New York Journal of Mathematics
New York J. Math. 20 (2014) 831–843.
Discrete Duhamel product, restriction of
weighted shift operators and related
problems
Mubariz T. Karaev, Suna Saltan and Tevfik Kunt
Abstract. By applying the discrete Duhamel product method we calculate the spectral multiplicity of the direct sum of some operators. In
particular, we prove that µ(T |Xi ⊕ A) = 1 + µ(A) and µ(S ⊕ A) = 2 for
the restriction of the weighted shift operator T |Xi , shift operator S and
some appropriate operators A on the Banach spaces.
Contents
1. Introduction and preliminaries
2. The results
References
831
835
843
1. Introduction and preliminaries
Recall that the classical Duhamel product of two analytic functions f (z) =
P b
P
gb (n) z n in Hol(D) is defined by
f (n)z n and g (z) =
n≥0
(1)
n≥0
d
(f ~ g)(z) =
dz
=
Zz
Zz
f (z − t)g(t)dt =
0
∞
∞
XX
n=0 m=0
f 0 (z − t) g(t)dt + f (0)g(z)
0
n!m! b
f (n) gb (m) z n+m ,
(n + m)!
where D := {z ∈ C : |z| < 1} is the unit disc of the complex plane C (see for
instance, Wigley [11, 12]). Beginning from these pioneering works of Wigley,
subsequently many interesting and important problems of analysis, operator
Received March 6, 2014.
2010 Mathematics Subject Classification. 47B38, 46B15.
Key words and phrases. Discrete Duhamel product, Weighted shift operator, Borel
transform, Banach algebra, Spectral multiplicity.
This work is supported by Suleyman Demirel University with Project 3465-YL1-13.
This paper was supported by King Saud University, Deanship of Scientific Research, College of Science Research Center.
ISSN 1076-9803/2014
831
832
MUBARIZ T. KARAEV, SUNA SALTAN AND TEVFIK KUNT
theory and Banach algebras theory were investigated namely by applying
the method of Duhamel products, see for example, [3, 6, 7, 10, 11, 12].
Here we will consider a generalization of the usual Duhamel product (1),
named as discrete Duhamel product, and use it to the study of spectral
multiplicity of direct sums of some operators.
Recall that a subspace E ⊂ X is called a cyclic subspace of an operator
A ∈ L (X) (Banach algebra of all bounded linear operators on X) if
span {An E : n = 0, 1, 2, . . . } = X,
where span stands for the closed linear hull. A vector x ∈ X is called cyclic
(x ∈ Cyc(A)) if
span {An x : n = 0, 1, 2, . . . } = X.
The spectral multiplicity µ(A) of the operator A is
µ(A) := inf {dim E : span {An E : n ≥ 0} = X} ,
a nonnegative integer or ∞. A is a cyclic operator (i.e., there exists a vector
x ∈ X such that x ∈ Cyc(A)) if and only if µ(A) = 1. For example, it follows
from the Weierstrass approximation theorem that µ(V ) = µ(Mx ) = 1 for
the classical Volterra integration operator V and multiplication operator Mx
defined in the space C [0, 1] by
Zx
V f (x) = f (t)dt,
0
Mx f (x) = xf (x),
respectively. Also, by the classical Beurling theorem [1] in the Hardy space
H 2 = H 2 (D) over the unit disc D, µ(S) = µ(S ∗ ) = 1, where S : H 2 → H 2 is
the classical unilateral shift operator in H 2 defined by Sf (z) = zf (z), and
S ∗ is the backward shift operator on H 2 defined by
f (z) − f (0)
.
z
(S ∗ is the simple co-analytic Toeplitz operator Tz̄ defined by
S ∗ f (z) :=
Tz̄ f (z) = P+ z̄f (z),
where P+ : L2 (∂D) → H 2 is the classical Riesz orthogonal projector.) However, it is well-known that µ(S n ) = n for any finite integer n, and µ(SE ) = ∞
for the shift operator SE acting in the vector valued Hardy space H 2 (E)
