# Interpolation by multipliers on certain spaces of analytic functions

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Vol. 23, No. 8 (2000) 547–554 S0161171200000958 © Hindawi Publishing Corp.

### SPACES OF ANALYTIC FUNCTIONS

K. MOSALEHEH and K. SEDDIGHI

Abstract.LetᏴbe a Hilbert space of analytic functions on a planar domainGsuch that, for eachλinG, the linear functionalof evaluation atλis bounded onᏴ. Furthermore,

assume thatzᏴandσ (Mz)=G¯is anM-spectral set forMz, the operator of multipli-cation byz. This paper is devoted to the study of interpolation by multipliers of the space Ᏼand, in particular, the Dirichlet space.

Keywords and phrases. Interpolation, multipliers, Dirichlet space, ﬁnitely connected do-mains.

2000 Mathematics Subject Classiﬁcation. Primary 47B38; Secondary 46E20.

1. Introduction. LetGbe a ﬁnitely connected domain in the complex planeC. The Bergman norm of a functionfanalytic onGis deﬁned by

f2 2=

G|f|

2dA, (1.1)

wheredAdenotes the usual Lebesgue area measure. The Bergman spaceL2

a(G)is the set of all functions analytic onGsuch thatf2<∞.

The Dirichlet spaceD(G) is the Hilbert space of analytic functions on G whose derivative lies inL2

a(G). The Dirichlet norm is deﬁned by

f2

D= |f (ω)|2+

G|f

|2dA (1.2)

forf∈D(G)and a ﬁxed pointωinG. Thisωis called the base point ofD(G). It is shown by Chan [5] that the norms obtained by ﬁxing diﬀerent points are equivalent. If

G=Uis the open unit disc andf (z)=∞n=0anzn, thenf2D= |a0|2+π∞n=1n|an|2. Iff∈D(U), then∞n=1|an|2<∞, and sof∈H2. Therefore,D(U)⊂H2.

An analytic functionϕinG is said to be a multiplier forD(G)if ϕf∈D(G)for allf∈D(G). The set of all multipliers of the Dirichlet space is denoted byM(D(G)). For each ϕ∈ M(D(G)), deﬁne the multiplication operator :D(G)→D(G)by

Mϕf=ϕf for allf∈D(G).Clearly, is a bounded operator onD(G). The norm of is the usual norm. If we identifyϕwith, thenM(D(G))can be thought of as a subspace of the space of bounded linear operators onD(G).

A sequence(ωn)inGis said to be an interpolating sequence forM(D(G)) if, for each bounded sequence(an)⊂C, there existsϕ∈M(D(G))such thatϕ(ωn)=an.

LetG1andG2be two conformally equivalent regions and leth:G1→G2be a confor-mal map which takes the base point that deﬁnes the norm ofD(G1)to that ofD(G2). Now, sinceG2|f|2dA=

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ontoD(G1)is an isometry. Therefore,D(G1)is unitarily equivalent toD(G2). We also observe that the mapM(D(G2))→M(D(G1))given byϕ→ϕ◦his an isomorphism. A circular domain is a domain that is obtained by removing a ﬁnite number of dis-joint closed subdisks from the open unit disk. We can assume that each removed closed subdisk has a positive radius since otherwise that point is a removable singu-larity of every function inD(G). We only prove this for the origin. Iff∈D(G), where

G containsB(0,R)\{0}, thenf (z)=n∞=−∞anzn, r <|z|< R for some 0< r < R. We haveπn≠0n|an|2(R2n−r2n)=r <|z|<R|f|2dA <

G|f|2dA. Now, ifr→0, then

an=0 forn <0. Therefore,fis analytic at 0.

