$k$th-order slant Toeplitz operators on the Fock space

Adv. Oper. Theory 2 (2017), no. 3, 318–333 http://doi.org/10.22034/aot.1703-1133 ISSN: 2538-225X (electronic) http://aot-math.org kTH-ORDER SLANT TOEPLITZ OPERATORS ON THE FOCK SPACE SHIVAM KUMAR SINGH 1∗ and ANURADHA GUPTA 2 Communicated by E. Ko Abstract. The notion of slant Toeplitz operators Bφ and kth-order slant Toeplitz operators Bφk on the Fock space is introduced and some of its properties are investigated. The Berezin transform of slant Toeplitz operator Bφ is also obtained. In addition, the commutativity of kth-order slant Toeplitz operators with co-analytic and harmonic symbols is discussed. 1. Introduction O. Toeplitz [16] in 1911 introduced the notion of Toeplitz operator Tφ for bounded measurable function φ with applications in prediction theory, wavelet analysis and differential equations. Later on Toeplitz operators on Hardy spaces have been studied extensively. In particular authors like Brown, Halmos [4] and Douglas [5] have done the remarkable study of this operators. Then, in the year 1995, Ho [8, 9, 10, 11], introduced slant Toeplitz operator having the property that its matrix with respect to standard orthonormal basis could be obtained by eliminating every alternate row of the matrix of the corresponding Toeplitz operator. After the introduction of class of slant Toeplitz operators, the study has gained voluminous importance due to its multidirectional applications as these class of operators have played major roles in wavelet analysis, dynamical system and in curve and surface modelling ( [6, 7, 13, 14, 17]). Many mathematicians (for e.g. [1, 2, 3]) generalized the notion of slant Toeplitz operators to different spaces such as Hardy spaces, Bergman space and studied its properties. Motivated by Copyright 2016 by the Tusi Mathematical Research Group. Date: Received: Mar. 2, 2017; Accepted: May 19, 2017. ∗ Corresponding author. 2010 Mathematics Subject Classification. Primary 47B35; Secondary 46E20. Key words and phrases. kth-order slant Toeplitz operator, Fock space, Berezin transform. 318 kTH-ORDER SLANT TOEPLITZ OPERATORS ON THE FOCK SPACE 319 the work of these researchers, here we introduce slant Toeplitz operators on the Fock space and study some of its algebraic properties. Let the space L2 (C, dµ) denote the Hilbert space of all Lebesgue measurable square integrable functions f on C with the norm 1/2 Z 2 |f (z)| dµ(z) kf k = C −|z|2 where the measure dµ(z) = e dA(z) and dA denotes the Lebesgue area measure on complex plane. We denote the Fock space by F2 which consists of all entire functions in L2 (C, dµ) and is a closed subspace of L2 (C, dµ). The space F2 is a Hilbert space with the inner product inherited from L2 (C, dµ) as Z f (z)g(z)dµ(z) where f and g are in F2 . hf, gi = (1.1) C The set of all polynomials in the complex variable z is denoted by P[z] which is contained in the Fock space and moreover it is dense. For n ≥ 0, let en (z) = n √z , then the set {en }n≥0 forms the orthonormal basis for