Prime and Semiprime Inner Functions_I.chalendar Et Al

8/13/2019 Prime and Semiprime Inner Functions_I.chalendar Et Al

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J. London Math. Soc. (2) 88 (2013) 779800 C2013 London Mathematical Societydoi:10.1112/jlms/jdt038

Prime and semiprime inner functions

Isabelle Chalendar, Pamela Gorkin and Jonathan R. Partington

Abstract

This paper studies the structure of inner functions under the operation of composition, andin particular the notions or primeness and semiprimeness. Results proved include the densityof prime finite Blaschke products in the set of finite Blaschke products, the semiprimeness offinite products of thin Blaschke products and their approximability by prime Blaschke products.An example of a nonsemiprime Blaschke product that is a Frostman Blaschke product is alsoprovided.

1. Introduction and notation

Let D be the open unit disk of the complex plane C. An analytic function u: D D is saidto be an inner function if it has radial limits of modulus 1 at almost every (a.e.) point of theunit circle T ={z C :|z|= 1}. Among the inner functions are the Blaschke products and thesingular inner functions. Moreover, each inner function is the product of such functions, withuniqueness of factorization up to a constant of modulus 1. If (an)n denotes the zero sequence(counting multiplicities) ofu, then the zeros satisfy the so-called Blaschke condition

n

(1 |an|)


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PRIME AND SEMIPRIME INNER FUNCTIONS 781

Table 1. Summary of known results for inner functions

Type of inner function Semiprime Prime

Singular If and only if finitely many singular-ities

Never

Omits more then 1 point ofD Never NeverOmits exactly 1 point ofD If and only if finitely many singular-

itiesNever

Surjective inner Thin Blaschke aB, B thin BlaschkezS a /E, Ea countable set

zS

From table 1, together with Urabes result [25] that an inner function with finitely manysingularities is semiprime, we see that an inner function with finitely many singularities can

omit at most one value inD

. For further results on the set of singularities of an inner functionomitting at least two points, we refer the reader to[5, 6].Looking at the table above raises several other questions. For example, is every surjective

inner function prime or semiprime? In addition to those appearing in the table, there are severalother examples of ways that singular inner functions can be multiplied by functions to becomeprime. For example, for the atomic singular inner function S(z) = exp((z+ 1)/(z 1)),multiplying by a Blaschke product with a simple zero at zero and satisfying |B(r)| 1 asr 1 changes the nonprime function S into the function BS, which is prime. Which Blaschkeproducts can be multiplied into the class of prime or semiprime Blaschke products?

A Blaschke product with zeros (an) is said to be a uniform Frostman Blaschke product if itsatisfies

supT 1 |an|

|an |


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782 I. CHALENDAR, P. GORKIN AND J. R. PARTINGTON

2. Background and preliminary results

In what follows, we let Pdenote the class of prime inner functions.Recall that a function f in H has radial limits a.e. We will usually denote the radial

limit function f byf. To obtain complete results, it is necessary to use maximal ideal spacetechniques and we may view our function f as a continuous function on M(H) via theGelfand transform off: f(x) :=x(f). We recall the relevant information here. We let M(H)denote the maximal ideal space of H; that is, the space of nonzero multiplicative linearfunctions on H equipped with the weak- topology. Identifying z D with point evaluationallows us to think of D as a subset of M(H) and Carlesons corona theorem tells us thatD is dense in M(H). We consider two decompositions of the maximal ideal space. First,we consider the fibers over points D defined by M = {y M(H) :y(z) =}, where zdenotes the identity function. Next, we consider the Gleason parts: given two points z, w Dthe pseudohyperbolic distance between them is denoted by and given by

(z, w) = z w

1 wz .

For two points in M(H), we have

(x, y) = sup{|f(y)|: fH, f 1, f(x) = 0}.

This definition agrees with the usual one whenx, y D; if we define two points to be equivalentwhen(x, y)< 1, then this defines an equivalence relation and the equivalence classes are calledGleason parts.

In this paper, one very important concept is the map Lxassociated with a pointx M(H):given a point x M(H) \ D, there is an associated map Lx that is a limit of the maps Lzdefined by

Lz(z) := z+ z

1 + zz

,

where (z) is a net in D converging to x in M(H). The function Lx maps the open unit

disk onto the Gleason partP(x) ofx,Lx(0) =x and, identifying a function fwith its Gelfandtransform, we have f Lx H whenever fH.

In general, we need to apply techniques developed by Hoffman[15] in his in-depth study ofthe Gleason parts in the maximal ideal space ofH. For completeness, we provide referencesfor the lemmas and propositions that we will use in the paper that are not readily availableand short proofs when possible. For the first result below, see also [3, p. 51; 12].

Lemma 2. Ifu is an inner function analytic in a neighborhood of0 T, then

limzD,z0

(1 |z|2)|u(z)|

1 |u(z)|2 = 1.

