Abstract
The theory of quaternionic slice regular functions was introduced in 2006 and successfully developed for about a decade over symmetric slice domains, which appeared to be the natural setting for their study. Some recent articles paved the way for a further development of the theory: namely, the study of slice regular functions on slice domains that are not necessarily symmetric. The present work is a panorama of geometric function theory in this new context, where new phenomena appear. For insta
Results & Lemmas (52)
Each result is auto-extracted from the paper via OCR - the math is shown verbatim and may carry occasional transcription artifacts. Click to expand the full statement.
Theorem 3.2
Theorem 3.2]. Section 12 presents a local version of the so-called spherical series expansions for slice regular functions. 2 Preliminaries…
Theorem 3.2]. Section 12 presents a local version of the so-called spherical series expansions for slice regular functions. 2 Preliminaries In this section devoted to preliminaries, we follow the presentation of [14, Chap- ter 1] (which derived from [2, 3, 10, 15, 16, 27]). The real algebra of quaternions will be denoted as H = R + iR + jR + kR; the real axis as R; the 2-sphere of quaternionic imaginary units as S; and the real subalgebra generated by 1 and by any I ∈S as LI := R + IR . If T ⊆H,
Theorem 2.4
Theorem 2.4 (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…
Theorem 2.4 (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 ΩI having an accumulation point in ΩI, then f = g in Ω. Definition 2.5. 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 + yS.
Theorem 2.6
Theorem 2.6 (Representation Formula). Let f be a slice regular function on a symmetric slice domain Ωand let x + yS ⊂Ω. For all I, J, K ∈S…
Theorem 2.6 (Representation Formula). Let f be a slice regular function on a symmetric slice domain Ωand let x + yS ⊂Ω. For all I, J, K ∈S with J ̸= K f(x + yI) = (J −K)−1 [Jf(x + yJ) −Kf(x + yK)] + (2) + I(J −K)−1 [f(x + yJ) −f(x + yK)] . Moreover, the quaternion b := (J −K)−1 [Jf(x + yJ) −Kf(x + yK)] and the quaternion c := (J −K)−1 [f(x + yJ) −f(x + yK)] do not depend on J, K but only on x, y. Another version of the Representation Formula was proven in [20, Theorem 2.4], yielding that every s
Lemma 2.7
Lemma 2.7 (Extension Lemma). Let Ωbe a symmetric slice domain and let I ∈S. If fI: ΩI →H is holomorphic then there exists a unique slice…
Lemma 2.7 (Extension Lemma). Let Ωbe a symmetric slice domain and let I ∈S. If fI : ΩI →H is holomorphic then there exists a unique slice regular function g : Ω→H such that gI = fI in ΩI. The function g is denoted by ext(fI) and called the regular extension of fI.
Theorem 2.8
Theorem 2.8 (Extension Formula). Let J, K be distinct imaginary units; let T be a domain in LJ, such that T + J is connected and T ∩R ̸= ∅;…
Theorem 2.8 (Extension Formula). Let J, K be distinct imaginary units; let T be a domain in LJ, such that T + J is connected and T ∩R ̸= ∅; let U := {x+yK : x + yJ ∈T }. Choose holomorphic functions r : T →H, s : U →H such that r|T ∩R = s|U∩R. Let Ωbe the symmetric slice domain such that Ω+ J = T + J , Ω∩R = T ∩R and set, for all x + yI ∈Ωwith x, y ∈R, y ≥0 and I ∈S, f(x + yI) := (J −K)−1 [Jr(x + yJ) −Ks(x + yK)] + (3) + I(J −K)−1 [r(x + yJ) −s(x + yK)] The function f : Ω→H is the (unique) slice
Lemma 2.11.
