# Product and Commutativity ofkth-Order  Slant Toeplitz Operators

Authors: Chaomei Liu; Yufeng Lu
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

The commutativity ofkth-order slant Toeplitz operators with harmonic polynomial symbols, analytic symbols, and coanalytic symbols is discussed. We show that, on the Lebesgue space and Bergman space, necessary and sufficient conditions for the commutativity of kth-order slant Toeplitz operators are that their symbol functions are linearly dependent. Also, we study the product of two kth-order slant Toeplitz operators and give some necessary and sufficient conditions.


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## Body

## 1. Preliminaries

Throughout this paper,k is a fixed positive integer, and k≥2. Let φ(z)=∑i=-∞∞‍aizi be a bounded measurable function on the unit circle 𝕋, where ai=〈φ,zi〉 is the ith Fourier coefficient of φ and {zi:i∈ℤ} is the usual basis of L2(𝕋), with ℤ being the set of integers. The kth-order slant Toeplitz operator Uφ with symbol φ in L∞(𝕋) is defined on L2(𝕋) as follows:
(1)Uφ(zl)=∑i=-∞∞‍aki-lzi.In the past several decades, slant Toeplitz operators have played outstanding roles in wavelet analysis, curve and surface modelling, and dynamical systems (e.g., see [1–10]). For instance, Villemoes [7] has associated the Besov regularity of solution of the refinement equation with the spectral radius of an associated slant Toeplitz operators and has used the spectral radius of the slant Toeplitz operators to characterize the Lp(1≤p≤∞) regularity of refinable functions;  Goodman et al.  [6] have shown the connection between the spectral radii and conditions for the solutions of certain differential equations that in the Lipschitz classes. However,  these mathematicians concentrated mainly on the applications, but these considerations serve as a source of motivation to introduce and study the properties of slant Toeplitz operators.In 1995, Ho [11–14] began a systematic study of the slant Toeplitz operators on the Hardy space. In [15–17], the authors discussed some properties of kth-order slant Toeplitz operator. In [18, 19], the authors defined the slant Toeplitz operator and kth-order slant Toeplitz operator on the Bergman spaces, respectively, and studied some properties of these operators.In this paper, properties ofkth-order slant Toeplitz operators with harmonic polynomial symbols, analytic symbols, and coanalytic symbols are discussed. We show that, on the Lebesgue space and Bergman space, the necessary and sufficient conditions for the commutativity of kth-order slant Toeplitz operators are that their symbol functions are linearly dependent. Meanwhile, we study the product of two kth-order slant Toeplitz operators and give some necessary and sufficient conditions.


## 2. Commutativity ofkth-Order Slant Toeplitz   Operators on L2(𝕋)

In [17], we investigated the properties of kth-order slant Toeplitz operators on L2(𝕋) and have obtained that forφ,ψ∈L∞(𝕋),Uφ and  Uψ commute (essentially commute) if and only if φ(zk)ψ-φψ(zk)=0.Immediately we come up with the following problem.Could the commutativity of twokth-order slant Toeplitz operators be fully characterized by their symbols?The partial answer to the pervious problem has been obtained in [17]: for ψ∈L∞(𝕋) and φ(z)=∑p=0k-1‍apzp or φ(z)=∑p=-k+10‍apzp, Uφ and Uψ commute (essentially commute) if and only if there exist scalars α and β, not both zero, such that αφ+βψ=0.In this section the commutativity ofkth-order slant Toeplitz operators with analytic symbols and harmonic symbols will be studied. First we discuss the commutativity of two kth-order slant Toeplitz operators with analytic symbols.Proposition 1.
Letφ,ψ∈H∞(𝕋), then the following statements are equivalent: (1.1)
φ(zk)ψ(z)=φ(z)ψ(zk);(1.2)
there exist scalarsα   andβ,  not both zero, such thatαφ+βψ=0.Proof.
Begin with the easy direction. First, suppose that (1.2) holds. Without loss of generality, letα≠0, so that φ=-(β/α)ψ. Thus, φ(zk)ψ(z)=φ(z)ψ(zk).
To prove the other direction of the proposition, suppose that (1.1) holds; that is,φ(zk)ψ(z)=φ(z)ψ(zk). Let φ(z)=∑p=0∞‍apzp  and ψ(z)=∑p=0∞‍bpzp, then
(2)∑p=0∞‍apzkp·∑p=0∞‍bpzp=∑p=0∞‍apzp·∑p=0∞‍bpzkp,
that is,
(3)∑p=0∞‍∑ki+j=p‍aibjzp=∑p=0∞‍∑i+kj=p‍aibjzp,
where i and j are both nonnegative integers. Hence, for all nonnegative integers i,  j,  and  p,
(4)∑ki+j=p‍aibj=∑i+kj=p‍aibj.
Now we would give the proof in four cases.
Case I. Suppose that a0b0≠0. Let λ=b0/a0 and we continue the proof by the induction.
Whenp=1, from (4) we get that a0b1=a1b0, which means that b1=λa1.
Whenp=2, from (4) we get that
(5)a0b2+a1b0=a2b0+a0b1,ifk=2,a0b2=a2b0,ifk>2,
which means that b2=λa2, since b1=λa1.
Now suppose thatbi=λai for all integers i with 0≤i≤m. Then, observe the connection between am+1 and bm+1. Let m+1=kl+r, where l and r are nonnegative integers with 0≤r≤k-1.
Whenp=m+1, from (4) we get
(6)a0bm+1+a1bm+1-k+a2bm+1-2k+⋯+albm+1-lk=b0am+1+b1am+1-k+b2am+1-2k+⋯+blam+1-lk.
From the assumption we have a0bm+1=b0am+1; that is; bm+1=λam+1.
Hence, from the above discussion we get thatbi=λai for all integers i with i≥0 by the induction, which means that ψ(z)=λφ(z). So, the required result holds.
Case II. Suppose that a0=0 and b0≠0. We want to show that ai=0 for all integers i with i≥0; that is, φ≡0. Suppose that there exists some ai which is not zero. Without loss of generality, let ai=0 for all integers i with 0≤i≤l-1 and al≠0, where l≥1 is an integer, then φ(z)=zlφ1(z) and φ1(z)=∑i=0∞‍ai+lzi=˙∑i=0∞‍ai′zi.
Becauseφ(zk)ψ(z)=φ(z)ψ(zk), we can get that z(k-1)lφ1(zk)ψ(z)=φ1(z)ψ(zk), which means that
(7)∑p=0∞‍∑ki+j=p‍ai′bjzp+(k-1)l=∑p=0∞‍∑i+kj=p‍ai′bjzp.
