sequence spaces de ned by fibonacci matrix

General Letters in Mathematics Vol. 6, No. 2, June 2019, pp.45-60

e-ISSN 2519-9277, p-ISSN 2519-9269

Available online at http:// www.refaad.com

https://doi.org/10.31559/glm2019.6.2.1

Sequence Spaces Defined by Fibonacci Matrix

M. KUCUKASLAN ∗1, B. ARIS 2

1 Mersin University Faculty of Science, Department of Mathematics 33343 Mersin, TURKEY.2 Istanbul University Faculty of Science, Department of Mathematics 34452 Istanbul, TURKEY.

[email protected],[email protected]

Abstract. In this paper, by using well known Fibonacci numbers, so far not described in the literature a new regular matrix

F = (fnk) is defined and compared with well known matrix transformations. By using this new matrix, Fibonacci sequence

space c0(F ), c(F ), l∞(F ) and lp(F ) (1 ≤ p <∞) are introduced. In addition to examining the properties of the new sequence

spaces some results which contains comparison of lp(F ) (1 ≤ p <∞) with other summability methods are given. Finally, α, β

and γ duals of c0(F ), c(F ), l∞(F ) are characterized.

Keywords: Fibonacci numbers, Matrix transformation, Regular matrix, Sequence space, Summability method.2010 MSC No: 40A05,40C05, 40D05

1 Introduction

Let w be the space of all real valued sequences. Each subspaces of w is called a sequence space. The subspacesc0, c, lp (1 ≤ p <∞) and l∞ are the sequence spaces of null, convergent, p-absolutely convergent series and bounded,respectively. That is,

c0 := {x = (xk) ∈ w : limk→∞

xk = 0},

c := {x = (xk) ∈ w : limk→∞

xk exists}

and

lp := {x = (xk) ∈ w :

∞∑k=1

|xk|p <∞}, (1 ≤ p <∞),

l∞ := {x = (xk) ∈ w : supk∈N|xk| <∞}.

∗Corresponding author. M. KUCUKASLAN and B. ARIS 1 [email protected]

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46 M. KUCUKASLAN and B. ARIS

The α, β and γ duals of a sequence space X ⊂ w are respectively defined as follow:

Xα := {a = (ak) ∈ w : |∑k

akxk| is convergent, ∀ x = (xk) ∈ X},

Xβ := {a = (ak) ∈ w :∑k

akxk is convergent, ∀ x = (xk) ∈ X},

Xγ := {a = (ak) ∈ w :∑k

akxk is bounded, ∀ x = (xk) ∈ X}.

Let X and Y be two sequence spaces and let A = (ank) be an infinite matrix such that ank ∈ R, for all n, k ∈ N.

We recall the matrix mapping by A : X 7→ Y , for every x = (xk) in X, Ax = (An(x)) in Y . The followingexpression

An(x) :=

∞∑k=1

ankxk, n ∈ N,

is called the A-transformation of the sequence x = (xk). Moreover, the domain XA of a matrix A is defined by

XA = {x ∈ w : Ax ∈ X}.

It is known that if A is an infinite triangle matrix, then the spaces XA and X are isometrically isomorphic [2].

Now, let A = (ank) be an infinite matrix and following conditions are given

supn∈N

∑k

|ank| <∞, (1)

limn→∞

ank = 0 for each k ∈ N, (2)

∃αk ∈ C, limn→∞

ank = αk for each k ∈ N, (3)

limn→∞

∑k

ank = 0, (4)

∃α ∈ C, limn→∞

∑k

ank = α, (5)

supK⊂F

∑n

|∑k∈K

ank| <∞, (6)

where F finite subset of N [5].

The following Lemma characterization of the matrix transformation was given between some sequence spaces.

Lemma 1.1. [15] The following statements hold:(a) A = (ank) ∈ (c0, c0) if and only if (1) and (2) hold.(b) A = (ank) ∈ (c0, c) if and only if (1) and (3) hold.(c) A = (ank) ∈ (c, c0) if and only if (1), (2) and (4) hold.(d) A = (ank) ∈ (c, c) if and only if (1), (3) and (5) hold.(e) A = (ank) ∈ (c0, l∞) if and only if (1) holds.(f) A = (ank) ∈ (c0, l1) if and only if (6) holds.

Lets pay attention that when α = 1 in (1), (2) and (5), we get the well known Silverman-Toeplitz Theorem in[15] which gives the necessary and sufficient conditions for a matrix to be regular.

Let {pn} be a sequence of non-negative numbers which are not all 0 and put

Pn := p1 + p2 + ...+ pn; p1 > 0.

