Our aimis to study bounded pluriharmoni fun tions by means of S-invariant operators

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DARIUSZ BURACZEWSKI, EWA DAMEK, AND ANDRZEJ HULANICKI

Abstra t. LetDbeanirredu ible,symmetri SiegeldomainandletSbeasolvable

groupwhi ha tssimplytransitivelyonD. WeexhibitthreeS-invariant,real,se ond

order,degenerateellipti operatorsL; L; Hsu hthataboundedfun tionF onDis

pluriharmoni ,ifandonlyifLF =LF =HF =0. Thethreeoperatorsarethesame

asin our former paper [DHMP℄, however there weneeded a onsiderably stronger

onditiononF toderivethesame on lusion.

1. Introdu tion

LetDbeasymmetri Siegeldomainand letS beasolvableLiegroupa tingsimply

transitivelyon D. Our aimis to study bounded pluriharmoni fun tions by means of

S-invariant operators. More pre isely, these are real S-invariant se ond order, ellipti

degenerate operators L annihilating holomorphi fun tions F and, onsequently, <F

and =F. Su h operators willbe alledadmissible.

The parti ular interest in restri ting our attention to the se ond order, degenerate

ellipti operators is aused by the fa t that for su h an operator there is a very well

understood potential theory. To wit, theory of bounded fun tions harmoni with re-

spe t to left-invariantoperators L satisfyingHormander ondition wasstudied in[D℄,

[DH℄, [DHP℄. The origin of this resear h goesba k toH.Furstenberg [F℄, Y.Guivar 'h

[G℄,andA.Raugi[Ra℄whodevelopedaprobabilisti approa htoboundedfun tionson

groups, harmoni with respe t to a probability measure ora se ond order degenerate

ellipti invariant operator. The fundamental result of the theory says that bounded

L-harmoni fun tionsarepre iselytheintegralsoftheirboundaryvaluesonanilpotent

subgroup N(L) of S against anappropriate Poisson kernel.

Thisobservation hanges onsiderablyourwayoflookingatfun tionsonsymmetri

domains. We study S-invariant obje ts instead of studying the ones invariant under

the whole group of isometries G. It turns out that they are more suitable as far as

hara terization of bounded pluriharmoni fun tions is the goal, though learly the

spa e of bounded pluriharmoni fun tionsis invariant underG.

ForanadmissibleLsatisfyingthe Hormander ondition theboundaryN(L)always

ontainsthe groupN() thata ts simplytransitivelyonthe Bergmann-Shilovbound-

ary. In this paper we are mostly interested in operators for whi h N(L) = N().

Ea h of the operators gives rise to a new Poisson kernel. Weexploit intensively those

Poisson kernels onN() toprove the following theorem:

The authors were partly supported by KBN grant 2P03A04316,Foundationfor Polish S ien es,

Subsidy3/99,andbytheEuropeanCommissionviatheTMRNetwork\Harmoni Analysis", ontra t

no.ERBFMRX{CT97{0159.

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Let D be an irredu ible symmetri Siegel domain. There are three admissible op-

erators L , L, H on D su h that if a real valued bounded fun tion F on D satises

LF = LF =HF =0, then F is pluriharmoni . If D is a tube domain, L and H are

suÆ ient and if D is biholomorphi ally equivalent to the omplex ball, L and H are

suÆ ient.

This theorem is an improvement of the result obtained in [DHMP℄, where a more

restri tive, not G invariant type of ondition was onsidered. Namely, the fun tions

studiedthere had the property

(H 2

) sup

z2D Z

N()

jF(vz)j 2

dv<1:

As it is proved in [DHMP℄ the onjugate fun tion e

F i.e. the fun tion su h that

F +i e

F is holomorphi , also satises (H 2

). But already for the upper half-plane the

onjugatefun tion e

F ofabounded realharmoni (=pluriharmoni )fun tion F isnot

bounded, in general. Consequently, the same is true for an arbitrary Siegel domain.

Ex ept for the last remark,the \bounded" result impliesthe H 2

result.

Inany aseoftheabovetheorem,thesumofoperatorsneededtoyieldthe on lusion

is an ellipti operator. The Poisson integral hara terization says that the spa e of

bounded fun tions harmoni with respe t to one ellipti operator L is always bigger

then the spa e of bounded holomorphi fun tions. But due to our theorem three of

them, in general, or two in spe ial ases, while appropriately hosen are suÆ ient to

obtainthe smallestpossiblespa e of ommonzeros -the pluriharmoni fun tions. For

otherglobal hara terizationsofpluriharmoni fun tionsonsymmetri Siegeldomains

we referto [L1℄, [L2℄, [BBG℄, [DH1℄, [J℄.

Let be the underlying symmetri one in the Jordan algebra V. Suppose V is

the enter of the step two nilpotent group N() = Z V. Let f(;u) = f

(u) be

the boundary value of F. If f 2 L 2

(N()), the basi strategy of the proof , used in

[DHMP℄, is toshow that for every 2Z,

supp

^

f

[

(1.1)

and, in the nontube ase, that the integrated representation U

f , 2

[

is zero

on the invariant subspa es whi h do not ontain the va uum state. The rest is then

pretty standard. If f 2 L 1

(N()), new diÆ ulties arise. One is aused by the fa t

that 0 may belong to the support of

^

f

. To deal with that we produ e a sequen e of

fun tions F

n

with boundaryvalues f

n

su h that

LF

n

=LF

n

=HF

n

=0;

supp f

n

(;^)\Zf0g=;;

(F lim

n!1 F

n

) isa onstant (1.2)

(Se tion4). Hen eweredu etothefun tionssatisfying(1.2). Toprovethatf

n

satises

(1.1) we use the operator H whi h is basi ally the Lapla e-Beltrami operator on the

produ tofupperhalfplanes(Se tion 5). Thebasi ideaisthesameasin[DHMP℄,but

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every one as for L -fun tions. All this requires more deli ate algebrai onsideration

(Lemma5.7) usingthe stru ture of .

