Volume 2007, Article ID 39819,11pages doi:10.1155/2007/39819
Research Article
On a Class of Composition Operators on Bergman Space
Namita Das, R. P. Lal, and C. K. Mohapatra
Received 4 May 2006; Revised 14 December 2006; Accepted 15 December 2006
Recommended by Manfred H. Moller
LetD= {z∈C:|z|<1}be the open unit disk in the complex planeC. LetA2(D) be the
space of analytic functions onDsquare integrable with respect to the measuredA(z)=
(1/π)dx d y. Givena∈Dand f any measurable function onD, we define the function Caf byCaf(z)=f(ϕa(z)), whereϕa∈Aut(D). The mapCais a composition operator onL2(D,dA) andA2(D) for alla∈D. Letᏸ(A2(D)) be the space of all bounded linear
operators fromA2(D) into itself. In this article, we have shown thatC
aSCa=Sfor all a∈Dif and only ifDS(ϕ a(z))dA(a)=S(z), where S∈ᏸ(A2(D)) andSis the Berezin
symbol of S.
Copyright © 2007 Namita Das et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
1. Introduction
LetD= {z∈C:|z|<1}be the open unit disk in the complex planeC. Let dA(z) be the area measure onDnormalized so that the area of the disk Dis 1. In rectangular and polar coordinates,dA(z)=(1/π)dx d y=(1/π)rd rdθ. LetL2(D,dA) be the Hilbert space
of Lebesgue measurable functions f onDwith
f2=
D
f(z)2
dA(z)
1/2
<+∞. (1.1)
The inner product is defined as
f,g =
forf,g∈L2(D,dA). The spaceL∞(D,dA) will denote the Banach space of Lebesgue
mea-surable functions f onDwith
f∞=ess supf(z):z∈D <∞. (1.3)
The Bergman spaceA2(D) is defined to be the subspace ofL2(D,dA) consisting of
ana-lytic functions. It is not so difficult to verify that (see [1])A2(D) is a closed subspace of
L2(D,dA). Since point evaluation atz∈Dis a bounded linear functional on the Hilbert
spaceA2(D), the Riesz representation theorem [2] implies that there exists a unique
func-tion KzinA2(D) such that
f(w)=
Df(z)Kz(w)dA(w) (1.4)
for allf inA2(D). LetK(z,w) be the function onD×Ddefined byK(z,ω)=K
z(w). The functionK(z,w) is called the Bergman kernel ofDand it can be verified that (see [3])
K(z,ω)= 1
(1−zw)2. (1.5)
Letka(z)=K(z,a)/
K(a,a)=(1− |a|2)/(1−az)2. The functionk
ais called the normal-ized reproducing kernel forA2(D). It is clear thatk
a2=1. Let Aut(D) be the Lie group
of all automorphisms(biholomorphic mappings) ofD. We can define for eacha∈Dan automorphismϕain Aut(D) such that
(i) (ϕa◦ϕa)(z)≡z, (ii)ϕa(0)=a,ϕa(a)=0,
(iii)ϕahas a unique fixed point inD.
In fact,ϕa(z)=(a−z)/(1−az) for allaandz inD. An easy calculation shows that the derivative ofϕa at z is equal to −ka(z). It follows that the real Jacobian determi-nant ofϕaatzis Jϕa(z)= |ka(z)|2=(1− |a|2)2/|1−az|4. Givenλ∈Dand f any
measur-able function onD, we define a functionUλf onDbyUλf(z)=kλ(z)f(ϕλ(z)). Notice that Uλ is a bounded linear operator on L2(D,dA) and A2(D) for all λ∈D. Further, it can be checked that U2
λ =I, the identity operator,Uλ∗=Uλ,Uλ(A2(D))⊂(A2(D)), andUλ((A2(D))⊥)⊂(A2(D))⊥for allλ∈D. ThusUλP=PUλfor allλ∈D, whereP is the orthogonal projection fromL2(D,dA) onto A2(D). Letφ:D→Dbe analytic.
