Arab Journal of Mathematical Sciences

Arab J Math Sci xxx (xx) (xxxx), xxx-xxx

The spectra and eigenvectors for the weighted mean matrix operator

Q1 E. Pazouki*,B.Yousefi

Department of Mathematics, Payame Noor University, P.O. Box 19395-3697, Tehran, Iran Received 28 February 2013; accepted 14 January 2015

Abstract. In this paper, first we give conditions under which the weighted mean matrix operator is bounded on the weighted Hardy spaces, and we characterize the spectrum of the weighted mean matrix operator acting on some sequence spaces. Then we investigate eigenvectors of weighted mean matrix operator.

Keywords: Weighted Hardy spaces; Spectrum; Weighted mean matrix operator; Eigenvalue; Eigenvector

2010 Mathematics Subject Classification: 47B37; 47A25

1. Introduction

Let{P(n)} be a sequence of positive numbers with P(0) = 1 and let 1 < p < to. We consider the space of sequences f = {f (n)}r=0 such that

f ir = uz iiß = E if (n)ip ß (n)p <

The notation f (z) = J2r=0 f (n)zn shall be used whether or not the series converges for any value of z. These are called formal power series and the set of such series is denoted by

Hp (P). Let f k (n) = 5k (n). So fk (z) = zk and then {fk }k is a basis such that \\fk 11 = P (k).

For 1 < p < to, Hp(P) = lpwhere ^ is the a-finite measure defined on the positive integers by p(K) = (P(n))p, K C No, (No = N U {0}).

* Corresponding author.

E-mail addresses: aha.pazoki@gmail.com (E. Pazouki), b_yousefi@pnu.ac.ir (B. Yousefi). Peer review under responsibility of King Saud University.

http://dx.doi.org/10.1016/jj.ajmsc.2015.01.001

1319-5166 © 2015 King Saud University. Published by Elsevier B.V. This is an open access article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/).

2 E. Pazouki, B. Yousefi

1 Let {an}r=0 be a sequence of positive numbers, and let An = Y1 n=o a-ift(i)p. The

2 weighted mean matrix operator on Hp (ft) is an infinite matrix A = [ank]n,k with

3 ank = \ An

[0 k > n.

4 Let X be a nontrivial complex Banach space of complex sequences. The spectrum

5 (a(A))x of a bounded operator A on X, is the set all of complex numbers A such that

6 the operator A — AI is not invertible on X. The resolvent (p(A))X of A is the complement

7 of (a(A))X. A complex number A is an eigenvalue of the operator A, whenever there exists

8 a nonzero complex sequence {f (n) }r=0 in X such that A{f (n)} = A{f (n)}. This nonzero

9 complex sequence {f (n)}r=0 is called eigenvector corresponding to A. For some another

10 sources on this topic see [1-9,11,10,21,13-15,19,20,16-18,12].

11 2. Boundedness of the weighted mean matrix operator

12 In this section we investigate the boundedness of the weighted mean operator matrix act-

13 ing on some sequences spaces. Consider the fixed measure space (N0, P(N0)where

14 ^(K) = SneK(ft(n))p. The collection of all complex sequences f = {f (n)} for which

15 Z^r=0 if (n)lft (n)p < to, will be denoted by l1 (^). In this case put f (z) = ^ r=0 f (n)zn £

16 l1 (^), and let

17 ||f II = £ if(n)ift(n)p.

18 By the usual way we can define lr (^) for r > 1.

19 Lemma 2.1. Suppose that f (z) = r=0 f (n)zn. Then f (z) £ Lr (^) if and only if there

20 exists M > 0 such that if (n) | < M for all n > 0.

21 Proof. Let f (z) =Y1 r=0 f (n)zn £ lr (^). Observe that |f (n)| < ||f ||r for all n (since by

22 definition of ^(S), it follows that ^(S) =0 if and only if S is an empty set). So

23 sup if (n)l < TO.

