On Slice Polyanalytic Functions of a Quaternionic Variable

Abstract

In this paper, we introduce the quaternionic slice polyanalytic functions and we prove some of their properties. Then, we apply the obtained results to begin the study of the quaternionic Fock and Bergman spaces in this new setting. In particular, we give explicit expressions of their reproducing kernels.Proposition 3.18. Let Ω be an axially symmetric domain and let f : Ω −→ H be a slice polyanalytic function. Assume that there exist J, K ∈ S, with J = K and U J , U K two subdomains of Ω + J and Ω + K respectively where Ω + J := Ω ∩ C + J and Ω + K := Ω ∩ C + K . If f = 0 on U J and U K , then f = 0 everywhere in Ω. Proof. Let f be a slice polyanalytic function on Ω such that f = 0 on U J and U K . Thus, since U J and U K are respectively subdomains of Ω + J and Ω + K . It follows, from the Splitting Lemma for slice polyanalytic functions combined with the classical complex analysis that f = 0 everywhere on Ω + J and Ω + K . Then, we just need to use the Representation Formula which allows to recover a slice function by its values on two semi-slices to complete the proof.

S-Polyregular Bargmann Spaces

“…Remark 2.13. Although Proposition 2.12 is Theorem 3.8 in [5], the proof we provide here is different.…”

“…5 Spectral realization of the S-polyregular Bargmann spaces 5.1 Discussion. In this section, we show that the S-polyregular Bargmann space SR 2 2,n (and therefore SR 2 1,n ) is closely connected to the concrete L 2 -spectral analysis of the slice differential operator 2 q in (5). To this end, we begin by considering the C ∞ -spectral properties of 2 q which requires to solve two problems.…”

“…The following result is a Splitting Lemma for the S-polyregular functions generalizing the standard one for the slice regular functions (see [5]).…”

“…The authors of [5] have introduced a star product for S-polyregular functions (for completness) without further results on it. In the sequel, we will review this notion and establish some related results.…”

“…Proof. Let f : H −→ H be a C ∞ -quaternionic-valued function which is solution of 2 q f = f µ on H. Then, f = f | H satisfies ∆ q f = f µ, where ∆ q denotes the restriction of the slice differential operator 2 q in (5)…”