J. Math. Anal. Appl. 401 (2013) 416–429

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Journal of Mathematical Analysis andApplications

journal homepage: www.elsevier.com/locate/jmaa

Hilbert spaces of entire Dirichlet series andcomposition operators

Hou Xiaolu, Hu Bingyang, Le Hai Khoi ∗Division of Mathematical Sciences, School of Physical and Mathematical Sciences, Nanyang Technological University (NTU), 637371 Singapore, Singapore

a r t i c l e i n f o

Article history:

Received 8 May 2012

Available online 23 December 2012

Submitted by Thomas Ransford

Keywords:

Hilbert space

Entire Dirichlet series

Ritt order

Logarithmic orders

Composition operator

a b s t r a c t

The aim of this paper is to introduce Hilbert spaces of entire Dirichlet series with real

frequencies and consider composition operators on these spaces. We establish necessary

and sufficient conditions for such series to have Ritt order zero, as well as to have finite

logarithmic orders. This allows us to apply the Polya theorem on composition of entire

functions to consideration of composition operators on theHilbert spaces of entireDirichlet

series. In particular, criteria for action, boundedness, compactness and compact difference

of such operators are obtained.

© 2012 Elsevier Inc. All rights reserved.

1. Introduction

Let us consider Dirichlet series with real frequencies

∞∑n=1

ane−λnz, an, z ∈ C, (1.1)

where 0 ≤ (λn) ↑ ∞ is a given sequence of real numbers.

As is well known, if we let

L = lim supn→∞

log n

λn

,

and set

D = lim supn→∞

log |an|λn

,

then in case L < ∞, the series (1.1) represents an entire function in C if and only if D = −∞ (see, e.g., [4,9]).

Throughout this paper, the condition L < ∞ is always supposed to hold.

Let H be a space of holomorphic functions on a set G ⊆ C. If an analytic function ϕ maps G into itself, the compositionoperator Cϕ is a linear operator defined by the rule

(Cϕ f )(z) = f ◦ ϕ(z), z ∈ G, f ∈ H .

∗ Corresponding author.

E-mail addresses: HO0001LU@e.ntu.edu.sg (X. Hou), BHU2@e.ntu.edu.sg (B. Hu), lhkhoi@ntu.edu.sg, lhkhoister@gmail.com (L.H. Khoi).

0022-247X/$ – see front matter© 2012 Elsevier Inc. All rights reserved.

doi:10.1016/j.jmaa.2012.12.036

X. Hou et al. / J. Math. Anal. Appl. 401 (2013) 416–429 417

Composition operators have been investigated on various spaces of holomorphic functions of one variable, as well asin higher dimensions. We refer the reader to the excellent monographs [2,14] for detailed information. In particular, anextensive study of composition operators was carried out on spaces of classical Dirichlet series, that is for the case when thefrequencies λn = log n, ∀n ≥ 1 (see, e.g., [1,3,5]).

The aim of the present paper is to study composition operators on Hilbert spaces of entire Dirichlet series. It should benoted that we make use of Pólya’s result [11], and therefore, the three notions of order for Dirichlet series (ordinary, Rittand logarithmic orders) play an important role in our approach. It seems this topic has never been treated before.

The structure of the papers is as follows. In Section 2we introduce some classes of Hilbert spaces of entire Dirichlet serieswhich arise in depending on the weight sequences. Several auxiliary properties needed in the sequel are provided. Section 3deals with a study of when the series of those Hilbert spaces have Ritt order zero, and finite logarithmic orders. The resultsobtained in this section allow to use the Polya result on composition of entire functions of finite ordinary order in the mainpart of the paper. In Section 4 we study various problems of composition operators on Hilbert spaces of entire Dirichletseries whose elements have Ritt order zero and finite logarithmic orders. The criteria for action, boundedness, compactnessand compact differences are obtained.

We note that some of our results were announced in [6].

2. Hilbert space of entire Dirichlet series

Let Λ := (λn), with 0 ≤ (λn) ↑ ∞ satisfying condition L < ∞, be given. Consider the normed space of entire Dirichletseries

H(E) :={f (z) =

∞∑n=1

ane−λnz : (an) ∈ E

},

where

E ={(an) : lim sup

n→∞log |an|

λn

= −∞, or limn→∞ |an|1/λn = 0

}.

The norm in this space is defined by the inner product

〈f , g〉 :=∞∑n=1

anbn, f (z) =∞∑n=1

ane−λnz, g(z) =

∞∑n=1

bne−λnz .

In [7], different properties of composition operators on a subspace H(E, ρ, 0) of the space H(E), such as action,boundedness, were considered.

It is important to note that the space H(E, ρ, 0) as well as H(E) are never complete with respect to the norm above, andhence other important properties of composition operators (compactness, compact differences, . . . ) cannot be considered.

