Finite rank intermediate Hankel operators on the Bergman space

© Hindawi Publishing Corp.

FINITE-RANK INTERMEDIATE HANKEL OPERATORS

ON THE BERGMAN SPACE

TAKAHIKO NAKAZI and TOMOKO OSAWA

(Received 11 January 1998)

Abstract.LetL2_=L2_{(D,r dr dθ/π)be the Lebesgue space on the open unit disc and} letL2

a=L2∩Ᏼol(D)be the Bergman space. LetPbe the orthogonal projection ofL2onto L2

aand letQbe the orthogonal projection onto ¯L2a,0= {g∈L2; ¯g∈L2a, g(0)=0}. Then I−P≥Q. The big Hankel operator and the small Hankel operator onL2

a are defined as: forφinL∞_,Hbig

φ (f )=(I−P)(φf )andHφsmall(f )=Q(φf )(f∈L2a). In this paper, the finite-rank intermediate Hankel operators betweenHbig_φ andHsmall_φ are studied. We are working on the more general space, that is, the weighted Bergman space.

2000 Mathematics Subject Classification. Primary 47B35.

1. Introduction. LetDbe the open unit disc inCand letdµbe the finite positive Borel measure onD. LetL2_=L2_(µ)=L2_{(D,dµ)andᏴol(D)be the set of all} holo-morphic functions onD. The weighted Bergman spaceL2

a=L2a(µ)is the intersection ofL2_{andᏴol(D). In general,L}2

a is not closed. In [6, Theorem 8], when(suppµ)∩D is a uniqueness set forᏴol(D), the first author and M. Yamada gave a necessary and sufficient condition for thatL2

ais closed. Throughout this paper, we assume thatL2a is closed. Whendµ=r dr dθ/π,L2

ais the usual Bergman space. Forµsuch thatL2

a(µ)is closed, whenᏹis the closed subspace ofL2(µ)andzᏹ⊆ᏹ, ᏹis called an invariant subspace. Suppose thatᏹ⊇zL2_a.Pᏹ_{denotes the orthogonal} projection fromL2_{ontoᏹ. Fo rφinL}∞_=L∞_(µ)=L∞_{(D,dµ), the intermediate Hankel} operatorHᏹ

φis defined by

Hᏹ φf=

I−Pᏹ_{(φf ) f∈L}2 a

. (1.1)

Whenᏹ=L2

a,Hᏹφis called a big Hankel operatorHbigφ and whenᏹ=

¯

z¯L2

a⊥,Hᏹφis called a small Hankel operatorHsmall

φ . Note thatHᏹφis called a little Hankel operator whenᏹ=¯L2

a⊥.

For arbitrary symbolφinL∞_{, in the case ofdµ=r dr dθ/π, bothH}big

φ andHsmallφ were studied when they are compact operators or Schatten class operators (see [12]). However it seems to have not been studied when they are finite-rank operators. When

¯

φis inL2

a, it is known (see [12, page 155]) that ifHbigφ is a finite-rank operator, then Hbigφ =0 and if ¯φis a polynomial, thenHsmallφ is a finite-rank operator. In this paper, for arbitrary symbolφinL∞_{we show that ifH}big

φ is a finite-rank operator, thenHbigφ =0, and we study whenHsmall

φ is a finite-rank operator. In fact, we study such problems for the intermediate Hankel operators Hᏹ

In [2, 7, 9, 10], intermediate Hankel operators were studied in special weights,dµ=

(α+1)(1−r2₎α_{r dr dθ/πfor−1< α <∞. In particular, Strouse [9] studied finite-rank} intermediate Hankel operators.

Letdµ=dσ (r )dθbe a Borel measure onD, wheredσ (r )is a positive measure on[0,1)with dσ ([0,1))=1/2π and dθ is the Lebesgue measure on∂D. L2

a(µ)is closed ifdσ ([t,1)) >0 for anyt >0 (see [6]). For this type measures, it is possible to study more precisely the intermediate Hankel operators. In fact,L2_{has the following} orthogonal decomposition:

L2₌ ∞ j=−∞

⊕ᏸ2_eijθ_{, (1.2)}

whereᏸ2_=L2_{(dσ )=L}2_{([0,1),dσ ). Set}

H2₌∞ j=0

⊕ᏸ2_eijθ_{, (1.3)}

thenL2

a ⊂H2⊂(z¯¯L2a)⊥ and L2=H2⊕e−iθH¯2. Ifᏹ=H2, it is easy comparatively to determine finite-rank Hankel operatorsHᏹ

φand we can do it completely in Section 5. We can expect thatHᏹ

φis close toHbigφ in caseᏹ⊆H2(see Section 5) andHᏹφis close toHsmall

φ in caseᏹ⊇H2(see Section 6).

