# Boundedness and Compactness of the Mean Operator Matrix on Weighted Hardy Spaces

Authors: Bahmann Yousefi; Ebrahim Pazouki
Journal: ISRN Mathematical Analysis
 (2012)

---

## Abstract

We investigate the boundedness and the compactness of the
mean operator matrix acting on the weighted Hardy spaces.


---

## Body

## 1. Introduction

First in the following, we generalize the definitions coming in [1]. Let β={β(n)} be a sequence of positive numbers with β(0)=1 and 1<p<∞. We consider the space of sequences f={f̂(n)}n=0∞ such that(1.1)‖f‖p=‖f‖βp=∑n=0∞|f̂(n)|pβ(n)p<∞.
The notation(1.2)f(z)=∑n=0∞f̂(n)zn
will be used whether or not the series converges for any value of z. These are called formal power series and the set of such series is denoted by Hp(β). Let f̂k(n)=δk(n). So fk(z)=zk and then {fk}k is a basis such that ∥fk∥=β(k). Recall that Hp(β) is a reflexive Banach space with norm ∥·∥β and the dual of Hp(β) is Hq(βp/q) where 1/p+1/q=1 and βp/q={β(n)p/q} [2]. For some other sources on this topic see [1–12].The study of weighted Hardy spaces lies at the interface of analytic function theory and operator theory. As a part of operator theory, research on weighted Hardy spaces is of fairly recent origin, dating back to valuable work of Allen Shields [1] in the mid- 1970s. The mean operator matrix has been the focus of attention for several decades and many of its properties have been studied. Some of basic and useful works in this area are due to Browein et al. [13–16], which are pretty large works that contain a number of interesting results and indeed they are mainly of auxiliary nature. Also, some properties of mean operator matrices have been studied recently by Lashkaripour on weighted sequence spaces [17–20]. In this paper, we have given conditions under which the mean operator matrix is bounded and compact as an operator acting on weighted Hardy spaces. More details of our works are as follows: the idea of Theorem 2.6 comes from [16]. In Theorem 2.9, we extend the method used in [20, Theorem 1.2] to show the boundedness of the mean operator matrix acting on the weighted Hardy spaces. Some inequalities are useful to find a bound for the mean operator matrix acting on weighted Hardy spaces [21–26]. For example the inequality proved in [26, Theorem 8] is used in the proof of Theorem 2.11.


