C O L L O Q U I U M M A T H E M A T I C U M
VOL. 80 1999 NO. 2
MAPPING PROPERTIES OF c 0
BY
PAUL L E W I S (DENTON, TEX.)
Abstract. Bessaga and Pe lczy´ nski showed that if c 0 embeds in the dual X ∗ of a Banach space X, then ℓ 1 embeds as a complemented subspace of X. Pe lczy´ nski proved that every infinite-dimensional closed linear subspace of ℓ 1 contains a copy of ℓ 1 that is complemented in ℓ 1 . Later, Kadec and Pe lczy´ nski proved that every non-reflexive closed linear subspace of L 1 [0, 1] contains a copy of ℓ 1 that is complemented in L 1 [0, 1]. In this note a traditional sliding hump argument is used to establish a simple mapping property of c 0 which simultaneously yields extensions of the preceding theorems as corollaries.
Additional classical mapping properties of c 0 are briefly discussed and applications are given.
All Banach spaces in this note are defined over the real field. The canon- ical unit vector basis of c 0 will be denoted by (e n ), the canonical unit vector basis of ℓ 1 will be denoted by (e ∗ n ), and a continuous linear transformation will be referred to as an operator. The reader is referred to Diestel [3] or Lindenstrauss and Tzafriri [8] for undefined notation and terminology.
Theorem 1. If T : c 0 → X is an operator and (x ∗ k ) is any bounded sequence in X ∗ so that
∞
X
k=1
|x ∗ k (T (e n k )) − 1| < ∞
for some subsequence (T (e n k )) of (T (e n )), then there is a sequence (w ∗ i ) in {x ∗ k − x ∗ j : k, j ∈ N} so that (w ∗ i ) is equivalent to (e ∗ i ) as a basic sequence and [w ∗ i ] is complemented in X ∗ .
P r o o f. Let (b k ) = T (e n k ) for k ∈ N, let C be a positive number so that C > 1 and kT (x)k ≤ Ckxk for all x, and choose B > 1 so that
2 sup kx ∗ n k < B.
Without loss of generality, suppose that
|x ∗ n (b n ) − 1| < 1 BC · 2 n+4
1991 Mathematics Subject Classification: Primary 46B20.
[235]
for each n. Further, since (b n ) is weakly null, suppose that (1)
n−1
X
i=1
|x ∗ i (b n )| < 1 BC · 2 n+5
for each n. Now let r 1 = 1, r 2 = 2, and choose r 3 and r 4 so that r 2 < r 3 < r 4
and
|(x ∗ r 4 − x ∗ r 3 )(b r 2 )| < 1 BC · 2 1+5 . Next choose r 5 and r 6 so that r 4 < r 5 < r 6 and
(2)
|(x ∗ r 6 − x ∗ r 5 )(b r 2 )| < 1 BC · 2 2+5 ,
|(x ∗ r 6 − x ∗ r 5 )(b r 4 )| < 1 BC · 2 2+5 .
An additional step clarifies the induction process. Choose r 7 and r 8 so that r 6 < r 7 < r 8 and
(3)
|(x ∗ r 8 − x ∗ r 7 )(b r 2 )| < 1 BC · 2 3+5 ,
|(x ∗ r 8 − x ∗ r 7 )(b r 4 )| < 1 BC · 2 3+5 ,
|(x ∗ r 8 − x ∗ r 7 )(b r 6 )| < 1 BC · 2 3+5 .
Continue this construction inductively, and let u n = b r 2n and z n ∗ = x ∗ r 2n for each n. Note that
|z ∗ i (u n ) − x ∗ r 2
i−1 (u n )| < 1 BC · 2 i+4 for n < i. Further,
|z n ∗ (u n ) − 1| < 1
BC · 2 n+4 and
n
X
i=1
|z ∗ i (u n+1 )| < 1 BC · 2 (n+1)+5 for each n.
Next let w ∗ n = z n ∗ − x ∗ r 2 n−1 = x ∗ r 2n − x ∗ r 2 n−1 for n ∈ N. Then
|w n ∗ (u n ) − 1| ≤ |z n ∗ (u n ) − 1| + |x ∗ r 2 n−1 (u n )|
< 1
BC · 2 n+4 + 1
BC · 2 n+5 = 3 BC · 1
2 n+1 · 1 2 4 . Also, kx ∗ n k ≤ B for each n.