We will also need a very useful characterization of thin Blaschke products, due toTolokonnikov[24, Theorem 2.3].

Theorem 3. A Blaschke productB is thin if and only if

lim|z|1

(|B(z)|2 + (1 |z|2)|B(z)|) = 1.

As a consequence of this characterization, it follows that thin Blaschke products are stableunder composition on the left by Mobius transformations, as well as composition on the right


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by finite Blaschke products (see[8, Proposition 4.3] for a proof of this result; for compositionon the left, the result is well known and follows easily from Tolokonnikovs characterization).We record this result for future reference.

Proposition 4. Let B be a thin Blaschke product, let a D andb be a finite Blaschkeproduct. Thena B andB b are still thin Blaschke products.

We now present two useful lemmas due to Hoffman.

Lemma 5. Let B be an infinite thin Blaschke product and let (zn)n be a subsequence ofzeros of B. If B Lzn converges tof on D for the weak- topology, then there exists Rsuch that f(z) =eiz.

Proof. Since B Lzn(0) =B(zn) = 0, it follows that f(0) = 0. Moreover, f 1. Since|(B Lzn)

(0)|= |B(zn)(1 |zn|2)| and since B is thin, we get |f(0)|= 1. The conclusionfollows from Schwarzs lemma.

We also need to analyze what happens whenever (zn)n is not necessarily a subsequence ofzeros ofB .

Lemma 6. LetB be an infinite thin Blaschke product and let(zn)n D such that |zn| 1.If B Lzn converges to f on D for the weak- topology, then there exists R such thatf(z) =ei orf is a Mobius transformation.

Proof. If |f(0)|= 1, since f 1, then the maximum principle implies that f is aconstant of modulus 1. Otherwise, note that f(0) B Lzn weak-converges tog = f(0) f,with g(0) = 0. Since B is thin, by Proposition 4, f(0) B is also thin. By Theorem 3,

limn

(1 |zn|2)|(f(0) B)

(zn)| + |(f(0) B)(zn)|2 = 1.

Since|(f(0) B)(zn)| 0, we get

limn

(1 |zn|2)|(f(0) B)

(zn)|= 1. (1)

Since |Lzn(0)|= 1 |zn|2 andLzn(0) =zn,(1) means |g

(0)|= 1. Now, Schwarzs lemma showsthat there exists R such that g(z) =eiz, and then f is a Mobius transformation.

We shall need the following elementary lemma from [8].

Proposition 7. Let I=U V for inner functions U and V, and I satisfying I(0) = 0.ThenI=U1 V1 whereU1(0) =V1(0) = 0.

The proof follows by writing I=U 1V(0) V(0) V.Given this proposition, if I(0)= 0, then we may compose with a Mobius transformation

to obtain I1= I(0) I with I1(0) = 0. Now I is prime if and only ifI1 is and, according toProposition 7,I

1is prime if and only if it has no factorization of the form U

1 V

1withU

1and

V1 vanishing at 0. Therefore, we may assume I(0) = 0 and U1(0) =V1(0) = 0.


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3. Preliminary density results

Every thin Blaschke productBhas the property that for almost all a D, the productaB Pfor almost all a D (see[8]). In fact, much more is true as we shall see below.

We show that we can always multiply a Blaschke product by a single Blaschke factor so thatthe resulting product is not of the formX Y withYa finite (nontrivial) Blaschke product. Inparticular, we can multiply every finite Blaschke product into the set of prime finite Blaschkeproducts. As a consequence, we will see that prime finite Blaschke products are dense in theset of finite Blaschke products, a result we were unable to find in the literature.

We will call a finite Blaschke product B monic if it is of the form

B(z) =zmn

j=1

z aj1 aj z

,

where the product is taken over the nonzero zeros ofB. Given a Blaschke product B, we letNB be the set of finite monic subproducts of B. Note that if C=B where B is a monic

Blaschke product and T

, then Cis the composition of two (possibly nonmonic) Blaschkeproducts if and only ifB is. We will rotate our Blaschke products, when necessary, to assumethat they are monic.

Theorem 8. For every Blaschke product B with B(0) = 0 and all but countably manya D,the productaB is not of the form U V withU inner andVa finite Blaschke productof degree greater than 1.

Proof. Using Proposition 7, we consider only factorizations for which B(0) =U(0) =V(0) = 0.

Let (zj ) denote the zero sequence ofB. For each n N, consider each Blaschke product Vof degree n with zero set contained in (zj ). Now, there are countably many choices for monicsubproducts ofB of degree n or subproducts B/V for which V is of degree n (choosing then indices of the zeros amounts to a bijection with N N N (n times)). Now, let NBdenote the set of monic subproducts V ofB with V(0) = 0, and consider

AV,n = {V(zj ) :V NB , j N, deg V =n}

and

DV,n = {(B/V)(zj ) :V NB, j N, deg V =n}.

