Complementability of c 0

The Classical Sequence Spaces

2.5 Complementability of c 0

Proof. If _∞

n=1x_n is WUC then _∞

n=1x^∗(x_n) is absolutely convergent for everyx^∗∈X^∗, which is equivalent to saying that_∞

k=1x^∗(x_nk) converges for each subseries_∞

k=1x_nkand eachx^∗∈X^∗. Hence_∞

k=1x_nkis weakly Cauchy and therefore weakly convergent by hypothesis. We deduce that _∞

n=1x_n converges unconditionally in norm by the Orlicz-Pettis theorem.

The Orlicz-Pettis theorem predates basic sequence techniques. It was first proved by Orlicz in 1929 [162] and referenced in Banach’s book [8]. He at- tributes the result to Orlicz in the special case whenX is weakly sequentially complete so that every WUC series has the property of the theorem. However, it seems that Orlicz did know the more general statement. Independently, Pet- tis published a proof in 1938 [178]. Pettis was interested in such a result as a by-product of the study of vector measures. If Σ is a σ-algebra of sets and µ : Σ→ X is a map such that for every x^∗ ∈ X^∗ the set function x^∗◦µ is a (countably additive) measure then the Orlicz-Pettis theorem implies thatµ is countably additive in the norm topology. Thus weakly countably additive set functions are norm countably additive.

This is an attractive theorem and as a result it has been proved, reproved, and generalized many times since then. It is not clear that there is much left to say on this subject! We will suggest some generalizations in the Problems.

2.5 Complementability ofc0 45 Proposition 2.5.2.The space_∞ is an isometrically injective space. Hence, if a Banach spaceX has a subspaceEisomorphic to_∞, thenE is necessarily complemented inX.

Proof. SupposeEis a subspace ofXandT :E→_∞is bounded. ThenT e= (e^∗_n(e))^∞_n=1for some sequence (e^∗_n)^∞_n=1 inE^∗; clearlyT= sup_ne^∗_n.By the Hahn-Banach theorem we choose extensions x^∗_n ∈X^∗ with x^∗_n =e^∗_n for eachn.By letting ˜T x= (x^∗_n(x))^∞_n=1 we are done.

c₀ is a subspace of _∞ (its bidual) and it is easy to see that c₀ will be injective if and only if it is complemented in _∞. Must a Banach space be complemented in its bidual? Certainly this is true for any space which is the dual of another space since for any Banach space X the space X^∗ is always complemented in its bidual, X^∗∗∗. To see this consider the natural embedding j : X → X^∗∗. Then j^∗ : X^∗∗∗ → X^∗ is a norm-one operator.

Denote by J the canonical injection of X^∗ into X^∗∗∗. We claim that j^∗J is the identity I_X∗ on X^∗. Indeed, suppose x^∗ ∈ X^∗ and that x ∈ X. Then x, j^∗J(x^∗)=jx, J x^∗=x, x^∗.Thusj^∗ is a norm-one projection ofX^∗∗∗

onto X^∗. If X is isomorphic (but not necessarily isometric) to a dual space we leave for the reader the details to check thatX will still be complemented in its bidual. So we may also ask ifc₀is isomorphic to a dual space.

As we will see next,c0isnotcomplemented in_∞. This was proved essen- tially by Phillips [180] in 1940 although first formally observed by Sobczyk [208] the following year. Phillips in fact proved the result for the subspacecof convergent sequences. The proof we give is due to Whitley [220] and requires a simple lemma:

Lemma 2.5.3.Every countably infinite set S has an uncountable family of infinite subsets {Ai}i∈I such that any two members of the family have finite intersection.

Proof. The proof is very simple but rather difficult to spot! Without loss of generality we can identifySwith the set of the rational numbersQ. For each irrational number θ, take a sequence of rational numbers (q_n)^∞_n=1 converging toθ.Then the sets of the formAθ={(q_n)^∞_n=1:q_n→θ}verify the lemma.

IfAis any subset ofNwe denote by_∞(A) the subspace of_∞given by

_∞(A) =

ξ= (ξ(k))^∞_k=1∈_∞:ξ(k) = 0 ifk∈A .