In Section 2, we describe an invariant subspace inL2

a whose codimension is of fi-nite. Moreover we show that there does not exist an invariant subspace which contains

L2

a properly and in whichL2ais of finite codimension. We also give a lot of examples of invariant subspaces which containL2

a and in which Hankel operators are studied in this paper. In Section 3, we describe finite-rank intermediate Hankel operators for arbitrary measureµ such that L2

a(µ)is closed. Moreover, we show that there does not exist any nonzero finite-rank Hankel operatorsHbigφ and there exists a nonzero finite-rank Hankel operatorHsmallφ . In fact, we give twonecessary and sufficient con-ditions for that ifHᏹ

φ is of finite rank≤, thenHᏹφ=0. In Sections 3, 4, and 5, we use the Fourier coefficients{ᏹj}∞j=−∞ofᏹand sowe assumedµ=dσ (r )dθ. Using the Fourier coefficients ofφandᏹ, we give a necessary and sufficient condition for thatHᏹ

φis of finite rank≤. Assuming thatφis a harmonic function, we can get a better necessary and sufficient condition. Whenᏹ⊆H2_{, using the Fourier coefficients}

{ᏹj}∞j=−∞, we give a necessary condition and a sufficient condition for that ifHᏹφis of finite rank≤, thenHᏹ

φ=0. Two conditions are very similar but are a little different. Applications are given to examples in Section 2.

2. Invariant subspaces. In this section, we assume that dµ = dσ (r )dθ and

dσ ([t,1)) >0 for anyt >0, except Propositions 2.1 and 2.2. For our purpose, the invariant subspaceᏹmust containzL2

a but kerHᏹφis an invariant subspace inL2a. If Hᏹ

φis of finite rank, then the codimension of kerHᏹφinL2ais finite. In order to study finite-rank intermediate Hankel operators, we need the generalization of a result of Axler and Bourdon [1] which determines finite codimensional invariant subspaces in

L2

a whendµ=r dr dθ/π. In Propositions 2.1 and 2.2, the measureµ is an arbitrary finite positive Borel measure such that L2

Proposition 2.3 is a corollary of [4, Theorem 1]. We will give several examples of in-variant subspaces which containzL2

a.

Proposition 2.1. Supposeᏹ is an invariant subspace in L2

a and is a positive integer. The codimension ofᏹinL2

ais, if and only ifᏹ=qL2a, whereq=

j=1(z−aj) andaj∈D (1≤j≤).

Proof. The proof is almost parallel to that in [1, Theorem 1]. We will give a sketch of it. Supposeᏹ⊥_=L2

aᏹand dimᏹ⊥=. Put

Szf=P(zf ) f∈ᏹ⊥, (2.1)

wherePis an orthogonal projection. Since <∞, there exists an analytic polynomial

b such thatb(Sz)=Sb(z)=0 and the degree of b is less than or equal to. Hence

bᏹ⊥_{⊆ᏹand sobL}2

a⊆ᏹ. We show that the zeros ofbare only inDand the degree ofb=. Thenᏹ=bL2

a. It is clear that the degree ofb=. In this direction, we did not need the condition such that(suppµ)∩Dis a uniqueness set.

Ifa∉D,(z−a)L2

a is dense inL2a. Assuminga≥1 and soa=1 without a loss of generality, ifε >0, then(z−1)L2

a=(z−1){z−(1+ε)}−1L2a. For anyf∈L2a, it is easy

tosee that

D

_z−z₍−₁1_+ε)f−f2dµ →0 (ε→0). (2.2) This implies that(z−1)L2

ais dense inL2a. Thus all zeros ofbmust be inD. The “if” part is clear because any pointa∈Dgives a bounded evaluation functional. Here we used the condition such that(suppµ)∩Dis a uniqueness set (see [6, (1) of Theorem 8]).

Proposition2.2. Suppose that(z−a)−1_{does not belong toL}2_{for eacha∈D. Ifᏹ} is an invariant subspace which containsL2

aproperly, then the codimension ofL2ainᏹ is infinite.

Proof. If dimᏹL2

a= <∞, by the proof of Proposition 2.1, there exists a poly-nomial b=j=1(z−aj) such that bᏹ⊆L2a and aj ∈D (1≤j ≤). Hence there exists a functionφinᏹsuch thatφ∉L2

aandg=bφ∈La2. Ifg(ak)=0 fo r so mek, theng/(z−ak)=φj=k(z−aj)cannot belong toL2because(z−ak)−1∉L2. Hence

g(aj)=0 for anyj. By [6, the proof in (1) of Theorem 8],g∈bL2a and soφ=g/b belongs toL2

a. This contradiction implies that dimᏹL2a= ∞. For an invariant subspaceᏹ, set

ᏹj=

fj∈ᏸ2; f∈ᏹ, f (z)= ∞

j=−∞

fj(r )eijθ

. (2.3)

