Global Journal of Science Frontier Research, F: Mathematics and Decision Science, Volume 21 Issue 4

© 2021 Global Journals 1 Global Journal of Science Frontier Research Volume XXI Issue IV Year 2021 32 ( F ) Version I Difference Sequence Spaces of Second order Defined by a Sequence of Moduli Let( x i ) be a cauchy sequence in ` ∞ ( u, 4 2 , F ) for each i ∈ N. Let r, x 0 be fixed.Then for each rx 0 > 0 there exists a positive integer N such that || x i − x j || 4 2 < rx 0 for all i, j ≥ N Using the definition of norm, we get sup k ≥ 0 f k ( | u k ( 4 2 x i k − 4 2 x j k ) | || x i − x j || 4 2 ) ≤ α, for all i, j ≥ N ie, f k ( | u k ( 4 2 x i k − 4 2 x j k ) | || x i − x j || 4 2 ) ≤ α, for all i, j ≥ N Hence we can find r > 0 with f k ( rx 0 2 ) ≥ α such that f k ( | u k ( 4 2 x i k − 4 2 x j k ) | || x i − x j || 4 2 ) ≤ f k ( rx 0 2 ) | u k ( 4 2 x i k − 4 2 x j k ) | || x i − x j || 4 2 ≤ rx 0 2 This implies that | u k ( 4 2 x i k − 4 2 x j k ) | ≤ rx 0 2 rx 0 = 2 Since u k 6 = 0 for all k, we have |4 2 x i k − 4 2 x j k | ≤ 2 for all i, j ≥ N Hence ( 4 2 x i k ) is a cauchy sequence in R. For each > 0 there exists a positive integer N such that |4 2 x i k − 4 2 x k | < for all i > N . Using the continuity of F = ( f k ) we can show that sup k ≥ N f k ( | u k ( 4 2 x i k − lim j →∞ 4 2 x j k ) | ) ≤ α, Thus sup k ≥ N f k ( | u k ( 4 2 x i k − 4 2 x k ) | ) ≤ α, since ( x i ) ∈ ` ∞ ( u, 4 2 , F ) and F = ( f k ) is continuous it follows that x ∈ ` ∞ ( u, 4 2 , F ) Thus ` ∞ ( u, 4 2 , F ) is complete. Proof. N otes