Absolutely continuous measures and compact composition operator on spaces of Cauchy transforms

PII. S0161171204310057 http://ijmms.hindawi.com © Hindawi Publishing Corp.

ABSOLUTELY CONTINUOUS MEASURES AND COMPACT

COMPOSITION OPERATOR ON SPACES

OF CAUCHY TRANSFORMS

YUSUF ABU MUHANNA and EL-BACHIR YALLAOUI

Received 5 October 2003

The analytic self-map of the unit diskD,ϕis said to induce a composition operatorCϕfrom the Banach spaceXto the Banach spaceYifCϕ(f )=f◦ϕ∈Yfor allf∈X. Forz∈Dand α >0, the families of weighted Cauchy transformsFαare defined byf (z)=TKαx(z)dµ(x), whereµ(x)is complex Borel measure,xbelongs to the unit circleT, and the kernelKx(z)= (1−xz)−1_{. In this paper, we will explore the relationship between the compactness of the}

composition operatorCϕacting onFαand the complex Borel measuresµ(x).

2000 Mathematics Subject Classification: 30E20, 30D99.

1. Background. LetTbe the unit circle andMthe set of all complex-valued Borel measures onT. Forα >0 andz∈D, we define the space of weighted Cauchy transforms Fαto be the family of all functionsf (z)such that

f (z)=

T Kα

x(z)dµ(x), (1.1)

where the Cauchy kernelKx(z)is given by

Kx(z)=

1

1−xz (1.2)

and whereµin (1.1) varies over all measures inM. The classFαis a Banach space with

respect to the norm

fFα=infµM, (1.3)

where the infimum is taken over all Borel measures µ satisfying (1.1). µ denotes the total variation norm ofµ. The familyF1has been studied extensively in the Soviet literature. The generalizations forα >0 were defined by MacGregor [7]. The Banach spacesFαhave been well studied in [2,3,5,7]. Among the properties ofFαwe list the

following:

(i) Fα⊂Fβwhenever 0< α < β,

(ii) Fαis Möbius invariant,

(iii) f∈Fαif and only iff∈F1+αandfF1+α≤αfFα,

(iv) ifg∈Fα+1, thenf (z)=0zg(w)dw∈FαandfFα≤(2/α)gF1+α.

The spaceFαmay be identified withM/H01, the quotient of the Banach spaceMof Borel measures byH1

whose conjugate belongs to the Hardy spaceH1_{. Hence,F}

αis isometrically isomorphic

toM/H1

0. Furthermore,Madmits a decompositionM=L1⊕Ms, whereMs is the space

of Borel measures, which are singular with respect to Lebesgue measure, andH1 0⊂L1. According to the Lebesgue decomposition theorem, anyµ∈Mcan be written asµ=

µa+µs, whereµais absolutely continuous with respect to the Lebesgue measure andµs

is singular with respect to the Lebesgue measure(µa⊥µs). Since|x| =1 in (1.1), if we

letx=eit_{, then we can decomposeµin such a way thatdµ(e}it_)=h(eit_)dt+dµ s(eit),

whereh(eit_)∈L1_{. Consequently, F}

α is isomorphic toL1/H01⊕Ms. Hence,Fα can be

written asFα=Fαa⊕Fαs, whereFαais isomorphic toL1/H01, the closed subspace of Mof absolutely continuous measures, andFαsis isomorphic toMs, the subspace ofM

of singular measures. Iff∈Fαa, then the singular part is null and the measureµfor

which (1.1) holds reduces todµ(t)=dµ(eit_)=h(eit_{)dt, whereh(e}it_)∈L1_{anddt is} the Lebesgue measure onT, see [1]. The functions inFαamay be then written as, for

x=eit_,

f (z)=

π

−πK α x(z)h

eitdt. (1.4)

Furthermore, ifh(eit_{)is nonnegative, then}

fFα=h

eit

L1. (1.5)

Remark 1.1. For simplicity, we will adopt the following notation throughout the

paper. We will reserveµfor the Borel measures ofM, and since in (1.1)|x| =1, we can writex=eit_{, wheret∈[−π , π ). We will reservedtfor the normalized Lebesgue of the}

unit circleT, anddσ for the singular part ofdµ. Hence, instead of writingdµ(eit₎₌

dµa(eit)+dµs(eit)=h(eit)dt+dµs(eit), we may simply writedµ(t)=hdt+dσ (t).