Thenᏹj is a subspace inᏸ2, rᏹj⊆ᏹj+1and hence dimᏹj+1≥dimᏹj. We call

{ᏹj}∞j=−∞the Fourier coefficients ofᏹ.ᏹjeijθmay not belong toᏹ. Ifᏹjeijθbelongs toᏹfor anyj, thenᏹhas the following decomposition:

ᏹ=

∞

j=−∞

This decomposition is called the Fourier decomposition ofᏹ. In general,ᏹdoes not have the Fourier decomposition but we can get an extension ˜ᏹofᏹwhich has the following Fourier decomposition:

˜ ᏹ=

∞

j=−∞

⊕(closure ofᏹj)eijθ. (2.5)

Proposition2.3. Ifᏹis an invariant subspace which containsL2

a andeiθᏹ⊆ᏹ, thenᏹ=χEq¯H2⊕χEcL2, whereχ_Eis a characteristic function inᏸ2andqis a

unimod-ular function inH2_{. Henceᏹ⊇H}2_{. If} ∞

j=0eijθᏹ= {0}, thenᏹ=q¯H2.

Proof. SupposeS0= ᏹeiθᏹ, then ᏹ=∞j=0⊕S0eijθ⊕ᏹ−∞, where ᏹ−∞= ∞

j=0eijθᏹ, andr S0⊂S0becauserᏹj⊆ᏹj+1. It is well known thatᏹ−∞=χGL2for a characteristic functionχF of some measurable subset inD. PutE=Gc then there exists a functionfinS0such that

|f|>0 o nE and f=0 o nF. (2.6)

Sincefis orthogonal tof eijθ_{for allj≥0.|f|}2_{belongs toᏸ}1_=L1_{(dσ )=L}1_{([0,1),dσ )} and so|f|belongs toᏸ2_{. Henceχ}

Ebelongs toᏸ2. Set

Fr eiθ₌       

fr eiθ _f

r eiθ iff≠0,

1 iff=0,

(2.7)

then F is a unimodular function inL2_{. Since r S}

0⊆S0, we can show thatχEF be-longs toS0and soS0=χEFᏸ2. Henceᏹᏹ∞=χEFH2. Since 1∈ᏹ,χEF¯∈H2and

q=F¯∈H2_,

Example2.4. (i) For 0< β <1, put

Tβ=spanznz¯m; βn≥m≥0. (2.8)

ThenTβ is an invariant subspace andTβ⊇L2a. Put Tβ=L2a forβ=0 and Tβ=H2 forβ=1. In general, L2

a ⊆Tβ ⊆H2 and Tβ (0≤β <1) has the following Fourier decomposition:

Tβ= ∞

j=0

⊕Tβjeijθ, (2.9)

where (Tβ)j = span{rjpj(r2); pjis a polynomial of degree at mostβj/(1− β)}. Janson and Rochberg [2] studiedHᏹ

φ whenᏹ=(T¯β)⊥. Then(T¯β)⊥=eiθH2⊕∞j=0⊕

{ᏸ2_(T¯

β)j}e−ijθ.

(ii) Fork≥0, putEk_=span{zm_z¯n_{; m=0,1,...,k;n=m,m+1,...}. ¯E}k_{is an} invari-ant subspace andL2

a⊆E¯k⊆H2. ¯Ekhas the following Fourier decomposition: ¯

Ek₌∞ j=0

⊕E¯k

jeijθ, (2.10)

where (E¯k₎

j =span{rj,...,rj+2k}. Strouse [9] studied Hᏹφ when ᏹ = (Ek)⊥. Then

(Ek₎⊥_=eiθ_H2_⊕∞

(iii) Fix a polynomialpof degreek, that is,p=kj=0ajzj. Put

Y (p)=spanzn_{, z}m_{p¯; n≥0, m≥0,}

Yk_=spanzz¯j_{;≥0,0≤j≤k.} (2.11)

BothY (p)andYk_{are invariant subspaces andL}2

a⊆Y (p)⊆Yk, andYkhas the fol-lowing Fourier decomposition:

Yk₌ ∞ j=−k

⊕Yk

jeijθ, (2.12)

where Yk

0 = span1,r2,...,r2k and (Yk)j = rj(Y0k) for j ≥ 0, and (Yk)−j = span{r2−j_{;j ≤≤k}for 1≤j≤k.(Y (p))j⊆(Y}k_{)j for any j but Y (p)does not} have a Fourier decomposition. Ifaj=0 fo r 1≤j≤k,(Y (p))j=(Yk)j for anyj and so ˜Y (p)=Yk_{. Peng, Rochberg, and Wu [7] and Wang and Wu [10] studiedH}ᏹ

φ when ᏹ=(Y¯k₎⊥_{. In general, we can defineY (g)for any functionginL}2_{. Usually,Y (g)does} not have the Fourier decomposition.

