Exercises and Further Results
1. Maximal Ideal Spaces
V
Some Uniform Algebra
This chapter develops the background from uniform algebra theory which will be needed for our analysis ofH∞below. Our treatment is quite brief. For a complete picture of the general theory the reader is referred to the books of Browder [1969], Gamelin [1969], and Stout [1971].
However, two topics special to H∞ will be covered in detail. In Section 2 we prove Marshall’s theorem that the Blaschke products generate H∞. In Section 5, three theorems on the predual ofH∞ are proved by representing linear functionals as measures on the ˇSilov boundary ofH∞.
Proof. Because m is linear, we only have to prove that m ≤1. If m is unbounded, or ifm>1, then there is f ∈ Asuch that f<1 but such thatm(f)=1. By (1.1) the series
∞ n=0
fn is norm convergent. Its sum satisfies
(1− f) ∞ n=0
fn =1, (1.2)
so that 1− f ∈A−1. But then
1=m(1)=m((1− f)−1)(m(1)−m(f))=0, a contradiction.
Theorem 1.2. Suppose M is a maximal(proper)ideal in A. Then M is the kernel of a complex homomorphism m: A→.
Proof. There are two steps. First we showMis closed. Now, if the closure M¯ ofMis proper, that is, if ¯M = A, then ¯Mis also an ideal inA. ThereforeM is closed if ¯M = A, becauseMis maximal. However, ifg∈ M, theng∈ A−1 and (1.2), applied to f =1−g, shows that1−g ≥1. Hence 1∈M¯ and M is closed.
The second step is to show the quotient algebraB= A/M satisfies B=·1,
where 1=1+Mnow denotes the unit inB. The quotient mapping will then define the complex homomorphism with kernelM. SinceMis maximal, B= A/Mis a field, and sinceMis closed,Bis complete in the quotient norm
f +M = inf
g∈Mf +g,
which also satisfies (1.1).
Suppose there exists f ∈B\·1. Then f −λ∈B−1for allλ∈, because Bis a field. On the disc|λ−λ0|<1/(f −λ0)−1, the series
∞ n=0
(λ−λ0)n((f −λ0)−1)n+1 (1.3)
converges in norm to (f −λ)−1, because of the identity 1
f −λ = 1 (f −λ0)
1
[1−(λ−λ0)/(f −λ0)].
Clearly f−1 =0, and by the Hahn–Banach theorem there is a bounded linear functionalLonBsuch thatL =1 and
L(f−1)=0.
(1.4)
Since (1.3) is norm convergent and sinceL =1, we have F(λ)=L((f −λ)−1)=∞
n=0
L(((f −λ0)−1)n+1)(λ−λ0)n,
when|λ−λ0|<1/(f −λ0)−1. Becauseλ0is arbitrary, this means thatF(λ) is an entire analytic function. Now for|λ|large, (1.2) yields
(f −λ)−1 = 1
|λ|
1− f λ
−1 ≤ 1
|λ|
∞ n=0
fn
|λ|n . Consequently,
|F(λ)| = |L((f −λ)−1)| ≤C/|λ|, |λ|large, and by Liouville’s theoremF ≡0. Hence
L(f−1)= F(0)=0, contradicting (1.4).
The setMA of complex homomorphisms ofAis called thespectrum or maximal ideal spaceofA. By Theorem 1.1,MAis contained in the unit ball of the dual Banach space A∗. Give MA the weak-star topology of A∗, in which a basic neighborhoodVofm0 ∈MA is determined by ε >0 and by
f1, f2, . . . , fn ∈A:
V = {m ∈MA:|m(fj)−m0(fj)|< ε,1≤ j ≤n}.
This topology onMAis called theGelfand topology. With the Gelfand topology MAis a weak-star closed subset of ball(A∗), because
MA= {m∈ball(A∗) :m(f g)=m(f)m(g), f,g∈A}.
By the Banach–Alaoglu theorem, which says ball(A∗) is weak-star compact, MAis a compact Hausdorff space. Writing
fˆ(m)=m(f), f ∈A, m ∈MA,
we have a homomorphism f → fˆfromAinto C(MA), the algebra of con- tinuous complex functions onMA. This homomorphism is called theGelfand transform. By Theorem 1.1, the Gelfand transfrom is norm decreasing:
fˆ = sup
m∈MA
|fˆ(m)| ≤ f.
