# Quaternionic slice regular functions and quaternionic Laplace transforms

Gang Han
School of Mathematics
Zhejiang University
<EMAIL_ADDRESS>

(June 12, 2010 )

###### Abstract

The functions studied in the paper are quaternion-valued functions of a
quaternionic variable. It is show that the left slice regular functions and
right slice regular functions are related by a particular involution. The
relation between left slice regular functions, right slice regular functions
and intrinsic regular functions is revealed.

The classical Laplace transform can be naturally generalized to quaternions in
two different ways, which transform a quaternion-valued function of a real
variable to a left or right slice regular quaternion-valued function of a
quaternionic variable. The usual properties of the classical Laplace
transforms are generalized to quaternionic Laplace transforms.

MAC2010: 30G35, 35A22

##  1 Introduction

The theory of slice regular functions (originally Cullen-regular functions) of
a quaternionic variable was born around 2005, and has rapidly developed since
then. Most of the known results and applications of slice regular functions
can be found in [GSS]. Many results in the theory of the complex analytic
functions are generalized to slice regular functions, such as the Cauchy
Integral Formula, the Maximum Modulus Principles, the Open Mapping Theorem,
the Laurent series, and etc.

The theory of slice regular functions has now been applied to develop a new
functional calculus in a noncommutative setting [CSS1]. The theory of regular
functions has also been generalized to real alternative algebras [GP].

The set of left (resp. right) slice regular functions are not closed under the
usual product of functions, but they are closed under the regular product
($*$-product). With respect to this regular product and the usual addition and
scalar multiplication, the set of left (resp. right) slice regular functions
(on some domain of $H$) becomes a real noncommutative associative algebra. The
theory of left slice regular functions and that of right slice regular
functions are parallel. Sometimes one may meet left slice regular functions
and right slice regular functions at the same time. The Cauchy formulas for
the left and for the right slice regular functions involve different Cauchy
kernels [CGS], unlike what happens in the classical case. The quaternionic
Cauchy kernel

$\phi(q,s)=-\left(|s|^{2}-2q\operatorname{Re}(s)+q^{2}\right)^{-1}(q-\bar{s})$

is a left slice regular function of $q$, as well as a right slice regular
function of $s$ [CGS].

In the paper we will demonstrate the relation between left slice regular
functions and right slice regular functions. They are related by an involution
on the space of quaternionic functions.

Let $\mathbb{H}$ be the real algebra of quaternions and $U\subseteq\mathbb{H}$
be an axially symmetric slice domain and $C(U,\mathbb{H})$ be the set of
continuous functions from $U$ to $\mathbb{H}$. Let $\mathcal{R}^{0}(U)$,
$\mathcal{R}^{l}(U)$ and $\mathcal{R}^{r}(U)$ be respectively the set of
intrinsic regular functions , left regular functions and right regular
functions on $U$. Let

$\eta:C(U,\mathbb{H})\rightarrow
C(U,\mathbb{H}),\eta(f)(q)=\overline{f(\overline{q})}.$

Then $\eta$ is an involution on $C(U,\mathbb{H})$, which maps left slice
regular functions to right slice regular functions and vice versa, which
induces an anti-isomorphism between the associative algebras
$\mathcal{R}^{l}(U)$ and $\mathcal{R}^{r}(U)$.

Another main result is the isomorphism of associative algebras

$\phi:\mathcal{R}^{0}(U)\bigotimes_{\mathbb{R}}\mathbb{H}\rightarrow\mathcal{R}^{l}(U),$

through which some known results, including the slice derivatives and the
regular conjugation of slice regular functions, can be better understood. See
Theorem 3.5 and Theorem 3.6.

Next we introduce the quaternionic Laplace transform.

The classical Laplace transform is an integral transform named after its
inventor Pierre-Simon Laplace. It transforms a complex-valued function of a
real variable to a complex-valued function of a complex variable. The
transform has many applications in mathematics, physics and engineering.

The classical Laplace transform can be naturally generalized to quaternions,
see [L1] and [L2]. As $\mathbb{H}$ is noncommutative, there are 2 types of
quaternionic Laplace transforms (see Section 4), which transform a quaternion-
valued function of a real variable to a left or right slice regular
quaternion-valued function of a quaternionic variable.

The quaternionic Laplace transform can be generalized to quaternionic linear
operators. The relation between quaternionic evolution operator and
quaternionic Laplace transform were introduced in [CS].