2. Introduction. IfXandYare Banach spaces, andLis a linear operator fromXtoY, we say thatLis bounded if there exists a positive constantAsuch thatL(f )Y≤AfX

for allf inX. We denote byC(X, Y )the set of all bounded linear operators fromXto Y. IfL∈C(X, Y ), we say thatLis a compact operator fromXtoY if the image of every bounded set ofXis relatively compact (i.e., has compact closure) inY. Equivalently, a linear operatorLis a compact operator fromXtoY if and only if for every bounded sequence{fn}of X, {L(fn)}has a convergent subsequence inY. We will denote by

K(X, Y )the subset ofC(X, Y )of compact linear operators fromXintoY.

LetH(D)denote the set of all analytic functions on the unit diskDand mapDinto D. IfXandY are Banach spaces of functions on the unit diskD, we say thatϕ∈H(D) induces a bounded composition operatorCϕ(f )=f (ϕ)fromXtoY, ifCϕ∈C(X, Y )

or equivalentlyCϕ(X)⊆Y, and there exists a positive constantAfor allf ∈X and Cϕ(f )Y≤AfX. In caseX=Y, then we say thatϕinduces a composition operator

Cϕ onX. Iff∈X, thenCϕ(f )=f (ϕ)∈X. Similarly, we say thatϕ∈H(D)induces a

compact composition operator ifCϕ∈K(X, Y ).

spaces of Cauchy transforms are defined in terms of Borel measures, it seems natural to investigate the relation between the behavior of the composition operator and the measure. The work in this paper was motivated by the work of Cima and Matheson in [1], who showed thatCϕ is compact onF1if and only ifCϕ(F1)⊂F1a. In our work, we

will generalize this result forα >1.

Now, ifCϕ∈C(Fα, Fα), thenCϕ(f )=(f◦ϕ)=f (ϕ)∈Fαfor allf ∈Fαand there

exists a positive constantAsuch that

Cϕ(f )Fα=f (ϕ)Fα=[µ]≤AfFα. (2.1)

SinceFαcan be identified with the quotient spaceM/H01, we can viewCϕ as a map,

Cϕ:M/H01 →M/H01,

f→f (ϕ). (2.2)

The equivalence class of a complex measureµwill be written as

[µ]=µ+H1 0=

µ+h:h∈H1 0

,

_[µ]=inf h

_µ+h. (2.3)

The spaceC(Fα, Fα)has been studied in [4], where Hibschweiler showed that

(1) ifα≥1, thenCϕ∈C(Fα, Fα)for any analytic self mapϕof the unit disc,

(2) Cϕ∈C(Fα, Fα)if and only if{Kxα(ϕ):|x| =1}is a norm bounded subset ofFα,

(3) ifCϕ∈C(Fα, Fα), thenCϕ∈C(Fβ, Fβ)for 0< α < β,

(4) ifCϕ∈C(Fα, Fα), then the operatorϕCϕ∈C(Fα+1, Fα+1).

In this paper, we will investigate necessary and sufficient conditions forCϕ to be

compact onFαforα≥1. SinceFαis Mobius invariant, then there is no loss of generality

in assuming thatϕ(0)=0.

3. Compactness and absolutely continuous measures. In this section, we will show that the compactness of the composition operatorCϕ onFα is strongly tied with the

absolute continuity of the measure that supports it. First, we state this lemma due to [6].

Lemma3.1. If0< α < β, thenF_α⊂F_βaand the inclusion map is a compact operator of norm one.

Next, we show that the bounded function ofFαbelongs toFαaforα≥1.

Proposition3.2. H∞∩F_α⊂F_αaforα≥1.

Proof. Suppose thatf∈H∞∩F_α, whereα≥1. Then, using the previous lemma and the fact thatH∞⊂F1a, we get that for anyz∈D,f (z)∈H∞∩Fα⊂F1a∩Fα⊆Fαa.

In the following theorem,gx(eit)isL1continuous function ofxmeans, _gx

eit−gy

gx(eit)isM+continuous, if

_xgxeitdµn(x)−

x

gxeitdµ(x)

_L1 →0, (3.2)

for any sequence{µn}of nonnegative Borel measures that are weak* convergent toµ.