(iv) For a unimodular functionqinH2_{, putᏹ=q¯H}2_{. Thenᏹis an invariant subspace} which containsH2_{. In general, ¯qH}2_{may not have the Fourier decomposition but for}

q=eiθ_{, fo r so me≥0,}

ᏹ=

∞

j=−

⊕ᏸ2_eijθ_{. (2.13)}

There are a lot of invariant subspaces betweenH2_ande−iθ_H2_{even if=1.}

(v) For arbitrary closed subspacesSinᏸ2_{, putᏹ=H}2_⊕Se−iθ_{. Thenᏹis an invariant} subspace betweenH2_ande−iθ_H2_.

3. Kronecker’s theorem. In this section, the measureµis an arbitrary finite posi-tive Borel measure such thatL2

ais closed. We will write

ᏹ∞_=ᏹ∩L∞ _(3.1)

and, for each positive integer,

ᏹ∞,_=φ∈L∞_{; φ(z)=g(z)} j=1

(z−aj)−1a.e.µonD,g∈ᏹ∞anda1,...,a∈D

.

(3.2) Thenᏹ∞_⊆ᏹ∞,1_⊆ᏹ∞,2_{⊆ ···.}

Kronecker (cf. [11, page 210]) described finite-rank Hankel operators on the Hardy space. Theorem 3.1 describes finite-rank intermediate Hankel operators on the (weighted) Bergman space. However the situation is very different from that of Kronecker becauseᏹ∞_=ᏹ∞,_{may happen for some >0. See Corollaries 3.3 and 3.4.}

Theorem3.1. Supposeᏹis an invariant subspace which containszL2

a, andφis a function inL∞_{. H}ᏹ

φis of finite rank≤if and only ifφbelongs toᏹ∞,. Proof. Note that kerHᏹ

φ= {f∈L2a; φf ∈ᏹ}. Sinceᏹis an invariant subspace, kerHᏹ

Theorem3.2. Supposeᏹis an invariant subspace which containsL2

a, andφis a function inL∞_{. Then the following are equivalent:}

(1)If Hᏹ

φis of finite rank, then Hᏹφ=0. (2)ᏹ∞_=ᏹ∞,_{for any >0.}

(3)Ifg∈ᏹ∞_{,a∈Dand(g(z)−g(a))/(z−a)∈L}∞_{, then(g(z)−g(a))/(z−a)} belongs toᏹ∞_.

(4)Ifᏹ_{is an invariant subspace and(ᏹ)}∞_ᏹ∞_{, then there does not exist a nonzero} polynomialbsuch thatb(ᏹ₎∞_⊆ᏹ∞_.

Proof. By Theorem 3.1, (1)(2) is clear.

(1)⇒(3). If there existsg∈ᏹ∞_{such that(g−g(a))/(z−a)∈L}∞_{does not belong to} ᏹ∞_{, putφ=(g−g(a))/(z−a), thenH}ᏹ

φis of rank 1 andHᏹφ=0.

(3)⇒(4). If (4) is not true, there existsψsuch thatψ∉ᏹ∞_,ψ∈(ᏹ)∞_andbψ∈ᏹ∞ for some polynomial:b=j=1(z−aj)andaj∈D(1≤j≤ <∞). We may assume thatφ=ψj−1=1(z−aj)∉ᏹ∞andg=(z−a)φ∈ᏹ∞. Then

g−ga

z−a =φ∈L

∞_{, φ∉ᏹ}∞_{. (3.3)}

(4)⇒(1). By Theorem 3.1, if Hᏹ

φ is of finite rank ≤, then φ∈ᏹ∞,. Ifφ∉ ᏹ∞, suppose ᏹ _{is an invariant subspace generated by φ and ᏹ, then (ᏹ)}∞_ᏹ∞ _but there does not exist a nonzero polynomialb such thatb(ᏹ₎∞_⊆ᏹ∞_{. Sinceφ∈ᏹ,} this contradicts thatφ∈ᏹ∞,_.

Corollary3.3. Suppose(suppµ)∩Dis a uniqueness set forᏴol(D). If Hbig_φ is of finite rank, then Hbigφ =0.

Proof. Theorem 3.2(3) implies the corollary. In fact, ifg∈L2

a∩L∞, theng(z)−

g(a)∈(z−a)L2

a by [6, the proof in (1) of Theorem 5.4]. Thus(g(z)−g(a))/(z−a) belongs toL2

a∩L∞.

Corollary3.4. Supposedµ=r dr dθ/π. LetD0be an open subset ofDandᏹ=

{f∈L2_{; f is analytic onD}

0}. Thenᏹis an invariant subspace and if Hᏹφ is of finite rank then Hᏹ

φ=0.

Proof. It is easy tosee thatᏹ∞_{satisfies Theorem 3.2(3).} Corollary3.5. Suppose that if Hᏹ

φis of finite rank then Hᏹφ=0. Ifᏹis an invariant subspace which contains ᏹproperly, then the codimension of ᏹin ᏹ _{is infinite or}

(ᏹ₎∞_=ᏹ∞_.

