STRUCTURE OF WEIGHTED HARDY SPACES IN THE PLANE NI˙HAT GO¨KHAN GO¨G˘U¨S¸1 ∗ 5 1 Abstract. We characterize certain weighted Hardy spaces on the unit disk 0 and completely describe their dual spaces. 2 g u A 1. Introduction and preliminaries 8 By a recent paper of Poletsky and Stessin [5] to each subharmonic function on 2 a bounded regular domain G which is continuous near the boundary corresponds V] a space Hp of analytic functions in G with a certain growth condition. These u C are namely Poletsky-Stessin Hardy spaces. They include and generalize the well- . known classical Hardy spaces. This new theory unifies the standpoints of various h analytic function spaces into one. t a The first generalizations in this direction of the theory of Hardy spaces on m hyperconvex domains in Cn was suggested and studied in [1]. More recently the [ theory is extended to hyperconvex domains in [5]. Boundedness and compactness 2 of the composition operators on these new Poletsky-Stessin Hardy and Bergman v 8 type spaces were investigated there. After this motivating work more investiga- 2 tion [2], [11], [12] revealed the structure and first examples of these Hardy type 3 4 spaces in the plane. . In [2] to understand the scale of weighted Hardy spaces u → Hp Alan and 1 u 0 the author completely characterized Hp spaces in the plane domains by their u 4 boundary values or by possessing a harmonic majorant with a certain growth 1 : (see also [11], [12]). Basically the version of the Beurling’s theorem proved in [2] v states that to each subharmonic exhaustion G corresponds an outer function ϕ i X which belongs to the class Hp so that Hp isometrically equals to M for p > 0, u u ϕ,p r a where M is the space ϕ2/pHp endowed with the norm ϕ,p kfk := kf/ϕ2/pk , f ∈ M . Mϕ,p p ϕ,p This result is especially useful to construct examples of analytic function spaces enjoyingcertaindesiredproperties. ThespaceM , whenkϕk ≤ 1,wasstudied ϕ,2 ∞ as a tool to understand certain sub-Hardy Hilbert spaces in the unit disk in [10]. Two problems were not answered in [2]: (1) Can we go back? That is, given analytic ϕ can one find a subharmonic exhaustion u so that Hp = M ? u ϕ,p Date: Received: xxxxxx; Revised: yyyyyy; Accepted: zzzzzz. ∗ Corresponding author. 2010 Mathematics Subject Classification. Primary 30H10, 30J99, Secondary 46E22. Key words and phrases. Weighted Hardy spaces, subharmonic exhaustion, dual space. 1 2 NI˙HATGO¨KHANGO¨G˘U¨S¸ (2) For the space Hp, consider the class of all representatives, i.e., subhar- u monic exhaustions v so that Hp = Hp. What kind of ”good” representa- u v tives are there? In this note we give answers for both questions. We show under certain growth conditions on the analytic function ϕ on the disk that it is possible to construct a subharmonic exhaustion u on the disk so that Hp equals to M . Moreover, u ϕ,p by the construction, u is real analytic and satisfies the bi-Laplacian in the unit disk. This is a new information related to the second question. In