A litmus test for unconditional bases

Special Types of Bases

3.5 A litmus test for unconditional bases

Proof. Clearly,J ={ξ∈ J^∗∗: lim_n→∞ξ(n) = 0}has codimension one in its bidual. To prove the fact that it is isometric to its bidual we observe that

ξJ^∗∗=(0, ξ(1), ξ(2), . . .)J, ξ∈ J^∗∗. Let

L(ξ) = lim

n→∞ξ(n), ξ∈ J^∗∗. We define

S(ξ) = (−L(ξ), ξ(1)−L(ξ), ξ(2)−L(ξ), . . .).

S maps J^∗∗ onto J and is one-to-one. Since · _J is a seminorm on J^∗∗

vanishing on constants,

S(ξ)_J =(0, ξ(1), . . .)_J =ξ_J^∗∗. ThusS is an isometry.

Corollary 3.4.7.J does not have an unconditional basis.

Proof. It follows immediately from the separability of J^∗∗, Theorem 3.3.3, and Theorem 3.4.6.

After the appearance of James’s example the termquasi-reflexivewas often used for Banach spacesX so thatX^∗∗/X is finite-dimensional.

The ideas of the James construction have been repeatedly revisited to pro- duce more sophisticated examples of similar type. For example, Lindenstrauss [130] showed that for any separable Banach spaceXthere is a Banach spaceZ with a shrinking basis such thatZ^∗∗/Z is isomorphic toX (see Section 13.1).

3.5 A litmus test for unconditional bases 67 Proof. Let (y_s) be a weakly Cauchy sequence in a closed subspace Y of X.

Since X has property (u), there is a WUC series _∞

i=1u_i in X so that the sequence (y_s−s

i=1u_i) converges to 0 weakly. By Mazur’s theorem there is a sequence of convex combinations of members of (y_s−s

i=1u_i) that converges to 0 in norm. Using the Cauchy condition we find integers (p_k), 0 = p₀ <

p₁ < p₂ < . . ., and convex combinations (p_k

j=p_k−1+1λ_j(y_j−j

i=1u_i))^∞_k=1 such that

p_k

j=p_k−1+1

λj(yj− j i=1

ui)≤2^−k for allk.

Putz0= 0, and for each integer k≥1 let z_k=

p_k

j=p_k−1+1

λ_jy_j ∈Y.

Then for anyx^∗∈X^∗,x^∗= 1, we have

|x^∗(z_k−z_k−1)| ≤2^−k+ 2^1−k +x^{∗ pk}

j=p_k−1+1

λj

j i=p_k−2+1

ui−

p_k−1

j=p_k−2+1

λj

j i=p_k−2+1

ui. Thus,

|x^∗(z_k−z_k−1)| ≤3·2^−k+ 2

p_k

j=p_k−2+1

|x^∗(u_j)|, which implies

∞ k=1

|x^∗(z_k−z_k−1)| ≤ 3 2+ 4

∞ j=1

|x^∗(u_j)|<∞.

Therefore,_∞

k=1(zk−zk−1) is a WUC series inY. Now one easily checks that the sequence

yn− n k=1

(zk−zk−1)_∞

n=1= (yn−zn)^∞_n=1 converges weakly to 0.

Proposition 3.5.3 (Pelczy´nski [168]). If a Banach space X has an un- conditional basis thenX has property (u).

Proof. Let (un)^∞_n=1 be aK-unconditional basis ofX with biorthogonal func- tionals (u^∗_n)^∞_n=1. If (x_n) is a weakly Cauchy sequence in X then for each k the scalar sequence (u^∗_k(x_n))^∞_n=1 converges, say, to α_k. Hence the sequence

(N

k=1t_ku^∗_k(x_n)u_k)^∞_n=1converges weakly toN

k=1t_kα_ku_k for eachN and any scalars (t_k). Therefore,

N k=1

kαkuk≤Ksup

n xn

for allN and any sequence of signs (k). Being weakly Cauchy, (xn) is norm- bounded thus_∞

k=1αkuk is a WUC series. Put y_n=x_n−

n k=1

α_ku_k.

(y_n) is weakly Cauchy. Also, lim_n→∞u^∗_s(y_n) = 0 for alls∈N. We claim that (y_n) converges weakly to 0. If not, there isx^∗∈X^∗so that lim_n→∞x^∗(y_n) = 1.