In the paper, we show that the classical properties of the classical Laplace
transforms can be generalized to quaternionic Laplace transforms. For example,
if $f(t)$ and $g(t)$ are quaternion-valued functions defined for all real
numbers $t\geq 0$, and $f\circ g$ is the convolution of $f$ and $g$. Then the
left Laplace transform of $f\circ g$ is the regular product of the left
Laplace transform of $f$ and the left Laplace transform of $g$ (Proposition
4.8).

Most of the notations in the paper are standard, and some notations will be
explained when they first appear in the paper.

Acknowledgements

I would like to heartily thank Hao Yu and Yeshun Sun for very helpful
discussions with them.

##  2 Some preliminary results on quaternions and slice regular functions

1\. The quaternions

The quaternions $\mathbb{H}$ were introduced by Hamilton in $1843,$ adding a
multiplicative structure to $\mathbb{R}^{4}.$ It is a real division algebra.
Let $\\{1,i,j,k\\}$ be the standard basis of $\mathbb{H}$. Let
$\alpha:\mathbb{H}\rightarrow\mathbb{H},q\rightarrow\overline{q}$ be the usual
conjugation, which is an involution of $\mathbb{H}$. In the paper, an
involution of an associative algebra means an anti-automorphism of the algebra
of order two. Let $\mathbb{R}\cdot 1$ and $\operatorname{Im}(\mathbb{H})$ be
respectively the $\mathbb{R}$-span of 1 and the $\mathbb{R}$-span of
$\\{i,j,k\\}$, which are the respective 1 and -1 eigenspace of $\alpha$. One
has $\operatorname{Im}(\mathbb{H})=\\{q\in\mathbb{H}|q^{2}\in\mathbb{R}\ and\
q^{2}\leq 0\\}$. One has the decomposition of $\mathbb{H}$:

$\mathbb{H}=\mathbb{R}\cdot 1\oplus\operatorname{Im}(\mathbb{H}).$

For $q=a+bi+cj+dk\in\mathbb{H}$, let $\operatorname{Re}(q)=a$ and
$\operatorname{Im}(q)=bi+cj+dk$, which are the respective real and imaginary
part of $q$. Let $|q|^{2}=q\overline{q}$, which makes $\mathbb{H}$ into an
4-dimensional Euclidean space with the inner product
$(p,q)=\operatorname{Re}(p\overline{q})$. The elements in $\mathbb{H}$ with
unit norm constitute a group under multiplication, denoted by $Sp(1)$.

Let $\text{Aut}(\mathbb{H})$ be the automorphism group of $\mathbb{H}$. One
knows that $\text{Aut}(\mathbb{H})\cong SO(3)$.

Let
$\widetilde{\text{Aut}}(\mathbb{H})=\text{Aut}(\mathbb{H})\bigsqcup\alpha\cdot\text{Aut}(\mathbb{H})$
be the group generated by $\text{Aut}(\mathbb{H})$ and $\alpha$, which
consists of all the automorphisms and anti-automorphisms of $\mathbb{H}$.
Since $\alpha$ restricts to minus identity on $\operatorname{Im}\mathbb{H}$,
$\widetilde{\text{Aut}}(\mathbb{H})\cong O(3)$, and we will identify
$\widetilde{\text{Aut}}(\mathbb{H})$ with $O(3)$.

Let

$\mathbb{S}=\\{q\in\mathbb{H}|q^{2}=-1\\},$

which is the 2-dimensional unit sphere of purely imaginary elements in
$\mathbb{H}$ with unit norm. It is the orbit of $i$ under
$\text{Aut}(\mathbb{H})$. For any $I\in S$, let

$\mathbb{C}_{I}=\\{x+Iy|x,y\in\mathbb{R}\\},$

which is a subalgebra of $\mathbb{H}$ isomorphic to $\mathbb{C}$. For any
$I,J\in\mathbb{H}$, if $J\neq\pm I$ then
$\mathbb{C}_{I}\cap\mathbb{C}_{J}=\mathbb{R}$. One has that
$\mathbb{C}_{I}=\mathbb{C}_{J}$ if and only if $J=\pm I$. One has
$\mathbb{H}=\bigcup_{I\in S}\mathbb{C}_{I}$. This is called the slice
structure of $\mathbb{H}$. For any open set $U$ in $\mathbb{H}$ and
$I\in\mathbb{S}$, let

$U_{I}=U\cap\mathbb{C}_{I}.$ (2.1)

For any $I\in\mathbb{S}$, let
$\tau_{I}:\mathbb{C}\rightarrow\mathbb{H},x+iy\mapsto x+Iy$ be the embedding,
whose image is $\mathbb{C}_{I}$. We also use $\tau_{I}$ to denote
$\mathbb{C}\rightarrow\mathbb{C}_{I},x+iy\mapsto x+Iy$. $\tau_{I}$ is
conjugation equivariant. As $\mathbb{C}_{I}=\mathbb{C}_{-I}$, there are 2
identifications $\mathbb{C}\rightarrow\mathbb{C}_{I}$, $x+iy\mapsto x+Iy$ and
$x+iy\mapsto x-Iy$.

