# Commuting Quasihomogeneous Toeplitz Operator and Hankel Operator on Weighted Bergman Space

Authors: Jun Yang
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We characterize the commuting Toeplitz operator and Hankel operator with quasihomogeneous symbols. Also, we use it to show the necessary and sufficient conditions for commuting Toeplitz operator and Hankel operator with ordinary functions.


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## Body

## 1. Introduction

LetdA denote Lebesgue area measure on the unit disk 𝔻, normalized so that the measure of 𝔻 equals 1. For α>-1, we denote by dAα the measure dAα(z)=(α+1)(1-|z|2)αdA(z). For 1≤p<+∞, the space Lp(𝔻,dAα) is a Banach space. The weighted Bergman space Aα2(𝔻) is the closed subspace of analytic functions in the Hilbert space L2(𝔻,dAα). For each z∈𝔻, the application Lz:Aα2(𝔻)→ℂ is continuous and can be represented as Lz(f)=f(z)=〈f,Kz(α)〉α, where
(1)Kz(α)(w)=1(1-wz-)2+α=∑n=0∞Γ(n+2+α)n!Γ(2+α)(wz-)n,hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhz,w∈𝔻.
This means that, if Pα is the orthogonal projection from L2(𝔻,dAα) onto Aα2(𝔻), then Pα can be defined by
(2)(Pαf)(w)=〈f,Kw(α)〉α=∫𝔻f(z)1(1-w-z)2+αdAα(z).
For a function f∈L∞(𝔻,dAα(z)), we define the Toeplitz operator Tf:Aα2(𝔻)→Aα2(𝔻) with symbol f by
(3)Tf(h)=Pα(fh).
It is well known that
(4)Tf(h)(z)=∫𝔻f(w)h(w)Kz(α)(w)¯dAα(w),z∈𝔻.LetU(α):L2(𝔻,dAα(z))→L2(𝔻,dAα(z)) be the unitary operator defined by U(α)f(z)=f(z)~=f(z-), where f belongs to L2(𝔻,dAα(z)). Let g be in L2(𝔻,dAα(z)); we define a bounded linear operator Mg on L2(𝔻,dAα(z)) as follows:
(5)Mg(f)=gf.
Then we can define the small Hankel operator as follows:
(6)Hg:Aα2(𝔻)→Aα2(𝔻)
as Hg=PαU(α)Mg.The study of commuting Toeplitz operators on the Bergman and Hardy spaces over various domains and related operator algebras has a long lasting history. On the Hardy space of the unit disk, Brown and Halmos [1] first showed that two Toeplitz operators are commuting if and only if either both symbols of these operators are analytic, or both symbols of these operators are coanalytic or a nontrivial linear combination of the symbols of these operators is constant. On the Bergman space, the situation is more complicated. Axler and Čučković obtained the analogous result for Toeplitz operators with bounded harmonic symbols on the Bergman space of the unit disk [2]. The problem of characterizing commuting Toeplitz operators with arbitrary bounded symbols seems quite challenging and is not fully understood until now. In [3], Čučković and Rao used the Mellin transform to characterize all Toeplitz operators on La2 which commute with Teipθrm for (m,p)∈N×N. Later in [4] Louhichi and Zakariasy gave a partial characterization of commuting Toeplitz operators on La2 with quasihomogeneous symbols. Recently, Lu and Zhang [5, 6] characterized the commuting Toeplitz operators and Hankel operators with quasihomogeneous symbols. There are also many other important results [7–13]. Motivated by those works, we study commuting Toeplitz operator and Hankel operator on the weighted Bergman space. In this paper, we obtain the necessary and sufficient conditions for commuting Toeplitz operator and Hankel operator.An operator that will arise in our study of Toeplitz operators is the Mellin transform, defined for any functionφ∈L1([0,1],rdr); from the formula, φα^(z) is the Mellin transform as follows:
(7)φα^(z)=∫01φ(r)(1-r2)αrz-1dr,
which is a bounded holomorphic function in the half plane {z:Rez>2}.Letφ∈L1(𝔻,dAα) be a radial function; that is, suppose that φ(z)=φ(|z|), z∈𝔻. In fact, if we define the function φr on [0,1] by φr(s)=φ(s), then a direct calculation shows that
(8)〈Tφ(zk),zl〉α={0fork≠l,2(α+1)φα^(2k+2)fork=l,
so that if k∈ℕ,
(9)Tφ(zk)=2(1+α)Γ(k+2+α)k!Γ(2+α)×∫01φ(r)r2k+1(1-|r|2)αdrzk=2(1+α)Γ(k+2+α)k!Γ(2+α)φα^(2k+2)zk.