Now suppose that q ∈ N and t i is a non-zero real number for 1 ≤ i ≤ q.
If ε i = sgn(t i w i ∗ (u i )), then
q
X
i=1
t i w ∗ i (ε i u 1 ) ≥ |t 1 w ∗ 1 (u 1 )| −
q
X
i=2
|w i ∗ t i (u 1 )|
≥ |t 1 |
1 − 3
BC · 1 2 2 · 1
2 4
−
|t 2 |
BC · 2 1+5 + . . . + |t q | BC · 2 (q−1)+5
. Further,
q
X
i=1
t i w ∗ i (ε 2 u 2 )
= |t 2 w ∗ 2 (u 2 )| −
q
X
i=1,i6=2
t i w ∗ i (ε 2 u 2 )
≥ |t 2 |
1 − 3
BC · 1 2 3 · 1
2 4
−
|t 1 | 1 BC · 1
2 2+5 + |t 3 | 1 BC · 1
2 2+5 + . . . + |t q | 1
BC · 1
2 (q−1)+5
. (Observe that
|w ∗ 1 (u 2 )| = |(x ∗ 2 − x ∗ 1 )(b r 4 )| ≤ |x ∗ 2 (b r 4 )| + |x ∗ 1 (b r 4 )|
< 1
BC · 2 r 4 +5 + 1
BC · 2 r 4 +5 < 1 BC · 2 2+5 from (1). Also,
|w ∗ 3 (u 2 )| = |(x ∗ r 6 − x ∗ r 5 )(b r 4 )| ≤ 1 BC · 2 2+5 from (2), and
|(x ∗ r 8 − x ∗ r 7 )(b r 4 )| < 1 BC · 2 3+5 from (3).)
In general, D X q
i=1
ε i u i ,
q
X
n=1
t n w ∗ n E
≥ |t 1 |
1 − 3
BC · 1 2 2 · 1
2 4
− |t 1 |
1
BC · 2 r 4 +5 + 1
BC · 2 r 6 +5 + . . . + 1 BC · 2 r 2q +5
+ |t 2 |
1 − 3
BC · 1 2 3 · 1
2 4
− |t 2 |
1
BC · 2 2+4 + 1
BC · 2 r 6 +5 + . . . + 1 BC · 2 r 2q +5
+ |t 3 |
1 − 3
BC · 1 2 4 · 1
2 4
− |t 3 |
2 1
BC · 2 3+4 + 1
BC · 2 r 8 +5 + 1
BC · 2 r 10 +5 + . . . + 1 BC · 2 r 2q +5
+ |t 4 |
1 − 3
BC · 1 2 5 · 1
2 4
− |t 4 |
3 1
BC · 2 4+4 + 1
BC · 2 r 10 +5 + . . . + 1 BC · 2 r 2q +5
+ . . . +
+ |t q |
1 − 3
BC · 1 2 q+1 · 1
2 4
− |t q |
q − 1 BC · 2 q+4
. Note that
3
BC · 2 2 2 4 + 1
BC · 2 r 4 +5 + . . . + 1
BC · 2 r 2q ≤ 2 BC · 2 4 , 3
BC · 2 3 2 4 + 1
BC · 2 2+4 + 1
BC · 2 r 6 +5 + . . . + 1
BC · 2 r 2q +5 ≤ 2 BC · 2 4 , . . .
3
BC · 2 q+1 2 4 + q − 1
BC · 2 q+4 ≤ 2 BC · 2 4 . Consequently,
D X q
i=1
ε i u i ,
q
X
n=1
t n w ∗ n E
≥ X q
i=1
|t i |
1 − 2
BC · 2 4
> 0.
Thus P q
i=1 ε i u i 6= 0, and
q
X
i=1
t i w ∗ i ≥
1 .
q
X
i=1
ε i u i
D X q
i=1
ε i u i ,
q
X
n=1
t n w n ∗ E
≥ X q
i=1
|t i |
1 − 1
BC · 2 3
c −1
. Hence
1 − 1
BC · 2 3
c −1
X q
i=1
|t i |
≤
q
X
i=1
t i w ∗ i ≤ B
q
X
i=1
|t i |, and (w ∗ i ) ∼ (e ∗ i ).