Finally, we annot use the operator U

f

dire tly and we have to repla e this by a

\weak" argument. We do this to prove our theorem for the Siegel upper half-plane

(N() being the Heisenberg group, Se tion 6). Then, for the general ase we use

Siegelupper-halfplanesimbeddedinS invarious dire tions. Thisnal step simplies

onsiderably the whole argument,also the one in [DHMP℄.

The authors are grateful to AlineBonami and Ryszard Szwar for valuable dis us-

sions and omments.

2. Preliminaries

2.1. Jordan algebras and irredu ible ones. Let be an irredu ible symmetri

one in a Eu lidean spa e (V;h;i). V has stru ture of a simple Eu lidean Jordan

algebra[FK℄, and

= intfx 2

: x2Vg

([FK℄,Theorem III.2.1).

We are going to use the language of Jordan algebras to des ribe solvable groups

a tingsimplytransitivelyon, and sowe re allbrie y some basi fa ts whi h willbe

needed later. The reader is referred to[FK℄ formore details.

We x a Jordan frame f

1

;:::;

r

g in V. This is a omplete system of orthogonal

primitiveidempotents:

2

i

=

i

;

i

j

=0 if i6=j;

1

++

r

=e;

su h that none of the

1

;:::;

r

is a sum of two nonzero idempotents. The length r is

independentof the hoi e of Jordan frameanit is alled the rankof V.

LetL(x) be the self - adjoint endomorphismof V given by multipli ationby x, i.e.

L(x)y=xy:

Let

V = M

1ijr V

ij (2.1)

bethe ommondiagonalizationofthe ommutingfamilyofself-adjointendomorphisms

L(

1

);:::;L(

r

). It is alled be the Peir e de ompositionof V ([FK℄, Theorem IV.2.1).

This means that V is the orthogonal dire t sum (2.1) and L(

k

) restri ted to V

ij has

the eigenvalue 1

2 (Æ

ik +Æ

jk

). Moreover,

V

ij

V

ij

V

ii +V

jj

;

V

ij

V

jk

V

ik

; if i6=k;

V

jk

V

jl

V

kl

; if k <l;

V

ij

V

kl

=f0g; if fi;jg\fk;lg=;: (2.2)

All V , i<j have the same dimension d and V =R ,j =1;:::;r.

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Forea hi<j wex on efor allanorthonormalbasis ofV

ij

whi h wenote e

ij with

1 d. To simplify the notation we write

i

= e 1

ii

. Then e

ij

is an orthonormal

basis of V.

Let G be the onne ted omponent of the group G() of all transformations in

GL(V)whi hleaveinvariant. Its Liealgebrawillbedenoted by G. G ontains L(x)

for allx2V.

The hoi e of the Jordan frame determines a solvable Lie group S

0

= N

0

A a ting

simplytransitivelyon. A =expA,whereAistheAbeliansubalgebraofG onsisting

of elements

H =L(a); a= r

X

j=1 a

j

j 2

M

i V

ii :

Forthe nilpotent part N

0

=expN

0

, we have

N

0

= M

i<jr N

ij

;

where, ( f. [FK℄),

N

ij

=fz2

i

:z 2V

ij g:

2.2. Symmetri Siegel domains. Let V C

= V +iV be the omplexi ation of V.

Weextend the a tion of G() to V C

.

In addition, suppose that we are given a omplex ve tor spa e Z and a Hermitian

symmetri bilinear mapping

: ZZ !V C

:

Weassume that

(;)2; 2Z;

and (;)=0 implies =0:

The Siegel domainasso iated with these data is dened as

D=f(;z)2Z V C

: =z (;)2g:

There isan representation :G()3g 7!(g)2Gl(Z) su h that

g(;!)=((g);(g)!); (2.3)

and allautomorphisms (a)2A admit ajointdiagonalization. The transformation

(;z)7!((g); gz)

isaholomorphi automorphismof D,[KW℄.The elements 2Z,x2V andg 2G()

a t on D inthe following way:

Æ(!;z)=(+!;z+2i(!;)+i(;));

xÆ(!;z)=(!;z+x);

gÆ(!;z)=((g)!;gz): (2.4)

The rst two a tions generate a two-step nilpotent(or abelian, if Z =0)group N()

of biholomorphi automorphismsof D:

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All three a tions (g restri ted to S

0

) generate a solvable Lie group S =N()S

0 , the

group N() being a normalsubgroup of S.

ForX 2S

0

, (2.3) implies that

X(;!)=((X);!)+(;(X)!); (2.6)

where denotes both representation of the group G() and the Lie algebra G. An

easy onsequen e of (2.6) is that the only possible eigenvalues for (H), H 2 A are

j (H)

2

, j =1;:::;r. Sowe may write

Z = r

M

j=1 Z

j

with the property that

(H) =

j (H)

2

for 2Z

j :

(A standard argument is ontained e.g. in[DHMP℄).

Finally,the Lie algebraS of S has the following de omposition

S =N()S

0

= r

M

j=1 Z

j

!

M

ij V

ij

!

M

i<j N

ij

!

A:

that orresponds to adiagonalizationof the adjoint a tion of A:

[H;X℄=

j (H)

2

X forX 2Z

j

;

[H;X℄=

i

(H)+

j (H)

2

X forX 2V

ij

;

[H;X℄=

j

(H)

i (H)

2

X forX 2N

ij

; (2.7)

where

j

(H)=

j (

P

r

k=1 a

k

k )=a

j

. If Z =0then N() =V and

D

T

=V +i

is a tube domain. To distinguish those two ases the solvable group for the tube will

be denoted S

T

=VS

0

. Clearly

S

T

= M

ij V

ij

!

M

i<j N

ij

!