addition, using the boundary value characterization from [2] we completely characterize the dual space of Hp and discuss the corresponding extremal and u dual extremal problems. After several months of submission of this paper there appeared a preprint [7]. Theorem 3.3 in this paper is similar to Theorem 2.1 below. In Theorem 2.1 we do not require any integrability condition on the subharmonic exhaustion u, however the authors in [7] require u to be intagrable. Let us start to recall basic definitions. A function u ≤ 0 on a bounded open set G ⊂ C is called an exhaustion on G if the set B := {z ∈ G : u(z) < c} c,u is relatively compact in G for any c < 0. When u is an exhaustion and c < 0, we set u := max{u,c}, S := {z ∈ G : u(z) = c}. c c,u Let u ∈ sh(G) be an exhaustion function which is continuous with values in R∪{−∞}. Following Demailly [3] we define µ := ∆u −χ ∆u, c,u c G\Bc,u where χ is the characteristic function of a set ω ⊂ G. We denote the class ω of negative subharmonic exhaustion functions on G by E(G). The class of all functions u ∈ E(G) for which ∆u < ∞ is denoted by E (G). 0 If u ∈ E(G), then the Demailly-Lelong-Jensen formula ([3]) takes the form R vdµ = (v∆u−u∆v)+c ∆v, (1) c,u ZSc,u ZBc,u ZBc,u where µ is the Demailly measure which is supported in the level sets S of u c,u c,u and v ∈ sh(G). Let us recall that by [3] if ∆u < ∞, then the measures µ G c,u converge as c → 0 weak-∗ in C∗(G) to a measure µ supported in the boundary R u ∂G. Following [5] we set sh (G) := sh := v ∈ sh(G) : v ≥ 0, sup vdµ < ∞ , u u c,u ( c<0 ZSc,u ) and Hp(G) := Hp := {f ∈ hol(G) : |f|p ∈ sh } u u u WEIGHTED HARDY SPACES 3 for every p > 0. We write kvk := sup vdµ = (v∆u−u∆v) (2) u c,u c<0 ZSc,u ZG for a nonnegative function v ∈ sh(G) and set 1/p kfk := sup |f|pdµ (3) u,p c,u c<0 ZSc,u ! for a holomorphic function f on G. We will use kfk = kfk when p = 1. u u,p By Theorem 4.1 of [5], Hp is a Banach space when p ≥ 1. It is clear that the u function f ≡ 1 belongs to Hp if and only if the Demailly measure µ has finite u u mass. If G is a regular bounded domain in C and w ∈ G, then the Green function v(z) = g (z,w) is a subharmonic exhaustion function for G. For example, when G G is the unit disk and v(z) = log|z|, then µ is the normalized arclength measure v on the unit circle. We denote by P (z,w) the Poisson kernel for the domain G. G The following Theorems are recollections from [2]. Theorem 1.1. [2, Theorem 2.3] Let G be a bounded domain, v ≥ 0 be a function on G, p > 0, and u ∈ E(G). The following statements are equivalent: i. v ∈ sh (G). u ii. The least harmonic majorant h = P (v) of ϕ in G belongs to the class G sh . u Furthermore, kvk = h∆u = khk . u u ZG We will denote by Hp(G) the space of analytic functions f in G for which |f|p has a harmonic majorant in G (see for example [4]). We always have Hp ⊂ Hp u by [5]. We will denote by ν the usual arclength measure on ∂G normalized so that ν(∂G) = 1. Theorem 1.2. [2, Theorem 2.10] Let G be a Jordan domain with rectifiable