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On quasi-similarity and reducing subspaces of multiplication operator on the Fock space
J. Math. Anal. Appl. 409 (2014) 899–905
Contents lists available at ScienceDirect
Journal of Mathematical Analysis and Applications journal homepage: www.elsevier.com/locate/jmaa
On quasi-similarity and reducing subspaces of multiplication operator on the Fock space Yucheng Li ∗ , Wenhua Lan, Jingli Liu Department of Mathematics, Hebei Normal University, Hebei Province Key Laboratory of Computational Mathematics and Applications, Shijiazhuang 050024, PR China
article
info
Article history: Received 6 May 2013 Available online 31 July 2013 Submitted by J.A. Ball
abstract Let Fα2 (α > 0) denote the Fock space which consists of all entire functions f in L2 (C, dλα ). n We prove that the multiplication operator Mz n is quasi-similar to M on Fα2 . Then the z 1 2 n reducing subspaces of Mz are characterized on Fα . © 2013 Elsevier Inc. All rights reserved.
Keywords: Fock space Multiplication operator Quasi-similarity Reducing subspaces
1. Introduction Let C be the complex plane and dA be ordinary area measure on C. Let Fα2 (α > 0) denote the Fock space which consists
of all entire functions f in L2 (C, dλα ), where dλα is the Gaussian measure, dλα (z ) = πα e−α|z | dA(z ). It is well known that Fα2 is a closed subspace of L2 (C), and Fα2 is a Hilbert space with the reproducing kernel Kα (z , w) = eα z w¯ . The normalized 2
α
K (z ,w) ¯ 2 |w| reproducing kernel at w is denoted by kw (z ) = √Kα (w,w) = eα z w− . If f , g ∈ Fα2 , the inner product of f and g is defined α by ⟨f , g ⟩α = C f (z )g (z )dλα (z ). In studying an operator on a Hilbert space, it is of interest to characterize the commutant of a given operator, for such a characterization should help in understanding the structure of the operator. From the information of the commutant, people research the similar equivalence and unitary equivalence of operators. The commutant of an analytic Toeplitz operator on the Hardy space and the Bergman space has been studied extensively in the literature. We mention here that the papers [2–4,7,10,11] and the books [5,13] include a lot of the knowledge of the corresponding operator theory. In [12], Zhu obtained a complete description of the reducing subspaces of multiplication operators induced by the Blaschke product with two zeros on the Bergman space. In 2007, Jiang and Li (see [7]) proved that each analytic Toeplitz operator MB(z ) is similar to n copies of the Bergman shift if and only if B(z ) is an n-Blaschke product. On the weighted Bergman space, Li (see [9]) proved that the n multiplication operator Mz n is similar to 1 Mz . In 2011, Ahmadi and Hedayatian (see [1]) generalized this result to bilateral shift operators. Jiang and Zheng in [8] showed that the main result in [7] holds on the weighted Bergman space. Recently, Douglas and Kim (see [6]) studied the reducing subspaces of an analytic multiplication operator Mz n on the Bergman space A2α (Ar ) of the annulus Ar , and they showed that Mz n has exactly 2n reducing subspaces.
∗
2
Corresponding author. E-mail addresses: [email protected] (Y. Li), [email protected] (W. Lan), [email protected] (J. Liu).
0022-247X/$ – see front matter © 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.jmaa.2013.07.059
900
Y. Li et al. / J. Math. Anal. Appl. 409 (2014) 899–905
n
In this paper, we first prove that the multiplication operator Mz n is quasi-similar to 1 Mz on the Fock space. We then characterize the commutant of the operator and show that Mz n has exactly 2n reducing subspaces on the Fock space. 2. The quasi-similarity of Mz n (n ≥ 2) Recall that en (z ) =
αn n z n!
(n = 0, 1, . . .) form an orthonormal basis of the Fock space. If we set Fj = span{enk+j } (j =
0, 1, . . . , n − 1), then we have the following lemma. Lemma 2.1. If Fj = span{enk+j }(j = 0, 1, . . . , n − 1), then (i) {enk+j }∞ an orthonormal basis of Fj . k= 0 form (ii) Fα2 = F0 F1 · · · Fn−1 . (iii) Fj is a reducing subspace of Mz n . Proof. (i) Note that
α nk+j nk+j α nm+j nm+j z¯ dλα ( z ) z ⟨enk+j , enm+j ⟩ = (nk + j)! (nm + j)! C nk+j α α nm+j nm+j α −α|z |2 = lim z nk+j z¯ e dA(z ). R→+∞ D (nk + j)! (nm + j)! π R
(2.1)
If k = m, we have
α nk+j R→+∞ (nk + j)! = 1.
⟨enk+j , enm+j ⟩ = lim
2π
dθ
R
0
r 2(nk+j)
0
α r −αr 2 e dr π (2.2)
If k = ̸ m, then ⟨enk+j , enm+j ⟩ = 0. (ii) It is easy to prove that Fj ⊥Ft , 0 ≤ j ̸= t ≤ n − 1. Next, for f ∈ Fα2 , we get that f has the form f =
∞
a0k enk + · · · +
∞
k=0
an−1, k enk+n−1 .
k=0
Suppose that f = 0. Then from
∞ n−1
ajk enk+j , el
= 0 (l = 0, 1, . . .),
(2.3)
k=0 j=0 n
we obtain that ajk = 0 (j = 0, . . . , n − 1, k = 0, 1, . . .). That is, 0 = 0 ⊕ 0 ⊕ · · · ⊕ 0. Therefore, Fα2 = F0 F1 · · · Fn−1 .
(iii) It is easy to see that both Fj and Fj⊥ are invariant subspaces of Mz n .
