](x); (3.2) R!+1 −R πξ that as we know exists in the L2 sense, exists too pointwise a.e., in fact for x 6= ±a. Indeed, by using residues we have that 1 Z R eiαξ lim dξ R!1 2πi −R ξ 1 eiαξ 1 1 equals 2 Res( ξ ; 0) = 2 if α > 0 and − 2 if α < 0, and this implies (3.2). For x = 0 we get that Z R sin 2πaξ lim dξ = 1; R!+1 −R πξ from which it follows that the integral in (3.1) equals π. Another example are the Fresnel integrals Z Z sin x2dx; cos x2dx: R R By the change of variable u = x2 one has 2 Z R 1 Z R sin u sin x2dx = p du; 0 2 0 u so the integral is not absolutely convergent. To compute the integral we will see more generally that Z R Z R −απx2 −2πixξ −απx2 1 − π ξ2 lim e e dx = 2 lim e cos 2πxξ dx = p e α : R!+1 −R R!+1 0 α (3.3) for Re α ≥ 0 and all ξ 2 C. For Re α > 0, the above is the Fourier transform −απx2 of fα(x) = e , a function in the Schwarz class. Since it is holomorphic π 2 p1 − α ξ in α and for α; ξ real equals α e , the same holds true by analytic continuation first in α and then in ξ, that is Z −απx2 −2πixξ 1 − π ξ2 F (α; ξ) = e e dx = p e α ; Re α > 0; ξ 2 C: R α 3 By making Re α ! 0 we can already conclude that the Fourier transform, in 2 π 2 −απx p1 − α ξ the sense of distributions, of e is α e also if Re α = 0. However, this does not imply (3.3). To compute the limit in (3.3) when Re α = 0 we will exploit the analyticity in x as follows. We consider the contour iπt 1 in the z-plane given by the segment [0;R], the arc t 7! Re ; 0 ≤ t ≤ 4 , i π and the segment from Re 4 to 0, and apply the Cauchy theorem to the entire function e−απz2 cos(2πzξ). The contribution of the arc is bounded by aRe−bR2+cR with some positive constants a; b; c depending on ξ, and so has limit zero when R ! 1. Hence, parametrizing with z = t 1+p i , we get 2 R R Z 2 1 + i Z 2 1 + i lim e−απx cos(2πxξ) dx = p lim e−απit cos(2π p tξ)dt: R!+1 0 2 R!1 0 2 If Re iα > 0, that is, Im α < 0 this shows that the limits exist and equals 1 + i 1 + i 1 − π ξ2 p F (iα; p ξ) = p e α ; 2 2 α p as claimed. For ξ = 0 and α = − i we get the value pπ for the Fresnel π 2 2 integrals. 3.2 The conjugate Poisson kernel and the Cauchy transform One method to study the summability of the Fourier integral is by the means method. The Abel method consists in using Z f^(ξ)e−2πtjξje2πix·ξ dξ: Rd The behaviour as t ! 0 is easily understood. Indeed, by Fubini's theorem the above equals Z Z −2πtjξj 2πiξ(x−y) f(y)( e e dξ)dy = (f ∗ Pt)(x); Rd Rd with Z −2πtjξj 2πix·ξ Pt(x) = e e dξ: Rd Let us compute explicitly Pt in d = 1: Z +1 Z 0 2πξ(ix−t) 2πξ(ix+t) Pt(x) = e dξ + e dξ = A + B; 0 −∞ with 1 1 A = ;B = A = : 2π(t − ix) 2π(t + ix) 4 This gives 1 t P (x) = : t π x2 + t2 1 −2πjξj Note that Pt(x) = t P (x=t), with P = P1, and that P1; e are a Fourier pair, P is positive with integral one. In dimension d > 1 the computation is −d harder, the result being Pt(x) = t P (x=t), with d+1 1 Γ( 2 ) P (x) = cd d+1 ; cd = d+1 ; (jxj2 + 1) 2 π 2 It is convenient to look as (f ∗ Pt)(x) as a function of (x; t) in the upper half space t > 0. Noticing that P is in all Lp-spaces we define p d Definition 1. For f 2 L (R ); 1 ≤ p ≤ +1, we define the Poisson trans- form u = P [f] as the function defined in the upper half-space t > 0 by Z Z 1 t ^ −2πtjξj 2πix·ξ P [f](x; t) = (f∗Pt)(x) = f(y) 2 2 dy = f(ξ)e e dξ π Rd (x − y) + t Rd Since we will need it again in a more general form we state again the main −d properties of convolutions with functions of the form ht(x) = t h(x=t), with 1 d h 2 L (R ). 1 d R −d Proposition 1. Let h 2 L (R ); m = h, and set ht(x) = t h(x=t).. For p d f 2 L (R ); 1 ≤ p ≤ +1, consider f ∗ ht. Then p d • kf ∗ htkp ≤ kfkpkhk1, and f ∗ ht ! mf as t ! 0 in L (R ); 1 ≤ p < +1. • If f is bounded and continuous at x0 then f ∗ ht(x0) ! mf(x0). • If f is bounded and uniformly continuous then f ∗ht ! mf uniformly.
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