The Carleson–Jacobs Theorem

Exercises and Further Results

2. The Carleson–Jacobs Theorem

To complete the proof of Corollary 1.8, notice thatG = f F/B0 is an H¹ function on the annulusr <|z|<1 ifr >|zj|. Using (1.10) and using the lemma locally, we see that G is analytic across T, and thatG is in fact a rational function. Since B0G is analytic acrossT, and since f/f∞ is an inner function, Theorem II.6.3 shows thatf is analytic acrossT. Consequently f/f∞ is a Blaschke product of finite degree, andFis a rational function.

NowB0hasnzeros inDandFhas a zero atz=0. By (1.10) and the argument principle, it follows thatf has at mostn−1 zeros.

In Corollary 1.8 the extremal functionf(z)=cB(z) can have fewer thann− 1 zeros. For example, supposegis the function of minimum norm interpolating

g(zj)=wj, 1≤ j ≤n−1,

and takewn =g(zn). Then f =ghas at mostn−2 zeros. Similarly, the ex- tremal function in Corollary 1.9 can have fewer thannzeros. However, if the in- terpolation (1.9) has two distinct solutions f1, f2withf1∞≤1,f2∞≤1, then the extremal function f0in Corollary 1.9 is a Blaschke product of degree n. For the proof, notice that (1.9) then has a solution f withf∞ <1, by the uniqueness asserted in Corollary 1.8. Then by Rouch´e’s theorem, f0andf0− f have the same number of zeros in|z|<1. Since f0(zj)= f(zj),1≤ j ≤n, it follows that f0has at leastnzeros in|z|<1.

Corollary 2.4 Chapter I, contains more information than we have obtained here, but the duality methods of this section apply to a wider range of problems.

f(θ)∈C such thatωf(δ)≤ω(δ)but such that the best approximation to f in H^∞is not continuous.

We prove Theorem 2.1 first. In doing so we can assume f −g∞ =dist(f,H^∞)=1. By Theorem 1.7, there isF∈ H₀¹,F1 =1, such that

(f −g)F= |F| a.e (2.1)

All information used in the proof will follow from (2.1) and the continuity of f. We need two lemmas.

Lemma 2.3. Let G=u+iv∈ H¹and let I be an arc on T such that almost everywhere on I

u>0, |v| ≤αu,

whereα >0. Let J be a relatively compact subarc of I and let V be the domain {r e^iθ :r0 <r <1,0∈ J}. Then G∈H^p(V)if

arctanα < π/2p.

The statementG ∈ H^p(V) means that|G|^phas a harmonic majorant inV.

Proof. Recall from Corollary 2.5 in Chapter III that an H¹ function whose boundary values lie in the sector S= {x >0,|y|< αx} is in H^p if arctan α < π/2p. EnlargingJ, we may assumeGhas a finite radial limit at the end- points ofJ. ThenM =sup{|G(z)|:z∈∂V,|z|<1}is finite, and

g= M(1+1/α)+G

has, at almost every point of∂^V, boundary value in the coneS(see Figure IV.1).

Figure IV.1. S= {|y|< αx,x>0}. On∂V,ghas values in the union of the disc and the shaded cone.

Becauseg∈ H¹(V), the Poisson integral formula (applied after conformally mappingVonto the unit disc) shows thatg(V)⊂S. By Corollary III.2.5 and the conformal invariance of the definition of H^p(V), we see thatg∈H^p(V) and hence thatG ∈ H^p(V).

Lemma 2.4. Let f(θ)∈C, let g(θ)∈H^∞and let F(θ)∈H₀¹ be functions such that(2.1)holds. Then

(a) F ∈H^p for all p<∞, and (b) ifτ ∈[0,2π]and if

fτ = f − f(τ), gτ =g− f(τ), then there isδ >0and r0 >0such that|gr(z)| ≥ ¹₂ on

W_τ = {r e^iθ :|θ−τ|< δ/2,r0 <r <1}.

Proof. To prove (a) let p<∞ and let ε >0 satisfy arctan(ε/(1−ε))<

π/2p. Chooseδso that|f(θ)− f(τ)|< εwhen|θ−τ|< δ. Then on I_τ = {θ :|θ−τ|< δ}, we have by (2.1)

−gτF=(f −g)F− fτF= |F| − fτF.

Consequently, Re(−gτF)>0 and |Im gτF|< ε/(1−ε) Re(−gτF) almost everywhere onIτ. By Lemma 2.3 we havegτF∈ H^p(Wτ). If we replaceWτ

by the intersection of two discs whose boundaries cross inside I_τ, a simple conformal mapping can be used to show

|θ−τ|<δ/2|gτF|^pdθ <∞.

Since|gτ| ≥ f −g∞−ω(δ)≥1−εonIτ, this means

|θ−τ|<δ/2|F|^pdθ <

∞, so thatF∈L^p. HenceF∈ H^p by Corollary 11.4.3.

