Boundedly-complete and shrinking bases

Special Types of Bases

3.2 Boundedly-complete and shrinking bases

By hypothesis, ifN ≤n_k<· · ·< n_k+l we have

k+l j=k+1

a_nje_nj≤K

n_k+l

j=n_k+1

a_je_j< , and so _∞

k=1a_nke_nk is Cauchy.

Definition 3.1.4.Let (e_n) be an unconditional basis of a Banach space X.

Theunconditional basis constant,K_u, of(e_n) is the least constantK so that equation (3.1) holds. We then say that (e_n) is K-unconditional whenever K≥K_u.

Remark 3.1.5.Suppose (e_n)^∞_n=1is an unconditional basis for a Banach space X. For each sequence of scalars (α_n) with|α_n|= 1, letT_(αn):X →X be the isomorphism defined byT_(αn)(_∞

n=1a_ne_n) =_∞

n=1α_na_ne_n. Then Ku= sup

T_(αn): (αn) scalars, |αn|= 1 for alln

.

If (e_n)^∞_n=1is an unconditional basis ofXandAis any subset of the integers then there is a linear projectionPAfromX onto [ek :k∈A] defined for each x=_∞

k=1e^∗_k(x)ek by

P_A(x) =

k∈A

e^∗_k(x)e_k.

P_A is bounded by the same argument used in the proof of Proposition 3.1.3.

{P_A : A ⊂ N} are the natural projections associated to the unconditional basis (e_n) and the number

Ks= sup

A

PA

(which is finite by the Uniform Boundedness principle) is called thesuppres- sion constantof the basis. Let us observe that in general we have

1≤Ks≤Ku≤2Ks.

In the older literature the term absolute basis is often used in place of unconditional basis, but this usage has largely disappeared. Unconditional bases seem to have first appeared in work of Karlin in 1948 [107]. In particular Karlin proved thatC[0,1] fails to have an unconditional basis. We will prove this later in this chapter.

3.2 Boundedly-complete and shrinking bases

Suppose (e_n)^∞_n=1is a basis for a Banach spaceXwith biorthogonal functionals (e^∗_n)^∞_n=1 ⊂X^∗. One of our goals in this section is to establish necessary and

sufficient conditions for (e^∗_n)^∞_n=1 to be a basis forX^∗. This is not always the case. For example, the coordinate functionals of the standard basis of₁cannot be a basis for^∗₁ since ^∗₁ is not separable. We will first prove that, at least, (e^∗_n)^∞_n=1 is a basic sequence inX^∗.

Proposition 3.2.1.Suppose that (e^∗_n)^∞_n=1 is the sequence of biorthogonal functionals associated to a basis(e_n)^∞_n=1 of a Banach spaceX. Then(e^∗_n)^∞_n=1 is a basic sequence in X^∗ with basis constant no bigger than that of(e_n)^∞_n=1. Proof. Given (e^∗_n)^∞_n=1, consider the subspaceH ofX^∗ given by

H=

x^∗∈X^∗ : S_N^∗(x^∗)−x^∗ →0

, (3.2)

where (S_N^∗)^∞_N=1 is the sequence of adjoint operators of the partial sum pro- jections associated to (en)^∞_n=1:

S_N^∗ :X^∗→X^∗, S_N^∗(x^∗) = N k=1

x^∗(e_k)e^∗_k.

Clearly (e^∗_n)^∞_n=1 is a basis forH, hence (e^∗_n)^∞_n=1 is basic. Notice that sup

N

S_N^∗|H_H→H≤sup

N

S_N^∗_X∗→X^∗ = sup

N

S_N, which gives the latter statement in the proposition.

Definition 3.2.2.Suppose thatXis a normed space and thatY is a subspace ofX^∗. Let us consider a new norm onX defined by

x_Y = sup

|y^∗(x)|:y^∗∈Y,y^∗= 1 . If there is a constantc≤1 such that for all x∈X we have

cx ≤ x_Y ≤ x, thenY is said to be ac-norming subspace forX inX^∗.

