Bi¸sgin Journal of Inequalities and Applications (2016)2016:304 DOI 10.1186/s13660-016-1252-4

R E S E A R C H Open Access The binomial sequence spaces which include the spaces p and ∞ and geometric properties Mustafa Cemil Bi¸sgin*

*Correspondence: [email protected] Abstract Department of Mathematics, r,s r,s Faculty of Arts and Sciences, Recep In this work, we introduce the binomial sequence spaces bp and b∞ which include r,s r,s Tayyip Erdogan˘ University, Zihni the spaces p and ∞, in turn. Moreover, we show that the spaces bp and b∞ are Derin Campus, Rize, 53100, Turkey BK-spaces and prove that these spaces are linearly isomorphic to the spaces p and ∞, respectively. Furthermore, we speak of some inclusion relations and give the r,s Schauder basis of the space bp . Lastly, we determine the α-, β-, and γ -duals of those r,s spaces and give some geometric properties of the space bp . MSC: 40C05; 40H05; 46B45 Keywords: matrix transformations; matrix domain; Schauder basis; α-, β-and γ -duals; matrix classes

1 The basic information and notations The set of all real (or complex) valued sequences is symbolized by w which becomes a vector space under point-wise addition and scalar multiplication. Any vector subspace of w is called a sequence space. The spaces of all bounded, null, convergent, and absolutely

p-summable sequences are denoted by ∞, c, c,andp,respectively,where≤ p < ∞.

A Banach sequence space is called a BK-space provided each of the maps pn : X −→ C

defined by pn = xn is continuous for all n ∈ N []. By considering the notion of BK-space,

one can say that the sequence spaces ∞, c,andc are BK-spaces according to their usual   | | sup-norm defined by x ∞ = supk∈N xk and p is a BK-space according to its p-norm defined by

   ∞ p   | |p x p = xk , k=

where  ≤ p < ∞.

For an arbitrary infinite matrix A =(ank) of real (or complex) entries and x =(xk) ∈ w, the A-transform of x is defined by

∞ (Ax)n = ankxk (.) k=

© The Author(s) 2016. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, pro- vided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. Bi¸sgin Journal of Inequalities and Applications (2016)2016:304 Page 2 of 15

and is supposed to be convergent for all n ∈ N []. In terms of the ease of use, we prefer that the summation without limits runs from  to ∞.

Given two sequence spaces X and Y , and an infinite matrix A =(ank), the sequence space

XA is defined by   XA = x =(xk) ∈ w : Ax ∈ X (.)

which is called the domain of an infinite matrix A.Also,by(X : Y), we denote the class of

all matrices such that X ⊂ YA.Ifank =fork > n and ann =forall n, k ∈ N, an infinite – matrix A =(ank) is called a triangle. Also, a triangle matrix A uniquely has an inverse A which is a triangle matrix.

Let the summation matrix S =(snk) be defined as follows:  ,  ≤ k ≤ n, snk = , k > n

for all k, n ∈ N. Then the spaces of all bounded and convergent series are defined by means

of the summation matrix such that bs =(∞)S and cs = cS,respectively. The theory of matrix transformation was set in motion by the theory of summability which was developed by Cesàro, Norlund, Riesz, etc. By taking into account this theory,

many authors have constructed new sequence spaces. For example, (∞)Nq and cNq in [], r r Xp and X∞ in [], ap and a∞ in []. Furthermore, many authors have used especially the r r r r Euler matrix for defining new sequence spaces. These are e and ec in [], ep and e∞ in r r r r (m) r (m) r (m) r (m) []and[], e(), ec()ande∞()in[], e( ), ec( )ande∞( )in[], e(B ), r (m) r (m) r r r r r ec(B ), and e∞(B )in[], e(, p), ec(, p), and e∞(, p)in[], e(u, p)andec(u, p) in []. r,s r,s In this work, we introduce the binomial sequence spaces bp and b∞ which include the r,s r,s spaces p and ∞, in turn. Moreover, we show that the spaces bp and b∞ are BK-spaces and prove that these spaces are linearly isomorphic to the spaces p and ∞,respectively. Furthermore, we speak of some inclusion relations and give the Schauder basis of the r,s space bp . Lastly, we determine the α-, β-, and γ -duals of those spaces and give some r,s geometric properties of the space bp .