Lemma 2.11. Let Y be an open subset of H and let J0 ∈S. Let C be a compact and path-connected subset of YJ0 such that C ∩R is a closed…
Lemma 2.11. Let Y be an open subset of H and let J0 ∈S. Let C be a compact and path-connected subset of YJ0 such that C ∩R is a closed interval and ∅̸= C \R ⊂Y + J0. Let q0 ∈C be such that maxp∈C | Im(p)| = | Im(q0)|. Then there exists ε > 0 such that Γ(C, ε) := [ p∈C\R B p, | Im(p)| | Im(q0)|ε ∪
Theorem 2.12
Theorem 2.12 (Local Extension). Let f be a slice regular function on a slice domain Ω. For every p0 ∈Ω, there exist a symmetric slice…
Theorem 2.12 (Local Extension). Let f be a slice regular function on a slice domain Ω. For every p0 ∈Ω, there exist a symmetric slice domain N with N ∩R ⊂Ω, a slice domain Λ with p0 ∈Λ ⊆Ω∩N, and a slice regular function ef : N →H such that ef coincides with f in N ∩R, whence in Λ. We will state and prove a stronger version of this result in Theorem 11.4.
Theorem 2.12
Theorem 2.12 has the following consequence. 5
Theorem 2.12 has the following consequence. 5
Corollary 2.13.
Corollary 2.13. Every slice regular function on a slice domain is real analytic. Motivated by Theorem 2.12, we give a new definition.…
Corollary 2.13. Every slice regular function on a slice domain is real analytic. Motivated by Theorem 2.12, we give a new definition. Definition 2.14. Let f be a slice regular function on a slice domain Ω. If ef is a slice regular function on a symmetric slice domain N with N ∩R ⊂Ωand if there exists a slice domain Λ ⊆Ω∩N where f and ef coincide, then ef, N, Λ is called an extension triplet for f. In [13], we proved the next result.
Theorem 2.15
Theorem 2.15 (Local Representation Formula). Let Ωbe a slice domain and let f: Ω→H be a slice regular function. For all J, K ∈S with J ̸= K…
Theorem 2.15 (Local Representation Formula). Let Ωbe a slice domain and let f : Ω→H be a slice regular function. For all J, K ∈S with J ̸= K and all x, y ∈R with y ≥0 such that x + yJ, x + yK ∈Ω, let us set b(x + yJ, x + yK) := (J −K)−1 [Jf(x + yJ) −Kf(x + yK)] , c(x + yJ, x + yK) := (J −K)−1 [f(x + yJ) −f(x + yK)] . For every p0 ∈Ω, there exists a slice domain Λ with p0 ∈Λ ⊆Ωsuch that the following properties hold for all x, y ∈R with y ≥0: • If U = (x+yS)∩Λ is not empty, then b, c are constant
Theorem 2.17
Theorem 2.17 (Extension). Let f be a slice regular function on a simple slice domain Ω. There exists a unique slice regular function ef:…
Theorem 2.17 (Extension). Let f be a slice regular function on a simple slice domain Ω. There exists a unique slice regular function ef : eΩ→H that extends f to the symmetric completion of its domain. In [13], we observed that the slice domain Ωof Example 2.10 is not simple. We also described, without giving a formal proof, a large class of examples of simple domains. We include a proof here.
Proposition 2.18.
Proposition 2.18. If an open subset Ωof H is starlike with respect to a point x0 ∈Ω∩R, then it is a simple slice domain.
Proposition 2.18. If an open subset Ωof H is starlike with respect to a point x0 ∈Ω∩R, then it is a simple slice domain.
Proposition 3.5.
Proposition 3.5. Let Ωbe a slice domain in H and let f: Ω→H be a slice regular function. Pick p ∈Ωand let dfp denote the real differential…
Proposition 3.5. Let Ωbe a slice domain in H and let f : Ω→H be a slice regular function. Pick p ∈Ωand let dfp denote the real differential of f at p. 1. If p ∈Ω∩R, then dfp(v) = vf ′ c(p) for all v ∈TpΩ≃H. As a consequence, dfp is singular if, and only if, f ′ c(p) = 0. 2. If p ∈Ω\ R, if I ∈S is such that p ∈ΩI and if we split TpΩ≃H as LI ⊕L⊥ I , then dfp(v + w) = vf ′ c(p) + wf ′ s(p) for all v ∈LI and w ∈L⊥ I . As a consequence, dfp is singular if, and only if, f ′
Theorem 4.3.