Proof. If dimᏹ_{/ᏹ< ∞, as in the proof of Proposition 2.2, then there exists} a nonzero polynomial b such that bᏹ _{⊆ ᏹ. Hence b(ᏹ)}∞ _⊆ᏹ∞_{. If (ᏹ)= ᏹ}∞_, by Theorem 3.2, this contradicts that ifHᏹ

φis of finite rank, thenHᏹφ=0.

Theorem 4.1. Supposeᏹ is an invariant subspace which containszL2

a andφ= ∞

j=−∞φj(r )eijθis a function inL∞. Then Hᏹφ is of finite rank≤if and only if there exist complex numbersb0,...,bsuch thatb=1and, for any integern,

j=0

bjrjφn−j(r )∈ᏹn. (4.1)

Ifis the minimum number of complex numbersb1,...,bsuch thatj=0bjrjφn−j(r )

∈ᏹnfor alln, then Hᏹφis of rank.

Proof. IfHᏹ

φis of rank≤, by Theorem 3.1 there exists a polynomialb=

j=0bjzj such thatbφ∈ᏹ. Then

  ∞

j=−∞

φj(r )eijθ    

j=0

bjrjeijθ  ₌ ∞

n=−∞  

j=0

φn−j(r )bjrj 

_einθ _(4.2)

and soj=0bjrjφn−j(r )∈ᏹnfor anyn. The converse and the second statement are clear by Theorem 3.2.

Corollary4.2. Letφ=φt(r )eitθ for some integertin Theorem 4.1. Then Hᏹφis of finite rank≤if and only if there exist complex numbersb0,...,b such thatb=1 and fort≤n≤+t, bn−trn−tφt(r )∈ᏹn.

Proof. Sinceφj(r )=0 fo rj=t, ifn < torn > +t, thenj=0bjrjφn−j(r )=0. Fort≤n≤+t,j=0bjrjφn−j(r )=bn−trn−tφt(r ), thus the corollary follows.

Corollary4.3. Letφ=∞j=1ajzj+∞j=0a−j¯zjin Theorem 4.1. Then Hᏹφis of rank

≤if and only if there exist complex numbersb0,...,b such thatb=1and for any nonpositive integernj=0bjan−jr2j−n∈ᏹn and, for 0< n < , j=nbjan−jr2j−n

∈ᏹn.

Proof. Ifn≥andn=0, then

j=0

bjrjφn−j(r )=

j=0

bjan−jrj+n−j=

j=0

bjan−j

rn _(4.3)

and hencej=0bjrjφn−j(r )∈ᏹn becausezL2a ⊆ᏹ. Now Theorem 4.1 implies the corollary.

Theorem 4.1 does not give an exact relation between the rank ofHᏹ

φand the num-ber of complex numbers b0,...,b such that b =1. However, we can show the following: ifHᏹ

φis of rank, then there exist complex numbersb0,...,b such that

b =1, j=0bjrjφn−j(r )∈ᏹn for anynandb=j=0bjzj has justzeros in D. That is, if=1, then|b0|<1.

By Theorem 4.1,Hᏹ

φ=0 if and only ifφn∈ᏹnfor anyn(i.e.,φ∈ᏹ). Moreover, Hᏹ

φis of rank≤1 if and only if there exist complex numbers(b0,b1)=(0,0)such that

5. Big Hankel operator andᏹ⊆H2. In this section, we assume thatdµ=dσ (r )dθ

anddσ ([t,1)) >0 for anyt >0. Hence we can define the Fourier coefficients{ᏹj}∞j=−k of ᏹ and we assumeᏹ=ᏹ˜. In this case, Hᏹ

φ is close to Hbigφ . Recall examples in Section 2, that is,Tβ,E¯k,Y (p), andYk.

Corollary5.1. Supposeᏹis an invariant subspace betweenzL2

aandH2, andφ= ∞

j=1ajzj+∞j=0a−jz¯j. Then Hᏹφis of finite rank≤if and only ifa−n=0forn > and there exists complex numbersb0,...,b such thatb=1andj=nbjan−jr2j−n∈ᏹn for0≤n≤andj=0bjan−jr2j−n=0for− < n <0.

Proof. Sinceᏹ⊆H2_{, by Corollary 4.3H}ᏹ

φis of finite rank≤if and only if there exist complex numbersb0,...,bsuch thatb=1 andj=0bjan−jr2j−n=0 fo rn <0 and j=nbjan−jr2j−n∈ᏹn for 0≤n≤. Ifj=0bjan−jr2j−n=0 fo rn <0, then

bjan−j=0 fo r 0≤j≤andn <0. Hence for eachj (0≤j≤), bja−t=0 ift > j. Thusa−t=0 ift > .