Using the Bessaga-Pelczy´nski selection principle (Proposition 1.3.10) we can extract a subsequence (y_nj) of (y_n) and find a block basic sequence (z_j) of (u_n) such that (z_j) is equivalent to (y_nj) andy_nj−z_j →0. We deduce that x^∗(z_j)→1 since

|x^∗(z_j)−1| ≤ |x^∗(z_j−y_nj)|+|x^∗(y_nj)−1| ≤ x^∗ z_j−y_nj

this tends to 0

+|x^∗(y_nj)−1|

this tends to 0

.

Without loss of generality we can assume that|x^∗(zj)|>1/2 for allj. Given (aj)∈c00, by letting j= sgn ajx^∗(zj) we have

∞ j=1

|a_j||x^∗(z_j)|= ∞ j=1

_ja_jx^∗(z_j)

=x^∗∞

j=1

_ja_jz_j

≤ x^∗K ∞ j=1

ajzj. Hence

∞ j=1

ajzj≥ 1 2Kx^∗

∞ j=1

|aj|. On the other hand we obtain an upper₁-estimate for_∞

j=1a_jz_jusing the boundedness of the sequence (z_j) and the triangle law. We conclude that (z_j) is equivalent to the standard₁-basis. This is a contradiction because (z_j) is weakly Cauchy whereas the canonical basis of₁ is not. Therefore our claim holds and this finishes the proof.

3.5 A litmus test for unconditional bases 69 Proposition 3.5.4.(i)J does not have property (u) and so cannot be em- bedded in any Banach space with an unconditional basis.

(ii) (Karlin [107])C[0,1] does not have an unconditional basis, and cannot be embedded in a space with unconditional basis.

Proof. (i) Assume that J has property (u). Since the sequence defined for eachnbysn =n

k=1ek is weakly Cauchy inJ, there exists a WUC series in J,_∞

k=1uk, so that the sequence (n

k=1ek−n

k=1uk)^∞_n=1converges weakly to 0. One easily notices that the series _∞

k=1uk cannot be unconditionally convergent inJ because that would force the sequence (s_n) to converge weakly to the same limit, when (s_n) is not weakly convergent inJ (it does converge weakly, though, to (1,1,1, . . . ,1, . . .)∈J˜). Therefore using Theorem 2.4.11, c₀ embeds in J, which implies that _∞ embeds in J^∗∗, contradicting the separability ofJ^∗∗.

That J does not embed into any space with unconditional basis follows immediately from Proposition 3.5.2 and Proposition 3.5.3.

(ii) This follows from (i) because J embeds isometrically into C[0,1] by the Banach-Mazur theorem (Theorem 1.4.3).

Thus we have seen that having an unconditional basis is very special and one cannot rely on the existence of such bases in most spaces. It is, however, true that most of the spaces which are useful in harmonic analysis or partial differential equations such as the spacesLp for 1< p <∞do have uncondi- tional bases (which we will see in Chapter 6). We will see also thatL1fails to have an unconditional basis. It is perhaps reasonable to argue that the reason the spaces L_pfor 1< p <∞seem to be more useful for applications in these areas is precisely because they admit unconditional bases!

From the point of view of abstract Banach space theory, in this context it was natural to ask:

The unconditional basic sequence problem.Does every Banach space contain at least an unconditional basic sequence?

This problem was regarded as perhaps the single most important prob- lem in the area after the solution of the approximation problem by Enflo in 1973. Eventually a counterexample was found by Gowers and Maurey in 1993 [71]. The construction is extremely involved but has led to a variety of other applications, some of which we have already met (see e.g. [115], [70], and [72]).

Problems

3.1.Let (u_n) be aK_u-unconditional basis in a Banach spaceX.

(a) Show that if (y_n) is a block basic sequence of (u_n) then (y_n) is an uncon- ditional basic sequence inX with unconditional constant≤K_u.

(b) Show that the sequence of biorthogonal functionals (u^∗_n) of (u_n) is an unconditional basic sequence inX^∗ with unconditional constant≤K_u.

3.2.Let (u_n) be an unconditional basis for a Banach spaceXwith suppression constantK_s. Prove that for allN, whenevera₁, . . . , a_N,b₁, . . . , b_N are scalars so that|a_n| ≤ |b_n|for all 1≤n≤N anda_nb_n>0 we have

N n=1

a_nu_n≤K_s N n=1

b_nu_n.

That is, the suppression constant can replace the unconditional constant in equation (3.1) when the sign of the coefficients in the linear combinations of the basis coincide.

3.3.Show that the sequence (e_n)^∞_n=1 of standard unit vectors is a monotone basic sequence forJ in both norms · J and · 0 (see Proposition 3.4.2).