The next result shows that if (e_n)^∞_n=1 is a basis for a Banach spaceX with basis constant Kthen the subspace [e^∗_n] =H ofX^∗ is reasonably big, in the sense that it is 1/K-norming forX.

Lemma 3.2.3.Let (e_n)^∞_n=1 be a basis for a Banach space X with basis con- stant K and biorthogonal functionals (e^∗_n)^∞_n=1. Then H = [e^∗_n] is a K⁻¹- norming subspace for X inX^∗. Thus the norm onX defined by

xH = sup

|h(x)| : h∈H,h ≤1 , satisfies

x

K ≤ x_H ≤ x (3.3)

for allx∈X.

3.2 Boundedly-complete and shrinking bases 55 Proof. Let x ∈ X. Since H ⊂ X^∗, it follows immediately that xH ≤ sup{|x^∗(x)| : x^∗ ∈ X^∗,x^∗ ≤ 1} = x. For the other inequality, pick x^∗∈S_X∗ so thatx^∗(x) =x. Then for eachN,

|(S_N^∗x^∗)x|

K ≤ |(S_N^∗x^∗)x|

S_N^∗x^∗ ≤sup

|h(x)|:h∈H,h ≤1

=xH. Now we letNtend to infinity and use that ifS_Nx−x →0 then|S_N^∗x^∗(x)|=

|x^∗(SNx)| → x.

Remark 3.2.4.The previous result can be interpreted as saying thatX em- beds isomorphically in H^∗ via the map x →j(x)|H, where j is the natural embedding ofX in its second dualX^∗∗. In the case that the basis (e_n)^∞_n=1 is monotone, equation (3.3) implies thatX embeds isometrically inH^∗. Definition 3.2.5.A basis (e_n)^∞_n=1 of a Banach spaceX is shrinking if the sequence of its biorthogonal functionals (e^∗_n)^∞_n=1is a basis forX^∗, i.e., if [e^∗_n] = X^∗.

Proposition 3.2.6.A basis(e_n)^∞_n=1 of a Banach spaceX is shrinking if and only if wheneverx^∗∈X^∗,

lim

N→∞x^∗|[e_n]_n>N= 0, (3.4)

where

x^∗|[e_n]_n>N= sup

|x^∗(y)|:y∈[e_n]_n>N

.

Proof. Suppose that (e^∗_n)^∞_n=1 is a basis forX^∗. Everyx^∗∈X^∗ can be decom- posed as (x^∗−S_N^∗x^∗) +S_N^∗x^∗ for eachN. Then the claim follows because

x^∗|[e_n]_n>N ≤x^∗−S_N^∗x^∗

|[e_n]_n>N+S_N^∗x^∗|[e_n]_n>N

this term is 0

≤ x^∗−S_N^∗x^∗

and we know that lim_N→∞x^∗−S_N^∗x^∗= 0.

For the converse, assume that (3.4) holds. LetK be the basis constant of (en)^∞_n=1 andx^∗ be an element inX^∗. Since for anyx∈X, (IX−SN)(x) is in the subspace [en]n>N, we have

|(x^∗−S_N^∗x^∗)(x)|=|x^∗(I_X−S_N)(x)|

≤x^∗|[e_n]_n≥N+1I_X−S_N x

≤(K+ 1)x^∗|[e_n]_n≥N+1x.

Hencex^∗−S_N^∗x^∗ ≤(K+1)x^∗|[e_n]_n≥N+1and so lim_N→∞x^∗−S_N^∗x^∗= 0.

ThusX^∗= [e^∗_n] and we are done.

Proposition 3.2.7.A basis(e_n)^∞_n=1 of a Banach spaceX is shrinking if and only if every bounded block basic sequence of(e_n)^∞_n=1 is weakly null.