De-fine the composition operator Cφ from A2(D) into itself as Cφf = f ◦φ. ThenCφ is a bounded linear operator onA2(D) andC
φ ≤(1 +|φ(0)|)/(1− |φ(0)|) (see [3] for a proof). Givena∈Dand f any measurable function onD, we define the function Caf byCaf(z)= f(ϕa(z)), where ϕa∈Aut(D). The mapCa is a composition opera-tor onA2(D). Letᏸ(A2(D)) be the space of all bounded linear operator fromA2(D)
into itself. For T∈ᏸ(A2(D)), define the map T on D asT(z)= Tk
z,kz. The map B:ᏸ(A2(D))→L∞(D) defined byB(T)=T is called the Berezin transform andT is
called the Berezin symbol of T.
Berezin transform are so far mainly in the study of Toeplitz and Hankel operators on the Bergman space. Regardless of the original motivation of Berezin for introducing it, the Berezin transform essentially provides a kind of “symbol” for certain natural operators on Hilbert spaces of analytic functions. Thus it is natural to ask the general question of how much information about the operator does its Berezin symbol carry. The problem is subtle and no general answer is known. In this work, we have shown that the Berezin symbol of a bounded linear operatorSfrom the Bergman space into itself satisfies cer-tain averaging condition if and only if the operatorS satisfy the intertwining relation CaSCa=Sfor alla∈D. Recently, the spectra of composition operators have attracted much attention (see [5–7]) from operator theorists. To this purpose, it is important to know what are the essential commutants of the invertible operators Ca,a∈D, and to characterize thoseS∈ᏸ(A2(D)) such thatC
aS−SCa=(CaSCa−S)Cais compact for all a∈D. In this work, we present a necessary and sufficient condition forCaSCa−S=0 to happen for alla∈Din terms of the Berezin symbol ofS. Related work in this area can be found in [5–8].
2. The unitary operatorUaand the Berezin transform
In this section, we will prove certain elementary properties of the unitary operatorUa and the Berezin transform.
Lemma 2.1. Forz,ω∈D,Uzkω=αkϕz(ω)for some complex constantαsuch that|α| =1.
Proof. Supposez,ω∈D. If f ∈A2(D), then
f,UzKω
=Uzf,Kω
=Uzf
(ω)= −f ◦ϕz
(ω)ϕz(ω)=f,−ϕ
z(ω)
Kϕz(ω)
. (2.1)
ThusUzKω= −ϕz(ω)Kϕz(ω). Rewriting this in terms of the normalized reproducing
kernels, we have
Uzkω=αkϕz(ω) (2.2)
for some complex constantα. SinceUz is unitary andkω2= kϕz(ω)2=1, we obtain
that|α| =1.
Lemma 2.2. For alla∈D,Uaka=1.
Proof. Ifa∈D, then first observe thatϕa(z)= −ka(z). Since (ϕa◦ϕa)(z)=zfor allz∈D, taking derivatives with respect tozin both sides, we obtain
Uaka
(z)=ka
ϕa(z)
ka(z)=1. (2.3)
Notice that for alla∈D, sinceUaka=1, henceka◦ϕa=1/kaandka−1∈H∞, the space of bounded analytic functions onD.
Proof. IfS(z) =T(z) for all z∈D, then
(S−T)kz,kz
=0 (2.4)
for allz∈D. This implies
(S−T)Kz,Kz
=K(z,z)(S−T)kz,kz
=K(z,z)·0=0. (2.5)
LetL=S−Tand define
F(x,y)=LKx,Ky. (2.6)
The functionFis holomorphic inxand yandF(x,y)=0 ifx=y. It can now be veri-fied that such functions must vanish identically. Letx=u+iv,y=u−iv. LetG(u,v)=
F(x,y). The functionGis holomorphic and vanishes ifuandvare real. HenceF(x,y)=
G(u,v)≡0. Thus evenLKx,Ky =0 for anyx,y. Since linear combinations ofKx,x∈D,
are dense inA2(D), it follows thatL=0. That is,S=T.