The Galois group of $\mathbb{C}$ over $\mathbb{R}$ is the order 2 cyclic group
generated by the complex conjugation $z\mapsto\overline{z}$, and will be
denoted by $C_{2}$.

One has the following homeomorphism of orbit spaces

$\mathbb{H}/{SO}_{3}\cong\mathbb{C}/{C}_{2}.$

The map

$\mathbb{C}\times\mathbb{S}\rightarrow\mathbb{H},(x+iy,J)\mapsto x+Jy,$

is a surjective continuous map, whose restriction
$(\mathbb{C}\setminus\mathbb{R})\times\mathbb{S}\rightarrow\mathbb{H}\setminus\mathbb{R}$
is a double covering.

2\. Definition of slice regular functions

Recall the $U_{I}$ defined in (2.1).

######  Definition 2.1.

Let $U$ be an open set in $\mathbb{H}$ and $f:U\rightarrow\mathbb{H}$ be real
differentiable. For every $I\in\mathbb{S},$ let
$f_{I}:U_{I}\rightarrow\mathbb{H}$ be the restriction of $f$ to $U_{I}$. The
function $f$ is said to be left slice regular (left regular) if for every
$I\in\mathbb{S},$

$\bar{\partial}_{I}f_{I}(x+Iy):=\frac{1}{2}\left(\frac{\partial}{\partial
x}+I\frac{\partial}{\partial y}\right)f_{I}(x+Iy)=0$

on $U_{I}.$ Analogously, a function is said to be right slice regular (right
regular) in $U$ if

$\bar{\partial}^{r}_{I}f_{I}(x+Iy):=\frac{1}{2}\left(\frac{\partial}{\partial
x}f_{I}(x+Iy)+\frac{\partial}{\partial y}f_{I}(x+Iy)I\right)=0$

on $U_{I}$.

The class of left (resp. right) regular functions on $U$ will be denoted by
$\mathcal{R}^{l}(U)$ (resp. $\mathcal{R}^{r}(U)$).

The left (slice) $I$ -derivative of $f\in\mathcal{R}^{l}(U)$ at a point
$q=x+Iy$ is given by

$\partial_{I}f_{I}(x+Iy)=\frac{1}{2}\left(\frac{\partial}{\partial
x}f_{I}(x+Iy)-I\frac{\partial}{\partial y}f_{I}(x+Iy)\right)$

Similarly, the right (slice) $I$ -derivative of $f\in\mathcal{R}^{r}(U)$ at
$q=x+Iy$ is given by

$\partial_{I}^{r}f_{I}(x+Iy)=\frac{1}{2}\left(\frac{\partial}{\partial
x}f_{I}(x+Iy)-\frac{\partial}{\partial y}f_{I}(x+Iy)I\right)$

Let us now introduce the slice derivative for regular functions.

######  Definition 2.2.

Let $U$ be an open set in $\mathbb{H}$. If $f\in\mathcal{R}^{l}(U)$, then the
left slice derivative, $\partial f$, of $f,$ is defined by:

$\partial f(q)=\left\\{\begin{array}[]{ll}\partial_{I}f_{I}(q)&\text{ if
}q=x+Iy,\quad y\neq 0\\\ \frac{\partial f}{\partial x}(x)&\text{ if
}q=x\in\mathbb{R}.\end{array}\right.$

If $f\in\mathcal{R}^{r}(U)$, then the right slice derivative, $\partial^{r}f$,
of $f,$ is defined by:

$\partial^{r}f(q)=\left\\{\begin{array}[]{ll}\partial_{I}^{r}f_{I}(q)&\text{
if }q=x+Iy,\quad y\neq 0\\\ \frac{\partial f}{\partial x}(x)&\text{ if
}q=x\in\mathbb{R}.\end{array}\right.$

Note that if $f$ is a left regular function, then its derivative is still left
regular because

$\bar{\partial}_{I}\left(\partial
f(x+Iy)\right)=\partial\left(\bar{\partial}_{I}f(x+Iy)\right)=0.$

Similar result holds for right regular functions.

