Hardy Spaces on the Polydisk

EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 9, No. 3, 2016, 292-304

ISSN 1307-5543 – www.ejpam.com

Hardy Spaces on the Polydisk

Khim R. Shrestha

University of Great Falls, 1301 20th St S, Great Falls, MT 59405

Abstract. In this paper we will study the boundary values properties of the functions in the Hardy spaces; generalize the F. and M. Riesz theorem to higher dimensions; discuss the existence of boundary values of the functions inHp(Dn_{)on non-distinguished boundary∂}Dn\Tn_{and the intersection of the} spacesH_up(Dn_).

2010 Mathematics Subject Classifications: 32A35, 32A40

Key Words and Phrases: Poisson integral, boundary values, exhaustion function

1. Introduction

This paper basically consists of two parts. In the first part, consisting of Sections 2, 3 and 4, we study the properties of the functions on the classical Hardy spaces ofn-harmonic functions and the Hardy spaces of holomorphic functions on the polydisk. In Section 2 we will show that the functions in the classical Hardy spaces can be restored by the Poisson integral of its radial limit. In Section 3 we will restate and prove the celebrated F. and M. Riesz theorem to higher dimensions. In Section 4 we will study the boundary values of the functions inHp(D₎ on the non-distinguished boundary,∂Dn\Tn_.

The second part of this paper consists of Section 5. In this section we study the Poletsky– Stessin Hardy spacesH_up(D2_{)on bidisk. We mainly establish two things - there are nontrivial} Poletsky–Stessin Hardy spaces and the intersection of the Poletsky–Stessin Hardy spaces over all exhaustion functions isH∞(D2), the space of bounded holomorphic functions onD2_.

2. Hardy Spaces and Poisson Integral Formula

Ann-harmonic functionuonDn_{is a function which is harmonic in each variable separately.} Denote byhp(Dn)the space of alln-harmonic functions satisfying

sup

0≤r<1

Z

Tn

|u_r(ζ)|p_{d m(ζ)<∞ (1)}

Email address:[email protected]

whereu_r(ζ) =u(rζ)andd mis the normalized Lebesgue measure onTn_{. Thep-th root of (1)} defines a norm onhp(Dn_)whenp≥_{1. With this normh}p₍Dn_{)is Banach.}

We will use the following notations:

z=(z₁, . . . ,z_n)

ζ=(ζ1, . . . ,ζn)

P(z,ζ) =P(z₁,ζ1). . .P(zn,ζn) where P(z,ζ)is the Poisson kernel and

P(z_j,ζj) =Re

ζj+zj ζj−zj

= 1− |zj| 2

|ζj−zj|2

, j=1, . . . ,n.

Theorem 1. Let u∈hp(Dn), p>1. Then there exists a function f ∈Lp(Tn)such that

u(z) =

Z

Tn

P(z,ζ)f(ζ)d m(ζ).

Proof. Takerjր1. Then (1) implies that there is a weakly convergent subsequence ofurj. We will write the subsequenceu_rj just to avoid the sub-subscript. Hence for g∈Lq(Tn₎

g7→ lim j→∞

Z

Tn

g(ζ)u_r

j(ζ)d m(ζ)

is a linear functional on Lq(Tn_{). By Riesz theorem there exists an f} ∈_Lp₍Tn_{)such that}

lim j→∞

Z

Tn

g(ζ)u_r

j(ζ)d m(ζ) =

Z

Tn

g(ζ)f(ζ)d m(ζ).

Now take g(ζ) =P(z,ζ). Then

u(z) = lim

j→∞urj(z) =_jlim_→∞

Z

Tn

P(z,ζ)u_r

j(ζ)d m=

Z

Tn

P(z,ζ)f(ζ)d m(ζ).

The second equality above follows from[7, Theorem 2.1.2].

What makes the above proof work is the duality of Lp spaces. Since L∞ is the dual of

L1, the same result holds with the same proof for p=∞. Of course we have to change the statement accordingly. But unfortunately L1 is not dual of anything, we don’t have the same result forp=1. Instead, since the space of finite signed measures on Tn _{is dual of the space} of continuous functionsC(Tn_{)we have the following result from[7, Theorem 2.1.3, (e)].} Theorem 2. If the hypothesis of Theorem 1 holds for p = 1 then there exists a finite signed measureµonTn_with

u(z) =

Z

Tn

So the functionu∈hp(Dn_{), p >1, is the Poisson integral of some function f} ∈ _Lp₍Tn_). Is there any other connection between uand f? We know, when n= 1, f is the boundary value function ofuand whenn> 1 the following theorem[7, Theorem 2.3.1] answers this question.