Theorem_3.3. _{For a holomorphic self-mapϕof the unit discDandα}≥_{1, ifCϕ is} compact onFα, then(Cϕ◦Kxα)(z)∈Fαaand

Cϕ◦Kxα

(z)=

π

−πgx

eitKαx(z)dt, (3.3)

wheregx(eit)is a function of two variablesx, with|x| =1andeit,gx(eit)L1≤a <∞,

gx(eit)is nonnegative, andM+is a continuous function ofx.

Proof. Assume thatC_ϕis compact and let{f_j}∞_j

=1be a sequence of functions such that

fj(z)=Kxα

ρjz= 1

1−ρjxz

α, (3.4)

where 0< ρj<1 and limj→∞ρj=1. Then, it is known from [2] thatfj(z)∈Fαfor every

j, andfj(z)Fα=1. Furthermore, there existµj∈M, such thatµj =1,dµj>0, and

fj(z)=

1

1−ρjxz α

=

T Kα

eit(z)dµjeit

=

T 1

1−e−it_zαdµj

eit.

(3.5)

SinceCϕ is compact onFα, then(Cϕ◦fj)∈Fα and Cϕ(fj) ≤ CϕfjFα= Cϕ for allj. Furthermore,Cϕ◦fj∈H∞, therefore using the previous result, we get that

(Cϕ◦fj)∈H∞∩Fα⊂Fαafor everyj. Therefore, there existsL1a nonnegative function

gjx(eit)such that

dµj

eit=gjxeitdt, gxj_L1≤Cϕ,

fj◦ϕ

(z)=Kα x

ρjϕ

= π

−π

gjx

eit_Kα eit(z)dt.

(3.6)

Now, becauseFαais closed andCϕis compact, the sequence{fj◦ϕ}∞j=1has a conver-gent subsequence{fj_k◦ϕ}that converges to(Kxα◦ϕ)(z)∈Fαa. Therefore,

lim

k→∞

fjk◦ϕ

(z)=lim

k→∞K

α x

ρjkϕ

=lim

k→∞

π

−π

gjk

xeitK_eαit(z)dt

= π

−π

gx

eit_Kα eit(z)dt

=Kxα◦ϕ

(z)= 1

1−xϕ(z)α∈Fαa,

where the functiongx(eit)is anL1nonnegative continuous function ofx, andgxL1 ≤ Cϕ. The continuity ofgx(eit)inM+ follows from the compactness and the fact

that iff (z)=TKxα(z)dµ(x), withdµ(x) >0, thenfFα= µ. This concludes the proof.

Corollary3.4. Letg_x(eit)be as in Theorem 3.3. Then, forh(x)∈H1₀, the operator

gx(eit)h(x)dx=u(eit)∈H01is bounded onH01.

Proof. For the operator to be well defined, (h(x)dx/(1−xϕ(z))α)=0 for all

h(x)∈H01. Hence,

gx(eit)h(x)dx=u(eit)∈H01.

In the opposite direction, we have the following lemma.

Lemma3.5. Letg_x(eit)be a nonnegativeM+continuous function such thatg_xL1≤

a <∞andgx(eit)defines a bounded operator onH01. Iff (z)=

(1/(1−xz)α_)dµ(x), letLbe the operator given by

Lf (z) =

_g

x

eit

1−e−it_zαdt dµ(x), (3.8)

thenLis compact operator onFα,α≥1.

Proof. _{First, note that the condition thatgx(e}it_{)defines a bounded operator onH0}1

implies that theLoperator is a well-defined function onFα. Let{fn(z)}be a bounded

sequence inFαand let{µn}be the corresponding norm bounded sequence of measures

inM. Since every norm bounded sequence of measures inMhas a weak* convergent subsequence, let{µn}be such subsequence that is convergent toµ∈M. We want to

show that{L(fn)}has a convergent subsequence inFα.

First, we assume that dµn(x) >0 for all n, and let wn(t)=

gx(eit)dµn(x) and

w(t)=gx(eit)dµ(x), then, asgx(eit)is anM+continuous function, we havewn(t),

w(t)∈L1_{for alln, andw}

n(t)→w(t)inL1. Now, sincegx(eit)is a nonnegative

con-tinuous function inxand{µn}is weak* convergent toµ, then

Lfn(z)

=

gx

eit_d(t)

1−e−it_zαdµn(x)=

wn(t)

1−e−it_zαdt,

Lf (z)=

gxeitd(t)

1−e−it_zαdµ(x)=

w(t)

1−e−it_zαdt.