Let $r>0$ and $B(0,r)$ be the open ball centered at $0$ with radius $r$. A
function $f:B(0,r)\rightarrow\mathbb{H}$ is left (resp. right) regular on
$B(0,r)$ if and only if $f$ has a series representation of the form
$f(q)=\sum_{n=0}^{\infty}q^{n}a_{n}$ (resp.
$f(q)=\sum_{n=0}^{\infty}a_{n}q^{n}$), where $\\{a_{n}\\},n\in\mathbb{N},$ is
a sequence of quaternions satisfying
$(limsup_{n\rightarrow\infty}|a_{n}|^{1/n})^{-1}\geq r$. See Theorem
2.1.5.Theorem 2.1.6. [CSS2].

3\. Intrinsic regular functions

Let $D\subseteq\mathbb{C}$ be a domain symmetric with respect to the real
axis. Then $D\cap\mathbb{R}\neq\emptyset$. Let $f:D\mapsto\mathbb{C}$ be a
complex function. Then $f$ is called intrinsic if $f$ is $C_{2}$-equivariant,
i.e.,

$f(\bar{z})=\overline{f(z)}.$ (2.2)

######  Definition 2.3.

Let $U\subseteq\mathbb{H}$. We say that $U$ is axially symmetric if $U$ is
$SO(3)$-invariant, i.e., for every $x+Iy\in U,$ all the elements
$x+\mathbb{S}y=\\{x+Jy|J\in\mathbb{S}\\}$ are contained in U. We say that $U$
is a slice domain if it is a connected open set whose intersection with every
complex plane $\mathbb{C}_{I}$ is connected.

If $U\subseteq\mathbb{H}$ is an axially symmetric slice domain, then
$U\cap\mathbb{R}\neq\o$.

Let $U\subseteq\mathbb{H}$ be an axially symmetric slice domain and
$C(U,\mathbb{H})$ be the set of continuous functions from $U$ to $\mathbb{H}$.
Let

$\eta:C(U,\mathbb{H})\rightarrow
C(U,\mathbb{H}),\eta(f)(q)=\overline{f(\overline{q})}.$ (2.3)

Then one has

$\eta^{2}=1$

and

$\eta(f\cdot g)=\eta(g)\cdot\eta(f),\ \ f,g\in C(U,\mathbb{H}),$

where $f\cdot g$ is the pointwise multiplication. So $\eta$ is an involution
on $C(U,\mathbb{H})$. This involution $\eta$ plays an important role in the
paper!

######  Definition 2.4.

Let $U$ be an axially symmetric slice domain. A function
$F:U\rightarrow\mathbb{H}$ is called slice-preserving, if
$F(U\cap\mathbb{C}_{I})\subseteq U\cap\mathbb{C}_{I}$ for each
$I\in\mathbb{S}$. If $F$ is slice-preserving and left regular, then $F$ is
called intrinsic regular. The set of intrinsic regular functions on $U$ will
be denoted by $\mathcal{R}^{0}(U)$.

The following result is clear.

######  Proposition 2.5.

Assume that $F:U\rightarrow\mathbb{H}$ is slice-preserving, where $U$ is an
axially symmetric slice domain in $\mathbb{H}$, then $F$ is left regular if
and only if $F$ is right regular if and only if $F$ is intrinsic regular.

By Remark $1.31$ of [GSS], one know that if $f$ is a left regular function on
$B(0,r)$ (for some $r>0$ ), then $f$ is slice preserving if, and only if, the
power series expansion $f(q)=\sum_{n\in\mathbb{N}}q^{n}a_{n}$ has real
coefficients $a_{n}\in\mathbb{R}$.

If $f$ is an intrinsic regular function, then
$\partial(f)(q)=\partial^{r}(f)(q)$, which will be denoted by
$f^{{}^{\prime}}(q)$, i.e., for intrinsic regular functions $f$,

$f^{{}^{\prime}}(q)=\partial(f)(q)=\partial^{r}(f)(q).$

4\. Extension from complex intrinsic holomorphic functions to quaternionic
intrinsic regular functions

The following construction is due to Feuter [F].

Let $f(x+iy)=u(x,y)+iv(x,y)$ be an intrinsic complex function on a domain
$D\subseteq\mathbb{C}$ symmetric with respect to the real axis, where $u$ and
$v$ are real-valued functions. Let $\widetilde{D}\subseteq\mathbb{H}$ be the
union of $SO(3)$-orbit of all the points in $D$, i.e.,
$\widetilde{D}=\\{x+\mathbb{S}y|x+iy\in D\\}$. Then $f$ induces a function,
$\widetilde{f}$, from $\widetilde{D}$ to $\mathbb{H}$, such that for any
$J\in\mathbb{S}$,

$\widetilde{f}:\widetilde{D}\rightarrow\mathbb{H},\widetilde{f}(a+Jb)=u(a,b)+Jv(a,b).$
(2.4)

This is well-defined because $f$ is intrinsic. This induced function
$\widetilde{f}$ is $O(3)$-equivariant and slice-preserving. This construction
is later generalized to regular functions.