2 Let {β(k)}∞ k=0 be a sequence of positive numbers with β(0) = 1. The Hilbert space H (β) consists of all formal power
series f (z ) =
∞
k=0 2
fˆ (k)z k (z ∈ C), and ∥f ∥2β =
is bounded on H (β) if and only if ∥Mz ∥ =
β(k) =
k!
. So ∥Mz ∥ = supk αk
k+1
α
∞
k=0 β(k+1) supk β(k)
|fˆ (k)|2 β(k)2 < ∞. It is known that the multiplication operator Mz < ∞. We view the Fock space Fα2 as a Hilbert space H 2 (β), where
= +∞.
Let H and K be complex Hilbert spaces. An operator X in L(H , K ) is said to be quasi-invertible if X has zero kernel and dense range. Recall that for A ∈ L(H ) and B ∈ L(K ), A is quasi-similar to B if there exist two quasi-invertible operators S and T in L(H , K ) and L(K , H ) respectively such that SA = BS and AT = TB. Theorem 2.2. The multiplication operator Mz n is quasi-similar to
n
1
Mz on the Fock space.
Proof. Note that
Mz ek = z
αk k z = k!
k+1
α
ek+1 .
(2.4)
Y. Li et al. / J. Math. Anal. Appl. 409 (2014) 899–905
901
Set Mj = Mz n |Fj (j = 0, 1, . . . , n − 1). Then
Mj enk+j = z
n
=
α nk+j nk+j z (nk + j)!
Γ (nk + n + j + 1) enk+n+j . Γ (nk + j + 1)α n
(2.5)
Define Xj : Fα2 → Fj such that Xj ek = akj enk+j . Then we have Xj Mz ek = Mj Xj ek . In fact,
Xj
k+1
α
ek+1 = Mj akj enk+j ,
and
k+1
α
ak+1,j enk+n+j = akj
Γ (nk + n + j + 1) enk+n+j . Γ (nk + j + 1)α n
From ak+1,j akj
=
Γ (nk + n + j + 1) , Γ (nk + j + 1)(k + 1)α n−1
(2.6)
we obtain
akj =
Γ (nk + j + 1) k!Γ (j + 1)α k(n−1)
,
(k ≥ 0).
(2.7)
In the following, we will analyze the limit of sequence akj as k → +∞. Set bkj =
(nk + j)! k!α k(n−1)
then akj =
bkj
Γ (j+1)
=
(nk + j)(nk + j − 1) · · · (k + 1) , α k(n−1)
(2.8)
.
Case 1. When 0 < α < 1, we have limk→∞ akj = +∞. Case 2. When α = 1, it is easy to see that limk→∞ akj = +∞. Case 3. When α > 1, we will consider the following equality ln bkj = (ln(nk + j) + ln(nk + j − 1) + · · · + ln(k + 1)) − k(n − 1) ln α
= Ak − Bk = Bk
Ak Bk
−1 .
(2.9)
Note that Bk is a monotone increasing sequence, and Bk → +∞ as k → ∞. By Stoltz’s theorem, we obtain lim
k→∞
Ak Bk
= lim
k→∞
Ak − Ak−1 Bk − Bk−1 ln
= lim
(nk+j)(nk+j−1)···(k+1) (nk−n+j)···k
(n − 1) ln α j ln n + k + ln((nk + j − 1) · · · (nk − n + j + 1)) = lim k→∞ (n − 1) ln α = +∞. k→∞
Therefore, there exists a positive integer k0 , such that when k > k0 , we have akj → +∞ as k → ∞. If we define Yj with Yj enk+j = ckj ek , then we have Yj Mj enk+j = Mz Yj enk+j . In fact, from
Yj
Γ (nk + n + j + 1) enk+n+j = Mz ckj ek , Γ (nk + j + 1)α n
(2.10) Ak Bk
> 2. Thus Ak − Bk > Bk , which implies that
902
Y. Li et al. / J. Math. Anal. Appl. 409 (2014) 899–905
we obtain ck+1,j ckj
=
Γ (nk + j + 1)(k + 1)α n−1 . Γ (nk + n + j + 1)
(2.11)
Thus,
ckj =
k!Γ (j + 1)α k(n−1)
Γ (nk + j + 1)
,
(k ≥ 0).
(2.12)
By the discussion, we deduce that limk→∞ ckj = 0. Hence, Xj and Yj are not invertible. Suppose that f ∈ ker Xj , and above ∞ ∞ f , enk+j ⟩ = ⟨ k=0 dk akj enk+j , enk+j ⟩, we deduce that dk = 0 (k = 0, 1, . . .), so f = j k=0 dk ek , dk ∈ C. Then from 0 = ⟨X ∞ ∞ ∗ ∗ ker Xj = {0}. Next, for g ∈ ker Xj , and g = k=0 mk enk+j , mk ∈ C, from 0 = ⟨ek , Xj g ⟩ = ⟨akj enk+j , k=0 mk enk+j ⟩, we have mk = 0 (k = 0, 1, . . .). So ker Xj∗ = (Ran Xj )⊥ = {0}. That is, RanXj = Fj . Similarly, we can obtain ker Yj = {0}, RanYj = Fα2 . M1 · · · Mn−1 , This implies that Xj and Yj are quasi-invertible. Therefore, Mj is quasi-similar to Mz . Since Mz n |Fα2 = M0 we conclude that Mz n is quasi-similar to
n
1
Mz .
On the weighted Bergman space, it is shown that MB(z ) is similarly equivalent to |a| < 1). In the following, we will investigate the situation on the Fock space. Lemma 2.3. Let fn (z ) =
αn (z − a)n ka (z ), where ka (z ) n!