Next we show that [w n ∗ ] is complemented in X ∗ . Suppose that v ∗ = P ∞
n=1 t n w ∗ n , and let U : X ∗ → [w ∗ n ] be defined by
U (x ∗ ) = X
n
x ∗ (u n )w ∗ n .
Since (u n ) is a subsequence of (b k ) and P b k is weakly unconditionally con- vergent, it is clear that U is well defined, continuous, and linear. Now observe that
kv ∗ − U (v ∗ )k
=
X t n w n ∗ − X
t n U (w n ∗ )
=
∞
X
n=1
t n w n ∗ −
∞
X
n=1
t n
X ∞
k=1
w ∗ n (u k )w ∗ k
=
∞
X
n=1
t n w n ∗ −
∞
X
n=1
t n w ∗ n (u n )w ∗ n −
∞
X
n=1
t n
X ∞
k=1, k6=n
w ∗ n (u k )w ∗ k
≤
∞
X
n=1
|t n | · |1 − w ∗ n (u n )| · kw ∗ n k +
∞
X
n=1
|t n | X ∞
k=1, k6=n
|w ∗ n (u k )|B
≤
∞
X
n=1
|t n | sup
k
n
|1 − w ∗ k (u k )| +
∞
X
i=1, i6=k
|w ∗ k (u i )| o
B.
Also,
∞
X
n=1
|t n | ≤ c 1 − BC·2 1 3
kv ∗ k.
Further,
sup
k
|1 − w k ∗ (u k )| ≤ 3 BC · 2 2 2 4 , and kw ∗ k k ≤ B for each k.
Next note that
∞
X
k=2
|w ∗ 1 (u k )| =
∞
X
k=2
|(x ∗ 2 − x ∗ 1 )(u k )|
= |(x ∗ 2 − x ∗ 1 )(T (e r 4 ))| + |(x ∗ 2 − x ∗ 1 )(T (e r 6 ))| + . . .
≤ (|x ∗ 2 T (e r 4 )| + |x ∗ 1 T (e r 4 )|) + (|x ∗ 2 T (e r 6 )| + |x ∗ 1 T (e r 6 )|) + . . .
< 1
BC · 2 r 4 +5 + 1
BC · 2 r 6 +5 + . . . < 1
BC · 2 r 4 +4 < 1 BC · 2 4 . A similar argument shows that
∞
X
i=1, i6=k
|w ∗ k (u i )| < 1
BC · 2 4
for each k. Thus
kv ∗ − U (v ∗ )k ≤ c 1 − BC·2 1 3
kv ∗ k
3
BC · 2 2 2 4 + 1 BC · 2 4
B < 1
7 kv ∗ k.
If U 1 = U |[w ∗
i ] , then kIdentity − U 1 k |[w ∗
i ] < 1, and U 1 is invertible on [w ∗ i ]. It is easy to see that U 1 −1 U is a projection from X ∗ onto [w ∗ i ].
Remark . (a) The operator T : c 0 → X satisfies the hypotheses of Theorem 1 if and only if lim inf kT (e n )k > 0. H. Rosenthal [11] has given a penetrating study of the situation in which T : ℓ ∞ (Γ ) → X is an operator so that inf γ∈Γ kT (e γ )k > 0.
(b) If (x ∗ k ) is w ∗ -null, the proof of Theorem 1 makes it clear that we may choose the sequence (w ∗ i ) in the conclusion of the theorem to be w ∗ -null.
As the following corollaries indicate, Theorem 1 unifies and extends sev- eral classical results.
Corollary 2 ([1, Thm. 4], [3, p. 48]). If c 0 embeds isomorphically in the dual X ∗ of the Banach space X, then X contains a copy of ℓ 1 which is complemented (in X ∗∗ and thus) in X.
P r o o f. If T : c 0 → X ∗ is an isomorphism, then let (x n ) be a bounded sequence in X (⊆ X ∗∗ ) so that P ∞
n=1 |x n (T (e n ))−1| = 0. Apply Theorem 1 to the sequence (x n ).
Corollary 3 ([10], [3, p. 72]). If ℓ 1 is a quotient of X, then X contains a copy of ℓ 1 which is complemented in X ∗∗ .
P r o o f. If T : X → ℓ 1 is a surjective operator, then T ∗ : ℓ ∞ → X ∗ is an isomorphism. Hence T |c ∗
0 is an isomorphism.
