A local representation formula for quaternionic slice regular functions
aa r X i v : . [ m a t h . C V ] M a y A local representation formula for quaternionicslice regular functions
Graziano Gentili, Caterina Stoppato
Dipartimento di Matematica e Informatica “U. Dini”, Universit`a degli Studi di FirenzeViale Morgagni 67/A, I-50134 Firenze, Italygraziano.gentili@unifi.it, [email protected]fi.it
Abstract
After their introduction in 2006, quaternionic slice regular functions have mostly beenstudied over domains that are symmetric with respect to the real axis. This choice was mo-tivated by some foundational results published in 2009, such as the Representation Formulafor axially symmetric domains.The present work studies slice regular functions over domains that are not axially sym-metric, partly correcting the hypotheses of some previously published results. In particular,this work includes a Local Representation Formula valid without the symmetry hypothesis.Moreover, it determines a class of domains, called simple, having the following property:every slice regular function on a simple domain can be uniquely extended to the symmetriccompletion of its domain.
Acknowledgements.
This work was partly supported by INdAM, through: GNSAGA; INdAM project“Hypercomplex function theory and applications”. It was also partly supported by MIUR, through theprojects: Finanziamento Premiale FOE 2014 “Splines for accUrate NumeRics: adaptIve models for Sim-ulation Environments”; PRIN 2017 “Real and complex manifolds: topology, geometry and holomorphicdynamics”.
The theory of quaternionic slice regular functions was introduced in [5, 6] as a possible quater-nionic analog of the theory of holomorphic complex functions. Let us denote the algebra ofquaternions as H ; the real axis as R ; the 2-sphere of quaternionic imaginary units as S ; and the2-plane spanned by 1 and by any I ∈ S as L I . If T ⊆ H , for each I ∈ S , let T I := T ∩ L I . Asusual, a domain in H is an open connected subset of H . Definition 1.1.
Let f be a quaternion-valued function defined on a domain Ω. For each I ∈ S ,let f I := f | Ω I be the restriction of f to Ω I . The restriction f I is called holomorphic if it hascontinuous partial derivatives and¯ ∂ I f ( x + yI ) := 12 (cid:18) ∂∂x + I ∂∂y (cid:19) f I ( x + yI ) ≡ . (1)The function f is called slice regular if, for all I ∈ S , f I is holomorphic.1hile the first works within the theory of slice regular functions focused on the case when Ωis a Euclidean ball centered at the origin, it soon became clear that there were larger classes ofquaternionic domains over which the theory was interesting and had useful applications. Indeed,the next definition and theorem were published in [1] and in [9], respectively. Definition 1.2.
Let Ω be a domain in H that intersects the real axis. Ω is called a slice domain if, for all I ∈ S , the intersection Ω I with the complex plane L I is a domain of L I . Theorem 1.3 (Identity Principle) . Let f, g be slice regular functions on a slice domain Ω . If,for some I ∈ S , f and g coincide on a subset of Ω I having an accumulation point in Ω I , then f = g in Ω . Furthermore, for slice regular functions over slice domains fulfilling the next definition, thework [1] proved a strong property called Representation Formula.
Definition 1.4.
