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

182 Hemen DuttaThen using (3), we getlim (f k(|∆ n (m)λ k x k |)) p k= 0k→∞Thus c 0 (F , ∆ n−1(m) , Λ, p) ⊂ c 0(F , ∆ n (m), Λ, p).The inclusion is strict follows from the following example.Example 3 Let m = 3, n = 2, f k (x) = x 10 , for all k ≥ 1 and x ∈ [0, ∞)and p k = 5 for all k odd and p k = 3 for all k even. Consider the sequencesΛ = ( 1k 3 ) and x = (k 4 ). Then ∆ 2 (3) λ kx k = 0, for all k ∈ N. Hencex ∈ c 0 (F, ∆ 2 (3) , Λ, p). Again we have ∆1 (3) λ kx k = −3, for all k ∈ N. Hencex does not belong to c 0 (F, ∆ 1 (3), Λ, p). Thus the inclusion is strict.Theorem 5 The spaces c 0 (F, ∆ n (m) , Λ, p), c(F, ∆n (m) , Λ, p) and l ∞(F, ∆ n (m) ,Λ, p) are not monotone and as such are not solid in general.Proof. The proof follows from the following example.Example 4 Let n = 2, m = 3, p k = 1 for all k odd and p k = 2 for all keven and f k (x) = x 4 , for all x ∈ [0, ∞) and k ∈ N. Then ∆ 2 (3) λ kx k =λ k x k −2λ k−3 x k−3 + λ k−6 x k−6 , for all k ∈ N. Consider the J th step space of asequence space E defined as, for (x k ), (y k ) ∈ E J implies that y k = x k for kodd and y k = 0 for k even. Consider the sequences Λ = (k 3 ) and x = ( 1k 2 ).Then x ∈ Z(F, ∆ 2 (3) , Λ, p) for Z = l ∞, c and c 0 , but its J th canonical preimagedoes not belong to Z(F, ∆ 2 (3) , Λ, p) for Z = l ∞, c and c 0 . Hence thespaces Z(F, ∆ 2 (3) , Λ, p) for Z = l ∞, c and c 0 are not monotone and as suchare not solid in general.