#### Hyponormality of generalised slant weighted Toeplitz operators with polynomial symbols

Arab. J. Math.
Hyponormality of generalised slant weighted Toeplitz operators with polynomial symbols
Munmun Hazarika 0
Sougata Marik 0
0 M. Hazarika (
For a sequence of positive numbers β = {βn}n∈Z, the space L2(β) consists of all f (z) = an ∈ C for which −∞∞ |an|2βn2 < ∞. For a bounded function ϕ(z) = ∞ operator A(ϕβ) is an operator on L2(β) defined as A(ϕβ) = W Mϕ(β), where−M∞ϕ(β) is the weighted multiplication operator on L2(β) and W is an operator on L2(β) such that W z2n = zn, W z2n−1 = 0 for all n ∈ Z. In this paper we show that for a trigonometric polynomial ϕ(z) = qn=− p an zn, A(ϕβ) cannot be hyponormal unless ϕ ≡ 0. We also show that, for k ≥ 2 the kth order slant weighted Toeplitz operator Uk(,βϕ) cannot be hyponormal unless φ ≡ 0. Also the compression of Uk(,βϕ) to H 2(β), denoted by Vk(,βϕ), cannot be hyponormal unless φ ≡ 0.
1 Introduction and preliminaries
The class of Toeplitz operators was first defined by Toeplitz in 1911 [11]. Since then this class of
non-selfadjoint operators has been widely studied. By eliminating every other row of a doubly infinite Toeplitz matrix,
Ho defined slant Toeplitz operators [7]. The spectral properties of this class of operators have a connection
to the smoothness of wavelets, and as such, appear frequently in wavelet analysis. Various properties of slant
Toeplitz operators have been studied in [7,12]. Motivated by the multidirectional applications of the slant
Toeplitz operators, Arora and Kathuria [2] introduced the notion of slant weighted Toeplitz operators. The
study of this new class has also gained momentum as it is expected to be meaningful not only to specialists
in the theory of Toeplitz operators, but would also be of interest to physicists, probabilists, and computer
scientists. Properties of these classes of operators are detailed in [2,3,5,6]. To define and understand slant
weighted Toeplitz operators we need to begin with a few preliminaries.
βn βn
Let β = {βn}n∈Z be a sequence of positive numbers with β0 = 1, r ≤ βn+1 ≤ 1 for n ≥ 0, and r ≤ βn−1 ≤ 1
for n ≤ 0, for some r > 0. Let f (z) = n∞=−∞ an zn, an ∈ C be the formal Laurent series (whether or not the
series converges for any values of z). Define f 2β = n∞=−∞ |an |2βn2. We consider the weighted sequence
space
zk
(L2(β), · β ) is a Hilbert space with an orthonormal basis given by {ek (z) = βk }k∈Z and with inner product
defined by
∞
n=−∞
L2(β) =
f (z) =
an zn | an ∈ C, f 2β =
|an |2βn2 < ∞
∞
n=−∞
∞
n=−∞
∞
n=−∞
2
anb¯n βn .
∞
n=−∞
an zn ,
∞
n=−∞
bn zn
=
Let L∞(β) denote the set of formal Laurent series ϕ(z) =
(i) ϕ L2(β) ⊆ L2(β), and
(ii) there exists some c > 0 satisfying ϕ f β ≤ c f β for each f ∈ L2(β).
For ϕ ∈ L∞(β), define the norm ϕ
∞ as
ϕ ∞ = inf{c > 0 : ϕ f β ≤ c f β for each f ∈ L2(β)}.
∞
n=−∞
L∞(β) is a Banach space with respect to · ∞.
We refer to [8,10] for details of the spaces L2(β) and L∞(β).
Let ϕ ∈ L∞(β) and ϕ(z) = an zn. Mϕ(β) is the weighted multiplication operator on L2(β) defined as
an zn having the following properties:
The slant weighted Toeplitz operator A(ϕβ) on L2(β) is defined [2] as
Mϕ(β)( f ) = ϕ f.
A(ϕβ) = W Mϕ(β)
where W is the operator on L2(β) given by
βn
W e2n (z) = β2n en (z), W e2n−1(z) = 0 for n ∈ Z.
