Generalized multiplication operators on weighted hardy spaces

ISSN 1995-0802, Lobachevskii Journal of Mathematics, 2011, Vol. 32, No. 4, pp. 298–303. c© Pleiades Publishing, Ltd., 2011.

Generalized Multiplication Operators on Weighted Hardy Spaces

Sunil Kumar Sharma1 and B. S. Komal2

(Submitted by F.G. Avkhadiev)1Indus International University, Una (H.P.), India

2M.I.E.T. Kot Bhalwal, Jammu, IndiaReceived June 17, 2011

Abstract—The main purpose of this paper is to study generalized multiplication operators onweighted hardy spaces.

DOI: 10.1134/S1995080211040226

Keywords and phrases: Generalized Multiplication operator, Compact operator, Normal oper-ator, Hermitian operator, Adjoint of an operator, Weighted Hardy spaces.

1. INTRODUCTION

Let {βn} be a sequence of positive real numbers with β(0) = 1. For 1 ≤ p < ∞, let Hp(β)be the space of formal series {f : f(z) =

∑∞n=0 fnzn}, where {fn}∞n=0 is a sequence of complex

number such that∑∞

n=0 |fn|pβpn < ∞. Then Hp(β) is a Banach space under the norm ||f ||pβ =

∑∞n=0 |f(n)|pβp(n) < ∞. For p = 2, the space H2(β) is a Hilbert space under the inner product defined

as 〈f, g〉 =∑∞

n=0 anbnβ2n, where f(z) =

∑∞n=0 anzn and g(z) =

∑∞n=0 bnzn are elements of H2(β).

Let Ω be a region in C and let H(Ω) denote the Banach space of holomorphic functions defined on Ω.If θ : Ω → C is a mapping such that θ.f ∈ H(Ω) whenever f ∈ Ω, then a multiplication transformationMθ : H(Ω) → H(Ω) is defined by Mθf = θf . In case Mθ is continuous, we call it a multiplicationoperator induced by θ. If d is the differential operator,then the generalized multiplication operatorMd

θ : H(Ω) → H(Ω) is defined by Mdθ f = θf ′, where f ′ is the derivative of f .

In this paper we initiate the study of generalized multiplication operators on weighted Hardy spaces.The symbol B(H) denote the Banach algebra of all bounded linear operator on H into itself and No

denote the set {0, 1, 2, 3, . . .}.

2. BOUNDED GENERALIZED MULTIPLICATION OPERATORSON WEIGHTED HARDY SPACES

Our main purpose in this section is to study bounded generalized multiplication operators onweighted Hardy spaces. We show that shift of multiplicity k can be obtained as a generalized multi-plication operator. The adjoint of a generalized multiplication operator is computed.

Theorem 2.1. Let {nβn}∞n=1 be a bounded sequence and let A ∈ B(H2(β)). Then A is a general-ized multiplication operator induced by e1 if and only if A is a diagonal operator with diagonal(1,2,...n,...).

Proof. Suppose A is a generalized multiplication operator induced by e1. Then A = Mdθ , where

θ = e1. Now

Aen = Mdθ en = nMθen−1 = ne1en−1 = nen.

Therefore A is a diagonal operator induced by the diagonal (1, 2, . . . , n, . . .).

298

GENERALIZED MULTIPLICATION OPERATORS 299

Conversely suppose that A is a diagonal operator with diagonal (1, 2, . . . , n, . . .).Clearly Aen = nen. We shall prove that A is a generalized multiplication operator.Consider

Aen = nen = ne1en−1 = e1den = Me1den = Mdθ en for some n ∈ N0

where θ = e1.Hence A is a generalized multiplication operator.

Theorem 2.2. Let θ : Ω → C be a map defined by θ(z) = zk. Then Mdθ : H2(β) → H2(β) is a

bounded operator if and only if

nβn+k−1

βn≤ M∀n ≥ 1 and k ≥ 2.

Proof. Suppose first that the condition is true. Then

(Mdθ f)(z) =

∞∑

n=1

nfnzn+k−1.

Therefore

||Mdθ f ||2 =

∞∑

n=1

n2|fn|2β2n+k−1 =

∞∑

n=1

|fn|2β2n

n2β2n+k−1

β2n

≤ M2||f ||2.

Hence ||Mdθ f || ≤ M ||f || for every f ∈ H2(β).

Therefore Mdθ is a bounded operator.

Conversely suppose that Mdθ is a bounded operator. Then there exists M > 0 such that

||Mdθ en|| ≤ M ||en||

or

||n en+k−1|| ≤ M ||en||

or nβn+k−1

βn≤ M for every n ∈ N and k ≥ 2.