Thus, we have a0′b0=0, and, so, since a0′=al≠0, we have b0=0. This leads to a contradiction.
Hence,φ≡0 and the required result holds.
Case III. Suppose that a0≠0 and b0=0. Similar to Case II, we can get that ψ≡0. So, the required result holds.
Case IV. Suppose that a0=0 and b0=0. If φ≡0 or ψ≡0, then the required result holds. Otherwise, without loss of generality, let ai=0 for all integers i with 0≤i≤l-1 and al≠0, and let bj=0 for all integers j with 0≤j≤m-1 and bm≠0, where l and m are positive integers. Then, φ(z)=zlφ1(z) and ψ(z)=zmψ1(z).
Ifl=m, then φ(zk)ψ(z)=φ(z)ψ(zk) is equal to φ1(zk)ψ1(z)=φ1(z)ψ1(zk). The proof is similar to Case I.
Ifl>m, then φ(zk)ψ(z)=φ(z)ψ(zk) is equal to zk(l-m)φ1(zk)ψ1(z)=φ1(z)ψ1(zk). The proof is similar to Case II.
Ifl<m, then φ(zk)ψ(z)=φ(z)ψ(zk) is equal to φ1(zk)ψ1(z)=zk(m-l)φ1(z)ψ1(zk). The proof is similar to Case III.From the preceding Proposition1, it is evident that Corollary 2 holds.Corollary 2.
Letφ¯,ψ¯∈H∞(𝕋), then the following statements are equivalent: (1.1)
φ(zk)ψ(z)=φ(z)ψ(zk);(1.2)
there exist scalarsα and β, not both zero, such that αφ+βψ=0.From Theorem 2.8 in [17], Proposition 1, and Corollary 2, we can obtain the following Theorem 3.Theorem 3.
Letφ,ψ∈H∞(𝕋) or φ¯,  ψ¯∈H∞(𝕋), the following statements are equivalent: (1.1)
Uφ and Uψ commute;(1.2)
Uφ and Uψ essentially commute;(1.3)
φ(zk)ψ(z)=φ(z)ψ(zk);(1.4)
there exist scalarsα and β, not both zero, such that αφ+βψ=0.Now we start to study the commutativity of twokth-order slant Toeplitz operators with harmonic symbols.Proposition 4.
Letφ(z)=∑p=-nn‍apzp and ψ(z)=∑p=-nn‍bpzp, where a-n2+b-n2≠0 and n is a positive integer, then the following statements are equivalent: (1.1)
φ(zk)ψ(z)=φ(z)ψ(zk);(1.2)
there exist scalarsα    andβ,  not both zero, such thatαφ+βψ=0.Proof.
We begin with the easy direction. First, suppose that (1.2) holds and letα≠0 without lost of generality, so that φ=-(β/α)ψ. Thus, φ(zk)ψ(z)=φ(z)ψ(zk).
To prove the other direction of the proposition, suppose that (1.1) holds. Sinceφ(z)=∑p=-nn‍apzp and ψ(z)=∑p=-nn‍bpzp, then φ(zk)=∑p=-nn‍apzkp, ψ(zk)=∑p=-nn‍bpzkp and
(8)φ(zk)ψ(z)=∑p=-nn‍apzkp·∑p=-nn‍bpzp=∑p=-(k+1)n(k+1)n‍∑ki+j=pn‍aibjzp,φ(z)ψ(zk)=∑p=-nn‍apzp·∑p=-nn‍bpzkp=∑p=-(k+1)n(k+1)n‍∑i+kj=pn‍aibjzp,
where both i and j are integers with -n≤i,j≤n. Because φ(zk)ψ(z)=φ(z)ψ(zk), we can get that
(9)∑ki+j=p‍aibj=∑i+kj=p‍aibj,
for any integers p with -(k+1)n≤p≤(k+1)n.
Now we start to investigate the connection betweenaj and bj (j=-n,…,n) by induction. Since a-n2+b-n2≠0, without loss of generality, let a-n≠0 and let λ=b-n/a-n, which also means that b-n=λa-n.
Whenp=-(k+1)n+1, then by (9), we can get that a-nb-n+1=a-n+1b-n, which means that b-n+1=λa-n+1, since a-n≠0.
Suppose thatb-n+j=λa-n+j for any integers j with 0≤j≤l-1, where 1≤l≤2n. Now we consider the connection between b-n+l and a-n+l. Let l=mk+r, where m and r are both nonnegative integers with 0≤r≤k-1.
Whenp=-(k+1)n+l, by (9) we get that
(10)a-nb-n+l+a-n+1b-n+l-k+⋯+a-n+mb-n+l-mk=b-na-n+l+b-n+1a-n+l-k+⋯+b-n+ma-n+l-mk.
From the assumption we get that a-nb-n+l=a-n+lb-n, which means that b-n+l=λa-n+l.
Hence, by the induction we obtain thatbj=λaj(j=-n,…,n); that is, ψ(z)=∑j=-nn‍λajzj=λ∑j=-nn‍ajzj=λφ(z). So, the required result holds.Lemma 5.
Letφ(z)=∑p=-nn‍apzp, ψ(z)=∑p=-mm‍bpzp, and bmb-m≠0, where n and m are integers and n>m≥1. If φ(zk)ψ(z)=φ(z)ψ(zk), then aj=0 for any integers j with m+1≤|j|≤n.Proof.
Sinceφ(z)=∑p=-nn‍apzp, ψ(z)=∑p=-mm‍bpzp, and φ(zk)ψ(z)=φ(z)ψ(zk), then
(11)∑p=-(kn+m)kn+m‍∑ki+j=p‍aibjzp=∑p=-(n+km)n+km‍∑i+kj=p‍aibjzp.
Since n>m≥1, we can get that for any integers p with n+km+1≤|p|≤kn+m,
(12)∑ki+j=p‍aibj=0,
for any integers p with 0≤|p|≤n+km,
(13)∑ki+j=p‍aibj=∑i+kj=p‍aibj.
We want to show thataj=0 for any integers j with m+1≤j≤n. Here are two cases: n<m+k and n≥m+k. Let m=lk+r, where l and r are nonnegative integers with 0≤r≤k-1.
First Case. If n<m+k, then k(m+1)+m=km+m+k>km+n. Now we, continue the discussion by induction.
Whenp=kn+m, by (12), we can get that anbm=0. So, an=0, since b-mbm≠0.
Whenp=kn+m-k, by (12) we can get that an-1bm+(m-k)+anbm-k=0, where (m-k)+=max{sgn(2m-k+1),0} and sgn is a sign function. So, an-1=0, since b-mbm≠0 and an=0.
Suppose thatan-j=0 for any integers j with 0≤j≤t<n-m-1. Now, consider the value of an-t-1.