A set T ⊆ H is called (axially) symmetric if, for all points x + yI ∈ T with x, y ∈ R and I ∈ S , the set T contains the whole sphere x + y S . Theorem 1.5 (Representation Formula) . Let f be a slice regular function on a symmetric slicedomain Ω and let x + y S ⊂ Ω . For all I, J, K ∈ S with J = Kf ( x + yI ) = ( J − K ) − [ Jf ( x + yJ ) − Kf ( x + yK )] + (2)+ I ( J − K ) − [ f ( x + yJ ) − f ( x + yK )] . Moreover, the quaternion b := ( J − K ) − [ Jf ( x + yJ ) − Kf ( x + yK )] and the quaternion c :=( J − K ) − [ f ( x + yJ ) − f ( x + yK )] do not depend on J, K but only on x, y . An alternative proof of the same result is included in the proof of [8, Theorem 2.4]. As aconsequence of the Representation Formula, every slice regular function on a symmetric slicedomain is real analytic, see [7, Proposition 7]. Another result proven in [1] allows to constructslice regular functions on symmetric slice domains, starting from H -valued holomorphic functions. Lemma 1.6 (Extension Lemma) . Let Ω be a symmetric slice domain and let I ∈ S . If f I : Ω I → H is holomorphic then there exists a unique slice regular function g : Ω → H such that g I = f I in Ω I . The function g is denoted by ext( f I ) and called the regular extension of f I . After the work [1], the development of the theory led to many interesting results valid oversymmetric slice domains. These results are collected in the monograph [4] and in a variety ofsubsequent works. On the other hand, the study of slice regular functions over slice domains thatare not symmetric has not been further developed for several years. A possible reason is that [1,Theorem 4.1] stated that every slice regular function on a slice domain Ω could be extended ina unique fashion to the symmetric completion e Ω of Ω, in accordance with the next definition.
Definition 1.7.
The (axially) symmetric completion of a set T ⊆ H is the smallest symmetricset e T that contains T . In other words, e T := [ x + yI ∈ T ( x + y S ) . (3)However, the recent work [2] disproved [1, Theorem 4.1] by means of a counterexample. Italso pointed out that the proof proposed in [1] implicitly applied the Identity Principle 1.3 overthe intersection of two sets, which was not a slice domain. The same work [2] then introduced thenotions of Riemann slice domain and (Riemann) slice domain of regularity , both being abstract2opological spaces. The Representation Formula 1.5 has been generalized to (Riemann) slicedomains of regularity in [2] and to Riemann slice domains in [3]. The work [3] also explored thealgebraic structure of slice regular functions over Riemann slice domains.In contrast with the approach of [2, 3], the present work furthers the study over slice domainsof H that are not symmetric. Our main result is a local version of Theorem 1.5 valid withoutthe symmetry hypothesis. Theorem 1.8 (Local Representation Formula) . Let Ω be a slice domain and let f : Ω → H bea slice regular function. For all J, K ∈ S with J = K and all x, y ∈ R with y ≥ such that x + yJ, x + yK ∈ Ω , let us set b ( x + yJ, x + yK ) := ( J − K ) − [ Jf ( x + yJ ) − Kf ( x + yK )] ,c ( x + yJ, x + yK ) := ( J − K ) − [ f ( x + yJ ) − f ( x + yK )] . For every p ∈ Ω , there exists a slice domain Λ with p ∈ Λ ⊆ Ω such that the following propertieshold for all all x, y ∈ R with y ≥ : • If U := ( x + y S ) ∩ Λ is not empty, then b, c are constant in U × U \ { ( u, u ) : u ∈ U \ R } . • If I, J, K ∈ S with J = K are such that x + yI, x + yJ, x + yK ∈ Λ , then f ( x + yI ) = b ( x + yJ, x + yK ) + Ic ( x + yJ, x + yK ) . (4)Moreover, we can exhibit a class of slice domains, called simple , for which the following resultholds. Theorem 1.9 (Extension) . Let f be a slice regular function on a simple slice domain Ω . Thereexists a unique slice regular function ˜ f : e Ω → H that extends f to the symmetric completion ofits domain. Section 2 states and proves a slightly corrected version of the General Extension Formula [1,Theorem 4.2], needed in the subsequent pages. Section 3 presents a Local Extension Theorem,which, besides its independent interest, is used to prove the aforementioned Local RepresentationFormula. The definition of simple domain is given in Section 4, which also presents a broad class ofexamples of simple domains, called slice convex domains . The domain used in the counterexampleof [2] is shown not to be simple. Section 4 also proves the aforementioned Extension Theorem.
