In this paper, we characterize the commutativity of H-Toeplitz operators with quasihomogeneous symbols on the Bergman space, which is different from the case of Toeplitz operators with same symbols on the Bergman space.
Citation: Jinjin Liang, Liling Lai, Yile Zhao, Yong Chen. Commuting H-Toeplitz operators with quasihomogeneous symbols[J]. AIMS Mathematics, 2022, 7(5): 7898-7908. doi: 10.3934/math.2022442
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In this paper, we characterize the commutativity of H-Toeplitz operators with quasihomogeneous symbols on the Bergman space, which is different from the case of Toeplitz operators with same symbols on the Bergman space.
Let D be unit disk in the complex plane C and dA(z)=1πrdrdθ be normalized Lebesgue area measure on D. Let L2(D,dA) denote the Hilbert space of all square integrable functions on D with the inner product
⟨f,g⟩=∫Df(z)¯g(z)dA(z), f,g∈L2(D,dA). |
Let H(D) be the set of analytic functions on D. The Bergman space L2a(D)=L2(D,dA)∩H(D) is the closed subspace of L2(D,dA). For nonnegative integer n, set en(z)=√n+1zn, z∈D. Then {en}n≥0 is an orthonormal basis of L2a(D). The Bergman space is a reproducing Hilbert space with the reproducing kernel Kz(w)=1(1−¯zw)2, z,w∈D. Let P be orthogonal projection from L2(D,dA) onto Bergman space L2a(D). For f∈L2(D,dA), it has the reproducing formula
P(f)(z)=⟨f,Kz⟩=∫Df(w)(1−zˉw)2dA(w). | (1.1) |
For ϕ∈L∞(D), the multiplication operator Mϕ is defined by Mϕ(f)=ϕf. The Toeplitz operator Tϕ:L2a(D)→L2a(D) and the Hankel operator Hϕ:L2a(D)→L2a(D) are defined respectively by
Tϕ=PMϕ, Hϕ=PMϕJ, |
where J:L2a(D)→¯L2a(D) is defined by J(en)=¯en+1(n≥0). It is clear that Tϕ and Hϕ are bounded operators on the Bergman space L2a(D).
Let L2h(D) be a set of all harmonic functions in L2(D,dA). The operator K:L2a(D)→L2h(D) is defined by
K(e2n)=en, K(e2n+1)=¯en+1, n=0,1,2,⋯. |
Obviously K is bounded on L2a(D), ‖K‖=1 and its adjoint operator K∗:L2h(D)→L2a(D) is given by
K∗(en)=e2n, K∗(¯en+1)=e2n+1, n≥0. |
For ϕ∈L∞(D), the H-Toeplitz operator Bϕ:L2a(D)→L2a(D) is defined by
Bϕ=PMϕK. |
It is easy to see that
B∗ϕ=K∗PhMˉϕ, |
where Ph is the orthogonal projection from L2(D,dA) onto L2h(D).
It is noted that the H-Toeplitz operator is closely related to the Toeplitz and Hankel operators. In fact, for each nonnegative integer n, we have
Bϕ(e2n)=PMϕK(e2n)=PMϕ(en)=Tϕ(en) | (1.2) |
and
Bϕ(e2n+1)=PMϕK(e2n+1)=PMϕJ(en)=Hϕ(en). | (1.3) |
It is known that Tϕ=0 if and only if ϕ=0, whence Bϕ=0 if and only if ϕ=0.
Recently, lots of study about Toeplitz and Hankel operators have been done on the Bergman space (see [1,6,7,9,11,13,14,15]). Various generalizations of Toeplitz and Hankel operators on spaces of analytic functions have been studied by many mathematicians. In 2007, Arora and Paliwal [2] have introduced and studied H-Toeplitz operators on the Hardy space, where they have clubbed the notion of Toeplitz and Hankel operators together. The importance of this notion is that it is associated with a class of Toeplitz operators and a class of Hankel operators on the Hardy space where the original operators are neither Toeplitz nor Hankel. Moreover, it can also be observed that an n×n H-Toeplitz matrix has 2n−1 degree of freedom rather than n2 and therefore for large n, it is comparatively easy to solve the system of linear equations where the coefficient matrix is an H-Toeplitz matrix.
