Structure of weighted Hardy spaces in the plane

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by  Nihat Gökhan Göğüş

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 semi-
continuous 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 exhaus-
tion 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)