Sequence Spaces Defined by Fibonacci Matrix 47

Definition 1.2. [14] The transformation

tn :=pnx1 + pn−1x2 + ...+ p1xn

Pn

is called the Norlund mean of (xk). Matrix representation of Norlund mean (N, p) is given as follow:

ank :=

{ pn−k+1

Pn, k ≤ n,

0, otherwise.

Definition 1.3. [14] The transformation given by

tn :=p1x1 + p2x2 + ...+ pnxn

Pn

is called the Riesz mean of (xk). Matrix representation of the (R, p) is given by

ank :=

{ pkPk, k ≤ n,

0, otherwise.

The transformations (N, p) and (R, p) are regular if and only if Pn →∞, (n→∞) [14].

The Cesaro mean (C, 1) is a special case of both the Norlund and the Riesz means with pn = 1 for all n and itsmatrix representation is

Cnk :=

{1n , k ≤ n,0, otherwise.

Definition 1.4. [1] Let λ = {λ(n)} be a strictly increasing sequence of positive integers. Cλ-transformation of asequence x = (xn) is defined by

tn :=x1 + x2 + ...+ xλ(n)

λ(n).

Definition 1.5. [14] A matrix A = (am,k) is called (M) matrix if A is triangular and the inequality

|n∑k=1

am,kxk| ≤ K|n∑k=1

an′,kxk|

holds for some n′, n′ = n′(n) (0 ≤ n′ ≤ n), (n = 1, 2, 3, ...) and for all m (m ≥ n).

The number n′ depends on n and (xn) but not m. The regular matrices are not contain to the class (M).

Consider Cesaro matrix, if k < n+ 1, then

1

n+ 1

k∑m=0

sm ≤1

k + 1

k∑m=0

sm

holds. So, Cesaro matrix is an (M) matrix [14].

Theorem 1.6. [14] Let A = (am,n) and B = (bm,n) be regular triangular matrices and A be an (M) matrix.Then, if

m∑n=1

| bm,nam,n

− bm,n+1

am,n+1| < M,

it follows that B is a stronger than A.


Theorem 1.7. [14] If the matrix A = (am,n) is triangular and satisfies the conditions:

am,k = 0, 0 ≤ am,kan,k

≤ K (7)

andam,kan,k

≥ am,k+1

an,k+1(8)

hold for 0 ≤ k ≤ n ≤ m, then A is an (M) matrix.

The sequence of Fibonacci numbers is

(fn) = (0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, ...).

It is clear from the elements of (fn) that f0 = 0, f1 = 1 and recurrence formula

fn := fn−1 + fn−2, n ≥ 2,

holds.

Some properties of Fibonacci numbers are given as follows:

limn→∞

fn+1

fn=

1 +√

5

2(golden ratio),

Fn :=

n∑k=1

fk = fn+2 − 1, n ∈ N,

∞∑k=1

1

fkconverges,

n∑k=1

fk2 = fnfn+1.

Fibonacci sequence have may applications in science, art and architecture. In recent years, due to the interestingstructure of the Fibonacci sequence, it has been found some application in the theory of summability and sequencespaces ([2], [6], [3], [4], [13]).

For example; M. Karakas in [12] by considering the Fibonacci matrix as follows:

fnk :=

{ f2kf2n+1−1 , 1 ≤ k ≤ n,

0, otherwise,(9)

established the sequence spaces c0(F ), c(F ), l∞(F ) and lp(F ) (1 ≤ p <∞).

Also, this type of sequence spaces were defined by E. E. Kara in [9] with a different type Fibonacci matrix asfollows:

fnk :=

− fn+1

fn, k = n− 1,

fnfn+1

, k = n,

0, 0 ≤ k < n− 1 or k > n.

(10)

In [9], Fibonacci difference sequence spaces lp(F ), (1 ≤ p <∞) and l∞(F ) are defined as:

lp(F ) := {x = (xk) ∈ w :∑n

| fnfn+1

xn −fn+1

fnxn−1|p <∞},

l∞(F ) := {x = (xk) ∈ w : supn∈N| fnfn+1

xn −fn+1

fnxn−1| <∞}.


Later, in [5], the spaces c0(F ) and c(F ) were defined by using Fibonacci difference matrix in (10) as:

c0(F ) := {x = (xk) ∈ w : limn→∞

(fnfn+1

xn −fn+1

fnxn−1) = 0},

and

c(F ) := {x = (xk) ∈ w : ∃l ∈ C such that limn→∞

(fnfn+1

xn −fn+1

fnxn−1) = l}.