boundary or a bounded domain with C2 boundary, p > 1, and u ∈ E(G). The following statements are equivalent: i. f ∈ Hp(G). u ii. f ∈ Hp(G) and |f∗| ∈ Lp(V ν), where u V (ζ) := P (z,ζ)∆u(z), ζ ∈ ∂G. (4) u G ZG iii. f ∈ Hp(G) and there exists a positive measure µ on ∂G such that |f∗| ∈ u Lp(µ ). Moreover, if E is any Borel subset of ∂G with measure ν(E) = 0, u then µ (E) = 0 and we have the equality u f f γdµ = P (γ)∆u (5) f u G Z∂G ZG for every γ ∈ L1(ν). f In addition, if f ∈ Hup(G), then kfku,p = kf∗kLp(µfu) and dµu = Vudν. f 4 NI˙HATGO¨KHANGO¨G˘U¨S¸ Remark 1.3. i. Theorem 1.2 is valid when p > 0 and G is the unit disk or more generally a Jordan domain with rectifiable boundary. In this case the Poisson integral of an Lp(dν) function u has non-tangential limits equal to u on ν-almost every boundary point. This is indeed what is needed in the proof of Theorem 1.2. ii. By replacing the function u by a suitable positive multiple tu, t > 0, we may assume that V ≥ 1 on ∂G. To do this it is enough to take a compact set u K ⊂ G so that ∆u(K) > r > 0. Let m := min min P (z,ζ). Then take ζ∈∂G z∈K G t := 1/(rm). We will use the assumption that V ≥ 1 when convenient. u iii. The weight function V is lower semicontinuous. To see this, suppose u ζ ∈ ∂G, ζ → ζ. By Fatou’s lemma j j liminfV (ζ ) = liminf P (z,ζ )∆u(z) ≥ P (z,ζ)∆u(z) = V (ζ). u j G j G u j j ZG ZG Note that V is the balayage of the measure ∆u on ∂G. u iv. Suppose G is a bounded domain with C2 boundary or a Jordan domain with rectifiable boundary and u ∈ E (G). Then 0 u(z) = g (z,w)∆u(w), z ∈ G. G ZG Since ∂g (ζ,w) G = P (w,ζ) G ∂n when ζ ∈ ∂G and w ∈ G, ∂u(ζ) exists for every ζ ∈ ∂G, where ∂ denotes the ∂n ∂n normal derivative in the outward direction on ∂G and ∂u(ζ) = V (ζ) = P (w,ζ)∆u(w), ζ ∈ ∂G. u G ∂n ZG By property (5) in Theorem 1.1 ∂u V (ζ)dν(ζ) = ∆u = (ζ)dν(ζ). u ∂n Z∂G ZG Z∂G To obtain Fatou’s type results we would like to compute the Radon-Nikodym derivative of the Demailly measures with respect to the usual arclength measure on the level sets. In the next result we provide this. Let ν denote the arclength c measure on S . Define c,u V (ζ) := P (z,ζ)∆u(z), ζ ∈ S , c,u Bc,u c,u ZBc,u where P (z,ζ) denotes the Poisson kernel for B . Bc,u c,u Proposition 1.4. Let u ∈ E(G), where G is a bounded regular domain. Suppose thatu is Lipschitz ineverycompactsubsetof G. Thenthe measures ν andµ are c c,u mutually absolutely continuous and µ = V ν with V ∈ L1(ν ). Moreover, c,u c,u c c,u c for each c < 0 there is a constant k > 0 so that V ≥ k on S . c c,u c c,u WEIGHTED HARDY SPACES 5 Proof. Let ϕ be a continuous function on S and let h(z) be the harmonic c,u function in B with boundary values equal to ϕ. By equality (1) we have c,u ϕ(ζ)dµ (ζ) = h(z)∆u(z) c,u ZSc,u ZBc,u = P (z,ζ)∆u(z) ϕ(ζ)dν (ζ) Bc,u c ZSc,u ZBc,u ! = ϕ(ζ)V (ζ)dν (ζ). c,u c ZSc,u Hence µ = V ν . Another observation using Fubini’s theorem gives c,u c,u c V (ζ)dν (ζ) = ∆u(z) = kµ k < ∞. c,u c c,u ZSc,u ZBc,u Thus V ∈ L1(ν ). Note that ν ≤ k′µ for some positive constant k′ by [3]. c,u c c c c,u c Hence V ≥ k on S for some k > 0. This completes the proof. (cid:3) c,u c c,u c Remark 1.5. The requirement that u is Lipschitz is only needed to write the harmonicmeasureonB oftheformP dν . Therearemuchweaker conditions c,u Bc,u c on domains for which the harmonic measure is absolutely continuous. The next auxiliary result allows one to compare the Demailly measures on S c,u with a measure on an arbitrary level set. Proposition 1.6. Let u be a subharmonic exhaustion function on a bounded regular domain G in C. Let G be relatively compact regular open sets in G so j that G ⊂ G and ∪G = G. Then for each j there is a u ∈ E(G ) and for each j j+1 j j j c < 0 there is a number s with c < s < 0 so that for any nonnegative function v ∈ sh(G), the integrals µ (v) are increasing and uj µ (v) ≤ µ (v) = kvk ≤ µ (v). c,u uj uj s,u This means kvk = lim µ (v) for every nonnegative subharmonic function v on u j uj G. Proof. Set u := u − P u. Clearly u ∈ E(G ). Take an integer j ≥ 1 and a j Gj j j 0 number s < 0 with c < s so that B ⊂ G ⊂ B . The comparison follows c,u j0 s,u from (1) and (2) if we note that c ≤ P u on B and P u ≤ s on B . (cid:3) Gj c,u Gj s,u If ϕ is a nonzero analytic function on D, let M denote the space ϕ2/pHp ϕ,p endowed with the norm kfk := kf/ϕ2/pk , f ∈ M . Mϕ,p p ϕ,p We will call a function ϕ ∈ H2 a u-inner function if |ϕ∗(ζ)|2V (ζ) equals 1 for u u almost every ζ ∈ ∂D. If, moreover, ϕ(z) is zero-free, we will say that ϕ is a singular u-inner function. The next result is Theorem 3.2 and Corollary 3.3 from [2]. 6 NI˙HATGO¨KHANGO¨G˘U¨S¸ Theorem 1.7. Let Y 6= {0} be a closed M -invariant subspace of H2(D). Then z u there exists a function ϕ ∈ H2 so that |ϕ∗(ζ)|2V (ζ) = 1 for almost every ζ ∈ ∂D u u and Y = M . In particular, there exists a u-innerand an outer function ϕ ∈ H2 ϕ,2 u so that H2 = M and these spaces are isometric. u ϕ,2 This function ϕ is determined uniquely up to a unit constant. Note that 1 1 1 V (eiθ) = = sgn , (6) u |ϕ(eiθ)|2 ϕ2(eiθ) ϕ2(eiθ) where we set sgnα := |α|/α for any complex number α 6= 0 and sgn0 := 0. If V ≥ 1 on ∂D, then |ϕ(ζ)| ≤ 1 for almost every ζ. Suppose that ∆u < ∞. Then the function 1 belongs to Hp. Hence ϕ−1 belongs to H2. Then it is an easy u R exercise to show that ϕ is an outer function. Theorem 1.8. The set Lp(V dθ) coincides with ϕ2/pLp(dθ) and the map f 7→ u ϕ−2/pf is an isometric isomorphism from the space Lp(V dθ) onto Lp(dθ). u Theorem 1.9. [2, Theorem 3.4] Suppose 0 < p < ∞, f ∈ Hp(D), f 6≡ 0, and u B is the Blaschke product formed with the zeros of f. Then there are zero-free ϕ ∈ H2 ∩H∞, S ∈ H∞ and F ∈ Hp so that ϕ is outer and singular u-inner, S u is singular inner, F is outer, and f = BSϕ2/pF. (7) Moreover, kfk = kFk and Hp(D) = M . p,u p u ϕ,p Corollary 1.10. The map f 7→ ϕ−2/pf is an isometric isomorphism from the space Hp onto Hp. u The following Lemma will be useful in the next section. Its proof is a simple calculation and we outline it