Then a natural question to ask is: how do we define subspaces of H(E) that can be Hilbert spaces? We are going toconsider this question.

To each sequence of real positive numbers β = (βn) we associate the following sequence space

�2β =

⎧⎨⎩a = (an) : ‖a‖ =

( ∞∑n=1

|an|2β2n

)1/2

< ∞⎫⎬⎭ ,

which is a Hilbert (sequence) space with the inner product

〈a, b〉 =∞∑n=1

anb̄nβ2n , ∀(an), (bn) ∈ �2

β, (2.1)

(see, e.g., [15]).First we note the following elementary result (partially stated earlier in [8] under the condition L = 0 with complex

frequencies), which is used very often in our discussion.

Lemma 2.1. Let 0 ≤ (λn) ↑ ∞ be given. The following conditions are equivalent:

(a) lim supn→∞log n

λn= L < ∞.

(b) ∀r > L : ∑∞n=1 e

−rλn < ∞.

(c) ∃ρ > 0 : ∑∞n=1 e

−ρλn < ∞.

(d) lim supn→∞log n

λn≤ ρ .

The above conditions imply

(e) ∀r < L : ∑∞n=1 e

−rλn = ∞.

418 X. Hou et al. / J. Math. Anal. Appl. 401 (2013) 416–429

Put

lim infn→∞

logβn

λn

= β∗, lim supn→∞

logβn

λn

= β∗.

In [8] the characterization of the relationship between �2β and E in terms of β∗ and β∗ was given. Analyzing the proof in [8]

we can prove, by a similar method, that such a characterization is true in a general case 0 ≤ L < +∞ for entire Dirichletseries (1.1). Namely, we have the following result.

Proposition 2.2. There are three alternative possibilities

(1) �2β ⊂ E, which is equivalent to β∗ = ∞.

(2) E ⊂ �2β , which is equivalent to β∗ < ∞.

(3) �2β \ E �= ∅ and E \ �2

β �= ∅, which is equivalent to β∗ < β∗ = ∞.

As a consequence, these spaces never coincide.

Proof. As the proof of (1) and (2) are quite similar, we prove (2). Combining (1) and (2) yields (3).

(2) Let E ⊂ �2β . Assume that β∗ = ∞. In this case there exist (Mp) ↑ ∞ and (np) ↑ ∞ such that

logβnp

λnp

> Mnp , ∀p ≥ 1.

Define a sequence (an) as follows

an ={e−Mpλnp , if n = np, p = 1, 2, . . .0, otherwise.

Then we have (an) ∈ E, but (an) clearly is not in �2β : a contradiction.

Conversely, suppose that β∗ < ∞. In this case there existM > 0 and N1 ∈ N such that

logβn

λn

≤ M, ∀n > N1.

Let ε > 0 be given. For an arbitrary (an) ∈ E, there is N2 ∈ N such that

|an|1/λn < ε < e−M−L−1.

Then for all n > max{N1,N2} we have∑n>N

|an|2β2n ≤

∑n>N

ε2λnβ2n ≤

∑n>N

ε2λne2Mλn

=∑n>N

(e−2(M+L+1)e2M)λn =∑n>N

(e−2(L+1))λn < ∞,

due to Lemma 2.1(b). That is, E ⊂ �2β . �

In the sequel the condition β∗ = ∞ is always supposed to hold. That is, the sequence of positive real numbers (βn)satisfies the condition

lim infn→∞

logβn

λn

= ∞, or the same, limn→∞

logβn

λn

= ∞. (E)

By Proposition 2.2, we always have �2β ⊂ E, and therefore, we can consider a subspace H(E, β) of the normed space

H(E), which consists of all entire Dirichlet series with coefficients from �2β , that is

H(E, β) ={f (z) =

∞∑n=1

ane−λnz : (an) ∈ �2

β

}. (2.2)

This is a Hilbert space with the inner product (2.1) above.

X. Hou et al. / J. Math. Anal. Appl. 401 (2013) 416–429 419

3. Ritt order and logarithmic orders

3.1. Different notions of orders for Dirichlet series

Let f (z) be an entire function in C. The ordinary order ρ of f is defined as the greatest lower bound of values of μ suchthat Mf (r) < er

μfor all sufficiently large r , where Mf (r) = max|z|=r |f (z)| , r > 0, is its maximum modulus (function).

This order can be computed by

ρ(f ) = ρ := lim supr→∞

log logMf (r)

log r,

(see, e.g., [10, II - Chapter 9]).For an entire Dirichlet series f (z) = ∑∞

n=1 ane−λnz , there exists the so-called Ritt order, which is defined, via the quantity

M(σ ) = supt∈R |f (σ + it)|, σ ∈ R, as follows

ρR = lim supσ→−∞

log logM(σ )

−σ.

