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Contents
Introduction iii
1 Preliminaries 1
2 Infinite Products 5
2.1 The Quaternionic Logarithm . . . 5
2.2 Infinite Products of quaternions . . . 9
2.3 Infinite Products of functions defined on H . . . 11
2.4 Infinite ∗-Products of regular functions . . . 15
3 The Weierstrass Factorization Theorem 19 3.1 Convergence-producing regular factors . . . 19
3.2 On the multiplicity of zeros of entire regular functions . . . 30
3.3 Factorization of zeros of an entire regular function . . . 32
4 The Borel Theorem 38 4.1 Growth of entire regular functions: Order and Type . . . 38
4.2 The convergence exponent . . . 41
4.3 Canonical Products and their Genus . . . 43
4.4 Estimates for Canonical Products . . . 46
Bibliografia 56
Introduction
The well known theory of Fueter regular functions, [9, 10, 29], is the first and most celebrated candidate among a few others, in the search for a quater-nionic analogue of the theory of holomorphic functions of one complex vari-able. 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, [16, 17]. Unlike Fueter’s, this theory includes the polynomials and the power series of the quaternionic variable q of the typeP
n≥0q na
n, with coefficients an∈ H. Furthermore, the
analogs (sometime peculiarly different) of many of the fundamental prop-erties of holomorphic functions of one complex variable can be proven in this new setting, like the Cauchy and Pompeiu Representation Formulas 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 [4, 5, 6, 7, 12, 13, 14, 16, 17, 28]). In fact the theory of slice regular functions is already rather rich and well estab-lished 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 + ix1 + jx2 + kx3 where the xl are real, and i, j, k are such that
i2 = j2 = k2 = −1, ij = −ji = k, jk = −kj = i, ki = −ik = j.
We set Re(q) = x0, Im(q) = ix1 + jx2 + kx3, |q| = px20+ x21+ x22+ x23
and call Re(q), Im(q) and |q| the real part, the imaginary part and the module of q, respectively. The conjugate of the quaternion q, defined as
¯
q = Re(q) − Im(q) = x0− ix1 − jx2 − kx3, satisfies |q| =
√
q ¯q = √qq and¯ allows the definition of the inverse of any element q 6= 0 as q−1 = |q|q¯2. If S
is the unit sphere of purely imaginary quaternions, i.e. S = {q ∈ H : q2 =
−1} = {q = ix1 + jx2 + kx3 : x21 + x22 + x23 = 1}, then every quaternion q
which is not real (i.e. with Im(q) 6= 0) can be written as q = x + Iy for x = Re(q), y = |Im(q)| and I = |Im(q)|Im(q) ∈ S. Moreover, using the same I ∈ S, we can write q = |q|eIϑ, for some ϑ ∈ R.
We can now state the following definition, see [5].
Definition 0.0.1. Let Ω be a domain in H. A function f : Ω → H is said to be slice regular if, for every I ∈ S, its restriction fI to the complex line
LI = R+RI passing through the origin and containing 1 and I has continuous
partial derivatives and satisfies ∂If (x + yI) := 1 2 ∂ ∂x + I ∂ ∂y fI(x + yI) = 0, in ΩI = Ω ∩ LI.
In what follows we will simply call regular a slice regular function. If Ω is a generic domain of H, then there are examples of regular functions defined on Ω which are not even continuous, see e.g. [5]. In [4], and in this same paper [5], a class of pathology-preventing domains of definition for regular functions were introduced, as follows:
Definition 0.0.2. Let Ω be a domain in H, intersecting the real axis. If ΩI = Ω ∩ LI is a domain in LI ' C for all I ∈ S then we say that Ω 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 0.0.3. 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 [5], any regular function defined on a slice domain Ω can be (uniquely) extended to the smallest symmetric slice domain which contains Ω, and for this reason we will always consider regular functions defined on symmetric slice domains.
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, 18, 19, 27].
In this thesis we study some properties of entire regular functions of a quaternionic variable. An entire regular function is a function that is regular on the whole of H, and it is represented by an everywhere convergent power series of the form:
f (q) = a0+ qa1+ q2a2+ · · · + qnan+ · · · .
These functions form a natural generalization of regular polynomials, and one of our intents is to extend to them some properties of quaternionic poly-nomials. Historically, in the classical theory of holomorphic functions, the central result about this generalization is the Weierstrass Factorization The-orem that allows us to represent an entire holomorphic function by an infinite product. At approximately the same time as this work of Weierstrass, La-guerre studied the connection between entire functions and polynomials and introduced the important concept of the genus of an entire function. Borel, Hadamard and Lindel¨of dealt with the connection between the growth of an entire holomorphic function and the distribution of its zeros. The rate of growth of a polynomial as the independent variable goes to infinity is deter-mined, of course, by its degree. On the other hand, the number of roots of a polynomial is equal to its degree. Thus, the more roots a polynomial has, the greater is its growth. This connection between the set of zeros of the function and its growth can be generalized to arbitrary entire holomorphic functions. The content of most of the classical theorems of the theory of entire holo-morphic functions consists in establishing relations between the distribution of the roots of the function and its asymptotic behavior as z → ∞.
The first important result that we present is a quaternionic version of the Weierstrass Factorization Theorem. The theorem that we obtain is in fact the following.
Theorem 0.0.4 (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∈Nbe 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 than 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 ∈ N, there exist cn ∈ Sn, δn ∈ San = Re(an) + |Im(an)|S,
rn, `n, mn∈ N such that f (q) = qm R(q) S(q) A(q) ∗ h(q) where R(q) = ∞ Y n=0 (1 − qb−1n )eqb−1n +...+rn1 qrnb−rnn ,
S(q) = ∞ Y n=0 q2 |cn|2 − 2qRe(cn) |cn|2 + 1 eq 2Re(cn) |cn|2 +...+ 1 `nq`n 2Re(c `n n ) |cn|2`n , A(q) = ∞ Y ∗ n=0 (1 − qδn−1) ∗ gn(q)
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
gn|Ln(z) = e
zδn−1+...+mn1 zmnδn−mn.
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.
As in the complex case, this theorem provided us the basic apparatus for the investigation of the properties of entire regular functions and was the starting point to study the connection between the growth of an entire regular function and the distribution of its zeros. In fact the second important result that we prove in this thesis is a quaternionic version of Borel’s Theorem. Once defined the order of a regular function, that measures its growth, and the convergence exponent of the sequence of its zeros, that characterizes the density of its zeros, we obtain a Borel Theorem for entire regular functions. Theorem 0.0.5. (Borel Theorem for regular functions) Let {δn}n∈N ⊆ H \
R be a sequence having finite exponent of convergence a and let A be the canonical product of non real type of genus p:
A(q) = ∞ Y ∗ n=0 G(q, δn−1, p).
Let {bn}n∈N ⊆ R \ {0} be a sequence having finite exponent of convergence b
and let R be the canonical product of real type of genus r: R(q) =
∞
Y
n=0
Let {cn}n∈N ⊆ H \ R be a sequence having finite exponent of convergence c
and let S be the canonical product of spherical type of genus `: S(q) = ∞ Y n=0 q2 |cn|2 − 2qRe(cn) |cn|2 + 1 eq 2Re(cn) |cn|2 +...+ 1 `q ` 2Re(c`n) |cn|2` .
Let Π(q) be the canonical product
Π(q) = R(q) S(q) A(q). Then ord(Π) = max{a, b, c}.
