Parametrization of K

Exercises and Further Results

5. Parametrization of K

maps this annulus into the domain bounded by the slit [−4,0] and an ellipse, and the circle |w| =1 is mapped onto the slit [−4,0]. Also ϕ(G(0))>0.

Therefore

g= ϕ◦G ϕ◦G(0)

hasg(0)=1 andgis real and negative almost everywhere onT\E.

Consequently

J(e^iθ)= F(e^iθ)Re(u₀(e^iθ))=₁

2λG(e¯ ^iθ)u₀(e^iθ)}{2 Reu₀(e^iθ) (5.3)

= ¹₂λG(e¯ ^iθ)(1+u²₀(e^iθ)), andJ ∈H¹. Almost everywhere we have

|F±J| = |F|(1±Reu0), so that (5.2) gives

F+J1 = F−J1 = F1 =1.

SinceFis not outer,u0is not constant, and J =0. Then F = F+J

2 + F−J 2 andFis not an extreme point. Furthermore,

J±F= ¹₂λG(1¯ ±2u0+u²₀)= ¹₂λG¯ (1±u0)² is an outer function, because 1±u0 is outer, and hence (5.1) holds.

Now assume F is an outer function, F1=1. If F=t F1+(1− t)F2,F1,F2 ∈ball(H¹),0<t <1, then

1= F1 =t F¯

|F| F1 dθ

2π +(1−t) F¯

|F| F2 dθ 2π

≤tF11+(1−t)F21≤1,

and equality must hold throughout these inequalities. HenceF11 = F21 = 1, and

F¯

|F|Fjdθ

2π = Fj1, j =1,2.

Since|F|>0 almost everywhere, that means Fj =kjF, wherekj >0 and tk1+(1−t)k2=1. But then by the subharmonicity of log|Fj|,

|Fj(0)| ≤ |F(0)|exp

logkjdθ 2π

,

becauseFis outer. Since|F(0)| ≤t|F1(0)| +(1−t)|F2(0)|, Jensen’s inequal- ity now yields

1≤t exp

logk₁dθ 2π

+(1−t) exp

logk₂dθ 2π

≤t

k1dθ

2π +(1−t)

k2dθ 2π =1.

Because the exponential is strictly convex we conclude thatkjis constant and kj = Fj1/F1 =1. HenceF1 =F2 =FandFis an extreme point.

It is unfortunately the case that anH¹ functionFof unit norm is not de- termined by its argument (which is defined modulo 2π almost everywhere, because|F|>0 almost everywhere). For example, ifFis not outer, then the two outer functions in (5.1) have the same argument asF(and as each other).

That follows from the construction ofF1 andF2or from the observation that by (5.1),

F¯

|F|Fjdθ

2π =1= Fj1, j =1,2,

which means ( ¯F/|F|)Fj = |Fj|almost everywhere. (See Example 1.4 above for another counterexample.) Whenh∈L^∞and|h| =1 almost everywhere, we define

S_h = {F∈ H¹:F1=1,F/|F| =ha.e}.

Geometrically,S_h is the intersection of ball(H¹) and the hyperplane

F:

h F¯ dθ 2π =1

, (5.4)

and so S_h is a convex set. Of course, sometimesS_h is empty, but we are interested in the caseS_h =∅^{. WhenS}h contains exactly one functionF, the hyperplane (5.4) touches ball(H¹) only at F, which means Fis an exposed point of ball(H¹). There is no good characterization of the exposed points of ball(H¹), or equivalently of those F ∈ball(H¹) such thatS_F/|F| = {F}.

However, ifS_hcontains two functions, thenS_his very large.

Theorem 5.2. Let h∈L^∞,|h| =1almost everywhere, and assumeS_hcon- tains at least two distinct functions. Let z0∈ D. Then{F(z0) :F∈S_h}con- tains a disc centered at the origin.

Proof. When|z0| ≤1,|z1| ≤1, (z−z1)(1−¯z1z)

(z−z0)(1−¯z0z) = (z−z1)z(¯z−z¯1) (z−z0)z(¯z−z¯0)

is real and nonnegative onT. If there exists F∈S_h having a zero of orderk atz0, then

Fz1(z)=

(z−z1)(1−¯z1z) (z−z0)(1−¯z0z)

k

F(z) is inS_h and the valuesFz1(z0) fill a disc about 0.

