The Weierstrass Factorization Theorem for Slice Regular Functions Over the Quaternions

Ann Glob Anal Geom DOI 10.1007/s10455-011-9266-0

ORIGINAL PAPER

The Weierstrass factorization theorem for slice regular functions over the quaternions

Graziano Gentili · Irene Vignozzi

Received: 8 November 2010 / Accepted: 5 April 2011 © Springer Science+Business Media B.V. 2011

Abstract The class of slice regular functions of a quaternionic variable has been recently introduced and is intensively studied, as a quaternionic analogue of the class of holomorphic functions. Unlike other classes of quaternionic functions, this one contains natural quaternionic polynomials and power series. Its study has already produced a rather rich theory having steady foundations and interesting applications. The main purpose of this article is to prove a Weierstrass factorization theorem for slice regular functions. This result holds in a formula- tion that reﬂects the peculiarities of the quaternionic setting and the structure of the zero set of such functions. Some preliminary material that we need to prove has its own independent interest, like the study of a quaternionic logarithm and the convergence of inﬁnite products of quaternionic functions.

Keywords Functions of a quaternionic variable · Weierstrass factorization theorem · Zeros of hyperholomorphic functions

Mathematics Subject Classiﬁcation (2000) 30G35 · 30C15 · 30B10

1 Introduction

The well-known theory of Fueter regular functions [9,10,25] is the ﬁrst and most celebrated candidate among a few others, in the search for a quaternionic analogue of the theory of holomorphic functions of one complex variable. This well-developed theory [3,19Ð21] inspired also the setting of Clifford algebras (see, e.g. [2]). Recently, Gentili and Struppa proposed a different approach, which led to a new notion of holomorphicity (called slice regularity) for quaternion-valued functions of a quaternionic variable [15,16]. Unlike Fueter’s, this

G. Gentili (B) · I. Vignozzi Dipartimento di Matematica “U. Dini”, Università di Firenze, Viale Morgagni 67/A, 50134 Firenze, Italy e-mail: [email protected]fi.it I. Vignozzi e-mail: [email protected]fi.it 123 Ann Glob Anal Geom theory includes the polynomials and the power series of the quaternionic variable q of the n ∈ H type n≥0 q an, with coefficients an . Furthermore, the analogs (sometime peculiarly different) of many of the fundamental properties of holomorphic functions of one complex variable can be proven in this new setting, like the Cauchy and Pompeiu representation formu- las and Cauchy inequalities, the maximum (and minimum) modulus principle, the identity principle, the open mapping theorem, the Morera theorem, the power and Laurent series expansion, the Runge approximation theorem, to cite only some of the most significant (see [4Ð7,12Ð16,24]). In fact, the theory of slice regular functions is already rather rich and well established on steady foundations, and appears to be of fundamental importance to construct a functional calculus in non commutative settings [8]. Let H denote the skew field of quaternions. Its elements are of the form q = x0 + 2 2 2 ix1 + jx2 + kx3 where the xl are real, and i, j, k are such that i = j = k =−1, ij =−ji = k, jk =−kj = i, ki =−ik = j. We set Re(q) = x0,Im(q) = ix1 + jx2 +kx3, | |= 2 + 2 + 2 + 2 ( ) ( ) | | q x0 x1 x2 x3 and call Re q ,Imq and q the real part,theimaginary part and the module of q, respectively. The conjugate√of the√ quaternion q,definedasq¯ = Re(q)−Im(q) = x0 −ix1 − jx2 −kx3, satisfies |q|= qq¯ = qq¯ and allows the definition ¯ inverse q = q−1 = q . S of the of any element 0as |q|2 If is the unit sphere of purely imaginary S ={ ∈ H : 2 =− }={ = + + : 2 + 2 + 2 = } quaternions, i.e., q q 1 q ix1 jx2 kx3 x1 x2 x3 1 , then every quaternion q which is not real (i.e. with Im(q) = 0) can be written as q = x + Iy = ( ), =| ( )| = Im(q) ∈ S ∈ S for x Re q y Im q and I |Im(q)| . Moreover, using the same I , we can write q =|q|eϑ I ,forsomeϑ ∈ R. We can now state the following definition, see [4]. Definition 1.1 Let be a domain in H. A function f : → H is said to be slice regular if, for every I ∈ S, its restriction fI to the complex line L I = R + RI passing through the origin and containing 1 and I has continuous partial derivatives and satisfies 1 ∂ ∂ ∂ I f (x + yI) := + I fI (x + yI) = 0, 2 ∂x ∂y in I = ∩ L I . In what follows, we will simply call regular a slice regular function. If is a generic domain of H, then there are examples of regular functions defined on which are not even continuous, see e.g. [4]. In [5], and in this same article [4], a class of pathology-preventing domains of definition for regular functions were introduced, as follows:

Deﬁnition 1.2 Let be a domain in H, intersecting the real axis. If I = ∩ L I is a domain in L I C for all I ∈ S then we say that is a slice domain. As it is well known, the domains of holomorphy are the most natural domains of definition for holomorphic functions of a complex variable. The same can be said, in the setting of quaternionic regular functions, for slice domains which have the following additional property:

Definition 1.3 A subset C of H is axially symmetric if, for all x + yI ∈ C with x, y ∈ R, I ∈ S, the whole set x + yS ={x + yJ : J ∈ S} is contained in C. For the sake of simplicity, we will call such a C a symmetric set. As it is proved in [4], any regular function defined on a slice domain can be (uniquely) extended to the smallest symmetric slice domain which contains , and for this reason we will always consider regular functions defined on symmetric slice domains. 123 Ann Glob Anal Geom

One of the interesting, basic features of regular functions on slice domains is the structure of their zero set: it consists of isolated points (zeros) and isolated spheres of type x + yS, with x, y ∈ R (spherical zeros). The fact that pointwise product does not preserve regularity led to the definition of a regular product (the ∗-product) which maintains regularity and allows the study of a factorization for quaternionic regular polynomials, as well as the construction of different notions of multiplicity for their roots (and for the zeros of regular functions), [11,17,18,23]. This article is devoted to the study of a quaternionic version of the well-known Weierst- rass factorization theorem, which will be valid for entire regular functions, i.e. for regular functions deﬁned on the whole of H. The main result that we obtain is in fact the following.

Theorem 1.4 (Weierstrass factorization theorem for regular functions) Let f be an entire regular function. Let: m ∈ N be the multiplicity of 0 as zero of f , {bn}n∈N ⊆ R \{0} be the sequence of the (non zero) real zeros of f , {Sn = xn + ynS}n∈N be the sequence of the spherical zeros of f ,and{an}n∈N ⊆ H \ R be the sequence of the non real zeros of f with isolated multiplicity greater then zero. If all the zeros listed above are repeated according to their multiplicities, then there exists a never vanishing, entire regular function h and, for all ∈ N ∈ δ ∈ = ( ) +| ( )|S , , ∈ N n ,thereexistcn Sn, n San Re an Im an , rn n mn such that f (q) = qm R(q) S(q) A(q) ∗ h(q) where, with obvious notations for the inﬁnite ∗-product, ∞ − − qb−1+ 1 q2b−2+···+ 1 qrn b rn R( ) = ( − 1) n 2 n rn n , q 1 qbn e n=0 ∞ ( 2) ( n ) 2 ( ) 2Re(cn ) 1 2 2Re cn 1 n 2Re cn q 2qRe cn q + q +···+ q S(q) = − + 1 e |cn |2 2 |cn |4 n |cn |2 n , |c |2 |c |2 n=0 n n ∞ A( ) = ∗ ( − δ−1) ∗ ( ) q 1 q n gn q n=0 and where, for all n ∈ N, the function gn is the never vanishing entire regular function whose restriction to the plane Ln = R + R[Im(δn)] is given by − zδ−1+ 1 z2δ−2···+ 1 zmn δ mn | ( ) = n 2 n mn n . gn Ln z e

In particular, we can choose rn = n = mn = n for all n ∈ N.

The above result is achieved in several steps. Some of them are inspired by, and similar to, the steps that lead to the complex version of the Weierstrass factorization theorem, while other are new and quite diverse, mainly due to the peculiarities of the quaternionic setting and the fact that the structure of the zero-set of the entire regular functions includes spherical zeros, which are not present in the case of the entire holomorphic functions. This article is organized as follows. After having presented some necessary preliminary results, in Sect.3, we study criteria to establish the convergence of infinite products of quaternions, and this latter necessity leads us to the (peculiar) definition of a quaternionic Logarithm. Then, in Sect.4, we investigate the uniform convergence of infinite products of regular functions. The results so found give us a tool to find, in Sect.5, conditions that can guarantee the uniform convergence of infinite products of regular functions to a regular function. In Sect. 6 we perform the study of different type of convergence-producing regular factors, some of 123 Ann Glob Anal Geom which arise only in the quaternionic setting. Section 7 is dedicated to give the definitions of spherical and isolated multiplicities for the zeros of a regular function; to do this, we follow the path which brought to the same definitions in the case of quaternionic regular polynomials. Eventually, in Sect.8, we prove the Weierstrass factorization theorem for entire regular functions. In all this procedure, the technical difficulties encountered are quite relevant, but we believe that the effort made to overcome them brought to a neat final result.