This paper is organized as follows. After having presented some necessary preliminary results, in Section 2.1 we introduce a (peculiar) definition of a quaternionic Logarithm that is necessary to study criteria to establish the convergence of infinite products of quaternions in Section 2.2. Then, in Section 2.3, we investigate the uniform convergence of infinite products of regular functions. The results so found give us a tool to identify, in Section 2.4, conditions that can guarantee the uniform convergence of infinite *-products of regular functions to a regular function. In Section 3.1 we perform the study of different type of convergence-producing regular factors, some of which arise only in the quaternionic setting. Section 3.2 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. In Section 3.3, we prove the Weierstrass Factorization Theorem for entire regular functions. In order to study the connection between the growth of an entire regular function and the distribution of its zeroes, in Section 4.1 we introduce the order of an entire regular function and the convergence exponent of a quaternionic sequence in Section 4.2. The Weierstrass Factorization theorem becomes considerably simpler if the zeroes of the various factors that are used to represent the regular function f satisfy certain conditions, that lead to the definition of canonical products and their genus, that we have introduced in Section 4.3. In Section 4.4 we prove the main results about estimates of canonical products of real, spherical and non real type. Finally we state and prove the Borel’s Theorem.
In all this procedure, the technical difficulties encountered are quite rel-evant, but we believe that the effort made to overcome them brought to a neat final result.
Part of this dissertation has appeared in [21] and in a chapter of the book [15].
Chapter 1
Preliminaries
This chapter 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., [17]) for all I ∈ S and for all J ∈ S such that I ⊥ J, we have that IJ ∈ S and 1, I, J, K = IJ form a basis for H with the same algebraic properties of the standard basis 1, i, j, k = ij. Moreover, the following result holds
Proposition 1.0.6. Let I and J be two elements in S, let hI, Ji ∈ 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 IJ can be decomposed
through the following formula:
IJ = −hI, J i + I × J.
One of the interesting features of the regular functions is a splitting prop-erty, which turns out to be a key tool in the theory:
Lemma 1.0.7 (Splitting Lemma). If f is a regular function on a slice do-main Ω then, for every I ∈ S and every J ⊥ I in S, there exist two holomor-phic functions F, G : ΩI → LI such that
fI(z) = F (z) + G(z)J
for all z ∈ ΩI.
The Splitting Lemma naturally leads to a version of the identity principle for regular functions (see [5]).
Theorem 1.0.8. (Identity Principle) Let f : Ω → H be a regular function on a slice domain Ω and let Zf = {q ∈ Ω : f (q) = 0} be the zero-set of f . If
there exists I ∈ S such that LI∩ Zf has an accumulation point, then f ≡ 0
on Ω.
Due to the non-commutativity of H, pointwise multiplication and compo-sition do not preserve slice regularity in general. Nevertheless, slice regularity is preserved for the following class of functions.
Definition 1.0.9. Let f : Ω → H be a slice regular function. We say that f is a slice preserving regular function if f (ΩI) ⊆ LI for all I ∈ S.
We recall moreover that a slice preserving regular function has a real power series expansion , see [15] for details.
We will now recall some results (proved in [5, 11, 17]) on the algebraic and topological structure of the zero-set of a regular function.
Theorem 1.0.10. Let f be a regular function on a symmetric slice domain Ω. If there exist 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 Zf consists of isolated points or isolated
2-spheres of the form S = x + yS, with x, y ∈ R, y 6= 0. It becomes now natural to give the following (see e.g. [11])
Definition 1.0.11. Let f be a regular function on a symmetric slice domain Ω. 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.
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 count-able. Moreover
Corollary 1.0.12. A zero q0 ∈ R of a regular function f defined on a sym-/
metric slice domain Ω 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 [14], that we will use in the sequel.
Proposition 1.0.13. Let f be a regular function on a symmetric slice do-main Ω. If f (LI) ⊆ LI for some I ∈ S then the zero-set Zf consists of
isolated points belonging to LI or isolated 2-spheres of type x + yS. As a
consequence, if f (q) 6= 0 for every q ∈ LI then f (q) 6= 0 for every q ∈ H.
Finally, if f (LI) ⊆ LI for all I ∈ S then the zero-set Zf consists of real
Since pointwise product of regular functions is not in general regular, inspired by the classic case of polynomials with coefficients in a non commu-tative algebra (see e.g. [25]), 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 [5] to the case of regular functions on a slice domain and it is based on the following result.
Lemma 1.0.14 (Extension Lemma). Let Ω be a symmetric slice domain and choose I ∈ S. If fI : ΩI → H is holomorphic, then setting
f (x + yJ ) = 1
2[fI(x + yI) + fI(x − yI)] + J I
2[fI(x − yI) − fI(x + yI)] extends fI to a regular function f : Ω → 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 Ω, let I, J ∈ S, with I ⊥ J, and choose holomorphic functions F, G, H, K : ΩI → LI such that for all z ∈ ΩI
fI(z) = F (z) + G(z)J, gI(z) = H(z) + K(z)J. (1.1)
Let fI ∗ gI : ΩI → 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. (1.2)
Using the Extension Lemma 1.0.14, the following definition is given in [5]: Definition 1.0.15. Let Ω ⊆ H be a symmetric slice domain and let f, g : Ω → H be regular. The function
f ∗ g(q) = ext(fI∗ gI)(q)
defined as the extension of (4.1.7) is called the regular product of f and g. We now recall a few properties of the regular multiplication, see [11] and[5]. Remark 1.0.16. Let Ω ⊆ H be a symmetric slice domain and let f, g : Ω → H be regular. If f (LI) ⊆ LI for all I ∈ S, then f ∗ g(q) = f (q) g(q). If
f (LI) ⊆ LI or g(LI) ⊆ LI for all I ∈ S, then f ∗ g(q) = g ∗ f (q).
An alternative expression of the regular product of two regular functions was introduced in [11]. In particular we have the next proposition proven in [5].
Theorem 1.0.17. Let f , g be regular functions on a symmetric slice domain Ω. For all q ∈ Ω, if f (q) = 0 then f ∗ g(q) = 0, else
f ∗ g(q) = f (q)g(f (q)−1qf (q))
Corollary 1.0.18. Let f , g be regular functions on a symmetric slice domain Ω. Then f ∗g(p) = 0 if and only if f (p) = 0 or f (p) 6= 0 and g(f (p)−1pf (p)) = 0.
Definition 1.0.19. Let f be a regular function on a symmetric slice domain Ω and suppose f splits on ΩI as in formula (1.1), fI(z) = F (z) + G(z)J . We
consider the holomorphic function
fIc(z) = F (¯z) − G(z)J (1.3) and define, according to the Extension Lemma 1.0.14, the regular conjugate of f by the formula
fc(q) = ext(fIc)(q) = ext(F (¯z) − G(z)J ).
Furthermore, the following definition is given under the same assumptions. Definition 1.0.20. The symmetrization of f is defined as
fs = f ∗ fc = fc∗ f = ext(fc
I ∗ fI)(q)) = ext(F (z)F (¯z) + G(z)G(¯z)). (1.4)
Remark 1.0.21. It turns out that fs(LI) ⊆ LI for all I ∈ S, and hence the
zero set of fs has the property described in Proposition 1.0.13. Moreover, if
f, g are regular functions on a symmetric slice domain, then is easy to verify that (f ∗ g)c = gc∗ fc and (f ∗ g)s = fsgs = gsfs.
The zero-sets of fc and fs are characterized in [11, 14] as follows.
Theorem 1.0.22. Let f be a regular function on a symmetric slice domain Ω. For all x, y ∈ R with x + yS ⊆ Ω, the zeros of the regular conjugate fc on x + yS are in one-to-one correspondence with those of f . Moreover,
the symmetrization fs 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 6= 0 and a real singleton {x} if y = 0.