Theorem 4.3. Let Ωbe a slice domain in H. Then +, ∗,c are operations on the set of slice regular functions on Ωand turn it into a real…
Theorem 4.3. Let Ωbe a slice domain in H. Then +, ∗,c are operations on the set of slice regular functions on Ωand turn it into a real ∗-algebra. Moreover, if f, g : Ω→H are slice regular functions, then the following properties hold: 1. if f is slice preserving, then f ∗g(q) = f(q)g(q) = g ∗f(q) in Ω; 2. if g is constantly equal to c, then f ∗g(q) = f(q)c for all q ∈Ω; 3. f s is a slice preserving regular function and f s = f ∗f c = f c ∗f .
Proposition 4.4.
Proposition 4.4. Let Ωbe a slice domain in H, let f: Ω→H be a slice regular function and pick p = x + yI ∈Ω. If f(p) = 0, then f ∗g(p) = 0.…
Proposition 4.4. Let Ωbe a slice domain in H, let f : Ω→H be a slice regular function and pick p = x + yI ∈Ω. If f(p) = 0, then f ∗g(p) = 0. Otherwise, the point f(p)−1pf(p) belongs to x + yS and f ∗g(p) = f(p) eg(f(p)−1pf(p)) where eg(q) = g◦ s(p) + Im(q)g′ s(p) for all q ∈x + yS (whence eg coincides with g in the connected component of (x + yS) ∩Ωthat includes p).
Theorem 4.3
Theorem 4.3 and Proposition 4.4 hold. 5 Zero sets This section studies the zero sets Z(f):= q ∈Ω: f(q) = 0 of slice regular functions f on…
Theorem 4.3 and Proposition 4.4 hold. 5 Zero sets This section studies the zero sets Z(f) := {q ∈Ω: f(q) = 0} of slice regular functions f on slice domains Ω. For the case of symmetric slice domains, see [14, Chapter 3] (which, in turn, derived from [3, 9, 11, 16]). If Ω is symmetric, then Z(f) consists of isolated points or isolated 2-spheres of type x + yS (unless it is the whole domain). Moreover, each 2-sphere of type x + yS cannot include several isolated zeros. In the case of slice domains
Proposition 5.1.
Proposition 5.1. Let Ωbe a slice domain in H, let f: Ω→H be a slice regular function and consider its zero set Z(f). Fix any 2-sphere S = x…
Proposition 5.1. Let Ωbe a slice domain in H, let f : Ω→H be a slice regular function and consider its zero set Z(f). Fix any 2-sphere S = x + yS (with x, y ∈R, y ̸= 0). For p ∈S, the equality f(p) = 0 is equivalent to f ◦ s (p) = −Im(p)f ′ s(p) . If the previous equality holds and if C denotes the connected component of Ω∩S that includes p, then either 1. f ′ s(p) ̸= 0 and Z(f) ∩C = {p}; or 2. f ′ s(p) = 0 and C ⊆Z(f). The former possibility is excluded if f is slice preserving.
Theorem 5.2.
Theorem 5.2. Let Ωbe a slice domain in H and let f: Ω→H be a slice regular function. Either f ≡0 or, for any S = x + yS intersecting Ωand…
Theorem 5.2. Let Ωbe a slice domain in H and let f : Ω→H be a slice regular function. Either f ≡0 or, for any S = x + yS intersecting Ωand for any connected component C of S ∩Ω, the set Z(f) \ C has no accumulation points in C.
Corollary 5.3.
Corollary 5.3. Let Ωbe a slice domain in H and let f: Ω→H be a slice regular function with f ̸≡0. Suppose p ∈Z(f), let S = x + yS include p…
Corollary 5.3. Let Ωbe a slice domain in H and let f : Ω→H be a slice regular function with f ̸≡0. Suppose p ∈Z(f), let S = x + yS include p and let C be connected component of S ∩Ωthat includes p. If either p ∈R or f ′ s(p) ̸= 0, then p is an isolated point in Z(f). If, instead, p ̸∈R and f ′ s(p) = 0, then Z(f) includes C and Z(f) \ C has no accumulation points in C.