(3.9)

Furthermore, becausewn(t)is nonnegative, then _L

fnFα=wnL1,

L(f )_Fα= wL1.

(3.10)

Now, sincewn−wL1→0, thenL(fn)−L(f )Fα→0, which shows that{L(fn)}has a convergent subsequence inFαand thusLis a compact operator for the case whereµ

is a positive measure.

In the case whereµis complex measure, we write

dµn(x)=

dµ1n(x)−dµ2n(x)

+idµ3n(x)−dµ4n(x)

where eachdµjn(x) >0,and definewnj(t)=

gx(eit)dµnj(x), then

wn(t)=

gxeitdµn(x)=wn1(t)−wn2(t)

+iw3

n(t)−wn4(t)

. (3.12)

Using an argument similar to the one above, we get that wnj(t), wj(t)∈L1, and wnj−wjL1 →0. Consequently, wn−wL1 →0, where w(t)= (w1(t)−w2(t))+

i(w3_(t)−w4_(t))=g

x(eit)dµ(x).

Hence,L(fn)−L(f )Fα≤ wn−wL1→0. Finally, we conclude that the operator is compact.

The following is the converse ofTheorem 3.3.

Theorem3.6. For a holomorphic self-mapϕof the unit discD, if

Kα x◦ϕ

(z)= 1

1−xϕ(z)α =

_g

xeit

1−e−it_zαdt, gx∈L

1 _(3.13)

is nonnegative,gxL1≤a <∞for allx∈Tandgxis anM+continuous function, then

Cϕis compact onFα.

Proof. We want to show thatC_ϕis compact onF_α. Letf (z)∈F_α. Then, there exists

a measureµinMsuch that for everyzinD,

f (z)=

1

1−xzαdµ(x). (3.14)

Using the assumption of the theorem, we get that

(f◦ϕ)(z)=

₁

1−xϕ(z)αdµn(x)=

_g

xeit

1−e−it_zαdt dµn(x), (3.15)

which by the previous lemma was shown to be compact onFα.

Now, we give some examples.

Corollary3.7. Letϕ∈H(D), withϕ_∞<1. Then,C_ϕ is compact onFα,α≥1.

Proof. (C_ϕ◦Kα_x)(z) = 1/(1−xϕ(z))α ∈ H∞∩F_α ⊂ F_αa and is subordinate to 1/(1−z)α_{, hence,}

Cϕ◦Kxα

(z)=

Kαx(z)gx

eitdt (3.16)

withgx(eit)≥0 and since 1=(Cϕ◦Kxα)(0)=

gx(eit)dt, we get thatgx(eit)1=1.

Remark3.8. In fact, one can show thatC_ϕ, as in the above corollary, is compact

Corollary 3.9. If C_ϕ is compact on Fα, α ≥ 1, and lim_r→1|ϕ(r eiθ)| = 1, then |1/ϕ(eiθ_{)| =0.}

Proof. IfC_ϕis compact, then

Cϕ◦Kxα

(z)=

Kαx(z)gx

eitdt. (3.17)

Hence, ifz=eiθ_andϕ(eiθ_{)=x, then}

lim

r→1

eiθ_{−r e}iθα

1−xϕr eiθα=0. (3.18)

Corollary_3.10. _IfCϕ∈_{K(Fα, Fα)forα}≥_{1, thenCϕ is a contraction.}

4. Miscellaneous results. We first start by giving another characterization of com-pactness onFα.

Lemma _4.1. _Letϕ ∈_{C(Fα, Fα), α >0, then ϕ} ∈_{K(Fα, Fα) if and only if for any} bounded sequence(fn)inFα, withfn→0uniformly on compact subsets ofDasn→ ∞, Cϕ(fn)Fα→0asn→ ∞.