with infinity dimensional Hilbert space E.
Note that as the norm, spectral radius, numerical radius, spectrum and
numerical range of operator, the spectral multiplicity is also an important
invariant of an operator. For this it is sufficient to remember, for example,
the spectral theorem for normal operators on the Hilbert space (see, for
example, Rudin [8]). Of course, the concept of cyclic subspace is very important in relation with the outstanding problem of existence of a nontrivial
DISCRETE DUHAMEL PRODUCT
833
invariant subspace. Namely, it is easy to see that an operator A : X → X has
no nontrivial invariant subspace if and only if x ∈ Cyc(A) for any nonzero
vector x ∈ X. Note that a subspace E ⊂ X is an invariant subspace for an
operator A, if AE ⊂ E, that is, Ax ∈ E for every x ∈ E.
Let A ⊕ B denote the direct sum of operators A ∈ L (X) and B ∈ L (Y ),
(A ⊕ B) (x ⊕ y) = Ax ⊕ By, x ⊕ y ∈ X ⊕ Y . It is known that (see, for
example, Halmos [2])
µ(A ⊕ B) ≤ µ(A) + µ(B)
for any operator A ⊕ B ∈ L (X ⊕ Y ). Here we will investigate the equality
µ(A ⊕ B) = µ(A) + µ(B) for some operators A and B (For basic facts on the
spectral multiplicity of direct sums of operators we recommend the papers
[4], [5] and references therein).
Let X be a Banach space with Schauder basis (en )n≥0 . Let (λn )n≥0 ⊂ C
be a bounded sequence of nonzero numbers λn . We set wn := λ0 λ1 . . . λn−1 ,
w0 := 1,
Xi := span (ek : k = i, i + 1, . . . ) , i = 0, 1, 2, . . . ,
for any two vectors x =
∞
P
xn en and y =
n=i
∞
P
yn en in Xi (i = 0, 1, 2, . . . ).
n=i
Then discrete Duhamel product (sometimes it is also called generalized
Duhamel product, see for instance [3] and references therein, and also Karaev
and Gürdal [6]) is defined by,
˜ y :=
x~
(2)
i
∞ X
∞
X
wn+m−i
xn yn en+m−i .
wn wm
n=i m=i
It is easy to see that the classical Duhamel product ~ (see formula (1))
1
corresponds to i = 0 and λn := n+1
, n ≥ 0, in (2). It is also easy to verify
˜
that the product ~ is commutative and associative.
i
Let T be the weighted shift operator acting in X by the formula
T en = λn en+1 , n = 0, 1, 2, . . . .
It can be easily shown that all subspaces Xi (i ≥ 0) are closed T -invariant
subspace (i.e., T Xi ⊂ Xi , i ≥ 0). Therefore the restricted weighted shift
operators T |Xi , i ≥ 0, are well-defined operators on the subspace Xi , i ≥ 0.
In this paper, we will develop a method of the paper [9] (see the proof of
Theorem 1 there) and investigate the spectral multiplicity of the operators
T |Xi ⊕ A, i ≥ 0; here ⊕ stands for the direct sum of operators on the direct
sum of Banach spaces. Our results also improve some results in [9] and [3].
Before giving the results of the paper, let us give some necessary definitions,
notations and preliminaries.
834
MUBARIZ T. KARAEV, SUNA SALTAN AND TEVFIK KUNT
The generalized Borel transform Bw from X onto the space of formal
series over the field of complex numbers C is defined as follows
!
∞
∞
X
X
1
xn en :=
xn e n ,
Bw
wn
n=0
n=0
where wn = λ0 λ1 . . . λn−1 , n ≥ 0. The inverse generalized Borel transform
is defined by
!
∞
∞
X
X
−1
Bw
xn en :=
wn x n e n .
n=0
n=0
Clearly, the classical Borel transform from Hol (D) (the space of all analytic
functions on the unit disc D) onto the space of formal power series C [[Z]]
1
over the field of complex numbers C corresponds to the case λn = n+1
,
n ≥ 0.
Recall that the class `pA (wn ) , wn ≥ 0, n ≤ 0, p ≥ 1 is defined by