Theorem 2.5.4.Let T : _∞ →_∞ be a bounded operator such that T ξ = 0 for allξ∈c₀.Then there is an infinite subsetAofNso thatT ξ= 0for every ξ∈_∞(A).

Proof. We use the family (Ai)_i∈Iof infinite subsets ofNgiven by Lemma 2.5.3.

Suppose that for every such set we can findξ_i∈_∞(Ai) withT ξ_i= 0.We can assume by normalization that ξ_i∞ = 1 for every i ∈ I. There must exist

n∈Nso that the setIn ={i∈ I : ξ_i(n)= 0}is uncountable. Similarly, there exists k ∈N so that the setIn,k ={i : |ξ_i(n)| ≥ k⁻¹} is also uncountable.

For each i∈ In,k choose α_i with|α_i|= 1 and α_iξ_i(n) =|ξ_i(n)|. LetFbe a finite subset of In,k.Consider y =

i∈Fα_iξ_i.Since the inter- section of the supports of any two distinctξ_i is finite we can writey=u+v whereu_∞≤1 andv has finite support. Thus

T y_∞=T u_∞≤ T, and so

e^∗_n(T y) =

i∈F

|ξ_i(n)| ≤ T.

It follows that if |F| =m we have mk⁻¹ ≤ T, i.e., m ≤kT. Since this holds for every finite subset ofIn,k we have shown thatIn,k is in fact finite, which is a contradiction.

Theorem 2.5.5 (Phillips-Sobczyk, 1940-1). There is no bounded projec- tion from_∞ ontoc₀.

Proof. If P is such a projection we can apply Theorem 2.5.4 to T =I−P, with I the identity operator on _∞, and then it is clear thatP ξ =ξ for all ξ∈_∞(A) for some infinite setA, which gives a contradiction.

Corollary 2.5.6.c₀ is not isomorphic to a dual space.

Proof. Ifc₀were isomorphic to a dual space then, by the comments that follow the proof of Proposition 2.5.2,c₀should be complemented inc^∗∗₀ , which would lead to contradiction with Theorem 2.5.5.

Several comments are in order here. Theorem 2.5.4 proves more than is needed for Phillips-Sobczyk’s theorem. It shows that there is no bounded, one-to-one operator from the quotient space _∞/c₀ into _∞; in other words the points of_∞/c₀ cannot be separated by countably many bounded linear functionals. (Of course, if E is a complemented subspace of a Banach space X, thenX/Emust be isomorphic to a subspace ofXwhich is complementary toE.)

Now we are also in position to note that c0 is not an injective space.

Actually there are no separable injective spaces, but we will see this later, when we discuss the structure of _∞ in more detail. For the moment let us notice the dual statement of Theorem 2.3.1.

Theorem 2.5.7.If X is a separable Banach space then X embeds isometri- cally into _∞.

2.5 Complementability ofc0 47 Proof. Let (x_n)^∞_n=1 be a dense sequence inX. For each integern pickx^∗_n ∈ X^∗ so that x^∗_n = 1 and x^∗_n(x_n) = x_n. The sequence (x^∗_n)^∞_n=1 ⊂ X^∗ is norming in X. Therefore the operator T : X →_∞ defined for each x in X byT(x) = (x^∗_n(x))^∞_n=1 provides the desired embedding.

Thus X separable can only be injective if it is isomorphic to a comple- mented subspace of_∞. Therefore classifying the complemented subspaces of _∞becomes important; we will see in Chapter 5 the (already mentioned) the- orem of Lindenstrauss [129] that_∞is a prime space and this will answer our question.

In the meantime we turn to Sobczyk’s main result in his 1941 paper, which gives some partial answers to these questions. The proof we present here is due to Veech [219].

Theorem 2.5.8 (Sobczyk, 1941). Let X be a separable Banach space. If E is a closed subspace ofX andT :E −→c₀ is a bounded operator then there exists an operatorT˜:X−→c₀ such that T˜|E =T andT˜ ≤2T.

