Banach Spaces of Continuous Functions
4.3 Isometrically injective spaces
Let us observe that if we consider A = ∞ (with the multiplication of two sequences defined coordinate-wise), Theorem 4.2.5 yields thatA=C(K) (isometrically) for some compact Hausdorff space K. This set K is usually denoted by βN. We also note that if (Ω,Σ, µ) is any σ-finite measure space thenL∞(Ω, µ) is again aC(K)-space. In each case the isomorphismpreserves order(i.e., nonnegative functions are mapped to nonnegative functions) since squares are mapped to squares.
80 4 Banach Spaces of Continuous Functions (ii) V(x+y)≤V(x) +V(y) for allx, y∈F.
A sublinear mapV :X → C(K) isminimal provided there is no sublinear mapU :X→ C(K) such thatU(x)≤V(x) for allx∈X andU =V. Lemma 4.3.3.LetX be a Banach space andF a linear subspace of X. Sup- pose V : X → C(K) and W : F → C(K) are sublinear maps such that W(y) +V(−y) ≥ 0 for all y ∈ F. If C(K) is order-complete then the map V ∧W :X → C(K)given by
V ∧W(x) = inf{V(x−y) +W(y) : y∈F}, is well defined and sublinear.
Proof. For each fixedx∈X we have
V(x−y) +W(y)≥V(−y)−V(−x) +W(y)≥ −V(−x)
for all y ∈ F. That is, −V(−x) is a lower bound of the set {V(x−y) + W(y) : y ∈ F}. Thus, by the order-completeness of C(K), we can define a mapV ∧W :F→ C(K) by
V ∧W(x) = inf{V(x−y) +W(y) : y∈F}. It is a straightforward verification to check thatV ∧W is sublinear.
Lemma 4.3.4.Let V : X → C(K) be a sublinear map. If C(K) is order- complete then there is a minimal sublinear mapW :X → C(K)withW(x)≤ V(x)for allx∈X.
Proof. Put S=
U :X → C(K) : U is sublinear andU(x)≤V(x) for allx∈X . S is nonempty (V ∈ S) and partially ordered. Let Ψ = (Ui)i∈I be a chain (i.e., a totally ordered subset) in S. Note that for each i ∈ I we have 0 = Ui(x+ (−x))≤Ui(x) +Ui(−x) for all x∈X, hence
Ui(x)≥ −Ui(−x)≥ −V(−x).
Thus, for eachx∈X, the set{Ui(x) :i∈I} ⊂ C(K) has a lower bound. By the order-completeness ofC(K), the map
UΨ(x) = inf
i∈IUi(x)
is well defined on X and sublinear. To see this, since Ψ is a totally ordered set, given i=j ∈I, without loss of generality we can assume thatUi ≤Uj. Then, for anyx, y∈X we have
UΨ(x+y)≤Ui(x+y)≤Uj(x) +Ui(y),
thereforeUΨ(x+y)−Uj(x)≤UΨ(y), which yieldsUΨ(x+y)−UΨ(y)≤UΨ(x).
Moreover,UΨ(x)≤V(x) for allx∈X. That is,UΨ ∈ S is a lower bound for the chain (Ui)i∈I. Using Zorn’s lemma we deduce the existence of a minimal elementW inS.
Lemma 4.3.5.Suppose thatC(K) is order-complete and let V :X → C(K) be a sublinear map. If V is minimal thenV is linear.
Proof. Given an element x ∈ X, let us call F its linear span, F = x. Then, W(λx) = −λV(−x) defines a linear map from F to C(K). Clearly, W(λx)≥ −V(−λx) for every realλ. Using Lemma 4.3.3 we can define onX the sublinear map
V ∧W(x) = inf
λ∈R
V(x−λx) +W(λx) .
By the minimality of V, V ∧W =V on X. ThereforeV ≤W onF, which implies that V(x) ≤ −V(−x). On the other hand,V(x) ≥ −V(−x) by the sublinearity of V, so V(−x) = −V(x). Since this holds for allx ∈ X, it is clear thatV is linear.
Theorem 4.3.6 (Goodner, Nachbin, 1949-1950). Let K be a compact Hausdorff space. Then C(K) is isometrically injective if and only ifC(K)is order-complete.
Proof. Assume, first, thatC(K) is order-complete. Let E be a subspace of a Banach space X and let S : E → C(K) be a linear operator with S = 1.