2 The binomial sequence spaces which include the spaces p and ∞ r,s r,s In this part, we define the binomial sequence spaces bp and b∞ which include the spaces p and ∞, respectively. Furthermore, we show that those spaces are BK-spaces and are

linearly isomorphic to the spaces p and ∞. Also, we show that the binomial sequence r,s ≤ ∞ space bp is not a Hilbert space except the case p =,where p < .  r,s r,s Let r, s be nonzero real numbers with r + s = . Then the binomial matrix B =(bnk)is defined as follows:    n sn–krk,≤ k ≤ n, br,s = (s+r)n k nk , k > n

for all k, n ∈ N.Forsr > , one can easily check that the following properties hold for the r,s r,s binomial matrix B =(bnk): (i) Br,s < ∞, Bi¸sgin Journal of Inequalities and Applications (2016)2016:304 Page 3 of 15

r,s lim →∞ ∈ N (ii) n bnk =(each k ), r,s (iii) limn→∞ k bnk =. Thus, the binomial matrix is regular whenever sr > . Here and in the following, unless stated otherwise, we suppose that sr >. r,s r,s By taking into account the binomial matrix B =(bnk), the binomial sequence spaces r,s r,s bp and b∞ are defined by   p   n n br,s = x =(x ) ∈ w : sn–krkx < ∞ ,≤ p < ∞, p k (s + r)n k k n k=

and   n r,s  n n–k k b∞ = x =(xk) ∈ w : sup s r xk < ∞ . ∈N (s + r)n k n k=

r,s r,s By considering the notation of (.), the binomial sequence spaces bp and b∞ can be r,s r,s redefined by the matrix domain of B =(bnk) as follows:

r,s r,s bp =(p)Br,s and b∞ =(∞)Br,s .(.)

Let us define a sequence y =(yk) as follows:

  k k Br,sx = y = sk–jrjx (.) k k (s + r)k j j j=

for all k ∈ N.ThissequencewillbefrequentlyusedastheBr,s-transform of x. We would like to touch on a point, if we take s + r =,weobtaintheEulermatrixEr = r r,s r,s (enk). So, the binomial matrix B =(bnk) generalizes the Euler matrix. Now, we want to continue with the following theorem which is needed in the next.

r,s r,s Theorem . The binomial sequence spaces bp and b∞ are BK-spaces according to their norms defined by

     ∞  p  r,s  r,s p x r,s = B x = B x bp p n n=

and     r,s  r,s   r,s x b∞ = B x ∞ = sup B x n , n∈N

where  ≤ p < ∞.

Proof We know that the sequence spaces p and ∞ are BK-spaces with their p-norm and sup-norm,respectively,where≤ p < ∞.Furthermore,(.) holds and the binomial r,s r,s matrix B =(bnk) is a triangle matrix. By taking into account these three facts and The- r,s r,s orem .. of Wilansky [], we conclude that the binomial sequence spaces bp and b∞ are BK-spaces, where  ≤ p < ∞. This completes the proof of the theorem.  Bi¸sgin Journal of Inequalities and Applications (2016)2016:304 Page 4 of 15

r,s r,s Theorem . The binomial sequence spaces bp and b∞ arelinearlyisomorphictothe sequence spaces p and ∞, in turn, where  ≤ p < ∞.

Proof To refrain from the usage of similar statements, we prove the theorem for only the r,s ≤ ∞ sequence space bp ,where p < . For the proof of the theorem, we need to show the r,s existence of a linear bijection between the spaces bp and p.LetL be a transformation r,s −→ r,s r,s such that L : bp p, L(x)=B x. By the definition of the binomial sequence space bp , ∈ r,s r,s ∈ we conclude that, for all x bp , L(x)=B x p. Furthermore, it is obvious that L is a linear transformation and x =  whenever L(x) = . Therefore, L is injective.