Assume further that $f$ is an intrinsic holomorphic function on $D$, then
$\widetilde{f}$ is left and right regular on $\widetilde{D}$ preserving each
$\widetilde{D}\cap\mathbb{C}_{I}$, thus is an intrinsic regular function on
$\widetilde{D}$. Let $H(D,\mathbb{C})$ be the set of intrinsic holomorphic
function on $D$ and define

$\text{ext}:H(D,\mathbb{C})\rightarrow\mathcal{R}^{0}(\widetilde{D}),f\mapsto\widetilde{f},$
(2.5)

which is an isomorphism of commutative and associative algebras. Let
$U=\widetilde{D}$. Then the inverse of this map is the restriction of
$g\in\mathcal{R}^{0}(U)$ to some single slice $U_{I}=U\cap\mathbb{C}_{I}$ (for
some $I\in\mathbb{S}$), where $g|_{U_{I}}\in H(U_{I},\mathbb{C}_{I})$. Write
$H(U_{I})=H(U_{I},\mathbb{C}_{I})$. Then (2.5) can also be written as

$\text{ext}:H(U_{I})\rightarrow\mathcal{R}^{0}(U).$ (2.6)

######  Example 2.6.

The exponential function

$\exp:\mathbb{H}\rightarrow\mathbb{H},\quad q\rightarrow
e^{q}=\sum_{n=0}^{\infty}\frac{q^{n}}{n!}$

is an intrinsic regular function induced from the usual complex exponential
function.

##  3 Main results on regular functions

In this section $U$ will always be an axially symmetric slice domain in
$\mathbb{H}$.

######  Proposition 3.1.

(Proposition 2.2.7 and Corollary 2.2.8 of [CSS2]) Let
$f\in\mathcal{R}^{l}(U).$ For any orthogonal unit vectors $I,J,K\in\mathbb{S}$
with $IJ=K$, there exist 4 uniquely determined intrinsic regular function
$h_{m}$ on $U$, such that $f(q)=h_{0}(q)+h_{1}(q)I+h_{2}(q)J+h_{3}(q)K$, $q\in
U$. Each $h_{m}$ is a real analytic function.

The following result is a corollary of the Identity Principle (Theorem 1.12 of
[GSS]).

######  Proposition 3.2.

Assume $f,g\in R^{l}(U)$ and $W\subseteq U\cap\mathbb{R}$ is a nonempty open
subset. If $f(q)=g(q)$ for any $q\in W$, then $f=g$.

For any $f\in\mathcal{R}^{l}(U)$, one knows that $f$ can be recovered from its
restriction to some single slice $U_{I}=U\cap\mathbb{C}_{I}$ with
$I\in\mathbb{S}$, see Corollary 1.16 of [GSS]. In fact, more is true.

Let $U$ be an axially symmetric slice domain of $\mathbb{H}$ and
$f\in\mathcal{R}^{l}(U)$. Then $f$ can be recovered from its restriction to
$U\cap\mathbb{R}$ as follows. By Proposition 3.1, one knows that
$f(q)=f_{0}(q)+f_{1}(q)i+f_{2}(q)j+f_{3}(q)k$ with
$f_{m}\in\mathcal{R}^{0}(U)$. Fix $I\in\mathbb{S}$. Each $f_{m}(q)$ is
determined by its restriction, $\bar{f_{m}}$, to $U_{I}$. As $\bar{f_{m}}$ is
holomorphic on $U_{I}$, $\bar{f_{m}}$ is determined by its restriction to
$U\cap\mathbb{R}$, which is denoted by $f_{m}|_{\mathbb{R}}$. In fact,
$\bar{f_{m}}$ is the analytic continuation of $f_{m}|_{\mathbb{R}}$ to
$U_{I}$, and $f_{m}=\text{ext}(\bar{f_{m}})$ (see (2.4)).

######  Proposition 3.3.

A left (resp. right) regular function $f:U\rightarrow\mathbb{H}$ is intrinsic
regular if and only if $f(U\cap\mathbb{R})\subseteq\mathbb{R}$.

###### Proof.

We will prove the result in the case $f\in\mathcal{R}^{l}(U)$.

If $f$ is intrinsic regular, then $f(U\cap\mathbb{C}_{I})\subseteq
U\cap\mathbb{C}_{I}$ for each $I\in\mathbb{S}$. Since
$\mathbb{C}_{I}\cap\mathbb{C}_{J}=\mathbb{R}$ for $J\neq\pm I$,
$f(U\cap\mathbb{R})\subseteq\mathbb{R}$.