## 2. Main Results

In this section we define an operator acting onHp(β) and then we will investigate its boundedness and compactness on Hp(β).Definition 2.1.
Let{an} be a sequence of positive numbers and define
(2.1)An=∑i=0naiβ(i)p.
The mean operator matrix associated with the sequence {an} is represented by the matrix A=[ank]n,k and is defined by
(2.2)ank={akβ(n)pAn,0≤k≤n,0,k>n.
From now on, byA we denote the mean operator matrix associated with the fixed sequence {an} as in Definition 2.1.Theorem 2.2 (see [12, Theorem 1]).
If0<an≤an+1 for all integers n≥0, then A is a bounded operator on Hp(β).Theorem 2.3 (see [12, Theorem 2]).
Let1/p+1/q=1 and bn>0  for n=0,1,… If
(2.3)M1=sup⁡n≥0⁡∑k=0nakβ(n)p+1Anβ(k)(bkbn)1/p<∞,M2=sup⁡k≥0∑n=k∞akβ(n)p+1Anβ(k)(bnbk)1/q<∞,
then A=[ank]n,k is a bounded operator on Hp(β) and ∥A∥≤M11/qM21/p.Recall that ifan,bn are two positive sequences, by an~bn, we mean that an/bn→1 whenever n→∞. Also, we write an=o(bn), if an/bn→0 as n→∞.Corollary 2.4.
Letlimn→∞nan/An be finite and 1/p+1/q=1. If
(2.4)sup⁡n≥0⁡∑k=0nakβ(n)p+1nanβ(k)(bkbn)1/p<∞,sup⁡k≥0∑n=k∞akβ(n)p+1nanβ(k)(bnbk)1/q<∞,
then A is a bounded operator on Hp(β).Proof.
Putlimn→∞nan/An=β. Then nan/β~An and so
(2.5)∑k=0nakβ(n)p+1Anβ(k)(bkbn)1/p~β∑k=0nakβ(n)p+1nanβ(k)(bkbn)1/pasn⟶∞,∑n=k∞akβ(n)p+1Anβ(k)(bnbk)1/q~β∑n=k∞akβ(n)p+1nanβ(k)(bnbk)1/qask⟶∞.
On the other hand
(2.6)sup⁡n≥0⁡∑k=0nakβ(n)p+1nanβ(k)(bkbn)1/p<∞,sup⁡k≥0⁡∑n=k∞akβ(n)p+1nanβ(k)(bnbk)1/q<∞,
thus Theorem 2.3 implies that A is a bounded operator on Hp(β).Lemma 2.5.
Suppose thatncan/β(n) is eventually increasing when the constant c>1-γ, and eventually decreasing when c<1-γ. Let
(2.7)S1(n)=1n∑k=1nakβ(n)anβ(k)(kn)-1/p,S2(k)=k1/q∑n=k∞akβ(n)anβ(k)1n1/(q+1).
If γ>1/p, then limn→∞S1(n)=limk→∞S2(k)=1/(γ-1/p).Proof.
Let1/p+1/q=1 and c2<1-γ<c1<1. Then in either case there is a positive integer N such that
(2.8)(kn)-c2<akβ(n)anβ(k)<(kn)-c1
for N≤k≤n. Suppose first that γ>1/p, then
(2.9)lim⁡n→∞⁡n1-1/panβ(n)=∞
and hence
(2.10)lim⁡n→∞1n∑k=1N-1akβ(n)anβ(k)(kn)-1/p=0.
Therefore
(2.11)lim⁡n→∞⁡sup⁡S1(n)≤lim⁡n→∞1n∑k=Nn(kn)-c1-1/p=∫01x-c1-1/pdx.
By calculus integral we get
(2.12)∫01x-c-1/pdx=11-c-1/p;c≠1q,
and so
(2.13)lim⁡n→∞⁡inf⁡S1(n)≥lim⁡n→∞1n∑k=Nn(kn)-c2-δ=∫01x-c2-1/pdx=11-c2-1/p.
Letting c1→1-γ from the right and c2→1-γ from the left, we have
(2.14)lim⁡n→∞S1(n)=1γ-1/p.
Also note that