F2 (see [18]). Let πn! P : L2 (C, dµ) −→ F2 be the orthogonal projection. Then for f ∈ L2 (C, dµ), we have P (f (z)) = hP f (w), Kz (w)i = hf (w), P (Kz (w))i Z 1 = hf (w), Kz (w)i = f (w)ezw̄ dµ(w), π C where Kz (w) = K(z, w) = π1 ewz is the Fock kernel and the normalized reproducing kernel is given by |z|2 1 K(w, z) Kz (w) = √ ewz̄− 2 . =p kz (w) = kKz k π K(z, z) Let L∞ (C) denote the set of all essentially bounded measurable functions in the entire complex plane, then for φ ∈ L∞ (C), the multiplication operator on L2 (C, dµ) is defined by Mφ (f ) = φ · f for all f ∈ L2 (C, dµ) and the Toeplitz operator Tφ on F2 is defined by Tφ (f ) = P (φ · f ) for all f ∈ F2 . In this paper, we have introduced the notion of slant Toeplitz operators Bφ and the kth-order slant Toeplitz operators Bφk on the Fock space and have studied its properties. In particular, we have given the explicit expression for Berezin transform of slant Toeplitz operator Bφ and also obtained the conditions for boundedness, compactness of Bφ . In addition, we have shown that the necessary and sufficient conditions for the commutativity of kth-order slant Toeplitz operators on the Fock space are that their symbols functions must be linearly dependent. 2. kth-order slant Toeplitz operators on the Fock space Lemma 2.1. For non-negative integers s and t, we have ( πs! if s = t . zs, zt = 0 otherwise 320 S.K. SINGH, A. GUPTA Proof. Indeed for s = t, using the equation (1.1) and the measure dµ(z), we get s z ,z t = Z s t −|z|2 Z ∞ z z̄ e dA(z) = r=0 C Z Z ∞ 2s+1 −r 2 r e dr = π = 2π r=0 Z 2π 2 rs+t+1 ei(s−t)θ e−r dθdr θ=0 ∞ s −l l e dl = πΓ(s + 1) = πs! l=0 ( πs! if s = t Then hz s , z t i = . 0 otherwise  Now in the next proposition we give the (m, n)th entry of the matrix of Tφ with respect to orthonormal basis {en }n≥0 on the Fock space. P P∞ i j Proposition 2.2. For the harmonic symbol φ(z) = ∞ i=0 ai z + j=1 bj z̄ , the (m, n)th entry of matrix of Tφ with respect to orthonormal basis {en }n≥0 of F2 is given by r  m!   am−n for m ≥ n hTφ en , em i = r n!  n!   bn−m for n > m m! where m and n are non-negative integers. Proof. Here the (m, n)th entry of matrix of Tφ with respect to orthonormal basis {en }n≥0 of F2 is given by 1 1 √ hTφ z n , z m i πn! πm! * ! + ∞ ∞ X X 1 1 = √ √ P ai z i+n + bj z̄ j z n , z m π n! m! i=0 j=1 *∞ + *∞ +! X X 1 1 ai z i+n , z m + bj z̄ j z n , z m = √ √ π n! m! i=0 j=1 ! ∞ ∞ X X 1 1 = √ √ ai z i+n , z m + bj z n , z j+m . π n! m! i=0 j=1 hTφ en , em i = √ For m ≥ n, by Lemma 2.1 and by equation (2.1), it follows that ∞ 1 1 X √ hTφ en , em i = √ ai z i+n , z m πn! πm! i=0 r 1 1 m! √ =√ am−n πm! = am−n . n! πn! πm! (2.1) kTH-ORDER SLANT TOEPLITZ OPERATORS ON THE FOCK SPACE For n > m, by Lemma 2.1 and by equation (2.1), it follows that ∞ 1 1 X √ hTφ en , em i = √ bj z n , z j+m πn! πm! j=1 r 1 n! 1 √ =√ bn−m . bn−m π(n)! = m! πn! πm! r  m!   am−n for m ≥ n Thus hTφ en , em i = r n!  n!   bn−m for n > m m! where m and n are non-negative integers. Hence the matrix of Tφ explicitly is given by  √ √ 2b a0 b1 2 √6b3 √  a1 2b a 1 √6b2 0 √  2a √2a 3b1 Tφ = √ 2 √ 1 √a0  6a 6a2 3a1 a0 3  .. .. .. .. . . . . and the adjoint of the matrix of Tφ is given by 321   ... . . .  . . .  . . .   √ √ a¯0 a¯1 √2a¯2 √6a¯3 . . .  b¯1 2a¯1 √6a¯2 . . .  √ √a¯0   ¯ ¯ ⋆ 2 b 2 b a ¯ 3 a ¯ . . . Tφ = √ 2 √ 1 √ 0 1    6b¯ ¯ ¯ 6 b 3 b a ¯ . . . 3 2 1 0   .. .. .. .. . . . . P∞ P ¯ j which is the matrix of Tφ̂ where φ̂(z) = i=0 āi z̄ i + ∞ j=1 bj z which is nothing but φ(z). So we get that Tφ⋆ = Tφ . Assume that k is an integer and k ≥ 2 and now consider an operator Wk on F2 given by ( n z k , if n is divisible by k . Wk (z n ) = 0 otherwise  Note here we denote the operator W2 by W . Proposition 2.3. The operator Wk is a bounded linear operator on F2 with norm 1. P n Proof. Let f (z) = ∞ n=0 an z be any arbitrary polynomial, then * ∞ + ∞ ∞ X X X akm z m , akm z m = kWk f k22 = |akm |2 πm! ≤π m=0 ∞ X m=0 m=0 |am |2 m! = kf k22 m=0 322 S.K. SINGH, A. GUPTA and since the set ofall polynomials is dense in F2 , so it follows that kWk k2 ≤ 1.  1 = 1. Hence kWk k2 = 1.  Also kWk k2 ≥ Wk √ π 2 Proposition 2.4. For the operator Wk , its adjoint Wk⋆ (z n ) = n = 0, 1, 2, 3 · · · and also kWk⋆ f k2 ≤ kf k2 , for all f ∈ F2 . n! z kn (kn)! for Proof. Since of polynomials is dense in F2 , therefore for any polynomial P∞ the set m f (z) = m=0 am z ∈ F2 , we have * !+ * + ∞ ∞ X X hWk⋆ z n , f (z)i = z n , Wk am z m = zn, akm z m m=0 m=0 πn! z kn , z kn = hz n , akn z n i = akn πn! = akn π(kn)!   n! kn z , f (z) . = (kn)! This implies that Wk⋆ (z n ) = kWk⋆ f k22 = = kWk⋆ ∞ X ∞ X m=0 |am | m=0 ∞ X =π m=0 We have kWk⋆ f k2 2 n! z kn . (kn)! am z m k22 =k  2 m! (km)! 2 |am |2 Since ∞ X am m=0 z km ,z m! km 2 z k2 (km)! km = ∞ X m=0 ∞ X |am |2 ( m! 2 ) π(km)! (km)! m! ≤π |am |2 m! = kf k22 , (km)! m=0 ≤ kf k2 for all f ∈ F2 .  Now we define slant Toeplitz operator and k th -order slant Toeplitz operator on the Fock space. Definition 2.5. For φ ∈ L∞ , we define the slant Toeplitz operator on the space F2 as an operator Bφ : F2 −→ F2 given by Bφ (f ) = W Tφ (f ) ∀f ∈ F2 . Definition 2.6. A k th -order slant Toeplitz operator Bφk for k ≥ 2 induced by a function φ ∈ L∞ on the space F2 is defined as Bφk = Wk Tφ . It is clear that Bφk is bounded linear operator on F2 for essentially bounded measurable function φ over C and for k = 2, the k th -order slant Toeplitz operator is simply the slant Toeplitz operator Bφ . Following proposition follows easily from the definition of k th -order slant Toeplitz operator Bφk . Proposition 2.7. Let φ1 , φ2 ∈ L∞ (C) and λ1 , λ2 be complex numbers. Then (1) Bλk1 φ1 +λ2 φ2 = λ1 Bφk1 + λ2 Bφk2 . kTH-ORDER SLANT TOEPLITZ OPERATORS ON THE FOCK SPACE 323 (2) kBφk1 k ≤ kφ1 k∞ . P P∞ i j th Proposition 2.8. For φ(z) = ∞ entry of the i=0 ai z + j=1 bj z̄ , the (m, n) k 2 matrix of Bφ with respect to orthonormal basis {en }n≥0 of F is given by r m!    akm−n for km ≥ n hBφk en , em i = p n! (m)!(n)!    