2 Preliminaries

This section is devoted to present some basic results that point out the main features of slice regular functions and of their domains of convergence. We begin with some technical properties of the quaternions, and recall that (see e.g. [16]) for all I ∈ S and for all J ∈ S such that I ⊥ J,wehavethatIJ ∈ S and 1, I, J, K = IJform a basis for H with the same algebraic properties of the standard basis 1, i, j, k = ij. Moreover, the following result holds Proposition 2.1 Let I and J be two elements in S,let I, J∈R be the Euclidean scalar product of I and J in R4, and let I × J ∈ S be their natural vector product in R3. Then the quaternionic product I J can be decomposed through the following formula: IJ =− I, J+I × J. One of the interesting features of the regular functions is a splitting property, which turns out to be a key tool in the theory: Lemma 2.2 (Splitting Lemma) If f is a regular function on a slice domain then, for every I ∈ S and every J ⊥ IinS, there exist two holomorphic functions F, G : I → L I such that

fI (z) = F(z) + G(z)J for all z ∈ I . The Splitting Lemma naturally leads to a version of the identity principle for regular functions (see [4]). Theorem 2.3 (Identity Principle) Let f : → H be a regular function on a slice domain and let Z f ={q ∈ : f (q) = 0} be the zero-set of f . If there exists I ∈ S such that L I ∩ Z f has an accumulation point, then f ≡ 0 on . We will now recall some results (proved in [4,11,16]) on the algebraic and topological structure of the zero-set of a regular function. Theorem 2.4 Let f be a regular function on a symmetric slice domain .Ifthereexist x, y ∈ R and distinct imaginary units I, J ∈ S such that f (x + yI) = f (x + yJ) = 0,then f (x + yL) = 0 for all L ∈ S. If f is not identically zero, its zero set Z f consists of isolated points or isolated 2-spheres of the form S = x + yS, with x, y ∈ R,y= 0. It becomes now natural to give the following (see e.g. [11]) Deﬁnition 2.5 Let f be a regular function on a symmetric slice domain . An isolated 2-sphere S = x + yS of zeros of f is called a spherical zero of f . Any point q0 belonging to a spherical zero is called a generator of the spherical zero. Any zero of f not belonging to a spherical zero is said isolated zero or non spherical zero. 123 Ann Glob Anal Geom

For a regular function deﬁned on a symmetric slice domain, both the set of spherical zeros and the set of isolated zeros have cardinality at most countable. Moreover

Corollary 2.6 Azeroq0 ∈/ R of a regular function f deﬁned on a symmetric slice domain is a generator of a spherical zero if and only if its conjugate q0 is a zero of f as well. For the sake of completeness, we state here a result proven in [13], that we will use in the sequel.

Proposition 2.7 Let f be a regular function on a symmetric slice domain .Iff(L I ) ⊆ L I for some I ∈ S then the zero-set Z f consists of isolated points belonging to L I or isolated 2-spheres of type x + yS. As a consequence, if f (q) = 0 for every q ∈ L I then f (q) = 0 for every q ∈ H. Finally, if f (L I ) ⊆ L I for all I ∈ S then the zero-set Z f consists of real isolated points or isolated 2-spheres of type x + yS.

Since pointwise product of regular functions is not in general regular, inspired by the classic case of polynomials with coefficients in a non commutative algebra (see e.g. [22]), the ∗-multiplication (or regular multiplication) between regular functions f and g defined on the open unit ball B of H was introduced in [11] by means of the power series expansion, to guarantee the regularity of f ∗ g. The definition of ∗-multiplication was extended in [4] to the case of regular functions on a slice domain and it is based on the following result. Lemma 2.8 (Extension Lemma) Let be a symmetric slice domain and choose I ∈ S.If fI : I → H is holomorphic, then setting 1 I f (x + yJ) = [ fI (x + yI) + fI (x − yI)] + J [ fI (x − yI) − fI (x + yI)] 2 2 extends fI to a regular function f : → H. The function f is the unique such extension and it is denoted by ext( fI ). In order to define the regular product of two regular functions f, g on a symmetric slice domain ,letI, J ∈ S, with I ⊥ J, and choose holomorphic functions F, G, H, K : I → L I such that for all z ∈ I

fI (z) = F(z) + G(z)J, gI (z) = H(z) + K (z)J. (1)

Let fI ∗ gI : I → H be the holomorphic function deﬁned by

fI ∗ gI (z) =[F(z)H(z) − G(z)K (z¯)]+[F(z)K (z) + G(z)H(z¯)]J. (2) Using the Extension Lemma 2.8, the following definition is given in [4]:

Deﬁnition 2.9 Let ⊆ H be a symmetric slice domain and let f, g : → H be regular. The function

f ∗ g(q) = ext( fI ∗ gI )(q) deﬁned as the extension of (2) is called the regular product of f and g.

We now recall a few properties of the regular multiplication, see [11]and[4].

Remark 2.10 Let ⊆ H be a symmetric slice domain and let f, g : → H be regular. If f (L I ) ⊆ L I for all I ∈ S,then f ∗ g(q) = f (q) g(q). If f (L I ) ⊆ L I or g(L I ) ⊆ L I for all I ∈ S,then f ∗ g(q) = g ∗ f (q). 123 Ann Glob Anal Geom

An alternative expression of the regular product of two regular functions was introduced in [11]. In particular we have the next proposition proven in [4].

Theorem 2.11 Let f , g be regular functions on a symmetric slice domain . For all q ∈ , if f (q) = 0 then f ∗ g(q) = 0,else

f ∗ g(q) = f (q)g( f (q)−1qf(q))

Corollary 2.12 Let f , g be regular functions on a symmetric slice domain .Then f ∗ g(p) = 0 if and only if f (p) = 0 or f (p) = 0 and g( f (p)−1 pf(p)) = 0.

Deﬁnition 2.13 Let f be a regular function on a symmetric slice domain and suppose f splits on I as in formula (1), fI (z) = F(z)+G(z)J. We consider the holomorphic function c( ) = (¯) − ( ) fI z F z G z J (3) and deﬁne, according to the Extension Lemma 2.8,theregular conjugate of f by the formula c( ) = ( c)( ) = ( (¯) − ( ) ). f q ext fI q ext F z G z J

Furthermore, the following definition is given under the same assumptions.

Deﬁnition 2.14 The symmetrization of f is deﬁned as s = ∗ c = c ∗ = ( c ∗ )( )) = ( ( ) (¯) + ( ) (¯)). f f f f f ext fI fI q ext F z F z G z G z (4)

s s Remark 2.15 It turns out that f (L I ) ⊆ L I for all I ∈ S, and hence the zero set of f has the property described in Proposition 2.7. Moreover, if f, g are regular functions on a symmetric slice domain, then is easy to verify that ( f ∗ g)c = gc ∗ f c and ( f ∗ g)s = f s gs = gs f s.

The zero-sets of f c and f s are characterized in [11,13] as follows.

Theorem 2.16 Let f be a regular function on a symmetric slice domain . For all x, y ∈ R with x + yS ⊆ , the zeros of the regular conjugate f c on x + yS are in one-to-one correspondence with those of f . Moreover, the symmetrization f s vanishes exactly on 2-spheres (or singleton) x + yS on which f has a zero. Note that x + yS is a 2-sphere if y = 0 and a real singleton {x} if y = 0.

3 Quaternionic logarithm and inﬁnite products of quaternions

We consider an infinite product of quaternions ∞ q0q1 ...qi ...= qi i=0 and, for n ∈ N, we denote by Qn = q0q1 ...qn the partial products. In analogy with the complex case (see [1]), we give the following definition. ∞ Definition 3.1 The infinite product i=0 qi is said to converge if and only if at most a finite number of the factors are zero, and if the partial products formed by the non vanishing factors tend to a finite limit which is different from zero. 123 Ann Glob Anal Geom

In what follows, we will always refer to an infinite product assuming that the vanishing factors are a finite number, and we will check its convergence just looking at the product of the non-vanishing terms. We point out that in a convergent product we have that limi→∞ qi = 1. = −1 In fact, this is clear by writing qi Qi−1 Qi . It is therefore preferable to write all infinite products in the form ∞ (1 + ai ) i=0 so that limi→∞ ai = 0 is a necessary condition for their convergence.

Remark 3.2In Definition 3.1, the requirement that the partial products of non vanishing ∞ ( + ) factors of i=0 1 ai tend to a finite limit different from zero finds its motivation in the ∞ ( + ) complex case. In fact, assuming this requirement, an infinite product i=0 1 ci with ∈ C ∞ ( + ) ci converges simultaneously with the series i=0 Log 1 ci , whose terms represent the values of the principal branch of the logarithm (see [1]). The convergence is not necessar- ily simultaneous if the limit of the infinite product is zero, as the following example shows. Let ∞ (1 − 1/i) (5) i=0 and consider the corresponding series ∞ Log(1 − 1/i). (6) i=0 S Denoting the partial sums of (6)bySn,wehavethatQn = e n . Since limn→∞ Sn =−∞, then ∞ Sn (1 − 1/i) = lim Qn = lim e = 0. n→∞ n→∞ i=0 Therefore, the infinite product (5) converges (to zero) while the series (6)diverges(to−∞).

To follow a similar approach to study the convergence of inﬁnite products in the quaternionic case, we need to introduce a logarithm on H.Theexponential function on H is naturally deﬁned as ∞ qn exp(q) = eq = n! n=0 and it coincides with the complex exponential function on any complex plane L I .

Deﬁnition 3.3 Let ⊆ H be a connected open set. We deﬁne a branch of the quaternionic logarithm (or simply a logarithm)on a function f : → H such that for every q ∈ e f (q) = q.

