Nevanlinna’s Proof

Exercises and Further Results

6. Nevanlinna’s Proof

Clearly|w⁽¹⁾_j | ≤1, because f1∈B . Moreover, the case|w⁽¹⁾_j | =1 for some j,2≤ j ≤n, is not possible, because then f1(z)≡w⁽¹⁾_j ,|z|<1, and by (6.2) E_n contains exactly one function. Thus we have

|w⁽¹⁾_j |<1, j =2,3, . . . .

Something has been gained, however, because (6.2) allows us to disregard the value f₁(z₁), andE_nis now described by then−1 equations (6.4), instead of by the originalnequations.

Repeating the above reasoning, we fixc2,|c2|<1, and write f1−w⁽¹⁾₂

1−w⁽¹⁾₂ f1

= f2+c2

1+c¯2f2

z−z2

1−z¯2z. (6.5)

Now f(z)∈E₂ if and only if f1(z2)=w⁽¹⁾₂ , and that happens if and only if (6.5) holds for some f₂ ∈B. Furthermore, whenn>2, f ∈E_n if and only if

f2(zj)=w⁽²⁾_j , 3≤ j ≤n,

wherew⁽²⁾_j is determined by (6.5). We also have|w⁽²⁾_j |<1, for the same reason that we had|w⁽¹⁾_j |<1.

Continue by induction, always assumingE_ncontains more than one function.

Fork≤n, f ∈E_n if and only if there are f0, f1, . . . , fk

inB such that f0 = f, such that

fk(zj)=w^(k)_j , k+1≤ j ≤n, where|w^(k)_j |<1, andw⁽⁰⁾_j =wj, and such that

f_k−1−w_k^(k−1)

1−w_k^(k−1)f_k−1 = fk+ck

1+c¯kfk

z−zk

1−z¯kz, (6.6)

where|ck|<1,ck to be determined. (The explicit values of thew^(k)_j are not important here.)

We now iterate (6.6) and obtain a one-to-one mapping of B onto E_n. Rewrite (6.6)

fk−1(z)= αk(z)+βk(z)fk(z) γk(z)+δk(z)fk(z), (6.7)

in which

αk(z)=w^(k−1)_k (1−z¯kz)+ck(z−zk), βk(z)=c¯kw^(k−1)_k (1−¯zkz)+(z−zk), γk(z)=(1−z¯kz)+ckw^(k−1)_k (z−zk), δk(z) =c¯k(1−z¯kz)+w^(k−1)_k (z−zk).

(6.8)

By induction, (6.3) and (6.7) yield

f(z)= An(z)+Bn(z)fn(z) Cn(z)+Dn(z)fn(z), (6.9)

in which

An =γnA_n−1+αnB_n−1, Bn =δnA_n−1+βnB_n−1, Cn =γnCn−1+αnDn−1, Dn =δnC_n−1+βnD_n−1 (6.10)

are polynomials of degree at mostninz. Then we have proved

Lemma 6.1. Suppose the polynomials An(z),Bn(z),Cn(z),Dn(z)are defined by(6.8)and(6.10). Then f(z)∈E_n if and only if f(z)satisfies(6.9)for some

fn(z)∈B.

The polynomials An,Bn,Cn, and Dn depend on the parameters c1,c2, . . . ,cn. Now fixck =z¯kw_k^(k−1). This makes

δn(0)= Dn(0)=0

in (6.8) and (6.10), which will facilitate the convergence argument below.

When|z|<1, (6.9) shows that{f(z) : f ∈E_n}is a closed discn(z) con- tained in ¯Dand defined by

An(z)+Bn(z)ζ

Cn(z)+Dn(z)ζ :|ζ| ≤1

. (6.11)

When|z| =1, this formula forn(z) makes sense, although some f ∈E_n are not defined atz. Taken(z)= {f(z) : f ∈ Ao∩E_n},Aothe disc algebra. The discn(z) degenerates into a point if and only if the determinant

An(z)Dn(z)−Bn(z)Cn(z)=0.