For given y =(yk) ∈ p,letusdefineasequencex =(xk)suchthat

 k k x = (–s)k–j(s + r)jy k rk j j j=

for all k ∈ N.Thenweget    r,s  x r,s = B x bp p

   ∞  p r,s p = B x n n=   ∞ p    n n p = sn–krkx (s + r)n k k n= k=   ∞ p    n n k k p = sn–k (–s)k–j(s + r)jy (s + r)n k j j n= k= j=

   ∞ p p = |yn| n=   = y p   = L(x) < ∞. p

∈ r,s Hence, we conclude that L is norm preserving and x bp ,namelyL is surjective. As a r,s consequence, L is a linear bijection. This means that the spaces bp and p are linearly r,s ∼ ≤ ∞ isomorphic, that is, bp = p,where p < . This completes the proof of the theorem. 

r,s Theorem . The binomial sequence space bp is not a Hilbert space except the case p =, where  ≤ p < ∞.

r,s Proof Let p = . Remembering Theorem .,onecansaythatb is a BK-space according to its -norm defined by

     ∞    r,s  r,s  x r,s = B x = B x . b  n n= Bi¸sgin Journal of Inequalities and Applications (2016)2016:304 Page 5 of 15

Moreover, this norm can be generated by an inner product such that

   r,s r,s x r,s = B x, B x  . b

r,s Therefore, b is a Hilbert space. Now, we assume that  ≤ p < ∞ and p = . We define two sequences y =(yk)andz =(zk) as follows:

–s + k(r + s) s k– s + k(r + s) s k– y = – and z =– – k r r k r r

for all k ∈ N.Thenweobtain

    p +   y + z r,s + y – z r,s == = y r,s + z r,s . bp bp bp bp

r,s Thus, the norm of the binomial sequence space bp does not satisfy the parallelogram equality. As a consequence, the norm cannot be generated by an inner product, that is, r,s  the binomial sequence space bp is not a Hilbert space whenever p =.Thiscompletesthe proof of the theorem. 

3 The inclusion relations and Schauder basis In this part, we speak of some inclusion relations and give the Schauder basis for the bi- r,s ≤ ∞ nomial sequence space bp ,where p < .

r ⊂ r,s r ⊂ r,s r r Theorem . The inclusions ep bp and e∞ b∞ strictly hold, where ep and e∞ are the Euler sequence spaces which include the spaces p and ∞, respectively.

r r,s r r,s Proof If r + s =,onecaneasilyseethatE = B . Therefore, the inclusion e∞ ⊂ b∞

holds. Suppose that  < r <ands = . Let us now consider a sequence x =(xk)suchthat   k ∈ N k ∈ ∞ r k ∈ ∞ xk =(–r ) for all k .Thenitisclearthatx =(xk) = ((– r ) )/ , E x = ((– – r) )/  r,s k ∈ ∞ ∈ r,s \ r and B x =((+r ) ) .Asaresultofthis,x =(xk) b∞ e∞. This shows that the in- r r,s clusion e∞ ⊂ b∞ is strictly. We can prove the other part of the theorem by using a similar technique. This completes the proof of the theorem. 

⊂ r,s ≤ ∞ Theorem . The inclusion p bp is strict, where  p < .

Proof First we assume that  < p < ∞. From the definition of the space p,wewrite  p |xk| < ∞ k

for all x =(xk) ∈ p.Forgivenanarbitrarysequencex =(xk) ∈ p, by taking into account the equality (.) and the Hölder inequality, we obtain

 k p p  k Br,sx = sk–jrjx k (s + r)k j j j=     p–  p k k k k ≤ |s|k–j|r|j × |s|k–j|r|j|x |p |s + r|k j j j j= j= Bi¸sgin Journal of Inequalities and Applications (2016)2016:304 Page 6 of 15

 k k = |s|k–j|r|j|x |p |s + r|k j j j=

k k j k s r p = |xj| , j s + r s j=

where  ≤ p < ∞. And

   k k j r,s p k s r p B x ≤ |xj| k j s + r s k k j=

 ∞ k j p k s r = |xj| j s + r s j k=j

 s + r p = |xj| . s j

r,s ∈ ∈ r,s If we consider the comparison test, we conclude that B x p,namelyx bp .Asacon- ⊂ r,s ∞ sequence p bp ,where

r,s Theorem . The spaces bp and ∞ overlap but these spaces do not include each other, where  ≤ p < ∞.