IfGis a ﬁnitely connected domain inC, there exists a conformal mapψofGonto some circular domainΩ, and soD(G)is isometric toD(). IfGis a ﬁnitely connected domain inCandK1,K2,...,KN are the bounded components ofC\G, thenG0=G∪

K1∪K2∪ ··· ∪KN and Gi=(C∪ {∞})\Ki, 1≤i≤N, are simply connected regions. By Cauchy integral formula, every functionf analytic onGcan be written as a sum

f=f0+f1+···+fN, where eachfiis analytic onGiandfk(∞)=0 ifk≠0. Moreover, this summation is unique up to the order.

We use the notationH(G)for the algebra of all analytic functions inGandH∞(G) for the algebra of all bounded analytic functions inG. IfGis unbounded,H0(G)is the space of all analytic functions inGthat vanish at inﬁnity. We say thatM(D(G))is rotation invariant if wheneverh∈M(D(G)),hθ∈M(D(G)), wherehθ(z)=h(e−iθz). Axler [1] considered the interpolation sequence for M(D(U)). In this paper, we consider the interpolation sequence for M(D(G)). Which we treat in Section 2. In Section 3, we consider the interpolation sequence for M(H), where H is a Hilbert space of analytic functions onGandM(H)is the set of all multipliers ofH.

2. Interpolation sequence forM(D(G)). We use Rosenthal-Dor theorem as proved by Rosenthal [11] and Dor [9].

Theorem2.1(Rosenthal-Dor theorem). Suppose thatEis a Banach space and(en) is a bounded sequence inE. Then there exists a subsequence(enk)such that either

(i) the map(ak)∞k=1akenk is an isomorphism ofl1intoE; or (ii) limk→∞ϕ(enk)exists for everyϕ∈E∗.

As we see in Rosenthal-Dor theorem, we must work with a Banach space. Therefore, we need to makeM(D()), whereΩis a circular domain into a Banach space. We can prove thatM(D())can be identiﬁed with E∗, whereE is a separable Banach space and each point evaluation atω∈Ωis aweak∗continuous linear functional on M(D())with norm 1.

Lemma2.2. If(ωn)is a sequence inr = {z:|z|< r},r <1, such that|ωn| →r, then there existsϕ1∈M(D(r))such thatlimn→∞ϕ1(ωn)does not exist andϕ1(0)=0, Furthermore, if(ωn)is a sequence in the annulusA= {z∈C:r <|z|<1}and|ωn| →

ror|ωn| →1, then there is aϕ∈M(D(A))such thatlimn→∞ϕ(ωn)does not exist.

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To prove the second part, suppose that|ωn| →r. Considering ωn inA1=(C

{∞})\r, sinceA1is the conformal image of{z:|z|< r1}for somer1>0, there exists

ϕ2∈M(D(A1))such that limn→∞ϕ2(ωn)does not exist and vanish at inﬁnity. Ifϕ3∈

M(D(U)), thenϕ3is analytic in the unit diskU. Therefore,ϕ3(ωn)is convergent. Now, we writeϕ=ϕ2+ϕ3. Then ϕ∈M(D(A)) becauseM(D(A))=M(D(U))+ M0(D(A1)), whereM0(D(A1))=H0(A1)∩M(D(A1)), (see [5, Theorem 3.2]). Hence, we ﬁnd a functionϕinM(D(A))such that limn→∞ϕ(ωn)does not exist.

If|ωn| →1, then there existsψ1∈M(D(U))such that limn→∞ψ1(ωn) does not exist. Ifψ2∈M0(D(A1)), thenψ2is analytic onA1. Hence,ψ2(ωn)is convergent. Therefore, ifψ=ψ1+ψ2, thenψis inM(D(A))and limn→∞ψ(ωn)does not exist.

Theorem2.3. Let(ωn) be a sequence in the circular domain=U\{K1∪K2∪

··· ∪KN}, whereKi= {z:|z−αi| ≤ri}, αi∈U,i=1,2,...,N, such that|ωn| →1or

|ωn| →ri,i=1,2,...,N. Then there is a subsequence of(ωn)that is interpolatingfor

M(D()).