!1/p
∞ 
p

X
b p
`pA (wn ) := f ∈ Hol (D) : kf k`p (wn ) :=
f (n) wn
< +∞ ,
A


n=0
(n)
where fb(n) := f n!(0) is the nth Taylor coefficient of the analytic function
∞
P
f (z) =
fb(n) z n on D. Note that:
n≥0
(a) Every bounded linear operator C on a Banach space X admits the
functional calculus from the class `1A (kC n k). Indeed, we can put
∞
def P b
f (C) =
f (n) C n for every function f ∈ `1 (kC n k), because in
A
n≥0
this case
∞
∞ X
X
b n
n
kf (C)k = fb(n) C ≤
f (n) kC k = kf k`1A (kC n k) ,
n=0
n=0
`1A (kC n k).
for every f ∈
(b) Every operator C ∈ L (X) (the Banach algebra of all bounded linear
∞
P
operators on X), satisfying the condition
kC n kq < +∞, admits
n=0
the functional calculus from the class
(
`pA
:=
`pA (D)
1
p
=
f ∈ Hol (D) :
kf kpkp
A
+
n=0
1
q
= 1, p ≥ 1. Indeed,
∞
X
kf (C)kL(X) = fb(n) C n where
)
∞ X
b p
=
f (n) < +∞ ,
n=0
L(X)
≤
∞ X
b n
f (n) kC kL(X)
n=0
DISCRETE DUHAMEL PRODUCT
≤
∞ X
b p
f (n)
!1/p
n=0
∞
X
835
!1/q
kC n kqL(X)
n=0
= kf k`p M (C, q) ,
A
where M (C, q) > 0 is a constant.
Let us define also the following (closed) subspace of the space `pA (D) :
(∞
)
X
`pA,i :=
fb(n) z n ∈ `pA : fb(k) = 0, k = 0, 1, 2, . . . , i − 1 , i = 1, 2, . . . .
n=0
2. The results
The following two lemmas can be proved by similar arguments used in
the proofs of Theorem 1 in [9], Theorem 2 of [6] and Theorem 16 in [5], and
therefore we omit their proofs.
Lemma 1. Let X be a Banach space with a Schauder basis (en )n≥0 , x, y be
two elements in Xi = span {en : n = i, i + 1, . . . } , i ≥ 0. Let T en = λn en+1 ,
λn 6= 0, n ≥ 0, be the weighted shift operator with bounded weights sequence
(λn )n≥0 continuously acting in X. Then we have:
X
wn+m−i
˜y =
(3)
en+m−i
x~
xn ym
wn wm
i
n,m≥i
= (Bw x) (T |Xi ) y = (Bw g) (T |Xi ) x,
where T |Xi is a restricted weighted shift operator and
(4)
def
(Bw x) (T |Xi ) y =
∞
X
1
xn (T |Xi )n y.
wn
n=i
Lemma 2. Let X be a Banach space with a Schauder basis (en )n≥0 continuously embedded in `p for some p ≥ 1. Let T en = λn en+1 , n ≥ 0,
be the weighted shift operator continuously acting in X. We put wn =
λ0 λ1 . . . λn−1 , w0 := 1. Suppose that for any integer i ≥ 1 there exists
an integer N ≥ i such that
X wn+m−i q
wn wm < +∞ for p > 1
n,m≥N
and
X wn+m−i wn wm < +∞ for p = 1,
n,m≥N
where p1 + 1q = 1. Suppose also that ken+m−i k ≤ Mi ken k kem k for all n, m ≥ i
and some Mi > 0. Then we have:
836
MUBARIZ T. KARAEV, SUNA SALTAN AND TEVFIK KUNT
(a)
x~
˜ y
(5)
i
Xi
≤ Ci kxkXi kykXi
˜
for all x, y ∈ Xi and some constant Ci > 0, i.e., Xi , ~ is a unital
i
Banach algebra with the unit element wi ei .
˜ -invertible if and only if xi 6= 0.
(b) An element x ∈ Xi is ~
i
The main result of the present paper is the following.
Theorem 1. Let X be a Banach space with a Schauder basis (en )n≥0 which