k ∈ k ∈ r,s r,s Proof It is obvious that v = ((–) ) ∞ and v = ((–) ) bp .So,thespacesbp and ∞ overlap, where  ≤ p < ∞. Here, we consider the sequences e = (,,,...)and u =(uk) s s k ∈ N | | ∈ ∞ defined by uk =(–r ) for all k ,where r > . Then we conclude that e but r,s ∈ ∈ r,s ∈ r,s ∈ ∈ r,s B e = e / p,thatis,e / bp and u / ∞ but B u =(,,,...) p,namelyu bp .As ∈ \ r,s ∈ r,s \ r,s aconsequence,e ∞ bp and u bp ∞. On account of this, bp and ∞ do not in- clude each other, where  ≤ p < ∞. This completes the proof of the theorem. 

⊂ r,s r,s ⊂ r,s ≤ ∞ Theorem . The inclusions ∞ b∞ and bp b∞ are strict, where  p < .

Proof The inequality

k  k k–j j   r,s ≤  x b∞ = sup s r xj x ∞ ∈N (s + r)k j k j=

r,s holds for all x ∈ ∞. In this way, the inclusion ∞ ⊂ b∞ holds. Now, we consider the se- s+r k ∈ N ∈ ∞ quence v =(vk)definedbyvk =(– r ) for all k . Then we conclude that v =(vk)/ r r,s k ∈ ∞ ∈ r,s ∞ ⊂ r,s but B v = ((– r+s ) ) ,namelyv =(vk) b∞. Therefore, the inclusion b∞ strictly holds. ∈ r,s ≤ ∞ For given x =(xk) bp ,where p < , by taking into account Theorem . and the ⊂ r,s ∈ ∈ r,s r,s ⊂ inclusion p ∞,weconcludethatB x ∞,namelyx b∞.Thus,theinclusionbp Bi¸sgin Journal of Inequalities and Applications (2016)2016:304 Page 7 of 15

r,s ∈ r,s \ r,s r,s ⊂ r,s b∞ holds. Also, it is clear that e b∞ bp . Hence, the inclusion bp b∞ is strict. This completes the proof of the theorem. 

Now, let us continue with the definition of the Schauder basis of a normed space. Let

(X, ·X)beanormedsequencespaceandd =(dk)beasequenceinX.Ifforeveryx ∈ X,

there exists a unique sequence of scalars λ =(λk)suchthat      n  lim x – λkdk = n→∞  k= X

then d =(dk)iscalledaSchauderbasisforX [].

r,s (k) Theorem . Let μk = {B x}k be given for all k ∈ N. We define the sequence g (r, s)= { (k) } r,s gn (r, s) n∈N of the elements of the binomial sequence space bp as follows:  ,  ≤ n < k, g(k)(r, s)=  n  n n–k k ≥ rn k (–s) (s + r) , n k

(k) for all fixed k ∈ N. Then the sequence {g (r, s)}k∈N is a Schauder basis for the binomial r,s ∈ r,s sequence space bp , and every x bp has a unique representation of the form  (k) x = μkg (r, s), k

where  ≤ p < ∞.

∈ r,s ≤ ∞ Proof Let x =(xk) bp be given, where  p < . For all non-negative integer m,we define

m [m] (k) x = μkg (r, s). k=

r,s r,s [m] Then, if we apply the binomial matrix B =(bnk)tox ,wewrite

m m  r,s [m] r,s (k) r,s (k) B x = μkB g (r, s)= B x ke k= k=

and     ≤ ≤ r,s [m] ,  n m, B x – x n = r,s (B x)n, n > m

for all m, n ∈ N.

