Some characterizations of slice regular Lipschitz type spaces

Abstract

We give some characterizations of Lipschitz type spaces of slice regular functions in the unit ball of the skew field of quaternions with prescribed modulus of continuity.

1 Introduction

The quaternionic valued functions of a quaternionic variable, often referred to as slice regular functions, was born in [1, 2]. This class of functions, which would somehow resemble the classical theory of holomorphic functions of one complex variable, has been studied extensively in the last years, see [2,3,4,5,6,7,8] and the references given there.

It was shown in [9, 10] some characterization of generalized Lipschitz type spaces of holomorphic functions with prescribed behavior near the unit circle centered at the origin, determined by a regular majorant in terms of the moduli of their members. Rather surprisingly, several authors attempted to extend the aforementioned characterizations to (holomorphic) smoothness spaces of both complex and vector-valued functions, see [11,12,13,14,15,16] and the references therein.

This paper is devoted to establish some analogous results of Dyakonov’s paper [9] for the theory of slice regular quaternionic functions. The main results are Proposition 3.5 and Corollary 3.15 of Proposition 3.14 as well as Corollary 3.18 of Proposition 3.17.

2 Preliminaries

2.1 Antecedents in standard complex analysis

For the convenience of the reader, we recall the relevant material from [9, 10] and provide some additional notations and terminology, thus making our exposition self-contained.

Let ${\mathbb {D}}$ stand for the unit disc in the complex plane ${\mathbb {C}}$, ${\mathbb {S}}^{1}$ be the unit circle and $\overline{{\mathbb {D}}}\,:=\,{\mathbb {D}}\cup {\mathbb {S}}^1$. The algebra $Hol({\mathbb {D}})$ consists of those holomorphic functions on ${\mathbb {D}}$ that are continuous up to ${\mathbb {S}}^1$.

A continuous function $\omega : [0, 2] \rightarrow {\mathbb {R}}_{+}$ with $\omega (0)=0$ will be called a regular majorant if $\omega (t)$ is increasing, $\displaystyle \frac{\omega (t)}{t}$ is decreasing for $t\in [0, 2]$ and such that

$$\begin{aligned} \int _0^x \frac{\omega (t)}{ t} dt + x \int _x^2 \frac{\omega (t)}{ t^2}dt \le C\omega (x), \quad 0<x<2. \end{aligned}$$

Here and subsequently C stands for a positive real constant, not necessarily the same at each occurrence. When necessary, we will use subscripts to differentiate several constants.

Given a regular majorant the Lipschitz type space, denoted by $\Lambda _\omega ({\mathbb {D}})$, consists (by definition) of all complex valued functions f defined on ${\mathbb {D}}$ such that

$$\begin{aligned} |f(z)-f(\zeta )| \le C \omega (|z-\zeta |) , \quad \forall z,\zeta \in {\mathbb {D}}. \end{aligned}$$

The class $\Lambda _{\omega } ({\mathbb {S}}^1)$, is defined similarly.

Let us state the main results of [9] as Theorems A and B, the proofs of which were considerably shorted in [10].

Theorem A

Let $\omega$ be a regular majorant. A function f holomorphic in ${\mathbb {D}}$ is in $\Lambda _{\omega } ({\mathbb {D}})$ if and only if so is its modulus |f|.

If $f\in Hol({\mathbb {D}})$, then |f| is a subharmonic function, hence the Poisson integral of |f|, denoted by P[|f|], is equal to the smallest harmonic majorant in ${\mathbb {D}}$. In particular, $P[|f|]-|f|\ge 0$ in ${\mathbb {D}}$.

Theorem B

Let $\omega$ be a regular majorant, $f\in Hol({\mathbb {D}})$, and assume the boundary function of |f| belongs to $\Lambda _{\omega }({\mathbb {S}}^1)$. Then f is in $\Lambda _w ({\mathbb {D}})$ if and only if

$$\begin{aligned} P[|f|](z)-|f(z)| \le C \omega (1-|z|). \end{aligned}$$

The following notation will be needed

$$\begin{aligned} \Vert f\Vert _{\Lambda _{\omega } ({\mathbb {D}})} = \sup \{ \frac{|f(z)- f(\zeta )|}{\omega (|z-\zeta |)} \ \mid \ z, \zeta \in {\mathbb {D}}, \ z\ne \zeta \}, \quad \forall f\in C(\overline{{\mathbb {D}}}, {\mathbb {C}}). \end{aligned}$$

The notation ${{\mathfrak {A}}} \asymp {{\mathfrak {B}}}$ means that there exist positive constants $C_1$ and $C_2$ such that $C_1 {{\mathfrak {A}}} \le {{\mathfrak {B}}} \le C_2 {{\mathfrak {A}}}$.

Let $\omega$ be a majorant and $f\in Hol({\mathbb {D}}) \cap C(\overline{{\mathbb {D}}}, {\mathbb {C}})$. We introduce the following notations:

$$\begin{aligned} N_1(f) :=&\Vert \ |f| \ \Vert _{\Lambda _{\omega } ({\mathbb {S}}^1)} + \sup \left\{ \frac{ P[|f|] (z) - |f|(z) }{ \omega ( 1- |z|) } \ \mid \ z\in {\mathbb {D}}\right\} , \\ N_2(f) :=&\Vert \ |f| \ \Vert _{\Lambda _{\omega } ({\mathbb {S}}^1)} + \sup \left\{ \frac{ | \ |f| (\zeta ) - |f|(r \zeta ) \ | }{ \omega ( 1- r) } \ \mid \ \zeta \in {\mathbb {S}}^1, \ 0\le r < 1 \right\} ,\\ N_3(f) :=&\Vert \ |f| \ \Vert _{\Lambda _{\omega } (\overline{{\mathbb {D}}} )}. \end{aligned}$$

In particular, we have:

• 1. If $\omega$ and $\omega ^2$ are regular majorants then

  $$\begin{aligned} \Vert f\Vert _{\Lambda _{\omega }({\mathbb {D}})} \asymp \sup \left\{ \frac{\left\{ P[|f|^2] (z) - |f(z) |^2\right\} ^{\frac{1}{2} } }{ \omega ( 1- |z|) } \ \mid \ z\in {\mathbb {D}}\right\} . \end{aligned}$$

  (2.1)

• 2. If $\omega$ is a regular majorant then

  $$\begin{aligned} \Vert f\Vert _{\Lambda _{\omega }({\mathbb {D}})} \asymp N_1(f) \asymp N_2(f) \asymp N_3(f), \end{aligned}$$

  (2.2)

for any $f\in Hol({\mathbb {D}}) \cap C(\overline{{\mathbb {D}}}, {\mathbb {C}})$.

2.2 Brief introduction to slice regular functions

A quaternion is given by $q=x_0 + x_{1} {e_1} +x_{2} e_2 + x_{3} e_3$ where $x_0, x_1, x_2, x_3$ are real values and the imaginary units satisfy: $e_1^2=e_2^2=e_3^2=-1$, $e_1e_2=-e_2e_1=e_3$, $e_2e_3=-e_3e_2=e_1$, $e_3e_1=-e_1e_3=e_2$. The skew field of quaternions is denoted by ${\mathbb {H}}$. The sets $\{e_1,e_2,e_3\}$ and $\{1,e_1,e_2,e_3\}$ are called the standard basis of ${\mathbb {R}}^3$ and ${\mathbb {H}}$, respectively. The vector part of $q\in {\mathbb {H}}$ is $\mathbf{{q}}= x_{1} {e_1} +x_{2} e_2 + x_{3} e_3$ and its real part is $q_0=x_0$. The quaternionic conjugation of q is ${\bar{q}}=q_0-\mathbf{q}$ and its norm is $\Vert q\Vert \,:=\,\sqrt{x_0^2 +x_1^2+x_2^3+x_3^2}= \sqrt{q{\bar{q}}} = \sqrt{{\bar{q}} q}$.

By abuse of notation, the unit open ball in ${\mathbb {H}}$ will be denoted by ${\mathbb {D}}^4\,:=\,\{q \in {\mathbb {H}}\ \mid \ \Vert q\Vert <1 \}$ so will the unit spheres in ${\mathbb {R}}^3$ (in ${\mathbb {H}}$) by ${\mathbb {S}}^2\,:=\,\{\mathbf{q}\in {\mathbb {R}}^3 \mid \Vert \mathbf{q}\Vert =1\}$ (${\mathbb {S}}^3\,:=\,\{{q}\in {\mathbb {H}} \mid \Vert {q}\Vert =1\}$), respectively.

The quaternionic structure allows us to see that $\mathbf{i}^{2}=-1$, for every $\mathbf{i}\in {\mathbb {S}}^2$. Then ${\mathbb {C}}(\mathbf{i})\,:=\,\{x+\mathbf{i}y; \ |\ x,y\in {\mathbb {R}}\}\cong {\mathbb {C}}$ as fields, and any $q\in {\mathbb {H}} \setminus {\mathbb {R}}$ may be rewritten by $x+ \mathbf{I}_q y$ where $x, y\in {\mathbb {R}}$ and $\mathbf{I}_q\,:=\,\Vert \mathbf{q}\Vert ^{-1}{} \mathbf{q}\in {\mathbb {S}}^2$; i.e., $q\in {\mathbb {C}}({\mathbf{I}_q})$. Note that $q\in {\mathbb {R}}$ belongs to every complex plane.

Given $u\in {\mathbb {S}}^3$, the mapping $\mathbf{q} \mapsto u\mathbf{q}{\bar{u}}$ for all $\mathbf{q}\in {\mathbb {R}}^3$ is a quaternionic rotation that preserves ${\mathbb {R}}^3$, see [17, 18]. For any $\mathbf{i} \in {\mathbb {S}}^2$ we will write

$$\begin{aligned} {{\mathbb {D}}}_\mathbf{i}\,:=\, {\mathbb {D}}^4 \cap {\mathbb {C}}(\mathbf{i}) \end{aligned}$$

and

$$\begin{aligned} {{\mathbb {S}}}_\mathbf{i}\,:=\, {\mathbb {S}}^2 \cap {\mathbb {C}}(\mathbf{i}). \end{aligned}$$

Now, we recall few aspects of the slice regular functions theory of [4,5,6, 8, 19].

Definition 2.1

Let $\Omega \subset {\mathbb {H}}$ be an open domain. A real differentiable function $f:\Omega \rightarrow {\mathbb {H}}$ is called (left) slice regular function on $\Omega$ if

$$\begin{aligned} {\overline{\partial }}_{\mathbf {i}}f\mid _{_{\Omega \cap {\mathbb {C}}(\mathbf {i})}}\,:=\,\frac{1}{2}\left( \frac{\partial }{\partial x}+{\mathbf {i}} \frac{\partial }{\partial y}\right) f\mid |_{_{\Omega \cap {\mathbb {C}}(\mathbf {i})}}=0 \ \ \text {on }\Omega _{\mathbf {i}}\,:=\,\Omega \cap {\mathbb {C}}(\mathbf {i}), \end{aligned}$$

for all $\mathbf{i}\in {\mathbb {S}}^2$ and its derivative, or Cullen derivative, see [1], is given by

$$\begin{aligned} f'=\displaystyle {\partial }_{\mathbf{i}}f\mid _{_{\Omega \cap {\mathbb {C}}(\mathbf{i})}} = \frac{\partial }{\partial x} f\mid _{_{\Omega \cap {\mathbb {C}}(\mathbf{i})}}= \partial _xf\mid _{_{\Omega \cap {\mathbb {C}}(\mathbf{i})}}. \end{aligned}$$

Let ${{\mathcal {S}}}{{\mathcal {R}}}(\Omega )$ denote the right linear space of slice regular functions on $\Omega$.