Proof. In either cases|ωn| →1 or|ωn| →ri, i=1,2,...,N, we can prove, in the same way as in [1, Theorem 1], that if case (i) of Rosenthal-Dor theorem is satisﬁed, then there exists some subsequence of(ωn)that is interpolating forM(D()). So, it is enough to prove that case (ii) of Theorem 2.1 does not hold.

Let|ωn| →rjfor somej=1,2,...,N. As above, there isϕj∈M0(D(j))such that limn→∞ϕj(ωn)does not exist. Ifϕ0∈M(D(U))andϕi∈M0(D(i)),i=1,2,...,N, thenϕ0(ωn),ϕi(ωn),i=1,2,...,N,ijare convergent. Therefore, ifϕ=ϕ0+ϕ1+

···+ϕN, thenϕ∈M(D())and limn→∞ϕ(ωn)does not exist because

MD()=MD(U)+M0D(Ω1)+···+M0D(N). (2.1) If|ωn| →1, there is anψ0∈M(D(U))such that limn→∞ψ0(ωn)does not exist and ifψi∈M0(D(i)),i=1,2,...,N, thenψi(ωn)are convergent. Hence,ψ=ψ0+ψ1+

ψ2+···+ψN is inM(D())and limn→∞ψ(ωn)does not exist. The proof is complete.

3. Interpolation sequence for the space of multipliers. LetHbe a Hilbert space whose elements are analytic functionsf:U→C. Assume that 1∈Hand the operator

Mzof multiplication byzmapsHinto itself and point evaluations are bounded linear functionals onH.

ConsiderDα,−∞< α <∞, the space of all analytic functions inUwith Taylor series

f (z)=∞n=0anznsuch that

f2

α=

n=0

(n+1|a

n|2<∞. (3.1)

If α= −1, then we have the Bergman space L2

a and if α=0, we obtain the Hardy spaceH2. But ifα=1, the spaceD1is the Dirichlet space. In this space,M

zis not a contraction because ifϕis a nonconstant function on ¯U, thenϕ∞<Mϕ(see [1]). For a nice treatment of this space (see [2, 3, 4]).

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multipliers ofHis denoted byM(H). It is well known thatM(H)⊂H∞(U). We also assume thatM(H)is rotation invariant. For an introduction to Hilbert spaces of an-alytic functions, see Seddighi [12, 13]. Shields and Wallen [15] have proved that if

H is a Hilbert space of analytic functions on a setS and if point evaluations are bounded linear functionals onH, also ifMzmapsHinto itself and is a contraction, thenMϕ = ϕ∞,ϕ∈M(H). Also, see Shields [14].

In this section, we assume thatMzis polynomially bounded in the sense that there is a constantC >0 such that

Mp ≤Cp∞ (3.2)

for every polynomialp. Now, letϕ∈M(H). Becauseϕ∈H∞(U), there is a sequence {pn}of polynomials such thatpn∞≤ ϕ∞and pn(z)→ϕ(z)for every z∈U. Now, it is easy to show that

Mpnf ,g → Mϕf ,g (3.3)

for everyf,g∈H. In other words,Mpn→Mϕ in the weak operator topology (WOT) ofB(H). In fact, if g=kλ for some λ∈U, then (3.3) is true. Ifg is a ﬁnite linear combination of the reproducing kernelskλ, then (3.3) is again true. BecauseH={kλ:

λ∈U}, we conclude that everyg∈H can be approximated by a sequence of ﬁnite linear combinations of thekλ,λ∈U. To prove that (3.3) holds for everyg∈H, we use an approximation process, inequality (3.2), and the fact that (3.3) is true for ﬁnite linear combinations. We can also prove that

Mϕ ≤Cϕ∞. (3.4)

This follows from (3.3) and the fact that

Mpnf ,g≤ Mpnfg ≤Cpnfg ≤Cϕfg. (3.5)

We recall that the spaceH(U)of holomorphic functions onUis equipped with the topology of uniform convergence on compact sets and because the point evaluations are continuous, the embedding H→H(U)is continuous. We also let H(U)denote the space of functions holomorphic on neighbourhoods of ¯U, with the inductive limit topology. In fact, a sequence{fk}inH(U)¯ converges to a functionf if and only if all the functions are analytic in some ﬁxed open setG containing ¯U, withfk→f uniformly on compact subsets ofG. We assume thatH(U)¯ ⊂M(H).