That is, for eachx∈E we have
−x ≤(Sx)(k)≤ x for allk∈K,
which, if we let 1 denote the constant function 1 onK, is equivalent to writing
−x ·1≤S(x)≤ x ·1. (4.5)
Thus, if we consider the sublinear map from X to C(K) given by V0(x) = x ·1, equation (4.5) tells us thatS(x)≥ −V0(−x) for allx∈E and so we can define onX the sublinear mapV =V0∧S as in Lemma 4.3.3:
V(x) = inf
V0(x−y) +S(y) :y∈E .
By Lemma 4.3.4 there exists T :X → C(K), a minimal sublinear map satis- fying T≤V. Lemma 4.3.5 yields thatT is linear.
OnE, we have T(x) ≤S(x) and T(−x) ≤ S(−x). Therefore, T|E = S.
Finally,T(x)≤ x ·1 andT(−x)≤ x ·1 for allx∈X, which implies that T ≤1.Thus, we have successfully extendedS from E toX.
82 4 Banach Spaces of Continuous Functions
Suppose, conversely, thatC(K) is isometrically injective. Then there is a norm-one projection P from ∞(K) onto C(K), where ∞(K) denotes the space of all bounded functions on K. Suppose that A, B are two nonempty subsets of C(K) such that f ∈A and g∈B impliesf ≤g.For eachs∈K, put a(s) = supf∈Af(s). Obviously,a∈∞(K). Let h=P(a). We will prove that f ≤h≤g for allf ∈A and allg∈B.
SinceP(1) = 1 andP has norm one, it follows that for each b ∈∞(K) withb >0 we have
P(1−λb) ≤1 for 0≤λ≤2/b.
We deduce thatP is a positive map, that is,P b≥0 wheneverb∈∞(K) and b≥0. Thus, iff ∈Athenf ≤aand, therefore,f ≤h. Analogously, ifg∈B we have g≥aand sog≥h.Hence,C(K) is order-complete.
The spaces K so that C(K) is order-complete are characterized by the property that the closure of any open set remains open; such spaces are called extremally disconnected. We refer the reader to the Problems for more infor- mation.
The natural question arises as to whether only C(K)-spaces can be iso- metrically injective. Both Nachbin and Goodner showed that an isometrically injective Banach spaceX is (isometrically isomorphic to) aC(K)-space pro- vided the unit ball ofX has at least one extreme point. The key here is that the constant function 1 is always an extreme point on the unit ball inC(K) and they needed to find an element in the space X to play this role. How- ever, two years later, in 1952, Kelley completed the argument and proved the definitive result:
Theorem 4.3.7 (Kelley, 1952). A Banach spaceX is isometrically injec- tive if and only if it is isometrically isomorphic to an order-complete C(K)- space.
Proof. We need only show the forward implication. For that, we are going to identifyX (via an isometric isomorphism) with a suitableC(K)-space which, by the isometric injectivity ofX, will be order-continuous appealing to The- orem 4.3.6.
The trick is to “find”K as a subset of the dual unit ball BX∗. Consider the set∂eBX∗ of extreme points ofBX∗ with the weak∗ topology. There is a maximal open subset,U, of∂eBX∗ subject to the property thatU∩(−U) =∅. This is an easy consequence of Zorn’s lemma again, as any chain of such open sets has an upper bound, namely, their union. LetKbe the weak∗closure of U inBX∗. K is, of course, compact and Hausdorff for the weak∗ topology.
Let us observe that K∩∂eBX∗ cannot meet −U since ∂eBX∗ \(−U) is relatively weak∗ closed in∂eBX∗. Then,K∩(−U) =∅.
We claim that ∂eBX∗ ⊂ (K∪(−K)). Indeed, suppose that there exists x∗ ∈ ∂eBX∗ \(K∪(−K)). Then there is an absolutely convex weak∗ open
neighborhood,V, of 0 such thatx∗∈/ V and (x∗+V)∩(K∪(−K)) =∅.Let U1 =U∪
(x∗+V)∩∂eBX∗
. Then U1 strictly containsU sincex∗ ∈U1. Supposey∗∈U1∩(−U1).Then eithery∗∈/ U or −y∗∈/ U; thus replacing y∗ by −y∗ if necessary we can assume y∗ ∈/ U. Then y∗ ∈ x∗+V; this implies thaty∗∈/ K∪(−K) and soy∗∈ −/ U.Hencey∗∈ −x∗−V and so 0∈2x∗+ 2V or x∗ ∈V yielding a contradiction. Thus U1∩(−U1) =∅, which contradicts the maximality ofU.
By the Krein-Milman theorem,BX∗ must be the weak∗closed convex hull ofK∪(−K) and, in particular, ifx∈X we have
x= sup
x∗∈BX∗
|x∗(x)|= max
x∗∈K|x∗(x)|.