Definition 2.2

A set $U\subset {\mathbb {H}}$ is called axially symmetric if $x+\mathbf{i}y \in U$ with $x,y\in {\mathbb {R}}$, then $\{x+\mathbf{j}y \ \mid \ \mathbf{j}\in {\mathbb {S}}^2\}\subset U$ and $U\cap {\mathbb {R}}\ne \emptyset$. A domain $U\subset {\mathbb {H}}$ is called slice domain, or s-domain, if $U_\mathbf{i} = U\cap {\mathbb {C}}(\mathbf{i})$ is a domain in ${\mathbb {C}}(\mathbf{i})$ for all $\mathbf{i}\in {\mathbb {S}}^2$.

Let $\Omega \subset {\mathbb {H}}$ an axially symmetric s-domain. A function $f\in {{\mathcal {S}}}{{\mathcal {R}}}(\Omega )$ is said to be intrinsic if $f(q)=\overline{f({\bar{q}})}$ for all $q\in \Omega$. The real linear space of intrinsic slice regular functions on $\Omega$ will be denoted by ${\mathcal {N}}(\Omega )$, see [7, 19]. We will denote by $Z_f$ the set of zeroes of function f.

Theorem 2.3

Let $\Omega \subset {\mathbb {H}}$ be an axially symmetric s-domain and $f\in {{\mathcal {S}}}{{\mathcal {R}}}(\Omega )$.

• 1. (Splitting Property) For every $\mathbf{i}, \mathbf{j}\in {\mathbb {S}}$, orthogonal to each other, there exist holomorphic functions $F, G:\Omega _\mathbf{i} \rightarrow {\mathbb {C}}({\mathbf{i}})$ such that $f_{\mid _{\Omega _\mathbf{i}}} =F +G \mathbf{j}$ on $\Omega _\mathbf{i}$, see [4].

• 2. (Representation Formula) For every $q=x+\mathbf{I}_q y \in \Omega$ with $x,y\in {\mathbb {R}}$ and $\mathbf{I}_q \in {\mathbb {S}}^2$ the following identity holds

  $$\begin{aligned} f(x+\mathbf{I}_q y) = \frac{1}{2}[ f(x+\mathbf{i}y)+ f(x-\mathbf{i}y)] + \frac{1}{2} \mathbf{I}_q \mathbf{i}[ f(x-\mathbf{i}y)- f(x+\mathbf{i}y)], \end{aligned}$$

  for all $\mathbf{i}\in {\mathbb {S}}^2$, see [5].

In the case of $\Omega ={{\mathbb {D}}^4}$ and given $f, g\in {{\mathcal {S}}}{{\mathcal {R}}}({\mathbb {D}}^4)$ there exist two sequences of quaternions $(a_n)$ and $(b_n)$ such that

$$\begin{aligned} f(q) = \sum _{n=0}^{\infty } q^n a_n,\ g(q) = \sum _{n=0}^{\infty } q^n b_n. \end{aligned}$$

The product $f*g$ is defined as $f*g(q)\,:=\,\sum _{n=0}^{\infty } q^n \sum _{k=0}^n a_k b_{n-k}$ for all $q\in {{\mathbb {D}}^4}$.

For $f(q)\ne 0$ the following property holds

$$\begin{aligned} f * g(q)= f(q)g(f(q)^{-1} qf(q)), \end{aligned}$$

see [4]. What is more, if $f^s$ has no zeroes, the $*$-inverse of f is given by

$$\begin{aligned} f^{-*} =\displaystyle \frac{1}{f^s} * f^c \end{aligned}$$

and

$$\begin{aligned} (f^{-*})'= - f^{-*} * f'* f^{-*}, \end{aligned}$$

where $f^c(q)\,:=\, \sum _{n=0}^{\infty } q^n \overline{a_n}$ for all $q\in {{\mathbb {D}}^4}$ and $f^s \,:=\, f* f^c = f^c * f$, see [4, 20, 21].

3 Main results

Definition 3.1

Let $\omega$ be a regular majorant and $\mathbf{i} \in {\mathbb {S}}^2$. The set of all functions $f:{\mathbb {D}}^4 \rightarrow {\mathbb {H}}$ such that

$$\begin{aligned} \Vert f(x)-f(y)\Vert \le C \omega (\Vert x - y \Vert ) , \quad \forall x,y \in {{\mathbb {D}}}_\mathbf{i} \end{aligned}$$

will be denoted by ${}_\mathbf{i} \Lambda _{\omega }({\mathbb {D}}^4).$

We write ${}_\mathbf{i} \Lambda _{\omega }({\mathbb {S}}^3)$ for the set of all functions $f:{\mathbb {S}}^3 \rightarrow {\mathbb {H}}$ such that

$$\begin{aligned} \Vert f(x)-f(y)\Vert \le C \omega (\Vert x - y \Vert ) , \quad \forall x,y \in {{\mathbb {S}}}_\mathbf{i}. \end{aligned}$$

The norm of a function $f\in {}_\mathbf{i}\Lambda _\omega ({\mathbb {D}}^4)$ is defined as

$$\begin{aligned} \Vert f\Vert _{ {}_\mathbf{i}\Lambda _\omega ({\mathbb {D}}^4)} = \sup \left\{ \frac{ \Vert f(x)- f(y) \Vert }{ \omega (\Vert x-y\Vert )} \ \mid \ x,y \in {\mathbb {D}}_\mathbf{i}, \ x\ne y\right\} . \end{aligned}$$

Definition 3.2

Let $\omega _1 ,\omega _2$ be regular majorants and $\mathbf{i}, \mathbf{j} \in {\mathbb {S}}^2$ orthogonal to each other. We write ${}_\mathbf{i} \Lambda _{\omega _1, \omega _2}({\mathbb {D}}^4)$ for the set of all functions $f:{\mathbb {D}}^4 \rightarrow {\mathbb {H}}$ such that

$$\begin{aligned} \Vert f_k(x)-f_k(y)\Vert \le C_k \omega _k (\Vert x - y \Vert ) , \quad \forall x,y \in {{\mathbb {D}}}_\mathbf{i}, \end{aligned}$$

for $k=1,2$, where $f\mid _{{\mathbb {D}}_\mathbf{i}} = f_1+f_2 \mathbf{j}$ with $f_1,f_2:{{\mathbb {D}}}_{\mathbf{i}}\rightarrow {\mathbb {C}}(\mathbf{i})$. The set ${}_\mathbf{i} \Lambda _{\omega _1, \omega _2}({\mathbb {S}}^3)$ consists of all $f:{\mathbb {S}}^3\rightarrow {\mathbb {H}}$ such that

$$\begin{aligned} \Vert f_k(x)-f_k(y)\Vert \le C_k \omega _k (\Vert x - y \Vert ) , \quad \forall x,y \in {{\mathbb {S}}}_\mathbf{i}, \end{aligned}$$

for $k=1,2$, where $f\mid _{{\mathbb {D}}_\mathbf{i}} = f_1+f_2 \mathbf{j}$ with $f_1,f_2:{{\mathbb {S}}}_{\mathbf{i}}\rightarrow {\mathbb {C}}(\mathbf{i})$. For $f\in {}_\mathbf{i}\Lambda _{\omega _1, \omega _2} ({\mathbb {D}}^4)$ we define

$$\begin{aligned} \Vert f\Vert _{ {}_\mathbf{i}\Lambda _{\omega _1, \omega _2}({\mathbb {D}}^4)}^2 = \sup \left\{ \frac{ \Vert f_1(x)- f_1(y) \Vert ^2 }{ \omega _1(\Vert x-y\Vert )^2} + \frac{ \Vert f_2(x)- f_2(y) \Vert ^2 }{ \omega _2(\Vert x-y\Vert )^2} \ \mid \ x,y \in {\mathbb {D}}_\mathbf{i}, \ x\ne y\right\} . \end{aligned}$$

Given $\mathbf{i}\in {\mathbb {S}}^2$, the $\mathbf{i}$-Poisson integral of $u\in C({{\mathbb {S}}}_\mathbf{i}, {\mathbb {R}})$ is

$$\begin{aligned} P_{\mathbf{i}}[u](q) =\frac{1}{2\pi } \int _0^{2\pi } u(e^{\mathbf{i} t}) \frac{ 1 -\Vert q\Vert ^2}{ \Vert q- e^{\mathbf{i} t} \Vert ^2} dt, \quad q\in {\mathbb {D}}^4. \end{aligned}$$

Remark 3.3

• 1. Let $f\in {\mathbb {D}}^4 \rightarrow {\mathbb {H}}$ and $f=f_1+f_2\mathbf{j}$ on ${{\mathbb {D}}}_\mathbf{i}$ with $f_1, f_2: {{\mathbb {D}}}_\mathbf{i} \rightarrow {{\mathbb {D}}}_\mathbf{i}$. Then

  $$\begin{aligned} 2f_1 = f-\mathbf{i}f \mathbf{i}, \ \ 2f_2 \mathbf{j} = f+\mathbf{i}f \mathbf{i}, \ \ \text {on } \ {{\mathbb {D}}}_\mathbf{i}, \end{aligned}$$

  where $\mathbf{i}, \mathbf{j}\in {\mathbb {S}}^2$ are orthogonal to each other. If $\mathbf{j}'$ is another orthogonal vector to $\mathbf{i}$ and $f=g_1+g_2\mathbf{j}'$ on ${{\mathbb {D}}}_\mathbf{i}$ then $f_1=g_1$, $f_2 =- g_2 \mathbf{j}'\overline{ \mathbf{j}}$ and due to the usage of the quaternionic norm in the previous definitions we see that theses do not depends of the choose of $\mathbf{j}$, since

  $$\begin{aligned} \Vert f_2(x)-f_2(y)\Vert = \Vert g_2(x)-g_2(y)\Vert , \quad \forall x,y\in {\mathbb {D}}_\mathbf{i}. \end{aligned}$$

• 2. Let $\omega , \omega _1 ,\omega _2$ regular majorants and $f:{\mathbb {D}}^4 \rightarrow {\mathbb {H}}$ with $f\mid _{{\mathbb {D}}_\mathbf{i}} = f_1+f_2\mathbf{j},$ where $f_1,f_2:{{\mathbb {D}}}_{\mathbf{i}}\rightarrow {\mathbb {C}}(\mathbf{i})$, with $\mathbf{i}, \mathbf{j} \in {\mathbb {S}}^2$ are orthogonal to each other. Due to inequalities

  $$\begin{aligned} \left. \begin{array}{l} \Vert f_1(x)-f_1(y)\Vert \\ \Vert f_2(x)-f_2(y)\Vert \end{array} \right\} \le \Vert f(x)-f(y)\Vert \le \Vert f_1(x)-f_1(y)\Vert + \Vert f_2(x)-f_2(y)\Vert , \end{aligned}$$

  for all $x,y\in {\mathbb {D}}_\mathbf{i}$, we get that

  $$\begin{aligned} {}_\mathbf{i} \Lambda _{\omega , \omega }({\mathbb {D}}^4) = {}_\mathbf{i} \Lambda _{\omega }({\mathbb {D}}^4) \end{aligned}$$

  and

  $$\begin{aligned} {}_\mathbf{i} \Lambda _{\omega _1, \omega _2}({\mathbb {D}}^4) \subset {}_\mathbf{i} \Lambda _{\omega _1+ \omega _2}({\mathbb {D}}^4). \end{aligned}$$

  Similar relationships are obtained for ${}_\mathbf{i} \Lambda _{\omega }({\mathbb {S}}^3)$ and ${}_\mathbf{i} \Lambda _{\omega _1, \omega _2}({\mathbb {S}}^3)$.

