Banach Spaces of Continuous Functions
4.5 Spaces of continuous functions on countable compact metric spaces
Rf(t) =˜ Rf1(t). . . Rfn(t),
so ˜Rf∈ C[0,1].The linear span of such functions is again dense by the Stone- Weierstrass theorem so ˜Rmaps into C[0,1].
Iff ∈ C([0,1]N) is of the formf1(t1). . . fn(tn) then it is clear that ˜Rf◦ψ˜= f.It follows that this equation holds for allf ∈ C([0,1]N).
Thus C([0,1]N) is isomorphic to a norm-one complemented subspace of C(∆N) or C(∆) as ∆ is homeomorphic to ∆N.
Now, supposeK is an uncountable compact metric space. ThenC(K) is isomorphic to a complemented subspace of C([0,1]N) by combining Proposi- tion 4.4.3 and Theorem 4.4.4. Hence, by the preceding argument,C(K) is iso- morphic to a complemented subspace ofC(∆). On the other handC(∆) is iso- morphic to a complemented subspace ofC(K) again by Proposition 4.4.3 and Theorem 4.4.4. We also have Proposition 4.4.5 which givesc0(C(∆))≈ C(∆).
We can apply Theorem 2.2.3 to deduce that C(K) ≈ C(∆). Of course, the same reasoning givesC[0,1]≈ C(∆).
4.5 Spaces of continuous functions on countable compact
96 4 Banach Spaces of Continuous Functions
IfKdoes not have finite index then its index can be defined as a countable ordinal. This was used by Bessaga and Pelczy´nski to give a complete classi- fication, up to linear isomorphism, of all C(K) for K countable. But we will not pursue this; instead we will give one result in the direction of classifying suchC(K)-spaces.
Theorem 4.5.2.LetK be a compact metric space. The following conditions are equivalent:
(i)K is countable and has finite Cantor-Bendixson index;
(ii)C(K)≈c0;
(iii)C(K)embeds in a space with unconditional basis;
(iv)C(K)has property (u).
Let us point out that this theorem greatly extends Karlin’s theorem (see Proposition 3.5.4 (ii)) thatC[0,1] has no unconditional basis.
Proof. (i)⇒ (ii). Let us suppose, first, that σ(K) = 1. Then K is a finite set, sayK ={s1, . . . , sn}. LetV1, . . . , Vn be disjoint open neighborhoods of s1, . . . , sn, respectively.V1, V2, . . . , Vn must also be closed sets since, for each j, no sequence in Vj can converge to a point which does not belong toVj. If we denote Vn+1 =K\(V1∪ · · · ∪Vn), Vn+1 must be a finite set of isolated points and is also clopen; we therefore can absorb it into, say, V1 without changing the conditions. Now,Ksplits inton-clopen setsV1, . . . , Vnand each Vj is homeomorphic toγN.Hence C(K) is isometric to the∞-product ofn copies ofc, thus it is isomorphic toc0.
The proof of this implication is completed by induction. Assume we have shown thatC(K)≈c0ifσ(K)< n,n≥2, and suppose thatσ(K) =n.Then C(K)≈c0.Consider the restriction map f →f|K.By Theorem 4.4.4,C(K) is isomorphic to C(K)⊕E, where E denotes the kernel of the restriction f → f|K. If U =K\K is the set of isolated points ofK then E can be identified with c0(U), which is isometric to c0.Hence C(K) is isomorphic to c0.
(ii)⇒(iii) is trivial, and (iii)⇒(iv) is a consequence of Proposition 3.5.3.
(iv)⇒(i) First observe that ifC(K) has property (u), then it immediately follows that K is countable by combining Theorem 4.4.8 with the fact that the space C[0,1] fails to have property (u). This means thatM(K) contains only purely atomic measures and that C(K)∗ = 1(K) is separable. Thus C(K)∗∗=∞(K).
Supposeh is an arbitrary element in ∞(K) with h ≤ 1. Then, since BC(K)is weak∗dense inB∞(K)by Goldstine’s theorem, andB∞(K)is weak∗ metrizable by Lemma 1.4.1, it follows that we can find a sequence (gn)∞n=1 in C(K) with gn ≤1 which converges weak∗ toh.(gn)∞n=1 is a weakly-Cauchy sequence inC(K), so by property (u) we can find a WUC series∞
n=1fn such that
gn −n k=1fk
n converges weakly to zero in C(K). This means that ∞
k=1fk =hfor the weak∗ topology. In particular we have that
∞ k=1
fk(s) =h(s), s∈K.
Since
fn is a WUC series, there is a constantM such that sup
N
sup
j=±1
N k=1
kfk(s)= ∞ k=1
|fk(s)| ≤M for everys∈K.
Putφ(s) =∞
k=1|fk(s)|andψ(s) =∞
k=1
|fk(s)| −fk(s)
=φ(s)−h(s).
Bothφandψare lower semicontinuous functions onK, that is, for everya∈R the setsφ−1(a,∞) andψ−1(a,∞) are open. We also have φ,ψ ≤M and h=φ−ψ.
Suppose thatKfails to have finite Cantor-Bendixson index. Then each of the setsEn =K(n−1)−K(n) is nonempty forn= 1,2, . . . (here, K(0) =K).
We pick a particularh∈∞(K) withh ≤1 so that h(s) = (−1)n, s∈En.
Since K fails to have finite index, the setK\ ∪∞n=1En is nonempty and we can definehto be zero on this set. Thus, we can writeh=φ−ψas above. If we put
an= sup
s∈E2n
φ(s), n= 1,2, . . . then|an| ≤M for alln.
Suppose > 0 and that n ≥ 1. Then, there exists s0 ∈ E2n so that φ(s0)> an−.Thus by the lower semicontinuity ofφthere is an open setU0 containings0so thatφ(s)> an−for everys∈U0.In particularU0∩K(2n−2) is relatively open in K(2n−2) and U0∩E2n−1 = ∅. Hence there exists s1 ∈ U0∩E2n−1so thatφ(s1)> an−. Thusψ(s1)> an+ 1−.Next we find an open set U1containings1 so thatψ(s)> an+ 1−fors∈U1.Reasoning as above we can finds2∈U1∩E2n−2 withψ(s2)> an+ 1−. But this implies φ(s2)> an+ 2−and soan−1≥an+ 2−.Since >0 is arbitrary we have:
an≤an−1−2, n= 1,2, . . . .
Clearly this contradicts the lower bound of−M on the sequence (an)∞n=1.The contradiction shows thatK has finite Cantor-Bendixson index.
If K and L are countable compact metric spaces with different but fi- nite Cantor-Bendixson indices thenK andL are not homeomorphic but the spaces C(K) and C(L) are both isomorphic to c0. Later we will see that, up to equivalence, there is only one unconditional basis ofc0, in the sense that any normalized unconditional basis is equivalent to the canonical basis.
Remark 4.5.3.Notice that sinceC(K)∗is isometric to1forevery countable compact metric spaceK, the Banach space1is isometric to the dual of many nonisomorphic Banach spaces.
98 4 Banach Spaces of Continuous Functions
Problems
4.1.LetKbe any compact Hausdorff space. Show that any extreme point of BC(K)∗ is of the form±δswhereδsis the probability measure defined on the Borel sets ofK byδs(B) = 1 ifs∈B and 0 otherwise.
4.2. The Banach-Stone Theorem.Suppose K and Lare compact Haus- dorff spaces such thatC(K) andC(L) are isometric. Show thatK andLare homeomorphic. [Hint: Argue that ifU :C(K)→ C(L) is any (onto) isometry, thenU∗ maps extreme points of the dual ball to extreme points.]