Inner Functions - Exercises and Further Results

Exercises and Further Results

6. Inner Functions

Theorem 5.5. Let f(z)∈N, f ≡0. Then

f(z)=C B(z)F(z)S1(z)/S2(z), |C| =1, (5.10)

where B(z)is a Blaschke product, F(z)is an outer function, and S1(z)and S2(z)are singular functions. Except for the choice of the constant C,|C| =1, the factorization(5.10)is unique. Every function of the form(5.10)is in N.

log|f(z)| ≤log|F(z)| +log|S1(z)| −log|S2(z)|,

so that log|f(z)| is majorized by a Poisson integral. It now follows from Theorem 5.1 that f(z)∈ N.

Corollary 5.6. Let f(z)∈N, f ≡0. Then in (5.10) the singular factor S2 ≡1if and only if f(z)∈N⁺.

Proof. S2(z)≡1 if and only if

dμ=log|f(θ)|(dθ/2π)−dμ1, withdμ1 ≥0, and this holds if and only if f(z)∈ N⁺.

For emphasis we state

Corollary 5.7. If f(z)∈H^p,p>0, then f(z)has a unique decomposition f(z)=C B(z)S(z)F(z),

where|C| =1,B(z)is a Blaschke product, S(z) is a singular function, and F(z)is an outer function in H^p.

Corollary 5.8. Let f(z)∈N⁺. Then f(z)∈ H^pif and only if the outer factor F(z)is in H^p.

The proofs of these corollaries are left to the reader.

where the measuredμ is positive and singular todθ. By the Factorization Theorem 5.5, every inner function has the form

f(z)=e^icB(z)S(z),

wherecis a real constant,Bis a Blaschke product, andSis a singular function. If f(z)∈N⁺and if|f(e^iθ)| =1 almost everywhere, thenf is an inner function, because N⁺∩L^∞= H^∞. However, if f(z)=1/S(z), whereSis a nonconstant singular function, then f ∈N and|f(e^iθ)| =1 almost everywhere, but f(z) is not an inner function. If it were, then forz∈ Dwe would have|f(z)| ≤1 and|1/f(z)| ≤1, but this is impossible.

Theorem 6.1. Let B(z)be a Blaschke product with zeros{zn}, and let E ⊂∂D be the set of accumulation points of{zn}. Then B(z)extends to be analytic on the complement of

E∪ {1/zn :n=1,2, . . .}

in the complex plane: In particular B(z)is analytic across each arc of(∂D)\E.

On the other hand, the function|B(z)|does not extend continuously from D to any point of E.

Proof. Let{Bn(z)}be the finite Blaschke products converging to B(z). Then Bn(1/z)=1/Bn(z)

by reflection, and lim_n→∞Bn(z) exists on{z:|z|>1}\{1/zn :n=1,2, . . .} and the limit is an analytic function on that region. Ifz0 ∈∂D\Eand ifδ >0 is small, then each Bn is analytic on =(z0, δ) and {Bn(z)} converges boundedly on∂\∂D. By the Poisson integral formula for, for example, this means{Bn(z)}converges on. HenceB(z) is analytic except on

E∪ {1/zn :n=1,2, . . .}.

Ifz0 ∈E, then

Dz→zlim 0

|B(z)| =0, while since|B(e^iθ)| =1 almost everywhere,

Dz→zlim 0

|B(z)| =1.

Thus|B|does not extend continuously toz0.

Aside from the Blaschke products, the simplest inner function is the singular function

S(z)=exp z+1 z−1

generated by the point mass at 1. A calculation shows thatSand all its derivatives have nontangential limit 0 ate^iθ =1. Under w=(1+z)/(1−z), the discDis conformally mapped to the right half plane{Rew >0}, so thatz=1 corresponds tow= ∞, and so that∂D\{1}corresponds to the imaginary axis.