Now both of these sets are countable and so we may choose a so that ifW NB , then

W(a) /V,n

AV,n and (B/W)(a) /V,n

DV,n ;

that is,

a /

{W:WNB}

V,n

W1(AV,n )

{B/W:WNB}

V,n

(B/W)1(DV,n )

.

Suppose

aB = U1 V1,

with V1 of degree m > 1. Since U1(0) = 0 =V1(0) and we suppose that V1 is monic, wehave

aB = U

1 V

1= V

1(U

2 V

1) with U

1(z) =zU

2(z). Then V

1 N

B or V

1(a) = 0 and

V1/a NB.


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IfV1 is a finite subproduct ofB , then since V1 is of degree greater than 1, there is a secondpointz0such thatV1(z0) =V1(a). SinceaBhas a zero of order 1 at a, it is not possible to havea= z0, so a=z0. By our choice ofa, we cannot havez0= zj for anyj . But thenB(z0)= 0 anda(z0)= 0, so (a B)(z0)= 0. But (aB)(z0) = (U1 V1)(z0) =U1(V1(a)) = (aB)(a) = 0, a

contradiction.IfV1= aB1 where B1 is a subproduct ofB , then

aB = V1(U2 V1)

implies

B2= U2 V1,

whereB2= aB/V1 = B/B1 and V1= aB1. By our choice ofa, we have B2(a)=B2(zj ) foranyj . On the other hand,V1(a) = 0 =V1(zj ) for somej , so B2(a) =U2(V1(a)) =U2(V1(zj )) =B2(zj ), a contradiction.

Corollary 9. The set of prime finite Blaschke products is dense in the set of all finiteBlaschke products and hence weak- dense in the set of all inner functions.

Proof. First note that for every finite Blaschke product, a B is a Blaschke product forevery a D, so we may assume B(0) = 0. Let B be a finite Blaschke product and supposethat D is a zero ofB. Consider C=B/. Then C is finite and therefore, for any > 0,we may use Theorem 8 to choose a close to so that aC cannot be factored as U V withVfinite, of degree greater than 1, and aC B< . Since C is finite, every factorizationaC=U V implies thatV is of finite degree, and therefore the degree of such a Vmust be 1.Thus, aCis prime.

Remark 10. The same method of proof shows that the set of (prime) finite Blaschkeproducts such that every subproduct is prime (one might call them hereditarily prime) is openand dense in the set of all finite Blaschke products.

We also obtain the following.

Corollary 11. The set of Blaschke products that are either prime or not of the form

X Y with Ya finite Blaschke product is dense in the set of all Blaschke products and hencedense in the set of all inner functions.

Proof. We know that prime Blaschke products are dense in the set of finite Blaschkeproducts, so assume that B is infinite. We show how to approximate B by a nonfactorizableBlaschke product: ifB(0) = 0, then the proof proceeds exactly as in Corollary 9. If not, thenconsider I=B(0) B. Then I(0) = 0. If I is Blaschke and not prime, then we can find asequence of Blaschke products (Cn) that cannot be factored as X Y withYa finite Blaschkeproduct converging uniformly to I. Therefore, (1B(0) Cn) converges uniformly to B and

these cannot be factored either, establishing the result. If1B(0) Cn is not a Blaschke productfor some n, then using Frostmans theorem we may replace B(0) by an appropriate Mobiustransformation to obtain a Blaschke product that cannot be factored.

To put our approximation methods into context, we show below that a small movement inthe zeros of a Blaschke product can produce a Blaschke product that is far from the original,


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even in the case of finite Blaschke products. Thus, to obtain prime Blaschke products thatapproximate our original function, other methods are needed.

Recall that the pseudohyperbolic metric is given by (z, w) =|(z w)/(1 wz)| forz, w D.

Proposition 12. (i) For each > 0, there is an integer N and two degree-N BlaschkeproductsB1, B2with zerosz1, . . . , zNandw1, . . . , wN, respectively,normalized to be positive atthe origin,such that the hyperbolic distances satisfy(zk, wk)< for eachkbut B1 B2=2. Moreover,we may takeNto grow as1/. Thus,uniform closeness of zeros does not guaranteeuniform closeness of the Blaschke products, even in the finite case.

(ii) For infinite Blaschke productsB1, B2 normalized to be positive at the origin, providedthat >0 is sufficiently small, the condition

k=1 (zk, wk)< impliesB1 B2< 3.

Proof. (i) Take a > 0 and note that for the functions f(z) =z and g(z) =

(a z)/(1 az), we have (f g)(z) =a(z2 1)/(1 az) so that f g is asymptotic to2a as a 0. It follows by the invariance of the pseudohyperbolic metric that two elementaryBlaschke factors b and b , with zeros at , and taking positive values at the origin, differinH norm by a quantity asymptotic to 2(, ) as(, ) 0. This implies also that thereis a point P(, ) on the circle at which the arguments ofb and b differ by asymptoticallythe same quantity, 2(, ).