Whenp=kn+m-k(t+1), by (12) we get that
(14)an-t-1bm+⋯+an-t-1+λbm-λk=0,
where
(15)λ=t+1,if2m≥(t+1)k,λ=[2mk],if2m<(t+1)k,
and [2m/k] is the biggest integer which is not bigger than 2m/k. Then, by the assumption and (14), we get that an-t-1=0, since bmb-m≠0.
Hence, from the above discussion we obtain thataj=0 for all integers j with m+1≤j≤n by the induction.
Second Case. If n≥m+k, then k(m+1)+m=km+m+k≤km+n. Now we continue the discussion by induction.
Whenp=kn+m, by (12), we can get that anbm=0. So, an=0, since b-mbm≠0.
Suppose thatan-j for any integers j with 0≤j≤t<n-m-1. Now, consider the value of an-t-1.
Ifp=kn+m-k(t+1)>km+n, by (12) we get that
(16)an-t-1bm+⋯+an-t-1+λbm-λk=0,
where
(17)λ=t+1,if2m≥(t+1)k,λ=[2mk],if2m<(t+1)k,
and [x] is the biggest integer which is not bigger than x. Then, by the assumption and (16), we get that an-t-1=0, since bmb-m≠0.
Ifp=kn+m-k(t+1)≤km+n, by (13), we get that
(18)an-t-1bm+⋯+an-t-1+λ1bm-λ1k=bmakn-km-kt-k+m+⋯+bm-λ2akn-km-kt-k+m+λ2k,
where
(19)λ1=t+1,if2m≥(t+1)k,λ1=[2mk],if2m<(t+1)k,λ2=2m,ifkn+km-kt-k+m≤n,λ2=[n+km+kt+k-kn-mk],ifkn+km-kt-k+m>n,
and [x] is the biggest integer which is not bigger than x. Then, by the assumption and (18), we get that an-t-1=0, since bmb-m≠0.
Hence, from the above discussion we obtain thataj=0 for all integers j with m+1≤j≤n by the induction.
Similarly, we could get thataj=0 for all integers j with -n≤j≤-(m+1).From Proposition4 and Lemma 5, it is evident that Proposition 6 holds.Proposition 6.
Letφ(z)=∑p=-nn‍apzp, ψ(z)=∑p=-mm‍bpzp and bmb-m≠0, where n,m are integers and n>m≥1, then the following statements are equivalent: (1.1)
φ(zk)ψ(z)=φ(z)ψ(zk);(1.2)
there exist scalarsα    andβ,   not both zero, such thatαφ+βψ=0.Theorem7 is obvious from Theorem 2.8   in [17] and Proposition 6.Theorem 7.
Letφ(z)=∑p=-nn‍apzp, ψ(z)=∑p=-mm‍bpzp, and bmb-m≠0, where n and m are integers and n>m≥1, and the following statements are equivalent:(1.1)
Uφ and Uψ commute;(1.2)
Uφ and Uψ essentially commute;(1.3)
φ(zk)ψ(z)=φ(z)ψ(zk);(1.4)
there exist scalarsα and β, not both zero, such that αφ+βψ=0.


## 3. Product of Twokth-Order Slant Toeplitz  Operators on L2(𝕋)

In [15, 17], the authors have investigated properties of the product of two kth-order slant Toeplitz operators on L2(𝕋) and have obtained the following result.Theorem 8 (see [15, 17]).
Letφ,ψ∈L∞(𝕋), then the following statements are equivalent: (1.1)
UφUψ is a kth-order slant Toeplitz operator;(1.2)
UφUψ is a zero operator;(1.3)
UφUψ is compact;(1.4)
φ(zk)ψ(z)=0.In this section, we will describe properties of the product of twokth-order slant Toeplitz operators with analytic symbols and harmonic symbols on L2(𝕋) by their symbols. First, we start to discuss properties of two kth-order slant Toeplitz product with analytic symbols.Proposition 9.
Letφ,ψ∈H∞(𝕋), then φ(zk)ψ(z)=0 if and only if φ=0 or ψ=0.Proof.
As we know, the “if” direction of the proposition is trivial.
Now suppose thatφ(zk)ψ(z)=0. Since φ,ψ∈H∞(𝕋), we have φ^(zk)ψ^(z)=0, where φ^ and ψ^ are the Poisson extensions of φ and ψ, respectively, and they are analytic on the unit disk 𝔻. Hence, we get that φ^(zk) is identically 0 or ψ^ is identically 0; that is, φ is identically 0 or ψ is identically 0.Similarly, we could obtain Corollary10.Corollary 10.
Letφ¯,ψ¯∈H∞(𝕋), then φ(zk)ψ(z)=0 if and only if φ=0 or ψ=0.It is obvious that Theorem11 holds from the preceding analysis.Theorem 11.
Letφ,ψ∈H∞(𝕋) or φ¯,ψ¯∈H∞(𝕋), then the following statements are equivalent: (1.1)
UφUψ is a kth-order slant Toeplitz operator;(1.2)
UφUψ is a zero operator;(1.3)
UφUψ is compact;(1.4)
φ(zk)ψ(z)=0;(1.5)
φ=0 or ψ=0.Now, we start to discuss the properties of twokth-order slant Toeplitz product with harmonic symbols.Proposition 12.
Letφ(z)=∑p=0n‍apzp and ψ(z)=∑p=0m‍b-pz-p, where n and m are both positive integers, then φ(zk)ψ(z)=0 if and only if φ=0 or ψ=0.Proof.
We begin with the easy direction. First, suppose thatφ=0 or ψ=0; then it is clear that φ(zk)ψ(z)=0.
Without loss of generality, letan2+b-m2≠0. Otherwise, an2+b-m2=0; then we can consider the value of ai and bj, where i and j are both integers with 0≤i≤n-1 and 0≤j≤m-1, since φ(z) and ψ are both polynomial functions. There are four cases: (1)ai=0 for all integers i with 0≤i≤n; (2)b-j=0 for all integers j with 0≤j≤m; (3)as≠0 and ai=0 for all integers i with s+1≤i≤n; (4)b-t≠0 and b-j=0 for all integers j with t+1≤j≤m, where s and t are both nonnegative integers with s≤n-1 and t≤m-1. If the first two cases hold, then the required result holds; if the latter two cases holds, then we have φ(z)=∑p=0s‍apzp and as2+b-m2≠0 or ψ(z)=∑p=0t‍b-pz-p and an2+b-t2≠0.