In 1964, Brown and Halmos [4] showed that on the Hardy space, two bounded Toeplitz operators Tϕ and Tψ commute if and only if: (i) Both ϕ and ψ are analytic, or (ii) both ˉϕ and ˉψ are analytic, or (iii) one is a linear function of the other. In [3] Axler and Cuckovic proved that if the two symbols are bounded harmonic functions, then the same result is also true for Toeplitz operators on the Bergman space. The situation with a general symbol is rather more complicated. Let ϕ,ψ∈L∞(D) be radial functions, i.e., ϕ(z)=ϕ(|z|), z∈D. It is well known and easy to see that two Toeplitz operators with radial symbols commute. In [10] Louhichi and Zakariasy showed that if p and s are integers such that ps≤0, then the Toeplitz operators with symbols eipθϕ and eisθψ commute only in certain trivial cases. This result is not true if both of the integers p and s satisfy ps>0. There are lots of examples of functions of positive quasihomogeneous degree which are the symbols of commuting Toeplitz operators (see [5]).
The H-Toeplitz operator on the Bergman space was first studied recentely by [8], where the commutativity of H-Toeplitz operators with analytic or harmonic symbols is discussed. Motivated by these works, in this paper we will characterize the commuting H-Toeplitz operators with quasihomogeneous symbols, nonharmonic ones.
The organisation of paper is as follows. In Section 2, we shall collect some notations and results as preliminaries. In Section 3, we will first discuss when the product of two H-Toeplitz operators with quasihomogeneous symbols is still an H-Toeplitz operator (see Theorem 3.1), as a byproduct, we get the characterization of semi-commuting H-Toeplitz operators with quasihomogeneous symbols (see Corollary 3.2). The remaining of Section 3 will characterize the commuting H-Toeplitz operators with quasihomogeneous symbols in terms of different degrees and same degrees respectively (see Theorems 3.4 and 3.6 respectively).
Let R be the space of functions which are square integrable in [0,1] with respect to the measure rdr. By using the fact that the trigonometric polynomials are dense in L2(D,dA) and that for k1≠k2, eik1θR is orthogonal to eik2θR, one sees that
L2(D,dA)=∑k∈ZeikθR. |
Thus, each function ϕ∈L2(D,dA) can be written as (see [5])
ϕ(reiθ)=∑k∈Zeikθφk(r), φk∈R. |
Moreover, if ϕ∈L∞(D)⊂L2(D,dA), then for each r∈[0,1),
|φk(r)|=|12π∫2π0ϕ(reiθ)e−ikθdθ|≤supz∈D|ϕ(z)|, k∈Z. |
Hence, the functions φk are bounded in the disk. We call every function in eikθR to be quasihomogeneous function of degree k.
The following lemma will be used frequently.
Lemma 2.1. Let p be an integer and φ a bounded radial function. Then for each nonnegative integer n,
Beipθφ(z2n)={2√n+12n+1(n+p+1)ˆφ(2n+p+2)zn+p,n+p≥0,0,n+p<0 |
and
Beipθφ(z2n+1)={2√n+22n+2(p−n)ˆφ(p+2)zp−1−n,n+1≤p,0,n+1>p. |
Proof. Note that Kz(w)=∑∞j=0(1+j)ˉzjwj, it follows from the reproducing formula (1.1) that, when n+p≥0,
Beipθφ(z2n)=PMeipθφK(z2n)=√2n+24n+2PMeipθφ(zn)=√n+12n+1∫Deipθφ(w)wn∞∑j=0(1+j)¯wjzjdA(w)=√n+12n+1∞∑j=0∫10∫2π0ei(p+n−j)θφ(r)rn+j+1(1+j)zj1πdθdr=√n+12n+1(2n+2p+2)zn+p∫10φ(r)r2n+p+1dr=2√n+12n+1(n+p+1)ˆφ(2n+p+2)zn+p. |
When n+p<0, the above calculation also shows Beipθφ(z2n)=0. Similarly, when n+1≤p,
Beipθφ(z2n+1)=PMeipθφK(z2n+1)=√2n+44n+4PMeipθφ(¯zn+1).=√n+22n+2∫Deipθφ(w)¯wn+1∞∑j=0(1+j)¯wjzjdA(w)=√n+22n+2∞∑j=0∫10∫2π0ei(p−n−1−j)θφ(r)rn+2+j(1+j)zjdθdrπ=√n+22n+2(2p−2n)zp−1−n∫10φ(r)rp+1dr=2√n+22n+2(p−n)ˆφ(p+2)zp−1−n. |
When n+1>p, the above computation also gives Beipθφ(z2n+1)=0. The proof is complete.