Proposition 5.1
Proposition 5.1 yields that Z(f) ⊇C. By Theorem 5.2, in all cases, Z(f) C has no accumulation points in C. The thesis follows. One novelty…
Proposition 5.1 yields that Z(f) ⊇C. By Theorem 5.2, in all cases, Z(f) \ C has no accumulation points in C. The thesis follows. One novelty with respect to the symmetric case is, that (x + yS) ∩Ωmay include several zeros of f without being entirely included in Z(f). Explicit examples will be provided in Section 7. Let us study the zero set of a regular product f ∗g and find that it consists of zeros of f, plus points that depend on g but are not necessarily zeros of g.
Proposition 5.4.
Proposition 5.4. Let Ωbe a slice domain in H and let f, g: Ω→H be slice regular functions. Then Z(f) ⊆Z(f ∗g). Moreover, Z(f ∗g) Z(f) = p…
Proposition 5.4. Let Ωbe a slice domain in H and let f, g : Ω→H be slice regular functions. Then Z(f) ⊆Z(f ∗g) . Moreover, Z(f ∗g) \ Z(f) = {p ∈Ω: g◦ s(p) = −f(p)−1 Im(p)f(p)g′ s(p)} .
Proposition 5.6.
Proposition 5.6. Let Ωbe a slice domain in H and let f: Ω→H be a slice regular function. The zero sets Z(f) and Z(f s) are related, as…
Proposition 5.6. Let Ωbe a slice domain in H and let f : Ω→H be a slice regular function. The zero sets Z(f) and Z(f s) are related, as follows: 1. Fix any 2-sphere S = x + yS (with x, y ∈R, y ̸= 0). If p ∈Z(f) and if C denotes the connected component of Ω∩S that includes p, then C ⊆Z(f s). 2. The equality Z(f) ∩R = Z(f s) ∩R holds. Thus, if f s ≡0, then f ≡0.
Proposition 5.6
Proposition 5.6 holds true, while property 2. is reduced to the equality Z(f) ∩ R = Z(f s) ∩R. Indeed, it may well happen that f s ≡0 while…
Proposition 5.6 holds true, while property 2. is reduced to the equality Z(f) ∩ R = Z(f s) ∩R. Indeed, it may well happen that f s ≡0 while f ̸≡0: Example 5.8. Consider the slice regular functions f0 and f1 defined in Exam- ple 2.10 and their difference D := f1 −f0 on the interior of the solid torus eC1. For x, y ∈R with y ≥0 and J ∈S such that x + Jy belongs to the interior of eC1, it holds D(x + Jy) = π(I + J) = πI + yJ π y . Thus, Dc(x + Jy) = πI −yJ π y and Ds(x + Jy) = π2 −y2 π2 y2 + 2yJ Re
Proposition 6.2.
Proposition 6.2. Let Ωbe a slice domain in H and let f: Ω→H be a slice regular function such that f ̸≡0. Then f −∗is a slice regular…
Proposition 6.2. Let Ωbe a slice domain in H and let f : Ω→H be a slice regular function such that f ̸≡0. Then f −∗is a slice regular function on Ω′ = Ω\ Z(f s) and the following equalities hold in Ω′: f −∗(q) = (f s)−∗∗f c(q) = f s(q)−1f c(q) , f ∗f −∗= f −∗∗f ≡1 . Moreover, if f is slice preserving, then f −∗is slice preserving and f −∗(q) = f(q)−1 for all q ∈Ω′.
Proposition 6.2
Proposition 6.2 allows us to make the following remark.
Proposition 6.2 allows us to make the following remark.
Proposition 6.6.
Proposition 6.6. Let Ωbe a slice domain in H and let f, g: Ω→H be slice regular functions. For each p = x + yI ∈Ω Z(f s), it holds f…
Proposition 6.6. Let Ωbe a slice domain in H and let f, g : Ω→H be slice regular functions. For each p = x + yI ∈Ω\ Z(f s), it holds f −∗∗g(p) = ef(Tf(p))−1 eg(Tf(p)) where Tf(p) := f c(p)−1pf c(p), while ef(q) = f ◦ s (p) + Im(q)f ′ s(p) and eg(q) = g◦ s(p) + Im(q)g′ s(p) for all q ∈x + yS. 16
Theorem 7.1.
