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 34 ( F ) Version I Difference Sequence Spaces of Second order Defined by a Sequence of Moduli Thus sup k ≥ N f k ( | u k ( 4 2 x i k − 4 2 x k ) | ) p k H ≤ α, since ( x i ) ∈ ` ∞ ( u, 4 2 , F, p ) and F = ( f k ) is continuous it follows that x ∈ ` ∞ ( u, 4 2 , F, p ) Thus ` ∞ ( u, 4 2 , F, p ) is complete. Let 0 < p k ≤ q k < ∞ for each k. Then we have c 0 ( u, 4 2 , F, p ) ⊆ c 0 ( u, 4 2 , F, q ) Let x ∈ c 0 ( u, 4 2 , F, p ) that is lim k →∞ ( f k ( | u k ( 4 2 x k ) | )) p k = 0 This implies that f k ( | u k ( 4 2 x k ) | ) ≤ 1 for sufficiently large k , since modulus function is non decreasing. Hence we get lim k →∞ ( f k ( | u k ( 4 2 x k ) | )) q k ≤ lim k →∞ ( f k ( | u k ( 4 2 x k ) | )) p k = 0 Therefore x ∈ c 0 ( u, 4 2 , F, q ) . (a) Let 0 < inf p k ≤ p k ≤ 1. Then we have c 0 ( u, 4 2 , F, p ) ⊆ c 0 ( u, 4 2 , F ) . (b) Let 1 ≤ p k ≤ sup k p k < ∞ . Then we have c 0 ( u, 4 2 , F ) ⊆ c 0 ( u, 4 2 , F, p ) . (a) Let x ∈ c 0 ( u, 4 2 , F, p ) , that is lim k →∞ ( f k ( | u k ( 4 2 x k ) | )) p k = 0 Since 0 < inf p k ≤ p k ≤ 1, lim k →∞ ( f k ( | u k ( 4 2 x k ) | )) ≤ lim k →∞ ( f k ( | u k ( 4 2 x k ) | )) p k = 0 Proof. Proof. Theorem 2.3. Theorem 2.4. N otes