Thus, Tφ is a diagonal operator on Aα2(𝔻) with coefficient sequence as follows:
(10)(2(1+α)Γ(k+2+α)k!Γ(2+α)φα^(2k+2))k=0∞.
This makes it relatively simple to work with the product of two operators with such radial symbols.Now, we define the “radialization” of a functionf∈L1(𝔻,dAα) by the following:
(11)rad(f)(z)=12π∫02πf(eitz)dt.
It is clear that a function f is a radial if and only if rad(f)=f.Letℜα be the space of weighted square integrable radial functions on 𝔻. By using that, trigonometric polynomials are dense in L2(𝔻,dAα) and that, for k1≠k2, eik1θℜα is orthogonal to eik2θℜα, we see that is
(12)L2(𝔻,dAα)=⊕k∈ℤeikθℜα.Definition 1.
Letφ be a function in L1(𝔻,dAα) which is of the form eikθf, where f is a radial function. Then one says that φ is a quasihomogeneous function of quasihomogeneous degree k.
A direct calculation gives the following lemmas which we will use often.Lemma 2.
Letp∈ℕ and let φ be an integrable radial function. Then,
(13)Teipθφ(zk)=2(α+1)Γ(k+p+2+α)(k+p)!Γ(2+α)×φα^(2k+p+2)zk+p,k≥0,Te-ipθφ(zk)={0if0≤k<p,2(α+1)Γ(k-p+2+α)(k-p)!Γ(2+α)×φα^(2k-p+2)zk-pifk≥p.Lemma 3.
Letφ be an integrable radial function. Then, for p∈ℤ+,
(14)Heipθφ(zk)=0,k≥0,
and for p∈ℕ,
(15)He-ipθφ(zk)={0ifk>p,2(α+1)Γ(-k+p+α+2)(-k+p)!Γ(2+α)×φα^(p+2)z-k+pif0≤k≤p.


## 2. Commuting of Toeplitz Operator and Hankel Operator

Theorem 4.
Leteipθf be a bounded function of quasihomogeneous degree p≥0 and g=∑k∈Z‍eikθgk,α(r)∈L∞(𝔻,dAα). Then TeipθfHg=HgTeipθf if and only if the following conditions holds(1)
f^α(2j+p+2)g^-j-k-p,α(k+j+p+2)=0, if 0≤k≤p-1 and j≥0;(2)
(Γ(k+2+α)/k!)f^α(2k+p+2)g^-j-k,α(k+j+2)=(Γ(j+p+α+2)/(j+p)!)f^α(2j+p+  2)g^-j-k-2p,α(k+j+2p+2), if k≥0 and j≥0.Proof.
Forj≥0,
(16)TeipθfHg(zj)=TeipθfH∑k∈Z‍eikθgk,α(r)(zj)=Teipθf(∑k=-j-∞‍2(α+1)Γ(-j-k+α+2)(-j-k)!Γ(2+α)×g^k,α(-k+2)z-j-k∑k=-j-∞‍2(α+1)Γ(-j-k+α+2)(-j-k)!Γ(2+α))=2∑k=-j-∞(α+1)Γ(-j-k+α+2)(-j-k)!Γ(2+α)×g^k,α(-k+2)Teipθf(z-j-k)=2∑k=-j-∞(α+1)Γ(-j-k+α+2)(-j-k)!Γ(2+α)×g^k,α(-k+2)2(α+1)·Γ(-j-k+p+α+2)(-j-k+p)!Γ(2+α)×f^α(-2j-2k+p+2)z-j-k+p=4(α+1)2∑k=0+∞‍Γ(k+α+2)k!Γ(2+α)×Γ(k+p+α+2)(k+p)!Γ(2+α)×f^α(2k+p+2)·g^-j-k,α(j+k+2)zk+p,HgTeipθf(zj)=H∑k∈Z‍eikθgk,α(r)Teipθf(zj)=2(α+1)Γ(j+p+α+2)(j+p)!Γ(2+α)f^α(2j+p+2)×H∑k∈Z‍eikθgk,α(r)(zj+p)=2(α+1)Γ(j+p+α+2)(j+p)!Γ(2+α)f^α(2j+p+2)×∑k=-j-p-∞2(α+1)·Γ(-j-k-p+α+2)(-j-p-k)!Γ(2+α)g^k,α(-k+2)z-j-p-k=∑k=-j-p-∞4(α+1)2Γ(j+p+α+2)(j+p)!Γ(2+α)×Γ(-j-k-p+α+2)(-j-p-k)!Γ(2+α)·f^α(2j+p+2)g^k,α(-k+2)z-j-p-k=4(α+1)2∑k=0+∞Γ(j+p+α+2)(j+p)!Γ(2+α)×Γ(k+α+2)(k)!Γ(2+α)f^α(2j+p+2)·g^-j-p-k,α(k+j+p+2)zk.
