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) = {F ∈ H(D): kF kHp(D) < ∞} where
kF kHp(D) = sup kFrkp. 0 X 2πinθ 2 f(θ) = lim ane ∈ H ([−/1, 2, 12]) N→∞ |n|≤N 2 is the unique L -limit of Fr as r → 1− with kF kH2 = kfk2. In the reverse direction, if f ∈ 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 ∈ h (D). Recall that there exists (an) ∈ ` (Z) such that for z = re X |n| 2πinθ U(z) = anr e , n∈Z and recall that the harmonic conjugate of U has the series representation X |n| 2πinθ V (z) = −i sgn(n)anr e . n∈Z This means of course that ∞ 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)n∈Z) = (−isgn(n)an)n∈Z. 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 ∞ 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))n∈Z defined by |n| an = −i sgn(n)r 2 2 is an element of ` (Z), hence there exists Qr ∈ L ([−1/2, 1/2]) such that |n| Qbr(n) = −i sgn(n)r for all n ∈ Z. We will therefore first compute Qr and then investigate what happens if r → 1−. We have ∞ X re2πiθ rne2πinθ = 1 − re2πiθ n=1 and −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)r|n|e2πinθ = . 1 − 2r cos(2πθ) + r2 n∈Z For fixed r < 1 the series on the left is uniformly and absolutely con- vergent for all θ ∈ [−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 ∈ 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 ∈ R. 3 2 Proof. Vr ∈ L ([−1/2, 1/2]) with |n| 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 ∈/ h (D). Indeed, Z 1/2 2 1 + r |Qr(t)|dt = 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 ∈ L2([−1/2, 1/2]), then there 2 exists U ∈ h (D) such that kUr − tk2 → 0 as r → 1−. The harmonic 2 conjugate V of U has the property that V ∈ h (D) with kV kh2(D) ≤ kUkh2(D) 2 (by Parseval), hence there exists v ∈ L ([−1/2, 1/2]) with kVr − vk2 → 0 as r → 1−. Moreover, X v(θ) = −i lim sgn(n)u(n)e2πinθ N→∞ b |n|≤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 |z| = 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 ε > 0. Consider the contour Γε 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 rε = 2 sin(ε/2). (Sketch this.) For ε ≤ |θ − t| ≤ π the contour is parametrized by ζ = eit. On the small circle segment it is parametrized by iθ i(t+θ) ζ = e + rεe , where |t| ≤ 2−1(π + ε). (Calculate angle sums in the triangle with inner angle ε and side lengths 1, 1, rε.) Cauchy’s theorem gives us Z iθ 1 F (ζ) F (e ) = iθ dζ. 2πi Γε ζ − e Call the large circle segment Γ1,ε and the small circle segment Γ2,ε. 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,ε ζ − e 2πi ε≤|t−θ|≤π e − e iθ i(t+θ) On the small circle segment we parametrize by ζ = e + rεe with −1 iθ i(t+θ) i(t+θ) |t| ≤ 2 (π + ε), so that ζ − e = rεe . Since dζ = rεie dt, we get Z 1 iθ i(t+θ) 1 iθ F (e + rεe )idt → F (e ) 2πi |t|≤2−1(π+ε) 2 as ε → 0. Multiplying by 2, letting ε → 0 in the original formula and subtracting F (eiθ) from both sides we obtain Z F (eit) F (eiθ) = lim dt. i(θ−t) ε→0 ε≤|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 Γε 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) ε→0 π ε≤|θ−t|≤π 1 − e Add both representations for F (eiθ) to obtain i Z F (eit) F (eiθ) = lim dt. ε→0 π θ−t ε≤|θ−t|≤π 2 tan( 2 ) Take imaginary parts on both sides. We obtain 1 Z Z u(t) v(θ) = lim dt. ε→0 ε≤|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. ε→0 ε≤|t|≤1/2 tan πt This is also called the Hilbert transform of f. Note that the limit as ε → 0 cannot combined with the integral, since the denominator has a zero if t = θ. In the situation described above this Hilbert transform gives the following result: When calculating U = Pr ∗ u, the conjugate V of U, and then the boundary function v of V , then v = f.e 6 It is therefore of immediate interest for us to be able to decide if this identity is always true. This requires first that we find out what the mapping properties of this transform actually are. The eventual goal is the following statement: if u ∈ L2([−1/2, 1/2, ]), 2 2 then ue ∈ L ([−1/2, 1/2]), and v = ue in L ([−1/2, 1/2]). The structure for this proof is as follows: We know that u ∈ L2 implies 2 v ∈ L . We will also show that v = ue a.e., and we will finally apply the statement that if two functions are pointwise a.e. equal and one is L2, then so is the other, i.e., the two functions are equal in L2. 