The Ritt order can also be expressed in terms of coefficients of the series (see [13]):

ρR = lim supn→∞

λn log λn

log 1|an|

.

For entire Dirichlet series with Ritt order 0, Reddy defined the following logarithmic orders (see [12])

ρR(L) = lim supσ→−∞

log logM(σ )

log(−σ), ρ∗(L) = lim sup

σ→−∞log logμ(σ)

log(−σ),

and especially,

ρc(L) = lim supn→∞

log λn

log

(1λn

log 1|an|

) .

He showed that ρR(L) = ρ∗(L) = ρc(L) + 1 ≥ 1.As noted in [7], for an entire Dirichlet series, if ρR > 0, then ρR(L) = ∞, and hence there is not much information we

can get from logarithmic orders.

3.2. Ritt order for series from H(E, β)

We are interested in the following question: when does each series from H(E, β) have Ritt order zero? The answer tothis question is given by the following characterization.

Theorem 3.1. Every function f ∈ H(E, β) has Ritt order zero if and only if the following condition (R) is satisfied:

lim infn→∞

logβn

λn log λn

= ∞. (R)

Proof (Necessity). Assume that condition (R) is false, that is

lim infn→∞

logβn

λn log λn

< ∞.

In this case, there existsM > 0 and (nk) ↑ ∞ such that

logβnk < Mλnk log λnk , ∀k ≥ 1.

Put

an =⎧⎨⎩

1

nλMλnn

, n ∈ {nk} (k = 1, 2, . . .)

0, otherwise.

Then we have

∞∑n=1

|an|2β2n =

∞∑k=1

|ank |2β2nk

≤∞∑k=1

1

n2k

< ∞,

which means that (an) ∈ �2β , that is, the function

∑∞n=1 ane

−λnz ∈ H(E, β).

420 X. Hou et al. / J. Math. Anal. Appl. 401 (2013) 416–429

On the other hand, since L = lim supn→∞log n

λn< ∞, there exists N > 0 such that ∀n > N

log n < (L + 1)λn,

which implies that for all k large enough (nk > N), we have

λnk log λnk

log nk + Mλnk log λnk

≥ λnk log λnk

(L + 1)λnk + Mλnk log λnk

.

However,

limn→∞

λnk log λnk

(L + 1)λnk + Mλnk log λnk

= 1

M.

Hence

lim supn→∞

λn log λn

log 1|an|

≥ lim supk→∞

λnk log λnk

log nk + Mλnk log λnk

≥ 1

M,

which shows that the Ritt order ρR(f ) ≥ 1M

> 0: a contradiction.

Sufficiency. Assume that there is a series∑∞

n=1 ane−λnz ∈ H(E, β) whose Ritt order is positive, that is

lim supn→∞

λn log λn

log 1|an|

> 0.

Then there exists a constantM > 0 and (nk) ↑ ∞ such that

λnk log λnk

log 1|ank |

>1

M,

or equivalently,

|ank | > λ−Mλnknk , ∀k ≥ 1.

Furthermore, by condition (R), there exists N > 0 such that for all n > N we have

logβn > Mλn log λn.

Then for all k large enough (nk > N), we have

|ank |βnk ≥ λ−Mλnknk λ

Mλnknk = 1,

which means that the series∑∞

n=1 |an|2β2n diverges: a contradiction. �

Remark 3.2. Condition (R) is obviously stronger than condition (E). Moreover, the example βn = λλnn (n = 1, 2, . . .), shows

that the inverse implication is not true.

Due to Theorem 3.1 we can introduce the following Hilbert space of entire Dirichlet series, whose elements have Rittorder zero

H(E, βR) :={ ∞∑

n=1

ane−λnz : (an) ∈ �2

β, β ∈ (R)

}.

Note that there is no information about whether the logarithmic orders of series in the space H(E, βR) are finite orinfinite. In the next subsection we study this problem.

3.3. Logarithmic orders for series from H(E, βR)

By Theorem 3.1, the elements of the space H(E, βR) all have Ritt order zero. Therefore, it makes sense to study thelogarithmic orders of the series of this space. More precisely, a question to ask is: under what condition do all these serieshave finite logarithmic orders? We have the following characterization.

Theorem 3.3. Every function from H(E, βR) has a finite logarithmic order if and only if the following condition (S) is satisfied

lim infn→∞

logβn

λ1+αn

= +∞, for some α > 0. (S)

X. Hou et al. / J. Math. Anal. Appl. 401 (2013) 416–429 421

Proof. As for Theorem 3.1, we use a method of contradiction. We omit the proof of sufficiency, as it is quite similar to thatof Theorem 3.1.