Furthermore, E. E. Kara and M. Ilkkan in [10] defined a matrix T = (tnk) as follows:

tnk :=

tn, 1 ≤ k = n,

− 1tn, k = n− 1,

0, 0 ≤ k < 1 or k > n.

For tn = fnfn+1

in the matrix T = (tnk), the matrix given in (10) has been obtained. Thus, a generalization of

Fibonacci difference sequence has been given. In [11], using the matrix T , the sequence spaces c0(T ) and c(T ) weredefined and α, β and γ duals of these spaces were determined.

In the light of these studies, here we define a new Fibonacci matrix F = (fnk) which is different (9) and (10) asfollows:

fnk :=

fk

fn+2−1 , 1 ≤ k ≤ n,

0, k > n.

(11)

It is clear from (11) that the Fibonacci matrix F = (fnk) is a lower triangle matrix.

In this paper, some relations between the matrix in F = (fnk) and some other known matrices has beencompared and some inclusion results are given. Sequence spaces c0(F ), c(F ), lp(F ) (1 ≤ p < ∞) and l∞(F ) aredefined.

It is shown that the sequence space X (= c0, c, lp (1 ≤ p <∞) or l∞) is isometrically isomorphic with thesequence space X(F ), respectively.

Finally, α, β and γ duals of the spaces c0(F ), c(F ), lp(F ) (1 ≤ p <∞) and l∞(F ) have been determined.

2 Inclusion results for the matrix F := (fnk)

In this section, F -transformation of x = (xk) is going to defined and its relationship with other well known matrixtransformations (such as Norlund, Riesz and Cesaro) will be examined.

Definition 2.1. Let F = (fnk) be a Fibonacci matrix given in (11). A real valued sequence y = (yn) is calledF -transform of a sequence x = (xn) if

yn := F (xn) =1

fn+2 − 1

n∑k=1

fkxk (12)

exists for all n ∈ N.

Definition 2.2. Let x = (xn) be a real valued sequence and l ∈ R. It is called F -convergent to l if (F (xn− l))n∈Nconvergent to 0.

Theorem 2.3. The Fibonacci matrix F = (fnk) is a regular summability method if and only if fn+2 − 1 → ∞when n→∞.


Proof Assume that F = (fnk) is a regular summability method. Then, from Silverman-Toeplitz theorem in[15].

limn→∞

fnk = limn→∞

fkfn+2 − 1

= 0

must be satisfied. So, fn+2 − 1→∞, n→∞.

Conversely, suppose fn+2 − 1 → ∞ when n → ∞. We should check the conditions of Silverman- ToeplitzTheorem. Then,

∞∑k=1

fkfn+2 − 1

=

n∑k=1

fkfn+2 − 1

= 1,

For every k ∈ N, by hypothesis

limn→∞

fnk = limn→∞

fkfn+2 − 1

= 0,

and

limn→∞

∞∑k=1

fnk = limn→∞

n∑k=1

fnk = limn→∞

n∑k=1

fkfn+2 − 1

= 1

hold. Thus, F = (fnk) is a regular summability method.

Theorem 2.4. The Fibonacci matrix F = (fnk) is an (M) matrix.

Proof It is enough that the matrix F satisfies (7) and (8). The following inequalities

0 ≤ fkfm+2 − 1

· fn+2 − 1

fk=fn+2 − 1

fm+2 − 1≤ fn+2

fm+2≤ 1

and

fk+1

fm+2−1fk+1

fn+2−1

=fn+2 − 1

fm+2 − 1· fkfk

=fk

fm+2 − 1· fn+2 − 1

fk=am,kan,k

holds. Thus, the Fibonacci matrix F is (M) matrix.

Definition 2.5. Let a = (an) and b = (bn) be two real valued sequences. They are called equivalent if there existpositive real numbers m and M such that inequality

m.an ≤ bn ≤M.an

holds for all n ∈ N. It is denoted by a � b.

The following Theorem gives a relation between F and (R, p):

Theorem 2.6. Let F = (fnk) be a Fibonacci matrix, x = (xn) be a real valued sequence. Then, xn → l (F ) ifand only if xn → l (R, p), for any sequence (pn) such that pn � fn for all n ∈ N.

Proof Suppose that xn → l (F ). So, we have

limn→∞

1

fn+2 − 1

n∑k=1

fk(xk − l) = 0.

Under the assumption on (pn), following inequality

1

Pn

n∑k=1

pk(xk − l) ≤1

Pn

n∑k=1

M.fk(xk − l) ≤M

m

1

fn+2 − 1

n∑k=1

fk(xk − l)


holds and by the same way we have also

m

M

1

fn+2 − 1

n∑k=1

fk(xk − l) ≤1

Pn

n∑k=1

pk(xk − l). (13)

Since xk → l (F ), then (13) and (13) give that

limn→∞

1

Pn

n∑k=1

pk(xk − l) = 0.