3. Main result and its applications
In this section, we will prove that a bounded linear operatorS fromA2(D) into itself
commutes with all the composition operatorsCa,a∈D, if and only ifSsatisfies certain averaging condition. We will also present some applications of this result.
Theorem 3.1. A bounded linear operatorS∈ᏸ(A2(D)) commutes with all the composition operatorsCa,a∈D, if and only if
S(z)=
DS
ϕa(z)
dA(a) (3.1)
for allz∈D.
Proof. SupposeS(z) =DS(ϕ a(z))dA(a) for allz∈D.
Then byLemma 2.1, there exists a constantαsuch that|α| =1 for allz∈D,
Skz,kz
=
D
Skϕa(z),kϕa(z)
dA(a)=
D
αSUakz,αUakz
dA(a)
=
D
UaSUakz,kz
dA(a)=
DUaSUadA(a)
kz,kz
=Skz,kz
,
(3.2)
whereS=DUaSUadA(a).
Thus byLemma 2.3,S=S. Hence for all f,g∈A2(D),S f,g = S f ,g.
That is,
D
SUaf,Uag
dA(a)=
The boundedness ofSand the antianalyticity ofK(z,a) inaimply that for eachz∈D, the function
S
f
K(·,a)
(z)K(z,a) (3.4)
is antianalytic ina. Therefore, by the mean value property of harmonic functions, we have
DS
f
K(·,a)
(z)K(z,a)dA(a)=S
f
K(·, 0)
(z)K(z, 0)=S f(z). (3.5)
Thus, from (3.5), it follows that
S f,g =
Dg(z)
DS
f
K(·,a)
(z)K(z,a)dA(a)dA(z). (3.6)
Using Fubini’s theorem, we obtain
S f,g =
D
DS
f
K(·,a)
(z)g(z)K(z,a)dA(z)dA(a). (3.7)
Now sinceka(z)=K(z,a)/
K(a,a) and (ka◦ϕa)(z)ka(z)=1 for allz,a∈D, the right-hand side of (3.7) is equal to
D DS f ka
(z)g(z)ka(z)dA(z)dA(a)
= D DS f ka
(z)g(z)ka
ϕa(z)ka(z)2dA(z)dA(a).
(3.8)
Finally, as (ϕa◦ϕa)(z)≡zand Jϕa(z)= |ka(z)|2, we obtain
S f,g = D DS f ka
ϕa(z)ka(z)gϕa(z)dA(z)dA(a). (3.9)
By hypothesis,S f,g =DSUaf,UagdA(a) and by usingLemma 2.2,
SUaf,Uag
=
S
f◦ϕ
a ka◦ϕa
,g◦ϕa
ka = S f
ka◦ϕa
,g◦ϕa
ka = DS f
ka◦ ϕa
(z)gϕa(z)
ka(z)dA(z).
(3.10)
Thus we obtain for all f,g∈A2(D),
DS
f
ka◦ϕa
(z)gϕa(z)
ka(z)dA(z)=
DS
f
ka
ϕa(z)
ka(z)g
ϕa(z)
Hence for all f,g∈A2(D),a∈D,
S
f
ka◦ϕa
,Uag
= S f ka
◦ϕa,Uag
. (3.12)
SinceUais unitary,Ua∈ᏸ(A2(D)), we get
S
f
ka◦ϕa
=S
f
ka
◦ϕa (3.13)
for all f ∈A2(D),a∈D.
That is, for all f ∈A2(D),a∈D,
SCa
f
ka
=CaS
f
ka
. (3.14)
Sincek−1
a ∈H∞, henceSCa=CaSfor alla∈D. ThusCaSCa=Sfor alla∈DasCa2=I, the identity operator inᏸ(A2(D)).