bn−km for n > km (km)! where m and n are non-negative integers. Proof. Here the (m, n)th entry of Bφk with respect to orthonormal basis {en }n≥0 of F2 is   zm zn k ,√ hBφ en , em i = Wk Tφ √ πn! πm! 1 = √ hP (φ · z n ), Wk⋆ z m i π n!m! + *∞ ∞ X X 1 m! z km = √ ai z i+n + bj z̄ j z n , (km)! π n!m! i=0 j=1   X !  ∞ ∞ X m! m! 1 bj z̄ j z n , z km + z km ai z i+n , = √ (km)! (km)! π n!m! i=0 j=1 ! r ∞ ∞ X m! X 1 . (2.2) = bj z n , z km+j ai z i+n , z km + π(km)! n! i=0 j=1 For km ≥ n, by Lemma 2.1 and by equation (2.2), it follows that r ∞ m! X 1 k hBφ en , em i = ai z i+n , z km π(km)! n! i=0 r r m! m! 1 1 akm−n π(km)! = akm−n . = π (km)! n! n! For n > km, by Lemma 2.1 and by equation (2.2), it follows that r ∞ 1 m! X k bj z n , z km+j hBφ en , em i = π(km)! n! j=1 r 1 m! 1 √ = bn−km π(n)! = m!n!bn−km . π(km)! n! (km)! Thus, r m!    akm−n for km ≥ n Bφk en , em = p n! (m)!(n)!    bn−km for n > km (km)! where m and n are non-negative integers.  324 S.K. SINGH, A. GUPTA Hence explicit form of the matrix of Bφk is given by  a0 b1 √ 2b2  1  ak √ ak−2 ak−1  2 √ √  2a2k 2a a 2k−1 √ √ √ 2k−2  6a 6a 3a3k−2 3k 3k−1  .. .. .. . . . ...   . . .   . . .  . . .  ··· For k = 2 we get the matrix of slant Toeplitz operator Bφ , which is given by  a0 b1   a2 a1  √ √  2a4 2a3 √  6a √6a 6 5  .. .. . . √ 2b2 1 √ a0 2 a √2 3a4 .. . ...   . . .   . . .  . . .  ... Following proposition follows from the matrix representation of Bφk . Proposition 2.9. Let φ ∈ L∞ be analytic function then Bφk = 0 if and only if φ = 0 almost everywhere. Remark 2.10. For an analytic function φ ∈ L∞ the correspondence φ −→ Bφk is one-one. It is evident that if Tφ is bounded on F2 then Bφ is always bounded. However reverse statement need not be true which is shown in the following example. Example 2.11. Bz and Bz̄ are bounded linear operators on F2 while Tz and Tz̄ are not bounded on F2 . For any non-negtive integer p, using the lemma 2.1 we have 2 kBz · z p k2 = kW Tz · z p k2 = kW z p+1 k ( p+1 2 kz 2 k , if p is odd = 0, otherwise  p+1  ( 2 )! p 2 kz k , if p is odd = p!  0, otherwise which shows Bz is a bounded linear operator. kTH-ORDER SLANT TOEPLITZ OPERATORS ON THE FOCK SPACE 325 Now for p ∈ N, consider kBz̄ · z p k2 = kW Tz̄ · z p k2 = kW P (z̄ · z p k2 ( p−1 2 2 p! p2 kz 2 k , if p is odd p−1 = = W z (p − 1)! 0, otherwise  p−1  ( 2 )! p 2 p kz k , if p is odd = (p − 1)!  0, otherwise which implies that Bz̄ is bounded. Now consider kTz z p k2 = z p+1 , z p+1 = π(p + 1)! = (p + 1) hz p , z p i = (p + 1)kz p k2 . √ This means kTz z p k = p + 1kz p k, thus Tz is not bounded. Similarly we have √ kTz̄ z p k = pkz p k and hence Tz̄ is also not bounded. Now we conclude this section by giving a result about the point spectra of kth-order slant Toeplitz operators. Theorem 2.12. Let φ be an invertible, analytic function in L∞ , then σp (Bφk ) = k k k σp (Bφ(z k ) ), where σp (Bφ ) denotes the point spectrum of Bφ . Proof. Suppose that λ ∈ σp (Bφk ). Then there exists a non-zero function f in F2 such that Bφk f = λf . Let M = φf , then by using the definition of Wk it follows that Bφk (z k )M = Wk Tφ(zk ) M = Wk P [φ(z k )(φf )] = Wk [φ(z k )(φf )] = φ(z)Wk (φf ) = φ(z) · Bφk (f ) = φ(z) · λf = λφf = λM . Since φ is invertible and f is non-zero, so M 6= 0 and therefore λ ∈ σp (Bφk ). k Conversely, let µ ∈ σp (Bφ(z Then there exist a non-zero function g in F2 k ) ). satisfying Bφk (z k )g = µg. Let N = φ−1 g. Then N ∈ F2 satisfies Bφk N = Wk Tφ (φ−1 g) = Wk P (φ · (φ−1 g)) = Wk g = φ−1 φWk g = φ−1 Wk (φ(z k )g) k = φ−1 Bφ(z k)g = φ−1 µg = µφ−1 g = µN Since φ is invertible and g is non-zero, so N 6= 0 and therefore, µ ∈ σp (Bφk ). k Thus σp (Bφk ) = σp (Bφ(z k ) ). 326 S.K. SINGH, A. GUPTA  Corollary 2.13. For an invertible, analytic function φ in L∞ , σp (Bφ ) = σp (Bφ(z2 ) ), where σp (Bφ ) denotes the point spectrum of Bφ . 3. Berezin Transform of Bφ Let H be any reproducing kernel Hilbert space on an open subset Ω of C. For a bounded operator S on H, the Berezin transform [15] denoted by S̃, is the complex valued function on Ω S̃(z) = hSkz , kz i for z ∈ Ω. For every bounded operator S on H, the Berezin transform S̃ is a bounded function on Ω. The normalized reproducing kernel in the Fock space is given by ! ∞ X |z|2 1 1 wz̄− |z|2 (wz̄)n 2 =√ e− 2 . kz (w) = √ e π π n=0 n! The following proposition gives the Berezin transform of the operator Wk : |z|2 P∞ Proposition 3.1. For the operator Wk , we have Wk (kz (w)) = √1π e− 2 m=0 and its Berezin transform is given by (wz̄)m (km)! ∞ 2m |z|2 X |z| 1 W̃k (z) = √ e− 2 . (km)! π m=0 Proof. By the definitions of operator Wk and normalized reproducing kernel kz , it follows that !   ∞ X 1 − |z|2 1 − |z|2 +wz̄ (wz̄)n = √ e 2 Wk Wk (kz (w)) = Wk √ e 2 n! π π n=0 and ∞ 1 − |z|2 X (wz̄)m =√ e 2 (km)! π m=0 + ∞ 1 − |z|2 X (wz̄)m , kz (w) W̃k (z) = hWk kz , kz i = √ e 2 (km)! π m=0  ∞  1 − |z|2 X (wz̄)m =√ e 2 , kz (w) (km)! π m=0 * ∞ ∞ 1 − |z|2 X |z|2m 1 − |z|2 X (z z̄)m =√ e 2 . =√ e 2 (km)! (km)! π π m=0 m=0  The explicit expression for the Berezin transform of the slant Toeplitz operator is given in the next result: kTH-ORDER SLANT TOEPLITZ OPERATORS ON THE FOCK SPACE 327 |z|2 Theorem 3.2. For an operator W its adjoint W ⋆ (kz (w)) = √1π e− 2 cosh(wz̄) and the Berezin transform of slant Toeplitz operator Bφ is Z Z 1 1 2 −|z−w|2 ˜ φ(w)e dA(w) + φ(w)e−|z−w| e−2(zw̄) dA(w). Bφ (z) = 2π C 2π C Proof. By the proposition 2.4 and the definition of normalized reproducing kernel kz , it follows that ! ∞ n 2 X |z| (wz̄) 1 W ⋆ (kz (w)) = √ e− 2 W ⋆ n! π n=0 ∞ 2n |z|2 X (wz̄) 1 = √ e− 2 (2n!) π n=0 |z|2 1 = √ e− 2 cosh(wz̄). π Now the Berezin transform of Bφ is given by B˜φ (z) = hBφ kz , kz i = hW Tφ kz , kz i = hTφ kz , W ⋆ kz i = hP (φkz ), W ⋆ kz i Z |z|2 1 2 =√ φ(w)kz (w)e− 2 cosh(wz̄)e−|w| dA(w) π C Z 1 2 2 = φ(w)ewz̄−|z| −|w| cosh(z w̄)dA(w) π C  zw̄  Z e + e−(zw̄) 1 wz̄−|z|2 −|w|2 φ(w)e dA(w) = π C 2 Z Z 1 1 2 2 wz̄+z w̄−(|z|2 +|w|2 ) = φ(w)e dA(w) + φ(w)ewz̄−zw̄−(|z| +|w| ) dA(w) 2π C 2π C Z Z 1 1 2 2 φ(w)e−|z−w| dA(w) + φ(w)e−|z−w| e−2(zw̄) dA(w). (3.1) = 2π C 2π C  R 2 In the expression (3.1), the term π1 C φ(w)e−|z−w| dA(w) is basically the Berezin transform of Tφ or simply the Berezin transform of φ. Also from (3.1) it is clear that B˜φ (z) → 0 as |z| → ∞. For φ ∈ L∞ , we observe that (1) B˜φ (z) ∈ L∞ (C, dλ). (2) kB˜φ (z)k∞ ≤ kφk∞ . (3) B̃λ1 φ1 +λ2 φ2 = λ1 B̃φ1 + λ2 B̃φ2 . Corollary 3.3. Let φ be essentially bounded measurable function on C. Then Bφ = 0 if and only if B˜φ = 0. 