First of all, since exp(q) never vanishes, we must suppose that 0 ∈/ .Ifweset Im(q)/|Im(q)| if q ∈ H \ R I = q any element of S otherwise 123 Ann Glob Anal Geom

θ we have that for every q ∈ H \{0} there exists a unique θ ∈[0,π] such that q =|q|e Iq . Moreover θ = arccos(Re(q)/|q|). The function arccos(Re(q)/|q|) will be called the principal quaternionic argument of q and it will be denoted by ArgH(q) for every q ∈ H \{0}. Hence, we are ready to define the principal quaternionic logarithm. Definition 3.4 Let ln be the natural real logarithm. For every q ∈ H \ (−∞, 0], we define the principal quaternionic logarithm (or simply principal logarithm)ofq as Re(q) Log(q) = ln |q|+arccos I . |q| q This same definition of principal logarithm was already given in [20], in the setting of Clif- ford algebras. In fact, Definition 3.4 can be obtained by specialising to the case of quaternions the paravector logarithm introduced in [20, Definition 11.24, p. 231]. It is also important to state that: Proposition 3.5 The principal logarithm is a continuous function on H \ (−∞, 0].

Proof The function q → ln |q| is clearly continuous on H \{0}. The function q → Re(q) Im(q) ∈ H \ R arccos |q| |Im(q)| is deﬁned and continuous for every q . Moreover, since for every strictly positive real r, and for every q ∈/ R,wehave Re(q) lim arccos Iq = 0 q→r |q| → Re(q) H \ (−∞, ] then q arccos |q| Iq is continuous on 0 . We observe that, if s, instead, is a strictly negative real number, then Re(q) lim arccos Iq q→s |q| does not exist. Remark 3.6 Our approach to the definiton of a quaternionic logarithm is inspired by the search for an inverse of the exponential map. While the principal logarithm obvioulsy satis- ﬁes the relation eLog(q) = q in H \ (−∞, 0], the equality Log(eq ) = q is valid only in the domain {q ∈ H :|Im(q)| <π}. As one may expect, we can prove the following nice result Proposition 3.7 The principal quaternionic logarithm coincides with the principal complex logarithm on any complex plane L I , with I ∈ S.

Proof Let I ∈ S. First of all notice that, since L I = L−I , every quaternion q ∈ L I \ R, can be written both as x + yI and as x − y(−I ), for suitable x, y ∈ R.If(x, y) ∈ R2 \ (−∞, 0], then by Definition 3.4,wehavethat x y Log(x + yI) = ln x2 + y2 + arccos I x2 + y2 |y| 123 Ann Glob Anal Geom and x (−y) Log(x − y(−I )) = ln x2 + y2 + arccos (−I ) x2 + y2 |y| coincide. Now, we can consider the restriction of the principal quaternionic logarithm to L I : 2 2 y Log(x + yI) = ln( x + y ) + ArgH(x + yI) I. |y| φ( + ) = ( + ) y ( + ) [ ,π) We set x yI ArgH x yI |y| . The function ArgH x yI , with values in 0 , is the not oriented angle from the positive real half line to the vector x + yI. When y > 0, then x + yI belongs to the upper half plane of L I ,andφ(x + yI) = ArgH(x + yI).On the other hand if y < 0, we obtain that x + yI belongs to the lower half plane of L I ,and φ(x + yI) =−ArgH(x + yI). Therefore, φ(x + yI) coincide with the imaginary part of the principal complex logarithm of L I , and the proof is complete.

In the complex case, it is well known that the argument of a product is equal to the sum of the arguments of the factors (up to an integer multiple of 2π). In the quaternionic case, we have the following lemma. ∈ N θ ,...,θ ∈[ ,π) n θ <π Lemma 3.8 Let n and let 1 n 0 be such that i=1 i . Then for every set {I1,...,In}⊆S, we have that n θ1 I1 θn In ArgH(e ...e ) ≤ θi . i=1 = θ ,θ ,...,θ ∈[ ,π) Proof By induction on n.Forn 1, the thesis is straightforward. Let 1 2 n 0 n θ <π φ ∈[,π) be such that i=1 i .Let 0 be the principal quaternionic argument of the θ θ θ θ φ product e 1 I1 ...e n−1 In−1 and let J ∈ S be such that e 1 I1 ...e n−1 In−1 = e J . Consider the φ θ product e J e n In :

φ J θn In e e = cos φ cos θn + cos φ sin θn In + sin φcos θn J + sin φ sin θn JIn.

From the formula (2.1), we have that JIn =− J, In+J × In, and hence we obtain that φ J θn In φ J θn In cos (ArgH(e e )) = Re e e = cos φ cos θn − sin φ sin θn J, In.

Since J, In≤|J||In|=1andsinφ sinθn ≥ 0 we obtain the inequality

φ J θn In cos (ArgH(e e )) ≥ cos φ cos θn − sin φ sin θn = cos (φ + θn). The function cos (x) decreases in [0,π], and hence we have

φ J θn In ArgH(e e ) ≤ φ + θn. Thanks to the induction hypothesis, we have the thesis.

∞ If {ai }i∈N ⊆ H is a sequence such that the series = |ai | converges, we know that the ∞ i 0 series i=0 ai converges too. With this in mind, we can prove the following result. { } ⊆ H ∞ | ( + )| Theorem 3.9 Letai i∈N be a sequence. If the series i=0 Log 1 ai converges, ∞ ( + ) then the product i=0 1 ai converges too. 123 Ann Glob Anal Geom ∞ | ( + )| { } Proof The convergence of the series i=0 Log 1 ai implies that the sequence ai i∈N tends to zero. We can therefore suppose that 1+ai ∈/ (−∞, 0].Foreveryi ∈ N,letθi ∈[0,π) θ I and let Ii ∈ S be such that 1 + ai =|1 + ai |e i i .ThenLog(1 + ai ) = ln |1 + ai |+θi Ii and therefore

|ln |1 + ai || ≤ |Log(1 + ai )| (7) and

|θi | = |θi Ii | ≤ |Log(1 + ai )| (8) for every i ∈ N. We have to prove that the sequence of the partial products n Qn = (1 + ai ) i=0 tends to a finite limit (which is different from zero). Since real number commute with quaternions, for every n ∈ N,wehavethat n n n θi Ii (1 + ai ) = |1 + ai | e . i=0 i=0 i=0 ∞ | + | By hypothesis and by inequality (7), the series i=0 ln 1 ai is convergent and hence the ∞ | + | infinite product i=0 1 ai is convergent too (see Remark 3.2). Then, it is sufficient to θ = n i Ii ⊆ ∂ ( , ) = { ∈ H :| |= } prove that the sequence Rn i=0 e B 0 1 q q 1 is convergent. We will show that {Rn}n∈N is a Cauchy sequence. For n > m ∈ N,wehave: n m m n m θ θ θ θ θ | − |= i Ii − i Ii = i Ii i Ii − i Ii Rn Rm e e e e e = = = = + = i 0 i 0 i 0 i m 1 i 0 m n n n θ θ θ θ = i Ii i Ii − = i Ii − ≤ i Ii , e e 1 e 1 ArgH e i=0 i=m+1 i=m+1 i=m+1 θ n i Ii = where the last equality holds due to the fact that i=m+1 e 1. Inequality (8) implies ∞ θ = n θ that the series i=0 i is convergent. Therefore the sequence Sn i=0 i of the partial sums is a Cauchy sequence. This yields that for all >0, there exists m0 ∈ N such that for all n > m > m0 n θi < . i=m+1 In particular for <π, by Lemma 3.8,wehavethat n n θi Ii ArgH e ≤ θi < . i=m+1 i=m+1 The assertion follows.

The following proposition will be useful in the sequel, while studying the convergence of the inﬁnite products. 123 Ann Glob Anal Geom

Proposition 3.10 Let Log be the principal quaternionic logarithm. Then − lim q 1Log(1 + q) = 1. (9) q→0

Proof Let {qn}n∈N ={xn + yn In}n∈N be a sequence such that qn → 0forn →∞.Wecan assume without loss of generality that yn > 0. We consider the following norm −1 |(xn + yn In) Log(1 + xn + yn In) − 1|. (10)

Easy calculations show that both the real part and the norm of the imaginary part of (xn + −1 yn In) Log(1 + xn + yn In) do not depend on the complex direction In. Therefore, we can ﬁx In = I0 for every n ∈ N. Hence, we consider −1 |(xn + yn I0) Log(1 + xn + yn I0) − 1|, (11) ( + )−1 ( + + ) ∈ ∈ N and we observe that xn yn I0 Log 1 xn yn I0 L I0 for all n . Since, we are in the complex plane L I0 , by the same arguments used in the complex case, the sequence (11) tends to zero. ∞ | ( + )| Corollary 3.11 The series n=0 Log 1 an converges at the same time as the simpler ∞ | | series n=0 an . ∞ | ( + )| ∞ | | Proof If either the series n=0 Log 1 an or the series n=0 an converges, we have limn→∞ an = 0. By Proposition 3.10,wehavethatforagiven >0

|an|(1 − ) ≤|Log(1 + an)|≤|an|(1 + ) for all sufﬁciently large n. It follows immediately that the two series are simultaneously convergent.

As a consequence of Theorem 3.9 and Corollary 3.11, we end this section by finding a useful sufficient condition for the convergence of quaternionic infinite products. { } ⊆ H Theorem 3.12 Let an n∈N . A sufficient condition for the convergence of the product ∞ ( + ) ∞ | | n=0 1 an is the convergence of the series n=0 an .