Necessity. Assume that condition (S) fails to hold. From this it follows, in particular, that for eachm = 1, 2, . . . with αm = 1m

we always have

lim infn→∞

logβn

λ1+αmn

= Lm < ∞.

Then there exists a sequence (n(m)k ) ↑ ∞ such that

logβn(m)k

λ1+αm

n(m)k

< Lm + 1,

which gives

βn(m)k

≤ e(Lm+1)λ

1+αm

n(m)k . (3.1)

Furthermore, as (λn) ↑ ∞, there exists N (m) > 0 such that for all k with n(m)k > N (m)

Lm + 1 ≤ λαm

n(m)k

,

or equivalently,

βn(m)k

≤ exp

(λ1+2αm

n(m)k

).

Now we apply the standard ‘‘diagonalization process’’ to choose n(1)k1

> N (1) and n(2)k2

> max{n(1)k1

,N (2)}. Having chosen

n(p)kp

, choose n(p+1)kp+1

> max{n(p)kp

,N (p+1)}. Just for a sequence (n(p)kp

)∞p=1 ↑ ∞, denoted for simplicity by lp := n(p)kp

, we have

βlp < exp

(λ1+2αp

lp

), αp = 1

p, ∀p ≥ 1.

We define a series f = ∑∞n=1 ane

−λnz as follows:

an =⎧⎨⎩

1

lpexp

(−λ

1+2αp

lp

), n ∈ (lp), p = 1, 2, . . .

0, otherwise.

Then ∑n

|an|2β2n =

∑p

|alp |2β2lp

≤∑p

1

lp2exp

(λ1+2αp

lp− λ

1+2αp

lp

)2 =∑p

1

lp2

< ∞,

which means that (an) ∈ �2β .

Now the most important step is to prove that ρc(L)(f ) = ∞. We have

ρc(L)(f ) = lim supn→∞

log λn

log

(1λn

log 1|an|

) = lim supp→∞

log λlp

log

(1

λlplog

(1

1lp

exp

(−λ

1+2αplp

)))

= lim supp→∞

log λlp

log

(1

λlp

(log lp + λ

1+2αp

lp

)) = lim supp→∞

log λlp

log

(log lp

λlp+ λ

2αp

lp

) .

Since

L = lim supn→∞

log n

λn

< ∞,

there exists P > 0 large enough such that for all pwith lp > L,

log lp

λlp

< L + 1

422 X. Hou et al. / J. Math. Anal. Appl. 401 (2013) 416–429

which implies that

lim supp→∞

log λlp

log

(log lp

λlp+ λ

2αp

lp

) ≥ lim supp→∞

log λlp

log

(L + 1 + λ

2αp

lp

) .

We prove that

limp→∞

log λlp

log

(L + 1 + λ

2αp

lp

) = ∞.

Take and fixM > 0. Since (λlp) ↑ ∞, there exists P ′ > 0 such that for all p > P ′

λlp > (L + 2)2M , and λlp > 1.

Put P ′′ = max{4M, P ′}, then ∀p > P ′′, sincelog λlp

log

(L + 1 + λ

2αp

lp

) > M ⇐⇒ λ1Mlp

> L + 1 + λ2αp

lp⇐⇒ λ

1Mlp

− λ2p

lp> L + 1,

we show that the last inequality holds. Indeed, as p > 4M , and p > P ′ we have

λ1Mlp

− λ2p

lp> λ

1Mlp

− λ12Mlp

= λ12Mlp

(λ

12Mlp

− 1

)> (L + 2)(L + 1) > L + 1.

This completes the proof. �

Remark 3.4. We can easily see that condition (S) is stronger than condition (R), and by the example βn = λλn log λnn (n = 1,

2, . . .), the inverse implication is not true.

By Theorem 3.3 we introduce the following Hilbert space of entire Dirichlet series, whose elements have Ritt order zeroand finite logarithmic order

H(E, βS) :={ ∞∑

n=1

ane−λnz : (an) ∈ �2

β, β ∈ (S)

}.

This is the main space we consider in this paper.

4. Composition operators

4.1. Composition operators on the space H(E, βS)

An analytic mapping ϕ : C → C induces the composition operator Cϕ defined by(Cϕ(f )

)(z) = f ◦ ϕ(z), z ∈ C, f ∈ H(E, βS).

We note that H(E, βS) is a functional B-space. Indeed, for every z ∈ C fixed, if f (z) = ∑n ane

−λnz ∈ H(E, βS), by theCauchy–Schwarz inequality, we have∣∣∣∣∣

∑n

ane−λnz

∣∣∣∣∣2

≤(∑

n

|an|2 β2n

)·(∑

n

e−2λnRe z

β2n

).