is embedded in `p for some integer p ≥ 1. Let T, T en = λn en+1 , n ≥ 0, be
the weighted shift operator with bounded weight sequence (λn )n≥0 , such that
T is continuous in X. We put wn := λ0 λn1 . . . λn−1 , w0 := 1. Suppose that
for any integer i ≥ 1 there exists an integer N ≥ i such that
X wn+m−i q
(6)
wn wm < +∞ for p > 1
n,m≥N
and
X wn+m−i wn wm < +∞ for p = 1,
(7)
n,m≥N
where p1 + 1q = 1. Suppose also that ken+m−i k ≤ Mi ken k kem k for all n, m ≥ i
and some Mi >0. Let Q : Y → Y be an operator on a Banach space Y such
q
∞
P
kQk k
that
< +∞. Then
|wk |
k=1
µ(T |Xi ⊕ Q) ≤ µ(T |Xi ) + µ(Q) = 1 + µ(Q).
Proof. First, note that the restricted operator T |Xi is cyclic for every i ≥ 1,
and therefore µ(T |Xi ) = 1 for every i ≥ 1. On the other hand, since
1 + µ(Q) ≥ µ(T |Xi ⊕ Q) ≥ max {1, µ(Q)} ,
it is clear that if µ(Q) = +∞, then µ(T |Xi ⊕ Q) = 1 + µ(Q). So, we
will assume that µ(Q) = n < +∞. For the proof, suppose in contrary that
µ(T |Xi ⊕Q) 6= n+1, that is µ(T |Xi ⊕Q) < n+1 or µ(T |Xi ⊕Q) = µ(Q) = n.
Then by the definition of the spectral multiplicity there exists n-dimensional
cyclic subspace for the operator T |Xi ⊕ Q. Let
n
o
x(1) ⊕ y (1) , x(2) ⊕ y (2) , . . . , x(n) ⊕ y (n)
be a cyclic tuple of vectors for the operator T |Xi ⊕ Q. Then
n
o
x(1) , x(2) , . . . , x(n)
is a cyclic tuple for the operator T |Xi .
DISCRETE DUHAMEL PRODUCT
837
Lemma 2, it follows from the conditions (6), (7) that
By considering
˜ is a Banach algebra. Consequently, for every x ∈ Xi the ”discrete
Xi , ~
i
˜ y, y ∈ Xi , is continuous in Xi and ,
Duhamel operator” Dx , Dx y := x~
i
kDx k ≤ Ci kxkXi (see inequality (5)). On the other hand, formula (2) shows
that
˜ y, m ≥ 0,
(T |Xi )m y = wi+m ei+m ~
(8)
i
and therefore
span {(T |Xi )m y : m ≥ 0} = closDy span {wi+m ei+m : m ≥ 0} ,
which implies that y ∈ Cyc (T |Xi ) if and only if Dy Xi = Xi . It is not
difficult to prove that Dy Xi = Xi if and
onlyif yi 6= 0, that is y is an
˜ , which is equivalent to the
invertible element of the Banach algebra Xi , ~
i
invertibility of the corresponding
discrete
Duhamel operator Dy . Thus, the
cyclicity of the tuple x(1) , x(2) , . . . , x(n) implies that there exists a number
(i )
i0 ∈ {1, . . . , n} such that xi 0 6= 0. We assume without loss of generality that
(1)
i0 = 1, that is xi 6=
this condition, as already mentioned above,
0. Under
x(1) is invertible in
˜
Xi , ~
(see the assertion (b) in Lemma 2). Therefore,
(1)
˜ such that z (1) ~
˜ x(1) = wi ei .
there exists a unique element z ∈ Xi , ~
i
i
(1)
˜ x(1) = zi(1) x(1)
6= 0. Let us consider the
Since z (1) ~
i , it follows that zi
i
i
following matrix:
i
z (1)
0
0
(2)
 −x ~
˜ z (1) wi ei
0

 (3) i (1)
˜z

0
wi e i
M :=  −x ~
i

..
..
..