4 Convergence of infinite products of functions defined on H ∞ ( ) Let i=0 gi q be an infinite product whose factors are functions of a quaternionic variable ⊆ H ∞ ( ) defined on an open set . We want to explain what we mean by saying that i=0 gi q { } converges uniformly on compact sets of . If all functions gi i∈N do not vanish on , then the definition is obviously given (according with Definition 3.1) by requiring that the sequence of partial products n Gn(q) = gi (q) i=0 converges uniformly on compact sets of to a never vanishing function. The problem which arises for the presence of factors with some zeros can be avoided by demanding that, for any compact set K , at most a finite number of factors gi vanish at some point of K : therefore, it will be sufficient to study the uniform convergence of the “residual” product formed by the factors gi that do not vanish on K . All this leads to the following: 123 Ann Glob Anal Geom

{ } ⊆ H Deﬁnition 4.1 Let gi i∈N be a sequence of functions deﬁned on an open set .We will say that ∞ gi (q) i=0 converges uniformly on (compact sets of) to a function P : → H if, for any compact K ⊆ :

(i) there exists an integer NK such that gi = 0onK for all i ≥ NK , (ii) the residual product ∞ gi (q) i=NK (whose factors do not vanish on K ) converges uniformly on K to a never vanishing

function PNK , (iii) we have, for all q ∈ K , NK −1 ( ) = ( ) ( ). P q gi q PNK q i=0 ∞ ( ) ⊆ H Remark 4.2 If i=0 gi q converges uniformly on compact sets of , by the previous definition we have that limi→∞ gi = 1 uniformly on compacts set of . Indeed, for any compact K ⊆ ,takingany j > NK ,wecanset, j G j (q) = gi (q) i=NK = −1 = −1 = −1 and write g j G j−1G j and hence lim j→∞ g j limi→∞ G j−1G j lim j→∞ G j−1 lim j→∞ G j = 1.

For this reason, we write the infinite products in the form ∞ (1 + fi (q)) i=0 { } so that the uniform convergence on compact sets of the sequence of functions fi i∈N to the zero function is a necessary condition for studying their convergence. ∞ ( + ( )) { } Remark 4.3 Let i=0 1 fi q be an infinite product. If the sequence of functions fi i∈N converges uniformly on compact sets of ⊆ H to the zero function, then condition (i) of Definition 4.1 is automatically fulfilled.

{ } ∞ When fi i∈N is a sequence of C function, the following result holds: { } ∞ ⊆ Proposition 4.4 Let fi i∈N be a sequence of C functions deﬁned on an open set H, converging uniformly to the zero function on compact sets of . If the inﬁnite product ∞ ( + ( )) ∞ i=0 1 fi q converges uniformly on compact sets of to a function P, then P is C on . 123 Ann Glob Anal Geom

Proof Set q ∈ and let K ⊆ be a compact neighborhood of q. By the assumption on the sequence { fi }i∈N, we have that (see Definition 4.1) there exists an integer NK such that the residual product ∞ (1 + fi (q)) i=NK converges (uniformly) on K to a never vanishing function, that we call PNK . Hence PNK is the limit of a uniformly convergent sequence of C∞ functions and so it is C∞. Then, we can write P as a finite product of C∞ functions: NK −1 ( ) = ( + ( )) ( ). P q 1 fi q PNK q i=0 Therefore P is C∞. Moreover, on K , the zero set of P coincides with the zero set of − NK 1( + ( )) i=0 1 fi q . Let ∞ ( − −1) 1 qai (12) i=0 { } ⊆ H \{ } ∈ H with ai i∈N 0 .Letq be a fixed quaternion. By Theorem3.12, a sufficient ∞ | |/| | condition for the convergence of (12) is the convergence of the series i=0 q ai .Asa consequence, a sufficient condition for the uniform convergence on compact sets of H of ∞ /| | (12) is the convergence of the series i=0 1 ai . For our purposes, we will need to introduce (as it is done in the complex case, see, e.g. [1]) suitable convergence-producing factors. The same arguments used in the complex case lead to the following statement. We will give the proof here, both for the sake of completeness, and to remark a few differences due to the quaternionic environment.

Theorem 4.5 Let {an}n∈N ⊆ H\{0} be such that limn→∞ |an|=∞. Then there exist certain integers {mn}n∈N such that the inﬁnite product ∞ − qa−1+ 1 (qa−1)2+···+ 1 (qa−1)mn ( − 1) n 2 n mn n 1 qan e (13) n=0 converges uniformly on compact sets of H. In particular we can choose mn = n for all n ∈ N. Proof Let n ∈ N.Letp (x) = x + 1 x2 +···+ 1 xmn where m ∈ N. If we prove the n 2 mn n absolute convergence of the series with general term − ( −1) ( ) = (( − 1) pn qan ), rn q Log 1 qan e (14)

(where Log is the principal logarithm so that rn(0) = 0) then, thanks to Theorem 3.9,we ∈ H − −1 ( −1) conclude the proof. For every q , the quaternions 1 qan and pn qan lie on the same complex plane. Thus, since rn(0) = pn(0) = 0, we can write rn(q) in the form ( ) = ( − −1) + ( −1). rn q Log 1 qan pn qan

Now, for a given R > 0, we consider only the terms with |an| > R. In the ball B(0, R) we | −1| < ( − −1) have that qan 1andsoLog1 qan can be expanded in a Taylor series

− − 1 − 1 − Log(1 − qa 1) =−qa 1 − (qa 1)2 − (qa 1)3 −··· . n n 2 n 3 n 123 Ann Glob Anal Geom

Then rn(q) has the representation

1 − + 1 − + ( ) =− ( 1)mn 1 − ( 1)mn 2 −··· rn q qan qan mn + 1 mn + 2 and we obtain easily the estimate + − 1 R mn 1 R 1 |rn(q)|≤ 1 − . mn + 1 |an| |an| As in the complex case, to conclude the proof, we are left to show the convergence of the series ∞ + − 1 R mn 1 R 1 1 − . (15) m + 1 |a | |a | n=0 n n n −1 The fact that limn→∞(1 − R/|an|) = 1 implies that the series (15) converges if the series ∞ + 1 R mn 1 m + 1 |a | n=0 n n converges. The latter, if for example we take mn = n, has a majorant geometric series with ratio strictly smaller than 1 and hence it is convergent.

A completely similar proof leads to the following generalization of the result. { } Theorem 4.6 Let fn n∈N be a sequence of functions that converges uniformly on compact sets of H to the zero function. Then the inﬁnite product ∞ 1 2 1 n fn (q)+ fn (q) +···+ fn (q) (1 − fn(q))e 2 n (16) n=0 converges uniformly on compact sets of H.

5 Convergence of an inﬁnite ∗-product of regular functions

The results proved in the last section (Theorems 4.5 and 4.6) concern infinite products that, in general, do not converge to a regular function. In order to study the factorization of zeros of a regular function we must consider the infinite regular products (or infinite ∗-products) denoted by ∗ . We will study the convergence of regular products on symmetric slice domains of H, and the reason will be clear in what follows. The next lemma is useful in the sequel. { } Lemma 5.1 Let fi i∈N be a sequence of regular functions defined on a symmetric slice domain . Let K be a symmetric compact set such that K ⊆ . We suppose that there exists an integer NK such that 1 + fi = 0 on K for all i ≥ NK .Set m ( ) = ∗ ( + ( )). FNK ,m q 1 fi q i=NK

Then for all m ≥ NK and for any q ∈ K m ( ) = ( + ( ( ))) = FNK ,m q 1 fi Ti q 0 i=NK 123 Ann Glob Anal Geom

( ) = ( )−1 ( ) > ( ) = = where Tj q FNK , j−1 q qFNK , j−1 q for j NK and Tj q qforj NK .

Proof We will prove the assertion by induction. Let q ∈ K . Clearly the assertion is true for = ( ) = + ( ) = m NK by hypothesis. In fact FNK ,NK q 1 fNK q 0. Suppose the assertion is true for m = NK ,...,n − 1. We can write ( ) = ( ) ∗ ( + ( )). FNK ,n q FNK ,n−1 q 1 fn q (17)

First of all we notice that Re(Tj (q)) = Re(q) and |Im(Tj (q)|=|Im(q)| for all j ≥ NK . Then, since K is a symmetric set, we have that Tj (q) ∈ K if and only if q ∈ K . By Theorem ( ) = 2.11,sinceFNK ,n−1 q 0 by induction hypothesis, we have that formula (17) becomes: ( ) = ( )( + ( ( ))). FNK ,n q FNK ,n−1 q 1 fn Tn q

By hypothesis the factor (1 + fn) never vanishes on K and hence, since Tn(q) ∈ K , , ( ) the function FNK n q never vanishes on K . By induction hypothesis again, we have that n−1 F , − (q) = (1 + f (T (q))), and hence NK n 1 i=NK i i ⎡ ⎤ n−1 n ( ) = ⎣ ( + ( ( )))⎦ ( + ( ( ))) = ( + ( ( ))). FNK ,n q 1 fi Ti q 1 fn Tn q 1 fi Ti q i=NK i=NK

In analogy with Definition 4.1 we give the following

{ } Definition 5.2 Let fi i∈N be a sequence of regular functions defined on a symmetric slice domain . The infinite ∗-product ∞ ∗ (1 + fi (q)) i=0 is said to converge uniformly on (compact sets of) to a function F : → H if, for any symmetric, compact set K ⊆ :