For any given  > , there exists a non-negative integer m such that

∞  p p  Br,sx ≤ n  n=m+ Bi¸sgin Journal of Inequalities and Applications (2016)2016:304 Page 8 of 15

for all m ≥ m.Thus,

     ∞  p  [m] r,s p x – x r,s = B x bp n n=m+    ∞  p ≤ r,s p B x n n=m+  ≤ <  

for all m ≥ m. This shows us that  (k) x = μkg (r, s). k

Lastly, we should show the uniqueness of this representation. For this purpose, assume that  (k) x = λkg (r, s). k

r,s Since the linear transformation L defined from bp to p in the proof of Theorem . is continuous, we have      r,s r,s (k) (k) B x n = λk B g (r, s) n = λken = λn k k

r,s for every n ∈ N, which contradicts the fact that (B x)n = μn for every n ∈ N. Therefore, ∈ r,s  every x bp has a unique representation. This completes the proof of the theorem.

r,s ≤ ∞ From Theorem .,weknowthatbp is a Banach space, where  p < . If we consider this fact and Theorem ., we can give the next corollary.

r,s ≤ ∞ Corollary . Thebinomialsequencespacebp is separable, where  p < .

4Theα-,β-, and γ -duals r,s In this part, we determine the α-, β-, and γ -duals of the binomial sequence spaces bp and r,s b∞,where≤ p < ∞. Now, we start with a definition. The multiplier space of the sequence spaces X and Y is denoted by M(X, Y)anddefinedby   M(X, Y)= y =(yk) ∈ w : xy =(xkyk) ∈ Y for all x =(xk) ∈ X .

By taking into account the definition of a multiplier space, the α-, β -,and γ -duals of a sequence space X are defined by

α β γ X = M(X, ), X = M(X, cs)andX = M(X, bs),

respectively. Bi¸sgin Journal of Inequalities and Applications (2016)2016:304 Page 9 of 15

For use in the next lemma, we now give some properties:  q sup |ank| < ∞,(.) ∈N n k

sup |ank| < ∞,(.) n,k∈N

lim ank = ak for each k ∈ N,(.) n→∞

  q

sup ank < ∞, (.) ∈F K k n∈K

  lim |ank| = lim ank ,(.) n→∞ n→∞ k k  sup |ank| < ∞, (.) ∈N k n

F N   ≤∞ where is the collection of all finite subsets of , p + q =and

Lemma . (see []) Let A =(ank) be an infinite matrix, then the following hold:

(i) A =(ank) ∈ ( : ) ⇔ (.) holds,

(ii) A =(ank) ∈ ( : c) ⇔ (.) and (.) hold,

(iii) A =(ank) ∈ ( : ∞) ⇔ (.) holds, ∈ ⇔   ≤∞ (iv) A =(ank) (p : ) (.) holds with p + q =and 

(viii) A =(ank) ∈ (∞ : ∞) ⇔ (.) holds with q =.

r,s r,s Theorem . Let v and v be defined as follows:     q r,s ∈ n n–k –n k ∞ v = a =(ak) w : sup (–s) r (r + s) an < ∈F k K k n∈K

and    r,s ∈ n n–k –n k ∞ v = a =(ak) w : sup (–s) r (r + s) an < . ∈N k k n

{ r,s}α r,s { r,s}α r,s ≤∞ Then b = v and bp = v , where 

Proof Let a =(an) ∈ w be given. Remembering the sequence x =(xn), which is defined in the proof of Theorem .,wehave

n n  a x = (–s)n–kr–n(r + s)ka y = Hr,sy n n k n k n k=

for all n ∈ N. Then, by considering the equality above, we deduce that ax =(anxn) ∈ ∈ r,s ∈ r,s r,s ∈  whenever x =(xk) b or x =(xk) bp if and only if H y  whenever y = Bi¸sgin Journal of Inequalities and Applications (2016)2016:304 Page 10 of 15