This complete the proof.Theorem 2.3. Let {n βn+k} be a bounded sequence. Let A be the shift of multiplicity k on Ω

defined by Aen = nen+k. Then A is a generalized multiplication operator.Proof. Suppose A is a shift of multiplicity k defined as above.Then

Aen = nek+n = ek+1nen−1 = Mzk+1den = Mdθ en

where θ(z) = zk+1.This complete the proof.Define A: H2(β) → H2(β) by

(A∧f)(n) =

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

0, if n = 0

n

β2n

[∞∑

k=n−1

fkθ(k − n + 1)β2k

]

, for n ≥ 1.

Theorem 2.4. Let Mdθ ∈ B(H2(β)). Then M∗d

θ = A.

Proof. Let f, g ∈ H2(β). Consider

〈f,Ag〉 =∞∑

n=1

f∧n (A∧g)(n)β2

n =∞∑

n=1

f∧n (A∧g)(n)β2

n

LOBACHEVSKII JOURNAL OF MATHEMATICS Vol. 32 No. 4 2011

300 SHARMA, KOMAL

=∞∑

n=0

f∧n+1

[

(n + 1)∞∑

k=n

g(k)θ(k − n)β2k

]

β2n+1 =

∞∑

k=0

g(k)

[k∑

n=0

(n + 1)f∧n+1θ(k − n)β2

n+1

]

β2k.

〈f,Ag〉 =∞∑

k=0

(Md∧θ f)(k)gkβ2

k = 〈Md∧θ f, g〉∀f, g ∈ H2(β).

Hence M∗dθ = A.

Theorem 2.5. Suppose Mdθ ∈ B(H2(β)) is compact operator. Then

limn→∞

limk→∞

n βn+k−1

βk= 0.

Proof. We first assume that Mdθ is a compact operator.

Then for every g =∑∞

n=0 g(n)en ∈ H2(β), we have

〈en, g〉 = g(n)βn → 0 as n → ∞,

since

||g||2 =∞∑

n=0

|g(n)|2β2n < ∞.

Further if the condition of the theorem is not satisfied then there exists ∈> 0 such that n βn+k−1

βk≥ ε for

infinitely many value of k and n.But

||Mdθ en||2 = n2

∞∑

k=0

|θ(k)|2β2n+k−1 ≥ ε

∞∑

k=0

|θ(k)|2β2k

= ε||θ|| for infinitly many value of k,

which contradict the fact that Mdθ is compact. Hence the condition must be satisfied.

This complete the proof.If

θ(z) =

∞∑

n=0

θ∧(n)zn,

write

τn(z) =∞∑

p=n

τ∧n (θ)(p)

βnzp,

whereτ∧n (θ)(p) = θ∧(p − n).

Theorem 2.6. If Mdθ is compact, then τ∧

n → 0 as n → ∞.

Proof. Suppose that Mdθ is a compact operator.

Then for every g =∑∞

k=0 g(k)ek ∈ H2(β), we have

〈en, g〉 = g(n)βn.

Since

||g||2 =

( ∞∑

n=0

|g(n)|2β2n

)

< ∞



It follows that g(n)βn → 0 as n → ∞.Thus en → 0 weakly as n → ∞.Hence Md

θ en → 0 strongly.

That is, ||Mdθ en|| → 0 as n → ∞ or τ∧

n → 0 as n → ∞.This complete the proof.Theorem 2.7. If generalized multiplication operator Md

θ commute with Cφ, then θoφ = θφ′.Proof. Suppose Md

θ commute with Cφ. Then

Mdθ Cφf = CφMd

θ f.

Taking f = e1, we have

CφMdθ e1 = θoφ,

where asMd

θ Cφe1 = θφ′.

Hence θoφ = θφ′.

3. NORMAL AND HERMITIAN GENERALIZED MULTIPLICATION OPERATORSON WEIGHTED HARDY SPACES

This section deals with normal and Hermitian generalized multiplication operators acting onweighted Hardy spaces. In this section we prove that a generalized multiplication operators is Hermitianif and only if θ(z) = az, where a is real.

Theorem 3.1: Let Mdθ ∈ B(H2(β)). Then Md

θ is normal if and only if θ(z) = az for every z ∈ D.

Proof: Let θ(z) =∑∞

n=0 anzn. Suppose Mdθ is normal. Then ||Md

θ eo|| = 0 and

||M∗dθ eo||2 =

∞∑

n=0

|〈M∗dθ eo, en〉|2 =

∞∑

n=1

1β2

n

|〈eo, nθen−1〉|2 = |ao|2β4

o

β21

.

Therefore ao = 0.Now

||Mdθ e1||2 =

∞∑

n=0

|〈Mdθ e1, en〉|2 =

∞∑

n=0

|〈θeo, en〉|2 =∞∑

n=0

|an|2β2n (1)

Also

||M∗dθ e1||2 =

∞∑

n=0

|〈M∗dθ e1, en〉|2 =

∞∑

n=1

1β2

n

|〈e1, nθen−1〉|2 = |a1|2β21(2). (2)

From (1) and (2) we geta2 = 0, a3 = 0, . . . , an = 0

∴ θ(z) = a1z.