If TeipθfHg=HgTeipθf, then we have
(17)∑k=0+∞Γ(k+α+2)k!Γ(2+α)Γ(k+p+α+2)(k+p)!Γ(2+α)×f^α(2k+p+2)g^-j-k,α(j+k+2)zk+p=∑k=0+∞‍Γ(j+p+α+2)(j+p)!Γ(2+α)Γ(k+α+2)(k)!Γ(2+α)×f^α(2j+p+2)g^-j-p-k,α(k+j+p+2)zk,
which is equivalent to
(18)∑k=0+∞Γ(k+α+2)k!Γ(2+α)Γ(k+p+α+2)(k+p)!Γ(2+α)×f^α(2k+p+2)g^-j-k,α(j+k+2)zk+p=∑k=0p-1Γ(j+p+α+2)(j+p)!Γ(2+α)Γ(k+α+2)(k)!Γ(2+α)×f^α(2j+p+2)g^-j-p-k,α×(k+j+p+2)zk+∑k=p+∞Γ(j+p+α+2)(j+p)!Γ(2+α)Γ(k+α+2)(k)!Γ(2+α)×f^α(2j+p+2)g^-j-p-k,α×(k+j+p+2)zk,
that is,
(19)∑k=0+∞‍Γ(k+α+2)k!Γ(2+α)Γ(k+p+α+2)(k+p)!Γ(2+α)×f^α(2k+p+2)g^-j-k,α×(j+k+2)zk+p=∑k=0p-1‍Γ(j+p+α+2)(j+p)!Γ(2+α)Γ(k+α+2)(k)!Γ(2+α)×f^α(2j+p+2)g^-j-p-k,α×(k+j+p+2)zk+∑k=0+∞‍Γ(j+p+α+2)(j+p)!Γ(2+α)Γ(p+k+α+2)(p+k)!Γ(2+α)×f^α(2j+p+2)g^-j-2p-k,α×(k+j+2p+2)zp+k.
From the aforementioned we get the following.
Case
1. For 0≤k≤p-1 and j≥0,
(20)f^α(2j+p+2)g^-j-p-k,α(k+j+p+2)=0.Case2. For k≥0 and j≥0,
(21)Γ(k+α+2)k!f^α(2k+p+2)g^-j-k,α(j+k+2)=Γ(j+p+α+2)(j+p)!f^α(2j+p+2)g^-j-2p-k,α×(k+j+2p+2).
As a special case of Theorem 4, we can have the following corollary.Corollary 5.
Letf be a bounded radial function and g=∑k∈Z‍eikθgk,α(r)∈L∞(𝔻,dAα). Then TfHg=HgTf if and only if
(22)Γ(k+2+α)k!f^α(2k+2)g^-j-k,α(k+j+2)=Γ(j+α+2)j!f^α(2j+2)g^-j-k,α(k+j+2),
for k≥0 and j≥0.Theorem 6.
Lete-ipθf be a bounded function of quasihomogeneous degree -p<0 and g=∑k∈Z‍eikθgk,α(r)∈L∞(𝔻,dAα). Then Te-ipθfHg=HgTe-ipθf if and only if the following conditions holds(1)
f^α(2k+p+2)g^-j-k-p,α(k+j+p+2)=0, if k≥0 and p>j≥0;(2)
(Γ(k+p+2+α)/(k+p)!)f^α(2k+p+2)g^-j-k-p,α(k+j+p+2)=(Γ(j-p+α+2)/(j-  p)!)f^α(2j-p+2)g^-j-k+p,α(k+j-p+2), if k≥0 and j≥p.Proof.