Conversely, assume $f(U\cap\mathbb{R})\subseteq\mathbb{R}$ and
$f(q)=f_{0}(q)+f_{1}(q)I+f_{2}(q)J+f_{3}(q)K$ with each
$f_{m}\in\mathcal{R}^{0}(U)$. Then $f_{m}(q)=0$ for $q\in U\cap\mathbb{R}$ for
$m=1,2,3$. Since $f_{m}(q)$ is complex holomorphic on each slice $U_{I}$,
$f_{m}(q)=0$ for $q\in U_{I}$ for $m=1,2,3$. As
$U=\cup_{I\in\mathbb{S}}U_{I}$, $f_{m}(q)=0$ for each $q\in U$ for $m=1,2,3$,
and $f=f_{0}$ is intrinsic regular.

∎

For any $f,g\in\mathcal{R}^{l}(U)$, the function $h(q)=f(q)g(q)$ may not be in
$\mathcal{R}^{l}(U)$. The left regular product $f*g$ of $f$ and $g$ (see
Definition 1.27 of [GSS]), is also in $\mathcal{R}^{l}(U)$. If
$f(q)=\sum_{n\in\mathbb{N}}q^{n}a_{n}$ and
$g(q)=\sum_{n\in\mathbb{N}}q^{n}b_{n}$ are left regular functions on $B(0,r)$.
Then the regular product of $f$ and $g$ is the regular function defined by

$(f*g)(q)=\sum_{n\in\mathbb{N}}q^{n}\sum_{k=0}^{n}a_{k}b_{n-k}$ (3.1)

on the same ball $B(0,r)$.

Assume $f\in\mathcal{R}^{l}(U)$. For the definition of its regular conjugate
$f^{c}$ and its symmetrization $f^{s}$, see Definition 1.33 of [GSS].

Assume $f\in\mathcal{R}^{l}(U)$ and $f\neq 0$. The regular reciprocal $f^{-*}$
of $f$ is the function defined on $U\backslash Z_{f^{s}}$, where
$Z_{f^{s}}=\\{q\in\mathbb{H}|f^{s}(q)=0\\}$, by

$f^{-*}=\frac{1}{f^{s}}f^{c}.$

See Definition 5.1 of [GSS].

######  Proposition 3.4.

Assume $f,g\in\mathcal{R}^{l}(U)$. One has

(1) The left regular product $f*g$ is the left regular function in
$\mathcal{R}^{l}(U)$ such that $(f*g)(q)=f(q)g(q)$ for $q\in U\cap\mathbb{R}$.

(2) The regular conjugate $f^{c}$ of $f$ is the function in
$\mathcal{R}^{l}(U)$ such that for $q\in U\cap\mathbb{R}$,
$f^{c}(q)=\overline{f(q)}$.

(3) The symmetrization $f^{s}$ of $f$ is the intrinsic regular function in
$\mathcal{R}^{0}(U)$ such that for $q\in U\cap\mathbb{R}$,
$f^{s}(q)=f(q)\overline{f(q)}$.

(4) The regular reciprocal $f^{-*}$ of $f$ is the function in
$\mathcal{R}^{l}(U)$ such that for $q\in(U\cap\mathbb{R})\backslash
Z_{f^{s}}$, $f^{-*}(q)=f(q)^{-1}$.

###### Proof.

By Proposition 3.2, we only need to show that the value of the regular
functions $f*g$, $f^{c}$, $f^{s}$ and $f^{-*}$ in $q\in U\cap\mathbb{R}$ is as
stated in the proposition.

(1) Without loss of generality one can assume $U\supseteq B(0,r)$ for some
$r>0$. Then by (3.1), $(f*g)(q)=f(q)g(q)$ for $q\in B(0,r)\cap\mathbb{R}$. By
Proposition 3.1, the function $U\cap\mathbb{R}\rightarrow\mathbb{H},q\mapsto
f(q)g(q)$ is a function of a real variable with each component real analytic,
thus there is a unique left regular function $h$ such that $h(q)=f(q)g(q)$ for
$q\in U\cap\mathbb{R}$. As $h(q)=(f*g)(q)$ for $q\in B(0,r)\cap\mathbb{R}$,
$h=f*g$ on $U$ by Proposition 3.2.

(2) It can be proved similarly as in (1).

(3) It follows from (1) and (2) as $f^{s}=f*f^{c}$.