Theorem 7.1. Let Ωbe a slice domain in H and let f: Ω→H be a slice regular function. Fix a 2-sphere S = x0 + y0S (with x0, y0 ∈R and y0 >…
Theorem 7.1. Let Ωbe a slice domain in H and let f : Ω→H be a slice regular function. Fix a 2-sphere S = x0 + y0S (with x0, y0 ∈R and y0 > 0) that intersects Ωand consider the slice domain Ω′ := Ω\ O(S). Then, there exists a slice regular function h : Ω′ →H such that f(q) = [(q −x0)2 + y2 0] ∗h(q) = [(q −x0)2 + y2 0]h(q) throughout Ω′. Moreover, h cannot be continuously extended to any open set intersecting O(S).
Theorem 7.2.
Theorem 7.2. Let Ωbe a slice domain in H and let f: Ω→H be a slice regular function. Fix p ∈Ωand consider the slice domain Ω′:= Ω O(p).…
Theorem 7.2. Let Ωbe a slice domain in H and let f : Ω→H be a slice regular function. Fix p ∈Ωand consider the slice domain Ω′ := Ω\ O(p). Then there exists a slice regular function g : Ω′ →H such that f(q) = (q −p) ∗g(q) throughout Ω′. Moreover, for any slice domain Λ that intersects O(p), there cannot exist a slice regular g : Λ →H such that the previous equality holds in Λ.
Proposition 7.3.
Proposition 7.3. Let Ωbe a slice domain in H, let f: Ω→H be a slice regular function and fix x ∈Ω∩R. It holds f(x) = 0 if, and only if,…
Proposition 7.3. Let Ωbe a slice domain in H, let f : Ω→H be a slice regular function and fix x ∈Ω∩R. It holds f(x) = 0 if, and only if, there exists a slice regular function g : Ω→H such that f(q) = (q −x) ∗g(q) = (q −x)g(q) throughout Ω.
Proposition 8.4.
Proposition 8.4. Let Ωbe a slice domain in H and let f, g: Ω→H be slice regular functions. Fix an S:= x+yS intersecting Ωand a connected…
Proposition 8.4. Let Ωbe a slice domain in H and let f, g : Ω→H be slice regular functions. Fix an S := x+yS intersecting Ωand a connected component C of the intersection S ∩Ω. 1. Z(f ∗g) is the union between Z(f) and the set of points p ∈Ω\ Z(f) such that q −f(p)−1pf(p) divides g(q) near p. 2. Z(f s) intersects C if, and only if, it includes it; this happens if, and only if, there exists ep ∈S such that q −ep divides f(q) near C. 3. There exists ep ∈S such that q −ep divides f(q) near C if, and
Theorem 8.6.
Theorem 8.6. Let Ωbe a slice domain in H and let f: Ω→H be a slice regular function with f ̸≡0. Fix an S:= x + yS (with x, y ∈R, y > 0)…
Theorem 8.6. Let Ωbe a slice domain in H and let f : Ω→H be a slice regular function with f ̸≡0. Fix an S := x + yS (with x, y ∈R, y > 0) intersecting Ωand a connected component C of the intersection S ∩Ω. There exist m ∈N such that [(q −x)2 + y2]m divides f(q) near C but [(q −x)2 + y2]m+1 does not. Moreover, for the function h such that f(q) = (q −x)2 + y2mh(q) near C, there exist a number n ∈N, points p1, . . . , pn ∈S (with pi ̸= ¯pi+1 for all i ∈{1, . . . , n −1}), and a slice regular fun
Theorem 8.8.
Theorem 8.8. Let Ωbe a slice domain in H and let f: Ω→H be a slice regular function. Fix p ∈Ωand set h:= f −f(p). 25
Theorem 8.8. Let Ωbe a slice domain in H and let f : Ω→H be a slice regular function. Fix p ∈Ωand set h := f −f(p). 25
Theorem 9.2.