Definition 3.4

The symbol $G \Lambda _{\omega }({\mathbb {D}}^4)$ stands for the set of all quaternionic-valued functions f defined on ${\mathbb {D}}^4$ such that

$$\begin{aligned} \Vert f(x)-f(y)\Vert \le C \omega (\Vert x- y \Vert ) , \quad \forall x,y \in {{\mathbb {D}}}^4. \end{aligned}$$

Meanwhile, $G \Lambda _{\omega }({\mathbb {S}}^3)$ denotes the set of all quaternionic-valued functions f defined on ${\mathbb {S}}^3$ such that

$$\begin{aligned} \Vert f(x)-f(y)\Vert \le C \omega (\Vert x- y\Vert ) , \quad \forall x,y \in {{\mathbb {S}}}^3. \end{aligned}$$

Proposition 3.5

Let $\omega _1 ,\omega _2$ regular majorants and $\mathbf{i} \in {\mathbb {S}}^2$. Then

$$\begin{aligned} {}_\mathbf{i} \Lambda _{\omega _1, \omega _2}({\mathbb {D}}^4)\cap {{\mathcal {S}}}{{\mathcal {R}}}({\mathbb {D}}^4) \ \subset \ G \Lambda _{\omega _1+ \omega _2}({\mathbb {D}}^4)\cap {{\mathcal {S}}}{{\mathcal {R}}}({\mathbb {D}}^4) \ \subset \ {}_\mathbf{i} \Lambda _{\omega _1 + \omega _2}({\mathbb {D}}^4)\cap {{\mathcal {S}}}{{\mathcal {R}}}({\mathbb {D}}^4) . \end{aligned}$$

Proof

The relationship $G \Lambda _{\omega _1+ \omega _2}({\mathbb {D}}^4)\cap {{\mathcal {S}}}{{\mathcal {R}}}({\mathbb {D}}^4)\subset {}_\mathbf{i} \Lambda _{\omega _1 +\omega _2}({\mathbb {D}}^4)\cap {{\mathcal {S}}}{{\mathcal {R}}}({\mathbb {D}}^4)$ is a direct consequence of Definition 3.4.

On the other hand, we shall see that given $f\in {}_\mathbf{i} \Lambda _{\omega _1, \omega _2}({\mathbb {D}}^4)\cap {{\mathcal {S}}}{{\mathcal {R}}}({\mathbb {D}}^4)$ there exists a constant $L>0$ such that

$$\begin{aligned} \Vert f (p)-f (q)\Vert \le L \sum _{k=1}^2 \omega _k (\Vert p-q\Vert ) , \quad \forall p,q\in {\mathbb {D}}^4. \end{aligned}$$

Given $p,q\in {\mathbb {D}}^4$ consider the following cases:

• 1. Suppose that $\mathbf{p}$ and $\mathbf{q}$ are both the zero vector. By the Splitting Property we get

  $$\begin{aligned} \Vert f (p)-f (q)\Vert \le \,&\Vert f_1 (p)-f_1 (q)\Vert + \Vert f_2 (p)-f_2 (q)\Vert \\ \le \,&C_3\left( \omega _1 (\Vert p - q \Vert ) + \omega _2 (\Vert p - q \Vert ) \right) , \end{aligned}$$

  where $C_3= \max \{C_1, C_2\}$.

• 2. Suppose $\mathbf{p}$ is not the zero vector while $\mathbf{q}$ is. Consider $z= p_0 + \mathbf{i} |\mathbf{p}|$ and $\zeta = q = {\bar{\zeta }}$. Combining the Representation Formula with Splitting Property we obtain

  $$\begin{aligned} 2\Vert f (p)-f (q)\Vert = \,&\Vert \left( 1-\frac{\mathbf{p}}{\Vert \mathbf{p}\Vert } \mathbf{i}\right) (f(z)- f(\zeta )) + \left( 1+\frac{\mathbf{p}}{\Vert \mathbf{p}\Vert } \mathbf{i}\right) (f({\bar{z}} ) - f({\bar{\zeta }})) \Vert \\ \le \,&2\Vert f(z)- f(\zeta )\Vert + 2 \Vert f({\bar{z}} ) - f({\bar{\zeta }}) \Vert \\ \le \,&2 \sum _{k=1}^2 ( \Vert f_k(z)- f_k(\zeta )\Vert + \Vert f_k({\bar{z}} ) - f_k({\bar{\zeta }}) \Vert )\\ \le \,&4C_3 \left( \omega _1 (\Vert p-q \Vert ) + \omega _2 (\Vert p - q \Vert ) \right) . \end{aligned}$$

  where $\Vert z-\zeta \Vert = \Vert p- q\Vert$ is used.

• 3. Consider $p, q\in {\mathbb {D}}^4$ such that neither $\mathbf{p}$ nor $\mathbf{q}$ is the zero vector. Set $z= p_0 + \mathbf{i} |\mathbf{p}|$ and $\zeta = q_0 + \mathbf{i} |\mathbf{q}|.$ Representation Formula gives

  $$\begin{aligned}&\ 2\Vert f (p)-f (q)\Vert \\&\quad = \left\| \left\{ \left( 1-\frac{\mathbf{p}}{\Vert \mathbf{p}\Vert } \mathbf{i}\right) f(z) + \left( 1+\frac{\mathbf{p}}{\Vert \mathbf{p}\Vert } \mathbf{i}\right) f({\bar{z}}) - \left( 1-\frac{\mathbf{q}}{\Vert \mathbf{q}\Vert } \mathbf{i}\right) f(\zeta ) - \left( 1+\frac{\mathbf{q}}{\Vert \mathbf{q}\Vert } \mathbf{i}\right) f({\bar{\zeta }}) \right\} \right\| \\&\quad = \left\| \left\{ \left( 1-\frac{\mathbf{p}}{\Vert \mathbf{p}\Vert } \mathbf{i}\right) (f(z)- f(\zeta ) ) + \left( 1+\frac{\mathbf{p}}{\Vert \mathbf{p}\Vert } \mathbf{i}\right) (f({\bar{z}}) - f({\bar{\zeta }}) ) \right. \right. \\&\qquad \left. \left. + \left( \frac{\mathbf{p}}{\Vert \mathbf{p}\Vert } \mathbf{i} -\frac{\mathbf{q}}{\Vert \mathbf{q}\Vert } \mathbf{i} \right) ( f({\bar{\zeta }}) -f(\zeta ) ) \right\} \right\| \\&\quad \le 2 \Vert f(z)- f(\zeta ) \Vert + 2 \Vert f({\bar{z}}) - f({\bar{\zeta }}) \Vert + \left\| \frac{\mathbf{p}}{\Vert \mathbf{p}\Vert } -\frac{\mathbf{q}}{\Vert \mathbf{q}\Vert } \right\| \Vert f({\bar{\zeta }}) -f(\zeta )\Vert \\&\quad \le 4 C_3 \sum _{k=1}^2 \omega _k (\Vert z - \zeta \Vert ) + \left\| \frac{\mathbf{p}}{\Vert \mathbf{p}\Vert } -\frac{\mathbf{q}}{\Vert \mathbf{q}\Vert } \right\| C_3 \sum _{k=1}^2 \omega _k (\Vert {\bar{\zeta }} - \zeta \Vert ). \end{aligned}$$

  Note that $\Vert z-\zeta \Vert = \sqrt{ (p_0-q_0)^2 + (\Vert \mathbf{p}\Vert - \Vert \mathbf{q} \Vert )^2 }\le \Vert p-q\Vert$ and as $\omega _1$ and $\omega _2$ are increasing functions then

  $$\begin{aligned} 2\Vert f (p)-f (q)\Vert \le&4 C_3 \left\{ \sum _{k=1}^2 \left( \omega _k (\Vert p-q\Vert ) + \frac{1}{4} \Vert \frac{\mathbf{p}}{\Vert \mathbf{p}\Vert } -\frac{\mathbf{q}}{\Vert \mathbf{q}\Vert } \Vert \omega _k (2 \Vert \mathbf{q} \Vert ) \right) \right\} . \end{aligned}$$

  (3.1)

  If $2\Vert \mathbf{q} \Vert \le \Vert p-q\Vert$ then $\omega _k (2 \Vert \mathbf{q} \Vert ) \le \omega _k (\Vert p-q\Vert )$, for $k=1,2$, and

  $$\begin{aligned} \displaystyle \Vert f (p)-f (q)\Vert \le 3 C_3 \sum _{k=1}^2 \omega _k (\Vert p-q\Vert ). \end{aligned}$$

  On the other hand, if $\Vert p-q\Vert < 2\Vert \mathbf{q}\Vert$, from (3.1), we get

  $$\begin{aligned} \frac{\Vert f (p)-f (q)\Vert }{\sum \nolimits _{k=1}^2 \omega _k (\Vert p-q\Vert ) } \le \,&2C_3 \left\{ 1 + \frac{1}{4} \left\| \frac{\mathbf{p}}{\Vert \mathbf{p}\Vert } -\frac{\mathbf{q}}{\Vert \mathbf{q}\Vert } \right\| \frac{\sum \nolimits _{k=1}^2 \omega _k (2 \Vert \mathbf{q} \Vert )}{ \sum \nolimits _{k=1}^2 \omega _k (\Vert p-q\Vert ) } \right\} \\ \\ \le \,&2C_3 \left\{ 1 + \frac{1}{4} \left\| \frac{\mathbf{p}}{\Vert \mathbf{p}\Vert } -\frac{\mathbf{q}}{\Vert \mathbf{q}\Vert } \right\| \sum _{k=1}^2 \frac{ \omega _k (2 \Vert \mathbf{q} \Vert )}{ \omega _k (\Vert p-q\Vert ) } \right\} \\ \\ \le \,&2C_3 \left\{ 1 + \frac{1}{4} \left\| \dfrac{\mathbf{p}}{\Vert \mathbf{p}\Vert } -\dfrac{\mathbf{q}}{\Vert \mathbf{q}\Vert } \right\| \dfrac{ 2\Vert \mathbf{q} \Vert }{ \Vert p-q\Vert } \sum _{k=1}^2 \dfrac{ \dfrac{ \omega _k (2 \Vert \mathbf{q} \Vert ) }{ 2 \Vert \mathbf{q} \Vert } }{ \dfrac{ \omega _k (\Vert p-q\Vert )}{ \Vert p-q\Vert } } \right\} \\ \end{aligned}$$