By (6.10) and induction (6.12)

AnDn−BnCn =(γnβn−αnδn)(A_n−1D_n−1−B_n−1C_n−1)

=ⁿ

k=1

(1− |zk|²|w^(k−1)_k |²)(1− |w_k^(k−1)|²)(z−zk)(1−z¯kz),

wherew⁽⁰⁾_k =wk. Thusn(z) reduces to a point if and only ifz =zj,1≤ j ≤ n. An elementary computation with (6.11) shows thatn(z) has radius

ρn(z)= |AnDn −BnCn|

|Cn|²− |Dn|² .

Lemma 6.2. IfE_n =∅, then when|z| =1, n(z)= D, ρ¯ n(z)=1, and

|Bn(z)| = |Cn(z)|,

|An(z)| = |Dn(z)|,

An(z)/Cn(z)=(Dn(z)/Bn(z))=λn(z), (6.13)

with|λn(z)|<1.

Proof. The conditions (6.13) follow from the assertion thatn(z)=D, by¯ the well-known characterization of the coefficients of the linear fractional transformations

ζ → A+Bζ C+Dζ

that map ¯Donto itself. Thatn(z)= D,¯ |z| =1, is proved by induction. Fix z=e^iθ andζ ∈ D. For¯ n=1 there exists a constant f1∈B such that

f1(e^iθ)+c1

1+c¯1f1(e^iθ) =

ζ−w1

1−w1ζ

e^iθ−z1

1−z¯1e^iθ (6.14)

and (6.2) then produces f ∈E₁such that f(e^iθ)=ζ. Forn>1 the set {f1∈B : f1(zj)=w⁽¹⁾_j , j =2, . . . ,n}

is nonempty, becauseE_n =∅, and by induction this set contains a function f1satisfying (6.14). As before, (6.2) then gives us f ∈E_nsuch that f(e^iθ)= ζ.

Lemma 6.3. Cn(z)has no zeros in|z| ≤1.

Proof. IfCn(z)=0, then by takingζ =0 in (6.11), we see that An(z)=0.

Then

An(z)Dn(z)−Bn(z)Cn(z)=0

and by (6.12),z =zj for some j =1,2, . . . ,n. Because by (6.12) the deter- minantAnDn−BnCn has only simple zeros, we also have

A_n(zj)Dn(zj)−Bn(zj)C_n(zj)=0,

whenCn(zj)= An(zj)=0. Still takingζ =0 in (6.11), we then obtain wj = lim

z→zj

An(z)/Cn(z)= lim

z→zj

A_n(z)/C_n(Z).

IfC_n(zj)=0, this means that A_n(zj)=0, while ifC_n(zj)=0, it means that A_n(zj)/C_n(zj)=wj. Thus we have

A_n(zj)=wjC_n(zj).

On the other hand, takingζ =1 in (6.11) gives Bn(zj)=wjDn(zj).

Therefore

A_n(zj)Dn(zj)−Bn(zj)C_n(zj)=0, a contradiction.

It will be convenient to renormalize (6.9). Let ψn(z)=

_n

k=1

|zk|

−¯zk

(1− |zk|²|w^(k−1)_k |²)(1− |w_k^(k−1)|²)(1−z¯kz)² _1/2

.

Sinceψn(z) has no zeros in ¯D, Pn = An

ψn

, Qn = Bn

ψn

, Rn = Cn

ψn

, and Sn = Dn

ψn

are rational functions analytic in|z| ≤1. From (6.9) we obtain f(z)= Pn(z)+Qn(z)fn(z)

Rn(z)+Sn(z)fn(z), fn ∈B, (6.15)

as our new parametrization ofE_n. The advantage of this normalization is that by (6.12) the determinant is now

Pn(z)Sn(z)−Qn(z)Rn(z)=n(z)=ⁿ

k=1

(−z¯k)

|zk|

z−zk

1−z¯kz,

the Blaschke product having zeros{z1,z2, . . . ,zn}and normalized byn(0)>

0. We also have

ρn(z)= |n(z)|

|Rn(z)|²− |Sn(z)|², so that by Lemma 6.2,

|Rn(z)|²− |Sn(z)|²=1, |z| =1. (6.16)

Now by Lemma 6.3,Rnhas no zeros in|z| ≤1, and (6.16) gives 1

|Rn(z)| ≤1, |z| ≤1.