Now suppose thatφ(zk)ψ(z)=0 and an2+b-m2≠0. Since φ(z)=∑p=0n‍apzp, ψ(z)=∑p=0m‍b-pz-p, then
(20)φ(zk)ψ(z)=∑p=0n‍apzkp·∑p=0m‍b-pz-p=∑p=-mkn‍∑ki-j=p‍aib-jzp,
where i and j are both integers with 0≤i≤n and 0≤j≤m. Because φ(zk)ψ(z)=0, we get, for any integers p with -m≤p≤kn,
(21)∑ki-j=p‍aib-j=0,
where i and j are both integers with 0≤i≤n and 0≤j≤m. Since an2+b-m2≠0, yet we only obtain either an≠0 or b-m≠0.
First Case. If an≠0. Now we want to show that ψ≡0 by the induction.
Whenp=kn, by (21), we get that anb0=0, which means that b0=0, since an≠0.
Whenp=kn-1, by (21), we get that anb-1=0, which means that b-1=0, since an≠0.
Now suppose thatb-l=0 for any integers l with 0≤l≤t, where t is an integer with 0≤t≤m-1. Considering the value of b-t-1, when p=kn-t-1, by (21), we get that
(22)anb-t-1+an-1b-t-1+k+an-2b-t-1+2k+⋯+an-λb-t-1+kλ=0,
where λ=min{n,[(t+1)/k]} and [x] is the biggest integer which is not bigger than x. By the assumption and the above equation, we get that anb-t-1=0; that is,  b-t-1=0, since an≠0.
From the preceding analysis, by the induction, we can obtain thatb-l=0 for any integers l with 0≤l≤m. Hence, ψ≡0.
Second Case.  If b-m≠0. Arguing as in the First case, we obtain that φ≡0.Proposition 13.
Letφ(z)=∑p=0n‍apzp and ψ(z)=∑p=-mm‍bpzp, where n and m are positive integers and an2+bm2+b-m2≠0, then the following statements are equivalent: (1.1)
φ(zk)ψ(z)=0;(1.2)
φ(z)ψ(zk)=0;(1.3)
φ=0    orψ=0.Proof.
We begin with the easy direction. First, suppose thatφ=0 or ψ=0; then it is clear that (1.1) and (1.2) hold.
Now suppose that (1.1) holds. Sinceφ(z)=∑p=0n‍apzp, ψ(z)=∑p=-mm‍bpzp, then
(23)φ(zk)ψ(z)=∑p=0n‍apzkp·∑p=-mm‍bpzp=∑p=-mkn+m‍∑ki+j=p‍aibjzp,
where i and j are both integers with 0≤i≤n and -m≤j≤m. Because φ(zk)ψ(z)=0, we get that for any integers p with -m≤p≤kn+m,
(24)∑ki+j=p‍aibj=0,
where i and j are both integers with 0≤i≤n and -m≤j≤m. Since an2+bm2+b-m2≠0, yet we can obtain that an≠0, b-m≠0, or bm≠0.
First Case. If an≠0. Now we start to continue the proof by induction.
Whenp=kn+m, by (24), we get that anbm=0, which means that bm=0, since an≠0.
Whenp=kn+m-1, by (24), we get that anbm-1=0, which means that bm-1=0, since an≠0.
Now suppose thatbl=0 for any integers l with t≤l≤m, where t is an integer with 0<t≤m. Considering the value of bt-1, when p=kn+t-1, by (24), we get that
(25)anbt-1+⋯+an-λbt-1+kλ=0,
where λ=min{n,[(m-t+1)/2]} and [x] is the biggest integer which is not bigger than x. By the assumption and the above equation, we get that anbt-1=0; that is, bt-1=0, since an≠0.
From the preceding analysis, by the induction we can obtain thatbl=0 for any integers l with 0≤l≤m. Then, by Proposition 12 we get that ψ≡0, since φ is not 0 identically.
Second Case. If bm≠0. In the following, we will continue the proof by induction.
Whenp=kn+m, by (24), we get that anbm=0, which means that an=0, since bm≠0.
Whenp=kn+m-k, by (24), we get that an-1bm+(m-k)+anbm-k=0, where (m-k)+=max{0,sgn(2m-k+1)} and sgn is a sign function. So, an-1=0, since an=0 and bm≠0.
Now, suppose thatal=0 for any integers l with t≤l≤n, where t is an integer with 0<t≤n. Considering the value of at-1, when p=m+kt-k, by (24), we get that
(26)bmat-1+⋯+bm-kλat-1+λ=0,
where λ=min{n-t+1,[2m/k]} and [x] is the biggest integer which is not bigger than x. By the assumption and the above equation, we get that bmat-1=0; that is, at-1=0, since bm≠0.
From the preceding analysis, by the induction we can obtain thatal=0 for any integers l with 0≤l≤n, which means that φ≡0.
Third Case. If b-m≠0. A computation analogous to the one done in the second case  from which we can get that φ≡0.
From the above analysis, we have that (1.3) holds.
Arguing as in the pervious discussion, we obtain that if (1.2) holds, then (1.3) is true.Proposition 14.
Letφ(z)=∑p=-nn‍apzp,  ψ(z)=∑p=-mm‍bpzp, where n and m are both positive integers and an2+bm2+a-n2+b-m2≠0, then φ(zk)ψ(z)=0 if and only if φ=0 or ψ=0.Proof.
We begin with the easy direction. First, suppose thatφ=0 or ψ=0; then it is clear that φ(zk)ψ(z)=0.
Now suppose thatφ(zk)ψ(z)=0. Since φ(z)=∑p=-nn‍apzp,  ψ(z)=∑p=-mm‍bpzp, then
(27)φ(zk)ψ(z)=∑p=-nn‍apzkp·∑p=-nn‍bpzp=∑p=-(kn+m)kn+m‍∑ki+j=p‍aibjzp,
where i and j are both integers with -n≤i≤n and -m≤j≤m. Because φ(zk)ψ(z)=0, we get that, for any integers p with -(kn+m)≤p≤kn+m,
(28)∑ki+j=p‍aibj=0,
where i and j are both integers with -n≤i≤n and -m≤j≤m.
Sincean2+bm2+a-n2+b-m2≠0, without loss of generality, suppose that an≠0. We want to continue the proof by induction.
Whenp=kn+m, by (28), we get that anbm=0, which means that bm=0, since an≠0, when p=kn+m-1, by (28), we get that anbm-1=0, which means that bm-1=0, since an≠0.
Now suppose thatbl=0 for any integers l with t≤l≤m, where t is an integer with 0<t≤m. Considering the value of bt-1. When p=kn+t-1, by (28) we get that
(29)anbt-1+⋯+an-λbt-1+kλ=0,
where λ=min{2n,[(m-t+1)/k]} and [x] is the biggest integer which is not bigger than x. By the assumption and the above equation, we get that anbt-1=0; that is,  bt-1=0, since an≠0.