The Extension Formula, published in [1, Theorem 4.2], provides a method to construct sliceregular functions starting from couples of holomorphic functions. We present here a slightlycorrected version of the same result. For all J ∈ S and all T ⊆ H , we use the notations L + J := { x + yJ : x, y ∈ R , y > } and T + J := T ∩ L + J . Theorem 2.1 (Extension Formula) . Let
J, K be distinct imaginary units; let T be a domainin L J , such that T + J is connected and T ∩ R = ∅ ; let U := { x + yK : x + yJ ∈ T } . Chooseholomorphic functions r : T → H , s : U → H such that r | T ∩ R = s | U ∩ R . Let Ω be the symmetric lice domain such that Ω + J = T + J , Ω ∩ R = T ∩ R and set, for all x + yI ∈ Ω with x, y ∈ R , y ≥ and I ∈ S , f ( x + yI ) := ( J − K ) − [ Jr ( x + yJ ) − Ks ( x + yK )] + (5)+ I ( J − K ) − [ r ( x + yJ ) − s ( x + yK )] The function f : Ω → H is the (unique) slice regular function on Ω that coincides with r in Ω + J ,with s in Ω + K and with both r and s in Ω ∩ R .Proof. Formula (5) yields f ( x + yI ) = [( J − K ) − J + I ( J − K ) − ] r ( x + yJ ) + − [( J − K ) − K + I ( J − K ) − ] s ( x + yK ) . Since ( J − K ) − J + J ( J − K ) − = | J − K | − [( K − J ) J + J ( K − J )] == [( J − K )( K − J )] − (2 + JK + KJ ) = 1and ( J − K ) − K + J ( J − K ) − = | J − K | − [( K − J ) K + J ( K − J )] == | J − K | − ( − − JK + JK + 1) = 0 , the function f coincides with r in Ω + J and in Ω ∩ R . Similarly, f coincides with s in Ω + K and inΩ ∩ R .We can prove that f is slice regular in Ω \ R by showing that, for each I ∈ S , f I is holomorphicin Ω + I . Indeed, a direct computation shows that¯ ∂ I f ( x + yI ) = [( J − K ) − J + I ( J − K ) − ] ¯ ∂ J r ( x + yJ ) + − [( J − K ) − K + I ( J − K ) − ] ¯ ∂ K s ( x + yK )for x + yI ∈ Ω + I . Since ¯ ∂ J r ( x + yJ ) ≡ + J and ¯ ∂ K r ( x + yK ) ≡ + K , it follows that¯ ∂ I f ( x + yI ) ≡ + I .Proving that f is slice regular near every point of Ω ∩ R requires a bit more work. Eachconnected component of Ω ∩ R admits an open neighborhood D in Ω that is a symmetric slicedomain and whose slice D J is included in the domain T of the function r . The domain U of s automatically includes D K . Let us consider the regular extension g = ext( r | DJ ) on D : g coincideswith the regular extension ext( s | DK ) by the Identity Principle 1.3, because r and s coincide in D ∩ R . In the symmetric slice domain D , we can apply the Representation Formula (2) to g .Taking into account that g J = r | DJ and g K = s | DK , we get g ( x + yI ) = ( J − K ) − [ Jr ( x + yJ ) − Ks ( x + yK )] ++ I ( J − K ) − [ r ( x + yJ ) − s ( x + yK )]for all x + yI ∈ Ω. Thus, g coincides with f | D . In particular, f is slice regular in D , as desired.Our final remark concerns the uniqueness of f . The Identity Principle 1.3 proves that, forany slice regular function h : Ω → H coinciding with r in Ω ∩ R , h coincides with f .The original statement of the Extension Formula set Ω := e T and used formula (5) for all x + yI ∈ Ω, without requiring y ≥
0. However, the following example proves that such an f maybe ill-defined. 4 xample 2.2. Let Ω be the slice domain constructed in [2, page 5]: it holds Ω ⊃ R ; moreover,for every J ∈ S , Ω + J := L + J \ ( h J ∪ a J ) where h J := ( −∞ , −