Note that by multiplyingandby a complex number of modulus 1, we may rotate P(, ),so that it equals 1, for example. Now for finite products of elementary Blaschke factors thedifferences in arguments add, and by taking O(1/) points we may obtain a total difference inargument of exactly radians at 1 (if it exceeds it slightly, then we can perturb the points toreduce it).

(ii) It follows in a standard way that if inner functions 1, 1, 2, 2 satisfy1 10. For an innerfunctionu, let us denote the change of argument ofu betweenand by arg u. By passingto a subarc, if necessary, we may assume 0 < arg I 0 such that arg V for all inner functions V withV(0) = 0.

By the above, we may choose n N such that znV maps (, ) around the circle morethan once for any inner function V with V(0) = 0. We claim that znI(z) is semiprime.

Suppose there exist U and V with U(0) =V(0) = 0 such that znI(z) =U(V(z)) =

V(z)U1(V(z)), where U1(0)= 0. Since I has distinct zeros, we know U(z) =zU1(z) withU1(0)= 0. Now write V(z) =z

kV1(z) with V1(0) = 0.We claim k= n. If not, then znkV2(z) =U1(V(z)) with n k >0. Then U1(0) =

U1(V(0)) = 0, a contradiction. So V(z) =znV1(z) and V1(0) = 0. By our choice of n, we

have T V(, ) and since znI(z) =U(V(z)) with I continuous on (, ) we must haveUcontinuous everywhere on T; that is, Uis a finite Blaschke product.

We can improve upon this result, multiplying an inner function with no repeated zeros into P.

Theorem 17. If an inner functionIis such that the Blaschke factor ofIhas no repeatedroots and is continuous on a subarc J of positive measure, then there is a nonempty open

subsetA D such that 2aI is semiprime for alla A.


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Proof. We may suppose without loss of generality that the change of argument ofIon theintervalJsatisfies 0< arg I < . Choosing the open setA such that


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We may also obtain a result for a(b B) where b is a finite prime Blaschke product (recallthat these are dense in the class of all finite Blaschke products, by Corollary 9), in particular,for all b of prime degree. This result will be of further use later.

Proposition 20. Let B be a thin Blaschke product, and b be a finite prime Blaschkeproduct of degree n >1. Then there is a countable set S D such that provided that a ischosen with aS and |(b B)(a)|> |b1(0)| for all the degree-(n 1) subproductsb1 ofb, wehavea(b B) prime.

Proof. Supposea(b B) =U V. By Theorem 8, we may exclude a countable set ofa andassume that V is infinite and U is finite, since a(b B) is a finite product of thin Blaschkeproducts and hence semiprime by Theorem 14.

As in the proof of Theorem 19, we write

a(b B) =U V = (aV2)U2 (aV2). (2)

Without loss of generality, given the way that the result is formulated, we may suppose b B=B(b1 B) and thatV2has infinitely many zeros in common with B. Taking a weak- limit alongsequences of common zeros ofV2 and B , and using Lemmas 5 and 6, we obtain

3b(1z) =v2(z)U2(2v2(z)), (3)

where1, 2, 3 Tand v2 is a finite Blaschke product withv2(0) = 0. Sinceb is prime, eitherU2 is constant or v2 is Mobius. Thus, we need only consider the latter case, and we havev2= 4z for some 4 T, since v2(0) = 0.

Note that by (2), we have |b(B(a))|= |V2(a)U2(0)| |U2(0)|. But we have |b1(0)|= |U2(0)| |b1(0)|, and the result follows.

In Section 7, we will present an example of another way to multiply Blaschke products thatare continuous on an interval into the class of prime functions.

6. Prime approximations of finite products of thin Blaschke products

Since we are interested in the density of prime and semiprime functions, it is natural to askwhat we can get as a uniform limit of thin Blaschke products and whether or not the primeBlaschke products are dense in the set of finite products of thin Blaschke products. To answerthese questions, we apply a result of Hedenmalm.

Let A denote the closed subalgebra generated by the thin Blaschke products and let E

denote the closed subalgebra of L

generated by H

and the complex conjugates of thefunctions inA.

Note that ifB is a Blaschke product invertible inE, then for allx in the maximal ideal spaceofE, we have 1 =x(1) =x(B B) =x(B) x(B) =|x(B)|2. Therefore, |x(B)|= 1.

Theorem 21 ([11, Theorem 2.6]). EveryE-invertible inner function is a finite product ofthin interpolating Blaschke products.

