By ω, the authors denote the space of all real or complex valued sequences. Any vector subspace of ω is called a sequence space.

We shall write l_∞ ,c and c₀ for the sequence spaces of all bounded, convergent and null sequences, respectively, which are BK- spaces with the same norm given by

for all k  ∈  N. Also, by l_p(0 < p < ∞) we denote the sequence space of all sequences associated with _p -absolutely convergent series. It is known that l_p is a complete p- normed space and a BK- space in the cases of 0 < p < 1 and 1 ≤ p < ∞ with respect to the usual p- norm and l_p - norm defined by

and

A sequence space X with a linear topology is called a K- space provided each of the maps p_n : X → C defined by p_n (x) = x_n is continuous for all n ∈ N, where C denotes the complex field and N = {0,1,2.....} A K- space X is called an FK- space provided X is a complete linear metric space. An FK- space whose topology is normal is called a BK- space.

Let X and Y be sequence spaces and A = (a_nk) be an infinite matrix of real or complex numbers (a_nk) , where n, k ∈ N. Then, we say that A defines a matrix mapping from X into Y if for every sequence x = (x_k)∈ X the sequence Ax = {A_n(x)}, the A - transform of x, exists and is in Y, where

(1)

By (X,Y), we denote the class of all infinite matrices that map X into Y. Thus, A ∈ (X,Y) if and only if the series on the right side of (1) converges for each n∈N and every x∈X  and A_x∈Y for all x∈X.

For a sequence space X, the matrix domain of an infinite matrix A in X is defined by

(2)

which is a sequence space.

We shall write e^(k) for the sequence whose only non-zero is a 1 in the k^th place for each k∈ N

Throughout this paper, let λ=(λ_k)^∞_k=0 be a strictly increasing sequence of positive reals tending to ∞ , that is

We shall use the convention that any term with a negative subscript is equal to zero, e.g., λ_-1=0 and x_-1 = 0.

We define the infinite matrix Δ=(λ_nk)^∞_n,k=0 by

(4)

for all k∈  N. Then, for any sequence x=(x_k)∈ω, the Λ - transform of x is the sequence Λ(x)={Λ_n(x)}, where

(5)

for all n∈ N. It is obvious that by (4) that the matrix Λ=(λ_nk)is a triangle, i.e., λ_nn ≠ 0 and λ_nn = 0 for all k > n(n∈N) . The idea of constructing a new sequence by means of the matrix domain of a particular limitation method has recently been studied by several authors, e.g., Altay and Başar [2], Mursaleen et all [3,14], Mursaleen and Noman [15,16,17], Malkowsky [12], Malkowsky and Savaş [13].

In this section the authors introduce a new sequence space, as the sets of all sequences whose Λ - transforms is in the space l_p(u) where 0 < p < ∞ and (u_n) is a sequence of positive real numbers, that is

With the notation of (2), we can redefine the space l^λ_p(u) as follows:

l^λ_p(u) = l_p(Λ, u).

• If 0 < p < 1, then  l^λ_p(u) is a complete p- normed space with the p- norm
• If 1 ≤ p < ∞, then l^λ_p(u) is a BK- space with the norm

In the present section, the authors give some inclusion relations concerning l^λ_p the space and l_p.

Lemma 2.1.

(6)

(7)

hold, where S(x)={S_n(x)} is the sequence defined by

Lemma 2.2.

It is obvious that Lemma 2.2 still holds if the sequence is replaced by

Let u = (u_n) be a bounded sequence of positive real numbers. Then, we have the following theorems, lemmas and n corollaries.

Theorem 2.3.

Then, we have for every n ∈ N that

This shows that Λ(y) = x, hence Λ(y)∈ l_p\ l_q and since u = (u_n) is a bounded sequence we have y is in l^λ_q(u), but not in  l^λ_p(u).  Therefore, the inclusion l^λ_p(u) ⊂ l^λ_q(u) is strict. This completes the proof.

Lemma 2.4.

S(x) (l_p)

So, we have S(x)∈ l_p.

Conversely, let x ∈ l^λ_p(u) Then, we have by the hypothesis that S(x)∈ l_p. It follows by (6) that

Lemma 2.5.