  As $\dfrac{\omega _k(t)}{t}$ is decreasing, for $k=1,2$, and $\Vert p-q\Vert < 2\Vert \mathbf{q} \Vert$ then

  $$\begin{aligned} \sum _{k=1}^2 \dfrac{ \dfrac{ \omega _k (2 \Vert \mathbf{q} \Vert ) }{ 2 \Vert \mathbf{q} \Vert } }{ \dfrac{ \omega _k (\Vert p-q\Vert )}{ \Vert p-q\Vert } } \le 2 \end{aligned}$$

  and

  $$\begin{aligned} \Vert f (p)-f (q)\Vert \le \,&2C_3 \left\{ 1 + \left\| \dfrac{\mathbf{p}}{\Vert \mathbf{p}\Vert } -\dfrac{\mathbf{q}}{\Vert \mathbf{q}\Vert } \right\| \dfrac{\Vert \mathbf{q} \Vert }{\Vert p-q\Vert }\right\} \displaystyle \sum _{k=1}^2 \omega _k (\Vert p-q\Vert ). \end{aligned}$$

  It is easily seen that

  $$\begin{aligned} \Vert \dfrac{\mathbf{p}}{\Vert \mathbf{p}\Vert } -\dfrac{\mathbf{q}}{\Vert \mathbf{q}\Vert } \Vert \dfrac{\Vert \mathbf{q} \Vert }{ \Vert p-q\Vert } \le \,&\left\| \dfrac{\mathbf{p}}{\Vert \mathbf{p}\Vert } -\dfrac{\mathbf{p}}{\Vert \mathbf{q}\Vert } + \dfrac{\mathbf{p}}{\Vert \mathbf{q}\Vert } -\dfrac{\mathbf{q}}{\Vert \mathbf{q}\Vert } \right\| \dfrac{\Vert \mathbf{q} \Vert }{ \Vert p - q \Vert } \\ \le \,&\dfrac{|\Vert \mathbf{q} \Vert -\Vert \mathbf{p} \Vert |}{\Vert \mathbf{p}\Vert \Vert \mathbf{q}\Vert } \dfrac{\Vert \mathbf{p}{} \mathbf{q} \Vert }{ \Vert p-q\Vert } + \dfrac{\Vert \mathbf{p} - \mathbf{q} \Vert }{ \Vert p-q\Vert } \le 2, \end{aligned}$$

  and

  $$\begin{aligned} \Vert f (p)-f (q)\Vert \le \,&6C_3 \sum _{k=1}^2 \omega _k (\Vert p-q\Vert ), \end{aligned}$$

which completes the proof by choosing $L=6C_3$. $\square$

Remark 3.6

We have proved more, namely that for $\omega _1=\omega _2= \omega$ we are lead to

$$\begin{aligned} {}_\mathbf{i} \Lambda _{\omega }({\mathbb {D}}^4)\cap {{\mathcal {S}}}{{\mathcal {R}}}({\mathbb {D}}^4) = G \Lambda _{\omega }({\mathbb {D}}^4)\cap {{\mathcal {S}}}{{\mathcal {R}}}({\mathbb {D}}^4). \end{aligned}$$

Therefore, every slice regular function space associated to a majorant on a fix slice, it is also associated to the same majorant on the four-dimensional unit ball and reciprocally.

We proceed to describe some algebraic properties of the previously introduced functions sets.

Proposition 3.7

Set $\mathbf {i} \in {{\mathbb {S}}}^2$.

• 1. Given a regular majorant $\omega$, the sets ${}_\mathbf{i} \Lambda _{\omega } ({\mathbb {D}}^4) \cap {{\mathcal {S}}}{{\mathcal {R}}}({\mathbb {D}}^4)$ and $G \Lambda _{\omega } ({{{\mathbb {D}}^4}})\cap {{\mathcal {S}}}{{\mathcal {R}}} ({\mathbb {D}}^4)$ are quaternionic right linear spaces.

• 2. Let $\omega _1,\omega _2$ be two regular majorants and let $f,g\in {}_\mathbf{i} \Lambda _{\omega _1, \omega _2} ({\mathbb {D}}^4)$. For every $a\in {\mathbb {H}}$ we have $f+g \in {}_\mathbf{i} \Lambda _{\omega _1, \omega _2} ({\mathbb {D}}^4)$ and

  $$\begin{aligned} fa \in {}_\mathbf{i} \Lambda _{\Vert a_1\Vert \omega _1 + \Vert a_2\Vert \omega _2, \ \Vert a_2\Vert \omega _1 + \Vert a_1\Vert \omega _2} ({\mathbb {D}}^4), \end{aligned}$$

  where $a=a_1+ a_2 \mathbf{j}$ with $a_1,a_2\in {\mathbb {C}}(\mathbf{i})$ and $\mathbf{j}$ is orthogonal to $\mathbf{i}$.

Proof

• 1. Given $f,g\in {}_\mathbf{i} \Lambda _{\omega }({\mathbb {D}}^4)\cap {{\mathcal {S}}}{{\mathcal {R}}}({\mathbb {D}}^4)$ and $a\in {\mathbb {H}}$ we see that

  $$\begin{aligned} \Vert (fa+ g)(x)-(fa+ g)(y)\Vert \le \,&\Vert a\Vert \Vert f(x)-f(y)\Vert + \Vert g(x)-g(y)\Vert \\ \le \,&C\omega (\Vert x - y \Vert ) ,\ \ \forall x,y \in {{\mathbb {D}}}_\mathbf{i}. \end{aligned}$$

  Similar inequalities are used to see that $G \Lambda _{\omega }({\mathbb {D}}^4)\cap {{\mathcal {S}}}{{\mathcal {R}}}({\mathbb {D}}^4)$ is a quaternionic right linear space.

• 2. Given $f,g\in {}_\mathbf{i} \Lambda _{\omega _1, \omega _2} ({\mathbb {D}}^4)$ we have $f+g \in {}_\mathbf{i} \Lambda _{\omega _1, \omega _2} ({\mathbb {D}}^4)$ following a similar computation to the above. Denote $a=a_1+ a_2 \mathbf{j}$, $f\mid _{{\mathbb {D}}_\mathbf{i}}=f_1+ f_2 \mathbf{j}$, $g\mid _{{\mathbb {D}}_\mathbf{i}}=g_1+ g_2 \mathbf{j}$ with $\mathbf{j}\in {\mathbb {S}}^2$ orthogonal to $\mathbf{i}$ and $a_1,a_2\in {\mathbb {C}}(\mathbf{i})$ and $f_1,f_2,g_1,g_2 \in Hol({\mathbb {D}}_\mathbf{i})$. We obtain that

  $$\begin{aligned} fa \mid _{{\mathbb {D}}_\mathbf{i}}\, =\, (f_1 a_1 - f_2 {\bar{a}}_2 ) + (f_1 a_2 + f_2 {\bar{a}}_1)\mathbf{j} \end{aligned}$$

  and

  $$\begin{aligned} \Vert (f_1 a_1 - f_2 {\bar{a}}_2 )(x)- (f_1 a_1 - f_2 {\bar{a}}_2 )(y) \Vert \le \,&C_1 (\omega _1(\Vert x-y \Vert ) \Vert a_1\Vert \\&+ \omega _2(\Vert x-y \Vert ) \Vert a_2\Vert ), \end{aligned}$$
  $$\begin{aligned} \Vert (f_1 a_2 + f_2 {\bar{a}}_1 )(x)- (f_1 a_2 + f_2 {\bar{a}}_1 )(y) \Vert \le \,&C_2 (\omega _1(\Vert x-y \Vert ) \Vert a_2\Vert \\&+ \omega _2(\Vert x-y \Vert ) \Vert a_1\Vert ). \end{aligned}$$

  for all $x,y\in {{\mathbb {D}}}_\mathbf{i}$. Note that picking out $\max \{\Vert a_1\Vert , \Vert a_2\Vert \}$ we can prove that

  $$\begin{aligned} {}_\mathbf{i} \Lambda _{\Vert a_1\Vert \omega _1 + \Vert a_2\Vert \omega _2, \ \Vert a_2\Vert \omega _1 + \Vert a_1\Vert \omega _2} ({\mathbb {D}}^4)\subset {}_\mathbf{i} \Lambda _{ \omega _1 +\omega _2} ({\mathbb {D}}^4). \end{aligned}$$

$\square$

Corollary 3.8

Let $\omega _1$, $\omega _2$ be two regular majorants and $f\in {}_\mathbf{i} \Lambda _{\omega } ({\mathbb {D}}^4) \cap {\mathcal {N}}({\mathbb {D}}^4)$. Then

$$\begin{aligned} \Vert f\Vert _{ {}_\mathbf{i}\Lambda _{\omega _1}({\mathbb {D}}^4)} = \Vert f\Vert _{ {}_\mathbf{k}\Lambda _{\omega _1}({\mathbb {D}}^4)} = \Vert f\Vert _{ {}_\mathbf{i}\Lambda _{\omega _1, \omega _2}({\mathbb {D}}^4)}, \end{aligned}$$

for all $\mathbf{k}\in {\mathbb {S}}^2$.

Proof

Note that given $f \in {\mathcal {N}}(\Omega )$ there exists a sequence of real numbers $(a_{n})_{n=0}^{\infty }$, see [7, 19], such that $f(q)= \sum _{n=0}^{\infty } q^n a_n$ for all $q\in {\mathbb {D}}^4$. Therefore for all $u\in {\mathbb {S}}^3$ one has that

$$\begin{aligned} \Vert f(x) - f(y)\Vert&= \Vert u \left( \sum _{n=0}^{\infty } x^n a_n - \sum _{n=0}^{\infty } y^n a_n \right) {\bar{u}} \Vert = \Vert \sum _{n=0}^{\infty } (ux{\bar{u}})^n a_n - \sum _{n=0}^{\infty } (uy{\bar{u}})^n a_n \Vert \\&= \Vert f(u x{\bar{u}}) - f(uy{\bar{u}})\Vert . \end{aligned}$$

Choosing $u\in {\mathbb {S}}^3$ such that $u\mathbf{i}{\bar{u}} = \mathbf{k}$ one obtains the first equality and for the second one we see that $f\mid _{{\mathbb {D}}_\mathbf{i}} = f\mid _{{\mathbb {D}}_\mathbf{i}} + 0 \mathbf{j}$, i.e., $f_1=f\mid _{{\mathbb {D}}_\mathbf{i}}$ and $f_2=0$ in Definition 3.2. $\square$

Remark 3.9

Note that if $f \in {}_\mathbf{i}\Lambda _{\omega }({\mathbb {D}}^4)\cap C({{\mathbb {D}}}_\mathbf{i}, {\mathbb {H}})$, then $f\mid _{{{\mathbb {D}}}_\mathbf{i}}$ can be extended to a continuous function on $\overline{{{\mathbb {D}}}_\mathbf{i}}$. Similarly, if $f \in G\Lambda _{\omega }({\mathbb {D}}^4)\cap C({\mathbb {D}}^4, {\mathbb {H}})$, then f can be extended to a continuous function on $\overline{{\mathbb {D}}^4}$.

Now, we shall extend [9, Theorems 1 and 2] to slice regular function theory.