Of course, every inner function that is a finite product of thin interpolating Blaschke productsis invertible in Eby definition. So, ifuis a uniform limit of a finite product of thin interpolatingBlaschke products, then each element of the sequence is an inner function invertible in E andtherefore u will be too. Therefore, u will be a finite product of thin interpolating Blaschke


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products. From this, it is easy to see that ifu is a finite product of thin interpolating Blaschkeproducts and a is a Mobius transformation, then a u is again a finite product of thininterpolating Blaschke products.

Lemma 22. IfB is a finite product of thin Blaschke products andC is an inner functionwithC B< 1, then Cis also a finite product of thin Blaschke products.

Proof. This result is a corollary of Theorem 21: letx M(E) and suppose x(C) = 0. Then|x(B)|< 1. By the comments preceding the proof of this lemma, x /M(E). So C is invertibleinA and therefore Cis a finite product of thin interpolating Blaschke products.

Lemma 22 may also be deduced using Rouches theorem on a pseudohyperbolic disk aroundthe zeros.

For Ca finite product of thin Blaschke products, the number of zeros ofC Ly in the disk

is bounded above by a constant Nindependent ofy, by Lemma 6. Therefore, we may choosea pointx M(H) such thatC Lx has the maximum number of zeros.

The next result suggests that finite products of thin Blaschke products can be semiprime inonly the obvious ways. To make this result work, we need to assume that our Blaschke productlooks prime on a Gleason part. This will always be approximately true, as we will see inTheorem 25, the main theorem in this section. The next result will be needed in the proof ofTheorem 25, but this representation is of sufficient interest to isolate this as a theorem.

Theorem 23. If C is an interpolating Blaschke product that is a finite product of thinBlaschke products satisfyingC Lx is prime for somex M(H) for which C Lx has themaximum number of zeros andC(x) = 0, then eitherC is prime, C= U V with V finite, orC=U V with U finite and V thin. Moreover, in the latter case, this factorization of C isunique up to composition with Mobius transformations.

Proof. Without loss of generality, we may supposeC(0) = 0. Suppose thatC is not primeand not of the form U V with V finite. Then C=U V with Ufinite and of degree greaterthan 1, since C is semiprime. Now we have x M(H) with C(x) = 0 and C Lx has themaximum number of zeros. We may suppose V(x) = 0. Note that then 0 = C(x) =U(V(x))implies U(0) = 0, so V is a subproduct of C. Since C(x) = 0, we know C Lx =

nj=1(cj

Lx) =b where b(z) =m

j=1 aj (z), cj Lx(z) =aj (z) for j = 1, . . . , m and the product ofthe remaining factors is for some T. Nowx is a nontrivial point and Cvanishes at 0 and

x so we may suppose c1(0) =c1(x) = 0. Note that since U is finite we must have |V(x)|< 1and (C Lx) =U(V Lx). Therefore, since we assume that C Lx is prime and the degree ofU is greater than 1, V Lx is degree 1. Thus, Vhas one zero in the part and we find that Uhas m >1 zeros and, by Lemma 5, U(z) =C(Lx(z)) for some T.

Consider any other y M(H) for which C(y) = 0. Then, since U is finite, we must have|V(y)|< 1 and C Ly = U (V Ly). If V Ly has more than one zero, then C Ly wouldhave more than m zeros. So if |V(y)|< 1, then we see that V has exactly one zero in thepart. Since V is a subproduct of C, it is a finite product of thin Blaschke products. ByGorkin, Lingenberg and Mortini [9, Lemma 3.1, Theorem 3.2], we see that V must be thin.Therefore,C= U V withUfinite andVthin, as desired. Now we claim that this factorizationis unique.

Suppose that we have a second factorization ofC=X Y. Then Xmust be finite, so wemust have |Y(x)|< 1. We may suppose C=X Y with C(x) =Y(x) = 0 and X(0) = 0. As


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above, we have that Y is thin and a subproduct ofC. Since Y(x) = 0, we have C Lx(z) =X(z) =U(z) for some z. So we may write X(z) =U(z) for some T. Thus,

U(Y(z)) =U(V(z)),

where Y and V are thin and Y(x) =V(x) = 0. If V(w) = 0 for some w D and Y(w)= 0,then U(Y(w)) = 0 implies Y(w) =aj where aj is a zero ofU.

Now we assume thatC is interpolating andY andVare both subproducts ofCvanishing atx. By Hoffman [14, Theorem 5.3], x is in the closure of the zero sequences (yn) and (vn) ofYandV, respectively, and there exist subnets (ynk) and (vnk) such that({ynk}, {vnk}) 0. ButC is interpolating and (C) := inf{n:C(zn)=0}

j=n (zj , zn)> 0 implies that if yn=vm and

C(yn) =C(vm) = 0, then (yn, vm) (C). Therefore,Y(zj ) =V(zj ) = 0 for infinitely manyj. Writing Y1= Y, we have

U(Y1(z)) =U(V(z)), Y1(zj ) = 0, V(zj ) = 0.