Theorem 9.2. Choose any sequence an n∈Z in H. Let R1, R2 ∈[0, +∞] be such that R1 = lim supm→+∞|a−m|1/m, 1/R2 = lim supn→+∞|an|1/n. For all…
Theorem 9.2. Choose any sequence {an}n∈Z in H. Let R1, R2 ∈[0, +∞] be such that R1 = lim supm→+∞|a−m|1/m, 1/R2 = lim supn→+∞|an|1/n. For all p ∈H the regular Laurent series centered at p associated to {an}n∈Z, namely, f(q) = X n∈Z (q −p)∗nan , converges absolutely and uniformly on the compact subsets of Σ(p, R1, R2) and it does not converge at any point of T (p, R1) nor at any point of H \ Σ(p, R2). Furthermore: if Ω(p, R1, R2) ̸= ∅, then the sum of the series defines a slice regular function f :
Theorem 9.3
Theorem 9.3 (Regular Laurent Expansion). Let Ωbe a domain in H, let f: Ω→H be a slice regular function and let p ∈H. There exists a…
Theorem 9.3 (Regular Laurent Expansion). Let Ωbe a domain in H, let f : Ω→H be a slice regular function and let p ∈H. There exists a sequence {an}n∈Z in H such that, for all 0 ≤R1 < R2 ≤+∞with Σ(p, R1, R2) ⊆Ω, f(q) = X n∈Z (q −p)∗nan (6) in Σ(p, R1, R2). If, moreover, Σ(p, R2) ⊆Ω, then an = 0 for all n < 0 and equality (6) holds throughout Σ(p, R2). We now give some new definitions and results. For the case of symmetric slice domains, see [14, §5.3] (derived from [27, 30]). Definition 9.4. Let f :
Theorem 9.7.
Theorem 9.7. Let Ωbe a slice domain in H and let f be semiregular in Ω. Fix p = x + yI ∈Ωand let C denote the connected component of (x +…
Theorem 9.7. Let Ωbe a slice domain in H and let f be semiregular in Ω. Fix p = x + yI ∈Ωand let C denote the connected component of (x + yS) ∩Ω that includes p. Let m := ordf(p) and n := maxep∈C ordf(ep). There exist a slice domain Λ with C ⊆Λ ⊆Ωand a slice regular function g : Λ →H such that f(q) = (q −x)2 + y2−n (q −p)∗(n−m) ∗g(q) in Λ \ (x + yS). Moreover, if n > 0 then g(p) ̸= 0.
Proposition 9.8.
Proposition 9.8. The set of semiregular functions on a slice domain Ωis a real ∗-algebra with respect to +, ∗,c. Moreover, it is a division…
Proposition 9.8. The set of semiregular functions on a slice domain Ωis a real ∗-algebra with respect to +, ∗,c. Moreover, it is a division ring.
Theorem 9.9.
Theorem 9.9. Let Ωbe a slice domain in H, let f be semiregular in Ωand suppose f ̸≡0. Fix an S:= x + yS intersecting Ωand a connected…
Theorem 9.9. Let Ωbe a slice domain in H, let f be semiregular in Ωand suppose f ̸≡0. Fix an S := x + yS intersecting Ωand a connected component C of S ∩Ω. There exist m ∈Z, n ∈N, p1, ..., pn ∈S (with pi ̸= ¯pi+1 for all i ∈{1, . . . , n −1}) such that f(q) = [(q −x)2 + y2]m(q −p1) ∗(q −p2) ∗... ∗(q −pn) ∗g(q) (7) in Ω, for some semiregular function g on Ωthat is slice regular near C and has gs |C ̸≡0.
Theorem 7.1
Theorem 7.1].
Theorem 7.1].
Theorem 10.1
Theorem 10.1 (Maximum Modulus Principle). Let Ωbe a slice domain in H and let f: Ω→H be a slice regular function. If |f| has a relative…
Theorem 10.1 (Maximum Modulus Principle). Let Ωbe a slice domain in H and let f : Ω→H be a slice regular function. If |f| has a relative maximum at p ∈Ω, then f is constant. We now prove the Minimum Modulus Principle over slice domains, exploiting the analogous result for symmetric slice domains, namely [14, Theorem 7.3].