Then (6.13) and the maximum principle yield

|Pn(z)| ≤ |Rn(z)|,

|Qn(z)| ≤ |Rn(z)|,

|Sn(z)| ≤ |Rn(z)| (6.17)

on|z| ≤1.

Theorem 6.4. AssumeE_∞ =

E_ncontains two functions with different val- ues at z=0. Then there are nj → ∞such that the limits

P(z)= lim

j Pnj(z), Q(z)=lim

j Qnj(z), R(z)= lim

j Rnj(z), S(z)= lim

j Snj(z)

all exist. The limits do not all vanish identically; in fact, P(z)S(z)−Q(z)R(z)=(z),

the Blaschke product with zeros{zn}. If f(z)∈B, then f ∈E_∞if and only if f(z)= P(z)+Q(z)f∞(z)

R(z)+S(z)f∞(z), |z|<1, (6.18)

for some f∞ ∈B.

Proof. The hypothesis that

E_ncontains two functions with different values atz=0 implies that

limn ρn(0)>0.

(This limit exists becauseρn+1≤ρn sinceE_n+1⊂E_n.) SinceDn(0)=0, by the choice of the constantscn in (6.6), we have

|Rn(0)|²= |n(0)|/ρn(0) and hence

n→∞lim |Rn(0)|<∞.

(6.19)

Choosenjso that the bounded sequence{1/Rnj(z)}converges onD. Because 1/Rn(z) has no zeros in|z|<1, the limit function either vanishes identically onDor it has no zeros onD. By (6.19) the limit does not vanish atz=0, and so the limit is zero free onD. Consequently Rnj(z) converges uniformly on compact subsets of|z|<1 to an analytic function R(z), which has no zeros

on|z|<1. By (6.17) we can take a finer subsequence{nj}so that limj Pnj(z)= P(z),

limj Qnj(z)= Q(z), and

limj Snj(z)=S(z) also all exist. Then

P(z)S(z)−Q(z)R(z)=lim

j nj(z)=(z), and so the limit functions are not all identically zero.

If f ∈

E_n, then by Lemma 6.1 there is fn ∈B such that f = Pn+Qnfn

Rn+Sn fn.

Refine the subsequence{nj}so that fnj(z)→ f∞(z),|z|<1, with f∞∈B. Then (6.18) follows. Conversely, if f∞∈B, then

f⁽ⁿ⁾= Pn +Qnf_∞ Rn +Snf_∞. is inE_n, and f^(nj)has limit f ∈E_∞.

It turns out that P,Q,R, andS do not depend on the subsequence{nj}.

They are uniquely determined by the original interpolation problem (6.1) and by the choice of the constantscn in (6.6). We shall not make use of this fact, however.

Before turning to Nevanlinna’s proof that E_∞ contains inner functions, we mention a simple consequence of (6.18). Because E_n+1 ⊂E_n, the discs n(z),|z|<1, decrease to a limit disc∞(z)= {f(z) : f ∈E_∞}. By (6.18),

∞(z)=

P(z)+Q(z)ζ

R(z)+S(z)ζ :|ζ|<1

.

Since the determinant of the coefficients for this mapping is P(z)S(z)− Q(z)R(z)=(z), ∞(z) is nontrivial wheneverz∈ {zj}. The radius of∞(z) is

ρ∞(z)=lim

n ρn(z)= (z)

|R(z)²| − |S(z)|².