By Theorem 1.2, f ∈ A−1if and only if ˆf(m) is nowhere zero. Indeed, if f ∈ A−1, then by Zorn’s lemma the ideal{f g:g∈ A}is contained in a (proper) maximal ideal.
The Banach algebraAis called auniform algebraif the Gelfand transform is an isometry, that is, if
fˆ = f, f ∈ A.
Theorem 1.3. The Gelfand transform is an isometry if and only if f2 = f2
(1.5)
for all f ∈ A.
Proof. Sincefˆis a supremum,fˆ2 = fˆ2and (1.5) holds for any uni- form algebra.
Now assume (1.5). By Theorem 1.1, we have fˆ ≤ f. To complete the proof we take f ∈ Awithfˆ =1, we fixε >0, and we showf ≤ 1+ε. By Theorem 1.2, f −λ∈A−1 when|λ|>1= fˆ. By (1.3) theA- valued function (f −λ)−1 is analytic on|λ|>1. This means that whenever L∈ A∗, the scalar function F(λ)=L((f −λ)−1) is analytic on|λ|>1. By compactness,
sup
|λ|=1+ε(f −λ)−1 = K is finite. By (1.2),
F(λ)=L −1
λ
1− f λ
−1
= −∞
n=0
L(fn)
λn+1 , |λ|>f. This series must also representF(λ) on|λ|>1. TakingL =1 we obtain from Cauchy’s theorem,
L(fn)= − 1
2πi |λ|=1+εF(λ)λndλ
≤(1+ε)n+1K. Consequently, by the Hahn–Banach theorem,
fn = sup
L=1|L(fn)| ≤(1+ε)n+1K. Settingn=2k and using (1.5), we conclude that
f ≤ lim
k→∞(K(1+ε)2k+1)1/2k =1+ε, as desired.
When Ais a uniform algebra, the range ˆAof the Gelfand transform is a uniformly closed subalgebra ofC(MA), and ˆAis isometrically isomorphic to
A. In that case we identifyfwith ˆf and write
f(m)=m(f)= fˆ(m), f ∈ A, m ∈M.
Thus we view Aas a uniformly closed algebra of continuous functions on MA. Note thatAseparates the points ofMAand thatAcontains the constant functions onMA.
Example 1. SupposeAis any algebra of continuous complex functions on a compact Hausdorff spaceY. IfAhas the uniform norm,f =supy∈Y|f(y)| and ifAis complete, thenAis a uniform algebra. IfAcontains the constant functions and separates the points ofY, thenYis homeomorphic to a closed subset ofMA, and we say thatAis auniform algebra on Y. This is the generic example, because any uniform algebraAis clearly a uniform algebra on its spectrumY =MA. If A=C(Y), thenMA=Y. (See Exercise 5.)
Example 2. Letl∞denote the space of bounded complex sequences. With the normx =supn|xn|and with the pointwise multiplication (x y)n =xnyn,l∞ is a uniform algebra, by Theorem 1.3. The maximal ideal space ofl∞has the special name β, theStone– ˇCech compactification of the positive integers .
The Gelfand transform ofl∞isC(β). To see this, note that ifx∈l∞is real, that is, ifxn ∈for alln, then ˆx(m) is real onMl∞ =β, because then (x−λ)−1 ∈l∞ whenever Imλ=0. It now follows from the Stone–Weier- strass theorem thatl∞ =C(β).
Since the functionalmn(x)=xnis multiplicative onl∞,can be identified with a subset ofβ, and the Gelfand topology is defined in such a way that is homeomorphic to its image inβ. Moreover,is dense inβ, because every function inC(β)=l∞is completely determined by its behavior on. The Stone– ˇCech compactificationβcan also be characterized functorially.
Theorem 1.4. Let Y be a compact Hausdorff space and letτ :→Y be a continuous mapping. Then the mappingτ has a unique continuous extension τ˜ :β→Y .
Ifτ()is dense in Y and if the images of disjoint subsets ofhave disjoint closures in Y, then the extensionτ˜is a homeomorphism ofβonto Y.