To prove (b) first takeεso small that onI_τ

−gτF=exp(u+iv), wherev∞< π/4. then

h=exp(−ivχIτ +vχIτ)

is inH²by Corollary III.2.6. TheH¹functiongτFhis real onIτ. By Lemma 1.10 and Theorem II.6.3, the inner factor ofgτFhis analytic acrossIt. Conse- quently, the inner factor ofgτ is analytic across Iτ. The representation for- mula for outer functions then shows |gτ(z)| ≥ ³₄ in a region {r0(τ)<r <

1,|θ−τ|< δ/2}. Sinceg_σ(z)=g_τ(z)−(f(σ)− f(τ)), we have|gσ(z)|> ¹₂ on the same region if|σ −τ|< δ. This means we may chooser0(τ) indepen- dent ofτ, and thus (b) is proved.

Proof of Theorem 2.1. By Lemma 2.4(b), g_τ(z) has a single-valued loga- rithm onW_τ, and the logarithm is defined by

logg_τ(z)= 1 2π

|θ−τ|<δlog|gτ(θ)|e^iθ +z

e^iθ −z dθ+R_τ(z), (2.2)

where R_τ(z) is the same integral over|θ−τ| ≥δ plus the logarithm of the inner factor of gτ(z). Since|gτ(z)| ≥ ¹₂ onWτ, the inner factor is bounded below on W_τ and it is analytic across {e^iθ :|θ−τ|< δ/2}. Hence there is r1>0 such thatRτ(z) is bounded and analytic onτ = {|z−e^iτ|<r1}. The radiusr1and the bound sup{|Rτ(z)|:z ∈τ}are independent ofτ.

Because|fτ−gτ| =1, we have

1− |gτ|² = |f_τ|²−2 Ref_τg¯_τ ≤ Aω(|θ−τ|), so that

|log|gτ(θ) ≤cω(|θ−τ|). Let(τ) be the truncated cone

(τ)=

z: |z−e^iτ|

1− |z| ≤2,r2 <|z|<1

,

wherer2 >0 is such that(τ)⊂τ. Forz ∈τ, e^iθ+z

e^iθ−z

≤ C

|θ−τ|, so that (2.2) yields

|loggτ(z)−Rτ(z)| ≤C _δ

0

ω(t)

t dt, z∈(τ).

By Schwarz’s lemma and the uniform bound on R_τ(z),|R_τ(z)−R_τ(w)| ≤ c|z−w|,z, w∈(τ). Therefore

|gτ(z)−g_τ(w)| ≤C|z−w| +η(δ), z, w∈τ, whereη(δ)→0(δ→0).

Now letτ andσ be so close together that there is a pointz∈(τ)∩(σ).

Taking nontangential limits, we then obtain (when 1− |z|is small)

|g(e^iτ)−g(e^iσ)| ≤ |g(e^iτ)−g(z)| + |g(z)−g(e^iσ)|

≤ |gτ(e^iτ)−gτ(z)| + |gσ(z)−gσ(e^iσ)|

≤2η(δ)+ |e^iτ−z| + |e^iσ −z|

≤2η(δ)+C|σ −τ|.

Fixingδ >0 with 2η(δ)< ε, we conclude that

σlim→τ|g(e^iσ)−g(e^iτ)|< ε, which means thatgis continuous.

Now letω(t) satisfy the hypotheses of Theorem 2.2.

Lemma 2.5. Letδ >0. Let f(e^it)=

ω(t), 0≤t ≤δ, 0, −δ≤t <0.

Extend f(e^it) to be smooth onδ <|t|< π and continuous on[−π, π]with f(−π)= f(π). Let g∈H^∞be the best approximation of f. If g is continuous, then

g(1)= ±if −g∞.

Proof. Since f is real, f ∈/ H^∞ and f −g∞>0. We may suppose f −g∞ =1, so that|f −g| =1 onT. We must prove Reg(1)=0. Since

|f −g| =1,log|g(t)| =0 on−δ <t <0. Suppose Re g(1)>0. Then for 0<t < δ,

log|g| = −log g− f

g

≤ −log

1− f Reg

|g|²

≤ −Cω(t).

Similarly, if Reg(1)<0, then on 0<t < δ,log|g| ≥Cω(t). Moreover,ghas a continuous logarithm on{|z−1|< δ,|z| ≤1}, again because|f −g| =1.

However, if Reg(1)=0, then by Lemma III.1.2,

r→1lim|argg(r)| = c+lim

ε→0

1 π

|t|>ε

log|g(t)| t dt

≤c+Clim

ε→0

1 π

|t|>ε

ω(t)

t dt = +∞.

This contradiction showsg(1)= ±i.

To prove Theorem 2.2 letδn =2⁻ⁿ, and let ωn(t)=ω

δn

2 −

δn+δn−1

2 −t

on [δn, δn−1], ωn(t)=0 off [δn, δn−1]. Let

f(e^it)=^∞

k=1

1

2ω4k+1(t)+i ∞ k=1

1

2ω4k+3(t).

Thenf²is real, so that f ∈/ H^∞. It is not hard to verify thatωf(δ)≤ω(δ). If gis the best approximation off, then Lemma 2.5 says that

g(e^iδ4k+1)= ±if −g∞, g(e^iδ4k+3)= ±f −g∞. Thusg(e^it) is not continuous att =0.