Theorem 3. If f ∈L1(Tn), ifσis a measure onTn_{which is singular with respect to d m, and if}

u=P[f +dσ], then u∗(ζ) = f(ζ)for almost everyζ∈Tn_.

Recall thatu∗(ζ) =lim_r→1u(rζ)is the radial limit. Thus anyn-harmonic function satisfy-ing the growth condition (1) forp>1 can be restored by the Poisson integral of its boundary value function.

Forp=1 we just saw in Theorem 2 thatu(z) =P[dµ](z). By the Lebesgue decomposition theorem

dµ= f d m+dσ

where σis singular with respect to m and f ∈ L1(Tn_{). Hence we haveu}∗_{(ζ) = f(ζ) butu} can not be restored by the Poisson integral of its boundary value function unless, of course,

P[dσ] =0.

Also in [7] it has been proved that if f ∈ Lp(Tn_{), 1} ≤ _{p <} ∞, and _{u = P[f] thenur} converges to f in the Lp-norm asr →1, i.e. lim_r→1kur− fkLp =0. But when p=1 we have the weak-∗convergence.

Theorem 4. Let f(z) = P[dµ](z) withµ a finite signed measure on Tn_{. Then frd m} → _dµ

weak-∗as r→1.

Proof. Letϕ∈C(Tn_{). Then}

Z Tn

ϕ(ζ)fr(ζ)d m(ζ)−

Z

Tn

ϕ(ζ)dµ(ζ)

= Z Tn

ϕ(ζ)

Z

Tn

P(rζ,η)dµ(η)

d m(ζ)−

Z

Tn

ϕ(η)dµ(η)

(∵_{P(rζ,η) =P(rη,ζ)) =}

Z Tn Z Tn

P(rη,ζ)ϕ(ζ)d m(ζ)

dµ(η)−

Z

Tn

ϕ(η)dµ(η)

= Z Tn Z Tn

P(rη,ζ)ϕ(ζ)d m(ζ)−ϕ(η)

dµ(η)

→0

because the inner integral goes to zero uniformly onη. Hence f_rd m→dµweak-∗as

r→1.

We define Hp(Dn), 0< p < ∞, to be the class of all holomorphic functions f ∈Dn _for which

sup

0≤r<1

Z

Tn

|fr(ζ)|pd m<∞

Since|f|p isn-subharmonic, sup in the definition can be replaced by lim asr→1. It is known that if f ∈Hp(Dn_{), 0< p <}∞, then _{f has a non-tangential limit at almost} all points of Tn [11, Ch. XVII, Theorem 4.8]. We denote this limit by f∗ as in [7] and call it a boundary value function. Moreover, we have the following results from Rudin (see[7, Theorem 3.4.2 and 3.4.3]).

Theorem 5. If f ∈Hp(Dn_),0<p<∞_{, then f}∗∈_Lp₍Tn_)and

(i) lim_r→1R_Tn|fr|pd m=

R

Tn|f

∗_|p_{d m}

(ii) limr→1

R

Tn|fr− f∗|pd m=0.

When p ≥ 1 the function in Hp(Dn_{) can be represented by the Poisson integral of its} boundary value function.

Theorem 6. If f ∈H1(Dn_{), then}

f(z) =

Z

Tn

P(z,ζ)f∗(ζ)d m.

(The casen=1 can be found in[6, Theorem 17.11].)

Proof. Sincez∈Dn_{, P(z,ζ)is bounded on}Tn_{and by(ii)of the theorem above}

Z

Tn

P(z,ζ)f_r(ζ)d m(ζ)−

Z

Tn

P(z,ζ)f∗(ζ)d m(ζ)

≤

Z

Tn

P(z,ζ)|f_r(ζ)−f∗(ζ)|d m(ζ)

→0.

Now by[7, Theorem 2.1.2]

f(z) =lim r→1fr(z)

=lim r→1

Z

Tn

P(z,ζ)fr(ζ)d m(ζ)

=

Z

Tn

f∗(ζ)d m(ζ).