Chapter 2
Infinite Products
2.1
The Quaternionic Logarithm
We consider an infinite product of quaternions q0 q1. . . qi. . . =
∞
Y
i=0
qi
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 2.1.1. The infinite productQ∞
i=0qi 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 nonvanishing factors tend to a finite limit which is different from zero.
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. In fact this is clear by
writing qi = Q−1i−1Qi. It is therefore preferable to write all infinite products
in the form
∞
Y
i=0
(1 + ai)
so that limi→∞ai = 0 is a necessary condition for their convergence.
Remark 2.1.2. In definition 2.1.1, the requirement that the partial products of non vanishing factors of Q∞
i=0(1 + ai) tend to a finite limit different from
zero finds its motivation in the complex case. In fact, assuming this require-ment, an infinite product Q∞
i=0(1 + ci) with ci ∈ C converges simultaneously
with the series P∞
i=0Log(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
∞
Y
i=0
(1 − 1/i) (2.1)
and consider the corresponding series
∞
X
i=0
Log(1 − 1/i). (2.2)
Denoting the partial sums of (2.2) by Sn, we have that Qn = eSn. Since
limn→∞Sn = −∞, then ∞ Y i=0 (1 − 1/i) = lim n→∞Qn = limn→∞e Sn = 0.
Therefore the infinite product (2.1) converges (to zero) while the series (2.2) diverges (to −∞).
To follow a similar approach to study the convergence of infinite prod-ucts in the quaternionic case, we need to introduce a logarithm on H. The exponential function on H is naturally defined as
exp(q) = eq = ∞ X n=0 qn n!
and it coincides with the complex exponential function on any complex plane LI.
Definition 2.1.3. Let Ω ⊆ H be a connected open set. We define a branch of the quaternionic logarithm (or simply a logarithm) on Ω a function f : Ω → H such that for every q ∈ Ω
ef (q) = q.
First of all, since exp(q) never vanishes, we must suppose that 0 /∈ Ω. If we set
Iq=
Im(q)/|Im(q)| if q ∈ H \ R any element of S otherwise
we have that for every q ∈ H \ {0} there exists a unique θ ∈ [0, π] such that q = |q|eθIq. Moreover θ = arccos(Re(q)/|q|). The function arccos(Re(q)/|q|)
will be called the principle 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 2.1.4. Let ln be the natural real logarithm. For every q ∈ H \ (−∞, 0], we define the principal quaternionic logarithm (or simply principal logarithm) of q as
Log(q) = ln |q| + arccos Re(q) |q|
Iq.
Proposition 2.1.5. The principal logarithm is a continuous function on H \ (−∞, 0].
Proof. The function q 7→ ln |q| is clearly continuous on H \ {0}. The function q 7→ arccosRe(q)|q| |Im(q)|Im(q) is defined and continuous for every q ∈ H \ R. Moreover, since for every strictly positive real r, and for every q /∈ R, we have lim q7→rarccos Re(q) |q| Iq = 0
then q 7→ arccosRe(q)|q| Iq is continuous on H \ (−∞, 0]. We observe that, if
s, instead, is a strictly negative real number, then lim q7→sarccos Re(q) |q| Iq
does not exist.
Proposition 2.1.6. The principal quaternionic logarithm coincides with the principal complex logarithm on any complex plane LI, with I ∈ S.
Proof. Let I ∈ S. First of all notice that, since LI = L−I, every quaternion
q ∈ LI \ 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 2.1.4 we have that Log(x + yI) = lnpx2+ y2+ arccos x
px2+ y2
! y |y|I and
Log(x − y(−I)) = lnpx2+ y2+ arccos x
px2+ y2
! (−y)
|y| (−I) coincide.
Now we can consider the restriction of the principal quaternionic logarithm to LI:
Log(x + yI) = ln(px2+ y2) + Arg
H(x + yI)
y |y|I.
We set φ(x + yI) = ArgH(x + yI)|y|y . The function ArgH(x + yI), with values in [0, π), 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 LI, and
φ(x + yI) = ArgH(x + yI). On the other hand if y < 0, we obtain that x + yI belongs to the lower half plane of LI, and φ(x + yI) = −ArgH(x + yI).
Therefore φ(x+yI) coincide with the imaginary part of the principal complex logarithm of LI, 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.
Lemma 2.1.7. Let n ∈ N and let θ1, ..., θn ∈ [0, π) be such that
Pn
i=1θi < π.
Then for every set {I1, . . . In} ⊆ S we have that
ArgH(eθ1I1· · · eθnIn) ≤
n
X
i=1
θi.
Proof. By induction on n. For n = 1 the thesis is straightforward. Let θ1, θ2, ..., θn ∈ [0, π) be such that
Pn
i=1θi < π. Let φ ∈ [0, π) be the principal
quaternionic argument of the product eθ1I1· · · eθn−1In−1 and let J ∈ S be such
that eθ1I1· · · eθn−1In−1 = eφJ. Consider the product eφJeθnIn:
eφJeθnIn = cosφcosθ
n+ cosφsinθnIn+ sinφcosθnJ + sinφsinθnJ In.
From the formula (1.0.6) we have that J In = −hJ, Ini + J × Inand hence we
obtain that
cos(ArgH(eφJeθnIn)) = Re eφJeθnIn = cosφcosθ
n− sinφsinθnhJ, Ini.
Since hJ, Ini ≤ |J||In| = 1 and sinφsinθn≥ 0 we obtain the inequality
cos(ArgH(eφJeθnIn)) ≥ cosφcosθ
n− sinφsinθn= cos(φ + θn).
The function cos(x) decreases in [0, π] and hence we have ArgH(eφJeθnIn) ≤ φ + θ
n.
2.2
Infinite Products of quaternions
If {ai}i∈N ⊆ H is a sequence such that the series
P∞
i=0|ai| converges, we know
that the series P∞
i=0ai converges too. With this in mind, we can prove the
following result.
Theorem 2.2.1. Let {ai}i∈N ⊆ H be a sequence. If the seriesP ∞
i=0|Log(1 +
ai)| converges, then the product
Q∞
i=0(1 + ai) converges too.
Proof. The function Log(1 + ai) is not defined if 1 + ai belongs to (−∞, 0].
The convergence of the series P∞
i=0|Log(1 + ai)| implies that the sequence
{ai}i∈N tends to zero. We can therefore suppose that 1 + ai ∈ (−∞, 0]. For/
every i ∈ N, let θi ∈ [0, π) and let Ii ∈ S be such that 1 + ai = |1 + ai|eθiIi.
Then Log(1 + ai) = ln |1 + ai| + θiIi and therefore
|ln |1 + ai|| ≤ |Log(1 + ai)| (2.3)
and
|θi| = |θiIi| ≤ |Log(1 + ai)| (2.4)
for every i ∈ N.
We have to prove that the sequence of the partial products Qn=
n
Y
i=0
(1 + ai)
tends to a finite limit (which is different from zero). Since real number commute with quaternions, for every n ∈ N we have that
n Y i=0 (1 + ai) = n Y i=0 |1 + ai| n Y i=0 eθiIi.