328 S.K. SINGH, A. GUPTA Proposition 3.4. If φ(z) is a non-negative bounded measurable function on C and if |z| → ∞ then Bφ is compact. Proof. If |z| → ∞ then B̃φ → 0 and so by the expression (3.1) it follows that R 2 φ(w)e−|z−w| dA(w) → 0 as z → ∞, that is, T̃φ (z) → 0 as z → ∞. But we C know that for a given φ, Tφ is compact on F2 if and only if T̃φ → 0 as z → ∞, ([15], Proposition 5.3). So it follows that Tφ is compact. Hence Bφ is compact on F2 .  4. Commutativity of kth-order slant Toeplitz operators In this section the commutativity of kth-order slant Toeplitz operators with coanalytic symbols and harmonic symbols has been studied. The following lemma follows from [12]. Lemma 4.1. Let f and g be analytic functions in L∞ (C), both of which are not identically zero and let q ≥ 0 be an integer. If f Wk⋆ g = gWk⋆ f , then the following are equivalent; (1) f (i) (0) = 0 for any integers i with 0 ≤ i ≤ q and f (q+1) (0) 6= 0 (2) g (i) (0) = 0 for any integers i with 0 ≤ i ≤ q and g (q+1) (0) 6= 0. Now in the next theorem we obtain the necessary and sufficient condition for the commutativity of kth-order slant Toeplitz operator with co-analytic symbols. Theorem 4.2. Let φ, ψ ∈ L∞ (C) be such that φ̄, ψ̄ are analytic functions then the following statements are equivalent: (1) Bφk and Bψk commute ; (2) there exist scalers α and β, not both zero, such that αφ + βψ = 0. Proof. Firstly suppose that (2) holds. Without loss of generality assume that α 6= 0, then φ = γψ where γ = −β/α. Then Bφk Bψk = Wk Tφ Wk Tψ = Wk Tγψ Wk Tψ = γWk Tψ Wk Tψ = Wk Tψ W Tγψ = Bψk Bφk . Conversely suppose that Bφk and Bψk commutes. Therefore we get that k⋆ k⋆ k⋆ k⋆ ⋆ ⋆ Bφ Bψ (1) = Bψ Bφ (1), or equivalently, φ̄Wk ψ̄ = ψ̄Wk φ̄. Now we have the following three cases. Case I. If φ ≡ 0 or ψ ≡ 0, then the result is obvious. φ(0) 6= 0 and ψ(0) 6= 0.PSince φ, ψ are analytic functions in L∞ (C), so Case II. IfP ∞ r s ⋆ let φ(z) = ∞ r=0 ar z and ψ(z) = s=0 bs z , then a0 6= 0, b0 6= 0 and Wk (φ(z)) = P∞ P r! kr s! ks z and Wk⋆ (ψ(z)) = ∞ z . Also r=0 ar s=0 bs (kr)! (ks)! ! ∞ ! ∞ ∞ ∞ X X X X r! r! ⋆ s kr ψWk (φ(z)) = bs z ar = z bs ar z s+kr (kr)! (kr)! s=0 r=0 s=0 r=0 and φWk⋆ (ψ(z)) = ∞ X r=0 ar z r ! ∞ X s! ks z bs (ks)! s=0 ! ∞ X ∞ X s! = ar bs z r+ks . (ks)! r=0 s=0 kTH-ORDER SLANT TOEPLITZ OPERATORS ON THE FOCK SPACE 329 Since φWk⋆ ψ = ψWk⋆ φ, therefore it follows that or equivalently, ∞ X ∞ ∞ X ∞ X X r! s! r+ks ar b s z = bs ar z s+kr (ks)! (kr)! r=0 s=0 s=0 r=0 ∞ X ∞ X ∞ ∞ X X r! s! ap−ks bs z p = ar bp−kr z p . (ks)! (kr)! s=0 p=ks r=0 p=kr (4.1) For p = 0, equation (4.1) gives that a0 b0 = b0 a0 , so b0 = (b0 /a0 )a0 , since a0 6= 0. Now take λ = b0 /a0 , so then b0 = λa0 . For 1 ≤ p ≤ k − 1, then by equation (4.1) it follows that ap b0 = bp a0 . This means bp = (b0 /a0 )ap , since a0 6= 0, that is bp = λap . 