3. The F. and M. Riesz Theorem

Now we want to generalize the F. and M. Riesz theorem. Theorem 7. Letµbe a complex Borel measure onTn_{. If}

Z

Tn

ei(kθ)dµ(θ) =0

for k= (k₁, . . . ,k_n)∈Zn _{with at least one kj, j=1, 2, . . . ,n positive, where}

(Whenn=1 see[6, Theorem 17.13].)

Proof. Define f(z) =P[dµ](z). Then, with the notations

z=(z₁, . . . ,z_n)withz_j=r_jeiθj_{,j=1, . . . ,n}

r|k|=r|k1|

1 . . .r

|kn| n

(k·θ) =k₁θ1+. . .+knθn

(k·t) =k₁t1+. . .+kntn

and using the series representation for the Poisson kernel, we get

f(z) =

Z

Tn

P(z,ei t)dµ(t)

=

Z

Tn

X

k∈Zn

r|k|ei(k·θ)e−i(k·t)

!

dµ(t)

=X

k∈Zn

Z

Tn

e−i(k·t)dµ(t)

r|k|ei(k·θ)

= X

k∈Zn +

ckzk

where c_k = R_Tne

−i(k·t)_dµ(t)andz

k = r|k|ei(k·θ). Notice that all other integrals in the above sum vanish by the hypothesis. Thus f(z)is holomorphic.

For 0≤r<1,

Z

Tn

|f_r(ζ)|d m(ζ) =

Z

Tn

Z

Tn

P(rζ,η)dµ(η)

d m(ζ)

≤

Z

Tn

Z

Tn

P(rζ,η)d|µ|(η)

d m(ζ)

=

Z

Tn

Z

Tn

P(rζ,η)d m(ζ)

d|µ|(η)

=kµk.

Thus f ∈H1(Dn_{)and hence f(z) =P[f}∗_{](z), where f}∗∈_L1₍Tn_{). Now the uniqueness of the} Poisson integral representation shows that

dµ= f∗d m

4. Boundary Values

Do the boundary values of functions inHp(Dn_{)exist on the non-distinguished boundary?} Now we want to look into this question.

Let{j₁, . . . ,j_k}and{i₁, . . . ,i_l}be disjoint sets of indices such that their union is{1, . . . ,n}

where j1< j2<. . .< jk andi1<i2<. . .<il. Define the sections ofDnas follows Dn

zj₁,...,z_jk ={(z1, . . . ,zn)∈

Dn_:zj

1, . . . ,zjk are fixed} and definefzj1,...,zjk = f|Dz jn₁,...,_{z j}

k

. We will write fzj1,...,zjk(zi1, . . . ,zil)instead offzj1,...,zjk(z1, . . . ,zn). We will see below that for f ∈Hp(Dn_{), 1}≤ _p< ∞, the non-tangential limit of _fz

j1,...,zjk exists at almost all points of the distinguished boundary of the section Dn

zj₁,...,z_jk which is

Tl

and the function f_z

j1,...,zjk can be restored by the Poisson integral of this limit. Theorem 8. Let f ∈Hp(Dn_),1≤_p<∞_{. Then fz}

j₁,...,z_jk ∈H p_(Dl_).

Proof. Without loss of generality we suppose that {j₁, . . . ,j_k}= {1, . . . ,k}. Let’s use the

following notations for the Poisson kernels

Pj(ζj) =

¨

P(z_j,ζj) j=1, . . . ,k

P(rξj,ζj) j=k+1, . . . ,n where|ξ_j|=1. Then, for 0<r<1, by Theorem 6

f_z1,...,z

k(rξk+1, . . . ,rξn) =

Z

Tn

P₁(ζ1). . .Pn(ζn)f∗(ζ1, . . . ,ζn)d mn.

By Hölder and Fubini

Z

Tl |f_z1,...,z

k(rξk+1, . . . ,rξn)| p_{d m}

l=

Z

Tl

Z

Tn

P₁(ζ1). . .Pn(ζn)f∗(ζ1, . . . ,ζn)d mn

p

d m_l

≤

Z

Tl

Z

Tn

P1(ζ1). . .Pn(ζn)|f∗(ζ1, . . . ,ζn)|pd mn

d m_l

=

Z

Tn

P1(ζ1). . .Pk(ζk)|f∗(ζ1, . . . ,ζn)|p

×

Z

Tl

Pk+1(ζk+1). . .Pn(ζn)d ml

d mn

≤ 2k

(1− |z₁|). . .(1− |z_k|)

Z

Tn

|f∗(ζ1, . . . ,ζn)|pd mn.