Proposition 5.2. Supposeᏹis an invariant subspace between zL2

a ande−ikθH2 wherek≥0, andφ=∞j=0φ−j(r )e−ijθ is a function inL∞. Then Hᏹφis of finite rank

≤if and only if

φ(z)=

j=−kψj(r )eijθ

j=0bjrjeijθ

, (5.1)

whereψn=j=0bjrjφn−j∈ᏹn, for−k≤n≤, and(b0,...,b)∈C. Proof. Note thatᏹ⊆e−ikθ_H2_andφ

j(r )=0 fo rj >0. IfHᏹφis of finite rank≤, then, by Theorem 4.1,

 

j=0

bjrjeijθ _∞

j=0

φ−j(r )e−ijθ  ₌

n=−k

ψn(r )einθ (5.2)

and ψn = j=0bjrjφn−j ∈ᏹn for −k≤ n≤ . The converse is also a result of Theorem 3.1.

Corollary 5.3. Suppose ᏹ is an invariant subspace in Proposition 5.2. If φ =

φ++φ−=∞j=1ajzj+∞j=0a−jz¯j andφ−∈L∞, then Hᏹφis of finite rank≤if and only if

φ(z)=φ++

j=−kψj(r )eijθ

j=0bjrjeijθ

, (5.3)

where ψn = j=0bjan−jrj+|n−j| ∈ ᏹn, for −k ≤ n ≤ , and (b0,...,b)∈ C. If

(b0,...,b)=(0,...,0), thenψn=0and soφ=φ+.

Theorem5.4. Supposeᏹis an invariant subspace betweenzL2

aande−ikθH2where

k≥0, andφ=∞j=1φ−j(r )eijθis a function inL∞.

(1)Ifᏹj∩rj+1ᏸ2= {0}for anyj≥0, then there does not exist any finite rank Hᏹφ except Hᏹ

φ=0.

(2)If there does not exist any finite rank Hᏹ

φexcept Hᏹφ=0, thenᏹ−(k−j)∩rj+1ᏸ∞=

Proof. (1) IfHᏹ

φis of finite rank, by Proposition 5.2,

ψn=

j=n

bjrjφn−j∈ᏹn, (5.4)

for 0≤n≤because φn−j(r )=0 fo r 0≤j ≤n−1. We may assume b =1. As

n=−1, r_φ

−1(r )∈ᏹ−1. Sinceᏹ−1∩rᏸ2= {0}, φ−1(r )=0. Asn=−2,

b−1r−1φ−1(r )+rφ−2(r )∈ᏹ−2. (5.5)

Sinceᏹ−2∩r−1ᏸ2= {0}andφ−1(r )=0, φ−2(r )=0. we can getφ−j(r )=0 fo r

j≤. In Proposition 5.2,ψn=0 fo r 0≤n≤and soφ≡0. (2) Ifrj+1_g∈ᏹ−

(k−j)∩rj+1ᏸ∞, then putφ=ge−i(k+1)θ. Ifg=0 thenφ∉ᏹand

zj+1_φ=rj+1_ge−i(k−j)θ_∈ᏹ−

(k−j)e−i(k−j)θ. (5.6) Sinceᏹhas the Fourier decomposition,ᏹjeijθ⊆ᏹand sozj+1φ∈ᏹ. Theorem 3.1 gives a contradiction.

We will apply results in this section to Example 2.4 in Section 2. Example5.5. (i) Supposeᏹ=Tβ(0≤β <1).

(1) Whenφ=∞j=1φ−j(r )e−ijθ is a function inL∞, there does not exist any finite rankHᏹ

φexceptHᏹφ=0 if and only ifβ=0.

(2) Whenφ=∞j=0ajzj+∞j=1a−j¯zjis a function inL∞, there does not exist any finite rankHᏹ

φexceptHᏹφ=0 if and only ifβ=0.

Proof. Recall that Tβ = ∞j=0⊕(Tβ)jeijθ and (Tβ)j = span{rjpj(r2); pjis a polynomial of degree at mostβj/1−β}.

(1) If β=0, then (Tβ)j∩rj+1ᏸ2= {0} for anyj ≥0 and if β=0, then (Tβ)j∩

rj+1_ᏸ∞_{= {0}for enough largej. Theorem 5.4 implies (1).}

(2) Ifβ=0, then there existsnsuch that 1−β≤β(n−1). Hence(Tβ)n−1rn+1. Sup-pose φ = ¯z, then zn_{φ = r}n+1_ei(n−1)θ _{and so z}n_{φ ∈ (T}

β)n−1ei(n−1)θ ⊂ Tβ. By Theorem 3.1,Hᏹ

φis of rank≤nandHᏹφ=0. (ii) Supposeᏹ=E¯m_{(0≤m <∞).}

(1) When φ=∞j=1φ−j(r )e−ijθ, there does not exist any finite rank Hᏹφ except Hᏹ

φ=0 if and only ifm=0.