If Σ is a σ-algebra, (µ n ) is a bounded sequence in cabv(Σ, X), and 0 < ε < δ, then (µ n ) is said to be (δ, ε)-relatively disjoint [11] if there is a pairwise disjoint sequence (A n ) in Σ so that
|µ n |(A n ) > δ and
∞
X
m=1, m6=n
|µ n |(A m ) < ε
for each n. Further, (µ n ) is said to be relatively disjoint if it is (δ, ε)-relatively disjoint for some pair (δ, ε). Rosenthal [11] and Kadec and Pe lczy´ nski [7]
showed that if (µ n ) is a relatively disjoint sequence in cabv(Σ, X), then (µ n ) ∼ (e ∗ n ) and [µ n ] is complemented in cabv(Σ, X).
If A is an algebra of subsets of Ω, then fabv(A, X) denotes the Banach
space (total variation norm) of all finitely additive set functions m : A → X
which have finite variation. Both [4] and [6] contain an extensive discussion
of spaces of measures. In addition, we note that [4] includes a detailed
presentation of results related to the Radon–Nikodym property. Note that
part (i) of Corollary 4 below contains an extension of Proposition 3.1 of [11]
to the setting of finitely additive set functions defined on an algebra of sets.
Further, we remark that in a classic paper Kadec and Pe lczy´ nski [7, Theo- rem 6] showed that if Y is any non-reflexive closed linear subspace of L 1 [0, 1], then Y contains a copy of ℓ 1 which is complemented in L 1 [0, 1]. Part (v) of the next corollary shows that if X and X ∗ have the Radon–Nikodym property, then any non-reflexive closed linear subspace of L 1 (µ, X) contains a copy of ℓ 1 which is complemented in L 1 (µ, X).
Corollary 4. (i) If (µ n ) is any bounded sequence in fabv(A, X) for which there is a pairwise disjoint sequence (A n ) in A and an ε > 0 so that
|µ n |(A n ) > ε
for each n, then there is a sequence (ν i ) in {µ n − µ k : k, n ∈ N} so that (ν i ) ∼ (e ∗ i ) and [ν i ] is complemented in fabv(A, X).
(ii) If K is a relatively weakly compact subset of fabv(A, X) and (A i ) is a pairwise disjoint sequence of members of A, then lim i |µ|(A i ) = 0 uni- formly for µ ∈ K.
(iii) If K is a relatively weakly compact subset of cabv(Σ, X), then {|µ| : µ ∈ K} is uniformly countably additive.
(iv) If µ is a finite positive measure on Σ and K is a relatively weakly compact subset of the space L 1 (µ, X) of Bochner integrable functions, then K is uniformly integrable.
(v) If Y is a closed linear subspace of fabv(A, X), Y is not reflexive, and X and X ∗ have the Radon–Nikodym property, then Y contains a copy of ℓ 1 which is complemented in fabv(A, X).
P r o o f. (i) For each n let (A n i ) k i=1 n be a partition of A n and (x ∗ n i ) k i=1 n be points in the unit ball of X ∗ so that
k n
X
i=1
x ∗ n i µ n (A n i ) > ε.
Now define the X ∗ -valued simple function s n by s n =
k n
X
i=1
χ A ni x ∗ n i , and observe that
T
s n dµ n > ε. Define T : c 0 → fabv(A, X) ∗ by T ((γ n )) = X
n
γ n s n .
Then T is an operator. Normalize and use Theorem 1 to conclude that some
sequence (ν i ) in {µ n − µ k : n, k ∈ N} is equivalent to (e ∗ n ) and that [ν n ] is
complemented in fabv(A, X).
(ii) Suppose that ε > 0 and (µ i ) is a sequence in K so that |µ i |(A i ) > ε for each i. Part (i) ensures that (e ∗ n ) is equivalent to some sequence in K −K.
However, this is impossible since K − K is relatively weakly compact.
(iii) Since each member of K is a countably additive measure on a σ-algebra, |K| = {|µ| : µ ∈ K} is uniformly countably additive if and only if lim i |µ|(A i ) = 0 uniformly for µ ∈ K whenever (A i ) is a pairwise disjoint sequence from Σ. Deny the uniform countable additivity of |K|, repeat the same construction as in (i), and obtain the same contradiction as in (ii).
(iv) If f ∈ L 1 (µ, X) and A ∈ Σ, put ν f (A) =
\