Now, let(ωn) be a sequence in U such that |ωn| →1. We show that there is a subsequence of(ωn)which is interpolating forM(H). Therefore, we must show that case (ii) of Rosenthal-Dor theorem does not hold. To prove this, we need the following lemmas.

Lemma3.1. Let ω∈U. Then there exists a function ψ analytic on U¯ such that

ψ(ω)=0; ψ(1)=1; andψ∞=1. In general, let ω1,ω2,...,ωn∈U. Then there existsψanalytic onU¯such that

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Proof. Letϕω(z)=eiθ(ω−z)(1−ωz)¯ 1 be an analytic automorphism of the discU. By a suitable choice ofθ, we haveϕω(ω)=0 andϕω(1)=1. For this choice ofθ, letψ=ϕω. Then ψ(ω)=0, ψ(1)=1, ψ∞=1 and becauseψ∈H(U)¯ , we conclude thatψis a multiplier.

For the second part, note that, for eachωi, there existsϕisatisfyingϕi(ωi)=0,

ϕi(1)=1, andϕi∞=1. Letψ=ϕ1ϕ2···ϕn. Thenψhas the desired properties.

Proposition3.2. Let(ωn)be a sequence in U such that|ωn| →1. Then there existsϕ∈M(H)such thatlimn→∞ϕ(ωn)does not exist.

Proof. First, we prove that if 0< r <1 andω1,ω2,...,ωn∈U, then there exists a functionϕanalytic on ¯Usuch that

ϕ(ω1)= ··· =ϕ(ωn)=1, ϕ(1)= −1, ϕ∞≤1r. (3.7) Consider

d(ω,z)= ω−z

1−ωz¯

. (3.8)

Lets >0 be chosen such thats=d(r ,−r )=2r /(1+r2). Thend(0,s)=d(r ,−r ), and

so there exists an analytic automorphismhofUsuch thath(0)=r,h(s)= −r. Ifψis the function given by Lemma 3.1 andϕ= 1rh◦(sψ), we haveϕ(1)=r1h◦(sψ)(1)=

1

rh(s)= −1,

ϕ(ω1)=ϕ(ω2)= ··· =ϕ(ωn)=r1h◦(sψ)(ω1)=r1h◦(sψ)(ω2)

= ··· =r1h◦(sψ)(ωn)=1rh(0)=1.

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Becausehis analytic on ¯U andh∞=1, we conclude thatϕis analytic on ¯U and ϕ∞≤1/r. Now, we let(rk)be a sequence in(0,1)such that ∞k=11/rk<∞and we ﬁnd a sequence satisfying conditions (11), (12), and (13) on [1, page 416], however, instead of condition (14), we assume that

ϕk∞≤r1

k. (3.10)

Using (3.4), we get

Mϕ

1ϕ2···ϕk≤Cϕ1ϕ2···ϕk∞≤C k

i=1

1

ri. (3.11) Hence,{Mϕ1ϕ2···ϕk}∞k=1is norm bounded inM(H). BecauseM(H)=E∗, whereEis a

separable Banach space, we conclude that there isϕ∈M(H)such that some subse-quence of{Mϕ1ϕ2···ϕk}is weakconvergent toMϕ. Now, if(ωn)is a sequence in

Usuch that|ωn| →1, thenϕsatisﬁesϕ(ωnk)=(−1)kand limk→∞ϕ(ωnk)does not exist.

