Some multiplier difference sequence spaces defined by a sequence ...

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178 Hemen DuttaProposition 1 The classes of sequences c 0 (F, ∆ n (m) , Λ, p), c(F, ∆n (m), Λ, p)and l ∞ (F, ∆ n (m), Λ, p) are linear spaces.Theorem 1 For Z = l ∞ , c and c 0 , the spaces Z(F, ∆ n (m), Λ, p) are paranormedspaces, paranormed bywhere H = max(1, sup p k ).k≥1g(x) = sup((f k (|∆ n (m)λ k x k |)) p k) 1 H ,k≥1Proof. Clearly g(x) = g(−x); x = θ implies g(θ) = 0.Let (x k ) and (y k ) be any two sequences belong to anyone of the abovespaces. Then we have,g(x + y) = sup(f k (|∆ n (m) λ kx k + ∆ n (m) λ ky k |)) p kHk≥1≤ sup(f k (|∆ n (m) λ kx k |)) p kH + sup(f k (|∆ n (m) λ ky k |)) p kHk≥1k≥1This implies thatg(x + y) ≤ g(x) + g(y).The continuity of the scalar multiplication follows from the followinginequality:in |α|.g(αx) = sup((f k (|∆ n (m) αλ kx k |)) p 1k ) Hk≥1= sup((f k (|α||∆ n (m) λ kx k |)) p 1k ) Hk≥1≤ (1 + [|α|])g(x), where [|α|] denotes the largest integer containedHence the spaces Z(F , ∆ n (m) , Λ, p), for Z = l ∞, c, c 0spaces paranormed by g.are paranormedTheorem 2 For Z = l ∞ , c and c 0 , the spaces Z(F, ∆ n (m), Λ, p) are completeparanormed spaces, paranormed by g as defined above.
Some multiplier difference sequence spaces 179Proof. We prove the result for the case l ∞ (F , ∆ n (m), Λ, p) and for the otherspaces it will follow on applying similar arguments.Let (x i ) be any Cauchy sequence in l ∞ (F , ∆ n (m), Λ, p). Let ε > 0 begiven and let ε 1 = sup(f k (ε)) p kH . Then there exists a positive integer n 0ksuch thatg(x i − x j ) < ε 1 , for all i, j ≥ n 0 .Using the definition of paranorm, we getsup((f k (|∆ n (m)λ k x i k − ∆ n (m)λ k x j k |))p k) 1 H < ε1 , for all i, j ≥ n 0 .k≥1It follows that|∆ n (m)λ k x i k − ∆ n (m)λ k x j k | < ε, for each k ≥ 1 and for all i, j ≥ n 0.Hence (∆ n (m) λ kx i k ) is a Cauchy sequence in C for all k ∈ N. This impliesthat (∆ n (m) λ kx i k ) is convergent in C for all k ∈ N. Let limi→∞ ∆n (m) λ kx i k = y kfor each k ∈ N.Let k = 1, we have(1) limi→∞∆ n (m)λ 1 x i 1 = limi→∞Similarly we have,n∑( ) n(−1) v λ 1−mv x i 1−mv = y 1vv=0(2) limi→∞∆ n (m)λ k x i k = limi→∞λ k x i k = y k , for k = 1, ..., nmThus from (1) and (2) we have lim x i 1+nm exists. Let lim x i 1+nm = x 1+nm .i→∞ i→∞Proceeding in this way inductively, we have lim x i k = x k exists for eachi→∞
Page 1 and 2: General Mathematics Vol. 17, No. 3
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