Properties of some new sets of sequences.

On the sets of sequences 19

2.3. Properties of some new sets of sequences.

In this subsection we shall characterize the sets E 1 µ, E 1 + µ for E ∈ s α , s ◦ α , s • α , and the sets w p α λ, w + p α λ and w ◦ + p α λ. In order to state some new results we need the following lemmas. First recall the well known result. L EMMA 1. A ∈ c , c if and only if A ∈ S 1 , lim n a nm = for each m ≥ 1. The next result has been shown in [11]. L EMMA 2. If 1 + is bijective from s α into itself, then α ∈ cs. We also need to state the following elementary result. L EMMA 3. We have 6 + 1 + X = X ∀ X ∈ c and 1 + 6 + X = X ∀ X ∈ cs. Put now w + p α λ = X ∈ s C + λ |X | p ∈ s α , w ◦ + p α λ = n X ∈ s C + λ |X | p ∈ s ◦ α o , see [11]. Letting β − = β n−1 n≥1 , with β = 1, for any β = β n n≥1 ∈ U +∗ , we can state the following results. T HEOREM 2. Let α ∈ U +∗ , λ, µ ∈ U and p 0. We successively have i a s α 1 µ = s α |µ| if and only if α ∈ c C 1 ; b s ◦ α 1 µ = s ◦ α |µ| if and only if α ∈ c C 1 ; c s • α 1 µ = s • α |µ| if and only if α ∈ b C. ii a s α 1 + µ = s α |µ| − if and only if α |µ| ∈ c C 1 ; b α n−1 α n µ n µ n−1 = o 1 implies s ◦ α 1 + µ = s ◦ α |µ| − ; c α |µ| ∈ c C + 1 if and only if s ∗ α 1 + µ = s α |µ| ; d α |µ| ∈ c C + 1 if and only if s ◦ ∗ α 1 + µ = s ◦ α |µ| . 20 B. de Malafosse iii a s α 6 + = s α if and only if α ∈ c C + 1 and s ◦ α 6 + = s ◦ α if and only if α ∈ c C + 1 . b α ∈ c C + 1 if and only if w + p α λ = s α|λ| 1 p , c if α ∈ c C + 1 , then w ◦ + p α λ = s ◦ α|λ| 1 p , d α |λ| ∈ c C 1 if and only if w p α λ = s α|λ| 1 p . e If α |λ| ∈ c C 1 , then w ◦ p α λ = s ◦ α|λ| 1 p . Proof. Assertion i has been proved in [10]. Throughout the proof of part ii we shall put β = α |µ| . Assertion ii a. First we have s α 1 + µ = s β 1 + . Indeed, X ∈ s α 1 + µ ⇔ D µ 1 + X ∈ s α ⇔ 1 + X ∈ s β ⇔ X ∈ s β 1 + . To get a, it is enough to show that β ∈ c C 1 if and only if s β 1 + = s β − . We assume that β ∈ c C 1 . From the inequality β n−1 β n ≤ 1 β n n X k=1 β k = O 1 , we deduce that β n−1 β n = O 1 and 1 + ∈ s β − , s β . Then for any given B ∈ s β the solutions of the equation 1 + X = B are given by x 1 = − u and 3 − x n = u + n−1 X k=1 b k , for n ≥ 2, where u is an arbitrary scalar. So there exists a real K 0, such that |x n | β n−1 = u + n−1 P k=1 b k β n−1 ≤ |u| + K n−1 P k=1 β k β n−1 = O 1 , since iii in Proposition 1 implies |u| β n−1 = O 1. So X ∈ s α and we conclude that 1 + is surjective from s β − into s β . Then β = α |µ| ∈ c C 1 implies s α 1 + µ = s α |µ| − . Conversely, assume that s α 1 + µ = s α |µ| − . If we take B = β, we get x n = x 1 − n−1 P k=1 β k , where x 1 is an arbitrary scalar and x n β n−1 = x 1 β n−1 − 1 β n−1 n−1 X k=1 β k = O 1 . On the sets of sequences 21 Putting x 1 = 0, we conclude that β ∈ c C 1 . ii b First we have 1 + ∈ s ◦ β − , s ◦ β because β n−1 β n = O 1. Let us show that 1 + is surjective from s ◦ β − into s ◦ β . For this, let B = b n n≥1 ∈ s ◦ β . The solutions X = x n n≥1 of the equation 1 + X = B are given by 3. We have x n β n−1 = o 1 − n−1 P k=1 b k β n−1 , because from Proposition 2.1, the condition β n−1 β n = o 1 implies β ∈ c C 1 and β → ∞ . Since B ∈ s ◦ β there is a sequence ν = ν n n≥1 ∈ c , such that b n = β n ν n . Then we have for a real M 0 n−1 P k=1 b k β n−1 ≤ 1 β n−1 n−1 X k=1 β k ν k for