To have a continuous point evaluation at z, it is sufficient that the second sum converges. By condition (S), there exists an

N1 ∈ N such that βn > eλ1+αn , ∀n > N1. Moreover, since L < ∞, there is an N2 ∈ N such that Re z + λα

n > L, ∀n > N2. Thenfor N = max{N1,N2}, we have∑

n>N

e−2λnRe z

β2n

≤∑n>N

e−2λn(Re z+λαn ) ≤

∑n>N

e−2Lλn < ∞,

by Lemma 2.1.By a similar argument, replacing λα

n by log λn, we can verify that the space H(E, βR) is also a functional Banach space.In [7], we considered the composition operators on the classes of entire Dirichlet series whose Ritt order is zero and

logarithmic order is finite. The criteria for an action and boundedness are obtained.The results obtained in the previous sections on Ritt order and logarithmic order of the series from H(E, β) allow us to

consider composition operators in the space H(E, βS). As this space is a Hilbert space, we can work also on other problems,such as compactness, compact difference, etc. which we could not have for [7].

X. Hou et al. / J. Math. Anal. Appl. 401 (2013) 416–429 423

4.2. Criterion for the action

We note that series from H(E, βS) have Ritt order zero and finite logarithmic orders. By [7, Theorem 3.2], the ordinaryorders of these series are finite. This important fact allows us, as in [7], to apply the Polya result [11].

Lemma 4.1 (Polya). If g(z) and h(z) are entire functions and g(h(z)

)is an entire function of finite ordinary order, then there are

only two possible cases:

(a) the internal function h(z) is a polynomial and the external function g(z) is of finite ordinary order;(b) the internal function h(z) is not a polynomial but a function of finite ordinary order, and the external function g(z) is of zero

ordinary order.

The first issue to study is the action of Cϕ on the space H(E, βS). We consider a problem: for which ϕ does f in H(E, βS)implies that f ◦ ϕ is also in H(E, βS)? We have the following criterion.

Theorem 4.2. Let ϕ(z) be an entire function on C. Let further, (βn) satisfy condition (S). The composition operator Cϕ(f ) = f ◦ϕmaps H(E, βS) into itself if and only if ϕ(z) is of the form

ϕ(z) = z + b, with Re b ≥ 0. (4.1)

• Proof of sufficiency. Suppose that

ϕ(z) = z + b, with Re b ≥ 0.

Take an arbitrary f (z) = ∑∞n=1 ane

−λnz ∈ H(E, βS), we have ‖f ‖2 = ∑∞n=1 |an|2β2

n < ∞. Then

Cϕ(f )(z) = f (ϕ(z)) =∞∑n=1

ane−λn(z+b) =

∞∑n=1

ane−λnbe−λnz,

and hence

‖Cϕ(f )‖2 =∞∑n=1

∣∣ane−λnb∣∣2 β2

n =∞∑n=1

|an|2e−2λnRe bβ2n ≤

∞∑n=1

|an|2β2n = ‖f ‖2 < ∞, (4.2)

as Re b ≥ 0, which means that Cϕ(f ) ∈ H(E, βS).Moreover, inequality (4.2) shows that Cϕ is bounded on H(E, βS).

• Proof of necessity. This is the most difficult part. For proving this, we need some notation and auxiliary results.The following result is simple.

Lemma 4.3. If f (z) ∈ H(E, βS) and am is the first nonzero coefficient of its representation, then

limRe z→+∞ eλmzf (z) = am.

Proposition 4.4. Suppose that the composition operator Cϕ(f ) = f ◦ ϕ maps H(E, βS) into itself. Then

ϕ(z) = az + b, (4.3)

where a ≥ 1 satisfies the condition

∀k ∈ N ∃N(k) such that a = λN(k)

λk

. (4.4)

Proof. Suppose Cϕ maps H(E, βS) into itself. We follows the proof in [7].Let f be an arbitrary Dirichlet series in H(E, βS) with nonzero ordinary order. Such function does exist, of course. For

example, all functions e−λkz (k ≥ 1) have ordinary order 1 [7]. Then f ◦ ϕ ∈ H(E, βS), and so by Polya Lemma 4.1, ϕ is apolynomial. We write

ϕ(z) = anzn + an−1z

n−1 + · · · + az + b.

As for each k ∈ N, the function qk(z) = e−λkz is in H(E, βS), we have Cϕ(qk) = qk(ϕ(z)

) ∈ H(E, βS), which means that

the function e−λkϕ(z) can be represented in the form

e−λkϕ(z) =∞∑

m=N(k)

b(k)m e−λmz, (4.5)

where N(k) is the index of the first nonzero coefficient.