Returning ton(z0),|z0|<1, and to its parametrization (6.11), we see that f ∈E_n solves the extremal problem f(z0)∈∂n(z0) if and only if

f(z)= An(z)+Bn(z)e^iϕ

Cn(z)+Dn(z)e^iϕ = Pn(z)+Qn(z)e^iϕ Rn(z)+Sn(z)e^iϕ

for some unimodular constante^iϕ. Thus f(z) is a rational function of degree at mostn. On two other occasions (Corollary 1.9 of this chapter and Corol- lary 2.4, Chapter I), we have seen that the extremal function f(z) is a finite Blaschke product. This fact also follows from the reasoning in this section, because by Lemma 6.2,|f(z)| =1 for|z| =1. Further analysis of the poly- nomialsAn(z),Bn(z),Cn(z), andDn(z) shows that the Blaschke product f(z) has degreen. (See Exercise 20.)

The preceding discussion suggests that we might find an inner function in E_∞ by setting f_∞ =e^iϕ in (6.18). That is how Nevanlinna first proved Theorem 4.1. The proof above of Theorem 4.4 was based on the same idea: The unimodular functionh0∈K was obtained by maximizing the linear functional Re

h dθ/2π,h∈K.

We return to Theorem 4.1, which we restate as follows.

Theorem 6.5. IfE_∞ contains two distinct functions, and if e^iϕ is any uni- modular constant, then

f(z)= P(z)+Q(z)e^iϕ R(z)+S(z)e^iϕ (6.20)

is an inner function inE_∞.

Proof. By Theorem 6.4, f ∈E_∞, and in particularf∞ ≤1. We suppose there existsE ⊂T,|E| ≡

χE(θ)dθ/2π >0, such that

|f(e^iθ)| ≤α <1, θ ∈E, (6.21)

and we argue toward a contradiction.

Reindex so that Rn(z)→ R(z),Pn(z)→ P(z),Qn(z)→ Q(z), and Sn(z)→S(z). Fix M >|R(0)|. Since by Lemma 6.3 Rn(z) has no zeros on D, we have¯

logM >log|Rn(0)| = 1 2π

log|Rn(e^iθ)|dθ for largen, sayn>n0. By (6.16), log|Rn(e^iθ)| ≥0, and hence

|{θ : log|Rn(e^iθ)|>2 logM/|E|}| ≤ |E|/2, n>n0. Therefore

En = {θ ∈ E:|Rn(e^iθ)| ≤M^2/|E|} satisfies

|En| ≥ |E|/2, n>n0. By (6.16) and (6.13), we also have

Pn(e^iθ) Rn(e^iθ)

² =1− 1

|Rn(e^iθ)|²,

which gives

Pn(e^iθ) Rn(e^iθ)

≤ {1−M^−4/|E|}^1/2 =β <1, (6.22)

forθ ∈En,n>n0.

Because f ∈E_n there exists fn ∈B such that f(z)= Pn(z)+Qn(z)fn(z)

Rn(z)+Sn(z)fn(z).

Since the rational functionsPn,Qn,Rn, andSnare continuous on ¯Dand since

|PnSn−QnRn| =1 on T, fn(z) has radial limit fn(e^iθ) whenever f(z) has radial limit f(e^iθ). Thus ate^iθ ∈ E, fk(e^iθ) exists and

f(e^iθ)= Pn(e^iθ)+Qn(e^iθ)fn(e^iθ) Rn(e^iθ)+Sn(e^iθ)fn(e^iθ). By Lemma 6.2, the mapping

ζ → Pn(e^iθ)+Qn(e^iθ)ζ Rn(e^iθ)+Sn(e^iθ)ζ,

which is an automorphism of D, preserves the pseudohyperbolic distance.

Hence

|fn(e^iθ)| =ρ(fn(e^iθ),0)=ρ(f(e^iθ),Pn(e^iθ)/Qn(e^iθ)). Forn>n0and fore^iθ ∈En, (6.21) and (6.22) then yield

|fn(e^iθ)| ≤ |f(e^iθ)| + |Pn(e^iθ)|/|Rn(e^iθ)|

1+ |f(e^iθ)||Pn(e^iθ)|/|Rn(e^iθ)|

≤ α+β

1+αβ =γ <1.