Proof. The mapping
T :C(Y)→l∞,
defined byT f(n)= f ◦τ(n), is a homomorphism fromC(Y) intol∞. Because Tis continuous, the adjoint mapping T∗: (l∞)∗→(C(Y))∗ is weak-star to weak-star continuous. Ifm∈(l∞)∗is a multiplicative linear functional, that is, ifm∈β, then sinceT(f g)=T(f)T(g),T∗(m) is also a multiplicative linear functional onC(Y), and by Exercise 5,T∗(m)∈Y =MC(Y). Restricting T∗ to β, we have a mapping ˜τ(m)=T∗(m) from β into Y, and by the definition ofT,τ(n)˜ =τ(n),n∈. Sinceβhas the weak-star topology of
(l∞)∗andY =MC(Y)has the weak-star topology of (C(Y))∗, the mapping ˜τ is continuous. Becauseis dense inβ,τ˜is the unique continuous extension ofτtoβ.
Now suppose thatτ(S1)∩τ(S2)=∅whereverS1⊂,S2⊂andS1∩ S2 =∅. Then there is f ∈C(Y) such that f =1 onτ(S1) and f =0 onτ(S2).
We can assume thatS2 =\S1, so thatT f =χS1. Because linear combinations of characteristic functions are dense inl∞, this means the range ofTis dense inl∞. If we further assume thatτ() is dense inY, then we have T f = f, f ∈C(Y), and the mappingTis an isometry. Consequently the range of Tis norm closed inl∞. Hence the range ofTis both dense and closed andT mapsC(Y) ontol∞. Now becauseT f = f, the homomorphismTis one- to-one, and soTis an algebra isomorphism fromC(Y) ontol∞. The adjoint T∗then defines a homeomorphism fromβontoY.
Theorem 1.4 determines the spaceβup to a homeomorphism, because if Zis another compact Hausdorff space and ifZcontains a dense sequence{zn} homeomorphic tosuch that Theorem 1.4 holds withZin place ofβ, then the correspondence
n↔zn
extends to a homeomorphism betweenβand Z.
The space β is extremely huge. It can be mapped onto any separable compact Hausdorff space. No point ofβ\can be exhibited concretely.
Example 3. The spaceL∞of essentially bounded, measurable functions on the unit circle is a uniform algebra when it is given the pointwise multiplication and the essential supremum norm
f =inf{α:|f| ≤αalmost everywhere}.
We fix the notationXfor the maximal ideal space of L∞, because this space will be reappearing from time to time. Under the Gelfand transform,L∞ is isomorphic toC(X), the algebra of continuous complex functions onX. This has the same proof as the corresponding result on l∞. If f ∈L∞ is real, then (f −λ)−1∈ L∞whenever Imλ=0, so that ˆf is real onX. The Stone- Weierstrass theorem then shows thatL∞=C(X).
Likeβ\, the spaceXis large and intractable. We cannot construct a single point ofX. Nevertheless, the spaceXis quite useful in the theory of bounded analytic functions. Some of the intricacies ofXare outlined in Exercise 8. See Hoffman’s book [1962a] for further details.
LetAbe a uniform algebra onMA. A closed subsetKofMA is called a boundary for Aif
f = sup
m∈K|f(m)|
for all f ∈ A.
Theorem 1.5. There is a smallest closed boundary K0, which is contained in every boundary K.
This smallest boundary is called the ˇSilov boundaryofA. Note thatAis a uniform algebra on its ˇSilov boundary.
Proof. Let K0 be the intersection of all boundaries. We must show K0 is a boundary forA.
Lemma 1.6. Let f1, f2, . . . ,fn ∈ A and set
U = {m :|fj(m)|<1,j =1, . . . ,n}.
Then either U∩K =∅for every boundary K, or else K\U is a boundary for every boundary K.
Accepting Lemma 1.6 for a moment, we prove Theorem 1.5. Suppose f ∈ A and|f|<1 on K0. Set J = {m:|f(m)| ≥1}. If we show J =∅for every suchf, then we will have proved that K0 is a boundary. Since J ∩K0 =∅, eachm ∈J has (by the definition ofK0) a neighborhoodUof the form in the lemma such thatU ∩Km =∅for some boundaryKm. CoverJby finitely many such neighborhoodsUi,1≤i ≤N. Then by the lemma,K\Uiis a boundary wheneverKis a boundary. Consequently, by induction
K1=MA\ N i=1
Ui =((MA\U1)\U2)\. . .\UN)
is a boundary. Since|f|<1 on K1, this meansf<1 andJ =∅, so that K0is a boundary forA.
