NORMED BARRELLED SPACES

Definition. Let X be a normed space. S ⊂ X is a bounding set if S ⊂ Sphere(X) and if (fn) ⊂ X ′ is an unbounded sequence in the dual space of X, there exists (xn) ⊂ S such that supn |fn(xn)| =∞. Theorem. Let S be a bounding set in a normed space X. If for any sequence (xn) ⊂ S, there exist a sequence (dk) ∈ Ball(l), dk 6= 0, integers Nk ≥ 0, C > 0, such that for every subsequence (xnk) of (xn) there is a further subsequence (xnkl ) and x ∈ X with x = ∑∞ j=1 tjxj and


Introduction.
In this paper we present a general "gliding hump" condition that implies the barrelledness of a normed vector space.Several examples of subspaces of l 1 are shown to be barrelled using the theorem.The barrelledness of the space of Pettis integrable functions is also implied by the theorem (this was first shown in [3]).
Results.The following theorem generalizes that given in [8].
Definition.Let X be a normed space.S ⊂ X is a bounding set if S ⊂ Sphere(X) and if (f n ) ⊂ X is an unbounded sequence in the dual space of X, there exists (x n ) ⊂ S such that sup n |f n (x n )| = ∞.
Theorem.Let S be a bounding set in a normed space X.If for any sequence (x n ) ⊂ S, there exist a sequence (d k ) ∈ Ball(l 1 ), d k = 0, integers N k ≥ 0, C > 0, such that for every subsequence (x n k ) of (x n ) there is a further subsequence (x n k l ) and x ∈ X with x = ∞ j=1 t j x j and then X is barrelled.(The condition needs to hold only for the index l greater than some integer).
Proof.Suppose X is not barrelled.Let (f n ) ⊂ X be a pointwise bounded sequence that is unbounded in norm, and let (x n ) ⊂ S satisfy sup n |f n (x n )| = ∞.We will use the notation The paper is in final form and no version of it will be published elsewhere. [205] C. E. STUART Note that f * ≤ f .Choose n 1 and x n1 such that We will also use the notation Note that M k is finite by the pointwise boundedness of (f n ), and that M k depends on Continue inductively to get and Now choose x ∈ X that satisfies the hypotheses of the theorem.Then using (1) and the fact that (t j ) ∈ Ball(l 1 ).Continuing, using ( ) on the third term.Simplifying we get Since t n k l x n k l will eventually be less than one we can write . This goes to infinity as l → ∞, which contradicts the assumption of pointwise boundedness.