2) + 2 J = { t + 2 J : t ∈ ( −∞ , − } is a half line with origin − J and a J is an appropriately defined arc, with endpoints − J and J , within the closed disk D J := { z ∈ L J : | z + 1 − J | ≤ } . The definition of a J for J ∈ S is the following: an imaginary unit I is fixed; for each J ∈ S , itis set T ( J ) := min {| J − I | , } and a J := − J + { (1 − T ( J )) e πJt + T ( J ) e − πJt : t ∈ [0 , / } . In particular, a I is the upper half of the circle ∂D I within L + I and, for every imaginary unit J with | J − I | ≥ (including J = − I ) , a J is the lower half of the circle ∂D J within L + J .Consider the planar domain T := Ω I = L I \ ( h I ∪ h − I ∪ a I ∪ a − I ) and its conjugate U := Ω I = L I \ ( h I ∪ h − I ∪ a − I ∪ a I ) . Consider the unique holomorphic functions r : T → L I and s : U → L I such that r ( x +2 I ) = ln( x ) = s ( x + 2 I ) for all x ∈ (0 , + ∞ ) . The intersection T ∩ U has three connectedcomponents: namely, the open disks that form the interiors of D I and D − I in L I and the set L I \ ( h I ∪ h − I ∪ D I ∪ D − I ) . In the second and third connected component, the functions r and s coincide; in the first component, they differ by a jump. If we apply Theorem 2.1 to r and s with J = I and K = − I , we get a slice regular function H \ ( f h I ∪ e a I ) → H . If, instead, we apply Theorem 2.1 to r and s with J = − I and K = I , we get a slice regularfunction H \ ( f h I ∪ g a − I ) → H . These slice regular functions coincide in the symmetric slice domain H \ ( f h I ∪ f D I ) but they differby a jump in the interior of f D I \ D − I . The function r used in the previous example is the restriction G I : Ω I → H of the slice regularfunction G : Ω → H constructed in [2, page 5]. This section is devoted to proving the Local Extension Theorem and the Local RepresentationFormula announced in the Introduction. We begin with a useful lemma. In the statement,the expression “closed interval” includes the degenerate interval consisting of a single point butexcludes the empty set. 5 emma 3.1.
Let Y be an open subset of H and let J ∈ S . Let C be a compact and path-connected subset of Y J such that C ∩ R is a closed interval and C \ R ⊂ Y + J . Let q ∈ C be suchthat max p ∈ C | Im( p ) | = | Im( q ) | . Then there exists ε > such that Γ( C, ε ) := [ p ∈ C \ R B (cid:18) p, | Im( p ) || Im( q ) | ε (cid:19) ∪ [ p ∈ C ∩ R B ( p, ε ) is a slice domain and C ⊂ Γ( C, ε ) ⊆ Y .Proof. By compactness, there exists an ε > p ∈ C , the Euclidean ball B ( p, ε ) is included in Y . Up to shrinking ε , it also holds ε < | Im( q ) | . Clearly, Γ = Γ( C, ε ) hasthe desired property C ⊂ Γ ⊆ Y . Let us prove that Γ is a slice domain.The union of open balls, each centered at one point of C , is a domain: it is obviously open;it is path-connected because any point can be joined to the center of a ball by a line segment,while centers are connected by paths within C .The domain Γ intersects R by construction.Moreover, for each I ∈ S , the slice Γ I is a domain in L I . Indeed, let ϑ := arcsin ε | Im( q ) | andlet ϑ ∈ [0 , π/