A:

In virtue of (2.4) we identity D with S. More pre isely lete=ie2D and let

:S 3s 7!(s)=sie2D (2.8)

Then isadieomorphismof S and D (or S

T

and D

T

inthe parti ular ase of atube

domain). identies also the spa es of smooth fun tionson S and D.

The Lie algebra S is then identied with the tangent spa e T

e

of D at e by the

dierentiald

e

. WethentransporttheBergmannmetri gandthe omplexstru tureJ

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tangent spa e T C

e

is identied with S C

and the de omposition T C

e

= T 1;0

e

T 0;1

e is

transported into

S C

=QP:

Now we pi k up a g-orthonormal basis of S and the orresponding basis of Q. For

j <k and 1d, we dene X

jk 2V

jk

;Y

jk 2N

jk

as the left-invariant ve tor elds

on S orresponding to e

jk

and 2e

jk 2

j

, respe tively. For ea h j we dene X

j

and H

j

as left-invariantve tor elds on S orrespondingto

j

and L(

j

) respe tively. Finally,

we hoose an orthonormal basis fe

j

g, = 1;:::;n related to 4h

j

;(;)i. For

j

=

x

j +iy

j

the orresponding oordinates, wedene X

j

;Y

j

asthe left-invariantve tor

elds onS whi h oin idewith

x

j

and

y

j

ate. ThenX

j

;X

jk

;H

j

;Y

jk

;X

j

;Y

j form

ag-orthonormalbasisof S andZ

j

=X

j iH

j ,Z

jk

=X

jk iY

jk ,Z

j

=X

j iY

j form

anorthonormalbasisofQwithrespe ttotheHermitians alarprodu t(;)= 1

2 g(;).

Forthe detailed al ulation of d

e

and J see [BBDHPT℄. In the oordinates

(;z)=

X

j;

j e

j

; X

ij

z

ij e

ij

;

Z

j , Z

jk ,Z

j

are left-invariantve tor elds orresponding to

z

j ,

z

jk ,

z

j at e.

2.3. Admissible operators. Under identi ation(2.8), holomorphi fun tions onD

are alled holomorphi fun tions on S. The left-invariant dierential operators on S

that annihilate holomorphi fun tions and are real se ond order and ellipti degener-

ate are alled admissible(see [DHP℄, [DHMP℄ and [BBDHPT℄ for more details about

admissibleoperators). In parti ular,these ondorderleft-invariantoperators

j ,

jk ,

L

j

with the property

j

(e) =

zj

(e);

jk

(e) =

z

jk

z

jk (e);

L

j

(e) =

z

j

z

j (e):

are su h. More expli itly,(see e.g. [DHMP℄):

j

=X 2

j +H

2

j H

j

;

L

j

=(X

j )

2

+(Y

j )

2

H

j

;

ij

=(X

ij )

2

+(Y

ij )

2

H

j : (2.9)

We willuse also left-invariant ve tor elds on N() that oin idewith

x

j ,

xij ,

x

j ,

y

j

ate. They willbedenoted by

f

X

j

; f

X

ij

; f

X

j

; e

Y

j

respe tively. Noti e that

XF(;u;y;a)=(Ad

ya f

X)F

ya (;u):

(2.10)

Then

j

F(;u;y;a) = a 2

j

(Ad

y f

X

j )

2

+ 2

a

j

F(;u;y;a);

L

F(;u;y;a) = a

j

(Ad

y g

X

j )

2

+(Ad

y g

Y

j )

2

a

F(;u;y;a):

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2.4. Poissonintegrals. Inthisse tion,were allsomegeneralresultsaboutbounded

harmoni fun tionsonS. LetLbease ondorderleft-invariantoperatoronSsatisfying

Hormander ondition. Denote by

A

(L) its image on A under the homomorphism

S ! A = S=N. The rst order part Z in

A

(L) determines the Poisson boundary

for L.

More pre isely,let W =f

1

;:::;

r

;

1

2

;:::;

r

2

;

i+j

2

;

j

i

2

;i<jg and let

W

1

=f2W : (Z)<0g;

N(L)=

2W

1 N

; N

+

(L)=

2WnW

1 N

;

N(L)=expN(L); N +

(L)=expN +

(L):

Thenthe bounded L-harmoni fun tionsonS are inone-one orresponden ewith L 1

fun tions onN(L)via the following Poisson integral

F(s)= Z

N(L)

f(sÆx)P

L

(x)dx;

(2.11)

where x!sÆx denotes the a tion onN(L) ason the quotientspa e N=N +

(L).

We will need this representation for the ase when N(L) = N() (the parti ular

ase N()=V in luded). Then (2.11)be omes

F(s)=F(xs

0 )=F

s0 (x)=

Z

N() f(xs

0 us

1

0 )P

L (u) du (2.12)

and f is *weak limit of F

yat

when

j (loga

t

) ! 0 for j = 1;:::;r. The fun tion f is

alled the boundary value of F.

In what follows we will onsider fun tions F(;u;y;a) harmoni with respe t to

an admissible operator L with the additionalproperty that for every , F

(u;y;a) =

F(;u;y;a) is annihilated by an admissible operator L on S

T

. Moreover, we assume

N(L)=N(), N(L )=V. Thus F has awelldened boundaryvaluef onN() su h

that for every 2 Z, f

(u)=f(;u) is the boundary value of F

on V. Then on top

of (2.12), wehave another representation of F. Namely,

F(;u;y;a)= Z

V f

(u v)P

ya

(v) dv;

(2.13)

where P

ya

(v) =deta 1

P

ya ((ya)

1

Æv) and deta is the determinantof the a tion V 3

v 7!aÆv 2V.

3. Main theorem

Let L be an ellipti admissible operator on S

T

. L may be onsidered also as a

left-invariantoperator onS.