Consequently

|fn(0)| ≤ 1 2π

|fn(e^iθ)|dθ ≤(γ|En| +(1− |En|))

≤

γ|E|

2 +

1−|E|

2

=η <1.

Take a subsequence{fnj}so that fnj(0)→ζ,|ζ| ≤η <1. Then f(0)=lim

j

Pnj(0)+Qnj(0)fnj(0)

Rnj(0)+Snj(0)fnj(0) = P(0)+Q(0)ζ R(0)+S(0)ζ .

Since P(0)S(0)−Q(0)R(0)=0, we conclude from (6.20) that ζ =e^iϕ, a contradiction.

An interesting apparently unsolved problem is to determine whether or when E_∞contains a Blaschke product. See page vii.

Notes

The first systematic treatment of dual extremal problems is the pa- per of Macintyre and Rogosinski [1950]. The functional analytic methods were introduced by Havinson [1949,1951] and by Rogosinski and Shapiro [1953]. Duren’s book [1970] contains a slightly different treatment and some references to the older literature. See also Goluzin [1952] and Landau [1916].

Section 2 is from the paper of Carleson and Jacobs [1972]. Kahane [1974] discusses a number of related questions. An interesting open prob- lem is to find intrinsic necessary and sufficient conditions for a function in L^∞ to have unique best approximation in H^∞. Exercise 17 gives a partial answer.

The primary references for Section 3 are Szeg¨o [1920] and Helson and Szeg¨o [1960]. Exercises 8 and 14 outline results similar to the Helson–Szeg¨o theorem.

Most of the material in Sections 4 and 5 originates with Adamyan, Arov, and Krein [1968], but the proofs in the text are considerably different from their spectral theory approach. In [1971] Adamyan, Arov, and Krein extend their results to the matrix valued case. Theorem 5.1 and Theorem 5.2 are from deLeeuw and Rudin [1958].

Section 6 is from Nevanlinna’s paper [1929], which includes a number of other classical results, all derived ultimately from Schwarz’s lemma. It is a fundamental paper long overlooked. Schur’s [1917] treatment of the coefficient problem is very similar. It is outlined in Exercise 21.

Exercises and Further Results

1. If f ∈L^p(),1≤ p≤ ∞,then

g∈Hinf^pf −gp = sup

G∈H^q Gq=1

f G d x .

2. Let f(z) be meromorphic on|z|<1, and suppose that on some annulusR<|z|<1, f(z) is analytic and of class H¹(i.e.|f(z)|possesses a harmonic majorant). Then f(e^iθ) exists almost everywhere. Prove f(e^iθ)≥0 almost everywhere if and only if f(z) is a rational function of the form

c n

j=1

(z−αj)(1−α¯jz) _n

j=1

(z−βj)(1−βjz), where|αj| ≤1,|βj| ≤1,c>0.

3. Letc0,c1, . . . ,cNbe given complex numbers and consider the max- imum problem

M = sup

f∈H^∞ f≤1

N j=0

cjaj

,

where f(z)= ajz^j.

(a) The dual extremal problem is M = inf

g∈H₀¹k−g1, wherek(z)=_n

j=0cjz^−j. It is equivalent to the minimum problem M =inf{h1:h∈ H¹:h=cN +c_N−1z+ · · · +c₀z^N + · · · }.

(b) The original extremal problem has unique extremal function f0and the dual problem has unique minimizing functiong0Moreover

f0(z)(k(z)−g0(z))= |k(z)−g0(z)|, |z| =1, so that

f0(z)(k(z)−g0(z))=cz^q n

j=1

(z−αj)(1−α¯jz)

z^N

withn+q= N,0<|αj| ≤1,c>0. Reindexingα1, . . . , αn, there iss,0≤ s≤nsuch that|αj|<1, j ≤s, and

f0(z)=γz^q s

j=1

z−αj

1−α¯jz

and

k(z)−g0(z)=cγ¯ⁿ

j=1

(1−α¯jz)² n s+1

z−αj

1−α¯jz

z^N, where|γ| =1.