2] be the angle between L I and L J within the 3-space R + I R + J R . There aretwo possibilities. • If ϑ ≥ ϑ then, for any p ∈ C \ R , the distance | Im( p ) | sin ϑ between p and L I is greaterthan, or equal to, | Im( p ) | sin ϑ = | Im( p ) || Im( q ) | ε . Thus,Γ I = [ p ∈ C ∩ R B ( p, ε ) ∩ L I is a union of open disks centered at points of the closed interval C ∩ R . • If ϑ < ϑ then every ball B (cid:16) p, | Im( p ) || Im( q ) | ε (cid:17) with p ∈ C \ R and every ball B ( p, ε ) with p ∈ C ∩ R intersects L I in an open disk centered at the orthogonal projection p I of p on L I . Such centers form a compact and path-connected subset of L I .In both cases, Γ I is a domain in L I by the argument we already used for Γ.We are now in a position to prove the first result we announced. Theorem 3.2 (Local Extension) . Let f be a slice regular function on a slice domain Ω . Forevery p ∈ Ω , there exist a symmetric slice domain N with N ∩ R ⊂ Ω , a slice domain Λ with p ∈ Λ ⊆ Ω ∩ N , and a slice regular function g : N → H such that g coincides with f in N ∩ R ,whence in Λ .Proof. If p ∈ Ω ∩ R then the thesis is obvious because Ω contains an open ball centered at p .We can therefore suppose p = x + y J with x , y ∈ R , y > J ∈ S .Since Ω is a slice domain, there exists a continuous path γ : [0 , → Ω J , γ ( t ) = α ( t ) + J β ( t ),such that γ (0) = p and γ (1) ∈ R . Up to restricting and reparametrizing γ , we can supposethe support of γ to only intersect the real axis at γ (1). By Lemma 3.1, there exists ε > M := Γ( γ ([0 , , ε ) is a slice domain with the property γ ([0 , ⊂ M ⊆ Ω. Let t ∈ [0 ,
1] be suchthat max [0 , | Im γ | = | Im γ ( t ) | .If we choose K ∈ S with 0 < | K − J | < ε | Im γ ( t ) | , then the support of the curve α + K β is included in M K . Indeed, the distance between α ( t ) + K β ( t ) and γ ( t ) = α ( t ) + J β ( t ) is | K − J | β ( t ), which is less than | Im γ ( t ) || Im γ ( t ) | ε for all t ∈ [0 ,
1) and is 0 for t = 1.6et N be the symmetric completion of the connected set M K . We point out the followingproperties of N : it includes the support of γ = α + J β ; it has N + K = M + K ; and it is a slicedomain. Moreover, N + J ⊆ M + J ⊆ Ω J . Indeed, for each x + yJ ∈ N + J it holds x + yK ∈ N + K = M + K . By direct computation, for all t ∈ [0 , | x + yK − γ ( t ) | − | x + yJ − γ ( t ) | = 2 yβ ( t )(1 − h K , J i ) ≥ . Thus, the distance between x + yJ and γ ( t ) is less than, or equal to, the distance between x + yK and γ ( t ). If B ( γ (1) , ε ) includes x + yK , then it also includes x + yJ . Similarly, ifthere exists t ∈ [0 ,
1) such that B (cid:16) γ ( t ) , | Im γ ( t ) || Im γ ( t ) | ε (cid:17) includes x + yK , then the same ball includes x + yJ . In both cases, x + yJ belongs to M + J .By the General Extension Formula 2.1 there exists a unique slice regular function g : N → H that coincides with f J in N + J ⊆ M + J ⊆ Ω J , with f K in N + K = M + K and with f in N ∩ R = M K ∩ R .Within the open set N ∩ Ω, the slice ( N ∩ Ω) J includes the support of γ . Lemma 3.1guarantees that there exists ε > γ ([0 , , ε ) has theproperty γ ([0 , ⊂ Λ ⊆ N ∩ Ω. Now, g and f coincide in N ∩ Ω ∩ R = N ∩ R , whence throughoutΛ by the Identity Principle 1.3.We can draw several consequences. Since any slice regular function g on a symmetric slicedomain is real analytic by [7, Proposition 7] and [8, Theorem 2.4], it follows that: Corollary 3.3.
Every slice regular function on a slice domain is real analytic.
The second consequence of Theorem 3.2 is the main result of this work.