From the preceding analysis, by the induction we can obtain thatbl=0 for any integers l with 0≤l≤m. Then, by Proposition 13 we get that ψ≡0, since φ is not 0 identically and φ(zk)ψ(z)=0 is equivalent to φ(zk)ψ(z)¯=φ(zk)¯·ψ(z)¯=0.Form Proposition 3.4 and Theorem 3.1 in [15, 17], we will obtain the following theorem which describes the product of two slant Toeplitz operators with harmonic symbols.Theorem 15.
Letφ(z)=∑p=-nn‍apzp, ψ(z)=∑p=-mm‍bpzp, where n and m are both positive integers and an2+bm2+a-n2+b-m2≠0, then the following statements are equivalent:(1.1)
UφUψ is a kth-order slant Toeplitz operator;(1.2)
UφUψ is a zero operator;(1.3)
UφUψ is compact;(1.4)
φ(zk)ψ(z)=0;(1.5)
φ=0 or ψ=0.


## 4. Commutativity ofkth-Order Slant Toeplitz  Operators on Bergman Spaces

Letℂ be the complex plane and let 𝔻 be the unit disk in ℂ. Let dA be the area measure on 𝔻 normalized so that ∫𝔻‍1dA=1. Let A2(𝔻) be the space of analytic functions in L2(𝔻,dA) which consists of Lebesgue measurable functions f on 𝔻 with
(30)∫𝔻‍|f(z)|2dA(z)<+∞.
It is well known that A2(𝔻) is a closed subspace of the Hilbert space L2(𝔻,dA) with the inner product 〈·,·〉 and {zi:i∈ℤ+} are the orthogonal basis in A2(𝔻), where ℤ+ is the set of nonnegative integers. Let L∞(𝔻) be the Banach space of Lebesgue measurable functions f on 𝔻 with
(31)∥f∥∞=esssup{|f(z)|:z∈𝔻}<+∞.LetP denote the orthogonal projection from L2(𝔻,dA) onto A2(𝔻). For f∈L∞(𝔻), the Toeplitz operator Tf on A2(𝔻) is defined by
(32)Tf(g)=P(fg),g∈A2(𝔻),
and the kth-order slant Toeplitz operator Bf on A2(𝔻) is defined by
(33)Bf=WkTf,
where Wk is a bounded linear operator on A2(𝔻) which is defined as
(34)Wk(zi)={zi/k,ifiisdivisiblebyk,0,otherwise.In this section we will investigate the commutativity ofkth-order slant Toeplitz operators with coanalytic symbols and harmonic symbols on Bergman space A2(𝔻). First, we study the commutativity of kth-order slant Toeplitz operators with coanalytic symbols.Lemma 16.
Letq≥0 be an integer, let f,g∈H∞(𝔻), both of which are not 0 identically. If fWk*g=gWk*f, the following statements are equivalent: (1.1)
f(i)(0)=0 for any integers i with 0≤i≤q and f(q+1)(0)≠0;(1.2)
g(i)(0)=0 for any integers i with 0≤i≤q and g(q+1)(0)≠0.Proof.
First, suppose that (1.1) holds. Sincef∈H∞(𝔻), we get that f(z)=zq+1f1(z) and (Wk*f)(z)=zkq+kf2(z), where f1(0)≠0, f2(0)≠0, and f1,f2∈H∞(𝔻). Because fWk*g=gWk*f and g∈H∞(𝔻), we get that
(35)f1(z)(Wk*g)(z)=z(k-1)(q+1)f2(z)g(z),
so g(0)=0. Since g is not 0 identically, without loss of generality, take g(j)(0)=0 for any integers j with 0≤j≤p and g(p+1)(0)≠0, then g(z)=zp+1g1(z) and (Wk*g)(z)=zk(p+1)g2(z), where g1(0)≠0, g2(0)≠0, and g1,g2∈H∞(𝔻); so, by (35) we get that
(36)f1(z)g2(z)z(k-1)(p+1)=z(k-1)(q+1)f2(z)g1(z),
so p=q. Otherwise, if p>q, then by the above equation we get that f1(z)g2(z)z(k-1)(p-q)=f2(z)g1(z), which means that f2(0)g1(0)=0, that leads a contradiction; if p<q, then by the above equation we get that f1(z)g2(z)=z(k-1)(q-p)f2(z)g1(z), which means that f1(0)g2(0)=0, that leads a contradiction. Hence (1.2) holds.
Similarly, we can obtain the other direction of the Lemma.Theorem 17.
Letφ¯,ψ¯∈H∞(𝔻), then the following statements are equivalent:(1.1)
Bφ and Bψ commute;(1.2)
there exist scalarsα and β, not both zero, such that αφ+βψ=0.Proof.
First suppose that (1.2) holds; then it is obvious thatBφ and Bψ commute.
Now suppose that (1.1) holds. So, we get thatBφ*Bψ*(1)=Bψ*Bφ*(1); that is, φ¯Wk*ψ¯=ψ¯Wk*φ¯.
Now we continue the discussion in three cases.
First Case. If φ≡0 or ψ≡0. It is obvious that the required result holds.
Second Case. If φ¯(0)≠0 and ψ¯(0)≠0. Since φ¯,ψ¯∈H∞(𝔻), let φ¯(z)=∑p=0∞‍apzp and ψ¯(z)=∑p=0∞‍bpzp, then a0≠0, b0≠0, and Wk*φ¯(z)=∑p=0∞‍ap((kp+1)/(p+1))zkp and Wk*ψ¯(z)=∑p=0∞‍bp((kp+1)/(p+1))zkp, so
(37)ψ¯Wk*φ¯(z)=∑p=0∞‍bpzp·∑p=0∞‍apkp+1p+1zkp=∑p=0∞‍∑ki+j=p‍ki+1i+1aibjzp,φ¯Wk*ψ¯(z)=∑p=0∞‍apzp·∑p=0∞‍bpkp+1p+1zkp=∑p=0∞‍∑i+kj=p‍kj+1j+1aibjzp.
Because φ¯Wk*ψ¯=ψ¯Wk*φ¯, we get that
(38)∑p=0∞‍∑ki+j=p‍ki+1i+1aibjzp=∑p=0∞‍∑i+kj=p‍kj+1j+1aibjzp,
that is, for any integers p≥0,
(39)∑ki+j=p‍ki+1i+1aibj=∑i+kj=p‍kj+1j+1aibj,
where i and j are both nonnegative integers. In the following, we want to continue the proof by the induction.
Whenp=0, by (39), we get that a0b0=a0b0, so b0=(b0/a0)a0, since a0≠0. Let λ=b0/a0, then b0=λa0.
Whenp=1, by (39), we get that a0b1=a1b0, so b1=λa1.