Proof. Without loss of generality we can assume thatT= 1. It is immediate to realize that the operatorT must be of the form

T x= (f_n^∗(x))^∞_n=1, x∈E

for some (f_n^∗) ⊂ E^∗. Moreover f_n^∗ ≤ 1 for all n and (f_n^∗) converges to 0 in the weak^∗ topology of E^∗. By the Hahn-Banach theorem, for eachn∈N there exists ϕ^∗_n ∈X^∗,ϕ^∗_n ≤1, such thatϕ^∗_n|E=f_n^∗.

X separable implies that (BX∗, w^∗) is metrizable (Lemma 1.4.1). Letρbe the metric on BX^∗ that induces the weak^∗ topology onBX^∗. We claim that lim_n→∞ρ(ϕ^∗_n, B_X∗ ∩E^⊥) = 0. If this is not the case, there would be some >0 and a subsequence (ϕ^∗_n

k) of (ϕ^∗_n) such that ρ(ϕ^∗_n

k, B_X∗ ∩E^⊥)≥for every k. Let (ϕ^∗_n

kj) be a subsequence of (ϕ^∗_n

k) such that ϕ^∗_n

kj

w^∗

−→ϕ^∗. Then ϕ^∗∈E^⊥∩B_X∗ since for eache∈E we have

ϕ^∗(e) = lim

j ϕ^∗_n

kj(e) = lim

j f_n^∗

kj(e) = 0.

Hence

ρ(ϕ^∗_n

kj, ϕ^∗)≥ for allj. (2.1) On the other hand

lim

j→∞ρ(ϕ^∗_n

kj, B_X∗∩E^⊥) =ρ(ϕ^∗, B_X∗∩E^⊥) = 0 (2.2) since the function ρ( ·, B_X∗∩E^⊥) is weak^∗ continuous onB_X∗. Clearly we cannot have (2.1) and (2.2) at the same time, so our claim holds.

Recall thatE^⊥is weak^∗closed, henceB_X∗∩E^⊥is weak^∗compact. There- fore for eachnwe can pickv_n^∗ ∈B_X∗∩E^⊥ such that

ρ(ϕ^∗_n, v_n^∗) =ρ(ϕ^∗_n, B_X∗∩E^⊥).

Letx^∗_n =ϕ^∗_n−v_n^∗ and define the operator ˜T onX by ˜T(x) = (x^∗_n(x)). Notice that ˜T(x)∈c₀ becausex^∗_n−→^w∗ 0. Moreover, for eachx∈X we have

T(x)˜ = sup

n |x^∗_n(x)|= sup

n

(|ϕ^∗_n(x)−v_n^∗(x)|)≤sup

n

(ϕ^∗_n+v^∗_n)x ≤2x, soT˜ ≤2.

Corollary 2.5.9.If E is a closed subspace of a separable Banach space X andE is isomorphic to c0, then there is a projection P fromX ontoE.

Proof. Suppose that T : E →c0 is an isomorphism and let ˜T : X → c0 be the extension of T given by the preceding theorem. Then P = T⁻¹T˜ is a projection fromX ontoE. (Note that sinceT˜ ≤2T, ifE isisometricto c₀thenP ≤2.)

Remark 2.5.10.It follows that if a separable Banach space X contains a copy ofc₀thenX is not injective.

We finish this chapter by observing that in light of Theorem 2.5.8 it is natural to define a Banach space Y to beseparably injective if whenever X is a separable Banach space,E is a closed subspace ofX and T :E →Y is a bounded operator then T can be extended to an operator ˜T :X →Y. It was for a long time conjectured that c₀ is the only separable and separably injective space. This was solved by Zippin in 1977 [225], who showed that, indeed,c₀is, up to isomorphism, the only separable space which is separably injective.

We also note that the constant 2 in Theorem 2.5.8 is the best possible (see Problem 2.7).

Problems

2.1.LetT :X →Y be an operator between the Banach spacesX,Y. (a) Show that if T is strictly singular then in every infinite-dimensional sub- spaceEofX there is a normalized basic sequence (x_n) withT x_n<2⁻ⁿx_n for alln.