If β ≡ 1, then A(ϕβ) is the slant Toeplitz operator as defined by Ho [7], and denoted simply as Aϕ .
(
1
)
A(ϕβ)(en ) =
2 Slant weighted Toeplitz operator
We begin with a few algebraic properties of A(ϕβ) that are found in [2,3]. For ϕ(z) =
we have:
∞
n=−∞
an zn ∈ L∞(β)
∞ zk
k=−∞ ββnk a2k−n ek for each n ∈ Z. Here {ek (z) = βk }k∈Z is an orthonomal basis for L2(β).
matrix of A(ϕβ) with respect to orthonormal basis {ek }k∈Z is
⎛ . . . ... ...
.
.
.
The matrix representation of A(ϕβ) is a two way matrix. If [·] denotes the central (0, 0)th entry, then the
In this paper we consider a trigonometric polynomial ϕ(z) = qn=− p an zn and show that A(ϕβ) cannot be
hyponormal unless ϕ ≡ 0. It may be mentioned that in Theorem 5 [12] it is shown that Aϕ is hyponormal iff
ϕ ≡ 0.
Let ϕ ∈ L∞(β) and ϕ(z) = n∞=−∞ an zn. The slant weighted Toeplitz operator A(ϕβ) is hyponormal if and
only if
[ A(ϕβ)∗, A(ϕβ)] = A(ϕβ)∗ A(ϕβ) − A(ϕβ) A(ϕβ)∗ ≥ 0
L2(β). Then for i, j ∈ Z,
zk
Let [λi, j ] be the matrix representation of [ A(ϕβ)∗, A(ϕβ)] with respect to orthonormal basis {ek (z) = βk }k∈Z of
λi,i are the diagonal entries of this matrix. If we denote λi,i by di , then for each i ∈ Z,
λi, j =
∞
k=−∞
di =
β2
βi βk j a¯2k−i a2k− j −
βi β j
βk2 a2i−k a¯2 j−k
∞
k=−∞
β2 2 β2
βk2 |a2k−i | − βi2 |a2i−k |
i k
2
such that di < 0, so that A(ϕβ) cannot be hyponormal.
As di = [ A(ϕβ)∗, A(ϕβ)]ei , ei , so di < 0 would imply that A(ϕβ) is not hyponormal. Hence for A(ϕβ) to be
hyponormal it is necessary that di ≥ 0 ∀i ∈ Z.
Here we will show that for a non-zero trigonometric polynomial ϕ(z) = qn=− p an zn, there will exist i ∈ Z
3 Hyponormality of A(ϕβ)
For ϕ ∈ L∞(β) given by ϕ(z) = n∞=−∞
the sequence of diagonal entries in the matrix representation of [ A(ϕβ)∗, A(ϕβ)] with respect to orthonormal basis
zk
{ek (z) = βk }k∈Z, then for each n ∈ Z we have,
an zn, we consider the operator A(ϕβ) on L2(β). If {dn}n∈Z represent
C (pn)|a p|2,
where
dn =
Let η = {(p,n) ∈ Z × Z | p = n}. Clearly, for (p,n) ∈ Z × Z − η, we have C(pn) = 0. For (p,n) ∈ η, we
define order of (p,n), denoted as o(p,n), and a (p,n)-induced set denoted by [p : n] as follows:
Definition 3.1 For (p,n) ∈ η, let u0 := n and um := p+u2m−1 ∀ m ∈ N. Let r be the smallest non-negative
integer such that p + ur is odd. Then o(p,n) := r and [p : n] := {uj : 0 ≤ j ≤ o(p,n)}.
Remark 3.2 Let (p,n) ∈ η, with u0 = n and um = p+u2m−1 for 0 < m ≤ o(p,n). Then from Definition 3.1
we have the following:
(i) If p < n then p < ui+1 < ui ≤ n ∀ 0 ≤ i < o(p,n),
(ii) If n < p then n ≤ ui < ui+1 < p ∀ 0 ≤ i < o(p,n).
βn2 < 0.