4 Pointwise radial limits of the conjugate function We note that since tan t is an odd function, we can write the Hilbert trans- form as Z 1/2 f(θ − t) − f(θ + t) fe(θ) = lim dt. ε→0 ε tan πt This is frequently preferrable, since the difference in the numerator may go to zero as t → 0 fast enough so that the zero in the denominator is being canceled. For much of the following discussion we will consider f ∈ L1([−1/2, 1/2]). Since L2([−1/2, 1/2]) ⊆ L1([−1/2, 1/2]), this includes the case that we are mostly interested in. Theorem 3. Assume f ∈ L1([−1/2, 1/2]). Let Vr(θ) = Qr ∗ f(θ). iθ Fix e ∈ T, and assume that Z 1/2 |f(θ − t) − f(θ + t)| dt < ∞. 0 tan(πt) Then limr→1− Vr and fe(θ) exist, and lim Vr(θ) = fe(θ). r→1− Proof. We note first that the assumptions imply that fe(θ) exists. Since Qr is also odd, we have Z 1/2 1 Vr(θ) − f(θ) = (Qr(t) − (f(θ − t) − f(θ + t))dt. 0 tan πt 7 We have with sin 2πt/(1−cos(2πt)) = 2 sin πt cos πt/(2 sin2 πt) = 1/ tan πt and a certain amount of algebraic manipulation that 1 (1 − r)2 1 Qr(t) − = . tan πt (1 − r)2 + 4r sin2 πt tan πt This gives Z 1/2 (1 − r)2 |f(θ − t) − f(θ + t)| V (θ) − f(θ) = dt. r e 2 2 0 (1 − r) + 4r sin πt tan(πt) The first factor is ≤ 1 for all r and converges to zero as r → 1−, so Lebesgue dominated convergence implies that the difference converges to zero. It follows that lim Vr(θ) exists, and that it equals fe(θ). In particular, if t 7→ f(θ − t) − f(θ + t) is differentiable at t = 0, then |f(θ − t) − f(θ + t)| |t| remains bounded near t = 0, i.e., the integrability assumption of the theorem is satisfied. 1 Lemma 1. Let u ∈ L ([−1/2, 1/2]), and let Vr = Qr ∗ u. Then for a.e. θ ! Z u(θ − t) lim V (re2πiθ) − dt = 0 r→1− 1−r<|t|<1/2 tan(πt) Proof. From the Lebesgue differentiation theorem (Folland, p. 98, Theorem 3.21), the assumption u ∈ L1 implies for almost every θ 1 Z δ lim |u(θ − t) − u(θ + t)|dt = 0. (1) δ→0 δ 0 We have Z 1/2 Vr(θ) = u(θ − t)Qr(t)dt −1/2 Z 1−r Z 1/2! = + Qr(t)(u(θ − t) − u(θ + t))dt. 0 1−r We obtain for the quantity that we want to estimate that Z 1/2 u(θ − t) − u(θ + t) Vr(θ) − dt =: I1 + I2, (2) 1−r tan πt 8 where Z 1−r I1 = Qr(t)(u(θ − t) − u(θ + t)dt 0 Z 1/2 1 I2 = Qr(t) − (u(θ − t) − u(θ + t))dt. 1−r tan πt (To see that this identity holds, add the integral on the left of (2) to I1 + I2, and verify that it simplifies to the integral representation of Vr.) The reason why this is a useful decomposition lies in the fact that 1 − r converges to zero, so the integration path in I1 converges to zero, hence Lebesgue differentiation may become applicable. For I2 we need to use the fact that Qr −1/ tan πt simplifies to an expres- sion with upper bound of the form constant times (1 − r)2/t3 (this will take a certain amount of algebra to establish), and then argue that the integral R 1/2 −2 1−r |u(θ − t) − u(θt)|dt is o((1 − r) ). We consider first I1. We have for 0 ≤ t ≤ 1 − r that there exists C > 0 so that 2r sin 2πt t 1 Q (t) = ≤ C ≤ C . r 1 − 2r cos 2πt + r2 (1 − r)2 1 − r Hence 1 Z 1−r |I1| ≤ |u(θ − t) − u(θ + t)|dt, 1 − r 0 and this goes to zero as r → 1 since θ is in the Lebesgue set of u. The estimate for I2 starts out as in the previous theorem by simplifying the first difference and estimating it above. We recall that 1 (1 − r)2 sin 2πt Qr(t) − = tan πt (1 − r)2 + 4r sin2 πt 2(1 − cos 2πt) The modulus of the right hand side has upper bound (1 − r)2 cos πt (1 − r)2 (1 − r)2 ≤ ≤ C , 2 sin(πt)((1 − r)2 + 4r sin2 πt) 8r sin3(πt) rt3 Hence (1 − r)2 Z 1/2 |u(θ − t) − u(θ + t)| |I2| . 