(yk) ∈  or y =(yk) ∈ p,respectively,where

   ∈ r,s α ⇔ n n–k –n k ∞ a =(an) b sup (–s) r (r + s) an < ∈N k k n

or

    q ∈ r,s α ⇔ n n–k –n k ∞ a =(an) bp sup (–s) r (r + s) an < , ∈F k K k n∈K

≤∞ { r,s}α r,s { r,s}α r,s ≤∞ respectively, where  < p . Therefore, b = v and bp = v ,where

r,s r,s r,s r,s r,s Theorem . Let v , v , v , v , and v be defined as follows:   ∞  j vr,s = a =(a ) ∈ w : (–s)j–kr–j(r + s)ka exists for each k ∈ N ,  k k j j=k  

n j r,s ∈ j–k –j k ∞ v = a =(ak) w : sup (–s) r (r + s) aj < , ∈N k n,k j=k   n r,s j j–k –j k v = a =(ak) ∈ w : lim (–s) r (r + s) aj  n→∞ k k j=k  ∞   j = (–s)j–kr–j(r + s)ka , k j k j=k   q n n j r,s ∈ j–k –j k ∞ ∞ v = a =(ak) w : sup (–s) r (r + s) aj < ,

n n j r,s ∈ j–k –j k ∞ v = a =(ak) w : sup (–s) r (r + s) aj < . ∈N k n k= j=k

Then the following equalities hold: { r,s}β r,s ∩ r,s (I) b = v v , { r,s}β r,s ∩ r,s ∞ (II) bp = v v , where 

Proof To avoid the repetition of similar statements, we give the proof of the theorem for r,s ∞ only the sequence space bp ,where

Let a =(ak) ∈ w be given. By considering the sequence x =(xk), which is used in the proof of Theorem .,weobtain

  n n  k k a x = (–s)k–j(r + s)jy a k k rk j j k k= k= j=   n n j = (–s)j–kr–j(r + s)ka y k j k k= j=k  r,s = G y n

∈ N r,s r,s for all n ,wherethematrixG =(gnk )isdefinedby   n j (–s)j–kr–j(r + s)ka ,≤ k ≤ n, gr,s = j=k k j nk , k > n

for all k, n ∈ N.Then: ∈ ∈ r,s r,s ∈ ∈ (II) ax =(akxk) cs whenever x =(xk) bp if and only if G y c whenever y =(yk) p, ∞ ∈{ r,s}β r,s ∈ where  < p < . This fact shows that a =(ak) bp if and only if G (p : c), where 

n n q j j–k –j k sup (–s) r (r + s) aj < ∞ (.) ∈N k n k= j=k

and

∞  j (–s)j–kr–j(r + s)ka exists for each k ∈ N, k j j=k

∞   { r,s}β r,s ∩ r,s where  < p < and p + q =.Asaresultofthis,weobtain bp = v v ,where 

r,s 5 Geometric properties of the binomial sequence space bp r,s In this part, we give some geometric properties of the binomial sequence space bp .Letus start with some notions.

Let (X, ·X ) be a Banach space. Then X is said to have the Banach-Saks property, if every bounded sequence u =(un)containsasubsequencev =(vn) such that the Cesàro  n means n+ k= vk are norm convergent []. X is said to have the weak Banach-Saks property, if every weakly null sequence u =(un)  n contains a subsequence v =(vn) such that the Cesàro means n+ k= vk are norm conver- gent []. Bi¸sgin Journal of Inequalities and Applications (2016)2016:304 Page 12 of 15

X is said to have Banach-Saks type p, if every weakly null sequence u =(un)hasasubse-

quence v =(vn)suchthat,forsomeM >,   n       ≤ p  vk M(n +) k= X

for all n ∈ N,where

where  ≤  ≤ [].

Theorem . (see []) A Banach space X has the weak fixed point property, if X provides the condition  

R(X)=sup lim inf xn + x <, n→∞

where the supremum is taken over all weakly null sequences (xn) of the unit ball and all points x of the unit ball.

r,s Theorem . The binomial sequence space bp is of the Banach-Saks type p.

r,s r,s Proof Let (un) be a weakly null sequence in the B(bp )unitballofbp .Wesupposethat ≤  (n) is a sequence of positive numbers provided n  .Constructv = u =andv = ∈ N un = u. Then we can find an m such that    ∞      (i)  v(i)e  < . r,s i=m+ bp

w By virtue of un −→ implyingun −→  coordinatewise, we can find an n ∈ N such that     m   (i)  un(i)e  < , r,s i= bp