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There are several examples of barrelledness in normed spaces that are implied by the theorem.
Corollary 1.Let S be a bounding set in a normed space X.Suppose that for any sequence (x n ) ⊂ S and any null sequence (t n ) of real numbers, we have for every subsequence Then we can show that We have t n k l x n k l = 1 k l ! and noting that t j = 0 for j = k l , it is easy to check that for k l ≥ 3, so the result follows from the theorem.
A topological vector space X is a K-space if for every null sequence (x n ) in X, every subsequence of (x n ) has a further subsequence ( X is an A-space if for every bounded sequence (x n ) in X and every null sequence of real (or complex) numbers (t n ), every subsequence of (t n x n ) has a subsequence (t The corollary above implies that normed A-spaces, and thus normed K-spaces, are barrelled.(See [9] for more information on these spaces).
It is shown in the paper [3] that the space of Pettis integrable functions defined on an atomless measure space satisfies property K with respect to a bounding set and so is barrelled.We refer the interested reader to the paper for details.
The following corollary is a general condition for a dense subspace of l 1 to be barrelled.Φ denotes the span of {e i : i ∈ N}, the canonical unit vectors.
Corollary 2. Let Φ ⊂ E ⊂ l 1 .E is barrelled if there exist a sequence (d k ) ∈ Ball(l 1 ) with d k = 0 for all k, C > 0, and integers N k ≥ 0 such that for every increasing sequence of integers (n k ), there is a subsequence (n k l ), and x ∈ E such that x j e j ≥ C|d k l |.
(Again, the condition need only hold for the index l greater than some integer).
Proof.The canonical unit vectors form a bounding set in l 1 , so the result follows directly from the theorem.
There are many examples of dense, barrelled subspaces of l 1 .See [6] for an application of these spaces.We next show that some of these are implied by the corollary.
C. E. STUART Corollary 3. Let Φ ⊂ E ⊂ l 1 .Suppose that E is monotone (that is, χ A • x ∈ E for all A ⊂ N and x ∈ E, where • stands for coordinatewise multiplication) and there is a fixed sequence (b k ) ∈ l 1 \Φ such that for any increasing sequence of integers (i k ) there is a subsequence (i k l ) and x ∈ E for which x i k l = b k l .Then E is barrelled.
Proof.We need to find a subsequence of (b k ) that satisfies the hypotheses of the Corollary 2. We can find So the result follows from Corollary 2.
The following result is due to Bennett [2].
Proof.l 0 is monotone and satisfies the hypotheses of Corollary 3 (say with b k = 2 −k ) so this result follows.
Bennett actually showed that scarce copies of l 0 (copies that satisfy a sparseness condition) are also barrelled.See [1] for details.This follows easily from the above result.
The following example is due to Ruckle [4].A sequence space is symmetric if x π(n) ∈ E for all (x n ) ∈ E and for any permutation π of N.
Proof.Actually, Ruckle's proof contains ideas similar to those used in the main Theorem.We will need to define N k to be something other than 0 for the first time.We essentially follow his construction and notation.
Let x ∈ E\Φ, and (h j ) an increasing sequence in N such that Let π be the permutation of the integers which interchanges h 2n−1 and h 2n and leaves the other integers the same.
Let {n 1 , n 2 , . ..} be an increasing sequence of integers for which Denote by θ the permutation which interchanges h 2nj −1 and h 2nj , and leaves the remaining integers unchanged.Let u = 1 2 (v − v θ ).Then u ∈ E, u j = 0 for j ∈ {h 2nj −1 , h 2nj , j = 1, 2, 3, . ..} We will define a sequence y that is a permutation of u and that satisfies the hypotheses of Corollary 2 with N k = 1 for all k.
Let (i k ) be any sequence of integers with i k > i k−1 + 1 and i 1 > 2. Let , and y j = 0 otherwise.
Then y is a permutation of u since it exhausts ±a k and has infinitely many zeros.We will show that y satisfies the hypotheses of Corollary 2.
We need to show that This follows easily from conditions (4) and ( 5) above.Note that we need N k = 1 for the result to follow.
The following corollary uses the idea of a modulus.A modulus is a non-negative, subadditive function q on [0, ∞) which is continuous and 0 at 0. Corollary 6. Subspaces of l 1 determined by a modulus q are barrelled.
Proof.Ruckle shows in [5] that the space of all sequences s in l 1 that satisfy j q(s j ) < ∞ is symmetric and properly contains Φ, so this result follows from the previous corollary.
The following result is due to Saxon [7].If b is any fixed sequence in l 1 with infinite support, then the dilation space E b is the span of Φ and the vectors i b i e ni as (n i ) ranges through all increasing subsequences of N.

Corollary 7 .
Dilation subspaces of l 1 that properly contain Φ are barrelled.Proof.Let b be a fixed sequence in l 1 with infinite support, and let b ij = d j be a subsequence of b that is non-zero, |d j | decreasing, and |d j | > 2 l>j |d l |.We can construct a sequence in E b that satisfies the hypotheses of Corollary 2 by a cancellation process similar to that used in the corollary above on symmetric spaces.In what follows, the subsequence (b ij ) is shown in brackets.We can define dilations of (b i ), denoted (c i ) and (f i ), as follows:(b i ) = b1 b 2 b 3 b 4 b 5 b 6 b 7 b 8 . . .(c i ) = b 1 0 b 2 b 3 b 4 0 b 5 b 6 0 b 7 b 8 . . .(f i ) = b 1 b 2 0 b 3 b 4 b 5 0 b 6 b 7 0 b 8 . . .(c i ) − (f i ) = 0 − b 2 b 2 0 0 − b 5 b 5 0 − b 7 b 7 . . .