The last quantity above is independent of r and is finite by Theorem 5. Thus the theorem is proved.

Corollary 1. If f ∈Hp(Dn_),1≤_p<∞_{, then the non-tangential limit f}∗

zj₁,...,z_jk of the function

f_z

j1,...,zjk exists almost everywhere onT

l _{and belongs to L}p_(Tl_).

The following theorems are the direct consequences of Theorems 5 and 6. Theorem 9. If1≤p<∞and f ∈Hp(Dn_{), then}

(i) lim r→1

R

Tl|(fzj₁,...,z_jk)r| p_{d m}

l =

R

Tl|f_z∗ j₁,...,z_jk|

p_{d m} l

(ii) lim r→1

R

Tl|(fzj1,...,zjk)r−f

∗

zj1,...,zjk| p_{d m}

l =0

where(f_z

j1,...,zjk)r(ζi1, . . . ,ζil) = fzj1,...,zjk(rζi1, . . . ,rζil). Theorem 10. If f ∈H1(Dn_{), then}

fzj₁,...,z_jk(zi1, . . . ,zil) =

Z

Tl

P(z_i

1,ζi1). . .P(zil,ζil)f

∗

zj₁,...,z_jk(ζi1, . . . ,ζil)d ml. Theorem 11. Let f be a holomorphic function inDn_{. If1}≤_p<∞_and

sup (zj₁,...,z_jk)

|zj1|=...=|zjk|

kfzj1,...,zjkkHp(Dn−k)=M<∞,

then f ∈Hp(Dn_).

Proof. For simplicity we take{j1, . . . ,jk}={1, . . . ,k}. And, of course, this theorem makes sense only whenk>0. Now for 0≤r<1,

Z

Tn

|f(rζ1, . . . ,rζn)|pd mn=

Z

Tk

Z

Tn−k

|f(rζ1, . . . ,rζn)|pd mn−k

d m_k

≤

Z

Tk

sup

0≤t<1

Z

Tn−k

|f(rζ1, . . . ,rζk,tζk+1, . . . ,tζn)|pd mn−k

d m_k

=

Z

Tk

kf_rζ1,...,rζkk p

Hp₍Dn−k₎d mk

≤Mp. Thus f ∈Hp(Dn_).

5. Poletsky–Stessin Hardy Spaces on the Bidisk

such thatu(z₁,z₂) →0 as(z₁,z₂) →(ζ1,ζ2) ∈∂D2. Following Demailly[2], for r < 0 we

define

S_u(r) =(z₁,z₂)∈D2_{:u(z1,z2) =r}

B_u(r) ={(z₁,z₂)∈D2_:u(z1,z2)<r}.

For convenience we will write z = (z₁,z2). Associated with this u we define the positive

measureµu,rcalled Monge-Ampère measures by µu,r= (d dcur)2−χD2_\B

u(r)(d d c_u)2

where ur = max{u,r}. These measures are supported by the level setsSu(r). Demailly has proved the following[2, Theorem 1.7].

Theorem 12(Lelong–Jensen Formula). For all r<0every plurisubharmonic functionϕonD2

isµu,r-integrable and

µu,r(ϕ) =

Z

Bu(r)

ϕ(d dcu)2+

Z

Bu(r)

(r−u)(d dcϕ)∧(d dcu).

Denote by E(D2_{) the set of all continuous negative plurisubharmonic functionsuon} D2 and equal to zero on∂D2 _{whose Monge–Ampère mass is finite, i.e.}

Z

D2

(d dcu)2<∞

and denote byE1(D2)the set of thoseu∈ E(D2)for which

R

D2d dcu=1.

Following[3]we define, what we call, the Poletsky–Stessin Hardy space Hup(D2), p> 0, as the space of all holomorphic functions onD2 _{for which}

lim sup

r→0− µu,r(|f|

p₎ <∞.

These new spaces are contained in the classical spaces, that is, Hup(D2) ⊂ Hp(D2). Since µu,r(|f|p) is an increasing function of r the lim sup in the definition can be replaced by lim. Forp≥1

kfkp

Hup

= lim

r→0−µu,r(|f|

p₎

is a norm and with this normH_up(D2_{)is Banach[3, Theorem 4.1]. The Poletsky–Stessin Hardy} spaces on the unit disk have been studied in detail in[1, 5, 8–10].