Theorem 3.4 (Local Representation Formula) . Let Ω be a slice domain and let f : Ω → H bea slice regular function. For all J, K ∈ S with J = K and all x, y ∈ R with y ≥ such that x + yJ, x + yK ∈ Ω , let us set b ( x + yJ, x + yK ) := ( J − K ) − [ Jf ( x + yJ ) − Kf ( x + yK )] c ( x + yJ, x + yK ) := ( J − K ) − [ f ( x + yJ ) − f ( x + yK )] . For every p ∈ Ω , there exists a slice domain Λ with p ∈ Λ ⊆ Ω such that the following propertieshold for all all x, y ∈ R with y ≥ : • If U = ( x + y S ) ∩ Λ is not empty, then b, c are constant in U × U \ { ( u, u ) : u ∈ U \ R } . • If I, J, K ∈ S with J = K are such that x + yI, x + yJ, x + yK ∈ Λ , then f ( x + yI ) = b ( x + yJ, x + yK ) + Ic ( x + yJ, x + yK ) . (6) Proof.
By Theorem 3.2, there exist a symmetric slice domain N , a slice domain Λ with p ∈ Λ ⊆ Ω ∩ N and a slice regular function g : N → H such that g coincides with f in Λ. If we applyTheorem 1.5 to g : N → H at x + yJ, x + yK ∈ N (with x, y ∈ R and y ≥ J − K ) − [ Jg ( x + yJ ) − Kg ( x + yK )] and ( J − K ) − [ g ( x + yJ ) − g ( x + yK )]do not depend on the choice of J, K ∈ S with J = K . When x + yJ, x + yK ∈ Λ, thesequaternions coincide with b ( x + yJ, x + yK ) and c ( x + yJ, x + yK ), respectively. Thus, thequaternions b ( x + yJ, x + yK ) , c ( x + yJ, x + yK ) do not depend on the choice of x + yJ, x + yK within U := ( x + y S ) ∩ Λ. By construction, f ( x + yI ) = g ( x + yI ) = b ( x + yJ, x + yK ) + Ic ( x + yJ, x + yK )when x + yI, x + yJ, x + yK ∈ Λ. 7n the previous statement and proof, for y = 0 we get well-defined b ( x, x ) , c ( x, x ) because,regardless of the choice of J, K ∈ S with J = K , it holds ( J − K ) − [ Jf ( x ) − Kf ( x )] = f ( x ) and( J − K ) − [ f ( x ) − f ( x )] = 0. This section proves the Extension Theorem announced in the Introduction. We begin with thefollowing definition.
Definition 4.1.
A slice domain Ω is simple if, for any choice of
J, K ∈ S , the setΩ + J,K := { x + yJ ∈ Ω + J : x + yK ∈ Ω + K } is connected.The next definition, proposition and remark provide many examples of simple domains. Definition 4.2.
A set T ⊆ H is slice convex if, for every I ∈ S , the slice T I is a convex subsetof L I . Proposition 4.3.
Let Ω be a slice domain. If Ω is slice convex, then it is a simple domain.Proof. If Ω is slice convex then each half-slice Ω + J is convex because it is the intersection of theconvex set Ω J with the (convex) half-plane L + J . Moreover, for each J, K ∈ S , the set A := { x + yJ ∈ L + J : x + yK ∈ Ω + K } is convex because it is a “copy” within the half-plane L + J of the convex set Ω + K . The set Ω + J,K ,which is the intersection of the convex sets Ω + J and A , is convex, whence connected.A similar technique proves what follows. Remark 4.4.
Let Ω be a slice domain. If Ω is starlike with respect to a point x ∈ Ω ∩ R , then Ω is a simple domain. Here is an example of slice domain that is not simple.
Example 4.5.
The slice domain Ω of Example 2.2 is not a simple domain. Indeed, Ω + I, − I hastwo connected components: one is the open disk that forms the interior of D I in L I ; the other is L + I \ ( h I ∪ D I ) . For simple slice domains, the following result holds.