Proof of Lemma 1.6. We suppose thatKis a boundary but thatK\U is not a boundary, and we show thatUintersects every boundary forA. By hypothesis there is f ∈ Asuch thatf =1 but such that supK\U|f(m)|<1. Replacingf by a power fn, we can assume that supK\U|f(m)|< ε, whereεfj<1,j = 1, . . . ,n, and where f1, . . . , fn are the functions definingU. Then|ffj|<1 onUby the definition ofU, while|f fj|<1 onK\Uby the choice ofε. Since Kis a boundary, that meansffj<1, j =1, . . . ,n. Hence
{m:|f(m)| =1} ⊂n
j=1
{m :|fj(m)|<1} =U.
Since{m:|f(m)| =1= f}meets every boundary, this implies thatUalso meets every boundary.
Of course, ifA=C(Y), thenAhas ˇSilov boundaryY. A pointx ∈MAis a peak pointforAif there is f ∈Asuch that
f(x)=1,
|f(y)|<1, y∈MA, y=x.
Clearly, every peak point forAis in the ˇSilov boundary ofA.
Example 4. Thedisc algebrais the algebra of functions continuous on the closed discDand analytic on the open discD. We reserve the notationAofor this algebra. With the supremum normf =supz∈D|f(z)|,Ao is a uniform algebra. Because analytic polynomials separate the points ofD,Aois a uniform algebra on D. By the maximum principle, the unit circleTis a boundary for Ao. If λ∈T, then f(z)=(1+λz)¯ /2 satisfies f(λ)=1,|f(z)|<1,z=λ. Thusλis a peak point forAoandTis the ˇSilov boundary ofAo. The maximal ideal space ofAo isD. (See Exercise 9.)
Example 5. H∞is a uniform algebra with pointwise multiplication and with the supremum norm
f =sup
z∈D|f(z)|.
We shall always writeMfor the maximal ideal space of H∞.
For each pointζ ∈D there existsmζ ∈Msuch that mζ(z)=ζ, where z denotes the coordinate function, because (z−ζ)∈/ (H∞)−1. Now whenever
f ∈ H∞, we have (f − f(ζ))/(z−ζ)∈H∞, and f = f(ζ)+(z−ζ)
f − f(ζ) z−ζ
. But then
mζ(f)= f(ζ)+mζ(z−ζ)mζ
f − f(ζ) z−ζ
= f(ζ),
so that the pointmζ ∈Mis uniquely determined by the conditionmζ(z)=ζ. Hence
ζ →mζ
defines an embedding ofDinto M. By the definition of the topology ofM, this embedding is a homeomorphism. We now identifyζ withmζ and regard Das a subset ofM. ThenDis an open subset ofMbecause
D= {m ∈M:|z(mˆ )|<1}.
From now on we identify the uniform algebraH∞with its Gelfand transform and we think ofH∞as a subalgebra ofC(M). There is no ambiguity in doing this because by the discussion above
fˆ(mζ)= f(ζ), ζ ∈D.
Now suppose|ζ| =1. Then (z−ζ)∈/(H∞)−1and there exist pointsm∈M such that ˆz(m)=ζ. As we shall see in a moment, however, the fiberMζ = {m : ˆz(m)=ζ}is very large whenζ ∈∂D.
The Gelfand transform of the coordinate functionzdefines a map ˆz:M→ D.¯
Having identified the open discDwith ˆz−1(D), we can write M= D∪
|ζ|=1
Mζ.
Thus we can imagineM as the open disc D with the large compact space Mζ =zˆ−1(ζ) lying aboveζ ∈∂D. The fibers Mζ over points ζ ∈∂D are homeomorphic to one another because the rotationτ(z)=ζz,|ζ| =1, induces an automorphism f → f ◦τ of H∞, and the adjoint of this automorphism mapsM1ontoMζ.
To see just how largeMζ is, takeζ =1 and consider the singular function S(z)=exp((z+1)/(z−1)). In Chapter II we showed that the cluster set of S(z) atζ =1 is the closed unit disc. That is, whenever|w| ≤1 there exists a sequence{zn}inDsuch thatzn →1 and S(zn)→w. By the compactness ofMthe sequence{zn}has a cluster pointm∈M1andS(m)=w. HenceS mapsM1onto the closed unit disc.