Thus, the mapJ that assigns to eachx∈X the function ˆx∈ C(K) given by ˆ
x(x∗) = x∗(x), x∗ ∈ K, is an isometry. We can therefore use the isometric injectivity ofX (extending the mapJ−1 :J(X)→X) to define an operator T :C(K)→X such thatT(ˆx) =xfor allx∈X withT= 1.
Let us consider the adjoint map T∗ : X∗ → M(K). If u∗ ∈ U, then T∗u∗=µ∈ M(K) withµ ≤1. LetV be any weak∗ open neighborhood of u∗ relative toK and putK0=K\V. We can define v∗∈X∗ by
v∗(x) =
V
x∗(x)dµ(x∗), x∈X, andw∗∈X∗ by
w∗(x) =
K0
x∗(x)dµ(x∗), x∈X.
Thenv∗ ≤ |µ|(V) andw∗ ≤ |µ|(K0).But,
K
x∗(x)dµ=x, Tˆ ∗(u∗)=x, u∗,
hencev∗+w∗=u∗.Sinceu∗= 1≥ µ, we must have|µ|(V)+|µ|(K0) = 1.
Thus, v∗+w∗ = 1 and so the fact that u∗ is an extreme point implies that v∗=v∗u∗ andw∗=w∗u∗.
Suppose|µ|(K0) =w∗=α >0.Then, u∗(x) =α−1
K0
x∗(x)dµ(x∗), x∈X, and, in particular,
|u∗(x)| ≤ max
x∗∈K0|x∗(x)|, x∈X.
This implies thatu∗is in the weak∗closed convex hull,C, ofK0∪(−K0).But u∗must be an extreme point inCalso, so by Milman’s theorem it must belong
84 4 Banach Spaces of Continuous Functions
to the weak∗closed setK0∪(−K0).Sinceu∗ ∈/ K0we have thatu∗∈(−K0), i.e.,−u∗∈K0.Thus,K0meets−U, soKmeets−U, which is a contradiction to our previous remarks.
Hence |µ|(K0) = w∗ = 0 and so |µ(V)| = 1 for every weak∗ open neighborhood V of u∗. By the regularity of µ we must have that µ =±δu∗
(δu∗ is the point mass at u∗). Thus µ =δu∗ for u∗ ∈ U. Since T∗ is weak∗ continuous we infer thatT∗(x∗) =δx∗ for allx∗∈K.We are done because if f ∈ C(K), then
T f, x∗=f(x∗),
soJ is ontoC(K). This shows that X is a C(K)-space.
At this point we have only one example where C(K) is order-complete, namely, ∞ (although, of course, ∞(I) for any index set I will also work).
There are, however, less trivial examples as the next proposition shows.
Proposition 4.3.8.
(i) If C(K) is (isometrically isomorphic to) a dual space, then C(K) is iso- metrically injective.
(ii) If(Ω,Σ, µ)is anyσ-finite measure space, thenL∞(Ω,Σ, µ)is isometrically injective.
(iii) For any compact Hausdorff space K the space C(K)∗∗ is isometrically injective.
Proof. For (i) we will first show that P ={f ∈ C(K) : f ≥0}, the positive cone ofC(K), is closed for the weak∗topology ofC(K) (regarded now as a dual Banach space by hypothesis). By the Banach-Dieudonn´e theorem it suffices to show that P∩λBC(K) is weak∗ closed for eachλ >0.ButP∩λBC(K) = {f : f −12λ·1 ≤ 12λ} is simply a closed ball, which must be weak∗ closed.
Let us see that C(K) is order-complete and then we will invoke Theo- rem 4.3.6 to deduce that C(K) is isometrically injective. Suppose A, B are nonempty subsets of C(K) such that f ∈ A, g ∈ B imply f ≤ g. For each f ∈A andg∈B, put
Cf,g={h∈ C(K) : f ≤h≤g}.
EveryCf,g is a (nonempty) bounded and weak∗ closed set. If f1, . . . , fn∈A and g1, . . . , gn ∈ B then ∩nk=1Cfk,gk is nonempty because it contains for example max(f1, . . . , fn). Hence, by weak∗ compactness, the intersection
∩{f∈A,g∈B}Cf,g is nonempty. If we pickhin the intersection we are done.
(ii) follows directly from (i) since L∞(µ) =L1(µ)∗.
(iii) Here we observe thatM(K) is actually a vast1-sum ofL1(µ)-spaces.
Precisely, using Zorn’s lemma one can produce a maximal collection (µi)i∈I of probability measures onKwith the property that any two members of the collection are mutually singular.