24 The converse is clear. Now the proof is complete. □

25 Thus lr(^) consists of all f (z) = Y^ r=0 f (n)zn such that ||f ||r = supn>0 if (n)| < to.

26 Corollary 2.2. Let 1 < p < q < to. Then the following statements holds

27 (i) If Er=0 ft(n) < TO, then lq (») C lp(¿).

28 (ii) If ft(n) > 1 for all n > 0, then lp (¿) C lq (¿).

29 Proof. Since r=0 ft(n) < to, there exists k such that ft(n) < 1 for all n > k. Thus

30 ft (n)p < ft (n) for all n > k, and so (N0, P(N0), ^) is a finite measure space. So by Theorem

31 25.13 in [2], part (i) is true.

The spectra and eigenvectors for the weighted mean matrix operator

Let f (z) = £r=0 f (n)zn e Lp(y), then ^r=0 If (n)|pP(n)p < to. So there exists k such that |f (n)|pP(n)p < 1 for all n > ki. Since P(n) > 1, we get

£ If (n)|qP(n)p = £ If (n)|qP(n)qP(n)p-q

n=0 n=0

£|/ (n)|p P (n)p

n)p <00.

Thus (ii) holds. This complete proof. □

Theorem 2.3. Let A be the weighted mean matrix operator. Assume that lim supn^r < 1. Then A is bounded on lr (y).

Proof. Let f (z) = Er=0 f (n)zn e lr (y), thus (Af)(z) = £ (£ f (k^ zn.

n=0 \k=0

Note that

l(A/ )(n)| =

ak P (n)p

< EE akprn)-1/(k)|

< E akP(n)p 11/1

for all n > 0. Since lim supn^r < 1, there exists N such that P (n) < P (k) for all

n > k > N. Thus

E akP(n)p < akP(n)p + ^^

ak P (k)p

k=0 N-1

akP(k)p Ak P(n)p Ak An P(k)p

k=0 N-1

P(n)p P(N)p P (N)p P (k)p

P (N )p

E. Pazouki, B. Yousefi

• — (- il ^ Z - 11 s-1)

3 for all n > 0. By Lemma 2.1, A is bounded on (ß). Now the proof is complete. □

4 Lemma 2.4. Suppose that B = [bnk ] (n,k = 0,1,...) is an infinite matrix of complex

5 numbers such that bnk = 0 for 0 < k < n, and bnk = 0 otherwise. If

Mi = sup l jbnfc \(^ ) < to, k>0^t \ß(k) '

then B is a bounded operator on l1(ß).

Proof. Let f (z) = £~=0 f (n)zn G l1 (ß). Then

(Bf)(z) = ^[ Y,bnkf(kM zn.

n=0 \k=0

We have

J2\(Bf)(n)\ß(n)p <Y,T,\bnkf (k)\ß(n)p

(n)\ß (n)

n>0 n>0k=0

ß(n)^ p

ll \bnk ||/ (k)\ß (k)> n>0 k=0 '

|\f (k)\ß (k)p(1 \bnk \( ß(n)X P

k>0 \n=k '

< M1| f| .

Thus A is bounded on l1 . Now the proof is complete. □

Theorem 2.5. Let lim infn^r ""a?* = 5, where 0 <5 < 1, and lim supn^r < 1. Thus the weighted mean matrix operator is bounded on l1 (^).

Proof. By Lemma 2.4, it is sufficient to show that

ak ft (n)p ( ft (n)

k>0^u An \ß (k)

Since lim infanß(n)P = S, there exists N1 such that an An) > 2

W - A tl^Qfil /V. "nf

1 _ q„.+i0(n -1)p\ < i 1 _ f

An+i / V 2

The spectra and eigenvectors for the weighted mean matrix operator

for all n > N1. For k > N1 and j > 1, we have

akP(k)p akP(k)p

ak+i P (k + i)p

. >1 - ^j Ak V 2

akP(k)

akP(k)p akP(k)p

TW V^afe

ihUs Z-j=0 Ak+j

< to for all k + j > N1. Since limsupn^r N2 such that P(n) < P(k) for all n > k > N2. If k > max{N1, N2}, then

< 1, there exists

ak P (k)p / P(n) An VP (k)

j=0 to

ak P (k)p f P (k + j ) P (k)

Ak+j akP(k)p

v- akP(n)VP(n)\ sup > -;- I ^TTW I < TO.

k>0 — n>k

An P(k)

Thus A is bounded on l1 (y). Now the proof is complete. □

3. The spectra for weighted mean matrix operator

In this section we investigate the spectra for the weighted mean matrix operator acting on some sequence spaces.