Now we will prove the converse. SupposeCaSCa=Sfor alla∈D. ThenCaS f=SCaf for alla∈D,f ∈A2(D). That is, for all f ∈A2(D),a∈D,
(S f)◦ϕa=S
f ◦ϕa
. (3.15)
ByLemma 2.2, (ka◦ϕa)ka=1 for alla∈D. Hence
SUaf =Skaf ◦ϕa=S
f ◦ϕ
a ka◦ϕa
=S
f
ka
◦ϕa
= S f ka
◦ϕa. (3.16)
Thus for f,g∈A2(D), sincek
a(ϕa(z))ka(z)=1, Jϕa(z)= |ka(z)|
2, andk
a(z)=K(z,a)/
K(a,a) for allz,a∈D, we obtain
SUaf,Uag
= D S f ka
ϕa(z)
g◦ϕa
(z)ka(z)dA(z)
= DS f ka
(z)g(z)ka◦ϕa
(z)ka(z)2dA(z)
= DS f ka
(z)g(z)ka(z)dA(z)
=
DS
f
K(·,a)
(z)g(z)K(z,a)dA(z).
(3.17)
Hence by using Fubini’s theorem, we obtain
D
SUaf,UagdA(a)=
D
DS
f
K(·,a)
(z)g(z)K(z,a)dA(z)dA(a)
= Dg(z)dA(z) DS f
K(·,a)
(z)K(z,a)dA(a).
We have already checked in the first part of the proof that for allz∈D,
DS
f
K(·,a)
(z)K(z,a)dA(a)=S
f
K(·, 0)
(z)K(z, 0)=S f(z). (3.19)
Thus
D
SUaf,Uag
dA(a)=
DS f(z)g(z)dA(z)= S f,g. (3.20)
When f =g=kz,z∈D, we obtain byLemma 2.1that
Skz,kz
=
D
SUakz,Uakz
dA(a)=
D
Skϕa(z),kϕa(z)
dA(a)=
DS
ϕa(z)
dA(a), (3.21)
and this concludes the proof.
LetPbe the orthogonal projection fromL2ontoA2(D). Forϕ∈L∞(D,dA), define the
Toeplitz operatorTϕfromA2(D) into itself asTϕf =P(ϕ f). Forϕ∈L∞(D,dA), let
ϕ(z)=
Dϕ
ϕa(z)
dA(a),
ϕ(z)=
Dϕ
ϕz(ω)
dA(ω).
(3.22)
Notice that
ϕ(z)=ϕkz,kz
. (3.23)
Corollary 3.2. Ifϕ∈L∞(D,dA), then there exists a constantcof modulus 1 such that
D
Dϕ
ϕϕa(z)(ω)
dA(ω)dA(a)=
D
Dϕ
cϕϕz(a)(ω)
dA(a)dA(ω). (3.24)
Proof. From (3.23), it follows that
DTϕ
ϕa(z)dA(a)=
D
Tϕkϕa(z),kϕa(z)
dA(a)=
D
ϕkϕa(z),kϕa(z)
dA(a)
=
Dϕ
ϕa(z)
dA(a)=
D
Dϕ
ϕϕa(z)(ω)
dA(ω)dA(a).
Given f,g∈A2(D), byLemma 2.2and Fubini’s theorem, we obtain
D
UaTϕUaf,gdA(a)
=
DdA(a)
Dϕ(z)
f◦ϕa
(z)ka(z)
g◦ϕa
(z)ka(z)dA(a)
=
DdA(a)
Dϕ
ϕa(ω)
f(ω)g(ω)ka◦ϕa
(ω)2ka(ω)2dA(ω)
=
DdA(a)
Dϕ
ϕa(ω)f(ω)g(ω)dA(ω)
=
Df(ω)g(ω)dA(ω)
Dϕ
ϕa(ω)
dA(a)
=
Dϕ(ω) f(ω)g(ω)dA(ω).