Given stri tly positive numbers

1

;:::;

r

and

1

;:::;

r let

H =

r

X

j=1

j

j (3.1)

L =

r

X

j=1

j L

j : (3.2)

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Main Theorem 3.3. Let D be an irredu ible symmetri Siegel domain, and let F be

a bounded fun tion on D su h that

(i) LF =0;

(ii) HF =0;

(iii) LF =0:

Then F is a pluriharmoni fun tion.

If the domain D is of type I, then ondition (iii) is void. If D is biholomorphi ally

equivalent tothe omplexballthenL maybe takentobe a multipleof H, so(i) isvoid.

Let us make some omments on erning the role of H. It turns out that for a

bounded fun tionF onS we have

Hf =0 () 8

j

F =0: (3.4)

Indeed, sin e V is a normal subgroup of S

T

, we may write S

T

= N

0

VA. In the

oordinates y2N

0

; x2V; a2A, for a smooth fun tiong on N

0

VA we have

j

g(yxa)=a 2

j (

2

x

j +

2

a

j

)g(yxa):

Therefore H= P

r

j=1

j a

2

j (

2

x

j +

2

a

j

) is the Lapla e-Beltrami operator on the produ t

of r-hyperboli half-planes

D =R

1

R +

1

:::R

r

R +

r

;

themetri being s aledby 1

j

onR

j

R +

j

:Thuseverybounded H-harmoni fun tion

onD isthe Poisson integral of aL 1

fun tionon

D =R

1

:::R

r

;

the Poisson kernel being the tensor produ t of the Poisson kernels for the hyperboli

half-planeRR +

, [Ko℄. Consequently, Hg =0 impliesthat g is annihilatedby ea h

of the operators

j

, j =1;:::;r.

This leads us tothe following

Proposition 3.5. Suppose F is a bounded fun tion annihilated by the operators (i)-

(iii). Then, adding to L and L appropriate linear ombination of

j

's with non-

negative oeÆ ients, by Proposition (2.1) in [DHMP℄, L and L an be repla ed by

operators su h that the maximal boundary of L is V and the maximal boundary for

L+L is N().

Consequently,from nowonweassumethatthe operators L andL inthe maintheorem

have their maximal boundaries V and N(), orrespondingly. We then have

lim

a!0

F(;x;y;a)=f(;x)=f

(x)

inthe

weak sense both on N() and V. Moreover, (2.13)holds.

Convolving (on N()) fromthe left by a S(N()) fun tion we may assumethat

f = e

f; 2S(N()); f 2L 1

(N()):

(3.6)

Throughout the rest of the paper we assume that F satises the assumptions of the

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Now the proof of the main theorem splits intofour parts.

Inthe rstpartwe dealwiththe distributionalpartialFouriertransformof f

along

V. We show that there is a sequen e

n

of fun tions on V su h that the Fourier

transform

^

n

vanishes in a neighbourhood of 0 in V, F

n

(;x;y;a) = f

V n

V

P

ya

satisfy (i)-(iii) and are *-weakly onvergent to F+ onstant (Se tion 4). For the

remainingthreepartswedealwithfun tionsF whi hsatisfy(i)-(iii)andsu hthatthe

support of the Fourier transform of the orresponding f

does not ontain 0. In the

se ondpart we prove the main theorem forthe tube domains(Se tion 5),inthe third

partforthetypeIIdomainswhi harebiholomorphi allyequivalenttothe omplexball

(Se tion6). Finallyinthefourthpartthe proofforthearbitraryirredu iblesymmetri

domainsof type II is redu ed tothe proof for the ases settledearlier (Se tion 7).

4. Partial Fourier transform

Let bea S hwartz fun tionon V su hthat

^

()= 8

<

:

1 for jj1

0 for jj2:

Fora given sequen e fk

n g

n=1;2;:::

of naturalnumbers tendingto innity, let

n (x)=

1

k d

n

x

k

n

!

: (4.1)

Given abounded fun tion g onV there exists fk

n g

n=1;2;:::

, k

n

!1 su h that

lim

n!+1 hg;

n

i exists :

We need a bit more here: we are going to sele t a sequen e fk

n g

n=1;2;:::

su h that for

all 2Z

lim

n!+1 hf

;

n

i exists and isindependent of :

The details are asfollows:

Lemma 4.2. Let 2S(N()) and f 2L 1

(N()): Then the mapping

Z 3 !f(;)2L 1

(V)

is ontinuous.

Proof. Indeed, weprove that for ea h ompa t subset K of Z

sup

2K ;x2V

jf(+h;x) f(;x)jC

K jhj:

By(2.5) we have

jf(+h;x) f(;x)j

kfk

L 1

(N()) Z

N() j

(h; 2=(h;))(!;u) 1

(!;u) 1

jd!du:

If h and are in a ompa t set, so are all elements (h; 2=(h;)) = exp [X℄ with

jXjCjhj. Moreover,

(h; 2=(h;))(!;u) 1

(!;u) 1

= Z

1

d

exp [rX℄(!;u) 1

dr:

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Hen e

jf(+h;x) f(;x)j kfk

L 1

(N()) Z

N()

X

(!;u) 1

d!du

Ckfk

L 1

(N()) jhj:

Lemma 4.3. Let g 2 L 1

(V) and let

n

be as in (4.1) be su h that lim

n!1 hg;

n i

exists. Then for every 2L 1

(V) su h that R

V

6=0 we have

lim

n!+1 hg

V ;

n

i= lim

n!+1 hg;

n i

Z

V :

Proof. Assume R

=1. We have h g;

n

i=hg;

n i and

jhg;

n

i hg;

n

ijkgk

L 1

(V) k

n n

k

L 1

(V) :

But

n (x)

n (x)=

Z

V

(y)

n

(x y)

n (x)

dy:

Hen e

jhg;

n

i hg;

n

ijkgk

L 1

(V) Z

V Z

V j (y)j

x y

k

n

!