Suppose thatbj=λaj for any integers j with 0≤j≤l, where l is a nonnegative integer. Now, consider the connection between al+1 and bl+1.
Whenp=l+1, by (39), we get that
(40)∑ki+j=l+1‍ki+1i+1aibj=∑i+kj=l+1‍kj+1j+1aibj,
that is,
(41)a0bl+1+⋯+kλ+1λ+1aλbl+1-λk=b0al+1+⋯+kλ+1λ+1bλal+1-λk,
where λ=[(l+1)/k] and [x] is the biggest integer which is not bigger than x. By this assumption, we can obtain that a0bl+1=b0al+1; that is, bl+1=λal+1, since a0≠0.
Hence, by the induction, we obtain thatbj=λaj for any nonnegative integers j from the pervoius discussion; that is, ψ(z)¯=∑p=0∞‍λapzp=λφ(z)¯. So, the required result holds.
Third Case. If φ and ψ are both not 0 identically, and φ¯(0)=0 or ψ¯(0)=0. Without loss of generality, take φ¯(i)(0)=0 for any integers 0≤i≤i1 and φ¯(i1+1)(0)≠0. By Lemma 16, we get that ψ¯(i)(0)=0 for any integers 0≤i≤i1 and ψ¯(i1+1)(0)≠0, and
(42)φ¯(z)=zi1+1φ1(z),ψ¯(z)=zi1+1ψ1(z),
where φ1,ψ1∈H∞(𝔻), and φ1(0)≠0, ψ1(0)≠0, so ψ¯Wk*(φ¯)=z(k+1)(i1+1)ψ1Wk*(φ1) and φ¯Wk*(ψ¯)=z(k+1)(i1+1)φ1Wk*(ψ1). Since ψ¯Wk*(φ¯)=φ¯Wk*(ψ¯), yet we can get that ψ1Wk*(φ1)=φ1Wk*(ψ1). Since φ1(0)≠0 and ψ1(0)≠0, yet by the second case we get that ψ1=λ1φ1, where λ1=ψ1(0)/φ1(0). So, ψ¯(z)=zi1+1ψ1(z)=λ1zi1+1φ1(z)=λ1φ1(z). The required result holds.Now we are in a position to discuss the commutativity of slant Toeplitz operators with harmonic symbols.Theorem 18.
Letφ(z)=∑p=0n‍apzp+∑k=1n‍a-pz-p and ψ(z)=∑p=0n‍bpzp+∑k=1n‍b-pz-p, where a-n2+b-n2≠0 and n≥1 is an integer, then the following statements are equivalent: (1.1)
Bφ and Bψ commute;(1.2)
there exist scalarsα and β, not both zero, such that αφ+βψ=0.Proof.
First suppose that (1.2) holds. It is obvious thatBφ and Bψ commute.
Now suppose that (1.1) holds. Letφ1(z)=∑p=0n‍apzp, φ2¯(z)=∑p=1n‍a-pz¯p, ψ1(z)=∑p=0n‍bpzp, and ψ2¯(z)=∑p=1n‍b-pz-p, then φ=φ1+φ2¯ and ψ=ψ1+ψ2¯. Since Bφ and Bψ commute, we have Tφ¯Wk*Tψ¯Wk*1=Tψ¯Wk*Tφ¯Wk*1, that is,
(43)φ2Wk*ψ2+ψ1(0)¯φ2+P(φ1¯Wk*ψ2)=ψ2Wk*φ2+φ1(0)¯ψ2+P(ψ1¯Wk*φ2).
Then, by (43), we get that for any integers p with kn+1≤p≤kn+n,
(44)∑i+kj=p‍a-i¯b-j¯kj+1j+1=∑ki+j=p‍a-i¯b-j¯ki+1i+1,
where i and j are positive integers which are not bigger than n. Since a-n2+b-n2≠0, without loss of generality, take b-n≠0. Now we continue the proof by the induction.
Whenp=kn+n, by (44), we get that a-n¯b-n¯((kn+1)/(n+1))=a-n¯b-n¯((kn+1)/(n+1)), so a-n¯=(a-n¯/b-n¯)b-n¯, since b-n≠0. Let λ=a-n¯/b-n¯, then a-n¯=λb-n¯.
Whenp=kn+n-1, by (44), we get that a-n+1¯b-n¯((kn+1)/(n+1))=a-n¯b-n+1¯((kn+1)/(n+1)), so a-n+1¯=λb-n+1¯, since b-n≠0.
Suppose thata-n+i¯=λb-n+i¯ for any integers i with 0≤i≤l<n-1. Now, consider the connection between a-n+l+1 and b-n+l+1.
Whenp=kn+n-l-1, by (44), we get that
(45)a-n+l+1¯b-n¯2n+1n+1+⋯+a-n+l+1-kγ¯b-n+γ¯kγ+1k+1=b-n+l+1¯a-n¯2n+1n+1+⋯+b-n+l+1-kγ¯a-n+γ¯kγ+1k+1,
where γ=[(l+1)/k] and [x] is the biggest integer which is not bigger than x. From the assumption we obtain that a-n+l+1¯b-n¯((2n+1)/(n+1))=b-n+l+1¯a-n¯((2n+1)/(n+1)), so a-n+l+1¯=λb-n+l+1¯, since b-n≠0.
From the pervous discussion, by the induction we obtain thata-n+i¯=λb-n+i¯ for any integers i with 0≤i≤n-1. Hence,  φ2(z)=∑p=1n‍a-pz¯p=∑p=1n‍λ¯b-pz¯p=λ¯ψ2(z).
Sinceφ2(z)=λ¯ψ2(z), by (43), we get that
(46)ψ1(0)¯φ2+P(φ1¯Wk*ψ2)=φ1(0)¯ψ2+P(ψ1¯Wk*φ2).
Then,
(47)〈ψ1(0)¯φ2+P(φ1¯Wk*ψ2),zkn〉=〈Wk*ψ2,zknφ1〉=b-na0¯1n+1=〈φ1(0)¯ψ2+P(ψ1¯Wk*φ2),zkn〉=〈Wk*φ2,zknψ1〉=a-nb0¯1n+1,
and we get that b-na0¯=a-nb0¯; that is, a0=λb0. So, by (46) we have that ψ1(0)¯φ2=φ1(0)¯ψ2 and P(φ1¯Wk*ψ2)=P(ψ1¯Wk*φ2), that is, P(λψ1-φ1¯Wk*ψ2)=0. Hence, for any integers l with kn-n≤l≤kn-1, we have
(48)0=〈zl,0〉=〈zl,P(λψ1-φ1¯Wk*ψ2)〉=〈zl,λψ1-φ1¯Wk*ψ2〉=〈(λψ1-φ1)zl,Wk*ψ2〉,
that is,
(49)〈∑p=0n‍(λbp-ap)zp+l,∑p=1n‍b-pkp+1p+1zkp〉=0.