Forj≥0, we have
(23)Te-ipθfHg(zj)=Te-ipθfH∑k∈Z‍eikθgk,α(r)(zj)=Te-ipθf2∑k=-j-∞‍(α+1)Γ(-j-k+α+2)(-j-k)!Γ(2+α)×g^k,α(-k+2)(z-j-k)=2∑k=-j-∞‍(α+1)Γ(-j-k+α+2)(-j-k)!Γ(2+α)×g^k,α(-k+2)Te-ipθf(z-j-k)=4(α+1)2∑k=-j-p-∞‍Γ(-j-k+α+2)(-j-k)!Γ(2+α)×g^k,α(-k+2)·Γ(-j-k-p+α+2)(-j-k-p)!Γ(2+α)×f^α(-2j-2k-p+2)z-j-k-p=4(α+1)2∑k=0+∞‍Γ(k+p+α+2)(k+p)!Γ(2+α)×g^-k-j-p,α(k+j+p+2)·Γ(k+α+2)k!Γ(2+α)f^α(2k+p+2)zk.
Then one has the following.
Case
1. For p>j≥0,
(24)HgTe-ipθf(zj)=H∑k∈Z‍eikθgk,α(r)Te-ipθf(zj)=0.
Case
2. For j≥p,
(25)HgTe-ipθf(zj)=H∑k∈Z‍eikθgk,α(r)Te-ipθf(zj)=2(α+1)Γ(j-p+α+2)(j-p)!Γ(2+α)×f^α(2j-p+2)H∑k∈Z‍eikθgk,α(r)(zj-p)=4(α+1)2∑k=p-j-∞Γ(j-p+α+2)(j-p)!Γ(2+α)×f^α(2j-p+2)×Γ(-j+p-k+α+2)(-j+p-k)!Γ(2+α)·g^k,α(-k+2)z-j+p-k=4(α+1)2∑k=0+∞Γ(j-p+α+2)(j-p)!Γ(2+α)×f^α(2j-p+2)×Γ(k+α+2)k!Γ(2+α)×g^-j-k+p,α(j+k-p+2)zk.
If Te-ipθfHg=HgTe-ipθf, then we have the following.
Case
1.For p>j≥0,
(26)∑k=0+∞Γ(k+p+α+2)(k+p)!Γ(2+α)g^-k-j-p,α(k+j+p+2)0i×Γ(k+α+2)k!Γ(2+α)f^α(2k+p+2)zk=0.Case2.For j≥p,
(27)∑k=0+∞Γ(k+p+α+2)(k+p)!Γ(2+α)g^-k-j-p,α(k+j+p+2)×Γ(k+α+2)k!Γ(2+α)f^α(2k+p+2)zk=∑k=0+∞Γ(j-p+α+2)(j-p)!Γ(2+α)f^α(2j-p+2)×Γ(k+α+2)k!Γ(2+α)g^-j-k+p,α×(j+k-p+2)zk.
That is one has the following.
Case
1. For p>j≥0,k≥0,
(28)g^-k-j-p,α(k+j+p+2)f^α(2k+p+2)=0.Case2. For j≥p,k≥0,
(29)Γ(k+p+α+2)(k+p)!g^-k-j-p,α×(k+j+p+2)f^α(2k+p+2)=Γ(j-p+α+2)(j-p)!f^α(2j-p+2)×g^-j-k+p,α(j+k-p+2).Theorem 7.
Lete-isθg be a bounded function of quasihomogeneous degree -s≤0 and f=∑k∈Z‍eikθfk,α(r)∈L∞(𝔻,dAα). Then TfHe-isθg=He-isθgTf if and only if g^α(s+2)=0 or (1)
f^k-j,α(k+j+2)=0, if s≥k≥0 and j>s;(2)
(Γ(-j+s+α+2)/(-j+s)!)f^k+j-s,α(-j+s+k+2)=(Γ(s-k+α+2)/(s-k)!)f^s-k-j,α(s-k+  j+2), if s≥k≥0 and s≥j≥0;(3)
f^k+j-s,α(-j+s+k+2)=0, if k>s and s≥j≥0.Proof.
Forj≥0, we have the following.