(x)

dxdy!0:

Lemma 4.4. Let f = e

f, with e

f 2 L 1

(N()), 2 S(N()) and let

n

be as in

(4.1). There is a subsequen e fk

n g

n=1;2;:::

su h that for every 2Z

lim

n!+1 hf

;

n

i=H() exists:

Proof. Let

1

;

2

;:::be a dense subset of Z. Using the above lemma and the diagonal

method,wesele tasubsequen e

n

su hthatlim

n!+1 hf

j

;

n

iexists forallj. Hen e,

by Lemma 4.2, the limitexists for all 2Z.

Lemma 4.5. Assume that F satises the assumptions of Theorem 3.3 and let its

boundary value be f = e

f, e

f 2L 1

(N()), 2S(N()). Thenthe fun tion

Z 3 !H()= lim

n!1 hf

;

n i

is bounded and harmoni with respe t to the Eu lidean Lapla ian on Z, so it is a

onstant fun tion.

Proof. Forpositivet let p

t

(;u)be the fundamental solutionof

e

L

t

= X

j;

( g

X

j )

2

+( g

Y

j )

2

t :

Werestri t our fun tion F to the submanifold

f(;x)exp[tH℄:(;x)2N(); H = r

X

j=1 H

j

; t2Rg:

More pre isely, leta

0

=expH and

e

F(;u;t)=F(;u;e;ta ); t>0:

(11)

Then

(LF)(;u;e;ta

0 )=t(

e

L

t )

e

F(;u;t):

Hen e

e

F(;u;t)=F(;u;e;ta

0

)=f p

t (;u):

On the otherhand,

F(;u;e;ta

0 )=f

V P

ta

0 (u)

and so by Lemmas 4.4and 4.3

H()= lim

n!1 hf

;

n

i= lim

n!1 hf

V P

ta

0

;

n i

= lim

n!1 hfp

t (;);

n i:

(4.6)

But

hf p

t (;);

n i =

Z

Z Z

V Z

V

f( !;u v 2=(;!))p

t

(!;v)dv

n

(u)dud!

= Z

Z hf

!

V p

t

(!;u+2=(;!));

n id!

Now

lim

n!+1 hf

!

V p

t

(!;u+2=(;!));

n

i=H( !) Z

V p

t

(!;u)du:

Therefore, by (4.6)

H()=H

Z p

t ();

(4.7)

where p

t () =

R

V p

t

(;u) du and

Z

denotes the Abelian onvolution on Z. Sin e p

t

is the usual heat kernel orresponding to the Lapla e operator on Z, (4.7) says that

H()is harmoni on Z, soit is onstant.

Tosummarize, the above lemmasallowtoformulate the mainresult of this se tion:

Theorem 4.8. Assume that F satises the assumptions of the main theorem and let

its boundary value be f = e

f, e

f 2L 1

(N()), 2 S(N()). Denote by f(;

^

) the

distributional partial Fourier transform of f along V. Let

n

(x) = k d

n

(k

n x) k

d

n

(k 1

n x);

f

n

=

n

V f;

F

n

(;x;y;a) = (f

n )

V P

ya (x):

Then

suppf

n

(;^)Z f2V : 2k

n

jj k 1

n g;

F

n

is annihilated by L, H and L and there is a onstant su h that the sequen e F

n

tends to F + .

Proof. We sele t fk

n g

n=1;2:::

as inLemma 4.4. Then by Lemmas 4.3and 4.5

lim

n!+1 n

V f

V P

ya

(x)= lim

n!+1 hf

;

n i= :

Sin e fk d

(k

n

)g

n=1;2;:::

is an approximate identity,the on lusion follows.

(12)

5. Tube domains

In this se tion we are going toprovethe main theorem for the tubedomains.

Theorem 5.1. LetF bea realboundedfun tiononatubedomainD

T

su hthatL F =

0 and HF =0. Then F is pluriharmoni .

Let f be the boundary value of F. In view of the previous se tion we may assume

that for an">0

supp

^

f f"<jj<"

1

g (5.2)

Sowe restri t ourselves to this lass of fun tions. We are going toshow that

supp

^

f [ : (5.3)

Assume (5.3) has been proved. From (5.2) and (5.3) itfollows that

f =f

1 +f

2

with

^

f

1

=

^

fj

;

^

f

2

( )=

^

f

1

() f

1

;f

2 2L

1

(V):

(Noti e that f

j

=

j

V

f for appropriate S hwartz fun tions

j

on V.

^

1

is 1 on a

neighbourhoodof supp

^

f \

,supp

^

1

\

=; and

^

2 ()=

^

1

( )). Then

F =F

1 +F

2

with F

1

(xya)=f

1

V P

ya

(x); F

2

(xya)=f

2

V P

ya (x);

F

1

is holomorphi ,F

2

isanti-holomorphi and F

2

=

F

1

. To see the latter, we write

f

n

=e i

1

n hx;ei

f

1

(x); g

n

=e i

1

n hx;ei

f

2 (x);

so

supp

^

f

n

; supp^g

n

when e

\

f

n

V P

ya ()=

^

f

n ()e

h;yaei

;

\

g

n

V P

ya

()=g^

n ()e

h;yaei

whi h implies that F

1

is a limit of holomorphi fun tions f

n

V P

ya , F

2

is a limit of

anti-holomorphi fun tions g

n

V P

ya .

Forthe proof of (5.3) we need few lemmas.

Lemma 5.4. For every , the fun tion ya7!

b

P

ya

() is smooth.