Since a0=λb0, we have
(50)〈∑p=1n‍(λbp-ap)zp+l,∑p=1n‍b-pkp+1p+1zkp〉=0.
When l=kn-1, by (50), we get that (λb1-a1)b-n¯(1/(n+1))=0, so a1=λb1, since b-n≠0. Then we have
(51)〈∑p=2n‍(λbp-ap)zp+l,∑p=1n‍b-pkp+1p+1zkp〉=0.
Suppose thataj=λbj for any integers j with 0≤j≤s, where 0≤s≤n-1. By (50), we get that
(52)〈∑p=s+1n‍(λbp-ap)zp+l,∑p=1n‍b-pkp+1p+1zkp〉=0.
Now consider the connection between as+1 and bs+1.
Whenl=kn-s-1, by (52), we get that (λbs+1-as+1)b-n¯(1/(n+1))=0, so as+1=λbs+1, since b-n≠0.
Form the above discussion, by the induction we can obtain thataj=λbj for any integers j with 0≤j≤n, so φ1(z)=∑p=0n‍apzp=∑p=0n‍λbpzp=λψ1(z).
Sinceφ2=λ¯ψ2, we have φ=φ1+φ2¯=λψ1+λψ2¯=λψ. Hence, the required result holds.Lemma 19.
Letφ(z)=∑p=0n‍apzp+∑p=1n‍a-pz-p and ψ(z)=∑p=0m‍bpzp+∑p=1m‍b-pz-p, where n and m are integers with n>m≥1 and b-m≠0. If Bφ and Bψ commute, then a-j=0 for any integers j with m+1≤j≤n.Proof.
Letφ1(z)=∑p=0n‍apzp, φ2¯(z)=∑p=1n‍a-pz¯p, ψ1(z)=∑p=0m‍bpzp and ψ2¯(z)=∑p=1m‍b-pz-p, then φ=φ1+φ2¯ and ψ=ψ1+ψ2¯. Since Bφ and Bψ commute, we have Tφ¯Wk*Tψ¯Wk*1=Tψ¯Wk*Tφ¯Wk*1; that is,
(53)φ2Wk*ψ2+ψ1(0)¯φ2+P(φ1¯Wk*ψ2)=ψ2Wk*φ2+φ1(0)¯ψ2+P(ψ1¯Wk*φ2),∑p=k+1km+n‍∑i+kj=p‍a-i¯b-j¯kj+1j+1zp+ψ1(0)¯∑p=1n‍a-p¯zp+∑p=0km‍∑kj-i=p‍ai¯b-j¯kj-i+1kj+1zp=∑p=k+1kn+m‍∑ki+j=p‍a-i¯b-j¯ki+1i+1zp+φ1(0)¯∑p=1m‍b-p¯zp+∑p=0kn‍∑ki-j=p‍bj¯a-i¯ki-j+1ki+1zp.
Sincen>m, let n=m+r, where r is a positive integer. Then, by the pervoius equation, we get that
(54)∑ki+j=kn+m‍a-i¯b-j¯ki+1i+1=0,thatis,a-n¯b-m¯kn+1n+1=0,
so a-n=0, since b-m≠0.
Suppose thata-n+t=0 for any integers t with 0≤t≤r-2. Now consider the value of a-n+t+1. By the assumption and the above equation we get that
(55)∑p=k+1km+n-t-1‍∑i+kj=p‍a-i¯b-j¯kj+1j+1zp+ψ1(0)¯∑p=1n-t-1‍a-p¯zp+∑p=0km‍∑kj-i=p‍ai¯b-j¯kj-i+1kj+1zp=∑p=k+1kn+m-kt-k‍∑ki+j=p‍a-i¯b-j¯ki+1i+1zp+φ1(0)¯∑p=1m‍b-p¯zp+∑p=0kn-kt-k‍∑ki-j=p‍bj¯a-i¯ki-j+1ki+1zp.
Since t+1<r, we can get that ∑ki+j=kn+m-kt-k‍a-i¯b-j¯((ki+1)/(i+1))=0; that is,
(56)a-n+t+1¯b-m¯k(n-t-1)+1n-t-1+⋯+a-n+t+1-λ¯b-m+kλ¯k(n-t-1+λ)+1n-t-1+λ=0,
where λ=min{t+1,[2m/k]} and [2m/k] is the biggest integer which is not bigger than 2m/k. Since b-m≠0, by assumption, we get that a-n+t+1=0.
From the preceding discussion, by the induction we can get thata-j=0 for any integers j with m+1≤j≤n.Theorem 20.
Letφ(z)=∑p=0n‍apzp+∑p=1n‍a-pz-p and ψ(z)=∑p=0m‍bpzp+∑p=1m‍b-pz-p, where a-n2+b-m2≠0, n and m are integers with n>m≥1, then the following statements are equivalent: (1.1)
Bφ and Bψ commute;(1.2)
there exist scalarsα and β, not both zero, such that αφ+βψ=0.Proof.
First suppose that (1.2) holds. It is obvious thatBφ and Bψ commute.
Now suppose that (1.1) holds. Sincea-n2+b-m2≠0, we can get that a-n≠0 or b-m≠0.
Ifa-n≠0, by Theorem 18, we can get the required result.
Ifb-m≠0, by Lemma 19, we can get that φ(z)=∑p=0n‍apzp+∑k=1m‍a-pz-p. Let φ1(z)=∑p=0n‍apzp, φ2¯(z)=∑p=1m‍a-pz¯p, ψ1(z)=∑p=0m‍bpzp, and ψ2¯(z)=∑p=1m‍b-pz-p, then φ=φ1+φ2¯ and ψ=ψ1+ψ2¯.
SinceBφ and Bψ commute, we have Tφ¯Wk*Tψ¯Wk*1=Tψ¯Wk*Tφ¯Wk*1; that is,
(57)φ2Wk*ψ2+ψ1(0)¯φ2+P(φ1¯Wk*ψ2)=ψ2Wk*φ2+φ1(0)¯ψ2+P(ψ1¯Wk*φ2).
By (57) we get that for any integers p with km+1≤p≤km+m,
(58)∑i+kj=p‍a-i¯b-j¯kj+1j+1=∑ki+j=p‍a-i¯b-j¯ki+1i+1,
where i and j are positive integers with -m≤i≤n and -m≤j≤m. Now we continue the proof by the induction.