Proposition 3.10

• 1. Set $\mathbf {i} \in {\mathbb {S}}^2$ and $f \in {{\mathcal {S}}}{{\mathcal {R}}}({\mathbb {D}}^4) \cap C(\overline{{\mathbb {D}}^4}, {\mathbb {H}})$. Let $\omega$ and $\omega ^2$ regular majorants. Then

  $$\begin{aligned} \Vert f\Vert _{{}_\mathbf{i}\Lambda _{\omega }({\mathbb {D}}^4)}^2 \asymp&\ \sup \left\{ \frac{P[\Vert f_1\Vert ^2] (x) - \Vert f_1(x)\Vert ^2 }{ \omega ( 1- \Vert x\Vert )^2 } \ \mid \ x\in {\mathbb {D}}_\mathbf{i}\right\} \\ \\&\ + \sup \left\{ \frac{ P[\Vert f_2\Vert ^2] (x) - \Vert f_2(x)\Vert ^2 }{ \omega ( 1- \Vert x\Vert )^2 } \ \mid \ x\in {\mathbb {D}}_\mathbf{i}\right\} , \end{aligned}$$

  where $f\mid _{{{\mathbb {D}}}_\mathbf{i}}=f_1+f_2 \mathbf{j}$ with $\mathbf{j}\in {\mathbb {S}}^2$ orthogonal to $\mathbf{i}$ and $f_1,f_2 \in Hol({{\mathbb {D}}}_\mathbf{i})$.

• 2. Set $\mathbf {i} \in {\mathbb {S}}^2$ and $f \in {{\mathcal {S}}}{{\mathcal {R}}}({\mathbb {D}}^4) \cap C(\overline{{\mathbb {D}}^4}, {\mathbb {H}})$. Let $\omega$ be a regular majorant. Then

  $$\begin{aligned} \Vert f\Vert _{{}_\mathbf{i}\Lambda _{\omega }({\mathbb {D}}^4)}^2 \asymp N_1(f_1)^2 + N_1(f_2)^2 \asymp N_2(f_1)^2 + N_2(f_2)^2 \asymp N_3(f_1)^2 + N_3(f_2)^2 , \end{aligned}$$

  where $f\mid _{{{\mathbb {D}}}_\mathbf{i}}=f_1+ f_2 \mathbf{j}$ with $\mathbf{j}\in {\mathbb {S}}^2$ orthogonal to $\mathbf{i}$ and $f_1,f_2 \in Hol({{\mathbb {D}}}_\mathbf{i})$.

• 3. Set $\mathbf {i}, \mathbf{k} \in {\mathbb {S}}^2$ and consider the regular majorants $\omega , \omega _1,\omega _2$.

  • (a) If $f\in {}_\mathbf{i} \Lambda _{\omega } ({\mathbb {D}}^4) \cap {{\mathcal {S}}}{{\mathcal {R}}}({\mathbb {D}}^4) \cap C(\overline{{\mathbb {D}}^4}, {\mathbb {H}})$, then

    $$\begin{aligned} \Vert f\Vert _{ {}_\mathbf{i}\Lambda _\omega ({\mathbb {D}}^4)}\le 2 \Vert f\Vert _{ {}_\mathbf{k}\Lambda _\omega ({\mathbb {D}}^4)}\le 4 \Vert f\Vert _{ {}_\mathbf{i}\Lambda _\omega ({\mathbb {D}}^4)}. \end{aligned}$$

  • (b) If $f \in {}_\mathbf{i} \Lambda _{\omega , \omega }({\mathbb {D}}^4)$, then

    $$\begin{aligned} \Vert f\Vert _{ {}_\mathbf{i}\Lambda _{\omega , \omega }({\mathbb {D}}^4)} = \Vert f\Vert _{ {}_\mathbf{i}\Lambda _{\omega }({\mathbb {D}}^4)} . \end{aligned}$$

    If $f \in {}_\mathbf{i} \Lambda _{\omega _1, \omega _2}({\mathbb {D}}^4)$ then

    $$\begin{aligned} \Vert f\Vert _{ {}_\mathbf{i}\Lambda _{\omega _1 + \omega _2}({\mathbb {D}}^4)} \le \Vert f\Vert _{ {}_\mathbf{i}\Lambda _{\omega _1, \omega _2}({\mathbb {D}}^4)}. \end{aligned}$$

Proof

• 1. By (2.1) and the fact that $\alpha \asymp \beta$ and $\delta \asymp \gamma$ imply $\alpha ^2 + \delta ^2 \asymp \beta ^2 + \gamma ^2$. Also, the application of inequalities

  $$\begin{aligned} \Vert f\Vert _{{}_\mathbf{i}\Lambda _{\omega }({\mathbb {D}}^4)}^2 \le \Vert f_1 \Vert _{{}_\mathbf{i}\Lambda _{\omega }({\mathbb {D}}_\mathbf{i})}^2 + \Vert f_2\Vert _{{}_\mathbf{i}\Lambda _{\omega }({\mathbb {D}}_\mathbf{i})}^2 \le 2 \Vert f\Vert _{{}_\mathbf{i}\Lambda _{\omega }({\mathbb {D}}^4)}^2. \end{aligned}$$

• 2. By (2.2) and the properties stated above.

• 3. Fact (a) follows from direct computations and the idea of the ensuing Fact (b) is the following:

  $$\begin{aligned} \Vert f\Vert _{ {}_\mathbf{i}\Lambda _{\omega , \omega }({\mathbb {D}}^4)}^2&= \sup \left\{ \frac{ \Vert f_1(x)- f_1(y) \Vert ^2 }{ \omega (\Vert x-y\Vert )^2} + \frac{ \Vert f_2(x)- f_2(y) \Vert ^2 }{ \omega (\Vert x-y\Vert )^2} \ \mid \ x,y \in {\mathbb {D}}_\mathbf{i}, \ x\ne y\right\} \\&= \sup \left\{ \frac{ \Vert f (x)- f (y) \Vert ^2 }{ \omega (\Vert x-y\Vert )^2} \ \mid \ x,y \in {\mathbb {D}}_\mathbf{i}, \ x\ne y\right\} = \Vert f\Vert _{ {}_\mathbf{i}\Lambda _{\omega }({\mathbb {D}}^4)}^2. \end{aligned}$$

$\square$

Corollary 3.11

• 1. Set $\mathbf {i} \in {\mathbb {S}}^2$ and $f \in {\mathcal {N}}({\mathbb {D}}^4) \cap C(\overline{{\mathbb {D}}^4}, {\mathbb {H}})$. If $\omega$ and $\omega ^2$ are regular majorant, then

  $$\begin{aligned} \sup \left\{ \frac{ P[\Vert f\mid _{{\mathbb {D}}_\mathbf{i}} \Vert ^2] (x) - \Vert f\mid _{{\mathbb {D}}_\mathbf{i}}(x)\Vert ^2 }{ \omega ( 1- \Vert x\Vert ) } \ \mid \ x\in {\mathbb {D}}_\mathbf{i}\right\} \asymp \Vert f\Vert _{{}_\mathbf{i}\Lambda _{\omega }({\mathbb {D}}^4)} . \end{aligned}$$

• 2. Suppose $\mathbf {i} \in {\mathbb {S}}^2$, $f \in {\mathcal {N}}({\mathbb {D}}^4) \cap C(\overline{{\mathbb {D}}^4}, {\mathbb {H}})$ and $\omega$ a regular majorant. Then

  $$\begin{aligned} \Vert f\Vert _{{}_\mathbf{i}\Lambda _{\omega }({\mathbb {D}}^4)}^2 \asymp N_1(f\mid _{{\mathbb {D}}_\mathbf{i}}) \asymp N_2(f\mid _{{\mathbb {D}}_\mathbf{i}} ) \asymp N_3(f\mid _{{\mathbb {D}}_\mathbf{i}}) , \end{aligned}$$

Proof

Both facts follow from $f_1=f_1\mid _{{\mathbb {D}}_\mathbf{i}}$ and $f_2=0$ in Definition 3.2 since $f \in {\mathcal {N}}({\mathbb {D}}^4)$ and $f(q)= \sum \nolimits _{n=0}^{\infty } q^n a_n$ for all $q\in {\mathbb {D}}^4$ iff $a_n\in {\mathbb {R}}$ for all n, see [7, 19]. $\square$

As the function sets given in Definitions 3.1, 3.2 and 3.4 depend of unit vectors the following proposition shows some relationships between them.

Proposition 3.12

Set $\mathbf {i} \in {\mathbb {S}}^2$ and consider the regular majorants $\omega , \omega _1,\omega _2$.

• 1. If $f\in {}_\mathbf{i} \Lambda _{\omega } ({\mathbb {D}}^4)$ then $\Vert f\Vert , \Vert f\pm \mathbf{i}f\mathbf{i} \Vert \in {}_\mathbf{i} \Lambda _{\omega } ({\mathbb {D}}^4)$.

• 2. ${}_\mathbf{i} \Lambda _{\omega }({\mathbb {D}}^4)={}_\mathbf{i}\Lambda _{\omega , \omega }({\mathbb {D}}^4)$.

• 3. ${}_\mathbf{i}\Lambda _{\omega _1, \omega _2}({\mathbb {D}}^4) \subset {}_\mathbf{i} \Lambda _{\omega _1+\omega _2}({\mathbb {D}}^4)$.

• 4. ${}_\mathbf{i} \Lambda _{\omega }({\mathbb {D}}^4) \cap {{\mathcal {S}}}{{\mathcal {R}}}({\mathbb {D}}^4) = {}_\mathbf{j} \Lambda _{\omega }({\mathbb {D}}^4) \cap {{\mathcal {S}}}{{\mathcal {R}}}({\mathbb {D}}^4)$.

Proof

• 1. Given $f\in {}_\mathbf{i} \Lambda _{\omega } ({\mathbb {D}}^4)$ set $f=f_1+f_2\mathbf{j}$ on ${{\mathbb {D}}}_\mathbf{i}$, where $\mathbf{j}\in {\mathbb {S}}^2$ is orthogonal to $\mathbf{i}$ and $f_1,f_2 : {{\mathbb {D}}}_\mathbf{i} \rightarrow {{\mathbb {D}}}_\mathbf{i}$. Then we see that

  $$\begin{aligned} 2f_1 = f-\mathbf{i}f \mathbf{i}, \ \ 2f_2 \mathbf{j} = f+\mathbf{i}f \mathbf{i}, \ \ \text {on } \ {{\mathbb {D}}}_\mathbf{i}. \end{aligned}$$

  (3.2)

  From inequalities

  $$\begin{aligned} |\ \Vert f(x)\Vert - \Vert f(y)\Vert \ | \le \Vert f(x)- f(y)\Vert , \\ \max \{ |\ \Vert f_1(x)\Vert - \Vert f_1(y)\Vert \ |, \ |\ \Vert f_2(x)\Vert -\Vert f_2(y)\Vert \ | \}\le \Vert f(x)- f(y)\Vert , \end{aligned}$$

  for all $x,y \in {{\mathbb {D}}}_\mathbf{i}$, it follows that $\Vert f\Vert , \Vert f\pm \mathbf{i}f\mathbf{i}\Vert \in {}_\mathbf{i} \Lambda _{\omega } ({\mathbb {D}}^4)$.