By hypothesis and by inequality (2.3) the seriesP∞
i=0ln |1 + ai| is convergent
and hence the infinite product Q∞
i=0|1 + ai| is convergent too (see Remark
2.1.2). Then it is sufficient to prove that the sequence Rn =
Qn
i=0eθiIi ⊆
∂B(0, 1) = {q ∈ H : |q| = 1} is convergent. We will show that {Rn}n∈N is a
Cauchy sequence. For n > m ∈ N, we have: |Rn− Rm| = n Y i=0 eθiIi − m Y i=0 eθiIi = m Y i=0 eθiIi n Y i=m+1 eθiIi − m Y i=0 eθiIi = = m Y i=0 eθiIi n Y i=m+1 eθiIi− 1 = n Y i=m+1 eθiIi − 1 ≤ ArgH n Y i=m+1 eθiIi ! ,
where the last equality holds due to the fact that Qn
i=m+1e θiIi
= 1. In-equality (2.4) implies that the series P∞
i=0θi is convergent. Therefore the
sequence Sn =
Pn
i=0θi 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
X
i=m+1
θi < .
In particular for < π, by lemma 2.1.7, we have that
ArgH n Y i=m+1 eθiIi ! ≤ n X i=m+1 θi < .
The assertion follows.
The following proposition will be useful in the sequel, while studying the convergence of the infinite products.
Proposition 2.2.2. Let Log be the principal quaternionic logarithm. Then lim
q→0q −1
Log(1 + q) = 1. (2.5)
Proof. Let {qn}n∈N = {xn+ ynIn}n∈N be a sequence such that qn → 0 for
n → ∞. We can assume without loss of generality that yn > 0. We consider
the following norm
|(xn+ ynIn)−1Log(1 + xn+ ynIn) − 1|. (2.6)
Easy calculations show that both the real part and the norm of the imaginary part of (xn+ynIn)−1Log(1+xn+ynIn) do not depend on the complex direction
In. Therefore we can fix In = I0 for every n ∈ N. Hence we consider
|(xn+ ynI0)−1Log(1 + xn+ ynI0) − 1| (2.7)
and we observe that (xn + ynI0)−1Log(1 + xn+ ynI0) ∈ LI0 for all n ∈ N.
Since we are in the complex plane LI0, by the same arguments used in the
complex case, the sequence (2.7) tends to zero. Corollary 2.2.3. The series P∞
n=0|Log(1 + an)| converges at the same time
as the simpler series P∞
Proof. If either the series P∞
n=0|Log(1 + an)| or the series
P∞
n=0|an|
con-verges, we have limn→∞an = 0. By Proposition 2.2.2 we have that for a
given > 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 2.2.1 and corollary 2.2.3, we end this section by finding a useful sufficient condition for the convergence of quaternionic infinite products.
Theorem 2.2.4. Let {an}n∈N ⊆ H. A sufficient condition for the
conver-gence of the product Q∞
n=0(1 + an) is the convergence of the series
P∞
n=0|an|.
2.3
Infinite Products of functions defined on
H
LetQ∞
i=0gi(q) be an infinite product whose factors are functions of a
quater-nionic variable defined on an open set Ω ⊆ H. We want to explain what we mean by saying that Q∞
i=0gi(q) converges uniformly on compact sets of
Ω. If all functions {gi}i∈N do not vanish on Ω, then the definition is
obvi-ously given (according with definition 2.1.1) by requiring that the sequence of partial products Gn(q) = n Y i=0 gi(q)
converges uniformly on compact sets of Ω 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:
Definition 2.3.1. Let {gi}i∈N be a sequence of functions defined on an open
set Ω ⊆ H. We will say that
∞
Y
i=0
gi(q)
converges uniformly on compact sets of Ω to a function P : Ω → H if, for any compact K ⊆ Ω:
(i) there exists an integer NK such that gi 6= 0 on K for all i ≥ NK,
(ii) the residual product
∞
Y
i=NK
gi(q)
(whose factors do not vanish on K) converges uniformly on K to a never vanishing function PNK,
(iii) we have, for all q ∈ K,
P (q) = "NK−1 Y i=0 gi(q) # PNK(q). Remark 2.3.2. IfQ∞
i=0gi(q) converges uniformly on compact sets of Ω ⊆ H,
by the previous definition we have that limi→∞gi = 1 uniformly on compacts
set of Ω. Indeed, for any compact K ⊆ Ω, taking any j > NK, we can set,
Gj(q) = j Y i=NK gi(q) and write gj = G−1j−1Gj.
Hence limj→∞gj = limi→∞G−1j−1Gj = limj→∞G−1j−1limj→∞Gj = 1.
For this reason we write the infinite products in the form
∞
Y
i=0
(1 + fi(q))
so that the uniform convergence on compact sets of the sequence of func-tions {fi}i∈N to the zero function is a necessary condition for studying their
convergence.
Remark 2.3.3. Let Q∞
i=0(1 + fi(q)) be an infinite product. If the sequence
of functions {fi}i∈N converges uniformly on compact sets of Ω ⊆ H to the
zero function, then condition (i) of definition 2.3.1 is automatically fulfilled. When {fi}i∈N is a sequence of C∞ function, the following result holds:
Proposition 2.3.4. Let {fi}i∈N be a sequence of C∞ functions defined on
an open set Ω, converging uniformly to the zero function on compact sets of Ω. If the infinite product Q∞
i=0(1 + fi(q)) converges uniformly on compact
Proof. Set q ∈ Ω and let K ⊆ Ω be a compact neighborhood of q. By the assumption on the sequence {fi}i∈N, we have that (see definition 2.3.1) there
exists an integer NK such that the residual product ∞
Y
i=NK
(1 + fi(q))
converges (uniformly) on K to a never vanishing function, that we call PNK.
Hence PNK is the limit of a convergent sequence of C
∞ functions and so it is
C∞. Then we can write P as a finite product of C∞ functions: P (q) = "NK−1 Y i=0 (1 + fi(q)) # PNK(q).
Therefore P is C∞. Moreover, on K, the zero set of P coincides with the zero set of QNK−1 i=0 (1 + fi(q)). Let ∞ Y i=0 (1 − qa−1i ) (2.8)
with {ai}i∈N ⊆ H \ {0}. Let q ∈ H be a fixed quaternion. By Theorem 2.2.4,
a sufficient condition for the convergence of (2.8) is the convergence of the series P∞
i=0|q|/|ai|. As a consequence, a sufficient condition for the uniform
convergence on compact sets of H of (2.8) is the convergence of the series P∞
i=01/|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 2.3.5. Let {an}n∈N ⊆ H \ {0} be such that limn→∞|an| = ∞.
Then there exist certain integers {mn}n∈N such that the infinite product ∞
Y
n=0
(1 − qa−1n )eqa−1n +12(qan−1)2+...+mn1 (qa−1n )mn (2.9)
converges uniformly on compact sets of H. In particular we can choose mn=
n for all n ∈ N.
Proof. Let n ∈ N. Let pn(x) = x + 12x2 + . . . +m1nxmn where mn∈ N. If we
prove the absolute convergence of the series with general term rn(q) = Log((1 − qa−1n )e
(where Log is the principal logarithm so that rn(0) = 0) then, thanks to
Theorem 2.2.1, we conclude the proof. For every q ∈ H, the quaternions 1 − qa−1n and pn(qa−1n ) lie on the same complex plane. Thus, since Log(1) =
pn(0) = 0, we can write rn(q) in the form
rn(q) = Log(1 − qa−1n ) + pn(qa−1n ).
Now, for a given R > 0, we consider only the terms with |an| > R. In the
ball B(0, R) we have that |qa−1n | < 1 and so Log(1 − qa−1
n ) can be expanded in a Taylor series Log(1 − qa−1n ) = −qa−1n − 1 2(qa −1 n ) 2− 1 3(qa −1 n ) 3 − . . . .
Then rn(q) has the representation
rn(q) = − 1 mn+ 1 (qa−1n )mn+1− 1 mn+ 2 (qa−1n )mn+2− . . .
and we obtain easily the estimate |rn(q)| ≤ 1 mn+ 1 R |an| mn+1 1 − R |an| −1 .