Theorem 3.3 If (p,n) ∈ η, then i∈[p:n] C(pi) = −β22n−p
If r > 0, then
Proof Let r = o(p,n) and [p : n] = {uj : 0 ≤ j ≤ o(p,n)} where u0 = n and uj = p+u2j−1 for 0 < j ≤ r.
βn2 < 0.
If r = 0, then C(pn) = −β22n−p
βu21 βn2 ,
C(pu0) = βu20 − β2n−p
2
2 2
C(puj) = βu2j βuj for 0 < j < r,
βuj+1 − βu2j−1
2
and C(pur) = −ββu2ur−r1 .
∴
i∈[p:n]
C(pi) = r C(puj) = − 2βn2 < 0.
j=0 β2n−p
Theorem 3.4 Let (p,q) ∈ η. Then np=q C(pn) < 0 if q < p, and qn=p C(pn) < 0 if p < q.
Proof Without loss of generality, we assume that p < q.
Letq0 = q. Fori > 0 letqi be thegreatest integer such that p < qi < qi−1 andqi ∈/ [p : q0]∪···∪[p : qi−1].
Thus there exist some finite q0 > q1 > ··· > qt such that [p : qi] ∩ [p : qj] = ∅ for i = j, and
t
∪i=0[p : qi] = {p + 1,...,q}.
By Theorem 3.3, n∈[p:qi] C(pn) < 0 ∀ 0 ≤ i ≤ t.
Therefore qn=p C(pn) = C(pp) + it=0( n∈[p:qi] C(pn)) < 0.
Theorem 3.5 Let ϕ(z) = n=m anzn where m,q ∈ Z and m ≤ q. If ϕ ≡ 0, then qn=m dn < 0.
q
Proof For n ∈ Z, dn = p=m C(pn)|ap|2. Therefore,
q
q
n=m
dn =
q q
p=m n=m
q p
p=m n=m
q
n=p
C(pn) |ap|2 =
C(pn) +
C(pn) |ap|2 < 0, by Theorem 3.4.
Theorem 3.6 Let ϕ(z) = qn=m anzn where m,q ∈ Z and m ≤ q. For ϕ ≡ 0, A(ϕβ) cannot be hyponormal.
Proof By Theorem 3.5, n=m dn < 0. Thus, there exist dn, m ≤ n ≤ q such that dn < 0. Hence A(ϕβ) can
q
not be hyponormal.
4 Generalised slant weighted Toeplitz operator
(5.1)
We now discuss the hyponormality of the generalised slant weighted Toeplitz operator which was first defined
in [4]. We consider the space L2(β) as introduced in Sect. 1 and for an integer k ≥ 2, let Wk : L2(β) → L2(β)
be defined as
Wk en (z) =
⎩⎪
0
⎧ β n
⎪⎨ βnk e nk (z) if n is divisible by k,
otherwise.
operator on L2(β), already mentioned in Sect. 1.
zi
The effect of Uk(,βϕ) on the orthonormal basis {ei (z) = βi }i∈Z can be given by
Uk(,βϕ)ei (z) = βi
1
n∞=−∞ ank−i βnen (z), where ϕ(z) = n∞=−∞
β j .
The adjoint of Uk(,βϕ), denoted by Uk(,βϕ)∗ is given by Uk(,βϕ)∗e j , ei = a¯k j−i βi
We assume that { ββknn }n∈Z is bounded so that Wk is bounded. For ϕ ∈ L∞(β) the kth order slant weighted Toeplitz
operator Uk(,βϕ) : L2(β) → L2(β) is defined as Uk(,βϕ) = Wk Mϕ(β), where Mϕ(β) is the weighted multiplication
an zn ∈ L∞(β).
For k = 2, Uk(,βϕ) is the slant weighted Toeplitz operator A(ϕβ) discussed earlier in Sects. 2 and 3. Properties of
the kth order slant weighted Toeplitz operator can be found in [4,6]. If βn = 1 ∀ n, then Uk(,βϕ) is the kth order
slant Toeplitz operator denoted by Uk,ϕ and discussed in [1,9].