.
.
.

˜ z (1)
0
0
−x(n) ~

i
...
...
...
..

0
0 


0 
.
.. 

. 
.
. . . wi ei
Then, by using formulas (3) and (4) in Lemma 1 and formula (8), we have:
 (1) 
x
 x(2) 


 x(3) 
(Bw M ) (T |Xi ) 

 . 
 .. 
x(n)
838
MUBARIZ T. KARAEV, SUNA SALTAN AND TEVFIK KUNT


(1) (T |X )
B
z
0
0
.
.
.
0
w
i

  (1) 
(1)
 Bw −x(2) ~
˜
x
z
(T |Xi ) IXi 0 . . . 0 


i

  x(2) 

 

(3) ˜ z (1)

 x(3) 
(T |Xi ) 0 IXi . . . 0 
=  Bw −x ~
.



i

  . 
.



..
..
..
.. 
..
. 

.
.
.
.
.

 x(n)


(n)
(1)
˜
Bw −x ~z
(T |Xi ) 0
0 . . . IXi
i


(1) (T |X ) x(1)
B
z
w
i


(1)
 Bw −x(2) ~
˜
z
(T |Xi ) x(1) + x(2) 


i




(1)
(1)
(3)
 Bw −x(3) ~
˜z
(T |Xi ) x + x 
=

i




.
..






(n)
(1)
(1)
(n)
˜z
Bw −x ~
(T |Xi ) x + x
i


˜ z (1)
x(1) ~
i



 (1)
(2)
(1)
(2)
x ~
˜ −x ~
˜z
+x 


i i



 (1)
˜ −x(3) ~
˜ z (1) + x(3) 
=
x
~


i
i




..


.



 (1)
(1)
(n)
(n)
˜
˜
+x
x ~ −x ~z
i
i


wi e i 

(1) ~
(1) + x(2) 
 −x(2) ~
˜
˜
x
z


i i




(3)
(1)
(1)
(3)

˜ x ~
˜z
+x 
=  −x ~

i
i




..


.



 (n)
(1)
(1)
(n)
˜
˜
+x
−x ~ x ~z
i
i


 

wi ei
wi e i
wi e i
˜ wi ei + x(2) 
 −x(2) ~
  −x(2) + x(2)   0 


 
 (3) i
 
(3)
˜ wi ei + x   −x(3) + x(3)   0 

=
=
=  −x ~

.



i
  .. 

 
.
.
.
 

  . 
..
.


(n)
(n)
(n)
(n)
0
−x + x
˜ wi e i + x
−x ~

i
DISCRETE DUHAMEL PRODUCT
839
So
 

x(1)
wi e i
 x(2)   0 
 


 (3)  

(Bw M ) (T |Xi )  x  =  0  .
 .   .. 
 ..   . 
0
x(n)

Since

(1)
Bw z
0
0
...
0



(1)
 Bw −x(2) ~

˜
z
e
0
.
.
.
0
i


i




(1)
 Bw −x(3) ~
˜z
0 ei . . . 0 
Bw M = 
,
i



..
.. .. . .
.. 

.
.


.
. .