Lemma 3.1 ([4, Lemma 1]). Let A be a weighted mean matrix operator, and define T = A — A/ where A e C such that ann = A for each n > 0. Then S (= [snk ]) = T-1 exists and,

f -1 akP(n)p

n (1 - jj

an ff(n)p a

0 < k < n

k = n k > n.

E. Pazouki, B. Yousefi

1 Theorem 3.2. Let — A

2 Then

^(n) = 5 where 0 <5 < 1 and let limn^r < 1.

KA))L1(M) Ç A :

1 - S 2- S

an ß(n)p

4 Proof. Let 0 < 5 < 1 and choose 0 < e < 5 — ^^. Since limn^r

N such that

ß(n+1) ß(n)

< 1, there exists

^ anß(n)p ß(n + 1) ^

S — e < -:-, -77-T— < 1

for all n > N. Notice that we have

1 - ^ anß(n)p

1 1 - S

2 S 2 S

9 for all n > N. Suppose A is a complex number such that A = A(n)P for all n > 0, and

10 IA — 2P51 > 1-1. If 5 < x < 1, then x e D, where

D = A :

Since A e D, thus |A — x| = 0. Define the function f by f (x) = ) for all 5 — e <

x < 1. By Lemma 4.9 in [4], f is continuous and 0 < f (x) < 1. So there exists 0 < M1 < 1

such that f (x) < M1 for all 5 — e < x < 1. If k > N, then

,ß(n)p

G D for all n > k. So

ß(n + 1)^ P S(n+l)k ß (n)

i A | (1 -

an+1 ß(n+1)p An+i

g«+iß(n+1)P A«+ i

an+i ß (n +1)p

Now by the relation |A -

,ß(n)p

i > |A - i- 2-f = M2 we get

£ |snfc1

n=fc+1

ß(n)P < (ß(k + 1A P i S(fc+1)fc | ß(k)p "V ß(k) J 1 - M

= 1 afc ß (k)p 1 1 - M Ak i A - akß(k)P i

1 / ß (k + 1)

a.k+iß(fc+1)pM ß(k)

i Ak+i i

(1 - M1 )M22

The spectra and eigenvectors for the weighted mean matrix operator 7

for all k > N. So by Lemma 2.4, S £ B(l1 (¿)), i.e. A £ (p(A))li(m). 1

Suppose that 5 = 1, and let A be a complex number such that A = A1 + iA2. Without loss 2

of generality we can say that 0 < A1 < 1 and A = a'n A(n)* for all n > 0, and iA — 1i > 0. 3

Then for e1 < iA — 1i and 1 — e1 < x < 1, we get |x — 1i < iA — 1i, i.e. iA — x| = 0. We know that ^rx and ^xx are two increasing functions, so there exists e2 > 0 such that 0 < A1 < 21X and A1 < for all 1 — e2 < x < 1. Thus we obtain

A: - 1

2-x 1x

Hence A £ Dx. If e < min{e1, e2}, then f is continuous and 0 < f (x) < 1 for all 1 — e < x < 1.Bya similar method used earlier we get S £ B(l1 (^)), i.e. A £ (p(A))ii(M). Thus the proof is complete. □

Lemma 3.3. Let C be a space of convergent sequences and limn^r < 1. Thus the

weighted mean matrix map A : C —► C is a bounded operator.