An operator that will arise in our study of H-Toeplitz operators is the Mellin transform, which is defined for any function φ∈L1([0,1],rdr), by the formula
ˆφ(z)=∫10φ(r)rz−1dr. |
It is clear that ˆφ is analytic in the half right plane {z:Rez>2}. It is important and helpful to know that the Mellin transform is uniquely determined by its value on an arithmetic sequence of integers. In fact, we have the following classical theorem (see [12], p. 102).
Lemma 2.2. Suppose that φ is a bounded analytic function on {z:Rez>0}. If φ vanishes at the pairwise distinct points {zk:k=1,2,…}, where inf{|zk|}>0 and ∑∞k=1(1zk)=∞, then φ=0.
As a simple application of the above lemma, we have the following fact which we will use frequently to prove our main results.
Corollary 2.3. For φ∈L1([0,1],rdr), if there exists a sequence of positive integers {nk}, such that ∑k1/nk=∞ and ˆφ(nk)=0 for all k, then φ=0.
In this section, we will characterize the commutativity of two H-Toeplitz operators with quasihomogeneous symbols.
In order to prove the semi-commuting H-Toeplitz operators, we first give the result for when the product of two H-Toeplitz operators with quasihomogeneous symbols is still an H-Toeplitz operator.
Theorem 3.1. Let p,s be integers and ϕ1,ϕ2 two bounded radial functions. If there is ϕ∈L∞(D) such that Beipθϕ1Beisθϕ2=Bϕ, then either ϕ1=ϕ=0 or ϕ2=ϕ=0.
Proof. Now for ϕ∈L∞(D), write
ϕ(reiθ)=∑k∈Zeikθφk(r), |
where each φk is bounded radial function. We show the conclusion by considering two cases.
Case 1. We assume s=2ℓ for some integer ℓ. Then using Lemma 2.1, direct calculations give that for nonnegative integer n, when n+ℓ≥0 and n+ℓ+p≥0,
Beipθϕ1Beisθϕ2(z4n)=A′n,ℓ,pˆϕ2(4n+2ℓ+2)ˆϕ1(2n+2ℓ+p+2)zn+ℓ+p, | (3.1) |
where
A′n,ℓ,p=4√(2n+1)(n+ℓ+1)(4n+1)(2n+2ℓ+1)(2n+2ℓ+1)(n+ℓ+p+1), |
and
Bϕ(z4n)=∞∑k=02√2n+14n+1(k+1)ˆφk−2n(k+2n+2)zk. | (3.2) |
Since Beipθϕ1Beisθϕ2(z4n)=Bϕ(z4n), then (3.1) and (3.2) give that
ˆφk−2n(k+2n+2)=0, k≠n+ℓ+p. | (3.3) |
Let j=k−2n, so k=j+2n. For each fixed interger j, when n>ℓ+p−j, then k=j+2n>n+ℓ+p. Hence, for each integer j, there is Nj, when n≥Nj, the Eq (3.3) implies
ˆφj(j+4n+2)=0. |
Obviously ∑∞n=Nj1/(j+4n+2)=∞, so by Corollary 2.3, we get φj=0 for every integer j, to obtain that ϕ=0.
The above has shown that Beipθϕ1Beisθϕ2=0. Now the Eq (3.1) gives that there is an integer N0>0,
ˆϕ2(4n+2ℓ+2)ˆϕ1(2n+2ℓ+p+2)=0, n≥N0. | (3.4) |
Set
E1={n≥N0:ˆϕ1(2n+2ℓ+p+2)=0}, |
E2={n≥N0:ˆϕ2(4n+2ℓ+2)=0}. |
If ∑n∈E11/n<∞, then ∑n∈E21/n=∞, thus by Corollary 2.3, we get ϕ2=0; similarly, if ∑n∈E21/n<∞, then it must be ϕ1=0.