(c) For|z|small write

(cN +cN−1z+ · · · +c0z^N)^1/2=^∞

j=0

λjz^j and set

PN(z)=^N

j=0

λjz^j.

AssumePN(z) has no zeros in|z|<1. Then there existα1, . . . , αn,n≤ N, such that 0<|αj| ≤1 and such that

PN(z)=λ0

n j=1

(1−α¯jz). Then

z^−NP_N² =cNz^−N n j=1

(1−α¯jz)² has the formk−g,g∈H¹. Setting

f0(z)= c¯N

|cN|z^N−n n

j=1

z−αj

1−α¯jz

,

we obtain f0P_N²/z^N ≥0 on|z| =1. Consequently f0is the extremal function, k(z)−g0(z)= P_N²(z)/z^N, and

M = P_N²1 =^N

j=0

|λj|².

(d) Landau [1913, 1916] determined the extremumMin the special case cj =1 by a different method. In this case PN =₀^Nλjz^j, where

λj =(−1)^j −¹₂

j

= (2j)!

4^j(j!)², and by Wallis’s formula

M∼logN/π (N → ∞).

ThusM has the same order of magnitude as the Lebesgue constantsDN1, whereDN =_−N^N e^{i jθ} is Dirichlet’s kernel for partial sums.

The dual problem in part (a) was first treated by F. Riesz [1920] using a variational argument.

4. A function f ∈C(T) is called badly approximable if its nearest functiongin H^∞ satisfiesg≡0. Showf is badly approximable if and only if|f(e^iθ)|is a positive constant and f(e^iθ) has negative winding number (see Poreda [1972] or Gamelin, Garnett, Rubel, and Shields [1976]).

5. Letg∈H^∞be the best approximation to f ∈C(T). It conjectured thatgis continuous if the conjugate function ˆf is continuous. (See page vii.) The converse of this is false. Use Exercise 4 above to find a badly approx- imable function whose conjugate function is not continuous (D. Sarason, un- published).

6. Let f ∈C(T), f ∈H^∞, letg∈H^∞be the best approximation tof, and letF∈ H₀¹be a dual extremal function, so that

(f −g)F = f −g∞|F|

almost everywhere.

(a) Fcan be chosen so that F(z)/zis outer.

(b) If F1∈ H₀¹ is another dual extremal function, thenF/F1 is a rational function.

(c) F is unique if and only ifz/F∈H¹. In that casez/F ∈H^p for all p<∞.

(d) More generally, if f ∈ L^∞\H^∞, ifFexists, and ifz/F ∈H¹, thenF is unique.

(See deLeeuw and Rudin [1958] and Carleson and Jacobs [1972].) 7. If f ∈ H^∞, then

dist(f,Ao)≤2 dist(f,C(T)),

where Ao = H^∞∩C(T) is the disc algebra. The constant 2 is sharp. (See Davie, Gamelin, and Garnett [1973].)

^{8. Let}dμ=wdθ/2π +dμsbe a positive measure on the circle. Then

finf∈F

|1−Ref|²dμ=

w⁻¹dθ 2π

₋₁ ,

whereF is the set of trigonometric polynomialsn>0ane^inθ (see Grenander and Rosenblatt [1957]). The result is due to Kolmogoroff [1941].

9. Find two closed subspacesF andGof a Hilbert space such that F ∩G= {0}

but such that

1=sup{|f,g |: f ∈F,g∈G,f = g =1}.

10. IfF andGare subspaces ofL²(μ), thenρ=1 if and only if 0=inf

|f −g|²dμ: f ∈F,g∈G;f ≥1,g ≥1

. Hence the propertyρ=1 is unchanged ifdμis replaced bywdμ, wbounded above and below.

11. If a weight functionw satisfies the Helson–Szeg¨o condition (3.3), then for someε >0, w∈ L^1+εandw⁻¹∈L^1+ε.

12. Letwbe a weight function. Assumeψ =(logw) is continuous except for jumps at a finite number of points. Thenwsatisfies (3.3) if and only if each jump is less thanπ (Helson and Szeg¨o [1960]).