(i) there exists an integer NK such that 1 + fi = 0onK for all i ≥ NK , (ii) the residual product ∞ ∗ (1 + fi (q)) i=NK converges uniformly on K (i.e. the sequence F , (q)= ∗ m (1 + f (q)) NK m i=NK i m≥NK converges uniformly on K ) to a never vanishing function FNK , (iii) we have, for all q ∈ K NK −1 ( ) = ∗ ( + ( )) ∗ ( ). F q 1 fi q FNK q i=0

It is important to establish that, for any given sequence of regular functions, the inﬁnite regular product and the inﬁnite product converge simultaneously. 123 Ann Glob Anal Geom

{ } Theorem 5.3 Let fn n∈N be a sequence of regular functions defined on a symmetric slice domain . The infinite ∗-product ∞ ∗ (1 + fn(q)) n=0 converges uniformly on compact sets of if and only if the infinite product ∞ (1 + fn(q)) n=0 converges uniformly on compact sets of . Proof It is enough to prove the convergence on symmetric compact sets of .LetK be such a compact set. Let NK be the integer such that for every n ≥ NK the factors (1 + fn(q)) do not vanish on K . By Lemma 5.1, the infinite ∗-product ∞ ∗ (1 + fi (q)) i=NK converges simultaneously with the infinite product ∞ (1 + fi (Tn(q))). i=NK

Recalling that Re(Tj (q)) = Re(q) and |Im(Tj (q)|=|Im(q)| for all j ≥ NK ,wehavethat Tj (q) ∈ K if and only if q ∈ K . This concludes the proof, since we are interested to shows the uniform convergence on K . Our next goal is to show that the limit of a uniformly convergent, regular inﬁnite product is a regular function. To this purpose, for every A ⊆ H and every f, g : A → H we set (with the usual notation)

f − gA = sup | f (q) − g(q)|. q∈A Thus, we have { } Lemma 5.4 Let hn n∈N be a sequence of regular functions deﬁned on a symmetric slice domain ⊆ H converging uniformly to a function h on compact sets of . Then h is regular on . Proof Given any pair of orthogonal vectors I and J in S, consider the orthogonal basis 1, I, J, IJ. We can write h I (x + yI) = h(z) as

h(z) = h0(z) + h1(z)I + h2(z)J + h3(z)IJ = F(z) + G(z)J with I = ∩ L I and F, G : I → L I . If we prove that F, G are holomorphic, then by Definition 1.1 we conclude the proof. By the Splitting Lemma, we have that, for all n ∈ N, the two functions Fn, Gn : I → L I such that

Consider any compact subset K ⊆ and set K I = K ∩ L I . Since by hypothesis, we have − = − = − = that limn→∞ hn h K 0, we obtain limn→∞ Fn F K I limn→∞ Gn G K I 0, too. Since the limit function of a sequence of holomorphic functions uniformly convergent on compact sets is itself holomorphic, we can conclude that the functions F and G are holomorphic on I . As we said, this completes the proof.

We are able to prove the announced result on the regularity of the limit function of a uniformly converging infinite regular product. { } Proposition 5.5 Let fn n∈N be a sequence of regular functions defined on a symmetric slice domain . If the infinite regular product ∞ ∗ (1 + fn(q)) n=0 converges uniformly on compact sets of to a function F, then F is regular on .

Proof By Definition 5.2, for any compact set K ⊆ H there exists an integer NK such that 1 + fn = 0onK if n ≥ NK . By Lemma 5.4, the residual ∗-product ∞ ∗ (1 + fn(q)) (18) n=NK converges uniformly on K to a regular function FNK that does not vanish on K . Therefore, the function F can be written as a finite product of regular functions NK −1 ( ) = ∗ ( + ( )) ∗ ( ), F q 1 fn q FNK q n=0 and hence it is regular on K . Moreover, by Corollary 2.12and Lemma 5.1, the zero set of F ∗ ∗ NK −1 ( + ( )) on K coincides with the zero set of the finite -product n=0 1 fn q . { } From now on, unless explicitly stated, we will always consider the case in which fn n∈N is a sequence of entire regular functions, i.e. the case in which = H. Consider an infinite ∗-product such as ∞ ∗ ( − −1) 1 qan (19) n=0 with {an}n∈N ⊆ H \{0}. In order to introduce convergence-producing regular factors that preserve the regularity of the product (19), we need a few results about series expansion of exponential type functions.

6 Convergence-producing regular factors

We want now to identify suitable convergence-producing regular factors (with respect to the ∗-multiplication) that will let certain classes of infinite ∗-products to converge. To do this, we need several technical results, that will be proven in the first part of this section. We warn the reader that our effort is to prove the analog of Theorem 4.5, when dealing with regular infinite ∗-products and when using convergence-producing regular factors. 123 Ann Glob Anal Geom

Lemma 6.1 Let p(x) ∈ C∞(R) and let f (x) = e p(x).Iff(k) and p(l) denote, respectively, the k-th derivative of f and the l-th derivative of p, then m ( + ) m ( ) ( + − ) f m 1 (x) = f i (x)p m 1 i (x). i i=0 Proof The proof is made by induction on m. ∈ N \{ } ( ) = + 1 2 +···+ 1 n (l) Lemma 6.2 For n 0 ,letpn x x 2 x n x .Ifp denotes the l-th derivative of p, then for all x ∈ R,wehave (i+ j−1)! j (n−1)! n−i ( ) (i − 1)!+i!x +···+ x +···+ x if i ≤ n p i (x) = j! (n−i)! n 0 if i > n. In particular ( ) (i − 1)! if i ≤ n p i (0) = n 0 if i > n. Proof By induction.

Using Lemma 6.1, in the particular case in which p is the polynomial pn which appears in Lemma 6.2, we are now able to ﬁnd an expression for all the derivatives of the function p (x) fn(x) = e n . ( ) ( ) ( ) = + 1 2 +···+ 1 n ∈ N \{ } ( ) = pn x k Lemma 6.3 Let pn x x 2 x n x with n 0 .Letfn x e .Iffn denotes the k-th derivative of fn,then (a) for m = 0,...,n ( ) f m (0) n = 1 m! (b) for all m ≥ n (m)( ) m−1 (i)( ) fn 0 = 1 fn 0 . m! m i! i=m−n Proof Thanks to Lemma 6.1,wehave m−1 ( ) m − 1 ( ) ( − ) f m (0) = f i (0)p m i (0) (20) n i n n i=0 and, by Lemma 6.2, ( − ) (m − i − 1)! if m− i ≤ n p m i (0) = n 0 if m− i > n.

To prove the assertion (a), we use ﬁnite induction on m.Ifm = 0, we have that fn(0) = ( ) e pn 0 = e0 = 1 = 0!, and hence the assertion is true. Suppose now that the assertion is true ( j)( ) = , ,..., − ≤ fn 0 = for m 0 1 i 1 with i n. By the induction hypothesis, we have that j! 1 for j < i and hence, by substitution, ( ) i− i− f i (0) 1 1 (i − 1)! 1 1 n = (i − j − 1)!= (i − 1)!=1. i! i! (i − 1 − j)! i! j=0 j=0 123 Ann Glob Anal Geom

Therefore (a) is proved. To prove (b) we consider m ≥ n + 1. By direct substitution in (20), we get

(m) m−1 f (0) 1 (m − 1)! ( ) n = f i (0)(m − i − 1)! m! m! i!(m − i − 1)! n i=m−n m−1 (i)( ) = 1 fn 0 m i! i=m−n and conclude the proof.

Remark 6.4 By Lemma 6.3, we also obtain that ( ) f m (0) n ≥ 0forallm ∈ N. m! ( ) ( ) = + 1 2 +···+ 1 n ∈ N \{ } ( ) = pn x Lemma 6.5 Let pn x x 2 x n x with n 0 .Letfn x e . Then, we have (a) for m = 0,...,n − 1 ( ) ( + ) f m (0) f m 1 (0) n − n = 0 m! (m + 1)! (b) for all m ≥ n (m)( ) (m+1)( ) (m−n)( ) fn 0 − fn 0 = 1 fn 0 . m! (m + 1)! m + 1 (m − n)!

Proof The statement (a) is true thanks to Lemma 6.3 (a). By Lemma 6.3 (b) we have , for all m ≥ n, (m+1)( ) m (i)( ) fn 0 = 1 fn 0 ( + )! + ! m 1 m 1 = + − i i m 1 n (m)( ) m−1 (i)( ) (m−n)( ) = 1 fn 0 + fn 0 − fn 0 + ! ! ( − )! m 1 m = − i m n i m n ( ) ( ) ( − ) 1 f m (0) f m (0) f m n (0) = n + m n − n = m + 1 m! m! (m − n)! (m)( ) (m−n)( ) = fn 0 − 1 fn 0 . m! m + 1 (m − n)! Hence, (m)( ) (m+1)( ) (m−n)( ) fn 0 − fn 0 = 1 fn 0 . m! (m + 1)! m + 1 (m − n)!