• 2. and 3. With the previous notation the identity

  $$\begin{aligned} \Vert f(x)- f(y)\Vert ^2 = \Vert f_1(x)-f_1(y)\Vert ^2 + \Vert f_2(x)-f_2(y)\Vert ^2 ,\quad \forall x,y\in {{\mathbb {D}}}_\mathbf{i}, \end{aligned}$$

  (3.3)

  holds, allowing us to see that

  $$\begin{aligned} {}_\mathbf{i} \Lambda _{\omega }({\mathbb {D}}^4)={}_\mathbf{i}\Lambda _{\omega , \omega }({\mathbb {D}}^4) \end{aligned}$$

  and

  $$\begin{aligned} {}_\mathbf{i}\Lambda _{\omega _1, \omega _2}({\mathbb {D}}^4) \subset {}_\mathbf{i} \Lambda _{\omega _1+\omega _2}({\mathbb {D}}^4). \end{aligned}$$

• 4. Given $f \in {}_\mathbf{i} \Lambda _{\omega }({\mathbb {D}}^4) \cap {{\mathcal {S}}}{{\mathcal {R}}}({\mathbb {D}}^4)$ and $\mathbf{j}\in {\mathbb {S}}^2$, according to the Representation Formula, we have for every $x_0+ x_1\mathbf{j}, y_0+ y_1\mathbf{j} \in {{\mathbb {D}}}_{\mathbf{j}}$ with $x_0,x_1, y_0,y_1 \in {\mathbb {R}}$

  $$\begin{aligned}&\Vert f(x_0+ x_1\mathbf{j}) - f(y_0+ y_1\mathbf{j})\Vert \\&= \frac{1}{2}\Vert ( 1- \mathbf{j} \mathbf{i})(f(x) - f(y)) + ( 1 + \mathbf{j} \mathbf{i}) (f({\bar{x}}) - f({\bar{y}} ) ) \Vert \\&\le \Vert f( x ) - f(y )\Vert + \Vert f({\bar{x}}) - f({\bar{y}}) \Vert \le 2 C \omega (\Vert x-y\Vert ) \\&\le 2 C \omega (\Vert (x_0+ x_1\mathbf{j}) - (y_0+ y_1\mathbf{j})\Vert ), \end{aligned}$$

  where $x = x_0+ x_1\mathbf{i}$ and $y=y_0+ y_1\mathbf{i}$.

$\square$

Remark 3.13

Repeating the computations presented in Propositions 3.7 and 3.12 enables us to see that ${}_\mathbf{i} \Lambda _{\omega }({\mathbb {S}}^3)$, ${}_\mathbf{i}\Lambda _{\omega _1, \omega _2}({\mathbb {S}}^3)$ and $G \Lambda _{\omega }({\mathbb {S}}^3)$ have similar properties of ${}_\mathbf{i} \Lambda _{\omega }({\mathbb {D}}^4)$, ${}_\mathbf{i}\Lambda _{\omega _1, \omega _2}({\mathbb {D}}^4)$ and $G \Lambda _{\omega }({\mathbb {D}}^4)$, respectively, and that is why they are omitted.

Let ${\omega }$, $\omega _1$ and $\omega _2$ be regular majorant. The next propositions characterize the elements of ${{\mathcal {S}}}{{\mathcal {R}}}({\mathbb {D}}^4)\cap {}_\mathbf{i} \Lambda _{\omega }({\mathbb {D}}^4)$ and of ${{\mathcal {S}}}{{\mathcal {R}}}({\mathbb {D}}^4)\cap {}_\mathbf{i} \Lambda _{\omega _1,\omega _2} ({\mathbb {D}}^4)$, which extend results contained in [9, 10].

Proposition 3.14

• 1. Set $f\in {{\mathcal {S}}}{{\mathcal {R}}}({\mathbb {D}}^4)$. Then $f\in {}_\mathbf{i} \Lambda _{\omega }({\mathbb {D}}^4)$ if and only if

  $$\begin{aligned} \Vert f'(x) \pm \mathbf{i} f'(x) \mathbf{i}\Vert \le C\frac{\omega (1-\Vert x\Vert )}{1-\Vert x\Vert }, \end{aligned}$$

  for C independent of $x\in {{\mathbb {D}}}_\mathbf{i}$.

• 2. Set $f\in {{\mathcal {S}}}{{\mathcal {R}}}({\mathbb {D}}^4)\cap C(\overline{{\mathbb {D}}^4}, {\mathbb {H}})$. Then

  $$\begin{aligned} \frac{1}{2}(1-\Vert x\Vert )\Vert f'(x) \pm \mathbf{i} f'(x) \mathbf{i}\Vert + \Vert f(x) \pm \mathbf{i} f(x) \mathbf{i}\Vert \le 2 M_x, \end{aligned}$$

  for C independent of $x\in {{\mathbb {D}}}_\mathbf{i}$, where

  $$\begin{aligned} M_x = \sup \{\Vert f(y)\Vert \ \mid \ \Vert y - x\Vert \le 1-\Vert x\Vert , \ y\in {{\mathbb {D}}}_\mathbf{i}\}. \end{aligned}$$

• 3. Set $f\in {{\mathcal {S}}}{{\mathcal {R}}}({\mathbb {D}}^4)$. Then $f\in {}_\mathbf{i} \Lambda _{\omega } ({\mathbb {D}}^4)$ if and only if

  $$\begin{aligned} \Vert f'(x)\Vert \le C\frac{\omega (1-\Vert x\Vert )}{1-\Vert x\Vert }, \end{aligned}$$

  for C independent of $x\in {{\mathbb {D}}}_\mathbf{i}$.

• 4. Set $f\in {{\mathcal {S}}}{{\mathcal {R}}}({\mathbb {D}}^4)\cap {}_\mathbf{i} \Lambda _{\omega _1, \omega _2} ({\mathbb {D}}^4)$. Then

  $$\begin{aligned} \Vert f '(x) \Vert \le C \left( \frac{ \sqrt{ \omega _1 (1-\Vert x\Vert )^2+\omega _2 (1-\Vert x\Vert )^2 } }{(1-\Vert x\Vert )}\right) , \end{aligned}$$

  for C independent of $x\in {{\mathbb {D}}}_\mathbf{i}$.

• 5. Set $f\in {{\mathcal {S}}}{{\mathcal {R}}}({\mathbb {D}}^4)\cap C(\overline{{\mathbb {D}}^4},{\mathbb {H}})$. Then

  $$\begin{aligned}&\frac{1}{4}(1-\Vert x\Vert )^2\Vert f'(x)\Vert ^2 + \Vert f(x)\Vert ^2 \\&\quad \le \, (\Vert x\Vert -1) (\ \Vert f_1'(x)\Vert \Vert f_1(x)\Vert + \Vert f_2'(x)\Vert \Vert f_2(x)\Vert \ ) + M_{1,x}^2+ M_{2,x}^2,\end{aligned}$$

  for $x\in {{\mathbb {D}}}_\mathbf{i}$, where $f\mid _{{{\mathbb {D}}}_\mathbf{i}}= f_1+ f_2\mathbf{j}$ with $f_1, f_2\in Hol({{\mathbb {D}}}_\mathbf{i}) \cap C(\overline{{\mathbb {D}}}_\mathbf{i}, {\mathbb {C}}(\mathbf{i}))$ and

  $$\begin{aligned} M_{k,x} = \sup \{\Vert f_k(y)\Vert \ \mid \ \Vert y-x\Vert \le 1-\Vert x\Vert , \ y\in {{\mathbb {D}}}_\mathbf{i}\}, \end{aligned}$$

  for $k=1,2$.

• 6. Consider $f\in {{\mathcal {S}}}{{\mathcal {R}}}({\mathbb {D}}^4)\cap C(\overline{{\mathbb {D}}^4}, {\mathbb {H}})$. Let us introduce one more piece of notation:

  $$\begin{aligned} M= \sup \{ \Vert f(w) \Vert \ \mid \ w\in {{\mathbb {D}}}_\mathbf{i}\}. \end{aligned}$$

  If there exists $\mathbf{i}\in {\mathbb {S}}^2$ and a regular majorant $\omega$ such that

  $$\begin{aligned} \Vert M^2- \overline{f(x)} f({\tilde{x}})\Vert \le C(1+\Vert x\Vert )\omega (1-\Vert x\Vert ), \end{aligned}$$

  for C independent of $x\in {{\mathbb {D}}}_\mathbf{i}\setminus Z_{f'}$, where

  $$\begin{aligned} {\tilde{x}} =\,&{(\overline{f (x)})}^{-1}T_g (f' (x)^{-1}x f' (x) ) \overline{ f (x)}, \\ g(x)= \,&1-\overline{f(x)} * f(x), \end{aligned}$$

  where $T_g (q) = (g^c(q))^{-1}q g^c(q)$ for all $q \in {\mathbb {D}}^4$ such that $g^ s(q)\ne 0$, then

  $$\begin{aligned} \Vert f'(x)\Vert \le \frac{C}{M} \frac{\omega (1-\Vert x\Vert )}{ 1-\Vert x\Vert }; \end{aligned}$$

  that is, $f\in {}_\mathbf{i}\Lambda _{\omega }({\mathbb {D}}^4)$, see Fact 3. of the present proposition.

Proof

• 1. With the Splitting Property in mind, consider $f_1,f_2\in Hol({{\mathbb {D}}}_\mathbf{i})$ such that $f\mid _{{{\mathbb {D}}}_\mathbf{i}} = f_1+f_2 \mathbf{j}$ where $\mathbf{j}\in {\mathbb {S}}^2$ is orthogonal to $\mathbf{i}$. From Fact 2. of Proposition 3.12 one has that $f\in {}_\mathbf{i} \Lambda _{\omega } ({\mathbb {D}}^4)$ if and only if $f_1,f_2\in \Lambda _{\omega , \omega } ({{\mathbb {D}}}_\mathbf{i})$, that is,

  $$\begin{aligned} \Vert f '(x) \pm \mathbf{i} f '(x) \mathbf{i}\Vert \le C\frac{\omega (1-\Vert x\Vert )}{1-\Vert x\Vert }, \end{aligned}$$

  for C independent of $x\in {{\mathbb {D}}}_\mathbf{i}$, where we use [10, Lemma 1]. The result is obtained from (3.2) applied to $f'$.

• 2. Splitting Property implies that $f\mid _{{{\mathbb {D}}}_\mathbf{i}} = f_1+f_2 \mathbf{j}$ where $f_1,f_2\in Hol({{\mathbb {D}}}_\mathbf{i})$ and $\mathbf{j}\in {\mathbb {S}}^2$ is orthogonal to $\mathbf{i}$. From Fact 2. of Proposition 3.12 one has that $f\in {}_\mathbf{i} \Lambda _{\omega }({\mathbb {D}}^4)$ if and only if $f_1,f_2\in \Lambda _{\omega , \omega }({{\mathbb {D}}}_\mathbf{i})$. Applying [10, Lemma 2] to the complex components of $f\mid _{{{\mathbb {D}}}_\mathbf{i}}$ and using (3.2) in f and $f'$ to finish the proof.