It remains to show there isF ∈S_h with F(z0)=0. If the convex setS_h contains two functions, then by Theorem 5.1 it contains a function F=uG

with nonconstant inner factoru(z). Let J(z) be the function defined in (5.3).

Then for 0<t <1,

F+t J =F(1+tReu0)

satisfiesF+t J1 =1, because of (5.2), andF+t J ∈S_h. By (5.3) we also have

F+t J = ¹₂λt G¯

1+ 2u0

t +u²₀

. When 0<t <1, the equation

ζ²+ 2ζ

t +1=0

has a rootζ(t),|ζ(t)|<1, and these roots fill the segment (−1,0). IfF+t J has no zero inDfor eacht ∈[0,1), then the range ofu0 is disjoint from the segment (−1,0]. But then by Corollary 4.8, Chapter II,u0 is both an inner function and an outer function, so thatu0is constant. Thus there existstsuch thatF+t J has a zero at some pointz1∈ D. Then

(z−z₀)(1−z¯₀z)

(z−z1)(1−z¯1z)(F+t J) is a function inS_h having a zero atz₀.

Returning to the topic of the previous section, we fix a coseth0+H^∞ of L^∞/H^∞and we assume

K = {h∈h0+H^∞:h ≤1}

contains more than one function. By Theorem 4.3,Kthen contains a function, which we callh0, such that|h0| =1 almost everywhere. We will need to recall that after a change of coordinatesh0is an extremal function:

Re

h0dθ 2π =sup

h∈K Re

hdθ 2π. (5.5)

(See the proof of Theorem 4.3.) The Adamyan, Arov, and Krein parametriza- tion ofKwhich we have been seeking is (5.7) below.

Theorem 5.3. There exists a unique outer function F ∈H¹,F1 =1, such that

h0 = F/|F|.

(5.6)

Defineχ ∈H^∞by

1+χ(z) 1−χ(z) = 1

2π

e^iθ+z

e^iθ−z|F(e^iθ)|dθ.

Then

K =

h0− F(1−χ)(1−w)

1−χw :w(z)∈H^∞,w∞≤1

. (5.7)

As a consequence of the parametrization (5.7) we see that, forz∈D,

h Pzdθ

2π :h∈ K

is a nondegenerate closed disc. Another corollary of (5.7) is the description by Nevanlinna [1929] of all solutions f ∈H^∞,f ≤1, of the interpolation problem

f(zj)=wj, j =1,2, . . . (see Section 6 below).

Condition (5.6) may seem to contradict Theorem 4.4, because whenh0+ H^∞<1, we have claimed that the extremal functionh0has the two forms

h₀ =F¯₁/|F1|, h₀ = F₂/|F₂|

withF1,F2 ∈H¹,Fj1=1. However, ifF1∈ H¹andg∈H^∞, and if F¯1

|F1|−g

∞ =α <1,

then |arg(g F1)| ≤sin⁻¹(1−α)< π/2, and Corollary III.2.5 shows that (g F1)⁻¹∈ H¹. Hence F₁⁻¹ =g(g F1)⁻¹ ∈H¹ and (5.6) holds with F2 = F₁⁻¹/F₁⁻¹1.

The proof of Theorem 5.3 requires three lemmas; the first uses an idea from Koosis [1973].

Lemma 5.4. There exists an outer function F∈ H¹,F1=1, such that h0 = F/|F|

almost everywhere.

Proof. We know |h0| =1 almost everywhere and we know there is g∈ H^∞,g≡0, such that h0−g∞ ≤1. Then |1−h¯0g| ≤1 almost every- where. Letα=arg ¯h0g. Then|α| ≤π/2 and

|g| = |h¯0g| ≤2 cosα

as shown in Figure IV.3. Letϕ=e^α−iα˜ . Thenϕ∈ H^p,p<1, by Theorem III.2.4. However, we also havegϕ∈H¹, because

|ϕ(e^iθ)g(e^iθ)| ≤2|ϕ(e^iθ) cosα(e^iθ)| =2 Reϕ(e^iθ), (5.8)

Figure IV.3. Why|g| ≤2 cosα.

and since Reϕ≥0, 1 2π

Reϕdθ ≤lim

r→1

Reϕ(r e^iθ)dθ

2π ≤ Reϕ(0).

Henceϕg∈H^p∩L¹= H¹. ThenF0 =ϕg/ϕg1is in ball(H¹) and ¯h0F0 ≥ 0, which means that

h₀ =F₀/|F0|.