We are now able to produce estimates for the derivatives of the function hn(x)= ( ) (1 − x)e pn x when x = 0. These estimates will be important to prove Proposition 6.7. 123 Ann Glob Anal Geom

( ) ( ) = + 1 2 + ...+ 1 n ∈ N \{ } ( ) = pn x Lemma 6.6 Let pn x x 2 x n x with n 0 .Letfn x e .Then for all m ≥ n we have

( ) ( + ) f m (0) f m 1 (0) 1 0 ≤ n − n ≤ . m! (m + 1)! n + 1

Proof By Remark 6.4,wehavethat

( − ) 1 f m n (0) n ≥ 0forallm ∈ N m + 1 (m − n)! and hence, by Lemma 6.5

( ) ( + ) f m (0) f m 1 (0) n − n ≥ 0forallm ∈ N m! (m + 1)! ( j) fn (0) Then, the sequence j! decreases, and hence j∈N

( ) ( ) f j (0) f 1 (0) n ≤ n = 1forallj ∈ N. j! 1! Therefore, we have that

( − ) 1 f m n (0) 1 0 ≤ n ≤ m + 1 (m − n)! m + 1 Since m ≥ n,weobtain

( − ) 1 f m n (0) 1 0 ≤ n ≤ . m + 1 (m − n)! n + 1 The thesis follows directly from Lemma 6.5.

As announced we have the following result.

Proposition 6.7 Let n ∈ N and x ∈ R.Weset

x+ 1 x2+···+ 1 xn hn(x) = (1 − x)e 2 n . (21)

Then, there exist ck ∈ R such that ∞ k+n+1 hn(x) = 1 − ck x (22) k=0 and 1 0 ≤ ck ≤ n + 1 for all k ∈ N. 123 Ann Glob Anal Geom

x+ 1 x2+···+ 1 xn Proof If we write formula (21) using the expansion of fn(x) = e 2 n ,weobtain that ∞ ( ) ∞ ( ) f m (0) f m (0) h (x) = n xm − x n xm n m! m! m=0 m=0 ∞ ( ) ( − ) f m (0) f m 1 (0) = 1 + n − n xm. m! (m − 1)! m=1 By Lemma 6.5 (a) we have that ∞ ( ) ( − ) f m (0) f m 1 (0) h (x) = + n − n xm n 1 ! ( − )! = + m m 1 m n1 ∞ (k+n+1) (k+n) f (0) f (0) + + = 1 + n − n xk n 1 (k + n + 1)! (k + n)! k=0 ∞ (k+n) (k+n+1) f (0) f (0) + + = 1 − n − n xk n 1. (k + n)! (k + n + 1)! k=0

(k+n)( ) (k+n+1)( ) = fn 0 − fn 0 + ≥ ∈ N If we set ck (k+n)! (k+n+1)! then, by Lemma 6.6,sincek n n for all k ,we obtain the thesis.

At this point, we have all the ingredients to prove an analog of Theorem 4.5 for the inﬁnite regular products.

Theorem 6.8 Let {an}n∈N ⊆ H \{0} be a sequence such that limn→∞ |an|=∞.For ∈ S ∈ ∈ N ∈ N all an,letIn be such that an L In . Then, for all n ,thereexistmn and a never vanishing entire regular function gn, whose restriction to L In is the function − za−1+ 1 z2a−2+···+ 1 zmn a mn e n 2 n mn n , such that the inﬁnite ∗-product ∞ ∗ ( − −1) ∗ ( ) 1 qan gn q (23) n=0 converges uniformly on compact sets of H. In particular we can choose mn = n for all n ∈ N.

∈ N V ( ) = ( − −1) ∗ ( ) Proof For all n ,letgn be as in the hypothesis and set n q 1 qan gn q . Theorem 5.3 yields that the ∗-product (23) converges simultaneously with the product ∞ Vn(q). (24) n=0 Thanks to Theorem 3.12, the product (24) converges uniformly on compact sets of H if the series ∞ |1 − Vn(q)| n=0 converges uniformly on compact sets of H.LetK ⊆ H be a compact set. Let R > 0besuch that K ⊆ B(0, R) and let n ∈ N be the integer such that |an| > R for all n ≥ n.Letvn(z) be V the restriction of n to the complex plane L In . Hence 123 Ann Glob Anal Geom

− −1+···+ 1 n −n v ( ) = ( − 1) zan n z an , n z 1 zan e and we can estimate the coefﬁcients of the power series Taylor expansion of vn at the point 0 by using Proposition 6.7. Since the regular extension of vn is unique by the identity principle (Theorem 2.3), the coefﬁcients of the power series expansion of Vn are the same of vn. Therefore, by Proposition 6.7,wehavethat: ∞ V ( ) = − k+n+1 −(k+n+1) n q 1 ckq an k=0 for every q ∈ H,whereck ∈ R are such that 1 0 ≤ ck ≤ n + 1 for all k ∈ N. Then ∞ ∞ ( + + ) |q| k n 1 | − V ( )| = k+n+1 −(k+n+1) ≤ 1 n q ckq an ck |an| k=0 k=0 ≤ 1 and since ck n+1 ,wehave ∞ + + 1 |q| n 1 k |1 − Vn(q)| ≤ (n + 1) |an| k=0 | | n+1 ∞ | | k ≤ 1 q q . n + 1 |an| |an| k=0 The factor on the right ∞ |q| k |an| k=0 is a geometric series and its ratio is strictly smaller than 1 when n ≥¯n,sinceq ∈ K ⊆ B(0, R) and |an| > R. Hence, the power series is convergent to the value − |q| 1 1 − , |an| and we obtain + − 1 |q| n 1 |q| 1 |1 − Vn(q)| ≤ 1 − . n + 1 |an| |an| − |q| 1 Since lim →∞ 1 − = 1 and the series n |an |

∞ + 1 |q| n 1 (25) n + 1 |a | n=0 n converges uniformly on K ,alsotheseries 123 Ann Glob Anal Geom

∞ |1 − Vn(q)| n=0 converges uniformly on K . To conclude, we remark that, for all n ∈ N, the function gn has no zeros. In fact gn is the (unique) regular extension of a holomorphic function on the complex plane L In that has no zeros. Thanks to Proposition 2.7, gn does not vanish on the whole of H. We will now present a geometrical result of particular significance for our purposes. This result is a consequence of a representation formula, that we state in a reduced form, and that is proven in its full generality in [5,4]. Theorem 6.9 Let f be a regular function on a symmetric slice domain ⊆ H. For all x, y ∈ R such that x + yS ⊆ ,thereexistb, c ∈ H such that f (x + yI) = b + Ic for all I ∈ S. Remark that c = 0 if and only if f is constant on x + yS. The announced geometrical result establishes the following: Proposition 6.10 Let f be a regular function on a symmetric slice domain ⊆ H.Let K ⊆ be a symmetric compact set. For every I ∈ S,p∈ H, and R > 0 such that ( ) ⊆ ( , ) f|L I K I B p R we have f (K ) ⊆ B(p, 2R). Proof Let I, p, R be as in the hypothesis. Remark that K symmetric means explicitly that K = x + yS x+yI∈K and hence that f (K ) = f (x + yS). x+yI∈K It is enough to prove that f (x + yS) ⊆ B(p, 2R) for x, y ∈ R such that x + yI ∈ K. Let x, y ∈ R be such that x + yS ⊆ K . By Theorem 6.9,thereexistb, c ∈ H such that f (x + yJ) = b + Jc ∈ S ( ) ⊆ ( , ) for all J .Since f|L I K I B p R ,wehavethat f (x + yI) = b + Ic ∈ B(p, R) and that f (x − yI) = b − Ic ∈ B(p, R). Since two antipodal points (i.e. b + Ic and b − Ic) of the 2-sphere f (x + yS) = b + Sc belong to B(p, R), then the center b of b + Sc also belongs to B(p, R) and the radius of b + Sc is smaller or equal than R. Hence f (x + yS) ⊆ B(p, 2R). 123 Ann Glob Anal Geom

We are now able to prove the following result, concerning the uniform convergence of sequences of regular functions.

Proposition 6.11 Let { fn}n∈N be a sequence of regular functions deﬁned on a symmetric slice domain ⊆ H. If the sequence converges uniformly on compact sets of to a regular function f , then { c} (a) the sequence of the regular conjugates fn n∈N converges uniformly on compact sets of to the regular conjugate f c; { s } (b) the sequence of symmetrizations fn n∈N converges uniformly on compact sets of to the symmetrization f s . Proof Let K ⊆ be a symmetric compact set. By hypothesis

lim fn − f K = 0. (26) n→+∞ We have to prove that c c s s lim f − f K = 0 and lim f − f K = 0. n→+∞ n n→+∞ n

Let I, J ∈ S be such that J ⊥ I . By the Splitting Lemma, there exist F, G : L I → L I holomorphic such that

fI (z) = F(z) + G(z)J and, for all n ∈ N,thereexistFn, Gn : L I → L I holomorphic such that

fn,I (z) = Fn(z) + Gn(z)J.