here. Lemma 1.11. Let c be a number with −1 < c < 0. Then there exists a function κ = κ defined on (−∞,0] with the following properties: c i. κ : (−∞,0] → (−∞,0] is non-decreasing, convex and C∞, ii. κ is real-analytic in (c,0], iii. κ(t) ≡ c when t ≤ c, κ(0) = 0, and κ′(0) = 1. Proof. Let a := −ln(−c), b := −1 , and e ln(−c) −a c+e(t−c)b, t > c, κ(t) := ( c, t ≤ c. Then 1 −a κ′(t) = e(t−c)b e(t−c)b+1 and 1 −a κ′′(t) = (1/e−(b+1)(t−c)b+1)e(t−c)b e(t−c)2b+2 for t > c. For t ≤ c, κ′(t) = κ′′(t) = 0. It can be checked that κ′′(t) > 0 for c < t ≤ 0, and κ satisfies all properties in i., ii. and iii. (cid:3) WEIGHTED HARDY SPACES 7 2. Finding subharmonic exhaustion Theorem1.2describes theweight functionV corresponding totheHardyspace u Hp when the Laplacian of u is known. In Theorem 1.9 we obtain a canonical u factorization for functions in Hp and we see that this space is a certain multiple u ϕ2/pHp of Hp. The singular u-inner function ϕ appearing in this factorization is related to the weight V by u 1 V (eiθ) = , a.e. θ. (8) u |ϕ(eiθ)|2 In this section we seek a converse to these results. Let G be a Jordan domain with rectifiable boundary and ψ be a given analytic function in H1(G). The problem is to find a subharmonic exhaustion u on G so that V (ζ) = |ψ(ζ)| when ζ ∈ ∂G. Taking a conformal map of G onto D we can u always suppose that G = D. This is a type of inverse balayage problem. We solve this next. Theorem 2.1. Let ψ be a lower semicontinuous function on ∂D so that ψ ≥ c for some constant c > 0. Then there exists a function u ∈ E so that ψ = V . u Moreover we have the following properties: a. u is the decreasing limit of functions in E ∩C∞(D) converging uniformly 0 to u on D. b. u ∈ E (D) if and only if ψ ∈ L1(dν). 0 c. If ψ is Ck, 0 ≤ k ≤ ∞, on ∂D, then u is Ck on D. If ψ is real-analytic, then there exists a compact K so that u is real-analytic on D\K. Proof. Suppose first that ψ is C2 on ∂D and let ρ(reiθ) := 1(r2 − 1)ψ(eiθ) for 2 reiθ ∈ D. Computing the Laplacian of ρ we get r2 −1d2ψ(eiθ) ∆ρ(reiθ) = 2ψ(eiθ)+ . 2r2 dθ2 By assumption ∆ρ(eiθ) = 2ψ(eiθ) ≥ 2c > 0. Hence there exists a compact B ⊂ D so that ∆ρ(z) > 0 on the open set Ω := D\B. Hence ρ is a non-positive subharmonc function on Ω and ρ|∂D ≡ 0. Since ρ is continuous on D, there exists a constant c < 0 so that the set B is relatively compact in D and S ⊂ Ω. c,ρ c,ρ Let κ = κ be the function proivided in Lemma 1.11. Define u(z) := κ(ρ(z)) c for z ∈ D. Now u = κ(ρ) is subharmonic in Ω, u ≡ c on B and u ≥ c on c,ρ D\B ⊂ Ω. Hence u ∈ E and V = ∂u = κ′(0)∂ρ = ψ on ∂D. c,ρ u ∂r ∂r Now let ψ be lower semicontinuous. There exists ψ , all C∞ on ∂D so that n c ≤ ψ (ζ) ≤ ψ (ζ), and ψ(ζ) = lim ψ (ζ) for every ζ ∈ ∂D. We let ψ ≡ 0. n n+1 n n 0 Replacing ψ by ψ −2−n we may assume that d := ψ −ψ ≥ 2−n−1. As in n n n n+1 n the first part of the proof we let ρ (z) := 1(r2−1)d (eiθ). There exists a compact n 2 n B ⊂ D so that ∆ρ (z) > 0 on the open set Ω := D\B . This time we choose n n n n constants −2−n ≤ c < 0 