This complete the proof.

Sufficiency of the Theorem 2.6 can be proven in the same analogy. So, it is omitted here.

Theorem 2.7. Let A = (ank) be any regular matrix and assume that

n∑k=1

|ank − fnk| → 0, (n→∞).

Then, xn → l (A) if and only if xn → l (F ), for any bounded sequence.

Proof Let x = (xn) be a bounded sequence. For any n, following inequality

|(Ax)n − (Fx)n| = |n∑k=1

ankxk −n∑k=1

fnkxk|

≤n∑k=1

|ank − fnk||xk|

≤ ||x||n∑k=1

|ank − fnk|

holds. Therefore, if xn → l (A), then we have

|(Fx)n − l| ≤ |(Fx)n − (Ax)n|+ |(Ax)n − l| → 0, n→∞. (14)

Similarly, if xn → l (F ), then we have

|(Ax)n − l| ≤ |(Ax)n − (Fx)n|+ |(Fx)n − l| → 0, n→∞. (15)

So, (14) and (15) complete the proof.

Now let’s define associate matrix F = (fnk) as follows:

fnk :=

fn−kfn+2−1 , k ≤ n,

0, k > n.

(16)

The matrix F = (fnk) can be expressed as a Norlund type Fibonacci matrix while the matrix F = (fnk) can beexpressed as a Riesz type Fibonacci matrix.

Next theorems is about F . But, firstly, we remind a Lemma in [14].

Lemma 2.8. If (N, pn) is a regular Norlund matrix, the series∑∞n=1 pnx

n−1 and∑∞n=1 Pnx

n−1 are convergentfor all x, |x| < 1.


Since the matrix F = (fnk) is a Norlund typed matrix, then the above lemma is valid for F = (fnk).

From the definition of the matrix F = (fnk), we can take Fn instead of Fn.

Since the following series are convergent for all |x| < 1,

f(x) :=

∞∑n=1

fnxn−1, F (x) :=

∞∑n=1

Fnxn−1,

then, the following series

k(x) =p(x)

f(x)=P (x)

F (x), k(x) :=

∞∑n=1

knxn−1,

h(x) =f(x)

p(x)=F (x)

P (x), h(x) :=

∞∑n=1

hnxn−1.

are also convergent for all x, |x| < 1.

Theorem 2.9. (N, pn) ⊆ (F ) if and only if there exists M > 0 such that

|k1|Pn + |k2|Pn−1 + ...+ |kn|P1 ≤M.Fn,

for every n holds and

limn→∞

knFn

= 0,

satisfied.

Proof The idea in [14] will be used in the proof. For sufficiency: Let (un) and (vn) be the (N, p) and (F )transformattion of a real valued sequence (sn), respectively. Then, we have following equality

∞∑n=1

Fnvnxn−1 =

∞∑n=1

Fn(fns1 + fn−1s2 + ...+ f1sn)

Fnxn−1

= (f1s1)x0 + (f2s1 + f1s2)x1 + (f3s1 + f2s2 + f1s3)x2 + ...

+ (fns1 + fn−1s2 + ...+ f1sn)xn−1 + ...

= s1(f1x0 + f2x

1 + f3x2 + ...) + s2(f1x+ f2x

2 + f3x3 + ...)

+ s3(f1x2 + f2x

3 + f3x4 + ...) + ...+ sn(f1x

n−1) + ...

= s1x0(f1x

0 + f2x1 + f3x

2 + ...+ fnxn−1) + s2x

1(f1x

0 + f2x1+

f3x2 + ...+ fn−1x

n−2)+ ...+ snxn−1(f1x

0) + ...

=( ∞∑n=1

snxn−1)( ∞∑

n=1

fnxn−1) = s(x)f(x). (17)

Using the same method, we get the equality

∞∑n=1

Pnunxn−1 = s(x)p(x). (18)

By the hypothesis, we know that

f(x) = k(x)p(x)

and

f(x)s(x) = k(x)p(x)s(x).


satisfied.By the Cauchy product of series and the equalities (17) and (18), we have

∞∑n=1

Fnvnxn−1 =

∞∑n=1

n∑m=1

kn−m+1Pmumxn−1

and the equality of these series gives that

Fnvn = knP1u1 + kn−1P2u2 + ...+ k1Pnun

for all n ∈ N. So,

vn =

∞∑n=1

anmum,

anm =

{ kn−m+1PmFn

, m ≤ n,0, m > n.