As in the complex case, to conclude the proof we are left to show the con-vergence of the series
∞ X n=0 1 mn+ 1 R |an| mn+1 1 − R |an| −1 . (2.11)
The fact that limn→∞(1 − R/|an|)−1 = 1 implies that the series (2.11)
con-verges if the series
∞ X n=0 1 mn+ 1 R |an| mn+1
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.
We can generalize this result as follows.
Theorem 2.3.6. Let {fn}n∈N be a sequence of functions that converges
uni-formly on compact sets of H to the zero function. Then the infinite product
∞ Y n=0 (1 − fn(q))efn(q)+ 1 2fn(q) 2+...+1 nfn(q) n (2.12) converges uniformly on compact sets of H.
2.4
Infinite ∗-Products of regular functions
The results proved in the last section (Theorem 2.3.5 and Theorem 2.3.6) concern infinite products that, in general, do not converge to a regular func-tion. In order to study the factorization of zeros of a regular function we must consider the infinite regular products (or infinite ∗−products). 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 2.4.1. Let {fi}i∈N be a sequence of regular functions defined on a
symmetric slice domain Ω. Let K be a symmetric compact set such that K ⊆ Ω. We suppose that there exists an integer NK such that 1 + fi 6= 0 on
K for all i ≥ NK. Set
FNK,m(q) = m Y ∗ i=NK (1 + fi(q)).
Then for all m ≥ NK and for any q ∈ K
FNK,m(q) = m Y i=NK (1 + fi(Ti(q))) 6= 0 where Tj(q) = FNK,j−1(q) −1qF
NK,j−1(q) for j > NK and Tj(q) = q for j =
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) 6=
0. Suppose the assertion is true for m = NK, · · · , n − 1. We can write
FNK,n(q) = FNK,n−1(q) ∗ (1 + fn(q)). (2.13)
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 1.0.17, since FNK,n−1(q) 6= 0 by induction
hypothesis, we have that formula (2.13) 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 FNK,n−1(q) = n−1 Q i=NK (1 + fi(Ti(q))) and hence FNK,n(q) = n−1 Y i=NK (1 + fi(Ti(q))) (1 + fn(Tn(q))) = n Y i=NK (1 + fi(Ti(q))).
In analogy with definition 2.3.1 we give the following
Definition 2.4.2. Let {fi}i∈N be a sequence of regular functions defined on
a symmetric slice domain Ω. The infinite ∗-product
∞
Y ∗
i=0
(1 + fi(q))
is said to converge uniformly on compact sets of Ω to a function F : Ω → H if, for any symmetric, compact set K ⊆ Ω:
(i) there exists an integer NK such that 1 + fi 6= 0 on K for all i ≥ NK,
(ii) the residual product
∞
Y ∗
i=NK
(1 + fi(q))
converges uniformly on K (i.e. the sequence n FNK,m(q) = Y ∗ m i=NK(1 + fi(q)) o m≥NK
converges uniformly on K) to a never vanishing function FNK,
(iii) we have, for all q ∈ K
F (q) = NK−1 Y ∗ i=0 (1 + fi(q)) ∗ FNK(q).
It is important to establish that, for any given sequence of regular functions, the infinite regular product and the infinite product converge simultaneously. Theorem 2.4.3. Let {fn}n∈N be a sequence of regular functions define on a
symmetric slice domain Ω. The infinite ∗−product
∞
Y ∗
n=0
(1 + fn(q))
converges uniformly on compact sets of Ω if and only if the infinite product
∞
Y
n=0
(1 + fn(q))
Proof. It is enough to prove the convergence on symmetric compact sets of Ω. Let K 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 2.4.1 the infinite
∗−product ∞ Y ∗ i=NK (1 + fi(q))
converges simultaneously with the infinite product
∞
Y
i=NK
(1 + fi(Tn(q))).
Recalling that Re(Tj(q)) = Re(q) and |Im(Tj(q)| = |Im(q)| for all j ≥ NK,
we have that 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 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)
kf − gkA= sup q∈A
|f (q) − g(q)|. Thus we have
Lemma 2.4.4. Let {hn}n∈N be a sequence of regular functions defined on
a symmetric slice domain Ω ⊆ H converging uniformly to a function h on compact sets of Ω. Then h is regular on Ω.
Proof. Given any pair of orthogonal vectors I and J in S, consider the or-thogonal basis 1, I, J, IJ . We can write hI(x + yI) = h(z) as
h(z) = h0(z) + h1(z)I + h2(z)J + h3(z)IJ = F (z) + G(z)J
with ΩI = Ω∩LI and F, G : ΩI → LI. If we prove that F, G are holomorphic,
then by definition 0.0.1 we conclude the proof. By the Splitting Lemma we have that, for all n ∈ N, the two functions Fn, Gn: ΩI → LI such that
hn(z) = Fn(z) + Gn(z)J
are holomorphic on ΩI. For z ∈ ΩI we have
Consider any compact subset K ⊆ Ω and set KI = K ∩ LI. Since by
hypoth-esis we have that limn→∞khn− hkK = 0, we obtain limn→∞kFn− F kKI =
limn→∞kGn − GkKI = 0, too. Since the limit function of a sequence of
holomorphic functions uniformly convergent on compact sets is itself holo-morphic, we can conclude that the functions F and G are holomorphic on ΩI. As we said, this completes the proof.
We are able to prove the announced result on the regularity of the limit function of a converging infinite regular product.
Proposition 2.4.5. Let {fn}n∈N be a sequence of regular functions defined
on a symmetric slice domain Ω. If the infinite regular product
∞
Y ∗
n=0
(1 + fn(q))
converges uniformly on compact sets of Ω to a function F , then F is regular on Ω.
Proof. By definition 2.4.2, for any compact set K ⊆ H there exists an integer NK such that 1 + fn 6= 0 on K if n ≥ NK. By Lemma 2.4.4 the residual
∗−product ∞ Y ∗ n=NK (1 + fn(q)) (2.14)
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 F (q) = "NK−1 Y ∗ n=0 (1 + fn(q)) # ∗ FNK(q)
and hence it is regular on K. Moreover, by corollary 1.0.18 and lemma 2.4.1, the zero set of F on K coincides with the zero set of the finite ∗−product
NK−1
Q ∗
n=0
(1 + fn(q)).
From now on, unless explicitly stated, we will always consider the case in which {fn}n∈Nis a sequence of entire regular functions, i.e. the case in which
Ω = H. Consider an infinite ∗-product such as
∞
Y ∗
n=0
(1 − qa−1n ) (2.15)
with {an}n∈N⊆ H\{0}. In order to introduce convergence-producing regular
factors that preserve the regularity of the product (2.15) we need a few results about series expansion of exponential type functions.
Chapter 3
The Weierstrass Factorization
Theorem
3.1
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 2.3.5, when dealing with regular infinite ∗-products and when using convergence-producing regular factors.
Lemma 3.1.1. Let p(x) ∈ C∞(R) and let f (x) = ep(x). If f(k) and p(l)
denote, respectively, the k-th derivative of f and the l-th derivative of p, then f(m+1)(x) = m X i=0 m i f(i)(x)p(m+1−i)(x).