Then, we have Λ(x)∉ l_p and since u = (u_n) is a bounded sequence we have x∈ l^λ_p(u). Thus, the sequence x is in l_p, but not in l^λ_p(u).

On the other hand, let 1 ≤ p < ∞ and define the sequence y=(y_k) by

for all k∈ N. Since 1/λ∉ l_p , we have y∉ l_p . Then, we have for every n ∈ N that

and hence

Then, we have Λ(y)∈ l_p and y∈ l^λ_p(u).  Thus the sequence y is in l^λ_p(u), but not in l_p where 1 ≤ p < ∞.

Lemma 2.6.

Proof.

Then, we have 1/λ∈ l_p and this completes the proof.

Lemma 2.7  If  1/λ∈ l₁ , then

Theorem 2.8.

Proof.

Conversely, suppose that  1/λ∈ l₁ Then M = by Lemma 2.7. Let x = (x_k)∈ l₁ be given. Then, we have

Since u = (u_n) is a bounded sequence we have X∈ l^λ_p(u) Hence, the inclusion l₁ ⊂  l^λ₁(u).

Corollary 2.9.If 1/λ∈ l₁ , then the inclusion l_p ⊂  l^λ_p(u) holds for 1 ≤  p < ∞.

Proof. The inclusion trivally holds for p = 1 which is obtained by Theorem 2.8. Let 1 < p < ∞ and take any x=(x_k)∈ l_p Then, for every n∈ N, we have by applying the Hölder's inequality that

Hence, we have that

where M= by Lemma 2.7. This shows that x∈  l^λ_p(u).

Corollary 2.10.

Proof.

Further, we have for every fixed k, n∈ N that

Thus, we obtain that

Therefore, we have e^(k)l^λ_p(u).

Lemma 2.11.

Theorem 2.12. The inclusion l_p⊂ l^λ_p(u) strictly holds if and only if 1/λ∈ l_p and where 1  ≤ p < ∞.

Proof. Suppose that the inclusion l_p ⊂ l^λ_p(u) is strict, where 1 ≤ p < ∞. Then, the necessity of the first condition is immediate by Lemma 2.6. Further, since the inclusion l^λ_p(u)⊂ l_p cannot be hold, Lemma 2.4 implies the existence of a sequence x∈ l^λ_p(u) such that Since x∈ l^λ_p(u), we have Thus, it follows by applying the Minkowski's inequality that  This means that and since {S_n(x)}∉ l_p it follows by the relation (7) that and hence Thus, we have by Lemma 2.2 to the necessity of the second condition

Conversely, since 1/λ∈ l_p we have by Corollary 2.10 that the inclusion l_p ⊂ l^λ_p(u) holds. Further, since we obtain by Lemma 2.2 that

Thus, it follows by Lemma 2.11 that there is a subsequence λ^(p) = λ(_kr)^∞_r=0 of the sequence λ = λ(_k) depend on p, such that k_r+1 - k_r ≥ 2  that for all r ∈ N  and 

(8)

Let us now define the sequence y = (y_k) for every k,rN by

 

Then, it is clear that y∉ l_p On the other hand, we have for every n,r∈ N that

 

This and (8) imply that Λ(y)∈ l_p and hence Y∈ l^λ_p(u). Thus, the sequence y is in l^λ_p(u) but not in l_p.

Theorem 2.13. The equality l^λ_p(u) = l_p holds if and only if where 1 ≤ p < ∞

Proof. The necessity follows from Theorem 2.12. For, if the equality holds, then the inclusion l_p ⊂  l^λ_p(u) holds and hence 1/λ∈l_p by Lemma 2.6. Further, since the inclusion l_p ⊂ l^λ_p(u) cannot be strict, we have Theorem 2.12 that and Hence

Conversely, suppose that     Then, there exists constant a >1   such that for all n∈ N. This shows that  1/λ∈ l₁ which leads us with Corollary 2.9 to the inclusion l_p⊂  l^λ_p(u) holds where 1 ≤ p < ∞ .   On the other hand, we have by Lemma 2.2 (b) that .

Let X ∈  l^λ_p(u), then  Thus, we obtain by the relation (7) that Therefore, we have by Lemma 2.4 that the inclusion l^λ_p(u)⊂ l_p. By combining the inclusion   we get the equality