Now, letHbe a Hilbert space of analytic functions on a circular domainΩsuch that 1∈H,zH⊂Hand point evaluations are bounded linear functionals onH. Clearly,

M(H)⊂H∞(). We also assume thatH()M(H)andM

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H∞()=H(U)+H

0(Ω1)+···+H0∞(N), (3.12) whereH∞

0(i)=H∞(i)∩H0(i),i=1,2,...,N (see [7, Theorem 3.1]), by the same technique as in Sections 2 and 3, we can prove that if(ωn)is a sequence inΩsuch that|ωn| →1 or|ωn| →ri,i=1,2,...,N, then there is a subsequence of(ωn)that is interpolating forM(H).

4. General domains and transformation to the unit disc. LetM >0 be a constant. We say that a compact setKinCis anM-spectral set for an operatorT ifσ (T )⊂ K

andf (T ) ≤MfKfor every rational functionfwith poles outsideK. Here,fK= max{|f (z)|:z∈K}. LetGbe an arbitrary bounded domain each point of which is a bounded point evaluation for a Hilbert space Ᏼ of functions analytic on G which contains the constant functions and admits multiplication by the independent variable

z,Mz, as a bounded operator. We denote such a Hilbert space by(,G). Also, we assume thatσ (Mz)=G¯is anM-spectral set forMzand we may assume thatᏹ()=

H∞(G) (or we may assume that each bounded component Gα of G is conformally equivalent to an open ballsuch that ifAα=C∪{∞}\Gα, thenH(A¯α)is contained in the space of multipliers onᏴ). However, we make a simpler assumption than this later. The idea is to transform the problem to the unit disc and to simplify the problem this way. Now, we indicate how our problem that concerns the operator of multiplication on (,G) can be transformed into a problem concerning an operator acting on a Hilbert space of functions analytic on a subset ofU.

Suppose thatH is a simply connected domain with G⊂H. Leth:U→H be the Riemann mapping corresponding toH and setΩ=h−1(G). This map enables us to

perform the desired transformation. First, we need our Hilbert space to consist of functions that are analytic on the corresponding subdomain of U. Therefore, we set

h=◦h= {f◦h:f∈}, (4.1)

with the inner product deﬁned by

f◦h,g◦h = f ,g (f ,g∈). (4.2)

It is easy to see thatᏴhis a Hilbert space. Furthermore, the functions inᏴhare analytic onΩ,Ω⊂U. Also, note that point evaluations are bounded onᏴhandᏴhcontains the constants. The operatorC:Ᏼhgiven byCf=f◦his clearly an isomorphism and satisﬁesLhC=CMz, whereLh:Ᏼh→hdeﬁned byLhg=hg,g∈h, is the operator of multiplication byhonᏴh. We also observe that the mapϕ→ϕ◦his an isometric isomorphism ofᏹ()→(h)just as the mapA→CAC−1ofᏮ()→(h)is.

To ﬁnd a simply connected domainHcontainingGwhich has some nice properties, we need the following lemma; whose proof is an application of transﬁnite induction (see Conway [8, Lemma 4.8, page 400]).

Lemma4.1. IfKis a compact subset of the plane, then there is a countable ordinal

α0such that, for everyα < α0, there is a componentVαofC\Ksuch that (a) if0is the ﬁrst ordinal,V0is the unbounded component ofC\K; (b) for each ordinalα,V−

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(c) ifVVαis a component ofC\Kand for anyα, thenV−∩[∪αVα]−= ∅. The enumeration of the components ofC\Kin Lemma 4.1 picks out those compo-nents that can bechainedto the unbounded component. So, ifKis ﬁnitely connected, the unbounded component is enumerated but there may be no others.

IfKis a compact set and{Vα:α < α0}are those components ofC\Kthat are picked out by Lemma 4.1 and ˆKis the polynomially convex hull ofK, then we set

L=Lα0=K\∪ˆ α<α0Vα∪W , (4.3)

whereW= ∪βVβis the union of all thoseVα,α < α0, such that∪Vβ meets the outer boundary of ˆK.