424 X. Hou et al. / J. Math. Anal. Appl. 401 (2013) 416–429

Multiplying both sides of (4.5) by eλN(k)z , by Lemma 4.3 we have

limRe z→+∞ e−λkϕ(z)+λN(k)z = b

(k)

N(k).

Since −λkϕ(z) + λN(k)z is entire, taking log as the principal branch of the logarithm function, there exists � ∈ Z such that

limRe z→+∞(−λkϕ(z) + λN(k)z) = log b

(k)

N(k) + 2π i�.

Dividing both sides by zλk we get

limRe z→+∞

(−

(anz

n−1 + an−1zn−2 + · · · + a + b

z

)+ λN(k)

λk

)= 0,

or equivalently,

limRe z→+∞

(anz

n−1 + an−1zn−2 + · · · + a2z + a + b

z

)= λN(k)

λk

.

By the assumption, λN(k) is a finite number, and so we must have

an = an−1 = · · · = a2 = 0,

which implies that

a = λN(k)

λk

, ∀k ∈ N,

and conditions (4.3)–(4.4) follow. In particular, for k = 1, we have a = λN(1)

λ1≥ 1, because N(1) ≥ 1 and the sequence (λk)

is increasing. �

Lemma 4.5. There are infinitely many n such that

βN(n) ≥ βn. (4.6)

Proof. – For the case a = 1, we have N(n) = n, ∀n ≥ 1, and the statement of the lemma is obvious.– Now consider the case a > 1. Assume that (4.6) is false, that is there are only finitely many n such that βN(n) ≥ βn. Then

there exists K > 0 such that

βN(n) < βn, ∀n ≥ K . (4.7)

Denote N (k)(n) = N(N (k−1)(n)

)for k ≥ 1, with N (0)(n) = n. We already noted above that N(n) > n, and hence

n < N(n) < N (2)(n) < · · · < N (n)(n) < · · ·In particular, for n = K we get an increasing sequence positive integers

K < N(K) < N (2)(K) < · · · < N (n)(K) < · · · ,for which, by (4.7),

βK > βN(K) > · · · > βN(n)(K) > · · · .Consequently,

logβN(n)(K)

[λN(n)(K)]1+α<

logβK

[λN(n)(K)]1+α,

where α is taken from condition (S).Letting n → ∞ in the last inequality, by condition (S), we get

limn→∞

logβN(n)(K)

[λN(n)(K)]1+α= ∞ ≤ lim

n→∞logβK

[λN(n)(K)]1+α= 0,

which is impossible. �

Now we introduce a new quantity which is useful in the sequel. Take and fix a positive integer k0, we denote

Bn := logβN(n)(k0)

λ1+α

N(n)(k0)

, n = 1, 2, . . . ,

where α > 0 taken from condition (S).

X. Hou et al. / J. Math. Anal. Appl. 401 (2013) 416–429 425

Lemma 4.6. There are infinitely many n such that

Bn ≤ Bn+1.

Proof. The proof of this lemma is similar to the previous one. �

Proposition 4.7. For the symbol ϕ(z) in formula (4.3), the coefficient a must equal to 1, that is

ϕ(z) = z + b, b ∈ C.

Proof. Assume that a > 1, which implies that N(k) > k for all k ≥ 1. By Lemma 4.6, there exists a sequence (np) ↑ ∞ suchthat Bnp ≤ Bnp+1, ∀p ≥ 1.

Setting N (np)(k0) = lp, ∀p ≥ 1, we have a sequence (lp) ↑ ∞, for which

N (np+1)(k0) = N(lp), p = 1, 2, . . .

Then the inequality Bnp ≤ Bnp+1, ∀p ≥ 1 is rewritten as follows

logβlp

λ1+αlp

≤ logβN(lp)

λ1+αN(lp)

, ∀p ≥ 1.

Note that λN(lp) = aλlp , ∀p ≥ 1, the last inequality becomes

βN(lp) ≥ βa1+α

lp, ∀p ≥ 1. (4.8)

Now we consider a sequence (an) defined by

an =⎧⎨⎩

1

nβn

, if n ∈ {lp} (p = 1, 2, . . .)

0, otherwise.

It is clear that (an) ∈ �2β , which means that the function f (z) = ∑∞

n=1 ane−λnz ∈ H(E, βS).

We are going to prove that g(z) = Cϕ(f )(z) �∈ H(E, βS).

Since lim supn→∞log n

λn= L < ∞, we have lim supp→∞

log lp

λlp= Lp ≤ L < ∞, which means that

∃P1 > 0 ∀p ≥ P1 : log lp

λlp

< Lp + 1,

or equivalently,

lp < eλlp (Lp+1)

, ∀p ≥ P1. (4.9)

Furthermore, from condition (S), it follows that

lim infp→∞

logβlp

λ1+αlp

= ∞,

which shows that

∃P2 > 0 ∀p ≥ P2 : βlp > eλ1+αlp . (4.10)

Note that

‖g‖2 =∞∑n=1

|an|2e−2λnRe bβ2N(n).

