Math 752 Fall 2015 1 Hardy Space Denote by H(D) the space of all analytic functions F : D ! C. We define the Hardy classes by p H (D) = fF 2 H(D): kF kHp(D) < 1g where kF kHp(D) = sup kFrkp: 0<r<1 p (This is the same norm as for h (D), but the space on which it is defined is p p different.) Evidently, H (D) ⊂ h (D). We define Hp([−1=2; 1=2]) to be the vector space of f 2 Lp([−1=2; 1=2]) such that fb(n) = 0 for n < 0. This definition makes sense since θ 7! e−2πinθ is in Lq([−1=2; 1=2]) for every 1 ≤ q ≤ 1, so that the integral defining the Fourier coefficient converges absolutely by H¨older'sinequality. 2 For p = 2 we have the following identifications. If F 2 H (D), then 1 1 X n X n 2πinθ F (z) = anz = anr e : n=0 n=0 Evidently 1 X n 2 janr j ≤ kF kH2(D) < 1; n=0 and the dominated convergence theorem shows that 1 X 2 janj < 1: n=0 2 This can be used (in the same way as the statements for h (D) and 2 L (R)) to show that X 2πinθ 2 f(θ) = lim ane 2 H ([−=1; 2; 12]) N!1 jn|≤N 2 is the unique L -limit of Fr as r ! 1− with kF kH2 = kfk2. In the reverse direction, if f 2 H2([−1=2; 1=2]), then F defined by 2πiθ F (re ) = Pr ∗ f(θ) 1 is an analytic function with kF kH2 = kfk2. (This can be seen directly from the Fourier series representation for Fr.) We summarize these statements in the following theorem. Theorem 1. The map that sends f to F as above defines an isometry (i.e., 2 2 a norm preserving bijection) between H ([−1=2; 1=2]) and H (D). 2 The Hilbert transform 2 2 2πiθ Let U 2 h (D). Recall that there exists (an) 2 ` (Z) such that for z = re X jnj 2πinθ U(z) = anr e ; n2Z and recall that the harmonic conjugate of U has the series representation X jnj 2πinθ V (z) = −i sgn(n)anr e : n2Z This means of course that 1 X n 2πinθ U(z) + iV (z) = a0 + 2 anr e n=0 2 is in H (D). The representation of the harmonic conjugate suggests the following definition. 2 2 Definition 1. define the map H : ` (Z) ! ` (Z) defined by H((an)n2Z) = (−isgn(n)an)n2Z: We will usually abuse notation by writing H(an) = −i sgn(n)an: We note that if a0 = 0, then kH(an)k2 = kank2. We note that the composition F −1 ◦ H ◦ F maps L2([−1=2; 1=2]) to L2([−1=2; 1=2]). Problem. Is there a more explicit representation of F −1 ◦ H ◦ F? Intuitively, H should be a convolution operator in the sense that 1 X F −1 ◦ H ◦ F(f)(x) = (−isgn(n)fb(n)e2πinx n=−∞ 2 looks like the Fourier series of a convolution. However −isgn(n) is not in 2 2 ` (Z), i.e., the function that we convolve with cannot be in L (R). However, if r < 1, then the sequence (an(r))n2Z defined by jnj an = −i sgn(n)r 2 2 is an element of ` (Z), hence there exists Qr 2 L ([−1=2; 1=2]) such that jnj Qbr(n) = −i sgn(n)r for all n 2 Z. We will therefore first compute Qr and then investigate what happens if r ! 1−. We have 1 X re2πiθ rne2πinθ = 1 − re2πiθ n=1 and −1 1 X X re−2πiθ r−ne2πinθ = rne−2πinθ = 1 − re−2πiθ n=−∞ n=1 which gives pointwise X 2r sin(2πθ) −i sgn(n)rjnje2πinθ = : 1 − 2r cos(2πθ) + r2 n2Z For fixed r < 1 the series on the left is uniformly and absolutely con- vergent for all θ 2 [−1=2; 1=2], hence it defines a continuous function on [−1=2; 1=2]. Since C[−1=2; 1=2] ⊂ L2([−1=2; 1=2]), we obtain 2r sin(2πθ) Q (θ) = : r 1 − 2r cos(2πθ) + r2 Theorem 2. Let u 2 L2([−1=2; 1=2]) and 2πiθ U(re ) = u ∗ Pr(θ) 2 the corresponding element in h (D). Then the harmonic conjugate V of U with V (0) = 0 has the representation 2πiθ V (re ) = u ∗ Qr(θ) for r < 1 and θ 2 R. 3 2 Proof. Vr 2 L ([−1=2; 1=2]) with jnj Vbr(n) = −i sgn(n)ub(n)r ; hence Vr = Pr ∗ u. Recall that Pr ≥ 0 and Z 1=2 Pr(θ)dθ = 1 −1=2 2πiθ 1 for all r < 1, i.e., P defined by P (re ) = Pr(θ) is in h (D). This implied in particular the solution of the Dirichlet problem. Unfortunately, Qr does not have the same nice property. 