• 3. It follows from Fact 1. commbining with the following consequence of the parallelogram identity:

  $$\begin{aligned} 4\Vert f'(x)\Vert ^2 = \Vert f'(x) + \mathbf{i} f'(x) \mathbf{i}\Vert ^2+ \Vert f'(x) - \mathbf{i} f'(x) \mathbf{i}\Vert ^2, \end{aligned}$$

  for all $x\in {{\mathbb {D}}}_\mathbf{i}$.

• 4. Kipping in mind the Splitting Property, set $f_1,f_2\in Hol({{\mathbb {D}}}_\mathbf{i})$ such that $f\mid _{{{\mathbb {D}}}_\mathbf{i}} = f_1+f_2 \mathbf{j}$ where $\mathbf{j}\in {\mathbb {S}}^2$ is orthogonal to $\mathbf{i}$. From Fact 2. of Proposition 3.12 one has that $f\in {}_\mathbf{i} \Lambda _{\omega }({\mathbb {D}}^4)$ if and only if $f_1,f_2\in \Lambda _{\omega , \omega }({{\mathbb {D}}}_\mathbf{i})$, i.e.,

  $$\begin{aligned} \Vert f_k'(x) \Vert \le C \frac{\omega _k (1-\Vert x\Vert )}{1-\Vert x\Vert }, \end{aligned}$$

  for C independent of $x\in {{\mathbb {D}}}_\mathbf{i}$ and for $k=1,2$, see [10, Lemma 1]. Applying (3.3) to $f'$ yields

  $$\begin{aligned} \Vert f'(x)\Vert ^2\le C^2 \left( \frac{\omega _1 (1-\Vert x\Vert )^2+\omega _2 (1-\Vert x\Vert )^2}{(1-\Vert x\Vert )^2}\right) . \end{aligned}$$

• 5. Given $f\mid _{{{\mathbb {D}}}_\mathbf{i}}= f_1+ f_2\mathbf{j}$ with $f_1, f_2\in Hol({{\mathbb {D}}}_\mathbf{i}) \cap C(\bar{{\mathbb {D}}}_\mathbf{i}, {\mathbb {C}}(\mathbf{i}))$ from [10, Lemma 2] we see that

  $$\begin{aligned} \frac{1}{2} (1- \Vert x\Vert ) \Vert f_k'(x)\Vert +\Vert f_k(x)\Vert \le M_{k,x}, \end{aligned}$$

  for $x\in {{\mathbb {D}}}_\mathbf{i}$, where $M_{k,x} = \sup \{ \Vert f_k (y)\Vert \ \mid \ \Vert y-x\Vert \le 1-\Vert x\Vert , \ y\in {{\mathbb {D}}}_\mathbf{i}\}$ for $k=1,2$. Therefore

  $$\begin{aligned} \frac{1}{4} (1- \Vert x\Vert )^2\Vert f_k'(x)\Vert ^2+ (1- \Vert x\Vert )\Vert f_k'(x)\Vert \Vert f_k(x)\Vert + \Vert f_k(x)\Vert ^2\le M_{k,x}^2, \end{aligned}$$

  for $k=1,2$. Adding terms in the previous inequalities and using (3.3) applied to f and $f'$, the main result follows.

• 6. Let $f\in {{\mathcal {S}}}{{\mathcal {R}}}({\mathbb {D}}^4)\cap C(\overline{{\mathbb {D}}^4}, {\mathbb {H}})$. With the notation $F(x)=\dfrac{f(x)}{M}$ for all $x\in {\mathbb {D}}^4$ rewrite the hypothesis as follows:

  $$\begin{aligned} \Vert 1- \overline{F(x)} F({\tilde{x}})\Vert \le \frac{C}{M^2}(1+\Vert x\Vert )\omega (1-\Vert x\Vert ), \end{aligned}$$

  for C independent of x, or equivalently

  $$\begin{aligned} \frac{\Vert 1- \overline{F(x)} F({\tilde{x}})\Vert }{1- \Vert x\Vert ^2} \le \frac{C }{M^2}\ \frac{\omega (1-\Vert x\Vert ) }{ 1- \Vert x\Vert }. \end{aligned}$$

  Equation (3.10) of [22, Theorem 3.7 (Schwarz-Pick lemma)] shows that if $f:{\mathbb {D}}^4 \rightarrow {\mathbb {D}}^4$ is a slice regular function and $q_0 \in {\mathbb {D}}^4$ implies that

  $$\begin{aligned} \Vert \partial _c f *(1 - f (q_0) *f (q))^{ -*} \Vert _{\mid _{q_0}} \le \frac{ 1}{1 - |q_0|^2} . \end{aligned}$$

  One can also see [23]. This finally yields

  $$\begin{aligned} \Vert F'(x)\Vert \le \frac{C}{M^2} \frac{\omega (1-\Vert x\Vert )}{ 1-\Vert x\Vert }, \end{aligned}$$

  that is

  $$\begin{aligned} \Vert f'(x)\Vert \le \frac{C}{M} \frac{\omega (1-\Vert x\Vert )}{ 1-\Vert x\Vert } \end{aligned}$$

  and Fact 6. is proved

$\square$

Corollary 3.15

• 1. Set $f\in {{\mathcal {S}}}{{\mathcal {R}}}({\mathbb {D}}^4)$. Then $f\in {}_\mathbf{i} \Lambda _{\omega }({\mathbb {D}}^4)$ if and only if

  $$\begin{aligned} \Vert f '(q) \Vert \le C\frac{\omega (1-\Vert q\Vert )}{1-\Vert q\Vert }, \end{aligned}$$

  for some C independent of the choice of $q\in {{\mathbb {D}}}^4$.

• 2. Set $f\in {{\mathcal {S}}}{{\mathcal {R}}}({\mathbb {D}}^4)\cap C(\overline{{\mathbb {D}}^4}, {\mathbb {H}})$ then

  $$\begin{aligned}&\frac{1}{2}(1-\Vert q\Vert )\Vert f'(q)\Vert + \Vert f(q) \Vert \\&\quad \le \ \ \sup \left\{ \Vert f(y)\Vert \ \mid \ \sqrt{ (\Re y - q_0)^2 +(\Im y - |\mathbf{q}| )^2} \le 1-\Vert q\Vert , \ y \in {{\mathbb {D}}}_\mathbf{i}\right\} \\&\qquad + \sup \left\{ \Vert f(y)\Vert \ \mid \ \sqrt{ (\Re y - q_0)^2 +(\Im y + |\mathbf{q}| )^2} \le 1-\Vert q\Vert , \ y \in {{\mathbb {D}}}_\mathbf{i}\right\} . \end{aligned}$$

• 3. If $f\in {{\mathcal {S}}}{{\mathcal {R}}}({\mathbb {D}}^4)\cap {}_\mathbf{i} \Lambda _{\omega _1, \omega _2} ({\mathbb {D}}^4)$ then

  $$\begin{aligned} \Vert f '(q) \Vert \le C \left( \frac{\omega _1 (1-\Vert q\Vert )+\omega _2 (1-\Vert q\Vert ) }{(1-\Vert q\Vert )}\right) , \end{aligned}$$

  for C independent of the choice of $q\in {{\mathbb {D}}}^4$.

Proof

• 1. The sufficient condition is a direct consequence of the previous proposition and

  $$\begin{aligned} \Vert f'(x) \pm \mathbf{i} f'(x) \mathbf{i}\Vert \le 2 \Vert f'(x) \Vert , \quad \forall x\in {\mathbb {D}}_\mathbf{i}. \end{aligned}$$

  On the other hand, let $f\in {}_\mathbf{i} \Lambda _{\omega }({\mathbb {D}}^4)\cap {{\mathcal {S}}}{{\mathcal {R}}}({\mathbb {D}}^4)$ and $q\in {\mathbb {D}}^4$ such that $\mathbf{q}$ is not the vector zero. Applying Representation Formula we deduce that

  $$\begin{aligned} 2\Vert f '(q)\Vert&\le 2\Vert f'(x)\Vert + 2\Vert f '({\bar{x}})\Vert \le \Vert f'(x) + \mathbf{i} f'(x) \mathbf{i}\Vert + \Vert f'(x) - \mathbf{i} f'(x) \mathbf{i}\Vert \\&\quad + \Vert f'({\bar{x}}) + \mathbf{i} f'({\bar{x}}) \mathbf{i}\Vert + \Vert f'({\bar{x}}) - \mathbf{i} f'({\bar{x}}) \mathbf{i}\Vert \\&\le 4 C\frac{\omega (1-\Vert x\Vert )}{1-\Vert x\Vert } = 4 C\frac{\omega (1-\Vert q\Vert )}{1-\Vert q\Vert }, \end{aligned}$$

  where $x= q_0 + \mathbf{i}|\mathbf{q}|\in {{\mathbb {D}}}_\mathbf{i}.$

• 2. For fixed $q\in {\mathbb {D}}^4$ such that $\mathbf{q}$ is not the zero vector, let $x=x_0 + \mathbf{i}|\mathbf{q}| \in {\mathbb {D}}_\mathbf{i}$. By the Representation Formula we obtain

  $$\begin{aligned}&\ \frac{1}{2}(1-\Vert q\Vert )\Vert f'(q)\Vert + \Vert f(q) \Vert \\&\quad \le \left( \ \frac{1}{2}(1-\Vert x\Vert ) \Vert f'(x)\Vert + \Vert f(x)\Vert \ \right) + \left( \ \frac{1}{2}(1-\Vert {\bar{x}}\Vert ) \Vert f'({\bar{x}})\Vert + \Vert f({\bar{x}})\Vert \ \right) \\&\quad \le \left( \ \frac{1}{4}(1-\Vert x\Vert )\Vert f'(x) + \mathbf{i} f'(x) \mathbf{i}\Vert + \frac{1}{2} \Vert f(x) + \mathbf{i} f(x) \mathbf{i}\Vert \Vert \ \right) \\&\qquad + \left( \ \frac{1}{4}(1-\Vert x\Vert ) \Vert f'(x) - \mathbf{i} f'(x) \mathbf{i}\Vert + \frac{1}{2} \Vert f(x) - \mathbf{i} f(x) \mathbf{i}\Vert \ \right) \\&\qquad + \left( \ \frac{1}{4}(1-\Vert {\bar{x}}\Vert ) \Vert f'({\bar{x}}) + \mathbf{i} f'({\bar{x}}) \mathbf{i}\Vert + \frac{1}{2} \Vert f ({\bar{x}}) + \mathbf{i} f({\bar{x}}) \mathbf{i}\Vert \Vert \ \right) \\&\qquad + \left( \ \frac{1}{4}(1-\Vert {\bar{x}}\Vert ) \Vert f'({\bar{x}}) - \mathbf{i} f'({\bar{x}}) \mathbf{i}\Vert + \frac{1}{2} \Vert f({\bar{x}}) - \mathbf{i} f({\bar{x}}) \mathbf{i}\Vert \ \right) \\&\quad \le \ M_x + M_{{\bar{x}}}, \end{aligned}$$

  where

  $$\begin{aligned} M_x =\left\{ \Vert f(y)\Vert \ \mid \ \sqrt{ (\Re y - q_0)^2 +(\Im y - |\mathbf{q}| )^2} \le 1-\Vert q\Vert , \ y \in {{\mathbb {D}}}_\mathbf{i}\right\} .\\ M_{{\bar{x}}} =\left\{ \Vert f(y)\Vert \ \mid \ \sqrt{ (\Re y - q_0)^2 +(\Im y + |\mathbf{q}| )^2} \le 1-\Vert q\Vert , \ y \in {{\mathbb {D}}}_\mathbf{i}\right\} . \end{aligned}$$

• 3. It follows in a similar way like Fact 1.