(2) Whenφ=∞j=0ajzj+∞j=1a−j¯zjis a function inL∞, there does not exist any finite rankHᏹ

φexceptHᏹφ=0 if and only ifm=0 o r 1. Proof. We recall that(E)¯ m₌∞

j=0⊕(E¯m)jeijθand(E¯m)j=span{rj,...,rj+2m}. (1) Ifm=0, then(E¯m₎

j∩rj+1ᏸ2= {0}for anyj≥0 and ifm=0, then(E¯m)j∩

rj+1_ᏸ∞_{= {0}for anyj≥0. Theorem 5.4 implies (1).} (2) Ifm=0, by (1) there does not exist any finite rankHᏹ

φexceptHᏹφ=0. Ifm=1, then(Em_)n=span{rn_,rn+2_{}forn≥0. WhenH}M

φis of finite rank, by Corollary 5.1,

a−n=0 fo rn > and if 0≤n≤,

j=n

for complex constantsc,d. Hence, for 0≤n≤,

bjan−j=0 fo rn+2≤j≤. (5.8) Sinceb=1, an−=0 fo r 0≤n≤and soa−j=0 fo r 0≤j≤. Whenm≥2, if

φ=z¯, thenzφ=r2_∈(E¯m₎

0=span{1,r2,...,r2m}andzφ∈E¯mbecause(E¯m)0⊂E¯m. HoweverHᏹ

φ=0. (iii) Supposeᏹ=Yk_.

(1) When φ=∞j=1φ−j(r )e−ijθ, there does not exist any finite rank Hᏹφ except Hᏹ

φ=0 if and only ifk=0.

(2) Whenφ=φ++φ−=∞j=0ajzj+∞j=1a−j¯zjandφ+are functions inL∞, there does not exist any finite rankHᏹ

φexceptHᏹφ=0 if and only ifk=0. Proof. Since Hᏹ

φ = Hᏹφ−, it is sufficient toprove (1). We recall that Yk =

∞

j=−k⊕(Yk)jeijθ, whereY0k=span{1,r2,...,r2k}and(Yk)j=rj(Yk)0forj≥0, and

(Yk₎

−j =span{r2−j, j≤ ≤ k} for 1≤ j ≤ k. If k=0, then Yk =L2a. If k≥1,

(Yk₎

−k=span{rk}. Theorem 5.4(2) implies that there exists a nonzero finite rankHᏹφ.

6. Small Hankel operator and ᏹ ⊇ H2_{. In this section, we assume that dµ =}

dσ (r )dθand dσ ([t,1)) >0 for anyt >0. Hence we can define the Fourier coeffi-cients{ᏹj}∞j=−∞ofᏹ. In this case,Hᏹφis close toHsmallφ and far fromHbigφ . Note that if ᏹ_{is an invariant subspace andᏹ⊆e}iθ_H2_{, thenᏹ=(ᏹ¯)}⊥_{is an invariant subspace} andᏹ⊇eiθ_H2_.

Proposition6.1. Supposeᏹ is an invariant subspace which containseikθ_H2_for some nonnegative integerk. Ifᏹ=L2_{, there exists at least a nonzero finite rank H}ᏹ

φ.

Proof. If ¯zn_{∈ᏹfor alln≥1, thenzz¯}n_{∈ᏹfor all≥1 becausezᏹ⊆ᏹ. LetᏱ} be the closed linear span of{z_z¯n_{;n≥1, ≥0}, thenᏱ⊆ᏹandgᏱ⊆Ᏹfor arbitrary} polynomialgofzand ¯z. It is well known thatᏱ=L2_{. This contradiction implies that} there exists at leastnsuch that ¯zn_{∉ᏹandn≥1. Ifφ=z¯}n_{, thenz}n+k_{φ∈ᏹ. Then} Hᏹ

φ=0 butHᏹφis of finite rank≤n+k, by Theorem 3.1.

Proposition6.2. Supposeᏹ is an invariant subspace which containseikθ_H2_for some nonnegative integerk. The following statements are valid.

(1)Ifφ=∞j=−∞φj(r )eijθis a function inL∞, then there exists a functionφinL2 such thatφ₌k−1

j=0φj(r )eijθ+∞j=1φ−j(r )e−ijθand Hᏹφ=Hᏹ_φ.

(2)Ifφ=∞j=kφj(r )eijθis a function inL∞, then Hᏹφ=0.

(3) If φ=∞j=−φj(r )eijθ is a function in L∞, then Hᏹφ is of rank ≤+k <∞. Conversely, if one of (1) or (2) is valid, thenᏹcontainseikθ_H2_.

Proof. Both (1) and (2) are clear becauseᏹ⊇eikθ_H2_{. (3) is a result of Theorem 3.1.} The converse is also clear.

We will consider Example 2.4 in Section 2.