1
n
1
−a n Mz , where B(z ) = ( 1z−¯ ) , (0 < az
= eαz a¯ − 2 α|a| is the normalized reproducing kernel at a (a ∈ C, a ̸= 0). 2
2 Then {fn }∞ n=1 form an orthonormal basis of Fα .
Proof. To prove the result, we will consider the inner product of fn and fm .
⟨fn , fm ⟩ =
C
αn (z − a)n n!
αm 1 1 2 2 m (z − a) eαz a¯ − 2 α|a| eα¯z a− 2 α|a| dλα (z ). m!
(2.13)
If we let z − a = w , then
α n n α m m α −α|w|2 ⟨fn , fm ⟩ = w w ¯ e dA(w) n! m! π C α n n α m m −α|w|2 α w w ¯ e dA(w). = lim R→+∞ D n! m! π R
(2.14)
From Lemma 2.1, we know that
⟨fn , fm ⟩ =
1, 0,
n = m, n ̸= m.
(2.15)
2 Therefore, {fn }∞ n=1 form an orthonormal basis of Fα .
Lemma 2.4. The multiplication operator M(z −a)n is unitarily equivalent to Mz n . Proof. Let Ba (z ) = z − a, and let CBa denote the composition operator, CBa : f (z ) −→ f (Ba ),
for any f ∈ Fα2 , a ∈ C.
The operator defined by X = Mka CBa : Fα2 → Fα2 is a unitary operator. In fact, for f , g ∈ Fα2 , by computation, we have
⟨Xf , Xg ⟩ =
f (z − a)g (z − a)eα z a¯ +α¯z a−α|a| dλα (z ) 2
C
= ⟨f , g ⟩. For f ∈ Fα2 , we obtain M(z −a)n Xf (z ) = (z − a)n Mka CBa f (z ) = (z − a)n ka (z )f (z − a),
(2.16)
XMz n f (z ) = Mka CBa Mz n f (z ) = ka (z )(z − a)n f (z − a), which implies that M(z −a)n X = XMz n . So M(z −a)n is unitarily equivalent to Mz n .
Y. Li et al. / J. Math. Anal. Appl. 409 (2014) 899–905
Theorem 2.5. The multiplication operator M(z −a)n is quasi-similar to
n
1
Mz on the Fock space.
Proof. Applying Theorem 2.2 and Lemma 2.4, we can obtain the desired result. If we set gn (z ) =
αn (z n!
+ a)n k−a (z ) and hn (z ) =
αn (a n!
903
− z )n ka (z ) (a ∈ C, n = 0, 1, . . .), then the following corollary
holds.
n
Corollary 2.6. (i) The multiplication operator M(z +a)n is quasi-similar to 1 Mz on the Fock space. n (ii) The multiplication operator M(a−z )n is quasi-similar to 1 Mz on the Fock space. Proof. The proof is similar to Lemmas 2.3, 2.4 and Theorem 2.5, and we omit it here.
3. The reducing subspaces of Mz n (n ≥ 2) From Lemma 2.1, we know that for any f ∈ Fα2 , there exists a unique orthogonal decomposition f = f0 + f1 + · · · + fn−1 , where fk ∈ Fk (0 ≤ k ≤ n − 1). In this section, we will characterize the reducing subspaces of Mz n on the Fock space. Lemma 3.1. A linear operator S on Fα2 commutes with Mz n if and only if there exist functions gk (0 ≤ k ≤ n − 1) in Fα2 such that Sf =
n−1 k=0
gk fk , where fk ∈ Fk .
2 2 2 Proof. Let Sf = k=0 gk fk , and Mgk : Fα → Fα is the multiplication operator defined by Mgk (h) = gk h for h ∈ Fα . Then for 2 any f ∈ Fα , we have
n−1
SMz n f = Sz n f = z n
n −1
gk fk = Mz n Sf .
(3.1)
k=0
That is, SMz n = Mz n S, as desired. Next, if we suppose that SMz n = Mz n S, then Mt∗n −z n S ∗ = S ∗ Mt∗n −z n , t ∈ C. From the property of the reproducing kernel, we know that ker Mt∗n −z n is generated by
2kπ Kα (z , t wk ) = eα zt wk |wk = e n i , 0 ≤ k ≤ n − 1 .
(3.2)
Note that S ∗ Kα (z , t ) ∈ ker Mt∗n −z n , and we obtain S ∗ Kα (z , t ) =
n−1
ak (t )Kα (z , t wk ),
ak (t ) ∈ C.
(3.3)
k=0
Therefore, Sf (z ) = ⟨Sf , Kα (ω, z )⟩
= ⟨f , S ∗ Kα (ω, z )⟩ =
n −1
ak (z )f (z wk )
k=0
=
n −1
ak (z )(f0 (z wk ) + f1 (z wk ) + · · · + fn−1 (z wk ))
k =0
=
n −1
ak (z )(f0 (z ) + wk f1 (z ) + · · · + wkn−1 fn−1 (z )),
(3.4)
k=0
where the last equality holds because wkn = 1 (0 ≤ k ≤ n − 1). Let gj =
n−1
j
ak (z )wk
(0 ≤ j ≤ n − 1), ak (z ) ∈ Fα2 .
(3.5)
k=0
Thus, we obtain Sf =
n−1 k=0
g k fk .
In the following, we will determine the reducing subspaces of the multiplication operator Mz n on the Fock space. 0 M1
···
0
0
·
·
0
0
· ···
··· ·
M Theorem 3.2. Suppose that B = (Bjk )n×n is a projection such that MB = BM, where M = Thenthere exist functions ϕjk ∈ Fα2 (0 ≤ j, k ≤ n − 1), such that Bjk = Mϕjk , and ϕjk =
cj , 0,
j = k, j ̸= k,
0
0
0
0 0
· Mn−1
where cj is a real number.
.