(b) Deduce thatT is strictly singular if and only if every infinite-dimensional closed subspace E contains a further infinite-dimensional closed subspace F so that the restriction ofT toF is compact.

2.2.Show that the sum of two strictly singular operators is strictly singular.

Show also that ifT_n :X →Y are strictly singular andT_n−T →0 thenT is strictly singular.

2.5 Complementability ofc0 49 2.3.Show that the set of all strictly singular operators on a Banach space X forms a closed two-sided ideal in the algebra L(X) of all bounded linear operators fromX toX.

2.4.Show that if 1< p < ∞and T :_p → _p is not compact then there is a complemented subspace E of_p so that T is an isomorphism of E onto a complemented subspaceT(E). Deduce that the Banach algebraL(_p) contains exactly one proper closed two-sided ideal (the ideal of compact operators).

Note that every strictly singular operator is compact in these spaces.

2.5.Show that L(_p⊕_r) for p =r contains at least two nontrivial closed two-sided ideals.

2.6.SupposeXis a Banach space whose dual is separable. Suppose that x^∗_n is a series inX^∗ which has the property that every subseries

x^∗_n

kconverges weak^∗. Show that

xn converges in norm. [Hint: Every x^∗∗ ∈ X^∗∗ is the limit of a weak^∗ converging sequence fromX.]

2.7.Letc be the subspace of_∞ of converging sequences. Show that for any bounded projectionP ofcontoc₀ we haveP ≥2.This proves that 2 is the best possible constant in Sobczyk’s theorem (Theorem 2.5.8).

2.8.In this exercise we will focus on the special properties of ₁ as a target space for operators and show itsprojectivity.

(a) SupposeT :X →1 is an operator from a Banach spaceX onto1. Show that then X contains a complemented subspace isomorphic to₁.

(b) Prove that ifY is a separable infinite-dimensional Banach space with the property that wheneverT :X →Y is a bounded surjective operator thenY is isomorphic to a complemented subspace ofX, thenY is isomorphic to₁. 2.9.LetX be a Banach space.

(a) Show that for any x^∗∗ ∈X^∗∗ and any finite-dimensional subspace E of X^∗ there existsx∈X such that

x<(1 +)x^∗∗, and

x^∗(x) =x^∗∗(x^∗), x^∗∈E.

(b) Use part (a) to deduce the following result of Bessaga and Pelczy´nski ([12]):

IfX^∗contains a subspace isomorphic toc₀ thenX contains a complemented subspace isomorphic to ₁, and henceX^∗ contains a subspace isomorphic to _∞. In particular, no separable dual space can contain an isomorphic copy of c₀. [This may also be used in Problem 2.6.]

2.10.For an arbitrary set Γ we definec₀(Γ) as the space of functionsξ: Γ→R such that for each >0 the set{γ : |ξ(γ)|> } is finite. When normed by ξ= max_γ∈Γ|ξ(γ)|, the spacec₀(Γ) becomes a Banach space.

(a) Show that c₀(Γ)^∗ can be identified with ₁(Γ) the space of functions η : Γ→Rsuch thatη∈c₀(Γ) andη=

γ∈Γ|η(γ)|<∞. (b) Show that1(Γ)^∗=_∞(Γ).

(c) Show, using the methods of Lemma 2.5.3 and Theorem 2.5.4, that c₀(R) is isomorphic to a subspace of_∞/c₀.

2.11.Let Γ be an infinite set and letPΓ denote its power setPΓ ={A: A⊂ Γ}.

(a) Show that ₁(PΓ) is isometric to a subspace of _∞(Γ). [Hint: For each γ∈Γ defineϕ_γ ∈_∞(PΓ) byϕ_γ = 1 whenγ∈Aand−1 whenγ /∈A.]

(b) Show that if₁(Γ) is a quotient of a subspace ofX then₁(Γ) embeds into X (compare with Problem 2.8).

(c) Deduce that if₁(Γ) embeds into X then₁(PΓ) embeds intoX^∗. (d) Deduce that^∗∗₁ contains an isometric copy of₁(PR).

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