Case
1.For j>s,
(30)TfHe-isθg(zj)=T∑k∈Z‍eikθfk,α(r)He-isθg(zj)=0.Case2.For s≥j≥0,
(31)TfHe-isθg(zj)=T∑k∈Z‍eikθfk,α(r)He-isθg(zj)=2(α+1)Γ(-j+s+α+2)(-j+s)!Γ(2+α)g^α(s+2)×T∑k∈Z‍eikθfk,α(r)(z-j+s)=4(α+1)2Γ(-j+s+α+2)(-j+s)!Γ(2+α)g^α(s+2)×∑k=j-s+∞Γ(s-j+k+α+2)(s-j+k)!Γ(2+α)f^k,α×(2s-2j+k+2)z-j+s+k,He-isθgTf(zj)=He-isθgT∑k∈Z‍eikθfk,α(r)(zj)=He-isθg(∑k=-j+∞2(α+1)×Γ(j+k+α+2)(j+k)!Γ(2+α)×f^k,α(2j+k+2)zj+k∑k=-j+∞2(α+1))=2∑k=0+∞(α+1)Γ(k+α+2)k!Γ(2+α)×f^k-j,α(k+j+2)He-isθg(zk)=4(α+1)2∑k=0sΓ(k+α+2)k!Γ(2+α)×Γ(-k+s+α+2)(-k+s)!Γ(2+α)f^k-j,α×(k+j+2)g^α(s+2)z-k+s.
Applying He-isθgTf=TfHe-isθg, we have the following.
Case
1. For j>s,
(32)4(α+1)2∑k=0s‍Γ(k+α+2)k!Γ(2+α)Γ(-k+s+α+2)(-k+s)!Γ(2+α)×f^k-j,α(k+j+2)g^α(s+2)z-k+s=0.Case2. For s≥j≥0,
(33)∑k=j-s+∞‍Γ(s-j+k+α+2)(s-j+k)!Γ(2+α)f^k,α(2s-2j+k+2)×Γ(-j+s+α+2)(-j+s)!Γ(2+α)g^α(s+2)z-j+s+k=∑k=0s‍Γ(k+α+2)k!Γ(2+α)Γ(-k+s+α+2)(-k+s)!Γ(2+α)×f^k-j,α(k+j+2)g^α(s+2)z-k+s.
If g^α(s+2)=0, then the equation holds.
Otherwiseg^α(s+2)≠0, we have the following.
Case
1. For j>s,
(34)∑k=0s‍Γ(k+α+2)k!Γ(2+α)Γ(-k+s+α+2)(-k+s)!Γ(2+α)×f^k-j,α(k+j+2)z-k+s=0.Case2. For s≥j≥0,
(35)∑k=j-s+∞‍Γ(s-j+k+α+2)(s-j+k)!Γ(2+α)f^k,α(2s-2j+k+2)×Γ(-j+s+α+2)(-j+s)!Γ(2+α)z-j+s+k=∑k=0s‍Γ(k+α+2)k!Γ(2+α)Γ(-k+s+α+2)(-k+s)!Γ(2+α)×f^k-j,α(k+j+2)z-k+s.
that is one has the following.
Case
1. For s≥k≥0,j>s,
(36)f^k-j,α(k+j+2)z-k+s=0.Case2. For s≥j≥0,
(37)∑k=0+∞‍Γ(s-j+α+2)(s-j)!Γ(2+α)f^k+j-s,α×(s-j+k+2)Γ(k+α+2)k!Γ(2+α)zk=∑k=0s‍Γ(k+α+2)k!Γ(2+α)Γ(-k+s+α+2)(-k+s)!Γ(2+α)×f^s-k-j,α(s-k+j+2)zk.
Then we get the following.
Case
1. For s≥k≥0, j>s, f^k-j,α(k+j+2)=0.
Case
2. For s≥j≥0,
(38)Γ(s-j+α+2)(s-j)!f^k+j-s,α(s-j+k+2)=Γ(-k+s+α+2)(-k+s)!f^s-k-j,α(s-k+j+2).Case3. For k>s,s≥j≥0,f^k+j-s,α(-j+s+k+2)=0.Corollary 8.
Letg be a bounded radial function and f=∑k∈Z‍eikθfk,α(r)∈L∞(𝔻,dAα). Then TfHg=HgTf if and only if  (1)
g^α(s+2)=0;(2)
f^k,α(k+2)=0 and f^-k,α(k+2)=0, for k>0.Finally, we will investigate the situation that both functions are ordinary functions.Theorem 9.
Letf=∑k∈Z‍eikθfk,α(r)∈L∞(𝔻,dAα) and g=∑l∈Z‍eilθgl,α(r)∈L∞(𝔻,dAα). Then TfHg=HgTf if and only if
(39)∑k=0+∞‍Γ(k+α+2)k!Γ(2+α)f^k-m,α(m+k+2)×g^-n-k,α(n+k+2)=∑k=0+∞‍Γ(k+α+2)k!Γ(2+α)f^n-k,α(n+k+2)×g^-k-m,α(k+m+2),
for m≥0 and n≥0.Proof.