Moreover, there exists a sequence{zn}inDsuch that limzn =1 and such that every interpolation problem
f(zn)=αn, n=1,2, . . . ,
{αn} ∈l∞, has solution f ∈ H∞. Such sequences, which are calledinterpo- lating sequences, will be discussed in Chapter VII, and a simple example of an interpolating sequence is given in Exercise 11. Here we only want to make this observation: If{zn}is an interpolating sequence, then by Theorem 1.4 the mapn→zn extends to define a homeomorphism fromβonto the closure of{zn}inM. Since limnzn =1 (as a sequence in the plane), we now see that M1contains a homeomorphic copy ofβ\.
By Fatou’s theorem, H∞ is a closed subalgebra of L∞. Now H∞ sepa- rates the points of X=ML∞, because every real L∞ function has the form u=log|f| with f ∈(H∞)−1 (see Theorem 4.5, Chapter II), and because u(m)=log|f(m)|whenuandfare viewed as elements ofC(X). Hence by compactness, the continuous map X→M, which is defined by restricting each multiplicative linear functional onL∞toH∞, is a homeomorphism. Ac- cordingly, we think ofXas a closed subset ofM. Since the injectionH∞⊂ L∞ is isometric,Xis a boundary forH∞. Moreover, ifKis a proper closed subset ofX, then sinceC(X)= L∞=log|(H∞)−1|, there is f ∈H∞such that
sup
K
log|f|<sup
X
log|f|,
and soKis not a boundary forH∞. We have proved the following theorem:
Theorem 1.7. The ˇSilov boundary of H∞is X =ML∞.
Every inner function has unit modulus onX, because it has unit modulus when viewed as an element ofL∞. So the singular functionS(z)=exp((z+ 1)/(z−1)) satisfies|S| =1 onX. NowS∈/(H∞)−1, butShas no zeros onD.
Therefore
M= D∪X.
Carleson’s corona theorem states thatDis dense inM. In other words, the coronaM\D¯ is empty. This famous result will be proved in Chapter VIII. For the present we only translate its statement into classical language.
Theorem 1.8. The open disc D is dense inMif and only if the following condition holds: if f1, . . . , fn ∈ H∞and if
1≤maxj≤n|fj(z)| ≥δ >0 (1.6)
for all z∈D, then there exist g1, . . . ,gn ∈ H∞such that f1g1+ · · · + fngn =1.
(1.7)
Proof. SupposeDis dense inM. Then by continuity we have
1≤maxj≤n|fj(m)| ≥δ
for allm∈M, so that{f1, . . . , fn}is contained in no proper ideal of H∞. Hence the idealJgenerated by{f1, . . . , fn}contains the constant 1. But
J = {f1g1+ · · · + fngn :gj ∈H∞}, and so (1.7) holds.
Conversely, supposeDis not dense inM. Then some pointm0∈Mhas a neighborhood disjoint fromD. This neighborhood has the form
V =n
j=1
{m:|fj(m)|< δ},
where δ >0, and where f1, . . . , fn ∈H∞, fj(m0)=0. The functions f1, . . . , fn satisfy (1.6) because V ∩D=∅, but they do not satisfy (1.7) withg1, . . . ,gn ∈H∞, because they all lie in the ideal{f : f(m0)=0}.
For the disc algebraAothe “corona theorem,” thatMA= D, is a very easy consequence of the Gelfand theory (see Exercise 9). Therefore, whenever (1.6) holds for f1, . . . , fn ∈ Ao, there existg1, . . . ,gn ∈ Ao such that (1.7) holds.
Now suppose we knew the “corona theorem with bounds” for Ao. In other words, suppose that whenever f1, . . . ,fn ∈ Ao satisfied (1.6), we could find g1, . . . ,gn ∈ Ao that solved (1.7) and in addition satisfied
gj ≤C
n, δ,max
j fj
. (1.8)
Then the corona theorem for H∞would follow by a simple normal families argument: Given f1, . . . , fn ∈ H∞ having (1.6), and givenr <1, we could
takeg(r)1 , . . . ,g(r)n in Aosuch that
fj(r z)g(rj)(z)=1, z ∈D,
and such thatg(rj)∞≤C(n, δ,maxjfj). For some sequencerk →1, gj(z)= lim
k→∞g(rjk)(z)
would then provideH∞solutions of (1.7). In other words, we would get the corona theorem forH∞if we had a proof of the easy “corona theorem” for Ao
that was constructive enough to include the bounds (1.8).