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1658 BRUNO DE MALAFOSSE (iv) If α ∈ Γ , there is an integer q ≥ 1 for which So there is a real M ′ > 0 for which k ≥ q +1 implies α k−1 α k ≤ χ
CALCULATIONS ON SOME SEQUENCE SPACES 1659 X = (x n,0 ) n≥1 in s α . Then there is a unique scalar x 1 such that n−1 ∑ x n,0 = x 1 − α k . (2.25) k=1 So ∣ ⎛ ⎞ ∣ xni , 0 n i −1 = 1 ∑ ⎝ α x1 − α k ⎠ ni ∣ α →∞ as i →∞, (2.26) ni ∣ k=1 and X ∉ s α , which is contradictory. We conclude that each of the properties e ∈ Ker∆ + ⋂ s α and e ∉ Ker∆ + ⋂ s α is impossible and ∆ + is not bijective from s α into itself. This proves the lemma. Lemma 2.6. For every X ∈ c 0 , Σ + (∆ + X) = X and for every X ∈ cs, ∆ + (Σ + X) = X. Proof. It can easily be seen that [ Σ + ( ∆ + X )] n = ∞ ∑ ( xm −x m+1 ) = xn ∀X ∈ c 0 , m=n [ ∆ + ( Σ + X )] ∞ n = ∑ ∞∑ x m − x m = x n m=n m=n+1 ∀X ∈ cs. (2.27) We can assert the following result, in which we put α + = (α n+1 ) n≥1 and s ◦∗ α (∆ + ) = s ◦ α(∆ + ) ⋂ s ◦ α. Note that from (2.5) we have s ∗ α ( ∆ + (e) ) = s ∗ α ( ∆ + ) = s α ( ∆ + ) ⋂ sα . (2.28) Theorem 2.7. (i) (a) s α (∆) = s α if and only if α ∈ Ĉ 1 , (b) sα(∆) ◦ = sα ◦ if and only if α ∈ Ĉ 1 , (c) s α (c) (∆) = s α (c) if and only if α ∈ Ĉ. (ii) (a) α ∈ Ĉ 1 if and only if s α +(∆ + ) = s α and ∆ + is surjective from s α into s α +, (b) α ∈ Ĉ+ 1 if and only if s∗ α (∆ + ) = s α and ∆ + is bijective from s α into s α , (c) α ∈ Ĉ+ 1 implies that s◦∗ α (∆ + ) = sα ◦ and ∆ + is bijective from sα ◦ into sα. ◦ (iii) α ∈ Ĉ+ 1 if and only if s α(Σ + ) = s α and s α (Σ + ) = s α implies sα ◦ (Σ+ ) = sα ◦ . Proof. (i) has been proved in [8]. (ii)(a) Sufficiency. If ∆ + is surjective from s α into s α +, then for every B ∈ s α + solutions of ∆ + X = B in s α are given by the n∑ x n+1 = x 1 − b k n = 1,2,..., (2.29) k=1
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