Therefore, as in Lemma 5.4,ifwesetK I = K ∩ L I , formula (26) implies

lim Fn − FK = 0 and lim Gn − GK = 0. (27) n→+∞ I n→+∞ I

Thanks to the continuity of the conjugate function on the complex plane L I , we obtain that

lim Fn − FK = 0 and lim Gn − GK = 0. (28) n→+∞ I n→+∞ I By Definition 2.13,wehavethat c( ) = (¯) − ( ) c ( ) = (¯) − ( ) fI z F z G z J and fn,I z Fn z Gn z J ∈ N. c c for all n By (27)and(28), we can easily prove that fn,I converges to fI uniformly on K I i.e c c lim f , − f K = 0. n→+∞ n I I I By Proposition 6.10 we have c − c ≤ c − c fn f K 2 fn,I fI K I for all n ∈ N, and hence c c lim f − f K = 0. n→+∞ n Since by Definition 2.14,wehavethat s( ) = ( ) (¯) + ( ) (¯) s ( ) = ( ) (¯) + ( ) (¯) fI z F z F z G z G z and fn,I z Fn z Fn z Gn z Gn z for all n ∈ N, the proof of (b) follows analogously. 123 Ann Glob Anal Geom

We now turn our attention back to the inﬁnite ∗-product, and state:

Proposition 6.12 Let {an}n∈N ⊆ H \{0} be such that limn→∞ |an|=∞.Let{gn}n∈N be the sequence of entire convergence-producing regular factors, defined as in Theorem 6.8,such that the infinite ∗-product ∞ ∗ ( − −1) ∗ ( ) 1 qan gn q n=0 converges uniformly on compact sets of H to a regular function h. Then the infinite product

∞ n 2 2Re(a ) 2Re(a2) 2Re(a ) q 2qRe(an) q n + 1 q2 n +···+ 1 qn n − + 1 e |an |2 2 |an |4 n |an |2n (29) |a |2 |a |2 n=0 n n converges uniformly on compact sets of H to the symmetrization hs of h.

Proof Set

k ( ) = ∗ ( − −1) ∗ ( ). hk q 1 qan gn q n=0

By hypothesis, the sequence {hk}k∈N converges uniformly on compact sets of H to h.By Remark 2.15, the symmetrization of a ∗-product is the product of the symmetrization of ∗-factors and hence, for all k ∈ N,

k s ( ) = ( − −1)s s ( ). hk q 1 qan gn q n=0 { s } H By Proposition 6.11, the sequence hk k∈N converges uniformly on compact sets of to the symmetrization hs of h, i.e. ∞ ∞ s ( − −1)s s ( ) = ∗ ( − −1) ∗ ( ) . 1 qan gn q 1 qan gn q (30) n=0 n=0 s ∈ N ∈ We want now to write explicitly the functions gn.Foralln ,ifan L In , the function gn −1+···+ 1 n −n ( ) = zan n z an ∈ H is deﬁned as the unique regular extension of fn z e (with z L In )to . ∈ Then for all z L In ,wehave

−1+···+ 1 n −n −1+···+ 1 n −n s ( ) = ( ) c( ) = zan n z an zan n z an fn z fn z fn z e e n 2Re(a ) 2Re(a2) 2Re(a ) − 1 n −n −1 1 n −n n + 1 2 n +···+ 1 n n za 1+···+ z a +za +···+ z a z 2 2 z 4 n z 2n = e n n n n n n = e |an | |an | |an | .

Hence, we obtain

( ) ( 2) ( n ) q 2Re an + 1 q2 2Re an +···+ 1 qn 2Re an s ( ) = |an |2 2 |an |4 n |an |2n gn q e 2 ( ) ∈ H ( − −1)s = q − 2qRe an + for all q . Observing that 1 qa 2 2 1 , we obtain the n |an | |an | thesis. 123 Ann Glob Anal Geom

7 On the multiplicity of zeros of entire regular functions

Let f be an entire regular function. If β is a real zero of f then, using in a “slicewise” manner the classical theory of holomorphic functions, it is easy to prove that there exists n ∈ N such that f (q) = (q − β)nh(q) where h is an entire regular function with h(β) = 0. For the non real zeros of f , a factorization result is not so trivial. The next proposition shows how to obtain it.

Proposition 7.1 Let f be an entire regular function. Let α ∈ H \ R beazerooff.Let Sα = Re(α)+|Im(α)|S.Iff ≡ 0 then there exist t ∈ N ∪{0}, t quaternions α1,...,αt ∈ Sα where αi = αi+1 for i = 1,...,t − 1 and m ∈ N ∪{0} such that t 2 2 m f (q) = q − 2qRe(α) +|α| g(q) ∗ ∗ (q − αi ) (31) i=1 where g is an entire regular function such that g(α) = 0.

Proof If f (α)¯ = 0 then, by Corollary 2.6, α is an isolated root, and hence we have m = 0. Otherwise, let I ∈ S be such that α = Re(α) +|Im(α)|I . By the Splitting Lemma, given J ∈ S, J ⊥ I , there exist two entire holomorphic functions R, S : L I → L I such that for all z ∈ L I

fI (z) = R(z) + S(z)J. Since f (α) = f (α)¯ = 0, we have that R(α) = R(α)¯ = 0andS(α) = S(α)¯ = 0. Then, there exist four integers p, q, r, s and two entire holomorphic functions U(z) e V (z) such that

R(z) = (z − α)p(z − α)qU(z) and S(z) = (z − α)r (z − α)s V (z) with U and V which do vanish neither in α nor in α¯ .Letm = min{p, q, r, s} and let p−m q−m r−m s−m U0(z) = (z − α) (z −¯α) U(z) and V0(z) = (z − α) (z −¯α) V (z).SinceL I is a commutative subspace of H,weobtain m m 2 2 m fI (z) = (z − α) (z −¯α) [U0(z) + V0(z)J]= z − 2zRe(α) +|α| [U0(z) + V0(z)J].

Since f is the unique regular extension of fI , it is of the form: m f (q) = q2 − 2qRe(α) +|α|2 h(q) where h is the regular extension of U0(z) + V0(z)J. We note that either α or α¯ is not a root of h and hence, by Corollary 2.6, the 2-sphere Sα is not a spherical zero of h.Ifh(α) = 0 then we have t = 0. Otherwise, we can retrace the process used, in [17], to factorize regular polynomials, and obtain that there exist α1 ∈ Sα and h1 entire regular function such that

h(q) = h1(q) ∗ (q − α1).

Recursively, if h1(α) = 0 the process ends with t = 1. Otherwise, if h1(α) = 0 we consider the function h1, and we obtain that there exist α2 ∈ Sα, α2 = α1,andh2 entire regular function such that

h(q) = h2(q) ∗ (q − α2) ∗ (q − α1). 123 Ann Glob Anal Geom

It is enough to prove that this process ends after a ﬁnite number of steps. We proceed by contradiction. Suppose that for all k ∈ N,thereexistk quaternions α1,...,αk lying onto the 2-sphere Sα and an entire regular function hk such that k h(q) = hk(q) ∗ ∗ (q − αi ) . i=1 If we consider the symmetrization of h,wehave s( ) =[ 2 − (α) +|α|2]k s ( ) h q q 2Re q hk q ∈ N ∈ S ∈ N s : → for all k . Hence, for any I and for all k , the holomorphic function h I L I L I is of the form s ( ) =[ 2 − (α) +|α|2]k s ( ) = ( − α)k ( −¯α)k s ( ) h I z z 2Re z hk z z z hk z s where z is any element of L I . As a consequence, the holomorphic function h I would have two roots with inﬁnite multiplicity. Contradiction.

We can at this point give the definition of multiplicity for the zeros of an entire regular function. This definition was already introduced in [17] for the case of polynomials and in [24] with the attention to the case of poles.

Deﬁnition 7.2 Let f be an entire regular function. If β is a real zero of f then its multiplicity is the largest natural number n such that f (q) = (q−β)n g(q) for some other entire regular function g.Ifx+yS is a spherical zero of f then its spherical multiplicity is 2m where m is the largest natural number such that f (q) =[q2−2qx+(x2+y2)]m h(q) for some other entire regular function h. Moreover, if α ∈ x + yS is such that h(α) = 0, then its isolated multiplicity is the largest natural number k such that there exist α1,...,αk ∈ x + yS with αi = αi+1 (for i = 1,...k−1) and an entire regular functionl with h(q) = l(q)∗(q−α1)∗(q−α2)∗···∗(q−αk).

8 Factorization of zeros of an entire regular function

Let us assume that f has a ﬁnite number of zeros, and let us set m ∈ N to be the multiplicity of 0 as a zero of f , let us denote the non zero, real zeros by bi for i = 1,...,k,the spherical ones by Sn = xn + ynS for n = 1,...,t, and the non real, isolated zeros by a j for j = 1,...,r. Assume also that the multiple zeros are repeated according to their multiplicities. By Proposition 7.1,forevery j = 1,...,r there exists δ j ∈ Re(a j ) +|Im(a j )|S such that k t 2 r − q 2qx − f (q) = qm (1 − qb 1) − n + 1 g(q) ∗ ∗ (1 − qδ 1) i x2 + y2 x2 + y2 j i=1 n=1 n n n n j=1 where g is a never vanishing entire regular function. If instead f has inﬁnitely many zeros the factorization problem is a much more delicate issue (as in the case of entire holomorphic functions). To study this situation, we begin by factorizing the real and spherical zeros.

Theorem 8.1 Let f be an entire regular function. Let m be the multiplicity of 0 as zero of f , let {bn}n∈N ⊆ R \{0} be the sequence of (non zero) real zeros of f where multiple zeros are 123 Ann Glob Anal Geom repeated according to their multiplicities. Then there exists a sequence {rn}n∈N ⊆ N such that ∞ − − qb−1+ 1 q2b−2+···+ 1 qrn b rn ( ) = m ( − 1) n 2 n rn n ( ) f q q 1 qbn e g q n=0 where g is an entire regular function without real zeros. In particular, we can choose rn = n for all n ∈ N.

Proof First of all, we note that

lim |bn|=∞ n→∞ for the real zeros of regular functions are isolated. Hence by Theorem 4.5, the inﬁnite product ∞ − −1+···+ 1 n −n R( ) = ( − 1) qbn n q bn q 1 qbn e n=0 converges uniformly on compact sets of H. By Lemma 5.4 R is an entire regular function and it has the same non zero real zeros with the same multiplicity as f . Hence there exists some entire regular function g without real zeros such that f (q) = qmR(q)g(q).