Example6.3. (ii) Suppose ᏹ=(Ek₎⊥_{(0≤k <∞)and φ=}∞

(1)Hᏹ

φ=0 if and only if 1

0φ−j(r )r

j+2t_{dσ=0 (j≥0,0≤t≤k). (6.1)}

(2)Hᏹ

φ is of rank≤1 if and only if there exist complex numbers(b0,b1)=(0,0) such that

b0 ₁

0φ−j(r )r

j+2t_{dσ= −b} 1

₁

0φ−j−1(r )r

j+2t+1_{dσ (6.2)}

forj≥0,0≤t≤k.

(3) Supposedσ=r dr /2π. Whenφ=∞j=0ajzj+∞j=1a−jz¯j, ifHᏹφis of rank≤1, thenHᏹ

φ=0.

Proof. From the remark in the last part of Section 4, (1) and (2) follows. (3) By (2), Hᏹ

φis of rank≤1 if and only if there exist complex numbers(b0,b1)=(0,0)such that

b0a−j _2j+1_2t+1= −b1a−j−1 _2j+1_2t+3 (6.3) forj≥0,0≤t≤k. Whenk=0, for eachj, ast=0,

b0a−j _2j1₊₁= −b1a−j−1 _2j1₊₃,

b0a−j _2j1₊₃= −b1a−j−1 _2j1₊₅.

(6.4)

This implies that a−j =a−j−1=0, for j ≥0, and so φ=∞j=1ajzj. When k=0, Corollary 3.3 implies (3)

(iv) Supposeᏹ=q¯H2_{for some unimodular functionqinH}2_{andφis a function} inL∞_.Hᏹ

φis of finite rankif and only if

φ=q¯ ∞

j=−

ψj(r )eijθ, (6.5)

whereψ−(r )=0.

Proof. Ifφ=q¯∞j=−ψj(r )eijθ, thenzφ∈ᏹand so, by Theorem 3.1,Hᏹφ is of finite rank≤. Sinceψ−(r )=0, bφ∉ᏹfor any polynomialbof degree≤−1 and soHᏹ

φis of finite rank. The converse is clear.

(v) Supposeᏹ=H2_⊕Se−iθ_{andSis a closed subspace inᏸ}2_{. Letφ=}∞

j=−∞φj(r )eijθ be a function inL∞_{. By Theorems 3.1 and 4.1,H}ᏹ

φ is of finite rank≤if and only if

φj(r )=0 fo rj≤ −(+2)and there exist complex numbersb0,...,bsuch thatb=1,

j=0

bjrjφn−j(r )=0 fo r −(+1)≤n <−1,

j=0

bjrjφ−1−j(r )∈S.

7. Restricted shift operator andᏹ⊆L2

a. In this section, we assumeµ=r dr dθ/π for simplicity. Letᏹbe an invariant subspace inL2

a and᏷=L2aᏹ. Fo rφinL∞a =

L2 a∩L∞,

S᏷ φf=

I−P᏷_{(φf ) (f∈᏷), (7.1)} whereP᏷ _{is the orthogonal projection from L}2

a to᏷. Sz᏷ is called a restricted shift operator. For any φin L∞

a, Sφ᏷ commutes with S᏷z. We do not know whether if the bounded linear operatorT on᏷commutes withS᏷

z, thenT =Sφ᏷ for someφinL∞a. If T S᏷

z =Sz᏷T and φ=T P᏷1 is bounded, then it is easy to see thatT =Sφ᏷ (cf. [5, page 784]). In the Hardy space instead of the Bergman space, Sarason [8] showed that this is true without any condition andT = φ∞.

We can define the Hankel operatorHᏹ

φ as in the introduction. HoweverHᏹφis not an intermediate Hankel operator. It is not so difficult to see the following: when᏷=

L2

aᏹandφinL∞a,

Hᏹ

φ = Sφ᏷. (7.2)

This is known for the Hardy space. In fact, forfinL2 a,

Hᏹ

φf=I−Pᏹφf=P᏷φP᏷f (7.3)

and soHᏹ

φf=Sφ᏷P᏷f forfinL2a. HenceHᏹφis of finite ranknif and only ifSφ᏷is of finite rankn. It is easy tosee thatS᏷

φis of finite rank≤nif and only if there exists an analytic polynomialp of degree ≤nsuch thatp(φ)∈ᏹ∞_{. Whenφis in L}∞_, Theorems 3.1 and 4.1 are true forHᏹ

φ. Supposeφis a function inL∞

a. (1)L2

a⊇kerHᏹφ⊇ᏹ.

(2) When the common zero setZ(ᏹ)ofᏹinDis empty, ifHᏹ

φis of finite rank then Hᏹ

φ=0. This is a result of (1) and Proposition 2.1.

(3) IfZ(ᏹ)is not empty, there exists a nonzero finite rankHᏹ φ.

Acknowledgement. This research was partially supported by Grant-in-Aid for scientific research, Ministry of Education.

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Takahiko Nakazi: Department of Mathematics, Hokkaido University, Sapporo060, Japan