all n ≥ 2. It remains to show that 1 β n−1 n−1 P k=1 β k ν k = o 1. For this consider any given ε 0. Since β → ∞ there is an integer N such that S n = 1 β n−1 N X k=1 β k ν k ≤ ε 2 for n N , and sup k≥N +1 |ν k | ≤ ε 2 sup n≥2 [C β β] n−1 . Writing R n = 1 β n−1 n−1 P k=N +1 β k ν k for n N + 2, we deduce that R n ≤ sup N +1≤k≤n−1 |ν k | [C β β] n−1 ≤ ε 2 . Finally, we obtain |x n | β n−1 = 1 β n−1 N X k=1 β k ν k + 1 β n−1   n−1 X k=N +1 β k ν k   ≤ S n + R n ≤ ε for n ≥ N, and X ∈ s ◦ β− . So we have proved ii b. 22 B. de Malafosse Assertion ii c. Necessity. Assume that β = α |µ| ∈ c C + 1 . Since we have s ∗ α 1 + µ = s ∗ β 1 + = s β , it is enough to show that 1 + is bijective from s β to s β . We can write that 1 + ∈ s β , s β , since 4 β n+1 β n ≤ 1 β n ∞ X k=n β k = O 1 n → ∞ . Further, from s β ⊂ cs, we deduce using Lemma 3 that for any given B ∈ s β , 1 + 6 + B = B. On the other hand 6 + B = ∞ P k=n b k n≥1 ∈ s β , since β ∈ c C + 1 . So 1 + is surjective from s β into s β . Finally, 1 + is injective because the equation 1 + X = O admits the unique solution X = O in s β , since K er 1 + = ue t u ∈ C and e t ∈ s β . Sufficiency. For every B ∈ s β the equation 1 + X = B admits a unique solution in s β . Then from Lemma 2, β ∈ cs and since s β ⊂ cs we deduce from Lemma 3 that X = 6 + B ∈ s β is the unique solution of 1 + X = B. Taking B = β, we get 6 + β ∈ s α that is β ∈ c C + 1 . As above to prove ii d it is enough to verify that β = α |µ| ∈ c C + 1 if and only if s ◦ ∗ β 1 + = s β . If β ∈ c C + 1 , 1 + is bijective from s ◦ β into itself. Indeed, we have D 1 β 1 + D β ∈ c , c from 4 and Lemma 1. Furthermore, since β ∈ c C + 1 we have s ◦ β ⊂ cs and for every B ∈ s ◦ β , 1 + 6 + B = B. From Lemma 1, we have 6 + ∈ s ◦ β , s ◦ β , so the equation 1 + X = B admits in s ◦ β the solution X = 6 + B and we have proved that 1 + is surjective from s ◦ β into itself. Finally, β ∈ c C + 1 implies that e t ∈ s ◦ β , so K er 1 + T s ◦ β = { 0} and we conclude that 1 + is bijective from s ◦ β into itself. iii a comes from ii, since α ∈ c C + 1 if and only if 1 + is bijective from s α into itself and is also bijective from s ◦ α into itself, and 6 + 1 + X = 1 + 6 + X = X for all X ∈ s α . b Assume that α ∈ c C + 1 . Since C + λ = 6 + D 1 λ , we have w + p α λ = n X ∈ s 6 + D 1 λ |X | p ∈ s α o = n X D 1 λ |X | p ∈ s α 6 + o ; and since α ∈ c C + 1 implies s α 6 + = s α , we conclude that w + p α λ = X ∈ s |X | p ∈ D λ s α = s α|λ| = s α|λ| 1 p . On the sets of sequences 23 Conversely, we have α |λ| 1 p ∈ s α|λ| 1 p = w + p α λ. So C + λ h α |λ| 1 p i p = ∞ X k=n α k |λ k | |λ k | n≥1 ∈ s α , i.e. α ∈ c C + 1 and we have proved i. We get iii c reasoning as above. iii d has been proved in [4]. iii e Assume that α |λ| ∈ c C 1 . Then w ◦ p α λ = n X ∈ s |X | p ∈ 1 λ s ◦ α o . Since 1 λ = 1D λ , we get 1 λ s ◦ α = 1 s ◦ α|λ| . Now, from i b we deduce that α |λ| ∈ c C 1 implies that 1 is bijective from s ◦ α|λ| into itself and w α λ = s ◦ α|λ| 1 p . We get e reasoning as above. As a direct consequence of Theorem 2 we obtain the following results given in [11]. C OROLLARY 1. Let r 0 be any real. We get r 1 ⇔ s r 1 = s r ⇔ s ◦ r 1 = s ◦ r ⇔ s r 1 + = s r . We deduce from the previous section the following. 3. Sets of sequences that are strongly α−bounded and α−convergent to zero with index p and generalizations.