Whenp=km+m, by (58), we get that a-m¯b-m¯((km+1)/(m+1))=a-m¯b-m¯((km+1)/(m+1)), so a-m¯=(a-m¯/b-m-)b-m¯, since b-m≠0. Let λ=(a-m¯/b-m-), then a-m¯=λb-m¯.
Whenp=km+m-1, by (58), we get that a-m+1¯b-m¯((km+1)/(m+1))=a-m¯b-m+1¯((km+1)/(m+1)), so a-m+1¯=λb-m+1¯, since b-m≠0.
Suppose thata-m+i¯=λb-m+i¯ for any integers i with 0≤i≤l<m-1. Now consider the connection between a-m+l+1 and b-m+l+1.
Whenp=km+m-l-1, by (58), we get that
(59)a-m+l+1¯b-m¯km+1m+1+⋯+a-m+l+1-rk¯b-m+r¯k(m-r)+1m-r+1=b-m+l+1¯a-m¯km+1m+1+⋯+b-m+l+1-rk¯a-m+r¯k(m-r)+1m-r+1.
From the assumption we obtain that a-m+l+1¯b-m¯((km+1)/(m+1))=b-m+l+1¯a-m¯((km+1)/(m+1)), so a-m+l+1¯=λb-m+l+1¯, since b-m≠0.
From the pervious discussion, by the induction we obtain thata-m+i¯=λb-m+i¯ for any integers i with 0≤i≤m-1. Hence,  φ2¯(z)=∑p=1m‍a-pz¯p=∑p=1m‍λ¯b-pz¯p=λ¯ψ2¯(z).
Sinceφ2(z)=λψ2(z), by (57), we get that
(60)ψ1(0)¯φ2+P(φ1¯W*ψ2)=φ1(0)¯ψ2+P(ψ1¯W*φ2).
Then,
(61)〈ψ1(0)¯φ2+P(φ1¯W*ψ2),zkm〉=〈W*ψ2,zkmφ1〉=b-m¯a0¯1m+1=〈φ1(0)¯ψ2+P(ψ1¯W*φ2),zkm〉=〈W*φ2,zkmψ1〉=a-m¯b0¯1m+1,
and we get that b-m¯a0¯=a-m¯b0¯; that is, a0=λ¯b0. So, by (60), we have that ψ1(0)¯φ2=φ1(0)¯ψ2 and P(φ1¯W*ψ2)=P(ψ1¯Wk*φ2); that is, P((λψ1¯-φ1¯)Wk*ψ2)=0. Hence, for any integers l with  km-m≤l≤km-1, we have
(62)0=〈zl,0〉=〈zl,P((λψ1¯-φ1¯)Wk*ψ2)〉=〈zl,(λψ1¯-φ1¯)Wk*ψ2〉=〈(λ¯ψ1-φ1)zl,Wk*ψ2〉,
that is,
(63)〈∑p=0m‍(λ¯bp-ap)zp+l-∑p=m+1n‍apzp+l,∑p=1m‍b-pkp+1p+1zkp〉=0.
Since a0=λb0, we have
(64)〈∑p=1m‍(λ¯bp-ap)zp+l-∑p=m+1n‍apzp+l,∑p=1m‍b-pkp+1p+1zkp〉=0.
Whenl=km-1, by (64), we get that (λ¯b1-a1)b-m¯((km+1)/(m+1))=0, so a1=λ¯b1, since b-m≠0. Then, we have
(65)〈∑p=2m‍(λ¯bp-ap)zp+l-∑p=m+1n‍apzp+l,∑p=1m‍b-pkp+1p+1zkp〉=0.
Suppose that aj=λ¯bj for any integers j with 0≤j≤s, where 0≤s≤m-1. By (64), we get that
(66)〈∑p=s+1m‍(λ¯bp-ap)zp+l-∑p=m+1n‍apzp+l,∑p=1m‍b-pkp+1p+1zkp〉=0.
Now consider the connection between as+1 and bs+1.
Whenl=km-s-1, by (66), we get that (λ¯bs+1-as+1)b-m¯((km+1)/(m+1))=0, so as+1=λ¯bs+1, since b-m≠0.
From the pervious discussion, by the induction, we can obtain thataj=λ¯bj for any integers j with 0≤j≤m. Hence,
(67)φ1(z)=∑p=0n‍apzp=∑p=0m‍λ¯bpzp+∑p=m+1n‍apzp=λ¯ψ1(z)+∑p=m+1n‍apzp.
Let φ3(z)=∑p=m+1n‍apzp, then φ=φ1+φ2¯=λ¯ψ1+φ3+λ¯ψ2¯=λ¯ψ+φ3, since φ2=λψ2.
Now we want to show thatφ3≡0. Because BφBψ=BψBφ, yet we get that Bφ3Bψ=BψBφ3, which means that
(68)Tφ3¯Wk*Tψ¯Wk*=Tψ¯Wk*Tφ3¯Wk*.
Since m≥1, we have Tφ3¯Wk*Tψ¯Wk*1=0; that is, Tφ3¯Wk*[ψ2¯+ψ1(0)¯]=0. Then, we get that for any integers l with 0≤l≤km-m-1, 〈Tφ3¯Wk*[ψ2¯+ψ1(0)¯],zl〉=0; that is,
(69)〈∑j=1m‍b-j¯kj+1j+1zkj+ψ1(0)¯,∑i=m+1n‍aizi+l〉=0.
By (69), we can get that for any integers l with max{0,km-n}≤l≤km-m-1, b-m¯akm-l¯(1/(m+1))=0; that is, akm-l=0, since b-m≠0.
Ifn≤km, then φ3≡0, which means that φ=λ¯ψ, so the required result holds.
Ifn≥km+1, then φ3(z)=∑p=km+1n‍apzp. So, by (68), we can get that Tφ3¯Wk*Tψ¯Wk*(zm)=0, that is, Tφ3¯Wk*[ψ2¯zkm+P(ψ1¯zkm)]=0. So we get that for any integers l with max{0,k2m+km-n}≤l≤k2m-1,
(70)〈Wk*[ψ2¯zkm+P(ψ1¯zkm)],∑i=km+1n‍aizi+l〉=0.
By (70), we can get that for any integers l with max{0,k2m+km-n}≤l≤k2m-1, b-m¯ak2m+km-l¯(1/(k2m+km+1))=0, that is, ak2m+km-l=0, since b-m≠0.
Ifn≤k2m+km, then φ3≡0, which means that φ=λ¯ψ, so, the required result holds.
Ifn≥k2m+km+1, then φ3(z)=∑p=k2m+km+1n‍apzp. Then successively by the pervoius method, we can get that ai=0 for all integers i with m+1≤i≤n, which means that φ=λ¯ψ. So, the required result holds.



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*Source: 473916-2013-04-14.xml*