ThusS(z)=e^−wis analytic across∂D\{1}andS(z) wraps∂D\{1}around∂D infinitely often. The vertical line Rew=α, α >0, comes from theorocycle

C_α =

1− |z|²

|1−z|² =α

This is a circle, with centerα/(1+α) and radius 1/(1+α), which is tangent to

∂^Dat 1. On this orocycle|S(z)| =e^−α, andSwrapsC_α\{1}around|ζ| =e^−α infinitely often. The functionS(z) has no zeros inD, but for everyζ,0, <|ζ|<

1,S(z)=ζ infinitely often in every neighborhood ofz =1.

Recall the notation

α(e^iθ)=

z∈D : |e^iθ−z|

1− |z| < α

, α >1 for the conelike region inDwith vertexe^iθ.

Theorem 6.2. Let S(z)be the singular function determined by the measure μ on ∂D, and let E ⊂∂D be the closed support of μ. Then S(z) extends analytically to\E. In particular S(z)is analytic across each arc of(∂D)\E.

On the other hand,|S(z)|does not extend continuously fromDto any point of E. For anyα >1and forμ-almost allθ

z→elim^iθS(z)=0, z∈ α(eⁱ^θ).

h→0lim

μ((θ−h, θ+h)) hlog 1/h = ∞, (6.1)

then every derivative S⁽ⁿ⁾(z)of S(z)satisfies

z→elim^iθS⁽ⁿ⁾(z)=0, z∈ α(eⁱ^θ).

(6.2)

Proof. For any measureμon∂D, the function eⁱ^θ+z

eⁱ^θ−zdμ(θ)

is analytic at all points not in the closed supportEofμ. HenceS(z) is analytic on\E.

Ifμis singular, then

h→0lim

μ((θ−h, θ+h))

2h = ∞

(6.3)

forμ-almost allθ. This follows from Lemma I.4.4 by repeating the proof of Lemma I.5.4, withd xanddμinterchanged. Forz ∈ α(e^iθ) and for|ϕ−θ|<

1− |z|²we have

Pz(ϕ)≥ c2

1− |z|². Settingh=1− |z|², we therefore obtain

−log|S(z)| ≥ c₂

hμ((θ−h, θ+h))→ ∞

asz→e^iθ,z∈ α(e^iθ), whenever (6.3) holds atθ. This shows|S|does not extend continuously at any point ofE.

If (6.1) holds atθ, then by similar reasoning lim

α(θ)z→e^iθ(−log|S(z)| +nlog(1− |z|²)= ∞ for everyn=1,2. . .. Hence we have

lim

α(θ)z→e^iθ

|S(z)|

(1− |z|²)ⁿ =0.

(6.4)

Now fixzin _α(θ) and consider two discs

1=(z,a(1− |z|²)), 2 =(z,¹₂a(1− |z|²)). If f(ζ) is analytic on1, and if

sup

|f(ζ)|

(1− |ζ|²)ⁿ < ε, then

sup2

|f⁽ⁿ⁾(ζ)|<C(a,n)ε,

where C(a,n) depends only on a and n. This is an easy consequence of Schwarz’s lemma on1, or of the Poisson integral formula for1. By (6.4) we therefore have

S⁽ⁿ⁾(z)→0 asz→e^iθ,z∈ α(θ), whenever (6.1) holds ate^iθ.

Theorem 6.3. Let f ∈H^p,p>0, and let be an open arc on∂D. If f(z)is analytic across , then its inner factor and its outer factor are analytic across . If f(z)is continuous across , then its outer factor is continuous across . Proof. Write f =BSF, whereBis a Blaschke product,Sis a singular function, andFis an outer function. We may suppose f ≡0. If f is analytic or continuous across , thenFis bounded on any compact subset of , because

|F| = |f|on .

Supposef is analytic across . If the zeros ofBhad an accumulation point on , thenf would have a zero of infinite order at some point of . This is impossible and soBis analytic across . Letμbe the measure determiningS.

Ifμhad a point charge ateîθ ∈ , thenS, and hence alsof, would have a zero of infinite order ateîθ, by Theorem 6.2. Thusμ({eîθ})=0 for allθ ∈ . Now ifμ(K)>0 for some compact subsetK of , thenK is uncountable and by Theorem 6.2, f(z) has infinitely many zeros onK. Thereforeμ(K)=0 andS is analytic across . HenceF = f/B Sis analytic across .