904
Y. Li et al. / J. Math. Anal. Appl. 409 (2014) 899–905 j
2 Proof. Applying Lemma 3.1, we have Bf = k=0 ak (z )wk , ϕj ∈ Fα . Set ak (z ) = j=0 ϕj fj , where ϕj = akl (z ) ∈ Fl (0 ≤ l ≤ n − 1). Then the operator B has the following matrix representation
n−1
n−1
n−1 ak0 k=0 n−1 ak1 B = k=0 · n −1
n −1
ak0 wk
k=0 n−1
k=0
ak1 wk
···
k=0
· n −1
ak,n−1
k=0
n−1
···
·
ak,n−1 wk
···
k=0
n−1 l =0
akl (z ), and
ak0 wkn−1
n −1 ak1 wk . k=0 · n −1 n−1 ak,n−1 wk n−1
(3.6)
k=0
It follows that B = (Mϕjk )n×n (0 ≤ j, k ≤ n − 1), and Mϕjk : Fk → Fj . Now we will analyze multiplication operators Mϕjk . If j = k, Mϕjj : Fj → Fj , suppose that ϕjj (z ) =
ϕjj (z ) =
∞
m=0
∞
l =0
jj
jj
al z l . Note that Fj = span{enk+j }, so al = 0 (l ̸= mn). Thus,
jj
amn z mn . Since B is a projection operator, we know that Mϕjj = Mϕ∗jj . From
⟨Mϕjj (z ) ej , emn+j ⟩ = ⟨ej , Mϕ∗jj (z ) emn+j ⟩ = ⟨ej , Mϕjj (z ) emn+j ⟩,
(3.7)
jj
jj
jj
we deduce that amn = 0 (m = 1, 2, . . .). Observe that Mϕjj is a self-adjoint operator, hence a0 is a real number. Set cj = a0 , we then have ϕjj = cj . If j ̸= k, Mϕjk : Fk → Fj , without loss of generality, assume that k < j, and set ϕjk (z ) =
∞
l =0
jk
al z l . Similar to the preceding
jk
discussion, we know that only the coefficient aj−k in the Taylor expansion of ϕjk (z ) is nonzero. Thus,
jk aj−k z j−k ek
⟨Mϕjk (z ) ek , ej ⟩ = ⟨
, ej ⟩ =
jk aj − k
j! . α j−k k!
(3.8)
On the other hand, since B is a projection operator, we have
⟨Mϕjk (z ) ek , ej ⟩ = ⟨ek , Mϕ∗jk (z ) ej ⟩ = ⟨ek , Mϕkj (z ) ej ⟩ = 0. jk
(3.9)
Hence aj−k = 0 and ϕjk = 0 (j ̸= k). The proof is complete.
Theorem 3.3. Suppose that Mz n : Fα2 → Fα2 . Then Mz n has 2n reducing subspaces. Proof. From Theorem 3.2, we know that B has the following form
a0 0
B= · 0
0 a1
· 0
··· ··· · ···
0 0
. ·
(3.10)
an − 1
B2 = B yields that 0 (0 ≤ k ≤ n − 1) or 1. Hence, ak = Mz n |Fα2 = M0 M1 · · · Mn−1 has 2n reducing subspaces a0 F0 ⊕ a1 F1 ⊕ · · · ⊕ an−1 Fn−1 ,
ak = 0 (0 ≤ k ≤ n − 1) or 1.
The minimal reducing subspaces are F0 , F1 , . . . , Fn−1 , as required.
(3.11)
Acknowledgments The authors would like to thank professor Chunlan Jiang for his helpful discussions in finishing the paper. The authors would like to express their sincere appreciation to the referee for his/her valuable suggestions and comments. The authors were supported by HBNSF (A2010000351, ZD200904, 140101). References [1] M.F. Ahmadi, K. Hedayatian, On similarity of powers of shift operators, Turkish J. Math. 35 (2011) 1–5. [2] J.A. Ball, Hardy space expectation operators and reducing subspaces, Proc. Amer. Math. Soc. 47 (1975) 351–357. [3] C.C. Cowen, The commutant of an analytic Toeplitz operator, Trans. Amer. Math. Soc. 239 (1978) 1–31.
Y. Li et al. / J. Math. Anal. Appl. 409 (2014) 899–905 [4] [5] [6] [7] [8] [9] [10] [11] [12] [13]
905
J.A. Deddens, T.K. Wong, The commutant of analytic Toeplitz operators, Trans. Amer. Math. Soc. 184 (1973) 261–273. R.G. Douglas, Banach Algebra Techniques in Operator Theory, Academic Press, New York, 1972. R.G. Douglas, Y.S. Kim, Reducing subspaces on the annulus, Integral Equations Operator Theory 70 (2011) 1–15. C.L. Jiang, Y.C. Li, The commutant and similarity invariant of analytic Toeplitz operators on Bergman space, Sci. China Ser. A 50 (5) (2007) 651–664. C.L. Jiang, D.C. Zheng, Similarity of analytic Toeplitz operators on the Bergman spaces, J. Funct. Anal. 258 (2010) 2961–2982. Y.C. Li, On similarity of multiplication operator on weighted Bergman space, Integral Equations Operator Theory 63 (1) (2009) 95–102. E. Nordgren, Reducing subspace of analytic Toeplitz operators, Duke Math. J. 34 (1967) 175–181. M. Stessin, K.H. Zhu, Generalized factorization in Hardy spaces and the commutant of Toeplitz operators, Canad. J. Math. 55 (2) (2003) 379–400. K.H. Zhu, Reducing subspaces for a class of multiplication operators, J. Lond. Math. Soc. 62 (2) (2000) 553–568. K.H. Zhu, Analysis on Fock Spaces, in: GTM, vol. 263, Springer-Verlag, New York, 2012.