Now, letMz act on(,G)in such a way thatσ (Mz)=G¯is anM-spectral set for

Mz. SetK=G¯in the preceding inductive process and ﬁndL. It turns out thatLis the smallest compact subset ofCthat containsKand hasL0simply connected.

Now, we setH=L0in our preceding argument and leth:UL0be the Riemann

mapping corresponding toL0. Setting=h1(G), we transform our problem

con-cerningMzon(,G)to similar problems concerningLhon(,), where᏷=h. The domainΩhas several nice properties such asΩ⊂U and∂U Ω¯. By Carathéodory theorem,hextends to a homomorphismh: ¯U→Lunder which the boundaries corre-spond to each other.

By [12, Theorem 3.6], we see thatσ (Lh)=G¯is a spectral set forLh. Another inter-esting property ofΩis that it contains a boundary annulus, i.e., there is 0< r <1 such that ¯A⊂Ω, where A= {z:r <|z|<1}. To see this, let {Vγ}be the bounded components ofC\Ω¯. Then∪Vγ∩∂U= ∅. Letr=dist(∪Vγ,∂U)and chooser < s <1 such thatB= {z:|z|< s}contains∪Vγ. Clearly,A= {z:s <|z|<1}satisﬁes ¯A⊂Ω.

Lemma4.2. Letbe an open connected subset of the open unit discUwith∂U⊂Ω¯. Suppose that{Vγ}are the bounded components ofC\Ω¯and assume that eachVγis the conformal image of some open disc with positive radius. Also we assumeH(Ω¯)⊂M(). Furthermore if{ωn}is a sequence insuch that|ωn| 1or|ωn|converges to a point on∂Vγfor someγthen there isϕ∈()such thatlimn→∞ϕ(ωn)does not exist.

Proof. Suppose|ωn| →1. We must show that if 0< r <1 andω1,ω2,...,ωn∈U, then there exists a functionϕinH(U)¯ such that

ϕ(ω1)= ··· =ϕ(ωn)=1, ϕ(1)= −1, ϕ ≤r1. (4.4)

Repeating the argument of Proposition 3.2, we conclude that if{rk}is a sequence in

(0,1)such that ∞k=11/rk<∞, then all the conditions for ﬁnding a multiplierϕsuch that limn→∞ϕ(ωn)does not exist hold provided a suitable form of (4) is in eﬀect.

Now, letϕ∈H(U)¯ . BecauseΩ⊂Uandσ (Lh)=G¯is anM-spectral set, we have

ϕ◦h1(L

h)≤Mϕ◦h−1G¯=Mϕ¯=Mϕ∞. (4.5) Sinceϕ◦h−1(L

h)=Lϕ, we haveLϕ ≤Mϕ∞. Hence,

1ϕ2···ϕk ≤Mϕ1ϕ2···ϕk∞≤M

 k

i=1

1

ri

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Becauseᏹ(h)is the dual of a separable Banach space, we conclude that there isϕ∈(h)such that{ϕ1ϕ2···ϕk}has a subsequence that converges pointwise onΩto

ϕ. Becauseϕp(ωnk)=1 wheneverp≥k, we haveϕ(ωnk)=limp→∞(ϕ1···ϕp)(ωnk)=

(ϕ1···ϕk−1)(ωnk)(−1)k−1.This shows that limk→∞ϕ(ωnk)does not exist. Now suppose that{ωn}is a sequence such that|ωn|converges to a point on∂Vγfor someγ. Fixγ. Letϕ0:Vγ→B(α,r )be a conformal mapping, whereα∈Candr >0. Letϕ1(z)=r /(z−α). Thenϕ2=ϕ1◦ϕ0maps ontoU. Because|ωn|converges to a point on∂Vγ, we conclude that|ϕ2(ωn)| →1 and we work with this sequence as above.

Acknowledgement. The research of the second author is partially supported by a grant from the Shiraz University Research Council.

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Mosaleheh: Department of Mathematics, Shiraz University, Shiraz, Iran E-mail address:mosaleheh@math.susc.ac.ir

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