Then for each p ≥ max{P1, P2}, by (4.8)–(4.10), we have

|alp |2e−2λlpRe bβ2N(lp)

≥ |alp |2e−2λlpRe bβ2a1+α

lp= 1

lp2e−2λlpRe bβ

2(a1+α−1)lp

≥ e−2λlp (Lp+1)−2λlpRe bβ

2(a1+α−1)lp

≥ e−2λlp (Lp+1)−2λlpRe be

2λ1+αlp

(a1+α−1)

= e2λlp [λα

lp(a1+α−1)−(Lp+1+Re b)]

.

426 X. Hou et al. / J. Math. Anal. Appl. 401 (2013) 416–429

Since λαlp(a1+α − 1) → ∞ as p → ∞, we have

∃P3 ∀p ≥ P3 : λαlp(a1+α − 1) − (Lp + 1 + Re b) > 0.

Thus we arrive at the fact that ∀p ≥ max{P1, P2, P3}|alp |2e−2λlpRe bβ2

N(lp)≥ e

2λlp [λαlp

(a1+α−1)−(Lp+1+Re b)] ≥ 1,

which shows that

∞∑n=1

|an|2e−2λnRe bβ2N(n) = ∞,

or equivalently, g �∈ H(E, βS). This completes the proof of the proposition. �

Proposition 4.8. For the symbol ϕ(z) = z + b, the real part of b must be non-negative, that is

Re b ≥ 0.

Proof. The proof of this proposition is similar to the previous one. �

Now the necessity part of Theorem 4.2 follows from Propositions 4.4, 4.7 and 4.8.

4.3. Criteria for boundedness, compactness and compact difference

4.3.1. Boundedness

From the proof of Theorem 4.2, it follows that when Cϕ acts from H(E, βS) into itself, this composition operator isautomatically bounded on H(E, βS). That is, we have the following result.

Theorem 4.9. Let ϕ(z) be an entire function on C. Let further, (βn) satisfy condition (S). The composition operator Cϕ(f ) = f ◦ϕis bounded on H(E, βS) if and only if ϕ(z) is of the form (4.3), that is

ϕ(z) = z + b, with Re b ≥ 0.

4.3.2. Compactness

Next we study the compactness of Cϕ .

Recall that a sequence of points (xn) in a Hilbert space H with the inner product 〈·, ·〉, is said to converge weakly to apoint x ∈ H if

〈xn, y〉 → 〈x, y〉, ∀y ∈ H.

The following criterion for compactness is used in our study.

Lemma 4.10 (See, e.g., [16]). Let H be a Hilbert space with the inner product 〈·, ·〉, T a linear operator on H. Then T is compactif and only if ‖T (xn)‖ → 0 whenever (xn) → 0 weakly in H.

Our second main result, which concerns compactness of Cϕ , is as follows.

Theorem 4.11. Let ϕ(z) be an entire function onC. Let further, (βn) satisfy condition (S). The composition operator Cϕ(f ) = f ◦ϕis compact on H(E, βS) if and only if

ϕ(z) = z + b, with Re b > 0. (4.11)

Proof (Necessity). Suppose Cϕ is compact on H(E, β). Then it is bounded and hence, by Theorem 4.9, ϕ(z) = z + b withRe b ≥ 0. It remains to prove that Re b > 0.

Consider a sequence of functions (qm) in H(E, β) of the following form

qm(z) = 1

βm

e−λmz, m = 1, 2, . . .

For any f = ∑∞n=1 bne

−λnz ∈ H(E, βS), since∑∞

n=1 |bn|2β2n < ∞, we have

limn→∞ |bn|βn = 0,

X. Hou et al. / J. Math. Anal. Appl. 401 (2013) 416–429 427

which implies that

limm→∞〈qm, f 〉 = lim

m→∞1

βm

b̄mβ2m = lim

m→∞ b̄mβm = 0.

This means (qm) converges to 0 weakly in H(E, β). By Lemma 4.10, the sequence (Cϕ(qm)) must converge to 0 in the norm

topology, that is∥∥Cϕ(qm)

∥∥ → 0 asm → ∞.

On the other hand, since

Cϕ(qm)(z) = qm(ϕ(z)) = qm(z + b) = 1

βm

e−λmbe−λmz,

we have∥∥Cϕ(qm)∥∥ = e−λmRe b,

whence (4.11) follows.

Sufficiency. Suppose ϕ(z) = z + b with Re b > 0. We prove that Cϕ is compact on H(E, βS).