1 For k ≤ p ≤ 2k − 1, then by equation (4.1) it follows that ap b0 + (k)! ap−k b1 = 1 bp a0 + k! bp−k a1 , but bm = λam for each m such that 0 ≤ m ≤ k − 1, therefore bp = λap , since a0 6= 0. So continuing in this manner it follows that bi = λai for eachP non-negative i, where λ = b0 /a0 . Therefore in this case we get Pinteger ∞ ∞ s ψ(z) = s=0 bs z = s=0 λas z s = λφ(z). Case III. If φ and ψ are both not zero identically and φ(0) = 0 or ψ(0) = 0. (m) Without loss of generality, suppose that φ (0) = 0 for any integer m such that (m+1) (0) 6= 0, where m1 is a non-negative integer. Therefore by 0 ≤ m ≤ m1 and φ (m) Lemma 4.1, it follows that ψ (0) = 0 for any integer m such that 0 ≤ m ≤ m1 (m+1) (0) 6= 0. Now we can write φ, ψ as and ψ φ(z) = z m1 +1 φ1 (z), ψ(z) = z m1 +1 ψ1 (z) (4.2) where φ1 , ψ1 are analytic functions in L∞ (C) and φ1 (0) 6= 0, ψ1 (0) 6= 0. Therefore by equation (4.2) it follows that, ψWk⋆ (φ(z)) = z (k+1)(m1 +1) ψ1 Wk⋆ (φ1 (z)) and φWk⋆ (ψ(z)) = z (k+1)(m1 +1) φ1 Wk⋆ (ψ1 (z)). Now as ψWk⋆ (φ) = φWk⋆ (ψ), so it gives ψ1 Wk⋆ (φ1 ) = φ1 Wk⋆ (ψ1 ). Since φ1 (0) 6= 0 and ψ1 (0) 6= 0, so by the second case it follows that ψ1 = λ1 φ1 , where λ1 = ψ1 (0)/φ1 (0). Therefore we have ψ(z) = z m1 +1 ψ1 (z) = λ1 z m1 +1 φ1 (z) = λ1 φ(z). Thus the symbol functions φ and ψ are linearly dependent.  In ([12]), C. Liu and Yufeng Lu discussed the commutativity of kth-order slant Toeplitz operators on Bergman space and gave the necessary and sufficient conditions for commutativity of kth-order slant Toeplitz operators. In the next theorem, we show that the two kth-order slant Toeplitz operators having harmonic symbols commute if and only if their symbol functions are linearly dependent. 330 S.K. SINGH, A. GUPTA Pn Pn Pn i i j Theorem 4.3. Let φ(z) = i=0 ai z + i=1 a−i z̄ and ψ(z) = j=0 bj z + Pn a−n j j=1 b−j z̄ , where b−n 6= 0 such that the ratio b−n is real and n ≥ 1 be an integer, then the following statements are equivalent: (1) Bφk and Bψk commute; (2) there exist scalers α and β, not both zero, such that αφ + βψ = 0. Proof. Suppose that (2) holds then Bφk and Bψk commute. Pn Pn i Now suppose that (1) holds. Let φ1 (z) = ai z i , φ2 (z) = i=1 a−i z̄ , i=0 Pn Pn j j ψ1 (z) = j=0 bj z and ψ2 (z) = j=1 b−j z̄ . Then φ = φ1 + φ2 and ψ = ψ1 + ψ2 . Since Bφk and Bψk commute, therefore it gives Tφ Wk⋆ Tψ Wk⋆ (1) = Tψ Wk⋆ Tφ Wk⋆ (1), that is φ2 Wk⋆ ψ2 + ψ1 (0)φ2 + P (φ1 Wk⋆ ψ2 ) = ψ2 Wk⋆ φ2 + φ1 (0)ψ2 + P (ψ1 Wk⋆ φ2 ) (4.3) or, equivalently, n n n X X X j! kj i ( a−i z )( b−j z ) + b0 a−i z i + P (kj)! i=1 i=1 j=1 n n n X X X i! ki j b−j z )( a−i z ) + a0 b−j z j + P =( (ki)! j=1 i=1 j=1 ! n n X X j! kj ( ai z̄ i )( b−j z ) (kj)! i=0 j=1 ! n n X X i! ( bj z̄ j )( a−i z ki ) . (ki)! j=0 i=1 Now for any integer r such that kn + 1 ≤ r ≤ kn + n, it follows that n n n n X X j! r X X i! r a−i b−j z = b−j a−i z (kj)! (ki)! j=1 i=1 i=1 j=1 (4.4) j+ki=r i+kj=r where i and j are positive integers not greater than n. Now we apply induction to complete the proof. When r = kn + n, then by equation (4.4) it follows that a−n b−n n! n! = b−n a−n , (kn)! (kn)!
