aa r X i v : . [ m a t h . F A ] F e b The subspace c is not complemented in ac N.N. Avdeev ∗† February 25, 2021
Abstract.
We prove that the subspace c of sequences that converge to zero is notcomplemented in the space ac of sequences that almost converge to zero. We proceedwith applying the same approach to inclusion chain c ⊂ A ⊂ ℓ ∞ . Let ℓ ∞ be the space of all bounded sequences endowed with the usual norm k x k = sup n ∈ N x n , (1)where N denotes set of all positive integers, and let c be the subspace of null-sequences.The famous and surprising result of Phillips [1] states that c is not complemented in ℓ ∞ : Theorem 1.1 ( [1]) . There is no bounded linear operator P : ℓ ∞ → c such that forevery x ∈ c the equality P x = x holds. The notion of almost convergence is a generalization of the notion of convergence.Before we proceed to it, we need to introduce the concept of Banach limits, which(expectably) provide a way to generalize the concept of the limit.
Definition 1.2.
A linear functional B ∈ ℓ ∗∞ is said to be a Banach limit if(i) B ≥ , that is Bx ≥ for x ≥ ,(ii) B I = 1 , where I = (1 , , . . . ) , ∗ This work was carried out at Voronezh State University and supported by the Russian Science Foundation grant 19-11-00197. † [email protected], [email protected] B ( T x ) = B ( x ) for x ∈ ℓ ∞ , where T is the shift operator T ( x , x , . . . ) =( x , x , . . . ) .The set of all Banach limits is denoted by B . The existence of Banach limits wasannounced by Mazur [ ? ] and proved by Banach [2]. Definition 1.3.
A sequence x ∈ ℓ ∞ is said to be almost convergent to t ∈ R if forevery B ∈ B the equality Bx = t (2)holds.The set of all sequences that are almost convergent to t is denoted by ac t ; the spaceof all almost convergent sequences is denoted by ac . Theorem 1.4 (Lorentz, [3]) . A sequence x = ( x , x , ... ) is almost convergent to t ∈ R iff lim n →∞ n m + n X k = m +1 x k = t (3) uniformly by m ∈ N . Alekhno gave an elegant proof [4, Theorem 8] that ac is not complemented in ℓ ∞ . The proof is based on the original Phillips’s proof of Theorem 1.1 and uses somelemmas from there. For more details on Banach limits, we refer the reader to [ ? , 5, 6].In Section 2 of the present article, we prove that c is not complemented in ac . So,all the three inclusions c ⊂ ℓ ∞ , ac ⊂ ℓ ∞ , and c ⊂ ac , (4)are not complemented. Our proof is inspired by Whitley’s approach [7] and the dis-cussion [8].In Sections 3 and 4 we proceed with applying the same approach to the inclusionchain c ⊂ A ⊂ ℓ ∞ , where A = { x ∈ ℓ ∞ : α ( x ) = 0 } , (5) α ( x ) = lim i →∞ max i Lemma 2.1. There exists an uncountable family of subsets { S i } i ∈ I , S i ⊂ N , such thateach S i is countable and for every i = j the intersection S i ∩ S j is finite.Proof. Consider a bijection N ↔ Q and let I = R . For i ∈ I we set S i = { q n } , where { q n } ⊂ Q is a sequence which converges monotonically to i .The following lemma is inspired by [8]. Lemma 2.2. For each linear operator Q : ac → ac such that c ⊆ ker Q , thereexists infinite subset S ⊂ N such that ∀ ( x ∈ ac : supp x ⊂ S )[ Qx = 0] . (7) Proof. Notice first that for any infinite subset S ⊂ N there exists x ∈ ac \ c with supp x ⊆ S . Indeed, we can find such x with < x ≤ χ S that contains infinitelymany ones, and the distance between the ones is enough for Lorentz’s criterion (3) tobe hold.Let { S i } i ∈ I be a family of subsets of N that satisfy the conditions of Lemma 2.1.Suppose to the contrary that ∀ ( infinite S ⊂ N ) ∃ ( x ∈ ac : supp x ⊂ S )[ Qx = 0] . (8)3n particular, ∀ ( i ∈ I ) ∃ ( x i ∈ ac : supp x i ⊂ S i )[ Q ( x i ) = 0] . (9)Note that x i / ∈ c because c ⊆ ker Q . Without loss of generality we can assumethat k x i k = 1 for all i ∈ I .Consider I n = { i ∈ I : ( Qx i ) n = 0 } , then I = S n ∈ N I n . Thus, we can find n suchthat I n is also uncountable (otherwise I would be countable as a countable union ofcountable sets, which contradicts to conditions of Lemma 2.1).