so that B is relatively compact in D, S ⊂ Ω , n cn,ρn cn,ρn n and B ⊂ B . Let u (z) := κ (ρ (z)) so that as proved in the first cn,ρn cn+1,ρn+1 n cn n part, V = d and u ∈ C∞(D). un n n 8 NI˙HATGO¨KHANGO¨G˘U¨S¸ Let ∞ u(z) := u (z). n n=0 X Since |u | ≤ |c | ≤ 2−n for all n, the sum converges uniformly on D. This shows n n that u ∈ E and properties in a. and c. are satisfied. Using (4) in Theorem 1.2, ∞ ∞ V (ζ) = P(z,ζ)∆u(z) = P(z,ζ)∆u (z) = d (ζ) = ψ(ζ). u n n D D Z n=0Z n=0 X X Due to an equality in Remark 1.3, ∆u(z) = V dν = ψdν. u ZD Z∂D Z∂D Hence u ∈ E if and only if ψ ∈ L1(dν). The proof is completed. (cid:3) 0 Wehave now the following converse to Theorem 1.9 to answer the first question in the introduction. Theorem 2.2. Let ϕ be a zero free analytic function on D so that |ϕ∗| equals ν-almost everywhere to an upper semicontinuous function on ∂D. Then there exists a u ∈ E(D) so that Hp = M and we have isometric isomorphism of two u ϕ,p spaces. Proof. It is enough to prove the theorem when p = 2. Since |ϕ∗| is upper semicontinuous on ∂D, there exists a constant m so that |ϕ∗| ≤ m. Hence the function ψ := 1/|ϕ∗|2 is lower semicontinuous and ψ ≥ 1/m. Let u ∈ E(D) be the exhaustion provided by Theorem 2.3 for the function ψ so that V = 1/|ϕ∗|2. If f ∈ H2, u u we write f = ϕf , where f = f/ϕ. Then 0 0 2π kfk2 = |f(eiθ)|2V (eiθ)dθ = kfk2 = kf k2 < ∞. 2,u u Mϕ,2 0 2 Z0 Thus f ∈ H2 and we have shown that H2 ⊂ M . Conversely, if f ∈ M , 0 u ϕ,2 ϕ,2 then clearly f ∈ H2 from the same equality above. The mapping f 7→ ϕf is u 0 clearly an isomorphism of H2 onto M which is an isometry. (cid:3) u ϕ,2 When the weight function V is smooth enough, there is a connection with the u corresponding subharmonic exhaustions and the bi-Laplacian equation ∆2u = 0. This is explained in the next result. Theorem 2.3. Let ψ ∈ C1(∂D) be a nonnegative function. Then there exists a function u and a constant M with the following properties: a. u ∈ E (D) and u is real analytic on D. 0 b. V (ζ) = ∂u(ζ) = ψ(ζ)+M for every ζ ∈ ∂D. u ∂n c. u satisfies the bi-Laplacian equation ∆2u = 0 on D. Proof. Let u(z) := 1(|z|2−1)[Pψ(z)+M], where Pψ(z) istheharmonicextension 2 of ψ on D. Then using polar coordinates ∆u(z) = 2[Pψ(z) + M] + 2|z|∂Pψ(z). ∂r Note that Pψ ∈ C1(D). Now take M large enough so that ∆u(z) ≥ 0 on D. WEIGHTED HARDY SPACES 9 Again taking the Laplacian it can be checked that ∆2u(z) = 0. Hence ∆u is harmonic on D and since ∆u = c(Pψ(0)+M) ≤ ckψk +cM < ∞, ∞ D Z u ∈ E . Clearly u is real analytic on D. On ∂D we have 0 ∂u V (ζ) = (ζ) = ψ(ζ)+M u ∂r for every ζ ∈ ∂D. (cid:3) Remark 2.4. Equality in (5) shows also that 2π ∂Pψ(eiθ) 2 log|z −eiθ| ψ(eiθ)+ +M dθ = log|1−wz|∆u(w) ∂r Z0 (cid:20) (cid:21) ZD for every z ∈ D. Therefore, in fact, u can be written as the difference of two potentials 1 1 2π ∂Pψ(eiθ) u(z) = log|z−w|∆u(w)dw− log|z−eiθ| ψ(eiθ)+ +M dθ 2π π ∂r ZD Z0 " # e for every z ∈ D. Here ∆u is harmonic. When v ∈ E(D), let R(v) denote the class of all functions u ∈ E(D) which generates