Here we consider a trigonometric polynomial ϕ(z) = qn=− p an zn and show that Uk(,βϕ) cannot be hyponormal
unless ϕ ≡ 0. It may be mentioned that if βn = 1 ∀ n then the kth order slant Toeplitz operator Uk,ϕ is
hyponormal iff ϕ ≡ 0, as shown in Theorem 5 [1].
5 Hyponormality of Uk(β,ϕ)
Let k ≥ 2 and ϕ(z) = ∞ an zn be in L∞(β). If {dn}n∈Z represents the sequence of diagonal entries in
the matrix representation no=f −[U∞k(,βϕ)∗, Uk(,βϕ)] with respect to orthonormal basis {ei (z) = βi }i∈Z, then for each
zi
n ∈ Z we have,
dn =
(
1
) If p, n ∈ Z are such that p = (k − 1)n, then C (pn) = 0.
(
2
) If ( p + n) is divisible by k and β p+n βkn− p = βn2, then C (pn) = 0.
k
Let ηk := {( p, n) ∈ Z × Z | p = (k − 1)n}. Then for ( p, n) ∈ Z × Z − ηk , we have C (pn) = 0.
Definition 5.1 For ( p, n) ∈ ηk , let u0 := n and um := p+ukm−1 ∀ m ∈ N. Let r be the smallest non-negative
integer such that p + ur is not divisible by k. We define order of ( p, n) as r , and denote it by o( p, n). Also we
define [ p : n] to be the set {u j : 0 ≤ j ≤ o( p, n)}.
For k = 3 we look at the following examples:
In view of the examples above, we first try to determine values a < b such that the following conditions will
hold:
(i) [p : n] = {uj : 0 ≤ j ≤ o(p,n)} is contained in the interval [a,b], and
(ii) for [p : n] ⊆ [a,b] we have either uj < uj+1 ∀ j, or uj+1 < uj ∀ j.
For this, we first define the functions ψg and ψs on R as follows:
For x ∈ R, let ψs(x) be the smallest integer greater than or equal to x. Thus, ψs(x) = m + 1 if m < x ≤
m + 1, m ∈ Z.
Again let ψg(x) be the greatest integer less than or equal to x. So, ψg(x) = m if m ≤ x < m + 1, m ∈ Z.
Hence for x ∈ R, ψg(x) < x < ψs(x) and ψs(x) − ψg(x) = 1 if x ∈/ Z, and ψg(x) = x = ψs(x) if x ∈ Z.
Thus for p ∈ Z and k ≥ 2, we have
p
p = (k − 1)ψs k − 1 − δ, 0 ≤ δ < k − 1.
p
p = (k − 1)ψg k − 1 + μ, 0 ≤ μ < k − 1.
Following these notations, we can now state the following results:
Theorem 5.2 For k ≥ 2, let (p,n) ∈ ηk, and p < n(k − 1). Also let [p : n] = {ui : 0 ≤ i ≤ o(p,n)} where
u0 = n and ui+1 = p+kui ∀ 0 ≤ i < o(p,n). Then
((
12
)) [upi+:1n<]⊆ui(∀k−p01,≤ni] ∩<Zo.(p,n).
Proof (
1
) As k−p1 < n, so ψs(k−p1) ≤ n.
Since ψs(k−p1) = kp−+1δ for 0 ≤ δ < k − 1, so
ψs(k−p1) ≤ n ⇒ kp−+1δ ≤ n ⇒ p+kn ≤ n − kδ ≤ n ⇒ u1 ≤ u0.
However, u1 = u0 ⇒ p+kn = n ⇒ p = (k − 1)n, which is not possible as (p,n) ∈ ηk. Thus we must
have u1 < u0.
Therefore, by induction, ui+1 < ui ∀ 0 ≤ i < o(p,n).
(
2
) If o(p,n) = 0 then [p : n] = {n} ⊂ (k−p1,n] ∩ Z.
If o(p,n) > 0 then for 0 ≤ i < o(p,n),
p
ui+1 < ui ⇒ kui+1 < kui ⇒ p + ui < kui ⇒ k − 1 < ui
In particular, if r = o(p,n), then k−p1 < ur−1 ⇒ ur = p+ukr−1 > k−p1.