so a−n = b−n a−n /b−n , since b−n 6= 0. Let γ = a−n /b−n , so we get a−n = γb−n . When r = kn + n − 1, then by equation (4.4) it follows that a−n+1 b−n n! n! = b−n+1 a−n , (kn)! (kn)! so a−n+1 = γb−n+1 , where γ = a−n /b−n , since b−n 6= 0. Suppose that a−n+s = γb−n+s for any integer s with 0 ≤ s ≤ l < n − 1. Now we consider the connection between a−n+l+1 and b−n+l+1 . When r = kn + n − l − 1, then by equation (4.4) we get that n! (n + λ)! + · · · + a−n+l+1−kλ b−n (kn)! (k(n + λ))! n! (n + λ)! + · · · + b−n+l+1−kλ a−n+λ = b−n+l+1 a−n (kn)! (k(n + λ))! a−n+l+1 b−n kTH-ORDER SLANT TOEPLITZ OPERATORS ON THE FOCK SPACE 331 i h and [x] denotes the greatest integer function not greater than where λ = (l+1) k x. Now from the assumption it follows that a−n+l+1 b−n n! n! = b−n+l+1 a−n , (kn)! (kn)! which implies that a−n+l+1 = γb−n+l+1 , where γ = a−n /b−n , since b−n 6= 0. Thus, from the induction we obtain that for any integer s such P−n+s Pn a−n+sr = γb n that 0 ≤ s ≤ n − 1. Hence φ2 (z) = r=1 a−r z = r=1 γb−r z r = γψ2 (z). Now since φ2 (z) = γψ2 (z), so by the equation (4.3) it follows that ψ1 (0)φ2 + P (φ1 Wk⋆ ψ2 ) = φ1 (0)ψ2 + P (ψ1 Wk⋆ φ2 ). (4.5) Then ψ1 (0)φ2 + P (φ1 Wk⋆ ψ2 ), z kn = φ1 (0)ψ2 + P (ψ1 Wk⋆ φ2 ), z kn , or equivalently, φ1 Wk⋆ ψ2 , z kn = ψ1 Wk⋆ φ2 , z kn , which further implies that b−n a0 π(kn)! = a−n b0 π(kn)! and hence a0 = γb0 , where γ = a−n /b−n , since b−n 6= 0. Now as a0 = γb0 , so from (4.5) it follows that ψ1 (0)φ2 = φ1 (0)ψ2 and P (φ1 Wk⋆ ψ2 ) = P (ψ1 Wk⋆ φ2 ),   that is, P (γψ1 − φ1 ) · Wk⋆ ψ2 = 0. Hence for any integer u such that kn − n ≤ u ≤ kn − 1, we have E D  = 0, z u , P (γψ1 − φ1 ) · Wk⋆ ψ2 or equivalently, h(γψ1 − φ1 )z u , Wk⋆ ψ2 i = 0, which on putting the values of φ1 , ψ1 and ψ2 gives that * n + n X X r! (γbr − ar )z r+u , b−r z kr = 0. (kr)! r=0 r=1 Since a0 = γb0 , so we have + * n n X X r! b−r z kr = 0. (γbr − ar )z r+u , (kr)! r=1 r=1 When u = kn − 1, then equation (4.6) gives that (γb1 − a1 ) b−n π(kn)! = 0, so this gives a1 = γb1 , since b−n 6= 0. So, then it follows + * n n X X r! b−r z kr = 0. (γbr − ar )z r+u , (kr)! r=2 r=1 (4.6) 332 S.K. SINGH, A. GUPTA Now suppose that aj = γbj for any integer j such that 0 ≤ j ≤ s, where 0 ≤ s ≤ n − 1.Then by equation (4.6), it follows that * n + n X X r! (γbr − ar )z r+u , b−r z kr = 0. (4.7) (kr)! r=s+1 r=1 Now consider the connection between as+1 and bs+1 . When u = kn − s − 1, then by equation (4.7), it follows that (γbs+1 − as+1 ) b−n π(kn)! = 0, which gives as+1 = γbs+1 , since b−n 6= 0. Thus from the induction we obtain that aj = γbjPfor any integers j with 0 ≤ j ≤ n. Therefore we get φ1 (z) = P n n r r r=1 ar z = r=1 γbr z = γψ1 (z). Also from above we have φ2 = γψ2 , so it follows that φ = φ1 + φ2 = γψ1 + γψ2 = γψ. Thus the symbol functions φ and ψ are linearly dependent.  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E-mail address: [email protected]