Case 2. We suppose s=2ℓ+1 for some integer ℓ. Then making use of Lemma 2.1, we get that for nonnegative integer n, when n+ℓ+1≥0 and n+ℓ+1+p≥0,
Beipθϕ1Beisθϕ2(z4n+2)=A″n,ℓ,pˆϕ2(4n+2ℓ+5)ˆϕ1(2n+2ℓ+p+4)zn+ℓ+1+p, |
where
A″n,ℓ,p=4√(2n+2)(n+ℓ+2)(4n+3)(2n+2ℓ+3)(2n+2ℓ+3)(n+ℓ+p+2), |
and
Bϕ(z4n+2)=∞∑k=02√2n+24n+3(k+1)ˆφk−2n−1(k+2n+3)zk. |
Using same arguments as done in Case 1, it follows from Beipθϕ1Beisθϕ2(z4n)=Bϕ(z4n) that the conclusion holds too. The proof is complete.
Note that the authors showed an example in [8] that BzBz≠Bz2. The following corollary is quickly, which answers that two H-Toeplitz operators with quasihomogeneous symbols semi-commute only in trivial case.
Corollary 3.2. Let p,s be integers and ϕ1,ϕ2 two bounded radial functions. Then the following are equivalent:
(a) Beipθϕ1Beisθϕ2=Bei(p+s)θϕ1ϕ2;
(b) Beipθϕ1Beisθϕ2=0;
(c) ϕ1=0 or ϕ2=0.
Theorem 3.1 also helps us to identify the H-Toeplitz operators which are idempotents. The following corollary is an immediate consequence.
Corollary 3.3. Let ϕ be a bounded quasihomogeneous function. Then B2ϕ=Bϕ if and only if ϕ=0.
Now we start to characterize when two H-Toeplitz operators with quasihomogeneous symbols commute. The first coming theorem tells that two H-Toeplitz operators with quasihomogeneous symbols such that the signs of their quasihomogeneous degrees are different commute only in the trivial case. Notice that in [8], the authors ever presented an example that BzBˉz≠BˉzBz.
Theorem 3.4. Let p,s be two distinct integers and ϕ1,ϕ2 two bounded radial functions. Then Beipθϕ1Beisθϕ2=Beisθϕ2Beipθϕ1 if and only if ϕ1=0 or ϕ2=0.
Proof. The sufficiency is obvious. We now show the necessity. Suppose that BeipθϕBeisθψ=BeisθψBeipθϕ and we deduce the conclusion by the following cases.
Case 1. p=2q and s=2ℓ for some integers q and ℓ, where q≠ℓ. Then when n+ℓ≥0 and n+ℓ+2q≥0, we have the Eq (3.1); and similarly, when n+q≥0 and n+q+2ℓ≥0, we have
Beisθϕ2Beipθϕ1(z4n)=A′n,q,sˆϕ1(4n+2q+2)ˆϕ2(2n+2q+s+2)zn+q+s, | (3.5) |
where
A′n,q,s=4√(2n+1)(n+q+1)(4n+1)(2n+2q+1)(2n+2q+1)(n+q+s+1), |
Note that n+ℓ+p≠n+q+s, thus by the Eqs (3.1) and (3.5), we conclude that there is an integer N0>0, the Eq (3.4) holds for all n≥N0. Thus done as the argument in Theorem 3.1 we may conclude that ϕ1=0 or ϕ2=0.
Case 2. p=2q+1 and s=2ℓ for some integers q and ℓ. For any nonnegative integer n,
Beipθϕ1Beisθϕ2(z4n)=Beisθϕ2Beipθϕ1(z4n). |
The left side of the above is (3.1) when n+ℓ≥0 and n+ℓ+p≥0, while the right side of the above is zero when n≥N for large enough positive integer N. It follows that (3.4) holds for N0=max(N,|ℓ|+|p|). So as discussed in Case 1, we see that ϕ1=0 or ϕ2=0.
Case 3. p=2q and s=2ℓ+1 for some integers q and ℓ. This case is similar to Case 2.
Case 4. p=2q+1 and s=2ℓ+1 for some integers q and ℓ with q≠ℓ. This case is similar to Case 1 when applying the equality
Beipθϕ1Beisθϕ2(z4n+2)=Beisθϕ2Beipθϕ1(z4n+2) |
for nonnegtive integer n. The detail is omitted and we finish the proof.