3 dt r 1−r t where . indicates the presence of some absolute positive constant. The ’obvious’ estimate (1 − r)2 ≤ t2 for t ≥ 1 − r is too weak to get the condition 9 that θ is in the Lebesgue set into the game, so we have to proceed more carefully. We proceed with an integration by parts. Define 1 Z x |u(θ − t) − u(θ + t)|dt if x 6= 0 F (x) = x 0 0 else. Since θ is in the Lebesgue set of u and u is periodic, we have that F is bounded on [0, ∞) and continuous at zero. With the previous estimate we have Z 1/2 0 2 (tF (t)) |I2| . (1 − r) 3 dt 1−r t F (t) 1/2 Z 1/2 F (t) = (1 − r)2 − 3(1 − r)2 dt. 2 3 t 1−r 1−r t The first term converges to zero as r → 1. In the integral we substitute t = (1 − r)τ. We obtain that the second term is bounded in modulus by Z ∞ F ((1 − r)τ) 3 3 dτ. 1 τ Since F is bounded, the integrand is bounded independently from r by an integrable function. Hence, Lebesgue dominated convergence may be used to move the limit r → 1 under the integral sign. The numerator converges pointwise to the function constant F (0), and this is zero as shown before. This is almost what we want, since it relates the limit of Vr and the R limit |t|≥ε with ε = 1 − r. The problem is the following: the lemma does not tell us if either of the limit actually exists, since we only established that the difference of the two limits goes to zero. In order to complete the investigation we need to show either that ue(θ) exists a.e., or that v(θ) exists a.e. The argumentation proceeds by showing that the second statement is true, i.e., it can be established that the following implication holds. 1 If u ∈ L ([−1/2, 1/2]), then lim Vr(θ) exists and is finite for a.e. θ. Hence, in the following we consider properties for u that imply pointwise radial convergence for Ur(θ) and Vr(θ). The first lemma is an extension of the solution of the Dirichlet problem. Recall that the Dirichlet problem has a continuous function on the unit circle as its input. The next lemma shows that if convergence everywhere is replaced by almost everywhere con- vergence, then the weaker assumption u ∈ L1 is sufficient. 10 1 2πiθ Lemma 2. Let u ∈ L ([−1/2, 1/2]) and define U(re ) = u ∗ Pr(θ). Then lim U(re2πiθ) = u(e2πiθ) r→1− for almost all θ. Proof. Assume u ∈ L1. We let θ be in the Lebesgue set of u. (As discussed before, the points θ not in the Lebesgue set have measure zero.) We need this in a slightly different form compared to the previous application; we note that 1 Z δ lim u(θ + t)dt = u(θ). (3) δ→0 2δ −δ Without loss of generality, we assume that θ = 0. (For general θ we apply the result to u1 defined by u1(t) = u(t + θ).) Set Z x ω(x) = u(t)dt. −1/2 We need to show that limr→1 U(r) exists and equals u(0). Integrate by parts and use that Pr(t) = Pr(−t) to obtain Z 1/2 U(r) = Pr(t)u(t)dt −1/2 1/2 Z 1/2 ∂ = ω(t)Pr(t) − [Pr(t)]ω(t)dt −1/2 −1/2 ∂t 1 − r Z 1/2 (1 − r2) sin 2πt = (ω(1/2) − ω(0)) + 4π 2 2 ω(t)dt 1 + r −1/2 (1 − 2r cos 2πt + r ) We have ω(0) = 0, and as r → 1, the first term converges to zero. Does the integral converge to u(0) ? We rewrite (3) in the form ω(t) − ω(−t) lim = u(0). t→0 2t If we call the remaining part of the integrand Fr(t), then we see that Fr(t) = −Fr(−t). Hence, adding the integral to itself, substituting t by −t in one of the intgrals, dividing by 2, and expanding by 2t/(2t) shows that the integral is equal to Z 1/2 ω(t) − ω(−2t) 2 2πtFr(t) dt. −1/2 2t 11 Happy coincidence: Tr defined by Tr(t) = 2π(1 + r)tFr(t) satifies Z 1/2 Tr(t)dt = 1 −1/2 and lim Tr(t) = 0 r→1 where the limit is uniform for |t| ≥ δ if δ > 0 is fixed, i.e., 2πt(1 + r)Fr(t) is an approximate identity. As in the previous proof, consider ω(t) − ω(−t) Ψ(t) = 2t for t 6= 0 and Ψ(0) = u(0). This is continuous at the origin. We have 2 Z 1/2 Z 1/2 lim (1+r)2πtFr(t)Ψ(t)dt−Ψ(0) = lim Tr(t)(Ψ(t)−Ψ(0))dt r→1 1 + r −1/2 r→1 −1/2 and with a proof analogous to the proof of the Dirichlet problem (split the integral into integration |t| ≤ δ and integration over |t| ≥ δ) it follows that this limit is zero, as required. We require this for Vr, and not for Ur. This is not possible