≥ as n n.Constructv = un . Then we can find an m > m such that    ∞      (i)  v(i)e  < . r,s i=m+ bp

If we use xn −→  coordinatewise one more time, we can find an n > n such that     m   (i)  un(i)e  < , r,s i= bp

as n ≥ n. Bi¸sgin Journal of Inequalities and Applications (2016)2016:304 Page 13 of 15

By continuing this method, we can constitute two increasing sequences (mk)and(nk) such that    m  k   (i)  un(i)e  < k r,s i= bp

for all n ≥ nk+ and    ∞      (i)  v(i)e  < k, r,s i=mk + bp

where vk = unk .Thus         m m ∞  n  n k– k      (i) (i) (i)   vk =  vk(i)e + vk(i)e + vk(i)e  r,s r,s k= bp k= i= i=mk–+ i=mk + bp     m  n k  n ≤  (i)   vk(i)e  + k r,s k= i=mk–+ bp k=

and    m p m p n k  n k  i i  v (i)e(i) = si–jrjv (j)  k  i k r,s (s + r) j k= i=mk–+ bp k= i=mk–+ j=

∞ p n   i i ≤ si–jrjv (j) ≤ n +. (s + r)i j k k= i= j=

Thus we obtain   n        ≤ p ≤ p  vk (n +) + (n +) . r,s k= bp

r,s As a consequence, the binomial sequence space bp is of the Banach-Saks type p.This completes the proof of the theorem. 

r,s We know from Theorem . that bp is linearly isomorphic to p.So,itisclearthat  r,s p R(bp )=R(p)= . By combining this fact and Theorem ., we can give the next theorem.

r,s Theorem . The binomial sequence space bp has the weak fixed point property, where 

  p Theorem . β r,s  ≤ p ≤  ≤ The inequality bp ( ) –[–( ) ] holds, where  .

Proof Let  ≤  ≤  be given. By assuming the inverse of the binomial matrix Br,s is D,we construct two sequences u and v as follows:

  p p  u = D – , D ,,,... ,   Bi¸sgin Journal of Inequalities and Applications (2016)2016:304 Page 14 of 15

  p p  v = D – , D – ,,,... .  

Then we obtain      r,s   r,s  B u = u r,s = and B v = v r,s =. p bp p bp

r,s r,s r,s ∈     r,s  This shows that u, v S(bp )and B u – B v p = u – v bp = . For  ≤ λ ≤ , we have      p  r,s r,s p λu +(–λ)v r,s = λB u +(–λ)B v bp p

 p  p =– + |λ –|  

and

  p  p  inf λu +(–λ)v r,s =– .(.) ≤λ≤ bp 

Thus, we obtain

    p p β r,s () ≤ – – . bp 

This completes the proof of the theorem. 

By using the equality (.), we find two more results.

r,s Corollary . β r,s  Since bp ( )=,the binomial sequence space bp is strictly convex.

r,s Corollary . β r,s  ≤  ≤ Since < bp ( ) , for < , the binomial sequence space bp is uni- formly convex.

6Conclusion r,s r,s r,s r,s By taking into account the binomial matrix B =(bnk), we conclude that B =(bnk)re- r r duces in the case r +s =toE =(enk) which is called the Euler matrix of order r. Therefore, r,s r,s our results obtained from the matrix domain of the binomial matrix B =(bnk) are more general and more extensive than the results on the matrix domain of the Euler matrix of r,s r,s order r. Furthermore, the binomial matrix B =(bnk) is not a special case of the weighed mean matrices. Thus, this paper has filled up a gap in the existent literature.

Competing interests The author declares that they have no competing interests.

Author’s contributions The author read and approved the final manuscript.

Acknowledgements I would like to express my thanks to the anonymous reviewers for their valuable comments.

Received: 13 August 2016 Accepted: 17 November 2016 Bi¸sgin Journal of Inequalities and Applications (2016)2016:304 Page 15 of 15

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