Proof. Let P(xya) =P

ya

(x). Then LP = 0 and so, there is an ellipti operator b

L in

variables ya su h that

[

(LP)

ya ()=

b

L b

P

ya ():

Lemma 5.5. For 2supp

^

f

b

P

ya

()=e P

r

i=1 a

i jW

i (;y)j

;

where W(;y)=h;Ad i:

(13)

Proof. By(3.4), for every j we have

0=

j

F(xya)=

j (f P

ya

(x)) =f (

j P)

ya (x);

where P(xya) = P

ya

(x) and (

j P)

ya

(x) =

j

P(xya). Now we use the following

theorem of Wiener [Ru℄: Let f 2 L 1

(V), g 2 L 1

(V) and f

V

g = 0. Then supp

^

f

f:g()^ =0g. Hen e if 2 supp

^

f we have

0=

\

(

j P)

ya

()=a 2

j

2

a

j

h;Ad

y e

j i

2

b

P

ya ():

Sin e j

^

P

ya

()j 1 for every 2 V, ya 2 S

0

, the uniqueness of the solution of the

above system of equations ompletes the proof of Lemma5.5.

Fora we write

ij

, 1j in for its oordinates.

Lemma 5.6 (DHMP). Let y2N. Then

Ad

y

j

=

j +y

j

+L(e

j )(y

j

y j

)

and so

W

j

(;y)=h;

j i+

X

k>j h

jk

;y

jk i+

X

j<kl <r h

kl

;L(e

j )(y

jk

y

jl )i:

In the set of pairs of natural numbers we introdu e the lexi ographi order:

(i;j)<(k;l),(i<k)_(i=k ^ j <l)

Lemma 5.7. If 2= [ , then there exist m and y

1

;y

2

2N su hthat

W

m (;y

1

)>0 and W

m (;y

2 )<0:

Proof. Let(k;l)=maxf(i;j):

ij

6=0)g. Then

Case 1: k =l=1.

This means that onlyone oordinateof is not equaltozero, when e 2[ .

Case 2: k <l.

Suppose y

ij

=0for (i;j)6=(k;l). Then by Lemma(5.6)

W

k

(;y)=

kk +h

kl

;y

kl i;

be ause L(e

k )(y

kl

y

kl )2V

l l and

l l

=0: Consequently, we an hoose y

kl

twi e in

su h away that W

k

hanges sign.

Case 3: 1<k =l.

Wemay assume

kk

<0. From 2= itfollows that there exists x2 su h that

h;xi>0:

Also,by Proposition(VI.3.5) [FK℄,there is y2N su h that

x=Ad

y (

r

X

j=1

j

j );

where

j

>0. Consequently,

0<h;xi=h;Ad

y (

r

X

j

j )i=

r

X

j h;Ad

y

j i:

(5.8)

(14)

Noti e that by Lemma (5.6), W

p

(;y)=0 for p>k and W

k

(;y)=

kk

:Indeed,

L(e

p )(y

p

y p

)2 M

p<l sr V

l s

so h;L(e

p )(y

p

y p

)i=0:

Wehave alsoh;y p

i=0for pk. It follows from(5.8) that thereis m<k su hthat

W

m

(;y)=h;Ad

y

m i>0

We are going to exhibit a y 0

su h that W

m (;y

0

) < 0. We assume that y

ij

=0 for

(i;j)6=(m;k). Then

W

m

(;y) =

mm +h

mk

;y

mk i+h

kk

;L(e

m )(y

mk

y

mk )i

=

mm +h

mk

;y

mk i+

kk hy

mk

;y

mk i:

But

kk

<0, sosele ting y

mk

suÆ iently large we obtainthe y 0

.

Corollary 5.9. Under assumptions of Theorem (5.1)

supp

^

f

[

:

Proof. If 2 supp

^

f \( [ )

, then on one hand side the fun tion y ! b

P

ya ()

is smooth, and on the other, one of the W

k

(;y)'s hanges sign, whi h ontradi ts

smoothnessof the fun tionand so(5.3) follows.

6. The Heisenberggroup

In this se tion we prove the main theorem in the ase when Z = C n

, V = R,

=R +

and

(;!)= 1

4

h;!i= 1

4 n

X

j=1

j

!

j :

Then

D=f(;z)2C n

C:=z>

1

4 jj

2

g

may be identied with N()A, where A = R +

and N() is the Heisenberg group

ZV:

(;u)(

0

;u 0

)=(+ 0

;u+u 0

+ 1

2

=h; 0

i):

In this ase we have onlytwo operators

H=

1

=X 2

+H 2

H

i.e.

1

F(;u;a)=a 2

(

2

u +

2

a

)F(;u;a); (6.1)

L=L

1

= X

X

) 2

+

Y

) 2

nH (6.2)

i.e.

LF(;u;a)=a(L

B n

a

)F(;u;a);

(6.3)

where

L

B

= X

f

X

2

+

e

Y

2

:

(15)

LetF beareal valuedfun tiononS su hthat

1

F =LF =0. Letf beitsboundary

value. By (6.1) and (6.2) wehave

F(;u;a)=f

H n

p

a

(;u)=f

R P

a (u):

(6.4)

where p

a

(;u) is the fundamental solution for L

B n

a

and P

a (u) =

1

a

a 2

+u 2

is the

Poisson kernel for the Lapla e operator 2

u +

2

a

. By Theorem (4.8), we may assume

that for apositive"

supp f(;

^

)C n

f:"<jj<"

1

g:

Now pro eedingas atthe beginningof se tion 5we take

f

j

=

j

R f

(u):

(6.5)

and

F

j

(;u;a)=f

j

H n

p

a

(;u)=(f

j )

R P

a (u):

(6.6)

Then

F =F

1 +F

2

with F

2

=

F

1

and it remainsto prove

Theorem 6.7. F

1

is holomorphi .