Proof. Since limn^r < 1, so there exists natural number N1 such that ft (n + 1) <

ft (n) for all n > N1 .If x = {xn} £ C and ||x||r < 1, then

\A( \ I ^ ^ akft(n)p I I iA(x)ni < -A-ixk i

ak P(n)p k=o An

Nl-1 ak P (n)p^ , , ^ ak P (k)p p (n)p

^ akP^^ akP\kr P\n I I

- £ "AT"|xk 1 + ^ m~plxk 1

k=0 n k=N1 n '

< N-1 akp (n)p+1

< ^ An

= N-1 akP(k)p Ak P(nL + ,

A. A. P(k)p +

k=0 N1-1

Ak An P(k)p P (n)p

- ^ p(k)p k=0

= v- P(n)p p(N1)p k=0P(N1 )p P(k)p +

ft (N1)

for all n > N1 . Therefore

< y wv +1

< ^ P(k)p

E. Pazouki, B. Yousefi

|A|U < [max E

ak P (m)p

N1-1 N1 -1

P (N1 )p

m I 0 k=0 P (k)2

m=0 k=0

+ 1 I .

This completes the proof. □

Theorem 3.4. Let A : C —► C be a linear map, and limn^r A(n) = 5 where

0 < S < 1. Assume that limr

5 (ct(a))c ^ A :

fi(n + 1)

< -< 2

< 1. Then

Proof. Suppose that |A — 11 > 1. This inequality is equivalent to a > —1, where

< 1 , we get

7 -I = a + iP. Since 0 < anAnZ

anP (n)p

> 1 + a

an P(n)p An

an P (n)p An

9 for all n > 0. Without loss of the generality suppose that P(n) is a decreasing sequence,

10 P (n + 1) < P(n) for all n > 0.

11 Consider 0 < 5 < 1, since

anP (n)

An-1 1 — an^(n)p ' is a monotone increasing sequence, thus there exists natural number N1 such that

anP(n)p < f) +1,

akP(k)p

An-1 \1 - S for all n > N1. We have

£ |Snk | < |Snn | + £ - | snn | +

k=0 An | A |2 n (1 + a jj^ ) j=k j

An|A|2(1+ a)(1+ a anAf^ ) 1 + An-1

| A | (1 + a anAnp ) An|A|2 (1+ a)(1+ a )

<u + a*^ M +

An-W V |A| |A|2(1 + a)

+ 1 ^ +

|A| | A |2 (1 + a)

The spectra and eigenvectors for the weighted mean matrix operator

If 5 = 1, then there is natural number N2 such that

1 anft (n)p

when 0 < a, we have 1 + a < 1+ a an< 1 + a, and if -1 < a < 0, then 1+ a < 3

1+ a anA(n)P < 1 + f, for all n > N2 .So 4

ak P (k)p

^T |snk | < |snn| + £ -

k=o An\A\2 n (1+ aajj)

- 1 snn \ +

An \ A \2 (1+ a)(1+ a ^^ ) +

Ln K*\ V-1- 1 <-"VV"L 1 <-x An 1 An-1

\A\(1+ a ) An \ A \2 (1 + a)(1+ a )

< M3 ( ^ +

\A\ \ A \2 (1+ a))'

where M3 = Max{1 + a, 1 + a}. Since

pin+11

An + 1 1 _ an+l^(n+1)p

11 _ an+ij(n+1)p I \1 AAn+i \

for all n,k > 0. Therefore S has bounded columns. By Proof of Theorem 3 in [5] we get A £ (p(A))c. Now the proof is complete. □

Theorem 3.5. Let A : C —► C be a linear map, and limn fci) < 1, then

anj(n) — S where

0 <S < 1. If limn^^ -jn"

" S A; A <

"l 2 - S <2

anP (n)p

Proof. Let A with = a + ift, be a fixed complex number and satisfy | A — ^r^ | > I-! Theorem 3.4 implies that | A — 11< 1 .If A = a"A(n)P for all n > 0, then a < —1. Since

aj P(j)p

1+n - 8 j

E. Pazouki, B. Yousefi

i we see

|snk| —

ak P (n)

An|A|2n |1 - jj| j=k

ak P (n)

Ak- 1 | A|2 n |1 + (1 - 1 ) aajp | j=k

Define

/ (t) =

1+11 - Alt

- 1 + 2(1 + a)t + ((1 + a)2 + P2 )t2,

for t > 0. It is clear that /(t) at t0 —

_ -2(1 + a)