Therefore [p : n] = {ui : 0 ≤ i ≤ o(p,n)} ⊆ (k−p1,n] ∩ Z.
Theorem 5.3 For k ≥ 2, let (p,n) ∈ ηk and p > n(k − 1). Also let [p : n] = {ui : 0 ≤ i ≤ o(p,n)} where
u0 = n and ui+1 = p+kui ∀ 0 ≤ i < o(p,n). Then
(
1
) ui < ui+1 ∀ 0 ≤ i < o(p,n).
(
2
) [p : n] ⊆ [n, k−p1) ∩ Z.
Proof (
1
) As n < k−p1 so n ≤ ψg(k−p1). Also ψg(k−p1) = pk−−μ1 for 0 ≤ μ < k − 1.
TBhuetrue1fo=reupk0−−μ1 ≥ n ⇒ p+kn ≥ n + μk ≥ n ⇒ u1 ≥ u0.
⇒ p = (k − 1)n, contradicting the fact that (p,n) ∈ ηk.
Thus u0 < u1, and so by induction we get ui < ui+1 ∀ 0 ≤ i < o(p,n).
(
2
) If o(p,n) = 0 then [p : n] = {n} ⊂ [n, k−p1) ∩ Z.
If o(p,n) > 0 then for 0 ≤ i < o(p,n),
Theorem 5.4 For k ≥ 2, let (p,m),(p,n) ∈ ηk. If m ∈/ [p : n] and n ∈/ [p : m], then [p : n] ∩ [p : m] = ∅.
Proof Let o(p,m) = r and [p : m] = {u0,u1,··· ,ur} where u0 = m and uj = p+uj−1 for 1 ≤ j ≤ r. As
k
m ∈/ [p : n], so u0 ∈/ [p : n].
Claim: For 0 ≤ j < r, uj ∈/ [p : n] ⇒ uj+1 ∈/ [p : n].
Let, if possible, the claim does not hold. Then there exists some 0 ≤ j < r such that uj+1 ∈ [p : n] but
uj ∈/ [p : n].
∴ either uj+1 = n, or uj+1 = p+ky for y ∈ [p : n].
Now uj+1 = n p : m], a contradiction.
Hence uj+1 = p+k⇒yfonr s∈om[e y ∈ [p : n].
But p+ky = uj+1 = p+kuj ⇒ uj = y ∈ [p : n], which is also a contradiction. Thus the claim is established.
Therefore,
u0 ∈/ [p : n] ⇒ uj ∈/ [p : n] ∀ 0 ≤ j ≤ r
⇒ [p : n] ∩ [p : m] = ∅
βn2 < 0.
Theorem 5.5 For k ≥ 2 if (p,n) ∈ ηk, then i∈[p:n] C(pi) = −βk2n−p
Proof being identical to that of Theorem 3.3, is omitted.
Theorem 5.6 For k ≥ 2, let (p,m) ∈ ηk and t = ψg(k−p1).
(
2
) If t ≥ m, then tnn==mt+C1C(pn()pn<)<0.0.
(
1
) If t < m, then m
Proof (
1
) Let t < m.
ICflak−ipm1 :∈pZ<th(ekn−t =1)mk−.p1, and so t < m ⇒ p < (k − 1)m.
p p
AAglsaoint i<fkm−1⇒∈/ tZ+th1en≤tm<. Tk−h1us<pt<+m1(.k − 1), and our claim is established.
Let q0 = m. By Theorem 5.2 we have [p : q0] ⊆ (k−p1,q0] ∩ Z = [t + 1,q0] ∩ Z. For i > 0 let qi be the
greatest integer such that t < qi < qi−1 and qi ∈/ ∪ij−=10[p : qj]. Clearly (p,qi) ∈ ηk, [p : qi] ⊆ [t + 1,qi],
and [p : qi] ∩ [p : qj] = ∅ ∀ 0 ≤ j < i.
Since there exist only finite number of integers in the interval [t + 1,q0], so there exist distinct integers
ξ
q0 > q1 > ··· > qξ such that ∪i=0[p : qi] = [t + 1,m] ∩ Z.