(2.15)lim⁡k→∞⁡sup⁡S2(k)≤lim⁡k→∞k1/q∑n=k∞(kn)-c11n1/q+1,lim⁡k→∞k1/q∑n=k∞(kn)-c11n1/q+1=11/q-c1.
If c1→1-γ, then 1/q-c1→γ-1/p and similarly we get
(2.16)lim⁡k→∞inf⁡S2(k)≥lim⁡k→∞k1/q-c2∑n=k∞(1n)1-1/q-c2,lim⁡k→∞k1/q-c2∑n=k∞(1n)1-1/q-c2=11/q-c2.
If c2→1-γ, then 1/q-c2→γ-1/p. This completes the proof.Theorem 2.6.
Letlimn→∞nanβ(n)p/An=γ, ncanβ(n)p be eventually monotonic for any constant c, and {β(n)} be bounded. Then A is a bounded operator if 1/γ<p.Proof.
Letδn=nanβ(n)p/An and suppose first that 0≤γ<∞. Then
(2.17)n(log⁡(An)-log⁡(An-1))=-nlog⁡(1-δnn)⟶γ
as n→∞, and hence
(2.18)log⁡(An)-log⁡(A1)=-n∑k=2nlog⁡(1-δkk)=ϵnlog⁡n,
where ϵn→γ. Consequently An=A1nϵn. Now suppose that γ=∞, then for n≥2,
(2.19)log⁡(An)-log⁡(An-1)=-log⁡(1-δnn)≥δnn
since δn→∞. If M>0, then there is N1∈ℕ such that δn≥M+1 for all n≥N1.
Without loss of the generality suppose that there is a positive real numbera>0 such that δn>a for n≤N1. Note that
(2.20)∑k=2N11k=log⁡N1+c+o(1)-1.
If n>N1, then
(2.21)∑k=2n1k=log⁡n+c+o(1)-1,∑k=N1+1n1k=log⁡n-log⁡N1.
Also,
(2.22)∑k=2nδk/klog⁡n≥a(∑k=2N11/k)+(M+1)(∑k=N1+1n1/k)log⁡n,∑k=2N11/k+∑k=N1+1n1/klog⁡n≥M1+1(log⁡n-log⁡N1)+a(log⁡N1+c+o(1)-1)log⁡n=M1+1+(a-M1-1)log⁡N1+a(c+o(1)-1)log⁡n,
for large amount of n last equality greater than M1. Hence
(2.23)log⁡An≥∑k=2nδkk=γnlog⁡n,
where γn→∞. It follows that, for any real number c, ncAn=nc+γn. Since
(2.24)nc-1An~1γncanβ(n)p,
thus ncanβ(n)p is eventually increasing for c>1-γ, and eventually decreasing for c<1-γ. But {β(n)}n is bounded, so there are M1,M2>0 such that M1<β(n)<M2, and
(2.25)ncanβ(n)=ncanβ(n)pβ(n)p+1,ncanβ(n)pβ(n)p+1≥ncanβ(n)pM1p+1.
This implies that ncan/β(n) is eventually increasing for c>1-γ. Similarly ncan/β(n) is eventually decreasing for c<1-γ. Thus
(2.26)∑k=1nakβ(n)p+1Anβ(k)(kn)-1/p~γn∑k=1nakβ(n)p+1nanβ(k)(kn)-1/p.
By Lemma 2.5(2.27)γn∑k=1nakβ(n)p+1nanβ(k)(kn)-1/p
is bounded and so
(2.28)∑k=1nakβ(n)p+1Anβ(k)(kn)-1/p
is bounded. We can see that
(2.29)∑n=k∞akβ(n)p+1Anβ(k)(kn)1/q
is also bounded. Now by Theorem 2.3, A is a bounded operator and so the proof is complete.Lemma 2.7.
Let{an},{tn} be nonnegative sequences with t-1=0. Then for all n∈ℕ one has
(2.30)∑k=0n(tkak)≤{max⁡0≤k≤n(1n-k+1∑j=Knaj)}(∑k=1n(n-k+1)(tk-tk-1)++t0(n+1)).Proof.
Employing the summation by parts, we get(2.31)∑k=0n(tkak)=∑k=0n(∑j=knaj)(tk-tk-1)≤∑k=0n(∑j=knaj1n-k+1)(tk-tk-1)+(n-k+1).
So
(2.32)∑k=0n(tkak)≤{max⁡0≤k≤n(1n-k+1∑j=knaj)}(∑k=1n(n-k+1)(tk-tk-1)++t0(n+1)),