(n)
(1)
˜
Bw −x ~z
0 0 . . . ei
i
we have
˜ ei ~
˜ ...~
˜ ei =
˜ − det (Bw M ) = Bw z (1) ~
~
i
i
i
i
1
Bw z
win−1
(1)
,
and hence
˜ − det (Bw M ) =
~
i
i
1
win−1
1 (1)
1
z = n z (1) 6= 0
wi i
wi
˜ -invertible element; see above). This implies that the
(because z (1) is ~
i
operators (Bw M ) (T |Xi ) and (Bw M ) (Q) are invertible in the spaces
X n := X
∗ · · · ∗ X}
| ∗ X {z
and
n
Y n := Y
∗ · · · ∗ Y},
| ∗ Y {z
n
respectively. Therefore, by Karaev’s result (about this result see in reference
of [5]), we have that
n
o
((Bw M ) (T |Xi ) x)j ⊕ ((Bw M ) (Q) y)j : j = 1, 2, . . . , n
is a cyclic tuple for the operator T |Xi ⊕ Q. So, we have a new cyclic tuple of
the form {wi ei ⊕ ȳ1 , 0 ⊕ ȳ2 , . . . , 0 ⊕ ȳn }, and therefore for every y ∈ Y there
exists a family of vector-polynomials {Pm,j }nj=1 such that
(9)
lim Pm,1 (T |Xi ) wi ei = 0 in Xi
m→∞
and
(10)
lim
m→∞
n
X
j=i
Pm,j (Q) ȳj = y in Y.
840
MUBARIZ T. KARAEV, SUNA SALTAN AND TEVFIK KUNT
We set
−1
qm,1 := Bw
Pm,1 =
X
wk (Pm,1 )k ek .
k≥0
It is clear that Pm,1 (T |Xi ) wi ei =
P
k≥0
wk (Pm,1 )k ek = qm,1 , and by consid-
ering this in (9) we obtain that lim qm,1 = 0 in X. Now by considering the
m→∞
condition
!
∞
Qk q
X
=: C < +∞,
|wk |
k=i
we have
kPm,1 (Q)kL(Y )
X
k
=
(Pm,1 )k Q k≥0
L(Y )
X
X
1 k
k
(Pm,1 ) Q
≤
Q
=
|w
|
(P
)
m,1 k
k
k |wk |
L(Y )
L(Y )
k≥0
k≥0
1 

p
X
X
wk (Pm,1 ) p  
≤
k
k≥0
k≥0
k
!q  1q
Q L(Y )