(T2h)eLreeftotre≥ mnm.=t+1 C(pn) = iξ=0( n∈[p:qi] C(pn)) < 0, by Theorem 5.5.
Claim: p > (k − 1)m.p
IBfukt−p(1p,∈mZ) t∈heηnkt = k−1 p ⇒ p ≥ m(k − 1).
⇒ p =⇒m(kk−−11≥). Smo p > m(k − 1).
Theorem 5.9 Let ϕ(z) =
then nw=l dn < 0.
ITfhku−ps1ou∈/r Zclathi menits e<stakb−l1is<hedt.+ 1
p
p
⇒ m < k−1
⇒ m(k − 1) < p.
Lemma 5.8 If p, s ∈ Z and ( p, s) ∈/ ηk , then s = ψg( k−p1 ).
Let q0 = m. By Theorem 5.3 we have [ p : q0] ⊆ [q0, k−p1 ) ∩ Z = [q0, t] ∩ Z. For i > 0 let qi be the smallest
integer such that qi−1 < qi ≤ t and qi ∈/ ∪ij−=10[ p : q j ]. Then there exist distinct integers q0 < q1 < · · · < qτ
such that [ p : qi ] ∩ [ p : q j ] = ∅ for i = j , [ p : qi ] ⊆ [qi , t] ∀ i , and ∪iτ=0[ p : qi ] = [m, t] ∩ Z.
Therefore tn=m C(pn) = iτ=0( n∈[p:qi ] C(pn)) < 0, by Theorem 5.5.
Theorem 5.7 If ϕ(z) = am zm then dn < 0 for each n ∈ Z such that m + n is not divisible by k.
Proof If ϕ(z) = am zm then for each n ∈ Z, dn = Cm(n)|am|2. If n ∈ Z such that m + n is not divisible by k,
then Cm(n) < 0 ⇒ dn < 0.
p
⇒ k−1 = s ∈ Z.
n=m an zn where m, q ∈ Z and m < q. If l := ψg( km−1 )−1 and w := ψg( k −q1 )+1,
q
Proof For p ∈ Z ∩ [m, q], let tp := ψg( k−p1 ), l p := tp − 1 and wp := tp + 1. Then l ≤ l p < tp < wp ≤ w.
Also by Lemma 5.8 we must have ( p, l) ∈ ηk and ( p, w) ∈ ηk .
Thus nw=l C(pn) = tnp=l C(pn) + nw=tp+1 C(pn) < 0 by Theorem 5.6. Therefore,
w
n=l
dn =
w
q
n=l p=m
q
w
p=m n=l
C(pn)|ap|2
=
C(pn) |ap|2 < 0.
Theorem 5.10 Let ϕ(z) =
n=m an zn where m, q ∈ Z and m ≤ q. For ϕ ≡ 0, Uk(,βϕ) cannot be hyponormal.
q
Proof By Theorems 5.7 and 5.9, there exist integers l ≤ w such that for some l ≤ n ≤ w.
Hence Uk(,βϕ) can not be hyponormal.
6 Hyponormality of Vk(,βϕ)
nw=l dn < 0. This implies that dn < 0
Let Vk(,βϕ) be the compression of the kth order generalised slant weighted Toeplitz operator Uk(,βϕ) to H 2(β). Here
H 2(β) = { f (z) = n∞=0 an zn| an ∈ C, f 2β = n∞=0 |an|2βn2 < ∞}. It is a Hilbert subspace of L2(β) with
zn
an orthonormal basis given by {en(z) = βn }n∈Z0 and with inner product defined by
∞
n=0
an zn,
∞
n=0
bn zn =
anb¯nβn2.
∞
n=0
Note that we use the notation N0 to represent the set {0, 1, 2, . . .}. For ϕ ∈ L∞(β) given by ϕ(z) =
n=−∞ an zn and k ≥ 2,
∞
∞ βi
i=0 βn
Vk(,βϕ)(en) =
aki−nei for each n ∈ N0.