Theorem 4.6 (Extension) . Let f be a slice regular function on a simple slice domain Ω . Thereexists a unique slice regular function ˜ f : e Ω → H that extends f to the symmetric completion ofits domain.Proof. For all J ∈ S , let us adopt the notation N ( J ) for the unique symmetric set such that N ( J ) + J = Ω + J and such that N ( J ) ∩ R = Ω ∩ R . For all J, K ∈ S , the intersection N ( J, K ) := N ( J ) ∩ N ( K ) is a slice domain because Ω is simple. We divide our proof into three steps.1. Let us prove that, for each J ∈ S , there exists a slice regular function g : N ( J ) → H that coincides with f in N ( J ) + J = Ω + J and in N ( J ) ∩ R = Ω ∩ R .8a) For each K ∈ S \ { J } , let g K : N ( J , K ) → H denote the slice regular function thatcoincides with f J in N ( J , K ) + J = Ω + J ,K , with f K in N ( K, J ) + K = Ω + K,J and with f in N ( J , K ) ∩ R = Ω ∩ R . Such a function exists by Theorem 2.1 and it is definedas g K ( x + yI ) := b ( x + yJ , x + yK ) + Ic ( x + yJ , x + yK ) with b ( x + yJ , x + yK ) := ( J − K ) − [ J f ( x + yJ ) − Kf ( x + yK )]and c ( x + yJ , x + yK ) := ( J − K ) − [ f ( x + yJ ) − f ( x + yK )]for all x, y ∈ R with y ≥ x + y S ⊂ N ( J , K ). For y = 0, we get b ( x, x ) = f ( x ) and c ( x, x ) = 0.(b) Fix p = x + y J , either in N ( J ) + J or in N ( J ) ∩ R . By Theorem 3.4, p has aneighborhood Λ such that, for Λ ′ := Λ \ Λ + J , the maps Λ ′ ∋ x + yK b ( x + yJ , x + yK )and Λ ′ ∋ x + yK c ( x + yJ , x + yK ) are constant with respect to K . Let us denotethese constants as b ( x + yJ ) , c ( x + yJ ), respectively, and let us set g ( x + y I ) := b ( x + y J ) + Ic ( x + y J )for all I ∈ S . We have thus constructed a function g : N ( J ) → H .(c) For each x + y S ⊂ N ( J ) (with x , y ∈ R , y ≥ x + y J such that the equality g ( x + yI ) = b ( x + yJ , x + yK ) + Ic ( x + yJ , x + yK ) = g K ( x + yI )holds for all x, y ∈ R with y ≥ I, K ∈ S such that x + yK ∈ Λ ′ . For each δ >
0, let us consider the following neighborhood of x + y S : T ( x + y S , δ ) := { x + yI : | x − x | + | y − y | < δ , I ∈ S } . There exist δ, ε > { x + yK ∈ T ( x + y S , δ ) : | K − J | < ε } . If we pick any K ∈ S such that 0 < | K − J | < ε , then g coincides with g K in T ( x + y S , δ ). It follows at once that g is slice regular in T ( x + y S , δ ) and that g ( x + y J ) = f ( x + y J ), as desired.2. Any two slice regular functions g : N ( J ) → H and g : N ( J ) → H that coincide with f in N ( J ) ∩ R = Ω ∩ R = N ( J ) ∩ R coincide with each other in the slice domain N ( J , J )by the Identity Principle 1.3.3. By steps 1. and 2., there exists a slice regular function g on [ J ∈ S N ( J ) = e Ωthat coincides with f in Ω ∩ R . By the Identity Principle 1.3, g coincides with f in theslice domain Ω. 9 eferences [1] F. Colombo, G. Gentili, I. Sabadini, and D. Struppa. Extension results for slice regularfunctions of a quaternionic variable. Adv. Math. , 222(5):1793–1808, 2009.[2] X. Dou and G. Ren. Riemann slice-domains over quaternions I. arXiv:1808.06994 [math.CV],2018.[3] X. Dou and G. Ren. Riemann slice-domains over quaternions II. arXiv:1809.07979 [math.CV],2018.[4] G. Gentili, C. Stoppato, and D. C. Struppa.
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