13. Letαbe real. There is a constantK_α such that _π

−π|p(θ˜ )|²|θ|^αdθ ≤K_{α π}

−π|p(θ)|²|θ|^αdθ

for all trigonometric polynomials if and only if−1< α <1 (see Hardy and Littlewood [1936]).

^{14. LetF}n be the set of all trigonometric polynomialsk≥nake^ikθ and letG_n =F_nbe the set of trigonometric polynomials of the formk≥nbke^−ikθ. Whenμis a positive finite measure onTwrite

ρn =sup

fg dμ¯ , where f ∈F_n,g∈G_n,

|f|²dμ≤1, and

|g|²dμ≤1. Wright dμ=wdθ/2π + dμs

withdμs, singular todθ.

(a) Ifdμs =0 or if logw /∈L¹, thenρn =1 for alln.

(b) Now assumedμ=wdθ/2π with ϕ=logw∈L¹. LetW be the set of weightswsuch that

n→∞lim ρn =0,

wheredμ=wdθ/2π. Thenw∈W if and only ife^−iϕ˜∈ H^∞+C. (c) H^∞+Cis a closed subalgebra ofL^∞.

(d) LetW₀ be the set of positive weights w such that, for everyε >0, ϕ=log w=r+s˜+t

withr∞< ε,s∞< ε, andt ∈C. ThenW₀ ⊂Wandw∈Wif and only if w= |p|²w0, wherew0∈W0andpis a trigonometric polynomial. (The set log W0 is the space VMO to be studied in Chapter VI.) (See Helson and Sarason [1967].)

15. Letube a unimodular function inL^∞and set S_u = {G ∈H¹ :G1=1,G/|G| =u}.

(a) There exist functionsufor whichS_u =∅^. (b) If F∈S_u, and if 1/F∈ H¹, thenS_u = {F}.

(c) If F ∈S_u = {F}, then for all α,|α| =1,(z−α)⁻²F(z)∈/ H¹. (See deLeeuw and Rudin [1958].)

16. Leth∈L^∞,|h| =1 almost everywhere.

(a) If dist(h,H^∞)<1, thenh= F/|F|for someF∈ H¹. Moreover, the outer factor ofFis invertible inH^1.(See Lemma 5.4.)

(b) If dist(h,H^∞)<1 but dist (h,H₀^∞)=1, thenh= F/|F|¯ for some F∈ H¹. (See Case 1 of the proof of Theorem 4.4.) It follows that dist( ¯h,H^∞)<1.

17.(a) Leth∈ L^∞. If the coset h+H^∞ is an extreme point of ball (L^∞/H^∞) then there isg∈ H^∞such that|h+g| =1 almost everywhere.

(b) Now suppose|h| =1 almost everywhere. Thenh+H^∞is an extreme point of ball(L^∞/H^∞) if and only if h+g>1 for all g∈ H^∞,g=0.

Thus the extreme points of ball(L^∞/h^∞) are the cosets containing exactly one function of unit modulus (see Koosis [1971]).

(c) If|h| =1 almost everywhere, thenh+g∞ >1 for allg∈ H^∞,g= 0, if and only ifh cannot be written as F/|F|,F∈ H¹. (See Lemmas 5.4 and 5.5.) This is a weak generalization of Exercise 5. If |h| =1, thenhis badly approximable byH^∞(h−g>1 ifg∈H^∞,g=0) if and only ifh is not the argument of an H¹ function. Example 4.2 shows there exist badly approximable functions not having constant modulus.

(d) L^∞/H^∞is a dual space. Hence ball(L^∞/H^∞) is the weak-star closed convex hull of the set of cosets{h+H^∞:|h| =1,hbadly approximable}.

18. LetF∈H¹,F1 =1. Then everyg∈H^∞such that F/|F| −g∞ ≤1

is of the form (5.10) if and only ifS_F/|F| = {F}.