If {dn}n∈N0 represent the sequence of diagonal entries in the matrix representation of [Vk(,βϕ)∗, Vk(,βϕ)] with respect
i
to orthonormal basis { βzi }i∈N0 , then for each n ∈ N0 we have,
dn =
∞
l=0
β2 2 β2 2
βl2 |akl−n| − βn2 |akn−l |
n l
=
∞
p=−∞
C(pn)|ap|2, where
⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪ β2βn+nk2βpn2 − βk2βnn2−p
⎪⎪⎪⎪⎩ − βk2n−p
if p > kn and (n + p) is not divisible by k,
if p > kn and (n + p) is divisible by k,
if − n ≤ p ≤ kn and (n + p) is not divisible by k,
if − n ≤ p ≤ kn and (n + p) is divisible by k,
if p < −n.
Let ζk = {( p, n) ∈ Z × N0 | p = (k − 1)n, and if p > kn then ( p + n) must be divisible by k}. Clearly, for
( p, n) ∈ Z × N0 − ζk , we have C (pn) = 0. For ( p, n) ∈ ζk , we define order of ( p, n), denoted as o( p, n), and
a ( p, n)-induced set denoted by [ p : n].
Definition 6.1 For ( p, n) ∈ ζk let u0 := n and um := p+ukm−1 ∀ m ∈ N. Then
(
1
) For p ≥ 0 we define o( p, n) to be the smallest non-negative integer r such that p + ur is not divisible by
k.
(
2
) For p < −n we define o( p, n) to be zero.
(
3
) For −n ≤ p < 0 let o( p, n) be the smallest non-negative integer for which either p < −ur or p + ur
is not divisible by k. Thus for each integer j with 0 ≤ j < r , we must have −u j ≤ p and p + u j is
divisible by k.
In all the above cases, [ p : n] := {u j : 0 ≤ j ≤ o( p, n)}.
Theorem 6.2 Let ( p, n) ∈ ζk . Then the following must hold:
(
1
) If kn < p, then
(
2
) If p ≤ kn, then
i∈[ p:n] C (pi) = 0.
i∈[ p:n] C (pi) = − βk2βnn2−p < 0.
Proof Let r = o( p, n) and [ p : n] = {u j : 0 ≤ j ≤ r } where u0 = n and u j = p+uk j−1 for 0 < j ≤ r .
(
1
) Let p > kn. Then by definition of ζk , we must have r > 0. Again,
C (pu0) = C (pn) =
(u j )
C p
=
2
βu j+1
βu2j
2
βu j
− βu2j−1
2
β n+p
k
β2
n
2
= ββuu201 ,
and C (pur )
2
βur .
= − βu2r−1
for 0 < j < r,
Therefore = rj=0 C (pu j ) = 0.
(
2
) This part io∈f[ tph:ne] pCr(poio)f is exactly similar to Theorem 3.3, and is therefore omitted.
Remark 6.3 For k ≥ 2, since ζk ⊆ ηk so Theorems 5.2–5.4 also hold for ( p, n) ∈ ζk .
Theorem 6.4 If ( p, n) ∈ ζk such that −n < p < 0, then [ p : n] ⊆ [0, n] ∩ Z.
f[Poprro:o1nf ≤]B⊆yj [T≤0h,enro].r∩TemhZe.n5.b2y(2D) ewfienihtiaovne 6[.p1(:3n),] −⊆u(jk−≤p1 ,pn∀]∩0 Z≤. Njo<w lre.t Arl=so,o−(pu,rn−)1, ≤u0 p= n ⇒anduurj ≥= 0p.+Tukhj−u1s
Theorem 6.5 If p ∈ Z and 0 ≤ p < q, then
PCrleoaorflyLaest tp =< ψ(kg−(k −1p)1q)0a,nsdo (qp0,=q0q).∈ ζk and by Theorem 5.2 [ p : q0] ⊆ [t + 1, q0]. Also by Theorem 6.2(
2
)
we have n∈[ p:q0] C (pn) < 0. If [t +1, q]∩Z = [ p : q0], then let q1 be the greatest integer such that t < q1 < q0
and q1 ∈/ [ p : q0]. As q1 ≥ t + 1 > k−p1 , so p < (k − 1)q1 and hence ( p, q1) ∈ ζk , [ p : q1] ⊆ [t + 1, q1]
and
n∈[ p:q1] C (pn) < 0. Moreover, by Theorem 5.4 [ p : q0] ∩ [ p : q1] = ∅. Continuing this process we
get finite distinct integers q0 > q1 > · · · > qξ in [t + 1, q0] such that [ p : qi ] ∩ [ p : q j ] = ∅ for i = j ,
n∈[ p:q j ] C (pn) < 0 ∀ j and ∪iξ=0[ p : qi ] = [t + 1, q] ∩ Z.