(3.26)
Thus
DTϕ
ϕa(z)dA(a)=
D
UaTϕUakz,kzdA(a)=
Dϕ(ω)
kz(ω)2
dA(ω)
=
Dϕ
ϕz(ω)
dA(ω)=
D
D
ϕ◦ϕa◦ϕz
(ω)dA(a)dA(ω). (3.27)
Thus byTheorem 3.1, we obtain
D
Dϕ
ϕϕa(z)(ω)
dA(ω)dA(a)=
D
Dϕ
ϕa◦ϕz
(ω)dA(a)dA(ω). (3.28)
Let
U=ϕa◦ϕz◦ϕϕz(a). (3.29)
Then U∈Aut(D) and U(0)=ϕa◦ϕz(ϕz(a))=ϕa(a)=0 and Uϕϕz(a)=ϕa◦ϕz.
We know that (see [9]) ifϕ∈Aut(D), then
ϕ(z)=eiθ z−b
1−bz (3.30)
for someθreal andb∈D. Furthermore,ϕ(0)=0 if and only ifϕ(z)=eiθz. Thus Uz= eiθz. Hence
ϕa◦ϕz=Uϕϕz(a)=eiθϕϕz(a)=cϕϕz(a), (3.31)
wherec=eiθ,θ∈R. Thus it follows that
D
Dϕ
ϕϕa(z)(ω)
dA(ω)dA(a)=
D
Dϕ
cϕϕz(a)(ω)
dA(a)dA(ω). (3.32)
Notice that we can defineUaonL2(D,dA) also. Supposeϕ∈L∞(D,dA), f,g∈L2(D, dA). Then by using Fubini’s theorem and making a change of variable, we obtain
D
ϕUaf,UagdA(a)=
DdA(a)
Dϕ(z)
f◦ϕa(z)ka(z)g◦ϕa(z)ka(z)dA(z)
=
DdA(a)
Dϕ
ϕa(ω)
f(ω)g(ω)dA(ω)
=
Df(ω)g(ω)dA(ω)
Dϕ
ϕa(ω)dA(a)
=
Dϕ(ω) f(ω)g(ω)dA(ω)= ϕ f ,g.
(3.33)
DefineJ:L2→L2 asJ f(z)= f(z). The mapJ is a unitary operator andJ∗=J. Let
A2(D)= {f :f ∈A2(D)}. Defineh
ϕ:A2(D)→A2(D) such thathϕf =P(ϕ f), wherePis the orthogonal projection fromL2ontoA2(D). The operatorh
ϕis called the little Hankel operator onA2(D). The following holds.
Corollary 3.3. Ifϕ∈L∞(D), f ∈A2(D),g∈A2(D), then
D
UahϕUaf,gdA(a)=hϕf,g
. (3.34)
Proof. From the above discussion it follows that for f ∈A2(D),g∈A2(D),
D
ϕUaf,UagdA(a)= ϕ f ,g. (3.35)
That is,
D
ϕUaP f,UaPg
dA(a)= ϕP f ,Pg. (3.36)
Notice thatP=JPJ. Therefore,
D
ϕUaP f,UaJPJg
dA(a)= ϕP f ,JPJg. (3.37)
SinceUaP=PUa, we obtain
D
UaJPJϕPUaf,gdA(a)=
D
ϕUaP f,JPJUagdA(a)= JPJϕP f ,g. (3.38)
Thus
D
UahϕUaf,g
dA(a)=hϕf,g
. (3.39)
Corollary 3.4. Ifϕ∈L∞(D,dA), then there exists a constantCsuch that
1
3Chϕ≤Supz∈D
D
UahϕUakz,kz
dA(a)≤C
3hϕ. (3.40)
Proof. From a result in [3], it follows that there exists a constantCsuch that
1
3Chϕ≤Supz∈Dhϕkz,kz≤C
3hϕ. (3.41)
ByCorollary 3.3,
Supz∈Dhϕkz,kz=Supz∈D
D
UahϕUakz,kz
dA(a), (3.42)
and the result follows.