The matrix (anm) is a regular matrix. Indeed,

limn→∞

anm = limn→∞

kn−m+1PmFn

= limn→∞

kn−m+1PmFn−m+1

= 0,

∞∑m=1

|anm| =|k1|Pn + ...+ |kn|P1

Fn≤M (n ∈ N)

and

limn→∞

n∑m=1

anm =k1Pn + ...+ knP1

Fn=FnFn

= 1.

Thus, from the regularity of (anm) the sufficiency of the Theorem 2.9 is proved. The necessary part of Theorem 2.9can also be shown easily with the conditions given in the theorem.

Now, let us consider a sequence (gn) defined by gn := gn−1 + gn−2, n ≥ 2 such that g0 = a and g1 = b for anypositive a, b ∈ R. This sequence can be called Fibonacci type sequence. So, Fibonacci typed matrix G = (gnk) isdefined as

gnk :=

gn−kGn

, k ≤ n ,

0 k > n

where Gn := g0 + g1 + ...+ gn = gn+2 − b and the matrix G = (gnk) is a regular matrix.

Theorem 2.10. For any two regular (F, fn) and (G, gn), there exists a regular (H, hn) such that (H, hn) ⊇(F, fn) and (H, hn) ⊇ (G, gn).

Proof Denote by {t1n} and {t2n} the (F, fn) and (G, gn) transformation of the sequence {sn}, respectively. Leta sequence {hn} as follows:

hn := fng0 + fn−1g1 + ...+ f0gn,


for all n ∈ N. The sequence {sn} is transformed into {tn} by (H, hn), where:

tn =h1sn + h2sn−1 + ...+ hns1

h1 + h2 + ...+ hn

=(f1g0 + f0g1)sn + (f2g0 + f1g1 + f0g2)sn−1 + ...+ (fng0 + ...+ f0gn)s1

(f1g0 + f0g1) + (f2g0 + f1g1 + f0g2) + ...+ (fng0 + ...+ f0gn)

=f1(g0sn + ...+ gn−1s1) + f2(g0sn−1 + ...+ gn−2s1)...+ fn(g0s1)

f1(g0 + ...+ gn) + ...+ fng0

=f1Gn( g0sn+...+gn−1s1

Gn) + ...+ fnG0( g0s1G0

)

f1Gn + ...+ fnG0

=f1Gnt

2n + ...+ fnG0t

20

f1Gn + ...+ fnG0

Clearly, (H, hn) mean of (sn) is the A = (amn) transformation of {t2n}, where

anm :=

fn−mGm∑nk=0 fn−kGk

, m ≤ n ,

0, m > n.

Since the equality

n∑m=0

anm =

n∑m=0

|anm| = 1,

satisfies and for∑nk=1 fn−kGk > KFn, (K ≥ g0 > 0) and for all fixed n, the inequality

0 ≤ limn→∞

fn−mGm∑nk=0 fn−kGk

< limn→∞

fn−kGkKFn

= 0

satisfies, then A = (anm) is a regular matrix. this implies, if {t2n} converges, then {tn} converges to the same value.This implies that (H, hn) ⊇ (G, gn) and by similar way, (H, hn) ⊇ (F, fn) can be shown.

Definition 2.11. [14] The matrices A and B are said to be equivalent if the inclusions A ⊇ B and B ⊇ A hold.

Theorem 2.12. The regular means (F, fn) and (G, gn) are equivalent if and only if the associated series∑∞n=1 |kn| and

∑∞n=1 |hn| are converge.

Proof Suppose that (F, fn) and (G, gn) are equivalent. Then, there is an M > 0 such that 0 < k0Fn < MGnand 0 < h0Gn < MFn hold for all n ∈ N and the sequence ( FnGn ) and (GnFn ) are both bounded. In continuation of theproof, if it is followed way as in [14], then the proof is completed.

Definition 2.13. Let λ = λ(n) be a strictly increasing sequence of positive integers. Fλ-transform of a sequencex = (xn) is defined by

tn :=f1x1 + f2x2 + ...+ fλ(n)xλ(n)

fλ(n)+2 − 1.

Let’s remind notation o(1) before giving next theorem:f(n) = o(g(n)) means for all c > 0 there exists some k > 0 such that 0 ≤ f(n) < cg(n) for all n ≥ k. The value of kmust not depend on n, but may depend on c.

Theorem 2.14. Let λ = {λ(n)} and µ = {µ(n)} be a strictly increasing sequences of N. If limn→∞fλ(n)+2−1fµ(n)+2−1

=

1, then Fλ is equivalent to Fµ on l∞.