Contents lists available at ScienceDirect
Journal of Mathematical Analysis and Applications journal homepage: www.elsevier.com/locate/jmaa
On quasi-similarity and reducing subspaces of multiplication operator on the Fock space Yucheng Li ∗ , Wenhua Lan, Jingli Liu Department of Mathematics, Hebei Normal University, Hebei Province Key Laboratory of Computational Mathematics and Applications, Shijiazhuang 050024, PR China
article
info
Article history: Received 6 May 2013 Available online 31 July 2013 Submitted by J.A. Ball
abstract Let Fα2 (α > 0) denote the Fock space which consists of all entire functions f in L2 (C, dλα ). n We prove that the multiplication operator Mz n is quasi-similar to M on Fα2 . Then the z 1 2 n reducing subspaces of Mz are characterized on Fα . © 2013 Elsevier Inc. All rights reserved.
Keywords: Fock space Multiplication operator Quasi-similarity Reducing subspaces
1. Introduction Let C be the complex plane and dA be ordinary area measure on C. Let Fα2 (α > 0) denote the Fock space which consists
of all entire functions f in L2 (C, dλα ), where dλα is the Gaussian measure, dλα (z ) = πα e−α|z | dA(z ). It is well known that Fα2 is a closed subspace of L2 (C), and Fα2 is a Hilbert space with the reproducing kernel Kα (z , w) = eα z w¯ . The normalized 2
α
K (z ,w) ¯ 2 |w| reproducing kernel at w is denoted by kw (z ) = √Kα (w,w) = eα z w− . If f , g ∈ Fα2 , the inner product of f and g is defined α by ⟨f , g ⟩α = C f (z )g (z )dλα (z ). In studying an operator on a Hilbert space, it is of interest to characterize the commutant of a given operator, for such a characterization should help in understanding the structure of the operator. From the information of the commutant, people research the similar equivalence and unitary equivalence of operators. The commutant of an analytic Toeplitz operator on the Hardy space and the Bergman space has been studied extensively in the literature. We mention here that the papers [2–4,7,10,11] and the books [5,13] include a lot of the knowledge of the corresponding operator theory. In [12], Zhu obtained a complete description of the reducing subspaces of multiplication operators induced by the Blaschke product with two zeros on the Bergman space. In 2007, Jiang and Li (see [7]) proved that each analytic Toeplitz operator MB(z ) is similar to n copies of the Bergman shift if and only if B(z ) is an n-Blaschke product. On the weighted Bergman space, Li (see [9]) proved that the n multiplication operator Mz n is similar to 1 Mz . In 2011, Ahmadi and Hedayatian (see [1]) generalized this result to bilateral shift operators. Jiang and Zheng in [8] showed that the main result in [7] holds on the weighted Bergman space. Recently, Douglas and Kim (see [6]) studied the reducing subspaces of an analytic multiplication operator Mz n on the Bergman space A2α (Ar ) of the annulus Ar , and they showed that Mz n has exactly 2n reducing subspaces.
∗
2
Corresponding author. E-mail addresses: [email protected] (Y. Li), [email protected] (W. Lan), [email protected] (J. Liu).
0022-247X/$ – see front matter © 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.jmaa.2013.07.059
900
Y. Li et al. / J. Math. Anal. Appl. 409 (2014) 899–905
n
In this paper, we first prove that the multiplication operator Mz n is quasi-similar to 1 Mz on the Fock space. We then characterize the commutant of the operator and show that Mz n has exactly 2n reducing subspaces on the Fock space. 2. The quasi-similarity of Mz n (n ≥ 2) Recall that en (z ) =
αn n z n!
(n = 0, 1, . . .) form an orthonormal basis of the Fock space. If we set Fj = span{enk+j } (j =
0, 1, . . . , n − 1), then we have the following lemma. Lemma 2.1. If Fj = span{enk+j }(j = 0, 1, . . . , n − 1), then (i) {enk+j }∞ an orthonormal basis of Fj . k= 0 form (ii) Fα2 = F0 F1 · · · Fn−1 . (iii) Fj is a reducing subspace of Mz n . Proof. (i) Note that
α nk+j nk+j α nm+j nm+j z¯ dλα ( z ) z ⟨enk+j , enm+j ⟩ = (nk + j)! (nm + j)! C nk+j α α nm+j nm+j α −α|z |2 = lim z nk+j z¯ e dA(z ). R→+∞ D (nk + j)! (nm + j)! π R
(2.1)
If k = m, we have
α nk+j R→+∞ (nk + j)! = 1.
⟨enk+j , enm+j ⟩ = lim
2π
dθ
R
0
r 2(nk+j)
0
α r −αr 2 e dr π (2.2)
If k = ̸ m, then ⟨enk+j , enm+j ⟩ = 0. (ii) It is easy to prove that Fj ⊥Ft , 0 ≤ j ̸= t ≤ n − 1. Next, for f ∈ Fα2 , we get that f has the form f =
∞
a0k enk + · · · +
∞
k=0
an−1, k enk+n−1 .
k=0
Suppose that f = 0. Then from
∞ n−1
ajk enk+j , el
= 0 (l = 0, 1, . . .),
(2.3)
k=0 j=0 n
we obtain that ajk = 0 (j = 0, . . . , n − 1, k = 0, 1, . . .). That is, 0 = 0 ⊕ 0 ⊕ · · · ⊕ 0. Therefore, Fα2 = F0 F1 · · · Fn−1 .
(iii) It is easy to see that both Fj and Fj⊥ are invariant subspaces of Mz n .