TheproofisbasedontheelementarytheoryofunitaryrepresentationsoftheHeisen-

berg group for whi h we referto [T℄. LetU

be the S hrodinger representation ofH n

,

([T℄, 1.2.1). In the underlying Hilbert spa e H

=L 2

(R n

) we onsider the basis on-

sisting of properly s aled Hermite fun tions

(1.4.18and se tion2.1of [T℄). Let

;

(;u)=(U

(;u)

;

):

Then

;

(;u)=(2) n=2

e iu

; (

q

jj);

(6.8)

where

;

are the spe ial Hermite fun tions, ([T℄, 1.4.19) These fun tions belong to

the S hwartz lass onC n

and

L

B

;

(;u)= (2j j+n)jj

; (;u):

(6.9)

Let

e

k

(;u)= X

jj=k

; (;u)

and

k

(;u)= Z

R e

k

(;u)()d:

(6.10)

for a 2C 1

(Rnf0g).

Then, k

2S(C n

R) and by (6.9)

L

B k

= (2k+n) k

0

;

0

(16)

Lemma 6.11. For every k 6=0 and 2C 1

(Rnf0g)

Z

H n

F

1

(;t;a) k

(;t)ddt=0:

(6.12)

Proof. By(6.5) and (6.6)

F

1 (;

^

;a)=f

1 (;

^

)e a

:

Hen e

(

u +i

a )F

1

(;u;a)=0

and so

(L

B in

u )F

1

(;u;a)=0:

(6.13)

Let 2C 1

(Rnf0g)and e

() = 1

(). By (6.13), (6.10)and (6.8),

0 = Z

H n

F

1

(;t;a)(L

B in

u )

k

e

(;t)ddt

= 2k Z

H n

F

1

(;t;a) k

(;t)ddt

and (6.12) follows.

UsingLebesgue dominated onvergen e theorem, by (6.12),we have

Z

H n

f

1 (;t)

k

(;t)ddt=0:

(6.14)

for k 6=0 and 2C 1

(Rnf0g).

Furthermore, f

1

translated by any element (;u) 2 H n

on the left is the boundary

value of F

1

translated onthe left by (;u). Therefore, the same proof giveus:

f

1

k

(;u)= Z

H n

f

1

((;u)(;t)) k

(;t)ddt=0:

(6.15)

for k 6=0 and 2C 1

(Rnf0g).

ForanL 2

(H n

)fun tionf

1

,(6.15)meansthatallthespe tralproje tionsf

1

e

k

vanish

for k 6= 0. So, learly (6.15) an be viewed as a weak version of that. For a good

fun tion f

1

, the next step would be an appli ation of the Fourier inversion formula

to F

1

. Sin e we are not in L 2

(H n

) we have to do it in a slightly more deli ate way.

Namely, we expand the fun tion

g(;u)=

1

R p

a (;u)

and we get

g(;u)=(2) n 1

1

X

k=0 Z

1

1 g

H n

e

k

(;u)jj n

d

(6.16)

(see theorem 2.1.1 [T℄), where the above series onverges inL 2

(H n

) norm.

Proposition 6.17.

Z

1

g

H n

e

k

(;u)jj n

d2S(H n

) (6.18)

(17)

and the series

1

X

k=0 Z

1

1 g

H n

e

k

(;u)jj n

d

(6.19)

onverges in L 1

(H n

).

Toprovethis propositionweneed moreinformationabout

;

. LetL

k

bethe k-th

Laguerrepolynomial,i.e.

L

k (t)e

t

= 1

k!

d

dt

!

k

e t

t k

:

Given amultiindex=(

1

;:::;

n

) and 2C n

let

L

()=L

1 (

1

2 j

1 j

2

):::L

n (

1

2 j

n j

2

):

(6.20)

Then

;

()=(2) n=2

L

()e

1

4 jj

2

; (6.21)

(see [T℄ 1.4.20)

We willuse the following well-known property of the Laguerrefun tions.

Lemma 6.22. For every l;p2N there exist = (l;p) and M =M(l;p) su h that

Z

1

0

(1+t) l

j

p

t L

k (t)j

2

e t

dt k M

: (6.23)

Proof. Toverify (6.23) we re all 5.1.13and 5.1.14in [Sz℄, whi h implythat

d

dt L

k (t)=

k 1

X

j=0 L

j (t):

Hen e, it suÆ es to have (6.23) for p = 0. But, for p = 0, (6.23) follows by the

orthogonalityand the re urren e relations(5.1.1,5.1.10 in [Sz℄)

Z

1

0 L

j (t)L

k (t)e

t

dt =Æ

j;k

;

tL

k

=(2k+1)L

k

(k+1)L

k+1

(k 1)L

k 1 :

Proof of proposition (6.17). For(6.18) we prove that

Z

1

1 g

H n

e

k

(;u)jj n

d = k

a

k (;u) (6.24)

with

a

k

()=e (

2k

n +1)jja

b

1 ()jj

n

:

Indeed,

g

H n

e

k

(;u)= X

jj=k Z

1

R p

a (;t)

; ((;t)

1

(;u))ddt

= X

(U

(;u)

;U

1

R pa

):

(6.25)

(18)

But

U

1

R pa

= b

1 ()U

pa

and

U

pa

=e (

2jj

n +1)jja

(6.26)

(To obtain (6.26) it is enough to solve the equation (L

B n

a )p

a

(;u) = 0 on the

Fourier transformside).

Now putting(6.26) into (6.25)we get

g

H n

e

k

(;u)= X

jj=k e

( 2k

n +1)jja

b

1 ()

; (;u)

whi h implies (6.24).

To estimateL 1

(H n

) norm of k

a

k

we write

k

a

k

(;u)=(2) n

2 Z

R e

iu

a

k ()

X

jj=k

; (

p

)d:

Hen e the S hwartz inequalityyields

Z

H n

j k

a

k

(;u)jdud Z

H n

(1+u 2

)j k

a

k (;u)j

2

dud

0

Z

C n

Z

R j

(

a

k ()

X

jj=k

; (

p

))j 2

dd+ Z

C n

Z

R j

a

k ()

X

jj=k

; (

p

)j 2

dd

1

A

1 e

2 (

2k

n +1)a

X

jj=k Z

C n

[";"

1

℄

j

(

; (

p

))j 2

+j

; (

p

)j 2

dd

!