Finally, by the remarks following Theorem 5.1 there is an outer functionF∈ ball(H¹) such that (5.6) holds.

Lemma 5.5. Suppose F(z)is any H¹ function, F≡0. Defineχ ∈H^∞by 1+χ(z)

1−χ(z) = 1 2π

e^iθ+z

e^iθ−z|F(e^iθ)|dθ.

(5.9)

Ifw(z)∈H^∞,w∞≤1,then

g(z)= F(z)(1−χ(z))(1−w(z)) 1−χ(z)w(z) (5.10)

is in H^∞and

F

|F| −g ∞≤1.

(5.11)

Proof. Note that

Re1+χ(z) 1−χ(z) =

Pz(θ)|F(θ)|dθ 2π >0. Let|w(z)|<1 and set

ϕ(z)= 1+χ(z)

1−χ(z)+ 1+w(z)

1−w(z) = 2(1−χ(z)w(z)) (1−χ(z))(1−w(z)). (5.12)

Thenϕis holomorphic and since

Re1+w(z) 1−w(z) ≥0, we have

Reϕ(z)≥

Pz(θ)|F(θ)|dθ

2π ≥ |F(z)| ≥0. (5.13)

A simple calculation from (5.10) and (5.12) gives g(z)=2F(z)/ϕ(z),

so thatg∈ H^∞. Whenw≡1,g=0, and whenw≡1, ϕhas nontangential limits almost everywhere satisfying

|ϕ(e^iθ)| ≥ |F(e^iθ)|.

Notice that

F(e^iθ)

|F(e^iθ)|−g(e^iθ) ≤1 if and only if

1

|F(e^iθ)| − g(e^iθ) F(e^iθ)

≤ 1

|F(e^iθ)|. (5.14)

The transformationζ →2/ζmaps the half plane Reζ ≥ |F(e^iθ)|onto the disc F(e^iθ)|⁻¹−w| ≤ |F(e^iθ)|⁻¹. By (5.13), Re (2F/g)= Re ϕ≥ |F| almost everywhere, so that we have (5.14) and consequently (5.11).

Lemma 5.6. The function F∈H¹,F1 =1, such that h0 = F/|F|

is unique.

Proof. Lemma 5.4 shows there exists at least one suchF. What we must show is thatS_h0does not contain two functions. But ifS_h0contains two functions, then by Theorem 5.2, there isF1 ∈S_h0 with ReF1(0)<0. SinceF11 =1, the functionχ associated withF1by (5.9) satisfiesχ(0)=0. Takingw≡ −1

in (5.10), we obtaing∈ H^∞such that Reg(0)<0, and by (5.11),h0−g∈K. Then

Re

(h0−g)dθ 2π > Re

h0dθ

2π, contradicting (5.5).

Conclusion of the proof of Theorem 5.3. By Lemmas 5.4 and 5.6, there is a unique outer functionF ∈H¹,F1 =1, such that (5.6) holds. By Lemma 5.5 every function of the form

h0− F(1−χ)(1−w)

1−χw ,

(5.15)

w∈H^∞,w∞≤1, lies inK. Now letg∈ H^∞,g≡0, be such thath0− g ≤1. We must showh0−ghas the form (5.15) forw∈ball(H^∞). The idea is in the proof of Lemma 5.4. Settingα=arg ¯h0g, andϕ=e^α−iα˜ , we have ϕg∈H¹ and, by its uniqueness,

F = ϕg ϕg1. Now Reϕ(z)>0 and by (5.8)

Re 2ϕ(e^iθ) ϕg1

≥ |ϕ(e^iθ)g(e^iθ)|

ϕg1

= |F(e^iθ)|.

The positive harmonic function Re 2ϕ(z)/ϕg1 is the Poisson integral of a positive measure with absolutely continuous part exceeding|F(e^iθ)|. Conse- quently

2ϕ(z) ϕg1

−1+χ(z) 1−χ(z) =k(z) has Rek(z)>0, so that

k(z)= 1+w(z) 1−w(z), w∈H^∞,w∞≤1. A calculation then gives

g= ϕg1F

ϕ =2F

2ϕ

|ϕg|1

₋₁

= 2F

1+χ 1−χ

1+w 1−w

=F(1−χ)(1−w) 1−χw , and (5.15) holds.