Proof Let x = (xn) be a bounded sequence and M(n) = max{{λ(n)}, {µ(n)}}, m(n) = min{{λ(n)}, {µ(n)}}.Since limn→∞

fλ(n)+2−1fµ(n)+2−1

= 1 and limn→∞fm(n)+2−1fM(n)+2−1

= 1, then, for any n,

∣∣∣∣(Fλx)n − (Fµx)n

∣∣∣∣ =

∣∣∣∣ 1

fµ(n)+2 − 1

µ(n)∑k=1

fkxk −1

fλ(n)+2 − 1

λ(n)∑k=1

fkxk

∣∣∣∣=

∣∣∣∣ 1

fM(n)+2 − 1

M(n)∑k=1

fkxk −1

fm(n)+2 − 1

m(n)∑k=1

fkxk

∣∣∣∣=

∣∣∣∣ 1

fM(n)+2 − 1

m(n)∑k=1

fkxk +1

fM(n)+2 − 1

M(n)∑k=m(n)+1

fkxk −1

fm(n)+2 − 1

m(n)∑k=1

fkxk

∣∣∣∣=

∣∣∣∣m(n)∑k=1

fkxk

(1

fM(n)+2 − 1− 1

fm(n)+2 − 1

)+

1

fM(n)+2 − 1

M(n)∑k=m(n)+1

fkxk

∣∣∣∣≤ ||x||∞

(m(n)∑k=1

fk|fm(n)+2 − fM(n)+2

(fM(n)+2 − 1)(fm(n)+2 − 1)|+

M(n)∑k=m(n)+1

fk|1

fM(n)+2 − 1|)

≤ ||x||∞(

(fM(n)+2 − fm(n)+2)(fm(n)+2 − 1)

(fM(n)+2 − 1)(fm(n)+2 − 1)+

(fM(n)+2 − fm(n)+2)

fM(n)+2 − 1

)≤ 2||x||∞

((fM(n)+2 − 1)− (fm(n)+2 − 1)

(fM(n)+2 − 1)

)≤ 2||x||∞

(1−

fm(n)+2 − 1

fM(n)+2 − 1

)= o(1)

holds. Thus, if x is Fλ-summable to L, then we have

0 ≤ |(Fµx)n − L| ≤ |(Fµx)n − (Fλx)n|+ |(Fλx)n − L| = o(1) + o(1) = o(1).

Similarly, if x is Fµ-summable to L, then we have

0 ≤ |(Fλx)n − L| ≤ |(Fλx)n − (Fµx)n|+ |(Fµx)n − L| = o(1) + o(1) = o(1).

This complete the proof of theorem.

Theorem 2.15. Cesaro matrix is stronger than Fibonacci matrix F = (fnk).

Proof We should check that the inequality in Theorem 1.6 holds. For this, if we take Cnk (Cesaro matrix)instead of B and F instead of A, we have

n∑k=1

|fn+2 − 1

nfk− fn+2 − 1

nfk+1| =

n∑k=1

fn+2 − 1

n(

1

fk− 1

fk+1)

=fn+2 − 1

n

n∑k=1

(1

fk− 1

fk+1).

It is known that n ≤ fn+2 − 1 holds for all n ∈ N. So, the following inequality

fn+2 − 1

n

n∑k=1

(1

fk− 1

fk+1) ≤

n∑k=1

(1

fk− 1

fk+1)

≤ (1

f1− 1

f2+

1

f2− 1

f3+ ...+

1

fn− 1

fn+1) ≤ 1− 1

fn+1

satisfied. This inequality gives that the condition of Theorem 1.6 is satisfied. Thus, F ⊂ Cnk.

Remark 2.16. The converse of Theorem 2.15 is not true, in general.


Let us consider x = (xn) as

xn :=(−1)n

n

The sequence (Cnx) = ( 1n

∑nk=1

(−1)kk ) is convergent, but if we consider F -transformation of (xn), then the sequence

(Fnx) = 1fn+2−1

∑nk=1

fk{−1}kk is not convergent. Thus, it seen that F * Cnk.

3 Fibonacci Sequence Spaces

In this section, Fibonacci sequence spaces will be defined and some properties of them will be given.

X denotes the sequence spaces c0, c, l∞ or lp (1 ≤ p <∞) and

(yn) := (Fx)n =( 1

fn+2 − 1

n∑k=1

fkxk)

is F -transform of x = (xn).

Definition 1. The set

X(F ) = {(xn) ∈ w : (yn) ∈ X}

is called Fibonacci sequence space.

Theorem 3.1. Let X be a sequence space. Then, X(F ) is a BK space with the following norms(i) If X = c0, c or l∞,

||x||X(F ) := sup{|yn| : n ∈ N}.