Ifν ∈ M(K), for each i ∈ I we define fi ∈L1(K, µi) to be the Radon- Nikodym derivativedν/dµi. Thus,dν =fidµi+γ, where γ is singular with respect toµi.Then it is easy to show (we leave the details to the reader) that for any finite setA⊂ I we have
i∈A
fiL1(µi)≤ ν.
Hence,
i∈I
fiL1(µi)≤ ν.
Notice that the last statement implies that only countably many terms in the sum are nonzero. Put
ν0=
i∈I
fidµi,
where the series converges in M(K). It is clear that the measure ν−ν0 is singular with respect to everyµiand, as a consequence, it must vanish onK.
It follows that the mapν →(fi)i∈Idefines an isometric isomorphism between M(K) and the1-sum of the spacesL1(µi) fori∈ I.
This yields thatC(K)∗∗ can be identified with the∞-sum of the spaces L∞(µi). Using (ii) we deduce thatC(K)∗∗ is isometrically injective.
Remark 4.3.9.We should note here that there are order-complete C(K)- spaces which are not isometric to dual spaces. The first example was given in 1951 (in a slightly different context) by Dixmier [43] and we refer to Prob- lem 4.8 and Problem 4.9 for details.
There is an easy but surprising application of the preceding proposition to the isomorphic theory [167]:
Theorem 4.3.10.L∞[0,1]is isomorphic to ∞.
Proof. First, observe that∞embeds isometrically intoL∞[0,1] via the map (ξ(n))∞n=1→
∞ n=1
ξ(n)χAn(t),
where (An)∞n=1 is a partition of [0,1] into sets of positive measure. Since∞ is an injective space, it follows that∞ is complemented inL∞[0,1].
On the other hand,L∞[0,1] also embeds isometrically into∞. To see this, pick (ϕn)∞n=1, a dense sequence in the unit ball ofL1, and mapf ∈L∞[0,1] to (1
0 ϕnf dt)∞n=1. Therefore, being an injective space,L∞[0,1] is complemented in ∞.
Furthermore,∞≈∞⊕∞and
86 4 Banach Spaces of Continuous Functions
L∞[0,1]≈L∞[0,1/2]⊕L∞[1/2,1]≈L∞[0,1]⊕L∞[0,1].
Using Theorem 2.2.3 (a) (the Pelczy´nski decomposition technique) we deduce that L∞[0,1] is isomorphic to∞.
We conclude this section by showing that a separable isometrically injec- tive space is necessarily finite-dimensional.
Proposition 4.3.11.For any infinite compact Hausdorff spaceK,C(K)con- tains a subspace isometric to c0. If K is metrizable this subspace is comple- mented.
Proof. Let (Un) be a sequence of nonempty, disjoint, open subsets ofK. Such a sequence can be found by induction: simply pick U1 so that K1 =K\U1 is infinite and then take U2 ⊂ K1 such that K2 = K1\U2 is infinite and so on. Next, pick a sequence (ϕn)∞n=1 of continuous functions on K so that 0≤ϕn≤1,maxs∈Kϕn(s) = 1 and{s∈K:ϕn(s)>0} ⊂Un, for alln∈N. Then for any (an)∈c00we have
∞ n=1
anϕn= max
n |an|.
Thus (ϕn)∞n=1 is a basic sequence isometrically equivalent to the unit vector basis ofc0.
IfKis metrizable, Theorem 4.1.3 implies thatC(K) is separable and we can apply Sobczyk’s theorem (Theorem 2.5.8) to deduce that the space [ϕn]∞n=1 is complemented by a projection of norm at most two.
Proposition 4.3.12.If C(K) is order-complete andK is metrizable then K is finite.
Proof. IfKis infinite,C(K) contains a complemented copy ofc0 by Proposi- tion 4.3.11. But if, moreover,C(K) is isometrically injective this would make c0injective, which is false because c0 is uncomplemented in∞ as we saw in Theorem 2.5.5.
Corollary 4.3.13.The only isometrically injective separable Banach spaces are finite-dimensional and isometric to n∞ for somen∈N.
Proof. If X is an isometrically injective Banach space, by Theorem 4.3.7, X can be identified with an order-complete C(K)-space for some compact HausdorffK. SinceX is separable, Theorem 4.1.3 yields thatK is metrizable and, by Proposition 4.3.12,Kmust be finite. ThereforeC(K) is (isometrically isomorphic to)|∞K|.
In fact, there are no infinite-dimensional injective separable Banach spaces (even dropping isometrically) but this is substantially harder and we will see it in the next chapter.