Proof. The proof is made by induction on m. If m = 0 we have f(1)(x) =
ep(x)p(1)(x) = f (x)p(1)(x) and hence the assertion is true. Suppose now that
the assertion is true for m = 0, · · · , n. Then f(n+1)(x) = d dxf (n)(x) = d dx n−1 X i=0 n − 1 i f(i)(x)p(n−i)(x) = = n−1 X i=0 n − 1 i f(i)(x)p(n−i+1)(x) + n−1 X i=0 n − 1 i f(i+1)(x)p(n−i)(x) = = n−1 X i=0 n − 1 i f(i)(x)p(n−i+1)(x) + n X j=1 n − 1 j − 1 f(j)(x)p(n+1−j)(x) = 19
= f (x)p(n+1)(x)+ n−1 X i=1 n − 1 i +n − 1 i − 1 f(i)(x)p(n−i+1)(x)+f(n)(x)p(1)(x) = =n 0 f (x)p(n+1)(x) + n−1 X i=1 n i f(i)(x)p(n+1−i)(x) +n n f(n)(x)p(1)(x) = = n X i=0 n i f(i)(x)p(n+1−i)(x).
Lemma 3.1.2. For n ∈ N \ {0}, let pn(x) = x + 12x2 + . . . + n1xn. If p(l)
denotes the l-th derivative of p, then for all x ∈ R we have
p(i)n (x) = ( (i − 1)! + i!x + . . . +(i+j−1)!j! xj+ . . . + (n−1)! (n−i)!x n−i if i ≤ n 0 if i > n. In particular p(i)n (0) = (i − 1)! if i ≤ n 0 if i > n.
Proof. By induction. The assertion is trivial to prove if i > n. To prove the assertion for 1 ≤ i ≤ n we use the finite induction. If i = 1 we have that p(1)n (x) = 1 + x + x2 + · · · + xn−1 and hence the assertion is true. Suppose
now that the assertion is true for i < n. Then we have p(i+1)n (x) = d dx(p (i) n (x)) = i!+· · ·+ (i + j − 1)! j! jx j−1+. . .+(n − 1)! (n − i)!(n−i)x n−i−1 = = i! + · · · + (i + (j − 1))! (j − 1)! x j−1+ · · · + (n − 1)! (n − (i + 1))!x n−(i+1).
Using lemma 3.1.1, in the particular case in which p is the polynomial pn
which appears in lemma 3.1.2, we are now able to find an expression for all the derivatives of the function fn(x) = epn(x).
Lemma 3.1.3. Let pn(x) = x + 12x2 + . . . + 1nxn con n ∈ N \ {0}. Let
fn(x) = epn(x). If f (k)
n denotes the k-th derivative of fn, then
(a) for m = 0, . . . , n
fn(m)(0)
(b) for all m ≥ n fn(m)(0) m! = 1 m m−1 X i=m−n fn(i)(0) i! . Proof. Thanks to lemma 3.1.1 we have
fn(m)(0) = m−1 X i=0 m − 1 i fn(i)(0)p(m−i)n (0) (3.1) and, by lemma 3.1.2, p(m−i)n (0) = (m − i − 1)! if m − i ≤ n 0 if m − i > n.
To prove the assertion (a) we use finite induction on m. If m = 0 we have that fn(0) = epn(0) = e0 = 1 = 0! and hence the assertion is true. Suppose now
that the assertion is true for m = 0, 1, . . . , i − 1 with i ≤ n. By the induction hypothesis we have that f
(j) n (0)
j! = 1 for j < i and hence, by substitution,
fn(i)(0) i! = 1 i! i−1 X j=0 (i − 1)! (i − 1 − j)!(i − j − 1)! = 1 i! i−1 X j=0 (i − 1)! = 1.
Therefore (a) is proved. To prove (b) we consider m ≥ n + 1. By direct substitution, we get fn(m)(0) m! = 1 m! m−1 X i=m−n (m − 1)! i!(m − i − 1)!f (i) n (0)(m − i − 1)! = 1 m m−1 X i=m−n fn(i)(0) i! and conclude the proof.
Remark 3.1.4. By lemma 3.1.3 we also obtain that fn(m)(0)
m! ≥ 0 f or all m ∈ N.
Lemma 3.1.5. Let pn(x) = x + 12x2 + . . . + 1nxn with n ∈ N \ {0}. Let
(a) for m = 0 . . . n − 1 fn(m)(0) m! − fn(m+1)(0) (m + 1)! = 0 (b) for all m ≥ n fn(m)(0) m! − fn(m+1)(0) (m + 1)! = 1 m + 1 fn(m−n)(0) (m − n)!.
Proof. The statement (a) is true thanks to lemma 3.1.3 (a). By lemma 3.1.3 (b) we have , for all m ≥ n,
fn(m+1)(0) (m + 1)! = 1 m + 1 m X i=m+1−n fn(i)(0) i! = = 1 m + 1 " fn(m)(0) m! + m−1 X i=m−n fn(i)(0) i! − fn(m−n)(0) (m − n)! # = = 1 m + 1 " fn(m)(0) m! + m fn(m)(0) m! − fn(m−n)(0) (m − n)! # = = f (m) n (0) m! − 1 m + 1 fn(m−n)(0) (m − n)!. Hence fn(m)(0) m! − fn(m+1)(0) (m + 1)! = 1 m + 1 fn(m−n)(0) (m − n)!. (3.2)
These estimates will be important to prove Proposition 3.1.7.
Lemma 3.1.6. Let pn(x) = x + 12x2 + . . . + 1nxn with n ∈ N \ {0}. Let
fn(x) = epn(x). Then for all m ≥ n we have
0 ≤ f (m) n (0) m! − fn(m+1)(0) (m + 1)! ≤ 1 n + 1. Proof. By Remark 3.1.4 we have that
1 m + 1
fn(m−n)(0)
and hence, by lemma 3.1.5 fn(m)(0)
m! −
fn(m+1)(0)
(m + 1)! ≥ 0 f or all m ∈ N Then, the sequence
n
fn(j)(0)
j!
o
j∈N decreases and hence
fn(j)(0)
j! ≤
fn(1)(0)
1! = 1 f or all j ∈ N. Therefore we have that
0 ≤ 1 m + 1 fn(m−n)(0) (m − n)! ≤ 1 m + 1 Since m ≥ n, we obtain 0 ≤ 1 m + 1 fn(m−n)(0) (m − n)! ≤ 1 n + 1. The thesis follows directly from lemma 3.1.5.
As announced we have the following result.
Proposition 3.1.7. Let n ∈ N and x ∈ R. We set hn(x) = (1 − x)ex+ 1 2x 2+...+1 nx n . (3.3)
Then there exist ck ∈ R such that
hn(x) = 1 − ∞ X k=0 ckxk+n+1 (3.4) and 0 ≤ ck≤ 1 n + 1 for all k ∈ N.
Proof. If we write formula (3.3) using the expansion of fn(x) = ex+
1 2x 2+...+1 nx n we obtain that hn(x) = ∞ X m=0 fn(m)(0) m! x m− x ∞ X m=0 fn(m)(0) m! x m =
= 1 + ∞ X m=1 fn(m)(0) m! − fn(m−1)(0) (m − 1)! ! xm. By lemma 3.1.5 (a) we have that
hn(x) = 1 + ∞ X m=n+1 fn(m)(0) m! − fn(m−1)(0) (m − 1)! ! xm = = 1 + ∞ X k=0 fn(k+n+1)(0) (k + n + 1)!− fn(k+n)(0) (k + n)! ! xk+n+1 = = 1 − ∞ X k=0 fn(k+n)(0) (k + n)! − fn(k+n+1)(0) (k + n + 1)! ! xk+n+1. If we set ck= f (k+n) n (0) (k+n)! − fn(k+n+1)(0)
(k+n+1)! then, by lemma 3.1.6 (b), since k + n ≥ n
for all k ∈ N, we obtain the thesis.