(6.1)
q
n=0 C (pn) =
∴
q
n=t+1
Caqse I: If p = 0 then t = 0. Also C0(0) = 0, since C (pn) = 0 if p = (k − 1)n. Therefore,
n=t+1 C (pn) < 0, by 6.1.
Case II: If p > 0, then assume s0 = 0. By Theorem 5.3(
2
), [ p : s0] ⊆ [s0, t ] ∩ Z. For i > 0 let si
be the smallest integer such that si−1 < si ≤ t and si ∈/ ∪ij−=10[ p : s j ]. So, there exist distinct integers
s0 < s1 < · · · < sτ such that [ p : si ] ∩ [ p : s j ] = ∅ for i = j , and ∪iτ=0[ p : si ] = [0, t ] ∩ Z. Also by Theorem
6.2 n∈[ p:si ] C (pn) ≤ 0 ∀ i .
Therefore tn=0 C (pn) = iτ=0( n∈[ p:si ] C (pn)) ≤ 0.
So, qn=0 C (pn) = tn=0 C (pn) + qn=t+1 C (pn) < 0.
Theorem 6.6 If p, q ∈ Z and p < 0 ≤ q, then
q
n=0 C (pn) < 0.
Hence,
n∈[ p:q j ] C (pn) < 0 ∀ j ∈ {0, 1, . . . , ξ }.
q
n=0 C (pn) < 0.
TPrhoeoreffo(1re) Leqtn=q0 <C (p−n) p<a0n.d 0 ≤ n ≤ q. Then p < −q ⇒ p < −n
(
2
) Suppose − p ≤ q.
Let q0 = q. As p < 0 < q, so ( p, q0) ∈ ζk and by Theorem 6.4 [ p : q0] ⊆ [0, q0] ∩ Z. Moreover, by
Theorem 6.2(
2
) n∈[ p:q0] C (pn) < 0. If [ p : q0] = [0, q0] ∩ Z, then qn=0 C (pn) = i∈[ p:q0] C (pi) < 0. If
[ p : q0] [0, q0]∩Z, then let q1 be the largest integer such that 0 ≤ q1 < q0 and q1 ∈/ [ p : q0]. Continuing this
process we get q0 > q1 > · · · > qξ ≥ 0 such that [ p : qi ] ∩ [ p : q j ] = ∅ for i = j, [0, q0] ∩ Z = ∪ξj=0[ p : q j ]
and
⇒ [ p : n] = {n} and C (pn) < 0.
q
n=m an zn where m, q ∈ Z and m ≤ q. If ϕ ≡ 0, then either
qn=+01 dn < 0, or
Theorem 6.7 Let ϕ(z) =
d0 < 0.
Proof (
1
) Let q ≥ 0
For n ≥ 0, dn = qp=m C (pn)|a p|2, and so
Again, for m ≤ p ≤ q,
(i) if p ≥ 0 then
(ii) if p < 0 then
qn=+01 C (pn) < 0 by Theorem 6.5,
q+1
n=0 C (pn) < 0 by Theorem 6.6.
qn=+01 dn =
qp=m (
q+1
n=0 C (pn))|a p|2.
2
= −β0
∞
l=0
(β)
Hence Vk,ϕ cannot be hyponormal.
qn=m an zn where m, q ∈ Z and m ≤ q . For ϕ ≡ 0, Vk(,βϕ) cannot be hyponormal.
qn=+01 dn < 0. Thus, there exists dn , 0 ≤ n ≤ q + 1 such that
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