(ii) If X = lp, (1 ≤ p <∞)

||x||X(F ) = (

∞∑k=1

|yk|p)1p .

Proof Since the matrix F is triangle, then the result of A. Wilansky in [16] gives that the space X(F ) is a BKspace.

Theorem 3.2. The space X(F ) is isometrically isomorphic to the space X.

Proof We should find an isometric isomorphism between the spaces X(F ) and X. For this, let us define afunction P by

P : X(F ) 7→ X, Px = y

where y = (yn) = (Fx)n = 1fn+2−1

∑nk=1 fkxk.

It is clear that P is a linear function. Since Px = 0 implies x = 0, then P is an injective function.

Now, we must show that P is surjective. If we consider a sequence x = (xn) as

xn =fk+2 − 1

fkyk −

fk+1 − 1

fkyk−1,

then for every y = (yn) ∈ X, there exists a sequence x = (xn) ∈ X(F ) such that Px = y. So, the equality

(Fx)n =1

fn+2 − 1

n∑k=1

fkxk =1

fn+2 − 1

n∑k=1

[(fk+2 − 1)yk − (fk+1 − 1)yk−1] = yn.

holds and it gives that x ∈ X(F ) as Fx ∈ X. Hereby, P is surjective.Also, P is norm preserving function, that is,

||Px||X = ||y||X = ||Fx||X = ||x||X(F ),

for any x ∈ X(F ). So, P is an isometry.


Theorem 3.3. Let X be any of the spaces c0, c and l∞. Then, the inclusion X ⊂ X(F ) holds.

Proof Since the matrix F is regular, then the inclusion is clear for X = c0 and c. Let us take x = (xn) ∈ l∞,there is a constant M > 0 such that |xn| ≤M for n ∈ N. Then, following inequality holds:

|Fn(x)| = | 1

fn+2 − 1

n∑k=1

fkxk| ≤1

fn+2 − 1

n∑k=1

fk|xk| ≤M

fn+2 − 1

n∑k=1

fk ≤M.

This inequality gives that x = (xn) ∈ l∞(F ).Now, if we take x = (xn) ∈ lp (1 ≤ p <∞), then there exists a constant M > 0 such that

||x||pp =

∞∑n=1

|xn|p < M.

We should show that x = (xn) ∈ lp(F ). From the Holder inequality and the definition of (fnk), we have[ n∑k=1

fkfn+2 − 1

|xk|]p≤

[( n∑k=1

(fk

fn+2 − 1)q) 1q

·( n∑k=1

|xk|p) 1p]p

≤( n∑k=1

(fk

fn+2 − 1)q) pq

·( n∑k=1

|xk|p)

where 1p + 1

q = 1.On the right side of the inequality, since( n∑

k=1

(fk

fn+2 − 1)q) pq

=

(f1q + f2

q + ...+ fnq

(f1 + f2 + ...+ fn)q

) pq

≤ 1,

then we have |Fn(x)|p ≤∑nk=1 |xk|p. It gives that

∞∑n=1

|Fn(x)|p ≤∞∑n=1

|x|p <∞.

Thus, it was shown that x ∈ lp(F ).

Theorem 3.4. Let F be Fibonacci matrix. Then, following inclusion

c0(F ) ⊂ c(F ) ⊂ l∞(F )

strictly holds.

Proof From Theorem 3.3, inclusion is obvious. Let us consider x = (xn) = (1, 1, 1, ...). Since the matrix F isregular, Fnx is convergent. So, (xn) ∈ c(F ). On the other hand, (Fnx) /∈ c0 and (xn) /∈ c0(F ).

If we take the sequence (xn) = ( (−1)n(fn+2+fn+1−1)fn

), for all n ∈ N. Then, we have

Fn(x) =1

fn+2 − 1

n∑k=1

fk(−1)k(fk+2 + fk+1 − 1)

fk.

We see that the sequence

Fnx =

1, n is even ,

0, n is odd

is in l∞. So, (xn) ∈ l∞. But, since (Fnx) /∈ c, then (xn) /∈ c(F ). and when n is odd, the sequence (Fn(x)) convergesto −1. We see that (xn) /∈ c(F ), but the sequence (xn) ∈ l∞(F ).