Proof. Assume (en)^∞_n=1 is not shrinking. ThenH =X^∗, hence there isx^∗ in X^∗\[e^∗_n],x^∗= 1, such that the series_∞

n=1x^∗(en)e^∗_nconverges tox^∗in the weak^∗ topology ofX^∗ but it does not converge in the norm topology ofX^∗. Using the Cauchy condition we can find two sequences of positive integers (p_n), (q_n) and δ > 0 such that p₁ ≤ q₁ < p₂ ≤ q₂ < p₃ ≤ q₃ < . . . and q_k

n=p_kx^∗(e_n)e^∗_n > δ for all k∈ N. Thus for each k there exists x_k ∈ X, x_k= 1, for whichq_k

n=p_kx^∗(e_n)e^∗_n(x_k)> δ. Put y_k=

q_k

n=p_k

e^∗_n(x_k)e_n, k= 1,2, . . .

(y_k)^∞_k=1 is a block basis of (e_n)^∞_n=1 which is not weakly null since x^∗(y_k)> δ for allk.

The converse implication follows readily from Proposition 3.2.6.

Definition 3.2.8.Let X be a Banach space. A basis (e_n)^∞_n=1 for X is boundedly-completeif whenever (a_n)^∞_n=1 is a sequence of scalars such that

sup

N

N n=1

anen<∞, then the series_∞

n=1a_ne_n converges.

Example 3.2.9.(a) The canonical basis of_pfor 1< p <∞is both shrinking and boundedly-complete. In 1 the canonical basis is obviously boundedly- complete, but 1 cannot have a shrinking basis because its dual, _∞, is not separable.

(b) As for c0, its natural basis is shrinking but not boundedly complete:

the series_∞

n=1e_n is not convergent inc₀despite the fact that sup

N

N n=1

e_n

∞= sup

N

(1,1, . . . ,1

N

,0,0, . . .)

∞= 1.

On the other hand, the summing basis ofc₀, (f_n)^∞_n=1, is not shrinking because the linear functionale^∗₁satisfiese^∗₁(f_n) = 1 for alln, so equation (3.4) cannot hold. (f_n)^∞_n=1 is not boundedly-complete either:

sup

N

N n=1

(−1)ⁿf_n

∞= 1, but the series_∞

n=1(−1)ⁿfn is not convergent.

3.2 Boundedly-complete and shrinking bases 57 Theorem 3.2.10.Let(e_n)^∞_n=1be a basis for a Banach spaceX with biorthog- onal functionals(e^∗_n)^∞_n=1. The following are equivalent:

(i)(e_n)^∞_n=1 is a boundedly-complete basis forX, (ii)(e^∗_n)^∞_n=1 is a shrinking basis forH,

(iii) The canonical mapj :X →H^∗ defined byj(x)(h) =h(x), for allx∈X andh∈H, is an isomorphism.

Proof. (i)⇒(iii) Using Remark 3.2.4 we need only show thatj is onto. For eachh^∗ ∈H^∗ there existsx^∗∗ ∈X^∗∗ so thatx^∗∗|H =h^∗. Let us consider the formal series_∞

n=1x^∗∗(e^∗_n)e_n inX. For eachN∈N, N

n=1

x^∗∗(e^∗_n)e_n=S_N^∗∗x^∗∗, whereS^∗∗_N is the double adjoint ofS_N. Hence

N n=1

x^∗∗(e^∗_n)e_n=S_N^∗∗x^∗∗ ≤sup

N

S^∗∗_N x^∗∗=Kx^∗∗.

(e_n)^∞_n=1 boundedly-complete implies that _∞

n=1x^∗∗(e^∗_n)e_n converges to some x∈X. Nowj(x) =h^∗ since for eachk∈Nwe have

j(x)(e^∗_k) =e^∗_k(x) =x^∗∗(e^∗_k) =h^∗(e^∗_k).

(iii) ⇒ (ii) Assume that j : X → H^∗ is an isomorphism onto. Then (j(e_n))^∞_n=1is a basis forH^∗and it is also the sequence of coordinate functionals for (e^∗_n)^∞_n=1. That means (e^∗_n)^∞_n=1 is a shrinking basis for H.