to do directly, since the conjugate Poisson kernel does not have the same nice integral prop- erties as the Poisson kernel. This is even more true for partial derivatives. The way around this consists in considering analytic functions instead. We know now that Ur converges to u pointwise, so if we can prove that the analytic function F defined by F = U + iV has the property that lim Fr(θ) r→1− exists and is finite for a.e. θ, then, since Vr = −i(Fr − Ur), we will be able to conclude that lim Vr(θ) r→1− exists and is finite for a.e. θ. The crucial ingredient in the following lemma is the fact that it makes no assumptions about behavior on the boundary. 12 Lemma 3. Let F be analytic in D and assume that 2πiθ G(re ) = Pr ∗ g(θ). 2 We proved this for h (D), and we looked at a sketch of a proof using the Riesz representation theorem. We indicated this for other hp-spaces as well, and the case p = ∞ is needed here. (There is a subtle point hidden ∞ above; we only need that G ∈ h (D). Why would the assumption that F is harmonic not be strong enough?) Evidently we also have g ∈ L1([−1/2, 1/2]). From Lemma 2 we obtain that lim Gr(θ) = g(θ) r→1− for almost all θ. Using the inverse M¨obiustransform we have for r < 1 1 − G(re2πiθ) F (re2πiθ) = . 1 + G(re2πiθ) Almost done; we want to conclude that F has a limit a.e., but we have to deal with those θ for which lim G(re2πiθ) = −1. r→1− 13 It follows from the next theorem that the set of θ with this property has measure zero, which finishes the proof. ∞ 1 Note that H (D) ⊆ H (D). We won’t give the proof of the following theorem in full detail. Intuitively, the boundary function of an element in 1 H (D) still behaves like an analytic function in the sense that if the boundary function is zero on a set with positive measure, then it already has to equal zero. 1 Theorem 4 (Uniqueness theorem). Let W ∈ H (D) with boundary values w ∈ H1([−1/2, 1/2]). Assume that the set {θ ∈ [−1/2, 1/2] : w(e2πiθ) = 0} has positive Lebesgue measure. Then W is identically zero on D. This is a deep theorem that describes a fundamental difference between analytic functions and harmonic functions in the unit disk. To see the difference, consider P defined by 2πiθ P (re ) = Pr(θ). 1 We know that P ∈ h (D), and lim P (re2πiθ) = 0 r→1− for all θ 6= 0. However, with z = re2πiθ we have 1 + z P (z) = < 1 − z which is evidently not the zero function in D. Before we discuss this theorem 2 and its proof, we finish the proof of v = ue for u ∈ L ([−1/2, 1/2]). Corollary 1. If F is analytic in D and is a (complex) linear combination of analytic functions with non-negative real part, then the conclusion of the lemma remains true. 1 2πiθ Theorem 5. Let u ∈ L ([−1/2, 1/2]) and let V (re ) = u ∗ Qr(θ). Then lim V (re2πiθ) r→1− exists and is finite for almost all θ. 14 1 Proof. If u ∈ L ([−1/2, 1/2]), then u = u+ − u− where u+, u− are nonneg- ative functions in L1. Since convolution with the Poisson kernel is a linear 2πiθ operation, the function U(re ) = Pr ∗ u(θ) may be written as 2πiθ U(re ) = u+ ∗ Pr(θ) − u− ∗ Pr(θ). The two terms on the right are both nonnegative (even positive) in the unit disk, and it follows that U is a linear combination of nonnegative func- tions. Forming F = U + iV we obtain from the previous corollary that lim Fr(θ) r→1− exists and is finite for a.e. θ. We had shown this property for Ur previously, and we obtain it therefore for Vr as well. 2 2 Theorem 6. Let u ∈ L ([−1/2, 1/2]). Then ue ∈ L ([−1/2, 1/2]). 2 2 Proof. u ∈ L implies that U defined by Ur = Pr ∗ u ∈ h (D). Hence the 2 conjugate function Vr ∈ h (D) (use the Fourier series representation of Vr to 2 see this). Let v be the L -limit of Vr as r → 1−. By the previous theorem we obtain that Vr converges pointwise a.e. to v as well. Lemma 1 implies that the limit defining ue exists for a.e. θ and equals 2 2 v(θ). Since v ∈ L , it follows that ue ∈ L as well. It remains to investigate the uniqueness theorem. 15