At this point we have all the ingredients to prove an analog of Theorem 2.3.5 for the infinite regular products.
Theorem 3.1.8. Let {an}n∈N⊆ H\{0} be a sequence such that limn→∞|an| =
∞. For all an, let In ∈ S be such that an ∈ LIn. Then, for all n ∈ N, there
exist mn ∈ N and a never vanishing entire regular function gn, whose
re-striction to LIn is the function e
za−1n +...+mn1 zmna−mnn , such that the infinite
∗-product ∞ Y ∗ n=0 (1 − qa−1n ) ∗ gn(q) (3.5)
converges uniformly on compact sets of H. In particular we can choose mn=
n for all n ∈ N.
Proof. For all n ∈ N, let gn be as in the hypothesis and set Vn(q) =
(1 − qa−1n ) ∗ gn(q). Theorem 2.4.3 yields that the ∗−product (3.5) converges
simultaneously with the product
∞
Y
n=0
Vn(q). (3.6)
Thanks to Theorem 2.2.4 the product (3.6) converges uniformly on compact sets of H if the series ∞
X
n=0
converges uniformly on compact sets of H. Let K ⊆ H be a compact set. Let R > 0 be such that K ⊆ B(0, R) and let n ∈ N be the integer such that |an| > R for all n ≥ n. Let vn(z) be the restriction of Vn to the complex
plane LIn. Hence
vn(z) = (1 − za−1n )e
za−1n +...+n1zna−nn
and we can estimate the coefficients of the power series Taylor expansion of vn at the point 0 by using Proposition 3.1.7. Since the regular extension of
vnis unique by the Identity Principle (Theorem 1.0.8), the coefficients of the
power series expansion of Vn are the same of vn. Therefore by Proposition
(3.1.7) we have that: Vn(q) = 1 − ∞ X k=0 ckqk+n+1a−(k+n+1)n
for every q ∈ H, where ck∈ R are such that
0 ≤ ck≤
1 n + 1 for all k ∈ N. Then
|1 − Vn(q)| = ∞ X k=0 ckqk+n+1a−(k+n+1)n ≤ ∞ X k=0 ck |q| |an| (k+n+1)
and, since ck ≤ n+11 , we have
|1 − Vn(q)| ≤ ∞ X k=0 1 (n + 1) |q| |an| n+1+k ≤ 1 n + 1 |q| |an| n+1 ∞ X k=0 |q| |an| k! . The factor on the right
∞ X k=0 |q| |an| k
is a geometric series and its ratio is strictly smaller than 1 when n ≥ ¯n, since q ∈ K ⊆ B(0, R) and |an| > R. Hence the power series is convergent to the
value
1 − |q| |an|
and we obtain |1 − Vn(q)| ≤ 1 n + 1 |q| |an| n+1 1 − |q| |an| −1 . Since limn→∞ 1 − |a|q| n| −1
= 1 and the series
∞ X n=0 1 n + 1 |q| |an| n+1 (3.7)
converges on K, also the series
∞
X
n=0
|1 − Vn(q)|
converges 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 LIn that has no zeros. Thanks to Proposition
1.0.13, 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 [4, 5]. Theorem 3.1.9. Let f be a regular function on a symmetric slice domain Ω ⊆ H. For all x, y ∈ R such that x + yS ⊆ Ω, there exist b, 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 3.1.10. Let f be a regular function on a symmetric slice do-main Ω ⊆ H. Let K ⊆ Ω be a symmetric compact set. For every I ∈ S, p ∈ H, and R > 0 such that
f|LI(KI) ⊆ B(p, R)
we have
Proof. Let I, p, R be as in the hypothesis. Remark that K symmetric means explicitly that
K = [
x+yI∈K
x + yS and hence that
f (K) = [
x+yI∈K
f (x + yS).
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 3.1.9 there exist b, c ∈ H such that
f (x + yJ ) = b + J c for all J ∈ S. Since f|LI(KI) ⊆ B(p, R) we have that
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).
We are now able to prove the following result, concerning the uniform con-vergence of sequences of regular functions.
Proposition 3.1.11. Let {fn}n∈N be a sequence of regular functions defined
on a symmetric slice domain Ω ⊆ H. If the sequence converges uniformly on compact sets of Ω to a regular function f , then
(a) the sequence of the regular conjugates {fnc}n∈N converges uniformly on compact sets of Ω to the regular conjugate fc;
(b) the sequence of symmetrizations {fs
n}n∈N converges uniformly on
com-pact sets of Ω to the symmetrization fs.
Proof. Let K ⊆ Ω be a symmetric compact set. By hypothesis lim
We have to prove that lim n→+∞kf c n− f ck K = 0 and lim n→+∞kf s n− f sk K = 0.
Let I, J ∈ S be such that J ⊥ I. By the Splitting Lemma there exist F, G : LI → LI holomorphic such that
fI(z) = F (z) + G(z)J
and, for all n ∈ N, there exist Fn, Gn: LI → LI holomorphic such that
fn,I(z) = Fn(z) + Gn(z)J.
Therefore, as in lemma 2.4.4, if we set KI = K ∩ LI, formula (3.8) implies
lim
n→+∞kFn− F kKI = 0 and n→+∞lim kGn− GkKI = 0. (3.9)
Thanks to the continuity of the conjugate function on the complex plane LI
we obtain that lim
n→+∞kFn− F kKI = 0 and n→+∞lim kGn− GkKI = 0. (3.10)
By definition 1.0.19, we have that
fIc(z) = F (¯z) − G(z)J and fn,Ic (z) = Fn(¯z) − Gn(z)J
for all n ∈ N. By (3.9) and (3.10) we can easily prove that fn,Ic converges to
fc I uniformly on KI i.e lim n→+∞kf c n,I− f c IkKI = 0. By Proposition 3.1.10 we have kfnc− fckK ≤ 2kfn,Ic − f c IkKI
for all n ∈ N and hence
lim n→+∞kf c n− f ck K = 0.
Since by definition 1.0.20, we have that
fIs(z) = F (z)F (¯z) + G(z)G(¯z) and fn,Is (z) = Fn(z)Fn(¯z) + Gn(z)Gn(¯z)
We now turn our attention back to infinite ∗-product, and state:
Proposition 3.1.12. 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 3.1.8, such that the infinite ∗-product
∞
Y ∗
n=0
(1 − qa−1n ) ∗ gn(q)
converges uniformly on compact sets of H to a regular function h. Then the infinite product ∞ Y n=0 q2 |an|2 − 2qRe(an) |an|2 + 1 eq 2Re(an) |an|2 +...+ 1 nq n 2Re(ann) |an|2n (3.11)
converges uniformly on compact sets of H to the symmetrization hs of h.
Proof. Set hk(q) = k Y ∗ n=0 (1 − qa−1n ) ∗ gn(q).
By hypothesis the sequence {hk}k∈N converges uniformly on compact sets of
H to h. By Remark 1.0.21 the symmetrization of a ∗-product is the product of the symmetrization of ∗-factors and hence, for all k ∈ N,
hsk(q) =
k
Y
n=0
(1 − qa−1n )s gsn(q).