(1 + a)2

. If 0 < S < 1, then we get t1 < ^ and

_ -(1 + a) (1 + a)2

5_ 1-5

has Minimum and / (t1 ) — 1 where

1 + 11-A 't

1 - S A(1 - S)

9 Let 0 < e < 1— — t1, thus there exists natural number N1 such that < 1 and

anP (n)^ S

< ^-? + 1,

An 1 1 S

for all n > N1. Note that f (t) is a continuous function so there is a positive number M1 such that

_, + ,_, 1\ an+1 P(n + 1)p

- A) AT

> M1 > 1,

for all n > N1. Using (*),

sn + 1k

P(n + 1)

p (n) y |1 + (1 -1 ) an+1 A(n+1)p |

The spectra and eigenvectors for the weighted mean matrix operator

for all n > N1, k > 0. Therefore S has bounded columns. From (*),

£|snk|

A„_ i

|A||1 + (1 - 1 ) ^AtT"" |

akP(n)p

k=0 Ak-1|A|^ |1 + (1 - 1 ) |

< 1 + 1-5 + 1 + P(N1 )p ^^

< 1,1 A0 | A |2 M2 M

k=0 ak P (k)p

n-N1 +1

k=Ni Ak-1 |A|2 n |1 + (1 - 1 ) jj|

< 1 + 1- + 1 + P(N1 )p ^1

k=0 |A|2M2MT-N1+1

| A|2 1 - S

M1n-k+1

k=N1 M1

where M2 - min{nN1k 11 + (1 - 1 ) jj|}N=0. By the proof of Theorem 3 of [5],

A G (p(A))c. 1

If S = 1 , then lim

ifi(n)" _

An — 1

= to. Since limt^r f (t) = to, there exists M2 > 0,

such that f (M2) > 1. It implies that there is natural number N2 such that

an fi(n)"

and / (i

) > 1 for all n > N2. Therefore

. + 1\ an+1 P(n +1)p

- A AT

> /(M2) > 1,

P (n + 1)

^p («W |1 + (1 -1 ) an+1 A(n+1)p |

and hence S has bounded columns. With relation

(1 + 2(1 + a)t + ((1 + a)2 + P2)t2)2 - ((1 + a)2 when t , we obtain

2a 21,

|1 + (1 - 1 ) | ((1+ a)2 + p2 )1

E. Pazouki, B. Yousefi

i as n

00. So there is a natural number N3 such that

|1 + (1 _ 1 ) OnMnr | ((1+ a)2 + p2) 1

for all n > N3 .If n > Max{N2, N3 }. Then by the similar way, y]lsnfc | < to. The Proof of Theorem 3 in [5], shows that A G (p(A))C. Now the proof is complete. □

Corollary 3.6. Let A : C

0 < S < 1. Assume that limn

C be a linear map, and lim

r((n)P _

((n + 1) ((n)

< 1. Then

(a(A))^ (M) Ci A;

anfi (n)p

= 5 where

Proof. By Theorem 2.3 A e B(l™ fa)). So A is an element of r(l™ the Banach algebra of bounded, infinite matrices over l™ fa). Let r (c) be the Banach algebra of bounded, infinite matrices over c, thus it is a closed subalgebra of r<(^) which contains identity matrices. By Lemma 4.8 of [4], (a(A))l<(^) C (a(A))c. Now the proof is complete. □

Proposition 3.7. Let A : C —► C be a linear map, and limn^™ A 0 < S < 1. If limn^™ < 1, then

((n)p _

= 5, where

(a(A))ip(M) C< A;

an Pfa)1

15 Proof. By Theorems 2.3 and 2.5 we get A e B(l1 fa)) n B(l™ fa)). If A e (p(A))li(m) n

16 (p(A))<(m), then the infinite matrix S = (A - Al)-1 e B(l1fa)) n B(l™ fa))- Thus

17 by the Riesz-Thorin Theorem, S e B(lpfa)). So A e (p(A))lp(M), i.e. (a(A))lP(M) C is (a(A))li (M) U (a(A))l< (M). Thus the proof is complete. □

19 4. Eigenvectors and diagonalization of the weighted mean matrix

20 OPERATOR

21 In this section first characterize eigenvectors and investigate diagonalizability of the

22 Q2 weighted mean matrix operator acting on weighted Hardy spaces.

Theorem 4.1. Let A : C

C be a linear map, and limn

an ((n)p _

((n+1)

= 1. If

< 1 , then

n^™ p(n)

(i) A is bounded on Hp fa) and (a(A))lp(m) = { an>™=0.