(4) For $q\in(U\cap\mathbb{R})\backslash Z_{f^{s}}$,
$f(q)\cdot\frac{1}{f^{s}(q)}f^{c}(q)=1$, thus
$f^{-*}(q)=\frac{1}{f^{s}(q)}f^{c}(q)=f(q)^{-1}$. Then (4) follows from
Proposition 3.2.

∎

Now let us review the Splitting Lemma before we prove the next result. Let
$f:U\rightarrow\mathbb{H}$ be a real differentiable function with $U$ an
axially symmetric slice domain. Then $f$ is left regular if and only if for
any $I\in\mathbb{S}$ and any $J\in\mathbb{S}$ with $J\perp I,$ there exist two
holomorphic functions $F,G:U_{I}\rightarrow\mathbb{C}_{I}$ such that for every
$z=x+yI,$

$f_{I}(z)=F(z)+G(z)J.$

And, $f$ is right regular if and only if for any $I\in\mathbb{S}$ and any
$J\in\mathbb{S}$ with $J\perp I,$ there exist two holomorphic functions
$F,G:U_{I}\rightarrow\mathbb{C}_{I}$ such that for every $z=x+yI,$

$f_{I}(z)=F(z)+JG(z).$

The space $\mathcal{R}^{l}(U)$ (resp. $\mathcal{R}^{r}(U)$) is a real
associative algebra equipped with the regular product. Recall the involution
$\eta$ defined in (2.3), let

$C(U,\mathbb{H})^{\eta}=\\{f\in C(U,\mathbb{H})|\eta(f)=f\\}.$

######  Theorem 3.5.

Let $U$ be an axially symmetric slice domain of $\mathbb{H}$.

(1) If $f\in\mathcal{R}^{l}(U)$ then $\eta(f)\in\mathcal{R}^{r}(U)$ and
viceversa. One has $\mathcal{R}^{0}(U)=\mathcal{R}^{l}(U)\cap
C(U,\mathbb{H})^{\eta}=\mathcal{R}^{r}(U)\cap C(U,\mathbb{H})^{\eta}$.

(2) The map

$\phi:\mathcal{R}^{0}(U)\bigotimes_{\mathbb{R}}\mathbb{H}\rightarrow\mathcal{R}^{l}(U),f(q)\otimes\lambda\mapsto
f(q)\lambda,$

is an isomorphism of real associative algebras, where the product of
$f_{0}\otimes I$ and $g_{0}\otimes J$ in
$\mathcal{R}^{0}(U)\bigotimes_{\mathbb{R}}\mathbb{H}$ is $f_{0}\otimes I\cdot
g_{0}\otimes J=f_{0}g_{0}\otimes IJ$. One has that $\mathcal{R}^{0}(U)$ is the
center of the associative algebra $\mathcal{R}^{l}(U)$.

The map $\phi$ is also an isomorphism of a left (free)
$\mathcal{R}^{0}(U)$-module and a right $\mathbb{H}$-module.

The results for $\mathcal{R}^{r}(U)$ are symmetric.

###### Proof.

(1)Let $f(q)\in\mathcal{R}^{l}(U)$. By the Splitting Lemma, for any
$I\in\mathbb{S}$ and any $J\in\mathbb{S}$ with $J\perp I,$ there exist two
holomorphic functions $F,G:\Omega_{I}\rightarrow L_{I}$ such that for every
$z=x+yI,$ $f_{I}(z)=F(z)+G(z)J$.

Let $g(q)=\overline{f(\overline{q})}$, then for every $z=x+yI,$
$g_{I}(z)=\overline{f_{I}(\overline{z})}=\overline{F(\overline{z})+G(\overline{z})J}=\overline{F(\overline{z})}+J(-\overline{G(\overline{z})})$.
Since $F,G:\Omega_{I}\rightarrow L_{I}$ are holomorphic functions,
$\overline{F(\overline{z})}$ and $-\overline{G(\overline{z})}$ are also
holomorphic, thus $g(q)=\overline{f(\overline{q})}$ is right regular.

One has $\mathcal{R}^{0}(U)\subseteq\mathcal{R}^{l}(U)\cap
C(U,\mathbb{H})^{\eta}$ by definition. Conversely, if $f\in
C(U,\mathbb{H})^{\eta}$ then for $q\in U\cap\mathbb{R}$,
$\overline{f(\overline{q})}=\overline{f(q)}=f(q)$ thus $f(q)$ is real. So, if
$f\in\mathcal{R}^{l}(U)\cap C(U,\mathbb{H})^{\eta}$, then
$f\in\mathcal{R}^{0}(U)$ by Proposition 3.3. So
$\mathcal{R}^{0}(U)\supseteq\mathcal{R}^{l}(U)\cap C(U,\mathbb{H})^{\eta}$.