In[4]Poletsky has proved that the intersection of all Poletsky–Stessin Hardy spacesHup(D),

p≥1, where Dis a strongly pseudoconvex domain with C2 boundary, isH∞(D), the space of bounded holomorphic functions. Hence it immediately follows that the intersection of all

Let ζ= (ζ1,ζ2)∈T2 andα= (α1,α2), 0< α1,α2 < π/2. Following[11] we define the

approach regionTα(ζ)as

Tα(ζ) =Tα1(ζ1)×Tα2(ζ2)

where Tαj(ζj)is the Stolz angle at ζj ∈Twith vertex angle 2αj. Here we will consider only the congruent symmetric approach regions meaning that the Stolz angles are symmetric with respect to the radius toζj and the vertex angles are equal, i.e. α1 = α2. Following [4] we

define the Green ball of radius 0<r<1 and center atwto be the set

G(w,r) ={z∈D2_:g(z,w)<logr}

where g(z,w) is the Green function for D2 _{with pole at w. The Green function for} D2 _is explicitly given by

g(z,w) =log max

z₁−w₁

1−w₁z₁

,

z₂−w₂

1−w₂z₂

. Hence it follows that

G(w,r) =

z₁∈D_:

z1−w1

1−w1z1

<

r

×

z₂∈D_:

z2−w2

2−w2z2

<

r

.

Lemma 1. Letζ= (ζ1,ζ2)∈T2 and 0< r <1. For any0< t <1there exists0< α < π/2 such that G(tζ,r)⊂Tα(ζ)where tζ= (tζ1,tζ2)and Tα(ζ) =Tα(ζ1)×Tα(ζ2).

Proof. Observe that

(

zj∈D:

zj−tζj

1−tζjzj

<r

)

is the image of the disk{|wj|<r} ⊂Cunder the conformal map

w_j7→ wj+tζj

1+tζjwj which is a disk contained inD_{with center at}

t(1−r2)

1−r2_t2 ζj

and radius equal to

r(1−t2)

1−r2_t2.

The tangents to this disk that pass throughζj make an angle of

α=arcsin

_r(1+t)

1+t r2

with the radius toζj. Hence

(

z_j∈D_:

z_j−tζj

1−tζ_jz_j

<r

)

⊂Tα(ζj)

for j=1, 2 andG(tζ,r)⊂Tα(ζ). Since for fixed 0<r<1

t7→ r(1+t)

1+t r2

is an increasing function of t∈[0, 1]we have

0< r(1+t) 1+t r2 ≤

2r

1+r2 <1.

From this it follows that

0< α≤arcsin

 ₂

r

1+r2

‹

< π 2.

Remark 1. For fixed0<r<1,

t7→ r(1−t 2₎

1−r2_t2

is a decreasing function of t∈[0, 1]that decreases to zero as t→1. Therefore we can make the size of the Green ball G(tζ,r)as small as we want simply by choosing t close enough to1.

The plurisubharmonic envelopeEφof a continuous functionφon a domainΩ⊂Cn_{is the} maximal plurisubharmonic function onΩless than or equal toφ. For a sequence of functions {u_j} ⊂ E(D2), we denote byE{u_j}the envelope of inf{u_j}. The following Lemma[4, Theorem 3.3]gives the estimate on the Monge–Ampère mass of the envelope.

Lemma 2. If Ω is a strongly hyperconvex domain and continuous plurisubharmonic functions

{uj} ⊂ E(Ω), then

Z

Ω

(d dcE{u_j})n≤X

Z

Ω

(d dcu_j)n.

Theorem 13. Let f be a holomorphic function onD2_{. Suppose that f has non-tangential limits}

at points{ζj} ⊂T2andlimj→∞|f∗(ζj)|=∞. Then for any p≥1there exists u∈ E1(D2)such that f ∈/Hup(D2).

The proof that Poletsky gave to this theorem in[4]in the case whenDis a strongly pseu-doconvex domain withC2boundary also works when the domain is a polydisk. We will mimic his proof in our context.