4 α, β and γ Duals of the Sequence Spaces X

In this section, we determine the α, β and γ duals of the sequence spaces c0(F ), c(F ), lp(F ) (1 ≤ p < ∞) for theFibonacci matrix F .The duals α, β and γ of X can be given as follows [16]:

Xα =⋂

(xn)∈X

[xn]α,

Xβ =⋂

(xn)∈X

[xn]β ,

Xγ =⋂

(xn)∈X

[xn]γ ,

where [xn]α, [xn]β and [xn]γ are sets of the sequences a = (an) ∈ w which satisfy

|∑

anxn| is convergent,∑anxn is convergent,∑anxn is bounded

respectively. Let us consider the sets

d1 = {a = (an) ∈ w : supK∈F

∑n

|∑k∈K

fkfn+2 − 1

ak| <∞}

d2 = {a = (an) ∈ w : supn∈N

n∑k=1

|n∑j=k

fjfn+2 − 1

aj | <∞}

d3 = {a = (an) ∈ w : limn→∞

n∑j=k

fjfn+2 − 1

aj exists for each n ∈ N}

d4 = {a = (an) ∈ w : limn→∞

n∑k=1

n∑j=k

fjfn+2 − 1

aj exists}.

Theorem 4.1. The following statements hold(a) {co(F )}α = {c(F )}α = d1.(b) {co(F )}β = d2 ∩ d3 and {c(F )}β = d2 ∩ d3 ∩ d4.(c) {co(F )}γ = {c(F )}γ = d2.

Proof As in [5] and [9], we define the matrix T = (tnk) as follows:

tnk :=

{ fkfn+2−1ak, k ≤ n ,

0, k > n .

for a given sequence a = (an) ∈ w. On the other hand, we know that since |∑anxn| < ∞ for every x = (xn) ∈

c0(F ), c(F ), then (yn) = (Fx)n ∈ c0, c(F ).By using the definition of α dual, we will show that (Ty)n ∈ l1 for (yn) ∈ c0, c.Then, we have

anxn =

n∑k=1

fkfn+2 − 1

ykak = (Ty)n (∀n ∈ N).

Thus, we see that (Ty)n ∈ l1 for (yn) ∈ c0, c if and only if (anxn) ∈ l1 for (xn) ∈ c0(F ), c(F ). This implies thata = (an) ∈ l1.

The other statements can be proved in a similar way.

For 1 < p ≤ ∞, let us find α, β and γ duals of the lp(F ).


Theorem 4.2. The α dual of the space lp(F ) is the set

d1 = {x = (xn) ∈ w : supK∈F

∑n

|∑k∈K

fkfn+2 − 1

xk|q <∞}

and β, γ duals are {lp(F )}β = d2 ∩ d3 and {l∞(F )}β = d3 ∩ d4 and {lp(F )}γ = d3, respectively.

Theorem 4.3. Let (xn) and (yn) be two sequence and (an) ∈ w be a dual sequence. If the sequences (xn) and(yn) are asymptotically equivalent,i.e, limn→∞

xnyn

= 1, then [xn]β = [yn]β

Proof Suppose that limn→∞xnyn

= 1 and (an) be arbitrary sequence in [xn]β . Then, for every (xn) ∈ X, the

sum∑∞n1anxn is convergent so that the partial sum sequence (sn) of converges to s ∈ R. For every ε > 0, there is

an n0 ∈ N such that ∀n ≥ n0, we have

|n∑k=1

akyk − s| = |n∑k=1

akxkykxk− s| < ε.

Thus, it is shown that (an) ∈ [yn]β . It can be easily proved the second part of the theorem.

Remark 4.4. The converse of Theorem 4.3 is not true, in general.

Let x = (xn) = ( 1n ), y = (yn) = ( 1

n2 ) and an = (−1)n, for ∀n ∈ N. Since the series∑∞n=1 (−1)n 1

n and∑∞n=1 (−1)n 1

n2 are convergent, then [xn]β = [yn]β . On the other hand, it seen that

limn→∞

1n1n2

6= 1.

5 Conclusion

Since F = (fnk) is a regular matrix, then the sequence (Fnx) is convergent for a convergent sequence (xn). Thus, asequence space containing the space c is obtained and many studies with regular matrices can be reexamined for theF matrix. Also, statistical convergence of the sequence (Fnx) can be studied .Lastly, if a arbitrary normed spaces X is taken instead of the space X = (c0 , c, lp (1 ≤ p < ∞), l∞), then it isobtained generalization of the Fibonacci sequence space X(F ). Given results in this paper can be investigated forthe general Fibonacci sequence space.

Also, in the paper [14], some properties of Fibonacci Frames, orthonormal bases and Fibonacci Riesz bases werestudied and dual Fibonacci Frames were characterized. Results given in [7] can be re-examined for the Fibonaccimatrix F = (fnk) defined in this paper.

ACKNOWLEDGEMENTS. The authors would like to thank the referees for their valuable comments andsuggestions that extend to the paper.

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sequence spaces de ned by fibonacci matrix

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