{ } Theorem 8.2 Let f be an entire regular function. Let Sn n∈N be the sequence of spherical zeros of f , where multiple zeros are repeated according to their spherical multiplicities. Then there exist {cn}n∈N ⊆ H \{R}, sequence of generators of the spherical zeros, and a sequence {n}n∈N ⊆ N such that ∞ ( 2) ( n ) 2 ( ) 2Re(cn ) 1 2 2Re cn 1 n 2Re cn q 2qRe cn q + q +···+ q f (q) = − + 1 e |cn |2 2 |cn |4 n |cn |2 n h(q) |c |2 |c |2 n=0 n n where h is an entire regular function without spherical zeros. In particular, we can choose n = n for all n ∈ N.

Proof For all n ∈ N let cn ∈ H be a generator of the spherical zero Sn. We note that limn→∞ |cn|=∞because the spherical zeros of regular functions are isolated. By Propo- sition 6.12, the inﬁnite product

∞ n 2 2Re(c ) 2Re(c2) 2Re(c ) q 2qRe(cn) q n + 1 q2 n +···+ 1 qn n S(q) = − + 1 e |cn |2 2 |cn |4 n |cn |2n |c |2 |c |2 n=0 n n converges uniformly on compact sets of H.SinceS is an entire regular function with the same spherical zeros as f , then there exists some entire regular function h, without spherical zeros, such that f (q) = S(q) h(q).

To reach the announced factorization, let us consider an entire regular function f without real or spherical zeros. 123 Ann Glob Anal Geom

Theorem 8.3 Let f be an entire regular function without real or spherical zeros. Let {an}n∈N be the sequence of (isolated, non real) zeros of f , where multiple zeros are repeated according to their multiplicities. Then there exists a never vanishing, entire regular function h and, ∈ N δ ∈ = ( ) +| ( )|S ∈ N for all n ,thereexist n San Re an Im an and mn such that ∞ c ( ) = ( ) ∗ ∗ ( − δ−1) ∗ ( ) f q h q 1 q n gn q (32) n=0 where, for all n ∈ N, the function gn is the never vanishing entire regular function whose restriction to the plane Ln = R + R[Im(δn)] is given by

− zδ−1+···+ 1 zmn δ mn | ( ) = n mn n . gn Ln z e

In particular, we can choose mn = n for all n ∈ N.

Proof Notice first of all that since f (an) = 0then f (an) = 0foralln ∈ N (in fact f has no spherical zeros). Moreover, since the zeros of f are isolated, the sequence {an}n∈N is such that limn→∞ |an|=∞.Now,byTheorem6.8, for every choice of the sequence {δn}n∈N ⊂ H, such that limn→∞ |δn|=∞, there exists a sequence of convergence-producing regular factors {gn}n∈N, defined as in the statement, such that the infinite ∗-product ∞ ∗ ( − δ−1) ∗ ( ) 1 q n gn q (33) n=0 converges uniformly on compact sets of H to an entire regular function. We will show how to choose the quaternions δn. In analogy with the case of polynomials (see [17]), we start to “add” the zero a0 to the function f . To this purpose we consider the function ( ) = ( ) ∗[ − ( )−1 −1 ( )]∗ ( ) f1 q f q 1 qf a0 a0 f a0 g0 q where g0 is defined as in the statement. The fact that f1(a0) = 0 is an immediate consequence δ = ( )−1 ( ) δ ∈ of Theorem 2.11.Weset 0 f a0 a0 f a0 and we note that 0 Sa0 . The function f1 { } has an n≥1 as sequence of isolated zeros and Sa0 as spherical zero (see [17] for details). We stress the fact that every an has multiplicity 1 and that all zeros are repeated according to their multiplicities. We can now factor the spherical zero Sa0 and obtain 2Re(a )q q2 f (q) = − 0 + f˜ (q) 1 1 2 2 1 |a0| |a0| ˜ ˜ with f1 an entire regular function having {an}n≥1 as sequence of zeros. Therefore f1(a1) = 0. By repeating this same procedure for f1 to “add” the zero a1, since the polynomial ( ) 2 − 2Re a0 q + q 1 2 2 has real coefficients, we obtain |a0| |a0| 2Re(a )q q2 f (q) = − 0 + f˜ (q) ∗[ − q f˜ (a )−1a −1 f˜ (a )]∗g (q) 2 1 2 2 1 1 1 1 1 1 1 1 |a0| |a0| ˜ −1 −1 ˜ = f1(q) ∗[1 − q f1(a1) a1 f1(a1)]∗g1(q)

˜ −1 ˜ where g1 is deﬁned as in the statement. We set δ1 = f1(a1) a1 f1(a1), and we note that δ ∈ { } 1 Sa1 . We also note that f2 has an n≥2 as sequence of isolated zeros and Sa0 and Sa1 as spherical zeros. Unlike the case of polynomials, this process has inﬁnitely many steps. For 123 Ann Glob Anal Geom this reason, at every step, we also add the convergence-producing regular factors. Iterating this process, we get ∞ ( ) ∗ ∗ ( − δ−1) ∗ ( ) = ( ) f q 1 q n gn q k q n=0 with k entire regular function having {δn}n∈N as a sequence of generators of its spherical { } zeros San n∈N and no other zeros. We can factor the spherical zeros of k and by Theorem 6.12, formula (30), we have that there exists an entire regular function h without zeros such that ∞ ∞ s ( ) ∗ ∗ ( − δ−1) ∗ ( ) = ( ) = ∗ ( − δ−1) ∗ ( ) ( ). f q 1 q n gn q k q 1 q n gn q h q n=0 n=0 Hence, by Remark 2.10 ∞ ( ) ∗ 1 ∗ ( − δ−1) ∗ ( ) = ( ) f q ∞ s 1 q n gn q h q (34) −1 = ∗ (1 − qδn ) ∗ gn(q) n 0 n=0 ∗ ∗ ∞ ( − δ−1)∗ By -multiplying on the right expression (34) by the regular conjugate of n=0 1 q n gn(q) ,weobtain ∞ ∞ c ∗ ( − δ−1) ∗ ( ) ∗ ∗ ( − δ−1) ∗ ( ) 1 q n gn q 1 q n gn q ( ) ∗ n=0 n=0 f q ∞ s −1 ∗ (1 − qδn ) ∗ gn(q) = n 0 ∞ c = ( ) ∗ ∗ ( − δ−1) ∗ ( ) h q 1 q n gn q n=0 and, by Definition 2.14,wehave ∞ c ( ) = ( ) ∗ ∗ ( − δ−1) ∗ ( ) . f q h q 1 q n gn q n=0

We are eventually ready to prove the announced result.

Theorem 8.4 [Weierstrass factorization theorem for regular functions] Let f be an entire regular function. Let: m ∈ N be the multiplicity of 0 as zero of f , {bn}n∈N ⊆ R \{0} be the sequence of the (non zero) real zeros of f , {Sn = xn + ynS}n∈N be the sequence of the spherical zeros of f , and {an}n∈N ⊆ H \ R be the sequence of the non real zeros of f with isolated multiplicity greater then zero. If all the zeros listed above are repeated according to their multiplicities, then there exists a never vanishing, entire regular function h and, for all ∈ N ∈ δ ∈ = ( ) +| ( )|S , , ∈ N n ,thereexistcn Sn, n San Re an Im an ,rn n mn such that f (q) = qm R(q) S(q) A(q) ∗ h(q) 123 Ann Glob Anal Geom where ∞ − − qb−1+ 1 q2b−2+···+ 1 qrn b rn R( ) = ( − 1) n 2 n rn n , q 1 qbn e n=0 ∞ ( 2) ( n ) 2 ( ) 2Re(cn ) 1 2 2Re cn 1 n 2Re cn q 2qRe cn q + q +···+ q S(q) = − + 1 e |cn |2 2 |cn |4 n |cn |2 n , |c |2 |c |2 n=0 n n ∞ A( ) = ∗ ( − δ−1) ∗ ( ) q 1 q n gn q n=0 and where, for all n ∈ N, the function gn is the never vanishing entire regular function whose restriction to the plane Ln = R + R[Im(δn)] is given by − zδ−1+ 1 z2δ−2+···+ 1 zmn δ mn | ( ) = n 2 n mn n . gn Ln z e

In particular, we can choose rn = n = mn = n for all n ∈ N.

Proof Thanks to Theorems 8.1 and 8.2, we obtain that there exist l entire regular function without real or spherical zeros such that f (q) = qm R(q) S(q) l(q) where R and S are as in the statement. Now we consider the regular conjugate lc of l.The function lc is an entire regular function whose (isolated) zeros on a 2-sphere S are in one-to- one correspondence with the (isolated) zeros of l (see Theorem 2.16). By Theorem 8.3,for ∈ N δ ∈ all n ,thereexist n San and gn never vanishing entire regular function such that ∞ c c( ) = c( ) ∗ ∗ ( − δ−1) ∗ ( ) l q h q 1 q n gn q (35) n=0 where hc is a never vanishing entire regular function. Taking the regular conjugate of both members of (35) and using Remark 2.15,wehavethat ∞ ( ) = ∗ ( − δ−1) ∗ ( ) ∗ ( ). l q 1 q n gn q h q n=0 Observing that h, like hc, is a never vanishing entire regular function, we conclude the proof.

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