Leth∞(D) be the space of bounded harmonic functions onD. Thenh∞(D)⊂L∞(D). It is well known (see [9]) that every harmonic function onDis the sum of an analytic function and the conjugate of another analytic function. Hence if f ∈h∞(D), thenf(z)= ∞
n=0anzn+
∞
n=0bnzn. It is not so difficult to verify (see [3]) that in this case f(z)= a0−(a1/2)z−(b1/2)z. Thus if f ∈h∞(D), thenf= f if and only iff is a constant.
Corollary 3.5. If f ∈h∞(D), thenCaTf =TfCafor alla∈Dif and only if f = f. That
is, if and only if f is a constant.
Proof. If f ∈h∞(D), thenTf is bounded linear operator onA2(D). Further, forz∈D,
Tf(z)=
Tfkz,kz
=f kz,kz
=f(z)=f(z). (3.43)
This is so because by the invariant mean value property [3], f ∈h∞(D) implies f = f. Hence
Tf
ϕa(z)
= fϕa(z)
(3.44)
for alla,z∈D. ByTheorem 3.1,CaTf =TfCafor alla∈Dif and only if for allz∈D,
DTf
ϕa(z)dA(a)=Tf(z). (3.45)
That is, if and only if for allz∈D,
Df
ϕa(z)dA(a)= f(z). (3.46)
HenceCaTf =TfCafor alla∈Dif and only if f= f. That is, if and only iff is a constant.
References
[1] J. B. Conway, A Course in Functional Analysis, vol. 96 of Graduate Texts in Mathematics, Springer, New York, NY, USA, 2nd edition, 1990.
[2] N. I. Akhiezer and I. M. Glazman, Theory of Linear Operators in Hilbert Space. Vol. I, vol. 9 of Monographs and studies in Mathematics, Pitman, Boston, Mass, USA, 1981.
[3] K. H. Zhu, Operator Theory in Function Spaces, vol. 139 of Monographs and Textbooks in Pure and Applied Mathematics, Marcel Dekker, New York, NY, USA, 1990.
[4] L. A. Coburn, “A Lipschitz estimate for Berezin’s operator calculus,” Proceedings of the American Mathematical Society, vol. 133, no. 1, pp. 127–131, 2005.
[5] D. D. Clahane, “Spectra of compact composition operators over bounded symmetric domains,” Integral Equations and Operator Theory, vol. 51, no. 1, pp. 41–56, 2005.
[6] M. Lindstr¨om and N. Palmberg, “Spectra of composition operators on BMOA,” Integral Equa-tions and Operator Theory, vol. 53, no. 1, pp. 75–86, 2005.
[7] B. D. MacCluer and R. J. Weir, “Linear-fractional composition operators in several variables,” Integral Equations and Operator Theory, vol. 53, no. 3, pp. 373–402, 2005.
[8] B. A. Cload, “Composition operators: hyperinvariant subspaces, quasi-normals and isometries,” Proceedings of the American Mathematical Society, vol. 127, no. 6, pp. 1697–1703, 1999. [9] J. B. Conway, Functions of One Complex Variable, vol. 11 of Graduate Texts in Mathematics,
Springer, New York, NY, USA, 1973.
Namita Das: P. G. Department of Mathematics, Utkal University, Vani Vihar, Bhubaneswar, Orissa 751004, India
Email address:[email protected]
R. P. Lal: Institute of Mathematics and Applications, 2nd Floor, Surya Kiran Building, Sahid Nagar, Bhubaneswar, Orissa 751007, India
Email address:[email protected]
C. K. Mohapatra: Institute of Mathematics and Applications, 2nd Floor, Surya Kiran Building, Sahid Nagar, Bhubaneswar, Orissa 751007, India