_ aj ¡3(j)P "

(ii) For every j > 0, Cj =

= r anp(n)^ ™ ~ { An }n=

A is an eigenvalue of A the weighted mean matrix

27 operator.

The spectra and eigenvectors for the weighted mean matrix operator

(iii) If Y^ fc=0 P (k) < to, then f (z) = ^ P (k)p zk e Lp(y) is the eigenvector for 1 eigenvalue of c0 = 1. 2

Proof. By the Proposition 3.7, A is bounded on Hp(P), and 3

(^(A)) r J anP(n)p (a(A))LP(M) Ç

_ an^(n)

, n > 0, then det(A/ — A) = 0. So (i) holds.

Theorem 3 of [3] implies that if A =

Suppose that Cj = A = a" A(j) , and j e N, we will show that there exists fA(z) = E°=0 5?(j )zj e Hp (P), such that

A(f?) = A(f?).

Define f?(i) = 0 for 0 < i < j, and f?(j) = a > 0. From (Af?))(j + 1) = Cj f? (j + 1), we have

(cj - cj+1 )/a(j + 1) =

aj P (j + 1)p

/a(j )

P(j + m p ajP(j)p P (j) / A

P (j + 1) P (j )

/a (j )

Cj (1 - Cj + 1 )/A (j):

/A(j + 1)=(' p /a(j).

P (j W (1 - j1 )

/a (j + k) =

P (j + k) P (j )

ïï^j I/a (j )

(1 - Cj+i )

(1 - )

vi=1 V c

(A (/a ))(j + k + 1) = Cj /a (j + k + 1),

it implies that

(cj - Cj+k+1 )/a(J + k + 1)

P(j + i)\ p aj+iP(j + k + 1)^^ (1 -P(j W A

j + k + 1 Vl=1 (1 - "j)

/A (j )

E. Pazouki, B. Yousefi

_ /ß(j + k + 1)\ p k+1 i=1

ß(j )

ß (j + k + 1)p

n (1 - Cj+i)

n (ci - Cj+i)

fA (j )

ß (j )P

(1 - Cj+i ) (1 - ^ )

fA(j).

Thus /a (z) = °=0 /a (j)zj, is a formal power series such that A(/a) = A(/a). For every j £ N, put e = 1 — 1+2> 0, so there exists natural number N such that

Cn > 1 - e =

1 + Cj'

ß (n + 1)p ß(n)p

for all n > N1. Let m > 0 such that j + m > N1, thus

Cj+m + 1 1 > 1 cj

1+ Cj'

1 - Cj

1 - Cj+m+1 < TT^J

Consider

ifA(j + m + 1)|pß(j + m + 1)p _ ß(j + m + 1)2p / 1 - Cj+m+1

ifA (j + m)|p ß(j + m)p

ß (j + m)2p

1 Cj+m + 1

Cj + m+1 Cj

cj + m+1

The formal power series ^fc=1 A(k)zk is in Hp(P). Therefore (ii) holds.

Note that £P(k) < to, it is clear that f0(z) = £P(k)pzk e Lp(y), thus

(Afo))(n) = £

akß(n)p

ß(k)p

fc=0 = ß (n)p,

so (iii) holds. Now the proof is complete. □

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The spectra and eigenvectors for the weighted mean matrix operator 15

[8] E. Pazouki, B. Yousefi, Compactness of the weighted mean operator matrix on weighted Hardy spaces, Int. J. 1 Appl. Math. 25 (4) (2012) 495-502.

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