Since C is interpolating, 0 is not a critical point ofU. So U is invertible on a neighborhood

around 0 and we conclude Y1(z) =V(z) on a neighborhood about zj. Therefore, Y1= Veverywhere.So for this Cthere is (essentially) only one factorization possible, namely C=U V where

C Lx(z) =U(z) for some T, establishing the result.

We now prove the main result of this section: prime Blaschke products are dense in the setof finite products of thin Blaschke products. In this proof, we will reduce the problem to aquestion about Blaschke products that are a product of two thin Blaschke products. To handlethe final case, we will need the following lemma.

Lemma 24. IfC= U V withU a degree-2Blaschke product having zeros at 0 anda D

andV thin, then for anyy M(H) \ D if C(y) = 0, then C Ly can have zeros only at 0andb where|b|= |a|.

Proof. Note that fora(z) = (z a)/(1 az),we have1a (z) = (z+ a)/(1 + az) and thatboth V and a Vare thin. We assume C=V(a(V)) and C(y) = 0.

If V(y) = 0, then V Ly(z) =z for some T and C Ly(z) =za(z). Therefore,C Ly has zeros at 0 and a.

If (a V)(y) = 0, then a (V Ly)(z) = (a V) Ly(z) =1z and

V Ly(z) =1a (1z) =

1z+ a

1 + a1z,

and the result follows.

Theorem 25. If B is a finite product of thin Blaschke products, then B can beapproximated by prime Blaschke products.

Proof. By composing with a Mobius transformation, we may suppose B(0) = 0. We dropone zero ofB to obtain the Blaschke product B1, choosing the zero to drop so that B1 stillvanishes at 0. Now we show how to move the zeros slightly so that we have an approximatingBlaschke product C that cannot be written as U V with V finite and such that C Lx isprime for some x M(H) for which C Lx has the maximum number of zeros.

We know that for >0,there exists an interpolating Blaschke product CwithC(0) = 0 andB1 C< < 1 (see[16]). By Lemma 22,Cis also a finite product of thin Blaschke products.


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WriteC=c1c2 cnwithcj thin. Choosex M(H) withC(x) = 0 and such thatC Lxhas

the maximum number of zeros. Since C(x) = 0, we know C Lx =n

j=1(cj Lx) =b whereb(z) =

mj=1 aj (z), cj Lx(z) =aj (z) for j = 1, . . . , m and the product of the remaining

factors is for some T. NowCvanishes atx so we may suppose c1(0) =c1(x) = 0. Thus,

for j= 2, . . . , n we may choose wj close to zero according to Theorem 8 and Corollary 9 sothatwj cj Lx = j wj forj T,wj cj cj is as small as we wish, and

nj=1 wj is

prime. In the case C Lx has three or more zeros, which we denote by {0, z1, . . . , zm1}, wemay (by adjusting the pseudohyperbolic distance between zeros of C Lx) assume that thezeros ofC Lx have the property that if, for T, the mapping w permutes the zeros ofC Lx, then w= 0. Therefore, the zeros of C Lx cannot be mapped to each other by anyMobius transformation other than the trivial one.

Using Theorem 8, we move one zero ofCslightly to assume that Cis not of the form U Vwith V finite. We note that this move can, at most, rotate C Lx, so it does not change thefact that C Lx is prime; this new C will still approximate B1, w will still not permutethe zeros of C Lx for any w= 0, T and we may still assume that C is interpolating

(see [16

] or [13

, Lemma 3.15] for an explicit statement and proof of how to approximate afinite product of interpolating Blaschke products by an interpolating Blaschke product). IfCis prime; then we are done. If not, then given our assumptions and Theorem 23 we knowC=U V with Ufinite and Vthin. Multiplying by a constant of modulus 1, we may assumeU(z) =z

nj=1((z aj )/(1 ajz)) and U(0) =V(0) = 0.

Now we are ready to choose D: we will choose close to the dropped zero ofB so that C B is as small as we want so that the following conditions hold.

(a) The functionCcannot be factored as X Y with Y finite.(b) Recalling that we have chosenx M(H) withC(x) =V(x) = 0 and such thatC Lx

has the maximum number of zeros, and the finitely many zeros of C Lx are denoted by0, z1, . . . , zm1, we choose D so that the subproducts Vj ofCwith at most finitely many

zeros different from those of V have the property |Vj ()| =|zk| for any j, k. This choice ispossible because there are only countably many such Vj and the Baire category theorem thenimplies that the set of such is dense.

(c) Additionally, chooseso that we also have |(C/Z)()| =|(C/Z)(vj )| for any zero, vj,of V, and Blaschke products Z such that V and Z are subproducts of C that have at mostfinitely many different zeros. There are, up to rotation, only countably many such Zsince thereare countably many ways to specify the differing zeros and the possible rotation constants infront are handled by the absolute values.