the same space Hp = Hp. We know a ”good” representative in R(v) v u for certain cases as a consequence of Theorem 2.3. Theorem 2.5. Let v ∈ E (D) so that V is bounded and PV + |z|∂PVv ≥ 0 on 0 v v ∂r D. Then R(v) contains a function u ∈ E which is real analytic and satisfies the 0 bi-Laplacian equation ∆2u = 0 on D. Moreover, V = V and the weight function u v V can be found by using the equation u 1 1 V (eiθ) = ∆u(seiθ)ds. u 2 Z0 Proof. Let u(z) := 1(|z|2 −1)PV (z). Then ∆u(z) = 2PV (z) +2|z|∂PVv(z) ≥ 0 2 v v ∂r by assumption. Hence u ∈ E , u is real analytic and satisfies the bi-Laplacian 0 equation ∆2u = 0 on D. Let h(z) := ∆u(z) and h (z) := h(sz) for 0 < s < 1. By s (4) of Theorem 1.2 V (eiθ) = P(z,eiθ)h(z)dz = lim P(z,eiθ)h (z)dz u s→1 s D D Z Z 1 2π 1 2π = lim r P(reit,eiθ) h (eiη)P(reit,eiη)dη dtdr s→1 s 2π Z0 Z0 (cid:20) Z0 (cid:21) 1 2π 1 2π = lim r h (eiη) P(reit,eiη)P(reiθ,eit)dt dηdr s→1 s 2π Z0 Z0 (cid:20) Z0 (cid:21) 1 2π = lim r h (eiη)P(r2eiθ,eiη)dηdr s→1 s Z0 Z0 1 1 1 = lim rh (r2eiθ)dr = ∆u(reiθ)dr. s→1 s 2 Z0 Z0 10 NI˙HATGO¨KHANGO¨G˘U¨S¸ (cid:3) 3. Representation of linear functionals First we describe the the space of annihilators of Hp in Lq(Vdθ), where 1/p+ u 1/q = 1. Let G ∈ Lq(dθ) and f = ϕ2/pF ∈ Hp. Define u 2π L (f) = L (ϕ2/pF) := F(eiθ)G(eiθ)dθ. G G Z0 Then L belongs to (Hp)∗ since |L (f)| ≤ kFk kGk = kfk kGk . We denote G u G p q u,p q by Hq the class of functions g in Hq with g(0) = 0. Then Hq is isometrically u,0 u u,0 isomorphic to Hq which is the space of functions g ∈ Hq with g(0) = 0. 0 Theorem 3.1. For 1 ≤ p < ∞, (Hp)⊥ is isometrically isomorphic to Hq which u 0 is isometrically isomorphic to Hq or Hq. u,0 u Proof. Suppose g ∈ Lq(Vdθ) is an annihilator of Hp. Then u 2π ϕ2/p(eiθ)g(eiθ)V(eiθ)einθdθ = 0 Z0 for every n = 0,1,2,.... Therefore ϕ2/pgV is the boundary function of some G ∈ H1 with G(0) = 0. In fact G is determined uniquely by g. From the equality |g|qV = |ϕ|2|G|qV = |G|q we see that G ∈ Hq and kgk = kGk . Take any f = ϕ2/pF ∈ Hp. Then u,q q u 2π 2π f(eiθ)g(eiθ)V(eiθ)dθ = F(eiθ)G(eiθ)dθ = 0. Z0 Z0 Conversely, take any G ∈ Hq. Now from [4, Sec. 7.2] if G ∈ Hq, then L ∈ 0 G (Hp)⊥. HencethemapG 7→ L fromHq onto (Hp)⊥ isanisometricisomorphism. u G 0 u (cid:3) Theorem 3.1 gives a canonical representation of (Hp)∗ as in the next statement u which can be compared to the classical case (see [4, Theorem 7.3] for example). Theorem 3.2. For1 ≤ p < ∞, (Hp)∗ isisometricallyisomorphicto Lq(Vdθ)/Hq. u u Furthermore, if 1 < p < ∞, for each L ∈ (Hp)∗ there exists a unique G ∈ Hq so u u that L(f) = L (f) for every f ∈ Hq. For each L ∈ (H1)∗ there exists a function G u u G ∈ H∞ so that L(f) = L (f) for every f ∈ H1. u G u The next theorem describes the preduals of Hp. u Theorem 3.3. Let u be a subharmonic exhaustion function on D. If 1 < p ≤ ∞ and 1/p+1/q = 1, then: i. Hp = Lq/Hq ∗. u u u,0 ii. Hp = (Lq/Hq)∗. u,0 (cid:0) u u(cid:1)

Structure of weighted Hardy spaces in the plane

Similar Structure of weighted Hardy spaces in the plane