Conversely suppose θ(z) = az, where a = a1.Let f =

∑∞n=0 fnen.

Then

||Mdθ f ||2 =

∣∣∣∣∣

∣∣∣∣∣

∞∑

n=0

fnMdθ en

∣∣∣∣∣

∣∣∣∣∣

2

=

∣∣∣∣∣

∣∣∣∣∣

∞∑

n=1

nfnaen

∣∣∣∣∣

∣∣∣∣∣

2

= |a|2∞∑

n=1

n2|fn|2β2n.

Now

||M∗dθ f ||2 =

∞∑

n=0

|〈M∗dθ fn, en〉|2 =

∞∑

n=1

1β2

n

|〈fn, nθen−1〉|2 = |a|2∞∑

n=1

n2|fn|2β2n.


302 SHARMA, KOMAL

Hence Mθd is normal.

Theorem 3.2. Mdθ is Hermitian if and only if θ(z) = az where a is real.

Proof: Suppose that the condition is true. Then for f, g ∈ H2(β) we have

〈Mdθ f, g〉 = 〈θf ′, g〉 = 〈ae1f

′, g〉 = a

∞∑

n=1

nfngnβ2n

and

〈M∗dθ f, g〉 = 〈f,Md

θ g〉 = a〈f, e1g′〉 = a

∞∑

n=1

nfngnβ2n as “a” is real.

Hence 〈M∗dθ f, g〉 = 〈Md

θ f, g〉 for every f, g ∈ H2(β). This proves that Mdθ is Hermitian. Conversely

suppose that Mdθ is Hermitian.

Let

θ =∞∑

k=0

θkzk.

Then

〈M∗dθ e1, e0〉 = 〈Md

θ e1, e0〉implies that

〈e1, 0〉 = 〈θ, e0〉or

θ(0)β20 = 0,

which implies that θ(0) = 0.Again

〈M∗dθ e2, e1〉 = 〈Md

θ e2, e1〉implies that

〈e2, θe0〉 = 0

orθ(2)β2

2 = 0

which implies

θ(2) = 0.

In general

〈M∗dθ en, e1〉 = 〈Md

θ en, e1〉implies that

〈en, θe0〉 = 0

orθ(n)β2

n = 0

which implies that

θ(n) = 0.

Thus θ(n) = 0 for all n = 0, 2, 3, . . ., that is θ(z) = θ(1)z or θ = θ(1)e1.Now

〈M∗dθ e1, e1〉 = 〈Md

θ e1, e1〉



or

〈e1, θ〉 = 〈θ, e1〉or

θ(1)β21 = θ(1)β2

1

so that

θ(1) = θ(1).

Hence θ(z) = az, where a = θ(1), is real.Example 3.3. Let θ : Ω → Ω be defined by θ(z) = 2z.Then

(Mdθ f)(w) = 〈θf ′, kw〉 =

∞∑

n=1

2nfnwn

and

(M∗dθ f)(w) = 〈M∗d

θ f, kw〉 = 〈f,Mdθ kw〉 =

∞∑

n=1

2nfnwn.

Therefore

(Mdθ f)(w) = (M∗d

θ f)(w).

This is true for each w ∈ Ω and f ∈ H2(β). Hence Mdθ = M∗d

θ . This shows that Mdθ is Hermitian.

REFERENCES1. C. C. Cowen, G. Gunatillake, and E. Ko, Hermitian weighted composition operators on weighted Hardy

spaces, Preprint.2. C .C. Cowen and B. D. MacCluer, Compostion operators on spaces of analytic functions (CRC Press,

Boca Raton, 1995).3. G. Gunatillake, Proc. Amer. Math. Soc. 136, 2895 (2008).4. B. S. Komal and R. S. Pathania, Bull. Cal. Math. Soc. 83, 515 (1991).5. B. D. MacCluer, X. Zong, and N. Zorboska, Illinios Journal of Mathematics 40 (4), 662 (1996).6. R. K. Singh and A. Kumar, Bull. Aust. Math. Soc. 16, 247 (1977).7. R. K. Singh and B. S. Komal, Bull. Austral. Math. Soc. 18, 439 (1978).8. S. K. Sharma and B. S. Komal, Int. Journal of Math Analysis, 5 (12), 1067 (2011).9. S. Sharma, B. S. Komal, and S. K. Sharma, Int. J. Contemp. Math. Sciences 5 (12), 579 (2010).

10. J. H. Shapiro and P. D. Taylor, Indiana. U. Math. J. 23, 471 (1973).11. A. L. Shields, Weighted shift operators and analytic function theory, Math. survey, A.M.S. Providence 13, 49

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Generalized multiplication operators on weighted hardy spaces

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