In what follows, fory M(H) \ D we let (y) =y T and we note that ifm M(H)

and y(z) =m(z), where z denotes the identity function, then m =y . Note that we maymake B C as small as we wish and so, ifCis prime, then we are again done.

If not, then we may apply Proposition 7 and assume C= X Y with X(0) =Y(0) = 0,and sinceCis not prime andCcannot be factored asX Y withYfinite, by Theorem 23we know thatY is thin. Since Cis a finite product of thin Blaschke products, C is semiprimeand Xmust be finite. So, since C(x) = 0 we must have |Y(x)|< 1. Composing with a Mobiustransformation, we may assume Y(x) = 0 (we may no longer assume Y(0) = 0, but our newX andY satisfyX(0) =X(Y(x)) =C(x) = 0 soY is still a subfactor ofC). So ( Lx)(CLx)(z) =X(z) for some T. Therefore,xU(z) =X(z) and so there exists Tsuchthat

xU(z) =X( z). (4)

Suppose that there are infinitely many zeros of V, call them {vj }, on which Y does notvanish. Note that V and Yare, up to one possible factor, subproducts of the interpolatingBlaschke productC. Then there is a pointy M(H) in the closure of{vj } at whichV(y) = 0.


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IfY(y) = 0, then we would have y in the closure of the zeros {yj } ofY . Therefore,y is in theclosure of{vj } (a subset on which Y does not vanish) and {yj} (a subset of the zeros ofY),which are two disjoint subsets of the zeros of the interpolating Blaschke product C. SinceC is interpolating, Hoffmans theorem [15, Theorem 6.1] tells us that disjoint subsets of an

interpolating sequence have disjoint closures, and we have obtained a contradiction. Thus,Y(y)= 0. So, using equation (4) we obtain

yU(Vz) = (( Ly)(U(V Ly))(z)

= ( C) Ly(z) =X(Y Ly)(z) =xU( Yw(z)), (5)

for some w D \ {0} and V, Y, y T. Replacing z byV z, we have for some 1 T and

w1= 0,

U(z) =yxU((1w1)(z)) =yxU(1w1(z)).

SoU(a) = 0 if and only ifU(1w1(a)) = 0 and 1w1 would permute the zeros ofU. By ourassumptions, this is impossible ifUhas degree 3 or higher. Therefore, ifUhas degree at least

3, then at most finitely many zeros ofYcan be different from the zeros ofV. We now showthat our choice of implies that this is impossible before handling the case in which U hasdegree 2.

We claim Y()= 0. If not, then write Y =Y1 and note that we have shown that thereare at most finitely many zeros of Y1 that are zeros of C but not zeros of V, and the sameargument shows that there can be at most finitely many zeros ofVthat are zeros ofCbut notzeros ofY1. Then

C=X Y =Y(X1 Y) = Y1(X1 Y).

Thus, C/Y1= X1 Y and (C/ Y1)() =X1(Y()) =X1(0) =X1(Y(vj )) = (C/ Y1)(vj ) forsome zero, vj , ofV, but we chose so that this is not possible.

Thus, Y()= 0 and our argument above shows that Y is a subfactor ofCthat differs from

Vby at most finitely many factors. Therefore,

0 =|( C)()|= |X(Y())|= |U( Y())|,

from equation (4). So, we have |Y()|= |zj| for some zero zj of U, again contradicting ourchoice ofin (b) above. Therefore, if the degree ofUis at least 3, then we conclude that Cis prime and approximates B .

So we may suppose that Uhas degree 2 or less. Then we know that Ucan be written

U(z) =z

z a

1 az

,

with a = 0.Suppose thatChas (at least) two points of discontinuity. Then there exist y1 and y2 in two

different fibers such that V(y1) =V(y2) = 0 and max{|Y(y1)|, |Y(y2)|}< 1. Composing withLy1 andLy2 , we have (V Lyj )(z) =j zand (Y Lyj )(z) =j wj (z) for somej , j T andwj D.

Thus, since

( (U V))(z) = ( C)(z) = (X Y)(z) =x(U( Y(z))),

we use equation(5) and write j = j to obtain

yj jz j z a

1 aj z =x

jwj (z)

j wj (z) a

1 aj wj (z)

.

Note that since Cis interpolating, a= 0. Now there are two possibilities: either wj

= 0 orwj =aj and each leads to the conclusion that y =x , which cannot happen on both


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the fiber overy1 andy2. ThereforeC, and hence B , can have at most one discontinuity, whichmust be in the fiber over x.

We will modify C slightly, obtaining for each > 0 a Blaschke product C that has theproperty that our new C approximatesB , C cannot be written as X0 Y0 withY0 finite and

C(0) = 0. We begin with C=X Y.We factorY =V1V2withV1and V2both infinite,V1(0) = 0 andV2(x) = 0,x M(H) \ D.