|wk |
= C kqm,1 k`p .
Since X ⊂ `p , we have kqm,1 k`p ≤ C̃ kqm,1 kX and thus
kqm,1 (Q)kL(Y ) ≤ C C̃ kqm,1 kX .
On the other hand, since qm,1 → 0 (m → ∞) in X, we obtain from this
inequality that qm,1 (Q) ⇒ 0, and therefore lim qm,1 (Q) ȳ1 = 0. Hence, it
m→∞
follows from (10) that
lim
m→∞
X
Pm,j (Q) ȳj = y.
j=2
Since y ∈ Y is arbitrary, this equality means that {ȳ2 , ȳ3 , . . . , ȳn } ∈ Cyc (Q),
which implies that µ(Q) ≤ n−1 (because card {ȳ2 , ȳ3 , . . . , ȳn } = n−1). But,
this contradicts to µ(Q) = n. The theorem is proven.
The following is an immediate corollary of Theorem 1.
Corollary 1. Let V , V f (z) =
Rz
0
f (t) dt, be a Volterra integration operator
on the space `pA (D), 1 ≤ p < ∞, and let Q be a bounded operator on a
separable Banach space Y satisfying
∞ X
k q
k! Q < +∞, for p > 1,
k=1
DISCRETE DUHAMEL PRODUCT
and
here
1
p
+
1
q
841
1
k
, for p = 1;
Q = O
k!
= 1. Then µ(V ⊕ Q) = 1 + µ (Q).
1
For the proof, it is sufficient to put in Theorem 1, X = `pA , λn = n+1
(n ≥ 0), en = z n (n ≥ 0) , i = 0, and to modify the proof of the theorem in
case p = 1.
Let S be the shift operator on `pA = `pA (D) defined by Sf (z) = zf (z).
Our next result calculates µ(S + C) for some C ∈ L (X).
Theorem 2. Let S be a shift operator on `pA , 1 < p < ∞, and C be a
cyclic bounded linear operator on a separable Banach space X such that
∞ P
C k q < +∞, where 1 + 1 = 1. Then µ(S ⊕ C) = 2.
p
k=0
q
Proof. As we have proved in Section 1,
kf (C)kL(X) ≤ M (C, q) kf k`p
A
∞
1
q
P
C k q , which means that C admits
for all f ∈ `pA , where M (C, q) :=
(11)
k=0
the functional calculus in the class `pA . Since µ(S) = 1 and µ(C) = 1, it
is clear that 1 ≤ µ(S ⊕ C) ≤ µ(S) + µ(C) = 2. Suppose in contrary that
µ(S ⊕ C) = 1. Then the operator S ⊕ C has a cyclic vector f ⊕ x ∈ `pA ⊕ X,
which implies that there exists polynomials pn such that
lim pn (S ⊕ C) (f ⊕ x) = 1 + 0 in `pA ⊕ X,
n→∞
or
lim (pn (S) ⊕ pn (C)) (f ⊕ x) = 1 ⊕ 0.
n→∞
Whence
lim pn (z) f (z) = 1 in `pA
(12)
n→∞
and
(13)
lim pn (C) x = 0 in X.
n→∞
It is obvious that pn f − 1 ∈ `pA . Let pn (z) f (z) − 1 =
∞
P
an,k z k . Then by
k=0
considering inequality (11) and (12), we have
k(pn f − 1) (C)kL(X) ≤ M (C, q) kpn f − 1k`p → 0 (as n → ∞) ,
A
which means that pn (C) f (C) ⇒ IX (as n → ∞). In particular,
lim pn (C) f (C) x = x in X.
n→∞
On the other hand, it follows from (13) that
0 = lim f (C) pn (C) x = lim pn (C) f (C) x = x,
n→∞
n→∞
842
MUBARIZ T. KARAEV, SUNA SALTAN AND TEVFIK KUNT
or x = 0, which is impossible, because x is a cyclic vector of the operator
C. This proves the theorem.
Let us denote
n
o
p
p
(n)
`p,n
:=
f
∈
`
(D)
:
f
∈
`
A
A
A
(
∞
X
= f ∈ `pA (D) : f (z) =
fb(k) z k and
k=0
∞ X
)
b p
k (k − 1) . . . (k − n + 1) f (k) < +∞ .
k=n
Theorem 3. Let S be a shift operator on the space `p,n
A , where p ≥ 1, and
C be a cyclic operator on the separable Banach space X satisfying
!q
k
∞
C X
< +∞,
(14)
k (k − 1) . . . (k − n + 1)
k=n
where
1
p
+
1
q
= 1. Then µ(S ⊕ C) = 2.
Proof. By using (14), we have for every f ∈ `p,n
A that
∞
∞ X
X
b k
k
b
kf (C)k = f (k) C ≤
f (k) C k=0 k=0
b b b
= f (0) + f (1) kCk + · · · + f (n − 1) C n−1 +
k
∞
C X
b k (k − 1) . . . (k − n + 1) f (k)
+
k (k − 1) . . . (k − n + 1)
k=n
≤ M1 (C, n) kf k`p,n +
∞ X
A
p
k (k − 1) . . . (k − n + 1) fˆ (k)
!1
p
·
k=n
·
∞
X
k=n
!q ! 1
k
q
C k (k − 1) . . . (k − n + 1)
= M2 (C, n) kf k`p,n ,
A
where M1 (C, n), M2 (C, n) > 0 are some numbers. Thus
kf (C)k ≤ M2 (C, n) kf k`p,n
(15)
A
`p,n
A .
for all f ∈
Note that µ(S) = 1. Now in order to complete the proof of
the theorem, it remains only to use inequality (15) and the arguments for
the proof of Theorem 2.
DISCRETE DUHAMEL PRODUCT
843
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(Mubariz T. Karaev) Department of Mathematics, College of Science, King Saud
University, P.O. Box 2455, Riyadh 11451, Saudi Arabia
mgarayev@ksu.edu.sa
(Suna Saltan) Suleyman Demirel University, Department of Mathematics, 32260,
Isparta, Turkey
sunasaltan@sdu.edu.tr
(Tevfik Kunt) Suleyman Demirel University, Department of Mathematics, 32260,
Isparta, Turkey
tevfik.kunt@gmail.com
This paper is available via http://nyjm.albany.edu/j/2014/20-42.html.