2 Let {β(k)}∞ k=0 be a sequence of positive numbers with β(0) = 1. The Hilbert space H (β) consists of all formal power
series f (z ) =
∞
k=0 2
fˆ (k)z k (z ∈ C), and ∥f ∥2β =
is bounded on H (β) if and only if ∥Mz ∥ =
β(k) =
k!
. So ∥Mz ∥ = supk αk
k+1
α
∞
k=0 β(k+1) supk β(k)
|fˆ (k)|2 β(k)2 < ∞. It is known that the multiplication operator Mz < ∞. We view the Fock space Fα2 as a Hilbert space H 2 (β), where
= +∞.
Let H and K be complex Hilbert spaces. An operator X in L(H , K ) is said to be quasi-invertible if X has zero kernel and dense range. Recall that for A ∈ L(H ) and B ∈ L(K ), A is quasi-similar to B if there exist two quasi-invertible operators S and T in L(H , K ) and L(K , H ) respectively such that SA = BS and AT = TB. Theorem 2.2. The multiplication operator Mz n is quasi-similar to
n
1
Mz on the Fock space.
Proof. Note that
Mz ek = z
αk k z = k!
k+1
α
ek+1 .
(2.4)
Y. Li et al. / J. Math. Anal. Appl. 409 (2014) 899–905
901
Set Mj = Mz n |Fj (j = 0, 1, . . . , n − 1). Then
Mj enk+j = z
n
=
α nk+j nk+j z (nk + j)!
Γ (nk + n + j + 1) enk+n+j . Γ (nk + j + 1)α n
(2.5)
Define Xj : Fα2 → Fj such that Xj ek = akj enk+j . Then we have Xj Mz ek = Mj Xj ek . In fact,
Xj
k+1
α
ek+1 = Mj akj enk+j ,
and
k+1
α
ak+1,j enk+n+j = akj
Γ (nk + n + j + 1) enk+n+j . Γ (nk + j + 1)α n
From ak+1,j akj
=
Γ (nk + n + j + 1) , Γ (nk + j + 1)(k + 1)α n−1
(2.6)
we obtain
akj =
Γ (nk + j + 1) k!Γ (j + 1)α k(n−1)
,
(k ≥ 0).
(2.7)
In the following, we will analyze the limit of sequence akj as k → +∞. Set bkj =
(nk + j)! k!α k(n−1)
then akj =
bkj
Γ (j+1)
=
(nk + j)(nk + j − 1) · · · (k + 1) , α k(n−1)
(2.8)
.
Case 1. When 0 < α < 1, we have limk→∞ akj = +∞. Case 2. When α = 1, it is easy to see that limk→∞ akj = +∞. Case 3. When α > 1, we will consider the following equality ln bkj = (ln(nk + j) + ln(nk + j − 1) + · · · + ln(k + 1)) − k(n − 1) ln α
= Ak − Bk = Bk
Ak Bk
−1 .
(2.9)
Note that Bk is a monotone increasing sequence, and Bk → +∞ as k → ∞. By Stoltz’s theorem, we obtain lim
k→∞
Ak Bk
= lim
k→∞
Ak − Ak−1 Bk − Bk−1 ln
= lim
(nk+j)(nk+j−1)···(k+1) (nk−n+j)···k
(n − 1) ln α j ln n + k + ln((nk + j − 1) · · · (nk − n + j + 1)) = lim k→∞ (n − 1) ln α = +∞. k→∞
Therefore, there exists a positive integer k0 , such that when k > k0 , we have akj → +∞ as k → ∞. If we define Yj with Yj enk+j = ckj ek , then we have Yj Mj enk+j = Mz Yj enk+j . In fact, from
Yj
Γ (nk + n + j + 1) enk+n+j = Mz ckj ek , Γ (nk + j + 1)α n
(2.10) Ak Bk
> 2. Thus Ak − Bk > Bk , which implies that
902
Y. Li et al. / J. Math. Anal. Appl. 409 (2014) 899–905
we obtain ck+1,j ckj
=
Γ (nk + j + 1)(k + 1)α n−1 . Γ (nk + n + j + 1)
(2.11)
Thus,
ckj =
k!Γ (j + 1)α k(n−1)
Γ (nk + j + 1)
,
(k ≥ 0).
(2.12)
By the discussion, we deduce that limk→∞ ckj = 0. Hence, Xj and Yj are not invertible. Suppose that f ∈ ker Xj , and above ∞ ∞ f , enk+j ⟩ = ⟨ k=0 dk akj enk+j , enk+j ⟩, we deduce that dk = 0 (k = 0, 1, . . .), so f = j k=0 dk ek , dk ∈ C. Then from 0 = ⟨X ∞ ∞ ∗ ∗ ker Xj = {0}. Next, for g ∈ ker Xj , and g = k=0 mk enk+j , mk ∈ C, from 0 = ⟨ek , Xj g ⟩ = ⟨akj enk+j , k=0 mk enk+j ⟩, we have mk = 0 (k = 0, 1, . . .). So ker Xj∗ = (Ran Xj )⊥ = {0}. That is, RanXj = Fj . Similarly, we can obtain ker Yj = {0}, RanYj = Fα2 . M1 · · · Mn−1 , This implies that Xj and Yj are quasi-invertible. Therefore, Mj is quasi-similar to Mz . Since Mz n |Fα2 = M0 we conclude that Mz n is quasi-similar to
n
1
Mz .
On the weighted Bergman space, it is shown that MB(z ) is similarly equivalent to |a| < 1). In the following, we will investigate the situation on the Fock space. Lemma 2.3. Let fn (z ) =
αn (z − a)n ka (z ), where ka (z ) n!
1
n
1
−a n Mz , where B(z ) = ( 1z−¯ ) , (0 < az
= eαz a¯ − 2 α|a| is the normalized reproducing kernel at a (a ∈ C, a ̸= 0). 2
2 Then {fn }∞ n=1 form an orthonormal basis of Fα .
Proof. To prove the result, we will consider the inner product of fn and fm .