Supposef is continuous across . LetK = {θ∈ : f(e^iθ)=0}. Then, as a function on ,Fis continuous at each point ofK, because|F| = |f|on and|F| =0 on K. On \K,|f|>0, so thatBandScannot tend to zero at any point of \K. ThenBandSare analytic across \K andFis continuous on . The Poisson integral representation now implies thatFis continuous on D∪ .

A compact setKin the plane has positivelogarithmic capacityif there is a positive measureσ onKwithσ =0 such that thelogarithmic potential

U_σ(z)=

log 1

|ζ−z|dσ(ζ)

is bounded on some neighborhood ofK. IfK ⊂D, thenKhas positive capacity if and only ifKsupports a positive massσ for whichGreen’s potential

G_σ(z)=

log 1−ζ¯z

ζ−z

dσ(ζ) (6.5)

is bounded onD, because the term

K log|1−ζ¯z|dσ(ζ) is always bounded on D. An arbitrary setEis said to have positive capacity if some compact subset ofEhas positive capacity. Since log 1/|ζ|is locally integrable with respect to area, any set of positive area has positive capacity. There are perfect sets of capacity zero, but these sets are very thin. For example, the Cantor ternary set on [0, 1] has positive capacity (see Tsuji [1959]).

The Green’s potential G_σ(z) in (6.5) clearly satisfies G_σ(z)≥0,z ∈D.

Sinceσ is finite and supported at a positive distance from∂^{D, we have} Gσ is continuous and zero at each point of∂D.

(6.6)

Further information about logarithmic capacity and potentials can be found in Tsuji’s book [1959], but we shall need only the facts cited above.

IfB(z) is a Blaschke product, let us agree to also call e^icB(z)

aBlaschke productwhencis a real constant.

Theorem 6.4(Frostman). Let f(z)be a nonconstant inner function on the unit disc. Then for allζ,|ζ|<1, except possibly for a set of capacity zero,

the function

f_ζ(z)= f(z)−ζ 1−ζ f(z) is a Blaschke product.

Proof. LetK be a compact set of positive capacity and letσ be a positive mass onKsuch thatG_σ(z) is bounded onD. We shall show

σ({ζ ∈K : fζ is not a Blaschke product})=0 and that will prove the theorem. Let

F(z)=Gσ(f(z))=

log

1−ζ f(z) ζ− f(z)

dσ(ζ).

ThenV ≥0 andVis bounded. Because f(z) is an inner function, (6.6) and dominated convergence imply that

r→1lim

V(r e^iθ)dθ=0.

Hence, by Fatou’s lemma,

K rlim→1

log|fζ(r^iθ)|dθ

2πdσ(ζ)=0. Becauseσ ≥0 and log|f_ζ| ≤0, this means

r→1lim

log|f_ζ(r e^iθ)|dθ 2π =0

forσ-almost everyζ. Theorem 2.4 then shows that f_ζ is a Blaschke product forσ-almost allζ.

Corollary 6.5. The set of Blaschke products is uniformly dense in the set of inner functions.

Proof. If f(z) is an inner function and if|ζ|is small, then f − fζ∞< ε.

By Frostman’s theorem f_ζ is a Blaschke product for many smallζ.

Corollary 6.5 should be compared to Carath´eodory’s theorem I.2.1, in which a weaker form of convergence, namely, pointwise bounded convergence, is obtained, but in which it is assumed only thatf∞≤1.

Let f ∈H^∞(D) and letz0 ∈∂D. Thecluster setoff atz0 is Cl(f,z0)=

r>0

f(D∩(z0,r)).

Thusζ ∈Cl(f,z0) if and only if there are pointszn inDtending toz0 such that f(zn)→ζ. The cluster set is a compact, nonempty, connected plane set.

It is a singleton if and only iff is continuous onD∪ {z0}. Therange setoff atz0is

R(f,z0)=

r>0

f(D∩(z0,r)),

so ζ ∈R(f,z0) if and only if there are points zn in D tending to z0 such that f(zn)=ζ,n=1,2, . . . In other words, the range set is the set of values assumed infinitely often in each neighborhood ofz0. The range setR(f,z0) is aGδset. ClearlyR(f,z0)⊂Cl(f,z0). If f(z) is analytic acrossz0, and not constant, then Cl(f,z0)= f(z0), andR(f,z0)=∅^.