Theorem 10.2
Theorem 10.2 (Minimum Modulus Principle). Let Ωbe a slice domain in H and let f: Ω→H be a slice regular function. If |f| has a relative…
Theorem 10.2 (Minimum Modulus Principle). Let Ωbe a slice domain in H and let f : Ω→H be a slice regular function. If |f| has a relative minimum at p ∈Ω, then either f(p) = 0 or f is constant.
Proposition 10.4.
Proposition 10.4. Let Ωbe a slice domain in H and let f: Ω→H be a slice regular function. Either f is constant or Df is a proper analytic…
Proposition 10.4. Let Ωbe a slice domain in H and let f : Ω→H be a slice regular function. Either f is constant or Df is a proper analytic subset of Ω\R. In the latter case, the dimension of Df cannot exceed 3.
Proposition 4.12
Proposition 4.12]) to the function ef on the symmetric slice domain N, we can conclude that ef is constant. Thus, f is constant in Λ,…
Proposition 4.12]) to the function ef on the symmetric slice domain N, we can conclude that ef is constant. Thus, f is constant in Λ, whence in Ωby the Identity Principle 2.4. We point out that, when Ωis not symmetric, the degenerate set Df needs not be symmetric. Indeed, the spherical derivative f ′ s may vanish identically on a connected component of (x + yS) ∩Ωwhile not vanishing elsewhere in (x + yS) ∩Ω. This happens, for instance, in Example 8.2. We are now ready for the announced theorem.
Theorem 10.5
Theorem 10.5 (Open Mapping). Let Ωbe a slice domain in H, let f: Ω→H be a nonconstant slice regular function and let Df be its degenerate…
Theorem 10.5 (Open Mapping). Let Ωbe a slice domain in H, let f : Ω→H be a nonconstant slice regular function and let Df be its degenerate set. Then f : Ω\ Df →H is open.
Lemma 11.1.
Lemma 11.1. Let f: Ω→H be a slice regular function, let I ∈S and let UI be a bounded Jordan domain in LI, with UI ⊂ΩI. If ∂UI is rectifiable…
Lemma 11.1. Let f : Ω→H be a slice regular function, let I ∈S and let UI be a bounded Jordan domain in LI, with UI ⊂ΩI. If ∂UI is rectifiable then f(z) = 1 2πI Z ∂UI ds s −z f(s) for all z ∈UI. 32
Proposition 11.2
Proposition 11.2 (Local Cauchy Formula I). Let Ωbe a slice domain in H, let f: Ω→H be a slice regular function and let ef, N, Λ be an…
Proposition 11.2 (Local Cauchy Formula I). Let Ωbe a slice domain in H, let f : Ω→H be a slice regular function and let ef, N, Λ be an extension triplet for f. If U is a bounded symmetric open subset of H with U ⊆N, if I ∈S and if the boundary ∂UI is a finite union of disjoint rectifiable Jordan curves, then f(q) = Z ∂UI (s −q)−∗(2πI)−1ds ef(s) for all q ∈U ∩Λ.
Proposition 11.3
Proposition 11.3 (Local Volume Cauchy Formula). Let Ωbe a slice domain in H, let f: Ω→H be a slice regular function and let ef, N, Λ be…
Proposition 11.3 (Local Volume Cauchy Formula). Let Ωbe a slice domain in H, let f : Ω→H be a slice regular function and let ef, N, Λ be an extension triplet for f. Let U be a bounded symmetric open subset of H with U ⊆N and assume the boundary ∂U to be C1. For w ∈∂U, let n(w) denote the outer normal versor of ∂U at w and let dσw denote the standard 3-volume form on ∂U. Then f(q) = Z ∂U C(q, w)n(w) ef(w)dσw for all q ∈U ∩Λ.