(ii)⇒(i) Let (a_n) be a sequence of scalars for which sup

N

N n=1

a_ne_n<∞. (3.5)

For eachNthe norm ofj(N

n=1a_ne_n) as a linear functional onH is equivalent to the norm ofN

n=1a_ne_n inX. Therefore, by (3.5), (N

n=1a_nj(e_n))^∞_N=1is a bounded sequence inX^∗∗. The Banach-Alaoglu theorem yields the existence of a weak^∗ cluster point, h^∗ ∈X^∗∗, of that sequence. In particular we have h^∗(e^∗_n) =an for eachn. Using the hypothesis we can write

h^∗= ∞ n=1

h^∗(e^∗_n)j(en) = ∞ n=1

anj(en),

where the series converges in the norm topology ofH^∗. Sincej is an isomor- phism, the series_∞

n=1a_ne_n converges in the norm topology ofX.

Corollary 3.2.11.c₀ has no boundedly-complete basis.

Proof. It follows from Theorem 3.2.10, taking into account that c₀ is not isomorphic to a dual space (Corollary 2.5.6).

Theorem 3.2.12.Let(e_n)^∞_n=1be a basis for a Banach spaceX with biorthog- onal functionals(e^∗_n)^∞_n=1. The following are equivalent:

(i)(en)^∞_n=1 is a shrinking basis forX,

(ii)(e^∗_n)^∞_n=1 is a boundedly-complete basis forH, (iii)H=X^∗.

Proof. (i)⇒(ii) Suppose that (a_n)^∞_n=1 is a sequence of scalars such that the sequence (N

n=1a_ne^∗_n)^∞_N=1 is bounded in X^∗ and let x^∗ ∈ X^∗ be a weak^∗ cluster point of this sequence. Since lim_N→∞(N

n=1a_ne^∗_n)(e_k) =a_k, it follows that x^∗(ek) =ak for eachk. Thus the series_∞

n=1ane^∗_n converges tox^∗. (ii)⇒(i) Suppose now that (e^∗_n)^∞_n=1is boundedly-complete. For anyx^∗in X^∗we know that the series_∞

n=1x^∗(en)e^∗_nconverges in the weak^∗topology of X^∗tox^∗. In particular, the sequence (N

n=1x^∗(e_n)e^∗_n)^∞_N=1is norm-bounded in X^∗. Hence, by the bounded-completeness of (e^∗_n)^∞_n=1, the series_∞

n=1x^∗(e_n)e^∗_n must converge to x^∗ in norm, so (e^∗_n)^∞_n=1 is a basis forX^∗.

(i)⇔(iii) is obvious.

Now we come to the main result of the section, which is due to James [80].

Theorem 3.2.13 (James, 1951).LetX be a Banach space. IfX has a basis (en)^∞_n=1 thenX is reflexive if and only if(en)^∞_n=1 is both boundedly-complete and shrinking.

Proof. Assume that X is reflexive and that (e_n)^∞_n=1 is a basis forX. Then X^∗ =H. If not, using the Hahn-Banach theorem, one could find 0 =x^∗∗ ∈ X^∗∗ such that x^∗∗(h) = 0 for all h ∈ H. By reflexivity there is 0 = x = _∞

n=1e^∗_n(x)e_n ∈ X such that x = x^∗∗. In particular we would have 0 = x^∗∗(e^∗_n) =e^∗_n(x) for alln, which would implyx= 0. Thus (e_n)^∞_n=1is shrinking.

Notice that (e_n)^∞_n=1is a basis forX^∗∗and is also the sequence of biorthogonal functionals associated to (e^∗_n)^∞_n=1. That implies that (e^∗_n)^∞_n=1 is a shrinking basis ofX^∗=H, hence by Theorem 3.2.10, (en)^∞_n=1 is boundedly-complete.

Conversely, (en)^∞_n=1 shrinking implies H = X^∗, and since (en)^∞_n=1 is boundedly-complete as well, the canonical mapj:X →H^∗in Theorem 3.2.10 (iii) is now the canonical embedding ofX ontoX^∗∗.

This theorem gives a criterion for reflexivity which is enormously useful, particularly in the construction of examples. Notice that the facts that the canonical basis of ₁ fails to be shrinking and that the canonical basis of c₀ fails to be boundedly-complete are explained now in the nonreflexivity of these spaces.