Сonsider now I n,k = { i ∈ I n : | ( Qx i ) n | ≥ /k } , then I n = S k ∈ N I n,k . Applying thesame argument as above, one can easily see that the set I n,k is uncountable for some k . Let us choose such I n,k and proceed with it.So, we have an uncountable set I n,k and ∀ ( i ∈ I n,k ) ∃ ( x i ∈ ac : supp x i ⊂ S i ) h k x i k = 1 and | ( Qx i ) n | ≥ /k i . (10)Consider a finite set J ⊂ I n,k with J > (here J stands for the cardinality ofthe set J ). Take y = X j ∈ J sign ( Qx j ) n · x j . (11)Since the intersection S i ∩ S j is finite for any i = j and supp x j ⊂ S j , the intersection T j ∈ J supp x j is also finite. Hence, y = f + z , where supp f is finite and k z k ≤ .On the other hand, ( Qy ) n = X j ∈ J (sign( Qx j ) n ) · ( Qx j ) n ≥ Jk . (12)Note, that f ∈ c and we have Qf = 0 , because c ⊆ ker Q . Thus, Qy = Q ( f + z ) = Qf + Qz = Qz and Jk ≤ ( Qy ) n ≤ k Qy k = k Qz k ≤ k Q k · k z k ≤ k Q k . (13)Due to (13), we obtain J ≤ k Q k k for every J ⊂ I n,k . This contradicts the fact that I n,k is uncountable, and we are done. Theorem 2.3. The subspace c is not complemented in ac . roof. Suppose to the contrary that there exists a continuous projection P : ac → c .Applying Lemma 2.2 to I − P we can find infinite subset S ⊂ N such that ∀ ( x ∈ ac :supp x ⊂ S )[( I − P ) x = 0] (such x ∈ ac \ c exists even if χ S / ∈ ac ). But then P x / ∈ c , which contradicts the fact that P is a projection onto c . A is not complemented in ℓ ∞ Lemma 3.1. For each linear operator Q : ℓ ∞ → ℓ ∞ such that c ⊆ ker Q , there existsinfinite subset S ⊂ N such that ∀ ( x ∈ ℓ ∞ : supp x ⊂ S )[ Qx = 0] . (14) Proof. Let { S i } i ∈ I be a family of subsets of N such that conditions of Lemma 2.1 arehold. Suppose to the contrary that ∀ ( infinite S ⊂ N ) ∃ ( x ∈ ℓ ∞ : supp x ⊂ S )[ Qx = 0] . (15)In particular, ∀ ( i ∈ I ) ∃ ( x i ∈ ℓ ∞ : supp x i ⊂ S i )[ Q ( x i ) = 0] . (16)The rest of the proof is similar to that of Lemma 2.2. Theorem 3.2. The subspace A is not complemented in ℓ ∞ .Proof. Suppose to the contrary that there exists a continuous projection P : ℓ ∞ → A .Then c ⊂ A ⊂ ker( I − P ) . Applying Lemma 3.1 to I − P we can find infinite subset S ⊂ N such that ∀ ( x ∈ ℓ ∞ : supp x ⊂ S )[( I − P ) x = 0] . Let S ′ ⊂ S be such that χ S contains infinite quantities of both ones and zeros. But then α ( P χ S ′ ) = α ( χ S ′ ) = 1 and P χ S ′ / ∈ A , which contradicts the fact that P is a projection onto A . c is not complemented in A To prove the fact, we need some auxiliary constructions from [9].Let us define linear operator F : ℓ ∞ → ℓ ∞ as following: ( F y ) k = y i +2 , for i < k ≤ i + 1 (17)5or example, F ( { , , , , , , ... } ) = { , , , , , , , , , , , , , , , , ... } It is easy to see that the equality α ( F x ) = lim k →∞ | x k +1 − x k | (18)holds.Let us also define linear operator M : ℓ ∞ → ℓ ∞ as following: M ω = (cid:18) , ω , , ω , ω , ω , , ω , ω , ω , ω , ω , , ..., , p ω p , p ω p , ..., p − p ω p , ω p , p − p ω p , ..., p ω p , p ω p , , p + 1 ω p +1 , ... (cid:19) . Note that due to (18) we have F M : ℓ ∞ → A . Lemma 4.1. For each linear operator Q : A → A such that c ⊆ ker Q , there existsinfinite subset S ⊂ N such that ∀ ( x ∈ A : supp x ⊂ S )[ Qx = 0] (19) and x ∈ A \ c such that supp x ⊆ S .Proof. Let { U i } i ∈ I be a family of subsets of N such that conditions of Lemma 2.1 arehold. Let { S i } i ∈ I be a family of subsets of N defined by the equality S i = supp F M χ U i .Obviously, the family of sets { S i } i ∈ I also satisfies the conditions of Lemma 2.1. More-over, for every i ∈ I we have x = F M χ U i ∈ A \ c .Suppose to the contrary that ∀ ( infinite S ⊂ N ) ∃ ( x ∈ A : supp x ⊂ S )[ Qx = 0] . (20)In particular, ∀ ( i ∈ I ) ∃ ( x i ∈ A : supp x i ⊂ S i )[ Q ( x i ) = 0] . (21)The rest of the proof is similar to that of Lemma 2.2. Theorem 4.2. The subspace c is not complemented in A . roof. Suppose to the contrary that there exists a continuous projection P : A → c .Applying Lemma 4.1 to I − P we can find infinite subset S ⊂ N such that ∀ ( x ∈ A : supp x ⊂ S )[( I − P ) x = 0] , and x ∈ A \ c such that supp x ⊆ S . But then P x = x / ∈ c , which contradicts the fact that P is a projection onto c . Author thanks Dr. Prof. E.M. Semenov and Dr. A.S. 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