Proof. SupposeC_ϕ∈K(F_α, F_α)and let(f_n)be a bounded sequence(f_n)inF_αwith

limn→∞fn→0 uniformly on compact subsets ofD. If the conclusion is false, then there

exists an >0 and a subsequencen1< n2< n3<···such that

Cϕ

fn_j

Fα≥, ∀j=1,2,3, . . . . (4.1)

Since(fn) is bounded andCϕ is compact, one can find another subsequence nj1 <

nj2< nj3<···andfinFαsuch that

lim

k→∞

Cϕ

fn_jk

−f

Fα=0. (4.2)

Since point functional evaluations are continuous inFα, then for anyz∈Dthere exists

A >0 such that

Cϕ

fn_jk

−f(z)≤ACϕ

fn_jk

−f_F

α →0 ask → ∞. (4.3)

Hence,

lim

k→∞

Cϕ

fn_jk

−f →0 (4.4)

uniformly on compact subsets ofD. Moreover, sincefn_jk→0 uniformly on compact

subsets ofD, thenf=0, that is,Cϕ(fn_jk)→0 on compact subsets ofFα. Hence,

lim

k→∞

Cϕ

fn_jkF

which contradicts our assumption. Thus, we must have

lim

n→∞Cϕ

fnFα=0. (4.6)

Conversely, let(fn)be a bounded sequence in the closed unit ball ofFα. We want to

show thatCϕ(fn)has a norm convergent subsequence. The closed unit ball ofFαis a

compact subset ofFαin the topology of uniform convergence on the compact subsets

ofD. Therefore, there is a subsequence(fnk)such that

fnk →f (4.7)

uniformly on compact subsets ofD. Hence, by hypothesis,

Cϕ

fnk

−Cϕ(f )_F

α →0 ask→ ∞, (4.8)

which completes the proof.

Proposition4.2. IfC_ϕ∈C(F_α, F_α), thenC_ϕ∈K(F_α, F_β)for allβ > α >0.

Proof. Let(f_n)be a bounded sequence in the closed unit ball ofF_α. Then,(f_n◦ϕ)

is bounded inFα and since the inclusion mapi:Fα→Fβ is compact, (fn◦ϕ)has a

convergent subsequence inFβ.

Proposition _4.3. _{Cϕ(f )} =_(f◦_{ϕ) is compact on Fα if and only if the operator}

ϕCϕ(g)=ϕ(g◦ϕ)is compact onFα+1.

Proof. Suppose thatC_ϕ(f )=(f◦ϕ)is compact onF_α. It is known from [4] that

ϕCϕ(g)=ϕ(g◦ϕ)is bounded on Fα+1. Let(gn)be a bounded sequence inFα+1,

with gn →0 uniformly on compact subsets ofD as n→ ∞. We want to show that

limn→∞ϕ(gn◦ϕ)F_α+1=0. Let(fn)be the sequence defined byfn(z)= z

0gn(w)dw.

Then,fn∈Fα and fnFα≤(2/α)gnF_α+1, thus(fn)is a bounded sequence inFα. Furthermore, using the Lebesgue dominated convergence theorem we get thatfn→0

uniformly on compact subsets ofD. Thus,

ϕgn◦ϕF_α+1=ϕ

_f

n◦ϕF_α+1

=fn◦ϕ

F_α+1

≤αfn◦ϕFα →0 asn→ ∞,

(4.9)

which shows thatϕCϕ(g)=ϕ(g◦ϕ)is compact onFα+1.

Conversely, assume thatϕCϕ(g)=ϕ(g◦ϕ)is compact onFα+1. Then, in particular,

ϕCϕ(f)=ϕ(f◦ϕ)=(f◦ϕ)is a compact for everyf∈Fα. Now, since(f◦ϕ)Fα≤ (2/α)(f◦ϕ)F_α+1. Let(fn)be a bounded sequence inFαwithfn→0 uniformly on compact subsets ofDasn→ ∞. We want to show that limn→∞(fn◦ϕ)Fα=0. Since any bounded sequence ofFαis also a bounded sequence ofFα+1, then(fn◦ϕ)Fα≤ (2/α)(fn◦ϕ)F_α+1→0 asn→ ∞and the proof is complete.

Acknowledgment. The authors would like to thank the referee for his many

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Yusuf Abu Muhanna: Department of Mathematics, American University of Sharjah, P.O. Box 26666, Sharjah, United Arab Emirates

E-mail address:[email protected]

El-Bachir Yallaoui: College of Arts and Sciences, Abu Dhabi University, P.O. Box 1790, Al Ain, United Arab Emirates