For the commuting of two H-Toeplitz operators with same degree quasihomogenous symbols, the situation becomes quite hard. We first give the following lemma.
Lemma 3.5. Let p be an integer and ϕ1,ϕ2 two bounded radial functions. Assume Beipθϕ1Beipθϕ2=Beipθϕ2Beipθϕ1.
(a) If p=2q for some integer q, then for any Rez>max(0,−2p),
^rp+2ϕ1(2z)^rp+2ϕ2(z+p)=^rp+2ϕ2(2z)^rp+2ϕ1(z+p). | (3.6) |
(b) If p=2q+1 for some integer q, then for any Rez>max(0,−2p−1),
^rp+4ϕ1(2z)^rp+4ϕ2(z+p−1)=^rp+4ϕ2(2z)^rp+4ϕ1(z+p−1). | (3.7) |
Proof. (a) Suppose p=2q for some integer q. So using
Beipθϕ1Beipθϕ2(z4n)=Beipθϕ2Beipθϕ1(z4n), |
together with (3.1) and (3.5) where s=p and integer n≥N=max(0,−3q), it follows that
ˆϕ1(4n+p+2)ˆϕ2(2n+2p+2)=ˆϕ2(4n+p+2)ˆϕ1(2n+2p+2) | (3.8) |
when n≥N, or equivalently,
^rp+2ϕ1(4n)^rp+2ϕ2(2n+p)=^rp+2ϕ2(4n)^rp+2ϕ1(2n+p) |
when n≥N. Set
Φ(z)=^rp+2ϕ1(2z)^rp+2ϕ2(z+p)−^rp+2ϕ2(2z)^rp+2ϕ1(z+p). |
It is easy to see that Φ is a bounded analytic function in the right half plane {z:Rez>max(0,−2p)}. The Eq (3.8) tells that Φ(2n)=0 when n≥N. Thus by Lemma 2.2 it concludes that Φ≡0 in the right half plane {z:Rez>max(0,−2p)}, which gives (3.6).
(b) Suppose p=2q+1 for some integer q. Similar to the previous case, it follows from
Beipθϕ1Beipθϕ2(z4n+2)=Beipθϕ2Beipθϕ1(z4n+2) |
that for big enough integer N, when n≥N,
ˆϕ1(4n+p+4)ˆϕ2(2n+2p+3)=ˆϕ2(4n+p+4)ˆϕ1(2n+2p+3), |
or equivalently,
^rp+4ϕ1(4n)^rp+4ϕ2(2n+p−1)=^rp+4ϕ2(4n)^rp+4ϕ1(2n+p−1) |
when n≥N. Hence same arguments used in (a) will give (3.7). The proof is complete.
Now we can characterize the commuting H-Toeplitz operators with same nonnegative degree quasihomogeneous symbols.
Theorem 3.6. Let p be a nonnegative integer and ϕ1,ϕ2 two bounded radial functions. Then Beipθϕ1Beipθϕ2=Beipθϕ2Beipθϕ1 if and only if there exist α,β∈C, |α|+|β|≠0 such that αϕ1+βϕ2=0.
Proof. We first show the sufficiency. If αϕ1+βϕ2=0 for α,β, not all zero, we may assume α≠0, then ϕ1=cϕ2, where c=−β/α. Hence Beipθϕ1=cBeipθϕ2, and so clearly Beipθϕ1Beipθϕ2=Beipθϕ2Beipθϕ1.
Now we show the necessity using Lemma 3.5. First consider the case p is even. Without loss of generality, we assume ϕ2≠0. Put
E={z: Rez>0, ^rp+2ϕ2(z)=0}. |
By (3.6) we get that for Rez>0,
^rp+2ϕ1(2z)^rp+2ϕ2(2z)=^rp+2ϕ1(z+p)^rp+2ϕ2(z+p), z+p,2z∉E. | (3.9) |
Case 1. Suppose p is positive. We claim that, there is z0∈(1+p,2+p) such that for any integer k≥0,
z0−p2k+2p∉E. | (3.10) |
In fact, on the one hand, we note that {z0−p2k+2p:k≥0} is a bounded sequence since
2p<z0−p2k+2p≤z0+p<2p+2. |
On the other hand, it is easy to check that for z1,z2∈(1+p,2+p) with z1≠z2,
{z1−p2k+2p: k≥0}⋂{z2−p2k+2p: k≥0}=∅. |
Now, if the claim is not true, then for each z∈(1+p,2+p), there is a nonnegative integer kz such that z−p2kz+2p∈E. It follows that the bounded infinite set
{z−p2kz+2p: z∈(1+p,2+p)}⊂E, |
which implies that the analytic function ^rp+2ϕ2≡0, and so ϕ2=0 by Corollary 2.3, a controdiction. Hence the claim holds.