$\square$

Here are some properties of $P_{\mathbf{i}}$ and its dependence on $\mathbf{i}$.

Proposition 3.16

Let $\mathbf {i} \in {\mathbb {S}}^2$. The following items hold

• 1. Given $r\in {\mathbb {S}}^3$ write $T_r({q})\,:=\,rq{\bar{r}}$ for all $q\in {\mathbb {D}}^4$. Then

  $$\begin{aligned} P_{T_r(\mathbf{i})} [u] (q) = P_\mathbf{i} [u\circ T_{r}] (T_{r}^{-1}(q)), \end{aligned}$$

  for all $u \in C^1 ( {\mathbb {S}}_{T_r(\mathbf{i})}, {\mathbb {R}})$.

• 2. Given $f \in C(\overline{{\mathbb {D}}^4}, {\mathbb {H}})$ we have on ${{\mathbb {D}}}_\mathbf{i}$ that

  $$\begin{aligned} \displaystyle P_\mathbf{i}[\Vert f\pm \mathbf{i} f\mathbf{i}\Vert ] \le 2 P_\mathbf{i}[\Vert f\Vert ] \le P_\mathbf{i}[\Vert f- \mathbf{i} f\mathbf{i}\Vert ] + P_\mathbf{i}[\Vert f+ \mathbf{i} f\mathbf{i}\Vert ]. \end{aligned}$$

• 3. Given $f\in {{\mathcal {S}}}{{\mathcal {R}}}({{\mathbb {D}}^4})\cap C(\overline{{\mathbb {D}}^4}, {\mathbb {H}})$ and $\mathbf{j}\in {\mathbb {S}}^2$ it follows that

  $$\begin{aligned} \displaystyle \frac{1}{2\pi }\int _{0}^{2\pi } \Vert (x- e^{\mathbf{j} t})^{-*2}* f(e^{\mathbf{j} t}) \Vert (1-\Vert x\Vert ^2) dt \le 2 P_\mathbf{i} [\Vert f\Vert ](x), \end{aligned}$$

  for all $x\in {{\mathbb {D}}}_\mathbf{i}$.

Proof

• 1. Since $e^{r\mathbf{i} {\bar{r}} t}= re^{ \mathbf{i} t} {\bar{r}}$ we have

  $$\begin{aligned} P_{T_r(\mathbf{i})} [u] (q) = \frac{1}{2\pi }\int _{0}^{2\pi } u(e^{r\mathbf{i} {\bar{r}} t} ) \frac{1-\Vert q\Vert ^2}{ \Vert q- e^{r\mathbf{i} {\bar{r}} t} \Vert ^2 } dt , \end{aligned}$$

  and it may be concluded that

  $$\begin{aligned} P_{T_r(\mathbf{i})} [u] (q) = \frac{1}{2\pi }\int _{0}^{2\pi } u\circ T_r (e^{\mathbf{i} t} ) \frac{1-\Vert {\bar{r}}q r \Vert ^2}{ \Vert {\bar{r}} q r- e^{ \mathbf{i} t} \Vert ^2}dt. \end{aligned}$$

• 2. It is due to (3.2) and (3.3)

• 3. Let $x\in {{\mathbb {D}}}_\mathbf{i}$ and $\mathbf{j}\in {\mathbb {S}}^2$. According to the Representation Formula and the established continuity we have

  $$\begin{aligned} (x- e^{\mathbf{j} t})^{-*2}) * f(e^{\mathbf{j} t})= & {} \frac{1}{2} [(1+\mathbf{j} \mathbf{i})(x- e^{ - \mathbf{i} t})^{- 2} f(e^{{-\mathbf{i}} t})\\&+ (1-\mathbf{j} \mathbf{i}) (x- e^{ \mathbf{i} t})^{- 2} f(e^{\mathbf{i} t})]. \end{aligned}$$

  Therefore

  $$\begin{aligned} \Vert (x- e^{\mathbf{j} t})^{-*2}) * f(e^{\mathbf{j} t}) \Vert (1-\Vert x\Vert ^2) =\,&\Vert x- e^{ - \mathbf{i} t}\Vert ^{- 2} \Vert f(e^{{-\mathbf{i}} t})\Vert (1-\Vert x\Vert ^2) \\&+ \Vert x- e^{ \mathbf{i} t}\Vert ^{- 2} \Vert f(e^{\mathbf{i} t}) \Vert (1-\Vert x\Vert ^2) \end{aligned}$$

  and

  $$\begin{aligned}&\frac{1}{2\pi }\int _{0}^{2\pi } \Vert (x- e^{\mathbf{j} t})^{-*2} * f(e^{\mathbf{j} t}) \Vert (1-\Vert x\Vert ^2) dt \\&\quad \le \, \frac{1}{2\pi }\int _{0}^{2\pi } \Vert x- e^{ - \mathbf{i} t}\Vert ^{- 2} \Vert f(e^{{-\mathbf{i}} t})\Vert (1-\Vert x\Vert ^2) dt\\&\qquad + \frac{1}{2\pi }\int _{0}^{2\pi } \Vert x- e^{ \mathbf{i} t}\Vert ^{- 2} \Vert f(e^{\mathbf{i} t}) \Vert (1-\Vert x\Vert ^2)dt \\&\quad = 2 P_\mathbf{i} [\Vert f\Vert ](x). \end{aligned}$$

$\square$

Proposition 3.17

Set $f\in {{\mathcal {S}}}{{\mathcal {R}}}({\mathbb {D}}^4)\cap C(\overline{{\mathbb {D}}^4})$.

• 1. Let $\omega$ be a regular majorant such that $\Vert f\Vert \in {}_\mathbf{i}\Lambda _{\omega }({\mathbb {S}}^3)$. Then $f\in {}_\mathbf{i}\Lambda _\omega ({\mathbb {D}}^4)$ if and only if

  $$\begin{aligned} P_\mathbf{{i}}(\Vert f\pm \mathbf{i} f \mathbf{i} \Vert )(x)-\Vert f (x)\pm \mathbf{i} f(x) \mathbf{i} \Vert \le C \omega (1-\Vert x\Vert ), \quad \forall x\in {{\mathbb {D}}}_{ \mathbf{i}}. \end{aligned}$$

• 2. Let $\omega _1$, $\omega _2$ be two regular majorant such that $\Vert f\Vert \in {}_\mathbf{i}\Lambda _{\omega _1,\omega _2}({\mathbb {S}}^3)$. Then $f\in {}_\mathbf{i}\Lambda _{\omega _1,\omega _2}({\mathbb {D}}^4)$ if and only if

  $$\begin{aligned} P_\mathbf{{i}}(\Vert f- \mathbf{i} f \mathbf{i} \Vert )(x)-\Vert f (x)- \mathbf{i} f(x) \mathbf{i} \Vert \le \,&C \omega _1 (1-\Vert x\Vert ),\\ P_\mathbf{{i}}(\Vert f+ \mathbf{i} f \mathbf{i} \Vert )(x)-\Vert f (x)+ \mathbf{i} f(x) \mathbf{i} \Vert \le \,&C \omega _2 (1-\Vert x\Vert ), \quad \forall x\in {{\mathbb {D}}}_{ \mathbf{i}}. \end{aligned}$$

Proof

• 1. Combining (3.2) with Proposition 3.12 and Theorem B of Section 2 completes the proof.

• 2. It follows from identities (3.2) and Theorem B of Section 2.

$\square$

Corollary 3.18

Consider $f\in {{\mathcal {S}}}{{\mathcal {R}}}({\mathbb {D}}^4)\cap C(\overline{{\mathbb {D}}^4})$.

• 1. Let $\omega$ be a regular majorant such that $\Vert f\Vert \in {}_\mathbf{i}\Lambda _{\omega }({\mathbb {S}}^3)$. If $f\in {}_\mathbf{i}\Lambda _{\omega } ({\mathbb {D}}^4)$ and $q\in {\mathbb {D}}^4$ satisfies

  $$\begin{aligned} \langle q, e^{\mathbf{i} t} \rangle \le q_0 \cos t \pm |\mathbf{q}| \sin t, \quad \forall t\in [0,2\pi ], \end{aligned}$$

  then

  $$\begin{aligned} P_\mathbf{{i}}(\Vert f \Vert )(q)- 2\Vert f (q_0 \pm \mathbf{i}|\mathbf{q}|) \Vert \le C \omega (1-\Vert q\Vert ). \end{aligned}$$

• 2. Let $\omega _1$, $\omega _2$ be two regular majorant such that $\Vert f\Vert \in {}_\mathbf{i}\Lambda _{\omega _1,\omega _2}({\mathbb {S}}^3)$. If $f\in {}_\mathbf{i}\Lambda _{\omega _1,\omega _2}({\mathbb {D}}^4)$ and $q\in {\mathbb {D}}^4$ holds

  $$\begin{aligned} \langle q, e^{\mathbf{i} t} \rangle \le q_0 \cos t \pm |\mathbf{q}| \sin t, \quad \forall t\in [0,2\pi ], \end{aligned}$$

  then

  $$\begin{aligned} P_\mathbf{{i}}(\Vert f \Vert )(q) - 2\Vert f (q_0 \pm \mathbf{i}|\mathbf{q}|) \Vert \le C \left( \omega _1 (1-\Vert q\Vert ) + \omega _2 (1-\Vert q\Vert ) \right) . \end{aligned}$$

Proof

• 1. Let $x= q_0 \pm \mathbf{i}|\mathbf{q}|$. It follows easily that $|q-e^{\mathbf{i}}| \ge |x- e^{\mathbf{i} t}|$ for all $t\in [0,2\pi ]$. Of course,

  $$\begin{aligned} 2P_\mathbf{{i}}(\Vert f \Vert )(q)&\le 2 P_\mathbf{{i}}(\Vert f \Vert )(x) \le P_\mathbf{{i}}(\Vert f+ \mathbf{i} f \mathbf{i} \Vert )(x) + P_\mathbf{{i}}(\Vert f- \mathbf{i} f \mathbf{i} \Vert )(x) \\&\le \Vert f (x) + \mathbf{i} f(x) \mathbf{i} \Vert + \Vert f (x)- \mathbf{i} f(x) \mathbf{i} \Vert + 2 C \omega (1-\Vert x\Vert ). \end{aligned}$$

  Therefore

  $$\begin{aligned} P_\mathbf{{i}}(\Vert f \Vert )(q)\le 2\Vert f (q_0 \pm \mathbf{i}|\mathbf{q}|) \Vert + C \omega (1-\Vert q\Vert ). \end{aligned}$$

• 2. Similar arguments to those above.

$\square$

4 Conclusions and future works

In summary, characterizations of the Lipschitz type spaces of slice regular functions in the unit ball of the skew-field of quaternions with prescribed modulus of continuity, despite the non-commutativity of quaternions, are presented. The main results go on to the global case. Importantly, the present findings suggest the possibility to extend the study to the theory of slice monogenic functions associated to Clifford algebras, as a good starting point for further research.