and at this time the proof is complete.Theorem 2.8 (see [26, Theorem 8]).
Let1/p+1/q=1, {xn} be a positive sequence, then
(2.33)∑j=0∞max⁡0≤i≤j(1j-i+1∑k=ijxk)p≤qp(∑k=0∞xkp).Theorem 2.9.
Let{an} be a positive sequence and
(2.34)M3=sup⁡n≥0(∑k=1nn-k+1An(akβ(k)-ak-1β(k-1))+β(n)p+1+(n+1)a0Anβ(0)β(n)p+1)
be finite. Then A is bounded and ∥A∥≤M3q.Proof.
Letf(z)=∑n=0∞f̂(n)zn∈Hp(β), thus
(2.35)A(f)(z)=∑n=0∞(∑k=0nakβ(n)pAnf̂(k))zn.
By definition of ∥·∥β, we have
(2.36)∑n≥0β(n)p|∑k=0nakβ(n)pAnf̂(k)|p≤∑n≥0(∑k=0nakβ(n)p+1Anβ(k)|f̂(k)|β(k))p.
In Lemma 2.7, consider tk=ak/β(k) and aj=|f̂(j)|β(j). Then
(2.37)∑n≥0(∑k=0nakβ(n)p+1Anβ(k)|f̂(k)|β(k))p≤∑n≥0{max⁡0≤k≤n1n-k+1∑j=kn|f̂(j)|β(j)}p×(∑k=1nn-k+1An(akβ(k)-ak-1β(k-1))+β(n)p+1+(n+1)a0Anβ(0)β(n)p+1)p.
Now, Theorem 2.8 implies that
(2.38)∑n≥0{max⁡0≤k≤n(1n-k+1∑j=kn|f̂(j)|β(j))}pM3p≤M3pqp∑k=1∞|f̂(k)|pβ(k)p,
and so we get ∥Af∥≤M3q∥f∥β for all f∈Hp(β). Thus A∈B(Hp(β)) and indeed ∥A∥≤M3q. This completes the proof.Corollary 2.10.
Let1/p+1/q=1, ak/β(k)≥ak-1/β(k-1) and
(2.39)M4=sup⁡n≥0∑k=0nakβ(n)p+1β(k)An<∞.
Then A is a bounded operator on Hp(β) and ∥A∥≤M4.Proof.
Note that(2.40)∑n≥0(∑k=0nakβ(n)p+1Anβ(k)|f̂(k)|β(k))p≤∑n≥0{max⁡0≤k≤n1n-k+1∑j=kn|f̂(j)|β(j)}p(∑k=0nakβ(n)p+1β(k)An)p.
Theorem 2.8 implies that
(2.41)∑n≥0{max⁡0≤k≤n(1n-k+1∑j=kn|f̂(j)|β(j))}pM4p≤M4pqp∑k=1∞|f̂(k)|pβ(k)p,
and so by Theorem 2.9 we obtain ∥Af∥≤qM4∥f∥β for all f∈Hp(β). Thus A∈B(Hp(β)) and indeed ∥A∥≤M4q. This completes the proof.Now, we characterize compactness of subsets ofHp(β) and then we will investigate compactness of the mean operator matrix on Hp(β).Theorem 2.11.
LetS be a nonempty subset of Hp(β). Then S is relatively compact if and only if the following hold:(i)
there existsM>0, such that for all ∑n=0∞f̂(n)zn∈S, |f̂(i)β(i)|≤M for all i∈ℕ∪{0};(ii)
givenϵ>0, there is n0∈ℕ such that ∑n=n0∞|f̂(n)|pβ(n)p<ϵp for all ∑n=0∞f̂(n)zn∈S.Proof.
LetS be relatively compact, thus there exist g1,…,gk∈Hp(β) such that
(2.42)S⊆⋃i=1kB(gi,1).
For every f(z)=∑n=0∞f̂(n)zn∈S, there is gi such that f∈B(gi,1). By Minkowski inequality we get
(2.43)∑n=0∞|f̂(n)|pβ(n)p≤[(∑n=0∞|f̂(n)-ĝi(n)|pβ(n)p)1/p+(∑n=0∞|ĝi(n)|pβ(n)p)1/p]p≤(‖f-gi‖+‖gi‖)p≤(1+‖gi‖)p≤(1+max⁡{‖gi‖:i=1,…,k})p.
Thus for every f∈S and n∈ℕ∪{0}, we get
(2.44)|f̂(n)β(n)|≤1+max⁡{‖gi‖:i=1,…,k}.
So (i) holds. Now suppose that ϵ is an arbitrary positive number. Since S is relatively compact, thus there exist h1,…,hk∈Hp(β) such that