⟨fn , fm ⟩ =
C
αn (z − a)n n!
αm 1 1 2 2 m (z − a) eαz a¯ − 2 α|a| eα¯z a− 2 α|a| dλα (z ). m!
(2.13)
If we let z − a = w , then
α n n α m m α −α|w|2 ⟨fn , fm ⟩ = w w ¯ e dA(w) n! m! π C α n n α m m −α|w|2 α w w ¯ e dA(w). = lim R→+∞ D n! m! π R
(2.14)
From Lemma 2.1, we know that
⟨fn , fm ⟩ =
1, 0,
n = m, n ̸= m.
(2.15)
2 Therefore, {fn }∞ n=1 form an orthonormal basis of Fα .
Lemma 2.4. The multiplication operator M(z −a)n is unitarily equivalent to Mz n . Proof. Let Ba (z ) = z − a, and let CBa denote the composition operator, CBa : f (z ) −→ f (Ba ),
for any f ∈ Fα2 , a ∈ C.
The operator defined by X = Mka CBa : Fα2 → Fα2 is a unitary operator. In fact, for f , g ∈ Fα2 , by computation, we have
⟨Xf , Xg ⟩ =
f (z − a)g (z − a)eα z a¯ +α¯z a−α|a| dλα (z ) 2
C
= ⟨f , g ⟩. For f ∈ Fα2 , we obtain M(z −a)n Xf (z ) = (z − a)n Mka CBa f (z ) = (z − a)n ka (z )f (z − a),
(2.16)
XMz n f (z ) = Mka CBa Mz n f (z ) = ka (z )(z − a)n f (z − a), which implies that M(z −a)n X = XMz n . So M(z −a)n is unitarily equivalent to Mz n .
Y. Li et al. / J. Math. Anal. Appl. 409 (2014) 899–905
Theorem 2.5. The multiplication operator M(z −a)n is quasi-similar to
n
1
Mz on the Fock space.
Proof. Applying Theorem 2.2 and Lemma 2.4, we can obtain the desired result. If we set gn (z ) =
αn (z n!
+ a)n k−a (z ) and hn (z ) =
αn (a n!
903
− z )n ka (z ) (a ∈ C, n = 0, 1, . . .), then the following corollary
holds.
n
Corollary 2.6. (i) The multiplication operator M(z +a)n is quasi-similar to 1 Mz on the Fock space. n (ii) The multiplication operator M(a−z )n is quasi-similar to 1 Mz on the Fock space. Proof. The proof is similar to Lemmas 2.3, 2.4 and Theorem 2.5, and we omit it here.
3. The reducing subspaces of Mz n (n ≥ 2) From Lemma 2.1, we know that for any f ∈ Fα2 , there exists a unique orthogonal decomposition f = f0 + f1 + · · · + fn−1 , where fk ∈ Fk (0 ≤ k ≤ n − 1). In this section, we will characterize the reducing subspaces of Mz n on the Fock space. Lemma 3.1. A linear operator S on Fα2 commutes with Mz n if and only if there exist functions gk (0 ≤ k ≤ n − 1) in Fα2 such that Sf =
n−1 k=0
gk fk , where fk ∈ Fk .
2 2 2 Proof. Let Sf = k=0 gk fk , and Mgk : Fα → Fα is the multiplication operator defined by Mgk (h) = gk h for h ∈ Fα . Then for 2 any f ∈ Fα , we have
n−1
SMz n f = Sz n f = z n
n −1
gk fk = Mz n Sf .
(3.1)
k=0
That is, SMz n = Mz n S, as desired. Next, if we suppose that SMz n = Mz n S, then Mt∗n −z n S ∗ = S ∗ Mt∗n −z n , t ∈ C. From the property of the reproducing kernel, we know that ker Mt∗n −z n is generated by
2kπ Kα (z , t wk ) = eα zt wk |wk = e n i , 0 ≤ k ≤ n − 1 .
(3.2)
Note that S ∗ Kα (z , t ) ∈ ker Mt∗n −z n , and we obtain S ∗ Kα (z , t ) =
n−1
ak (t )Kα (z , t wk ),
ak (t ) ∈ C.
(3.3)
k=0
Therefore, Sf (z ) = ⟨Sf , Kα (ω, z )⟩
= ⟨f , S ∗ Kα (ω, z )⟩ =
n −1
ak (z )f (z wk )
k=0
=
n −1
ak (z )(f0 (z wk ) + f1 (z wk ) + · · · + fn−1 (z wk ))
k =0
=
n −1
ak (z )(f0 (z ) + wk f1 (z ) + · · · + wkn−1 fn−1 (z )),
(3.4)
k=0
where the last equality holds because wkn = 1 (0 ≤ k ≤ n − 1). Let gj =
n−1
j
ak (z )wk
(0 ≤ j ≤ n − 1), ak (z ) ∈ Fα2 .
(3.5)
k=0
Thus, we obtain Sf =
n−1 k=0
g k fk .
In the following, we will determine the reducing subspaces of the multiplication operator Mz n on the Fock space. 0 M1
···
0
0
·
·
0
0
· ···
··· ·
M Theorem 3.2. Suppose that B = (Bjk )n×n is a projection such that MB = BM, where M = Thenthere exist functions ϕjk ∈ Fα2 (0 ≤ j, k ≤ n − 1), such that Bjk = Mϕjk , and ϕjk =
cj , 0,
j = k, j ̸= k,
0
0
0
0 0
· Mn−1
where cj is a real number.
.