Proof. Let us take a sequence{aj}of positive numbers such that

∞

X

j=1

a_j <∞and

∞

X

j=1

For 0< t_j <1 we writeG_j =G(t_jζj,e−1). By Lemma 1 there exists 0< αj < π/2 such that

Gj ⊂ Tαj(ζj). Now we inductively construct a sequence {tk}, 0 < tk <1, satisfying certain conditions. Choose any 0<t1<1. Suppose thatt1, . . . ,tk−1have already been chosen. Now

chose 0<t_k<1 so that the following conditions are satisfied: (i) |f|>|f∗(ζ_k)|/2 onG_k

(ii) G_k∩G_j=φ

(iii) g(z,t_kζk)>−aj/2k+1 onGj (iv) ajg(z,tjζj)>−ak/2j+1onGk

for 1≤ j≤k−1. The conditions (i) and (ii) can be achieved simply by takingt_kclose enough to 1. SinceGj, j <k, and Gkare disjoint, g(z,tkζk)→0 uniformly on Gj as tk →1. Hence (iii) can be achieved for t_k close enough to 1. Since g(z,t_jζj) = 0 whenz ∈ ∂D2, we can chooset_k so close to 1 that

Gk⊂ k−1

\

j=1

z∈D2_{:ajg(z,tjζj)>}−_ak/2j+1 _.

Thus (iv) can be achieved. Define

u_j(z) =a_jmax{g(z,t_jζj),−2}. Note that ifF is an open set inD2_{containingG(tjζj,e}−2_)then

Z

F

(d dcu_j)2=a2_j.

Letu=E{u_j}. Since the series v=P∞_j=1u_j converges uniformly onD2_{,v∈ E(}D2_{). Sou}≥_v is a continuous plurisubharmonic function onD2 _{equal to 0 on∂}D2_{. By Lemma 2,}

Z

D2

(d dcu)2≤

∞

X

j=1

Z

D2

(d dcu_j)2=

∞

X

j=1

a2_j <∞.

Henceu∈ E(D2). Now we evaluateR_G

k(d d

c_u)2_{. Observe thatu}

k ≥u≥ v onD2. By the conditions on the choices oftj, on∂Gk we get

−ak≥u≥ − k−1

X

j=1 a_k

2j+1 −ak−

∞

X

j=k+1 a_k

2j+1 ≥ −

3 2ak.

Henceu+3ak/2≥0 on ∂Gk and the set Fk = {6(u+ 32ak)<uk} compactly belongs toGk. Moreover, ifz∈∂G(t_kζk,e−2)then

6



u(z) +3

2ak

‹

≤6



u_k(z) +3

2ak

‹

ThusG(t_kζk,e−2)⊂Fk. By the comparison principle

36

Z

Gk

(d dcu)2=

Z

Gk

(d dc6(u(z) +3

2ak))

2_≥

Z

Fk

(d dcu_k)2=a2_k.

Hence by Lelong–Jensen formula

kfkp

Hup ≥

Z

D2

|f|p(d dcu)2≥

∞

X

k=1

Z

Gk

|f|p(d dcu)2≥ 1

36·2p

∞

X

k=0

|f∗(ζk)|pak2=∞.

Hence f ∈/Hp(D2).

The following corollary shows the existence of nontrivial Poletsky–Stessin Hardy spaces on the bidisk.

Corollary 2. For every p≥1there exists a function u∈ E₁(D2_{)such that Hu}p₍D2₎6⊆_Hp₍D2_).

Proof. Take f ∈Hp(D2_{)that is unbounded. Then the non-tangential limit f}∗_on T2 _must be unbounded because otherwise

f(z) =

Z

T2

P(z,ζ)f∗(ζ)d m

would imply that f(z)is bounded. So there exists a set of points{ζj} ∈T2 such that limj→∞|f∗(ζj)|=∞. Hence the corollary follows from Theorem 13.

Now we prove the most important theorem of this section. Theorem 14. Let p≥1. Then

\

u∈E1(D2)

H_up(D2_{) =H}∞₍D2_).

Proof. Let f ∈ T_u∈E

1(D2)H

p

u(D2). Then the non-tangential limit f∗ on T2 is bounded because otherwise by Theorem 13 there would exist au∈ E₁(D2_{)such thatf} ∈_/Hup₍D2_{). Thus,} since f∗is bounded,

f(z) =

Z

T2

P(z,ζ)f∗(ζ)d m

implies that f ∈H∞(D2).

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