Take an arbitrary sequence (fk) ⊂ H(E, βS) which converges to 0 weakly. Denote

fk(z) =∞∑

m=1

a(k)m e−λmz .

Then we have

gk(z) := (Cϕ(fk))(z) =∞∑

m=1

a(k)m e−λmbe−λmz

in H(E, β).

Moreover,

‖gk‖2 =∞∑

m=1

∣∣a(k)m

∣∣2 e−2λmRe bβ2m.

We prove that ‖gk‖2 → 0 as k → ∞.

Note that by the weak convergence of (fk) to 0, for each function qm(z) = 1βm

e−λmz ∈ H(E, βS), we have

limk→∞〈fk, qm〉 = lim

k→∞ a(k)m βm = 0 (m = 1, 2, . . .). (4.12)

Furthermore, since (fk) converges weakly to 0, it is bounded (by the uniform boundedness principle), which means thatthere is a positive constant C such that ‖fk‖ ≤ C, ∀k ≥ 1.

Note that conditions Re b > 0 and (λm) ↑ ∞ give limm→∞ e−2λmRe b = 0. Then for an arbitrary ε > 0 given, there existsM > 0 such that for allm > M

e−2λmRe b <ε

2C2. (4.13)

By condition (4.12), for eachm = 1, 2, . . . ,M there is Km ∈ N such that ∀k > Km∣∣a(k)m

∣∣2 β2m <

ε

2M. (4.14)

From these facts, (‖gk‖) is clearly seen to converge to 0, as k → ∞. By Lemma 4.10, Cϕ is compact. �

4.3.3. Compact difference of composition operators

Finally, we study a compact difference of composition operators.

Theorem 4.12. Suppose that ϕ1 = z + b1, ϕ2 = z + b2 with Re b1 ≥ 0, Re b2 ≥ 0 (that is composition operators Cϕ1and Cϕ2

act fromH(E, βS) into itself). Then the difference Cϕ1−Cϕ2

is compact onH(E, βS) if and only if either of the following conditionsis satisfied:

(1) Both Cϕ1and Cϕ2

are compact on H(E, βS) (that is Re b1 > 0, Re b2 > 0).

(2) Re b1 = Re b2 = 0 and limn→∞ cos[λn(Im b1 − Im b2)] = 1.

428 X. Hou et al. / J. Math. Anal. Appl. 401 (2013) 416–429

Proof (Necessity). Consider the sequence of functions

qm(z) = 1

βm

e−λmz, m = 1, 2, . . . ,

in H(E, βS). We have noted that this sequence converges weakly to zero. Then by Lemma 4.10, we have

limm→∞

∣∣∣∣(Cϕ1− Cϕ2

)(qm)∣∣∣∣ = 0. (4.15)

However,∥∥(Cϕ1− Cϕ2

)(qm)∥∥ =

∥∥∥∥ 1

βm

e−λm(z+b1) − 1

βm

e−λm(z+b2)

∥∥∥∥=

∥∥∥∥ 1

βm

e−λmb1e−λmz − 1

βm

e−λmb2e−λmz

∥∥∥∥= ∣∣e−λmb1 − e−λmb2

∣∣ .Consequently, for (4.15) to hold, we must have

limm→∞

∣∣e−λmb1 − e−λmb2∣∣ = 0. (4.16)

There are three cases.

Case 1. If Re b1 > 0 and Re b2 > 0, then both composition operators are compact and hence their difference is necessarilycompact.

Case 2. If Re b1 > 0 and Re b2 = 0 (the case when Re b1 = 0 and Re b2 > 0 is similar), by triangle inequality we have∣∣e−λmb1 − e−λmb2∣∣ ≥ |e−λmb2 | − |e−λmb1 | = 1 − e−λmRe b1 .

Since Re b1 > 0, limm→∞(1 − e−λmRe b1

) = 1 > 0, which, by (4.16), is impossible.

Case 3. If Re b1 = Re b2 = 0, let us denote Im bi = ti, we then have∥∥(Cϕ1− Cϕ2

)(qm)∥∥2 = |e−λmb1 − e−λmb2 |2 = |e−iλmt1 − e−iλmt2 |2

= | cos(λmt1) − i sin(λmt1) − cos(λmt2) + i sin(λmt2)|2= 2 − 2 [cos(λmt1) cos(λmt2) − sin(λmt1) sin(λmt2)]

= 2 − 2 cos [λm(t1 − t2)] .

For (4.15) to hold, we must have

limm→∞ cos [λm(t1 − t2)] = 1.

Sufficiency. The proof of this part is similar to that of Theorem 4.11. �

Acknowledgments

The authors thank the referee for pointing out that H(E, βS) is a functional Banach space, as well as for useful remarksand comments that led to the improvement of this paper.

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