Nowthe relation(6.21)and lemma(6.22)implythat thelastintegralis dominatedby

k M

and so (6.19) follows.

Now we are able to expandF

1

. By(6.15) and proposition(6.17) we have

F

1

(;u;a) = f

1

H n

(

1

R p

a )(;u)

= X

k f

1

H n

k

a

k (;u)

= f

1

H n

0

a

0 (;u);

where

0

a

0

(;u)=(2) n

2 Z

1

0 e

a

0;0 (;u)

b

1 ()

n

d:

Soit suÆ es toprovethat

G(;u;a)=e

a

0;0 (;u)

is holomorphi . We have,

G(;u;a)=(2) n=2

e

a

e ix

e 1

4

jj 2

=(2) n=2

e

i(x+ia+

1

4 ijj

2

)

=e iz

(19)

7. Type II domains

LetF beafun tion onS that satises(3.6) and theassumptions of Theorem(3.3).

In viewof Theorem(4.8) and (5.3) wemay assume thatits boundaryvaluef satises

supp f

1

Z(\f:"

1

>jj>"g):

Then

F(;u;y;a)=F

1

(;u;y;a)+F

2

(;u;y;a);

where

{ LF

1

=LF

2

=0,

{ F

2

=F

1 ,

{ for every xed ,F

1

isholomorphi onS

T

and F

2

anti-holomorphi on S

T ,

{ the boundary value f

j of F

j is f

j

=

j

V

f for

j

asin Se tion5.

We are goingto prove the following

Theorem 7.1. Let F be a bounded fun tion on S su h that

(X

j +iH

j

)F = 0 for j =1;:::;r;

(7.2)

(X

ij +iY

ij

)F = 0 for 1i<j r; =1;:::;d;

(7.3)

LF = 0:

(7.4)

Then F is holomorphi . If instead of (7.2), (7.3), (7.4) we have

(7:2 0

) (X

j iH

j

)F = 0 for j =1;::: ;r

(7:3 0

) (X

ij iY

ij

)F = 0 for 1 i<j r; =1;:::;d

(7:4) LF = 0

then F is anti-holomorphi .

Clearly F

1

satises (7.2)-(7.4), while F

2

satises (7.2'), (7.3') and (7.4). The proof

of the rst part and the se ond part are identi al, so we show only the rst one. We

doit intwo steps formulatedin the followinglemmas.

Lemma 7.5. Let

L

j

= X

L

j :

Assume that a bounded fun tion F satises (7.2)-(7.4). Then for every j, L

j

F =0.

Lemma 7.6. Let F be a bounded fun tion on S satisfying (7.2), (7.3) and

L

j

F =0 for j =1;:::;r:

(7.7)

Then F is holomorphi .

Proof of Lemma (7.5). Let

F

ya

(;u)=F(;u;y;a):

By(7.2) and (2.10)

a

F(;u;y;a)=iAd

y (

f

X

j )F

ya (;u):

(20)

Therefore, (7.4) implies

X

j;

a

j

Ad

y (

f

X

j )

2

+

Ad

y (

e

Y

j )

2

i

Ad

y (

f

X

j )

2

!

F

ya

=0:

(7.8)

We x y and j and we take a

j

= t, a

k

= t 2

for k 6= j. Then dividing (7.8) by t and

letting t tend tozero we obtain

D

j;y f =

X

Ad

y (

f

X

j )

2

+

Ad

y (

e

Y

j )

2

i

Ad

y (

f

X

j )

2

!

f =0:

(7.9)

We do this for every j. D

j;y

is a left-invariant operator onN(). We shall show that

(7.9) implies

D

j;y F

s

=0:

(7.10)

for every s

0 2S

0

; y2N

0

. Then taking s=ya in (7.9) weobtain (7.5).

Sin e F

s

(;u)=f

P

s

(u), it remainstoprove the followingstatement:

Let D be a left-invariant dierential operator on N() su h that Df =0. Then for

every s2S

0

DF

s

(;u)=D(f

V P

s

(u))=0:

(7.11)

We have

f

V P

s (x)=

Z

V

f(;x u)P

ya

(u)du= Z

V

f((0; u)(;x))P

ya (u)du:

Hen e

D(f

V P

s (x)) =

Z

V

(Df)((0; u)(;x))P

ya (u)du;

(7.12)

provided we an justify the hange of the order of integration and dierentiation.

But, sin e f = e

f, right-invariantdierentialoperators on N() applied tof yield

bounded fun tionsonN(). Therefore, for every left-invariantoperatorD,jDf(;x)j

is dominated by a polynomial depending only on . This proves (7.12), (7.11) and

ompletes the proof of Lemma (7.5).

Proof of Lemma (7.6). We xj and we onsider the group

S

j

=exp h

linfH

j

;X

j

;X

j

;Y

j

=1;:::;ng i

:

S

j

a ts simply transitively onthe Siegel upper-half plane des ribed in Se tion 6. As-

sume that F satisesthe assumptionsof Lemma (7.6) and letF

j

=Fj

Sj

. Then

(X

j +iH

j )F

j

=0 and L

j F

j

=0:

Consequently, by Se tion6, F

j

is holomorphi onS

j , i.e.

(X

j +iY

j )F

j

=0 for every =1;:::;n:

(7.13)

If insteadof F we take F

x

1

=F(x

1 x),x

1

;x2S,then (7.14) implies

(X

j +iY

j

)F =0:

(7.14)

(21)

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Instytut Matematy zny, Uniwersytet Wro lawski, Pla Grunwaldzki 2/4, 50-384

Wro law, Poland

E-mailaddress: dburamath.uni.wro .pl

same address inWro law

E-mailaddress: edamekmath.uni.wro .pl

same address inWro law

E-mailaddress: hulani kmath.uni.wro .pl