Theorem 11.4
Theorem 11.4 (Local Extension). Let f be a slice regular function on a slice domain Ωand let J0 ∈S. Let C be a symmetric, compact and…
Theorem 11.4 (Local Extension). Let f be a slice regular function on a slice domain Ωand let J0 ∈S. Let C be a symmetric, compact and path-connected subset of H such that ∅̸= C+ J0 ⊂Ω+ J0 . If C ∩R is not empty, suppose it is a closed interval included in Ω. Then there exist an extension triplet ef, N, Λ for f and a real number δ > 0 such that [ |J−J0|≤δ C+ J ⊂Λ .
Theorem 11.5
Theorem 11.5 (Local Cauchy Formula II). Let Ωbe a slice domain in H and let f: Ω→H be a slice regular function. Let U be a symmetric open…
Theorem 11.5 (Local Cauchy Formula II). Let Ωbe a slice domain in H and let f : Ω→H be a slice regular function. Let U be a symmetric open subset of H such that U is compact and path-connected. If U ∩R ̸= ∅, we assume the same intersection to be a closed interval of R included in Ω. Suppose, for some J0 ∈S, that the boundary ∂UJ0 is a finite union of disjoint rectifiable Jordan curves and that U + J0 ⊂Ω+ J0 . Then, there exists a real number ε > 0 such that, for all I ∈S and for all q ∈ [ |J−J0|<ε
Theorem 12.1.
Theorem 12.1. Let Ωbe a slice domain in H and let f: Ω→H be a slice regular function. If x0 + y0S intersects Ωand if C is a connected…
Theorem 12.1. Let Ωbe a slice domain in H and let f : Ω→H be a slice regular function. If x0 + y0S intersects Ωand if C is a connected component of the intersection, then there exist an open neighborhood U of C and quaternions {ak}k∈N such that f(q) = X n∈N [(q −x0)2 + y2 0]n[a2n + qa2n+1] for all q ∈U.
Theorem 12.2.
Theorem 12.2. Let Ωbe a slice domain in H and let f: Ω→H be a slice regular function. Suppose x0, y0, r1, r2 ∈R to fulfill the inequalities…
Theorem 12.2. Let Ωbe a slice domain in H and let f : Ω→H be a slice regular function. Suppose x0, y0, r1, r2 ∈R to fulfill the inequalities 0 < r1 < r2 ≤y0 and set U := U(x0 + y0S, r1, r2). If there exists J0 ∈S such that U + J0 ⊂Ω, then there exist a real number ε > 0 and quaternions {ak}k∈Z such that f(q) = X n∈Z [(q −x0)2 + y2 0]n[a2n + qa2n+1] for all q belonging to the open neighborhood [ |J−J0|<ε U +
Theorem 12.4.
Theorem 12.4. Let Ωbe a slice domain in H and let f: Ω→H be a slice regular function. Suppose x0, y0, r2 ∈R to fulfill the inequalities 0 <…
Theorem 12.4. Let Ωbe a slice domain in H and let f : Ω→H be a slice regular function. Suppose x0, y0, r2 ∈R to fulfill the inequalities 0 < r2 ≤y0 and set V := U(x0 + y0S, 0, r2) = U(x0 + y0S, r2) \ (x0 + y0S) . If there exist I0 ∈S and δ > 0 such that [ |J−I0|<δ V + J \ {x0 + y0J} ⊂Ω, then there exist quaternions {ak}k∈Z such that f(q) = X n∈Z [(q −x0)2 + y2 0]n[a2n + qa2n+1]
Theorem 9.9
Theorem 9.9 and Definition 9.10 to observe what follows. Suppose, for all n ≤0, there exists k < 2n such that ak ̸= 0: then every point of…
Theorem 9.9 and Definition 9.10 to observe what follows. Suppose, for all n ≤0, there exists k < 2n such that ak ̸= 0: then every point of C , except at most one, is an essential singularity for f. Suppose, instead, there exists n ≤0 such that ak = 0 for all k < 2n and let m be the maximum such n. Notice that, by construction, bΩ:= Ω∪C is a slice domain. By Proposition 9.8, f is semiregular in bΩ. Moreover, if C denotes the connected component of (x0 + y0S) ∩bΩthat includes C then: ordC f (x0 + y