Theorem 6.6. Let f(z) be an inner function on D, and let z0 ∈∂D be a singularity of f(z)(that is, a point at which f(z)does not extend analytically).

Then

Cl(f,z0)=D¯ and

R(f,z0)= D\L, where L is a set of logarithmic capacity zero.

Theorem 6.6 shows that, despite Fatou’s theorem, the boundary behavior of anH^∞function can be rather wild. For example, if f(z) is a Blaschke product whose zeros are dense on∂^{D, or if} ^f(z) is the singular function determined by a singular measure with closed support∂D, then the conclusion of Theorem 6.6 holds at every z0 ∈∂D, even though f(z) has nontangential limits almost everywhere.

Proof. Since sets of capacity zero have no interior, the assertion about the range sets implies the assertion about the cluster set.

We are assuming f(z) is not analytic across any arc containingz0. If f(z) is a Blaschke product, thenz0 is an accumulation point of the zeros of f(z).

Thus 0∈R(f,z0) iff is a Blaschke product. In general fζ(z)= f(z)−ζ

1−ζ¯ f(z)

is a Blaschke product whenζ ∈L, a set of capacity zero. Since f_ζ also has a singularity atz₀, we see that forζ ∈L,z₀ is an accumulation point of the zeros of f_ζ, thereforez0is an accumulation point of theζ-points of f(z).

That proves Theorem 6.6, but by using Theorem 6.2 we can get more precise information. Suppose f(z) is an inner function, supposez0is a singularity of f(z) and supposeζ ∈ D\R(f,z0). Then the inner function fζis not a Blaschke product. Moreover, the proof above shows its Blaschke factor cannot have zeros

accumulating atz0. Write f_ζ = B_ζS_ζ. We have just showed thatB_ζis analytic acrossz0. HenceS_ζ has a singularity atz0. Two cases now arise.

Ifz0is an isolated singularity ofS_ζon∂D, then the singular measureμgiving rise toS_ζ contains an atom atz0. In this case S_ζ and all its derivatives tend nontangentially to 0 atz0. It follows readily that f_ζ(z) and all its derivatives tend nontangentially to 0, so that nontangentially f(z) tends toζ, while all the derivatives of f(z) tend to 0. These conclusions also hold ifμsatisfies (6.1) atz0. It is quite clear that for fixed z0, there can be at most one point ζ ∈D\R(f,z0) at which these conclusions can hold.

The alternative case is thatμ({z0})=0. Then by Theorem 6.2, z0 is the limit of a sequence of points{eîθⁿ}at each of whichSζ has nontangential limit 0, and therefore at each of which f(z) has nontangential limitζ. In this case Theorem 6.2 tells us even more. Eitherμis continuous on some neighborhood ofz₀, orμassigns positive mass to each point in a sequenceeîθⁿ tending to z0. Ifμis not continuous, there areeîθⁿ →z0such that at each pointeîθⁿ, f(z) tends nontangentially toζ, and each derivative f^(k)(z) tends nontangentially to 0. Ifμis continuous, then by Theorem 6.2 each neighborhood ofz0contains uncountably manyeîθ at which f(z) tends nontangentially toζ.

The above reasoning can be summarized as follows:

Theorem 6.7. Let f(z) be an inner function on D and let z0∈∂D be a singularity of f(z). For|ζ|<1at least one of the following holds.

(a) ζ is in the range set of f at z0.

(b) f(z) has nontangential limit ζ at z₀, and each derivative f⁽ⁿ⁾ has nontangential limit 0 at z₀.

(d) Each neighborhood of z0on∂D contains uncountably many points at which f(z)has nontangential limitζ.

Considerably more about cluster theory can be found in the interesting books of Noshiro [1960] and of Collingwood and Lohwater [1966].

Dalam dokumen Bounded Analytic Functions (Halaman 82-89)