Now we fix a z0∈(1+p,2+p) such that (3.10) holds for each integer k≥0. By (3.9), we have
^rp+2ϕ1^rp+2ϕ2(z0−p2k+2p)=^rp+2ϕ1^rp+2ϕ2(2⋅(z0−p2k+1+p))=^rp+2ϕ1^rp+2ϕ2(z0−p2k+1+2p), k≥0. |
It induces that
^rp+2ϕ1^rp+2ϕ2(z0−p2k+2p)=^rp+2ϕ1^rp+2ϕ2(z0+p)=:c, k≥0. |
Notice that z0−p2k+2p→2p as k→∞, so the above implies that the analytic function
^rp+2ϕ1^rp+2ϕ2(z)≡c, Rez>0, |
which means that the Mellin transformation of rp+2(ϕ1−cϕ2) is identically zero in the right half plane {z:Rez>0}, hence we get that ϕ1=cϕ2 by Corollary 2.3, as desired.
Case 2. Suppose p=0. In this case, (3.6) becomes
ˆϕ1(2z+2)ˆϕ2(z+2)=ˆϕ2(2z+2)ˆϕ1(z+2), Rez>0. |
Replacing z by z−1 in the above we get
ˆϕ1(2z)ˆϕ2(z+1)=ˆϕ2(2z)ˆϕ1(z+1), Rez>1. |
Applying the same arguments done in Case 1, we will obtain the desired conclusion.
When p is positive and odd, the proof is similar by using (3.7). We omit the detail and finish the proof.
We don't know whether the previous theorem is true when p is negative. But for special symbols, it is still the case.
Theorem 3.7. Let j,k,s,t be integers. Then BzjˉzkBzsˉzt=BzsˉztBzjˉzk if and only if j=s and k=t.
Proof. The sufficiency is clear. Now we show the necessity. Let ϕ1=zjˉzk=rj+kei(j−k)θ and ϕ2=zsˉzt=rs−tei(s−t)θ. Since ϕ1≠0 and ϕ2≠0, so Theorem 3.4 tells that j−k=s−t:=p. It is left to show that j+k=s+t.
We only consider the case when p=2q for some integer q (the case p=2q+1 is similar). So by (3.6) we have
^rp+2rj+k(2z)^rp+2rs+t(z+p)=^rp+2rs+t(2z)^rp+2rj+k(z+p) |
when Rez>max(0,−2p). By the definition of the Mellin transformation, the above yields that
(p+2+j+k+2z)(2p+2+s+t+z)=(p+2+s+t+2z)(2p+2+j+k+z) |
when Rez>max(0,−2p). Thus it is easy to get that j+k=s+t, and which together with j−k=s−t induces j=s and k=t. The proof is complete.
In this research, it obtains the following characterizations for the commuting H-Toeplitz operators with quasihomogeneous symbols on the Bergman space.
(1) Let p,s be two distinct integers and ϕ1,ϕ2 two bounded radial functions. Then Beipθϕ1Beisθϕ2=Beisθϕ2Beipθϕ1 if and only if ϕ1=0 or ϕ2=0.
(2) Let p be a nonnegative integer and ϕ1,ϕ2 two bounded radial functions. Then Beipθϕ1Beipθϕ2=Beipθϕ2Beipθϕ1 if and only if there exist α,β∈C, |α|+|β|≠0 such that αϕ1+βϕ2=0.
(3) Let p,s be integers and ϕ1,ϕ2 two bounded radial functions. Then Beipθϕ1Beisθϕ2=Bei(p+s)θϕ1ϕ2 if and only if ϕ1=0 or ϕ2=0.
The authors would like to thank the referees for many helpful comments and suggestions. The research was supported by the National Natural Science Foundation of China (Grant Nos. 11771401).
The authors declare that there is no conflict of interest.
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