Forj≥0, we have
(40)HgTf(zj)=H∑l∈Z‍eilθgl,α(r)T∑k∈Z‍eikθfk,α(r)(zj)=H∑l∈Z‍eilθgl,α(r)(∑k=-j+∞‍2(α+1)0×Γ(j+k+α+2)(j+k)!Γ(2+α)0×f^k,α(2j+k+2)zj+k∑k=-j+∞‍)=∑k=-j+∞‍2(α+1)Γ(j+k+α+2)(j+k)!Γ(2+α)×f^k,α(2j+k+2)·(∑l=-(j+k)-∞‍2(α+1)Γ(-j-k-l+α+2)(-j-k-l)!Γ(2+α)×g^l,α(-l+2)z-k-j-l∑l=-(j+k)-∞‍)=∑k=0+∞‍2(α+1)Γ(k+α+2)k!Γ(2+α)×f^k-j,α(j+k+2)·(∑l=0+∞‍2(α+1)Γ(l+α+2)l!Γ(2+α)×g^-l-k,α(l+k+2)zl∑l=0+∞‍),TfHg(zj)=T∑k∈Z‍eikθfk,α(r)H∑l∈Z‍eilθgl,α(r)(zj)=T∑k∈Z‍eikθfk,α(r)(∑l=-j-∞‍2(α+1)×Γ(-j-l+α+2)(-j-l)!Γ(2+α)×g^l,α(-l+2)z-j-l∑l=-j-∞‍)=∑l=-j-∞‍2(α+1)Γ(-j-l+α+2)(-j-l)!Γ(2+α)×g^l,α(-l+2)·(∑k=j+l+∞‍2(α+1)Γ(-j-l+k+α+2)(-j-l+k)!Γ(2+α)×f^k,α(-2j-2l+k+2)z-j-l+k∑k=j+l+∞‍)=∑k=0+∞‍2(α+1)Γ(k+α+2)k!Γ(2+α)×g^-k-j,α(k+j+2)·(∑l=0+∞‍2(α+1)Γ(l+α+2)l!Γ(2+α)×f^l-k,α(l+k+2)zl∑l=0+∞‍).
From the aforementioned we get, for m≥0 and n≥0,
(41)(HgTf(zm),zn)=(∑k=0+∞‍2(α+1)Γ(k+α+2)k!Γ(2+α)f^k-m,α×(m+k+2)·∑l=0+∞‍2(α+1)Γ(l+α+2)l!Γ(2+α)g^-l-k,α×(l+k+2)zl,zn∑k=0+∞‍)=(∑k=0+∞‍2(α+1)Γ(k+α+2)k!Γ(2+α)×f^k-m,α(m+k+2)·2(α+1)Γ(n+α+2)n!Γ(2+α)g^-n-k,α×(n+k+2)zn,zn∑k=0+∞‍)=4(α+1)2∑k=0+∞‍Γ(k+α+2)k!Γ(2+α)f^k-m,α×(m+k+2)g^-n-k,α×(n+k+2),(TfHg(zm),zn)=(∑k=0+∞‍2(α+1)Γ(k+α+2)k!Γ(2+α)×g^-k-m,α(k+m+2)·∑l=0+∞‍2(α+1)Γ(l+α+2)l!Γ(2+α)×f^l-k,α(l+k+2)zl,zn∑k=0+∞‍)=(∑k=0+∞‍2(α+1)Γ(k+α+2)k!Γ(2+α)×g^-k-m,α(k+m+2)·2(α+1)Γ(n+α+2)n!Γ(2+α)f^n-k,α×(n+k+2)zn,zn∑k=0+∞‍)=4(α+1)2∑k=0+∞‍Γ(k+α+2)k!Γ(2+α)×f^n-k,α(n+k+2)×g^-k-m,α(k+m+2).
If HgTf=TfHg, then we get
(42)∑k=0+∞‍Γ(k+α+2)k!Γ(2+α)f^k-m,α(m+k+2)×g^-n-k,α(n+k+2)=∑k=0+∞‍Γ(k+α+2)k!Γ(2+α)f^n-k,α(n+k+2)×g^-k-m,α(k+m+2),
for m≥0 and n≥0.
The converse is easy to get.



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*Source: 408168-2013-07-28.xml*