19. Let f ∈H^∞,f∞=1. Then f is an extreme point of ball(H^∞) if and only if

log(1− |f(e^iθ)|)dθ= −∞

(see deLeeuw and Rudin [1958]).

20. In the notation of Section 6, and withck=z¯kw^(k−1)_k ,An(z) has degree at mostn−1 andBn(z) has degreen. Whene^iϕis a unimodular constant, the polynomials An(z)+Bn(z)e^iϕ andCn(z)+Dn(z)e^iϕ have no common zero.

Thus

f(z)= An(z)+Bn(z)e^iϕ Cn(z)+Dn(z)e^iϕ is a Blaschke product of degreen.

21. Fix complex numbersc₀,c₁, . . . and set

E_n = {f ∈B : f(z)=c0+c1z+ · · · +cnzⁿ+ · · · }.

Writeγ0 =c0 and assume|γ0| ≤1.

(a) If |γ0| =1 thenE₀ contains exactly one function, f =γ0. Suppose

|γ0|<1 and write f ∈E₀

f = z f1+γ0

1+γ¯0z f₁, (E.1)

with f₁ ∈B. SoE₀ is in one-to-one correspondence withB. Then f ∈E₁if and only if f₁(0)=γ1 =c₁/(1− |c0|²). In particularE₁ =∅if and only if

|γ1| ≤1, andE₁consists of exactly one function if and only if|γ1| =1.

(b) Continue by induction. We obtain γ0, γ1, . . . , γn, . . .. If |γ0|<

1,|γ1|<1, . . . ,|γn|<1, thenE_n is in one-to-one correspondence with B through the formulas f0= f,

fk−1= z fk+γk−1

1+γ¯k−1z fk, 1≤k≤n,

fk ∈β. If|γk| =1 but|γj|<1, j <k,thenE_kconsists of exactly one func- tion, a Blaschke product of degreek.

(c) SupposeE_n =∅. Then the (n+1)th coefficientscn+1of the functions inE_nfill a closed disc with radius

ωn =(1− |γ0|)· · ·(1− |γn|). (E.2)

(Use (E.1) and induction.) (d) Suppose

E_n =∅. Then of course

E_n = {f}, f =₀^∞cnzⁿ. Prove

n→∞lim ωn =exp 1 2π

log(1− |f|²)dθ.

(Hint: By (E.1) and (E.2) we have, on|z| =1, 1− |f|²= (1− |γ0|²)(1− |f1|²)

|1+γ¯0f1|² = ωn(1− |f_n+1|²) _n

j=0|1+γ¯jz f_j+1|². Because the denominator has no zeros onD, this gives

exp 1 2π

log(1− |f|²)dθ =ωnexp 1 2π

log(1− |f_n+1|²)dθ ≤ωn. For the reverse inequality use Szeg¨os theorem. For ε >0 there is P(z)= 1+b1z+ · · · +bNz^N such that

|P|²(1− |f|²)dθ

2π ≤ε+exp 1 2π

log(1− |f|²)dθ.

Let f^∗∈E_n be obtained by setting fn+1 =0. Then by Szeg¨os theorem ωn =exp

log(1− |f^∗|²)dθ 2π ≤

|P|²(1− |f|^∗2)dθ 2π.

By the Parseval relation,

n→∞lim

|P|²(1− |f^∗|²)dθ 2π ≤

|P|²(1− |f|²)dθ 2π. (e) Let Anbe the matrix

⎡

⎢⎢

⎣

c₀ c₁ · · · cn

0 c₀ c_n−1

... ...

0 · · · 0 c0

⎤

⎥⎥

⎦

and letInbe the (n+1)×(n+1) identity matrix. ThenE_n =∅if and only if In− A^∗_nAn is nonnegative definite. If In−A^∗_nAn is positive definite, thenE_n is infinite. More precisely,

det(In− A^∗_nAn)=ω0ω1· · ·ωn.

(See Schur [1917], except for part (d), which is due to Boyd [1979].)

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^∞.