1 Remark 1. Qr 2= h (D). Indeed, Z 1=2 2 1 + r jQr(t)jdt = log ; −1=2 π 1 − r and this diverges to infinity as r ! 1−. This means in particular that the boundary representation of the harmonic conjugate has additional technical difficulties. Summary of what we know so far: If u 2 L2([−1=2; 1=2]), then there 2 exists U 2 h (D) such that kUr − tk2 ! 0 as r ! 1−. The harmonic 2 conjugate V of U has the property that V 2 h (D) with kV kh2(D) ≤ kUkh2(D) 2 (by Parseval), hence there exists v 2 L ([−1=2; 1=2]) with kVr − vk2 ! 0 as r ! 1−. Moreover, X v(θ) = −i lim sgn(n)u(n)e2πinθ N!1 b jn|≤N for a.e. θ. Goal. Find a representation for v in terms of the starting function u without having to calculate Fourier coefficients. The first question that we need to consider is how such a representation should look like. We need a certain amount of complex analysis to answer this question. We start with a function F that is analytic on D(0;R), where R > 1. The strategy is as follows. We will use Cauchy's theorem to express F as an integral over a suitably chosen contour. We then will deform the contour to the circle jzj = 1. Finally, we will take real and imaginary parts to obtain a representation of V in terms of u. 4 3 Hilbert transform on the unit circle iθ Let F be analytic on D(0;R) with R > 1 and F (0) = 0. Fix z0 = e on the unit circle. Fix " > 0. Consider the contour Γ" that is the unit circle traced counter clockwise except about z0, where we change it outwards by a circle segment with center z0 and radius r" = 2 sin("=2): (Sketch this.) For " ≤ jθ − tj ≤ π the contour is parametrized by ζ = eit: On the small circle segment it is parametrized by iθ i(t+θ) ζ = e + r"e ; where jtj ≤ 2−1(π + "). (Calculate angle sums in the triangle with inner angle " and side lengths 1; 1; r".) Cauchy's theorem gives us Z iθ 1 F (ζ) F (e ) = iθ dζ: 2πi Γ" ζ − e Call the large circle segment Γ1;" and the small circle segment Γ2;". On the large segment we parametrize by ζ = eit and obtain Z Z it 1 F (ζ) 1 F (e ) it it dζ = it iθ ie dt: 2πi Γ1;" ζ − e 2πi "≤|t−θ|≤π e − e iθ i(t+θ) On the small circle segment we parametrize by ζ = e + r"e with −1 iθ i(t+θ) i(t+θ) jtj ≤ 2 (π + "), so that ζ − e = r"e . Since dζ = r"ie dt, we get Z 1 iθ i(t+θ) 1 iθ F (e + r"e )idt ! F (e ) 2πi jt|≤2−1(π+") 2 as " ! 0. Multiplying by 2, letting " ! 0 in the original formula and subtracting F (eiθ) from both sides we obtain Z F (eit) F (eiθ) = lim dt: i(θ−t) "!0 "≤|t−θ|≤π 1 − e 5 We would like to have a kernel that is purely imaginary on the unit circle, and we can simplify the integrand further. To do so, note that if we 0 use the contour Γ" that is obtained by tracing along the inner part of the small circle around eiθ rather than the outer part, and applying Cauchy's theorem to eiθF (ζ) ζ 7! ζ which is also analytic in D(0;R) due to the assumption that F (0) = 0, we obtain with a similar reasoning that 1 Z ei(θ−t)F (eit) F (eiθ) = lim dt: i(θ−t) "!0 π "≤|θ−t|≤π 1 − e Add both representations for F (eiθ) to obtain i Z F (eit) F (eiθ) = lim dt: "!0 π θ−t "≤|θ−t|≤π 2 tan( 2 ) Take imaginary parts on both sides. We obtain 1 Z <F (eit) =F (eiθ) = lim dt "!0 2π θ−t "≤|θ−t|≤π tan( 2 ) Change of variable by replacing θ by 2πθ and substituting t = 2πτ. Set also F (e2πiθ) = u(θ) + iv(θ). We then get Z u(t) v(θ) = lim dt: "!0 "≤|θ−t|≤1=2 tan π(θ − t) Based on this example we formally define the integral transform f 7! fe by Z f(θ − t) fe(θ) = lim dt: "!0 "≤|t|≤1=2 tan πt This is also called the Hilbert transform of f.