(2.45)S⊆⋃i=1kB(hi,ϵ2).
Since hi∈Hp(β), there exists Ni∈ℕ such that
(2.46)∑n=Ni∞|ĥi(n)|pβ(n)p<ϵp2p
for i=1,…,k. Put
(2.47)N0=max⁡{Ni:i=1,…,k},
and consider f∈S. Then there exists i∈{1,…,k}, such that f∈B(hi,ϵ/2). Hence we get
(2.48)∑n=N0∞|f̂(n)|pβ(n)p≤[(∑n=N0∞|f̂(n)-ĥi(n)|pβ(n)p)1/p+(∑n=N0∞|ĥi(n)|pβ(n)p)1/p]p≤(‖f-hi‖+ϵ2)p≤ϵp.
So (ii) holds.
Conversely, assume thatϵ>0 be given and let (i) and (ii) hold. By condition (ii), there exists n0∈ℕ such that
(2.49)∑n=n0∞|f̂(n)|pβ(n)p<ϵp2,
for all f∈S. Let Mn0 be the closed linear span of the set {1,z,…,zn0-1} in Hp(β). Consider ℂn0 and Mn0 with norms
(2.50)‖(z1,…,zn0)‖=(∑n=1n0|zn|pβ(n)p)1/p,
for all (zi)i=1n0∈ℂn0, and
(2.51)‖∑i=0n0-1aizi‖=(∑i=0n0-1|ai|pβ(i)p)1/p
for all ∑i=0n0-1aizi∈Mn0. Define L:Mn0→ℂn0, by
(2.52)L(∑i=0n0-1aizi)=(a0,…,an0-1).
Clearly, we can see that L is a bounded linear operator. Now, consider the compact subset
(2.53){(zi)i=1n0:∑i=1n0|zi|pβ(i)p≤n0Mp}
in ℂn0. Then we have
(2.54){∑i=0n0-1f̂(i)zi:∑n=0∞f̂(n)zn∈S}⊆L-1{(zi)i=1n0:∑i=1n0|zi|pβ(i)p≤n0Mp}.
Since
(2.55)L-1{(zi)i=1n0:∑i=1n0|zi|pβ(i)p≤n0Mp}
is a compact subspace of Mn0, so there exist g1,…,gk∈Mn0 such that
(2.56)L-1{(zi)i=1n0:∑i=1n0|zi|pβ(i)p≤n0Mp}∈⋃i=1kB(gi,ϵ21/p).
Hence for every
(2.57)f∈{∑i=0n0-1f̂(i)zi:f(z)=∑n=0∞f̂(n)zn∈S}
there is i∈{1,…,k} satisfying
(2.58)∑n=0n0-1|f̂(n)-ĝi(n)|pβ(n)p≤ϵp2.
Also, we have
(2.59)(‖f-gi‖β)p≤∑n=0n0-1|f̂(n)-ĝi(n)|pβ(n)p+∑n=n0∞|f̂(n)|pβ(n)p≤ϵp2+ϵp2≤ϵp.
Thus, S is relatively compact and so the proof is complete.Theorem 2.12.
Let the mean matrix operatorA be bounded on Hp(β), and
(2.60)lim⁡m→∞⁡(∑n=m∞β(n)p2+pAnp)1/p(∑k=0m(akβ(k))q)1/q=0,
where 1/p+1/q=1. Then A is a compact operator on Hp(β).Proof.
LetBHp(β) be the closed unit ball of Hp(β). Define S=A(BHp(β)) and note that S is a bounded subset of Hp(β). Put rn=|f̂(n)|an, un=β(n)p2+p/Anp, vk=(β(k)/ak)p, and
(2.61)Em=(∑n=m∞un)1/p(∑k=0mvk1-q)1/q.
Note that limm→∞Em=0. So for every ϵ>0, there exists m0∈ℕ such that Em<ϵ/(qp-1p)1/p for all m≥m0. Note that if
(2.62)f(z)=∑k=0∞f̂(k)zk∈BHp(β),
then
(2.63)Af(z)=∑n=0∞(∑k=0nakβ(n)pf̂(k)An)zn∈S.
Since ∥f∥βp≤1, we have
(2.64)∑n=m∞|Âf(n)|nβ(n)p≤∑n=m∞β(n)p2+pAnp(∑k=0nak|f̂(k)|)p=∑n=m∞un(∑k=0nrk)p≤ϵp∑k=0∞(rk)pvk≤ϵp∑k=0∞|f̂(k)|pβ(k)p≤ϵp.
Thus by Theorem 2.11, S is compact and so the proof is complete.



---

*Source: 945741-2012-05-08.xml*