By Proposition 3.1.11, the sequence {hsk}k∈N converges uniformly on compact
sets of H to the symmetrization hs of h, i.e. ∞ Y n=0 (1 − qa−1n )s gns(q) = " ∞ Y ∗ n=0 (1 − qa−1n ) ∗ gn(q) #s . (3.12)
We want now to write explicitly the functions gs
n. For all n ∈ N, if an ∈
LIn, the function gn is defined as the unique regular extension of fn(z) =
eza−1n +...+1nzna−nn (with z ∈ L
In) to H. Then for all z ∈ LIn we have
fns(z) = fn(z) fnc(z) = e za−1n +...+1nzna−nn ezan−1+...+n1znan−n = eza−1n +...+n1zna −n n +zan−1+...+1nznan−n = ez 2Re(an) |an|2 +...+ 1 nz n 2Re(ann) |an|2n . Hence we obtain gns(q) = eq 2Re(an) |an|2 +...+ 1 nq n 2Re(ann) |an|2n
for all q ∈ H. Observing that (1 − qa−1n )s =
q2 |an|2 − 2qRe(an) |an|2 + 1 we obtain the thesis.
3.2
On the multiplicity of zeros of entire
reg-ular 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) 6= 0. For the non real zeros of f , a factorization result is not so trivial. The next proposition shows how to obtain it.
Proposition 3.2.1. Let f be an entire regular function. Let α ∈ H \ R be a zero of f . Let Sα = Re(α) + |Im(α)|S. If f 6≡ 0 then there exist
t ∈ N ∪ {0}, t quaternions α1, ..., αt∈ Sα where αi 6= αi+1 for i = 1, . . . , t − 1
and m ∈ N ∪ {0} such that
f (q) = q2− 2qRe(α) + |α|2m g(q) ∗ t Y ∗ i=1 (q − αi) ! (3.13) where g is an entire regular function such that g(α) 6= 0.
Proof. If f ( ¯α) 6= 0 then, by corollary 1.0.12, 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 : LI → LI such that for all z ∈ LI
fI(z) = R(z) + S(z)J.
Since f (α) = f ( ¯α) = 0 we have that R(α) = R( ¯α) = 0 and S(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)sV (z) with U and V which do vanish neither in α nor in ¯α. Let m = min{p, q, r, s} and let U0(z) = (z − α)p−m(z − ¯α)q−mU (z) and V0(z) = (z − α)r−m(z −
¯
α)s−mV (z). Since L
I is a commutative subspace of H we obtain
fI(z) = (z−α)m(z− ¯α)m[U0(z)+V0(z)J ] = z2− 2zRe(α) + |α|2
m
[U0(z)+V0(z)J ].
Since f is the unique regular extension of fI it is of the form:
f (q) = q2− 2qRe(α) + |α|2m 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 1.0.12, the 2-sphere Sα is not
a spherical zero of h. If h(α) 6= 0 then we have t = 0. Otherwise we can retrace the process used, in [19], 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) 6= 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 6= α1,
and h2 entire regular function such that
h(q) = h2(q) ∗ (q − α2) ∗ (q − α1).
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, there exist k quater-nions α1, ..., αk lying onto the 2-sphere Sα and an entire regular function hk
such that h(q) = hk(q) ∗ k Y ∗ i=1 (q − αi) ! . If we consider the symmetrization of h we have
hs(q) = [q2− 2Re(α)q + |α|2]khs k(q)
for all k ∈ N. Hence, for any I ∈ S and for all k ∈ N, the holomorphic function hsI : LI → LI is of the form
hsI(z) = [z2− 2Re(α)z + |α|2]khs
k(z) = (z − α)k(z − ¯α)khsk(z)
where z is any element of LI. As a consequence the holomorphic function hsI
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 [19] for the case of polynomials and in [28] with the attention to the case of poles. Definition 3.2.2. 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)ng(q) for some other entire regular function g. If x + 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)]mh(q) for some other entire
regular function h. If α ∈ x + yS is a non real zero of f then its isolated multiplicity is the largest natural number k such that there exist α1, · · · , αk∈
x + yS with αi 6= αi+1 (for i = 1, · · · k − 1) and an entire regular function l
3.3
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 aj for j = 1, · · · , r. Assume also that the
multiple zeros are repeated according to their multiplicities. By Proposition 3.2.1, for every j = 1, · · · , r there exists δj ∈ Re(aj) + |Im(aj)|S such that
f (q) = qm " k Y i=1 (1 − qb−1i ) # " t Y n=1 q2 x2 n+ yn2 − 2qxn x2 n+ yn2 + 1 # g(q)∗ r Y ∗ j=1 (1−qδ−1j )
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 3.3.1. Let f be an entire regular function. Let m be the multiplic-ity 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 repeated according to their multiplicities. Then there exists a sequence {rn}n∈N⊆ N such that
f (q) = qm " ∞ Y n=0 (1 − qb−1n )eqb−1n +...+rn1 qrnb−rnn # g(q)
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
n→∞|bn| = ∞
for the real zeros of regular functions are isolated. Hence by Theorem 2.3.5 the infinite product
R(q) =
∞
Y
n=0
(1 − qb−1n )eqb−1n +...+1nqnb−nn
converges uniformly on compact sets of H. By lemma 2.4.4 R is an entire regular function and it has the same real zeros (different from zero) with
the same multiplicity as f . Moreover the function R is a slice preserving regular function and so it has a real power series expansion. Hence we can consider it as an holomorphic function and conclude that there exists some entire regular function g without real zeros such that
f (q) = qmR(q)g(q).
Theorem 3.3.2. 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
f (q) = " ∞ Y n=0 q2 |cn|2 −2qRe(cn) |cn|2 + 1 eq 2Re(cn) |cn|2 +...+ 1 `nq`n 2Re(c `n n ) |cn|2`n # h(q)
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 Proposition 3.1.12 the infinite product S(q) = ∞ Y n=0 q2 |cn|2 − 2qRe(cn) |cn|2 + 1 eq 2Re(cn) |cn|2 +...+ 1 nq n 2Re(cnn) |cn|2n
converges uniformly on compact sets of H. The function S is an entire regular function with the same spherical zeros as f . Moreover the function R is slice preserving and so it has a real power series expansion. Hence we can consider it as an holomorphic function and conclude that 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.
Theorem 3.3.3. 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
a never vanishing, entire regular function h and, for all n ∈ N, there exist δn ∈ San = Re(an) + |Im(an)|S and mn ∈ N such that
f (q) = h(q) ∗ ∞ Y ∗ n=0 (1 − qδn−1) ∗ gn(q) !c (3.14) 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
gn|Ln(z) = e
zδn−1+...+mn1 zmnδn−mn.
In particular we can choose mn= n for all n ∈ N.
Proof. Notice first of all that since f (an) = 0 then f (an) 6= 0 for all n ∈ 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, by Theorem 3.1.8, 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
∞
Y ∗
n=0
(1 − qδn−1) ∗ gn(q) (3.15)
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 [19]) we start to “add” the zero a0 to the function f . To
this purpose we consider the function
f1(q) = f (q) ∗ [1 − qf (a0)−1a0−1f (a0)] ∗ g0(q)
where g0 is defined as in the statement. The fact that f1(a0) = 0 is an
immediate consequence of Theorem 1.0.17. We set δ0 = f (a0)−1a0f (a0) and
we note that δ0 ∈ Sa0. The function f1 has {an}n≥1 as sequence of isolated
zeros and Sa0 as spherical zero (see [19] 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
f1(q) = 1 − 2Re(a0)q |a0|2 + q 2 |a0|2 ˜ f1(q)
with ˜f1 an entire regular function having {an}n≥1as sequence of zeros.
There-fore ˜f1(a1) 6= 0. By repeating this same procedure for f1 to “add” the zero
a1, since the polynomial
1 − 2Re(a0)q |a0|2 + q2 |a0|2
has real coefficients, we obtain f2(q) = 1 −2Re(a0)q |a0|2 + q 2 |a0|2 ˜ f1(q) ∗ [1 − q ˜f1(a1)−1a1−1f˜1(a1)] ∗ g1(q) =