904
Y. Li et al. / J. Math. Anal. Appl. 409 (2014) 899–905 j
2 Proof. Applying Lemma 3.1, we have Bf = k=0 ak (z )wk , ϕj ∈ Fα . Set ak (z ) = j=0 ϕj fj , where ϕj = akl (z ) ∈ Fl (0 ≤ l ≤ n − 1). Then the operator B has the following matrix representation
n−1
n−1
n−1 ak0 k=0 n−1 ak1 B = k=0 · n −1
n −1
ak0 wk
k=0 n−1
k=0
ak1 wk
···
k=0
· n −1
ak,n−1
k=0
n−1
···
·
ak,n−1 wk
···
k=0
n−1 l =0
akl (z ), and
ak0 wkn−1
n −1 ak1 wk . k=0 · n −1 n−1 ak,n−1 wk n−1
(3.6)
k=0
It follows that B = (Mϕjk )n×n (0 ≤ j, k ≤ n − 1), and Mϕjk : Fk → Fj . Now we will analyze multiplication operators Mϕjk . If j = k, Mϕjj : Fj → Fj , suppose that ϕjj (z ) =
ϕjj (z ) =
∞
m=0
∞
l =0
jj
jj
al z l . Note that Fj = span{enk+j }, so al = 0 (l ̸= mn). Thus,
jj
amn z mn . Since B is a projection operator, we know that Mϕjj = Mϕ∗jj . From
⟨Mϕjj (z ) ej , emn+j ⟩ = ⟨ej , Mϕ∗jj (z ) emn+j ⟩ = ⟨ej , Mϕjj (z ) emn+j ⟩,
(3.7)
jj
jj
jj
we deduce that amn = 0 (m = 1, 2, . . .). Observe that Mϕjj is a self-adjoint operator, hence a0 is a real number. Set cj = a0 , we then have ϕjj = cj . If j ̸= k, Mϕjk : Fk → Fj , without loss of generality, assume that k < j, and set ϕjk (z ) =
∞
l =0
jk
al z l . Similar to the preceding
jk
discussion, we know that only the coefficient aj−k in the Taylor expansion of ϕjk (z ) is nonzero. Thus,
jk aj−k z j−k ek
⟨Mϕjk (z ) ek , ej ⟩ = ⟨
, ej ⟩ =
jk aj − k
j! . α j−k k!
(3.8)
On the other hand, since B is a projection operator, we have
⟨Mϕjk (z ) ek , ej ⟩ = ⟨ek , Mϕ∗jk (z ) ej ⟩ = ⟨ek , Mϕkj (z ) ej ⟩ = 0. jk
(3.9)
Hence aj−k = 0 and ϕjk = 0 (j ̸= k). The proof is complete.
Theorem 3.3. Suppose that Mz n : Fα2 → Fα2 . Then Mz n has 2n reducing subspaces. Proof. From Theorem 3.2, we know that B has the following form
a0 0
B= · 0
0 a1
· 0
··· ··· · ···
0 0
. ·
(3.10)
an − 1
B2 = B yields that 0 (0 ≤ k ≤ n − 1) or 1. Hence, ak = Mz n |Fα2 = M0 M1 · · · Mn−1 has 2n reducing subspaces a0 F0 ⊕ a1 F1 ⊕ · · · ⊕ an−1 Fn−1 ,
ak = 0 (0 ≤ k ≤ n − 1) or 1.
The minimal reducing subspaces are F0 , F1 , . . . , Fn−1 , as required.
(3.11)
Acknowledgments The authors would like to thank professor Chunlan Jiang for his helpful discussions in finishing the paper. The authors would like to express their sincere appreciation to the referee for his/her valuable suggestions and comments. The authors were supported by HBNSF (A2010000351, ZD200904, 140101). References [1] M.F. Ahmadi, K. Hedayatian, On similarity of powers of shift operators, Turkish J. Math. 35 (2011) 1–5. [2] J.A. Ball, Hardy space expectation operators and reducing subspaces, Proc. Amer. Math. Soc. 47 (1975) 351–357. [3] C.C. Cowen, The commutant of an analytic Toeplitz operator, Trans. Amer. Math. Soc. 239 (1978) 1–31.
Y. Li et al. / J. Math. Anal. Appl. 409 (2014) 899–905 [4] [5] [6] [7] [8] [9] [10] [11] [12] [13]
905
J.A. Deddens, T.K. Wong, The commutant of analytic Toeplitz operators, Trans. Amer. Math. Soc. 184 (1973) 261–273. R.G. Douglas, Banach Algebra Techniques in Operator Theory, Academic Press, New York, 1972. R.G. Douglas, Y.S. Kim, Reducing subspaces on the annulus, Integral Equations Operator Theory 70 (2011) 1–15. C.L. Jiang, Y.C. Li, The commutant and similarity invariant of analytic Toeplitz operators on Bergman space, Sci. China Ser. A 50 (5) (2007) 651–664. C.L. Jiang, D.C. Zheng, Similarity of analytic Toeplitz operators on the Bergman spaces, J. Funct. Anal. 258 (2010) 2961–2982. Y.C. Li, On similarity of multiplication operator on weighted Bergman space, Integral Equations Operator Theory 63 (1) (2009) 95–102. E. Nordgren, Reducing subspace of analytic Toeplitz operators, Duke Math. J. 34 (1967) 175–181. M. Stessin, K.H. Zhu, Generalized factorization in Hardy spaces and the commutant of Toeplitz operators, Canad. J. Math. 55 (2) (2003) 379–400. K.H. Zhu, Reducing subspaces for a class of multiplication operators, J. Lond. Math. Soc. 62 (2) (2000) 553–568. K.H. Zhu, Analysis on Fock Spaces, in: GTM, vol. 263, Springer-Verlag, New York, 2012.