Thus, since Y is thin, V1 Lx(z) =1 for all z D, V2 Lx(z) =2z and Y Lx(z) =12zwhere |j|= 1. For >0 (to be chosen later), consider

Y= V1(V2).

We may need to change one zero of V1 slightly, but this will only rotate V1 on M(H) \ D.

So ourC will look like

C= W( V2)

Y a

1 aY

, (6)

where W has one zero that differs from a zero ofV1 and Y =V1V2. We will choose later so

that C is also prime.Now, we know V1 Lx = 1 and 1 is independent of . We may change the nonzero zero

of X (that is, a) very slightly, if necessary, to assume (1a)= 0. As mentioned above, ifnecessary, we move one zero of V1 to obtain W, so that we may still assume that our newBlaschke product,

C = W( V2)(a Y),

cannot be factored as X1 Y1 with Y1 finite. Note that W Lx = , say, and all otherconstants remain the same. Thus, if C= X Y with X(0) =Y(0) = 0, then Y is an(infinite) thin Blaschke product, X has degree 2 and so X(z) =z(z). We may do thisso that C is still interpolating (see [13, Lemma 3.15] or [16]).

We show ||= |a|: by Lemma 24, ify M(H

) \D

and C(y) = 0, then C Ly(w) = 0implies that either w= 0 or |w|= ||. On the other hand, if we choose y so that C(y) =V1(y) = 0, then

V1 Ly(z) =1z, V2 Ly(z) =2 and Y Ly(z) =12z.

Thus, W Ly(z) =z and we have

C Ly(z) =2z

12z a

1 12az

.

Thus, |a|= |12a|= ||, as claimed.Now recall that we chosexso thatV1 Lx = 1and V2 Lx(z) =2z. Therefore, there exists

y in the Gleason part ofx with

V2(y) = 0.

So,

V1 Ly =1,

sinceV1is constant on the Gleason part ofx, and we also know that ( V2) Ly(z) =2,z,with|2,|= 1; also, W Lx = , and

Y Ly(z) = ((V1 Ly)(V2 Ly)(z)) =11 (2,z) =1

2,z+

1 + 2,z.

Therefore,

C Ly(z) =2,z 1((2,z+ )/(1 + 2,z)) a

1 a1((2,z+ )/(1 + 2,z))

.


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But we know C Ly(w) = 0 implies w = 0 or |w|= ||= |a|. So ifw = 0, then

12,w+

1 + 2,w=a.

Since|w|= |a|, we obtain |2,w 1a2,w|= |1a |, whence|a||1 1a|= |1a |.

Squaring both sides and solving for , we find = 2(1a)/(1 + |a|2). By our choice of a,

we know (1a)= 0, so we may choose arbitrarily small if(1a)< 0 and we may choose < (1a)/(1 + |a|

2) otherwise and we see that C cannot be factored as X Y.Thus, our choice of yields C prime and we may approximate B usingC.

7. Examples

In the table provided in the introduction, there is no example of a surjective inner function

that is not semiprime. We present such an example below.

Example 26. There exists a surjective Blaschke product that is not semiprime.

Proof. LetB andCbe thin Blaschke products. Since w B is thin, we know thatw Bis always a Blaschke product for everyw D and so we can always find a solution toB(z) =w.Thus,B is surjective. Consequently,B Cis surjective. IfB Cis not a Blaschke product, thenwe can compose with a Mobius transformation via Frostmans theorem to obtain a surjectiveBlaschke product. Since B Cis not semiprime, we have the desired example.

Recall that a Blaschke product is called a uniform Frostman Blaschke product if its zerosequence satisfies

supT

n=1

1 |an|

| an|


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conjugate pairs.) Sincev has a single singularity, there is a countable sequence of angles withvertex on T at which v has radial limit 1. We will choose u with finite angular derivativeeverywhere, continuous at all points ofT \ {1} and satisfying u(0) = 0. Since we assume thatv is real on the real axis and since v satisfies condition (FA), we know that v(r) must tend to

1 or1, asr +1 orr 1. We claim thatu v has finite angular derivative at every pointof the circle.

First note that when rei ei for = 0, we know that v(rei) tends to v(ei) and v isanalytic aboutei. Therefore,v is conformal (angle preserving) atei. Since we assume that uhas finite angular derivative everywhere, we see that u v has finite angular derivative at ei

for = 0.Now suppose rn 1. Then, from our choice ofv , we know that v(rn) tends to 1. Since u

has finite radial limit everywhere, we know (u v)(rn) tends to a point T. We also have

(1 |v(rn)|)/(1 |rn|)v < .

Since we assume that u has finite angular derivative everywhere, we also know

limn

1 |u(v(rn))|1 |v(rn)|

exists and is finite. Thus,

1 |(u v)(rn)|

1 |rn|


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|b(1)|= |d(1)|, by the JuliaCaratheodory theorem (see[3, Theorem 2.44]). Since 0 < |d(1)|

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