(2) Proposition 3.1 implies that

$\mathcal{R}^{l}(U)=\mathcal{R}^{0}(U)\oplus\mathcal{R}^{0}(U)i\oplus\mathcal{R}^{0}j\oplus\mathcal{R}^{0}(U)k,$

which implies that $\phi$ is an $\mathbb{R}$-linear isomorphism of
$\mathcal{R}^{0}(U)\otimes_{\mathbb{R}}\mathbb{H}$ and $\mathcal{R}^{l}(U)$.
It is directly verified that $\phi$ preserves multiplications in the two
associative algebras. So $\phi$ is an isomorphism of real associative
algebras. It is easy to see that $\mathcal{R}^{0}(U)$ is the center of the
associative algebra $\mathcal{R}^{l}(U)$.

The left $\mathcal{R}^{0}(U)$-module structure and right $\mathbb{H}$-module
structure on $\mathcal{R}^{l}(U)$ and
$\mathcal{R}^{0}(U)\otimes_{\mathbb{R}}\mathbb{H}$ are the obvious ones. It is
clear that $\phi$ is an isomorphism of the left $\mathcal{R}^{0}(U)$-module
and the right $\mathbb{H}$-module.

∎

In the rest, we will identify $\mathcal{R}^{l}(U)$ with
$\mathcal{R}^{0}(U)\bigotimes_{\mathbb{R}}\mathbb{H}$ by $\phi$.

######  Theorem 3.6.

Let $U$ be an axially symmetric slice domain of $\mathbb{H}$.

(1)One has $\eta(f*g)=\eta(g)*\eta(f)$ for $f,g\in\mathcal{R}^{l}(U)$, where
the 1st (resp. the 2nd) product is left (resp. right) regular product. So
$\eta:\mathcal{R}^{l}(U)\rightarrow\mathcal{R}^{r}(U)$ is an anti-isomorphism
of associative algebras. Similarly,
$\eta:\mathcal{R}^{r}(U)\rightarrow\mathcal{R}^{l}(U)$ is also an anti-
isomorphism of associative algebras.

(2) The map

$\mathcal{R}^{0}(U)\bigotimes_{\mathbb{R}}\mathbb{H}\rightarrow\mathcal{R}^{0}(U)\bigotimes_{\mathbb{R}}\mathbb{H},f\otimes\mu\mapsto
f\otimes\overline{\mu}$

induces the regular conjugation on $\mathcal{R}^{l}(U)$, i.e.,
$(f\otimes\mu)^{c}=f\otimes\overline{\mu}$. And similar result holds for
$\mathcal{R}^{r}(U)$.

(3)Assume $f\in\mathcal{R}^{l}(U)$. Write $f=\sum_{m=0}^{3}f_{m}(q)\otimes
J_{m}$, with $\\{J_{m}|m=0,1,2,3\\}$ a basis of $H$. Then

$\partial(\sum_{m=0}^{3}f_{m}(q)\otimes
J_{m})=\sum_{m=0}^{3}f_{m}^{{}^{\prime}}(q)\otimes J_{m}.$

(4) Let $I\in\mathbb{S}$ and recall the map ext in (2.6). Let
$\psi:H(U_{I})\bigotimes_{\mathbb{R}}\mathbb{H}\rightarrow\mathcal{R}^{l}(U)$
be the composition of

$\text{ext}\otimes
1:H(U_{I})\bigotimes_{\mathbb{R}}\mathbb{H}\rightarrow\mathcal{R}^{0}(U)\bigotimes_{\mathbb{R}}\mathbb{H}$

with
$\phi:\mathcal{R}^{0}(U)\bigotimes_{\mathbb{R}}\mathbb{H}\rightarrow\mathcal{R}^{l}(U)$.
Then $\psi$ is an isomorphism of associative algebras.

If one identifies any element $g\otimes J\in
H(U_{I})\bigotimes_{\mathbb{R}}\mathbb{H}$ with the function $U_{I}\rightarrow
H,q\mapsto g(q)J$, then $\psi(g\otimes J)$ is the regular extension of
$g\otimes J$. (For the definition of regular extension, see Lemma 1.22 of
[GSS].) Similar result holds for $\mathcal{R}^{r}(U)$.

###### Proof.

(1)Assume $f,g\in\mathcal{R}^{l}(U)$ with $f(q)=f_{0}(q)\otimes I$ and
$g(q)=g_{0}(q)\otimes J$. Then

$\displaystyle\eta(f*g)$ $\displaystyle=\eta(f_{0}\otimes I\cdot g_{0}\otimes
J)$ $\displaystyle=\eta(f_{0}g_{0}\otimes IJ)$
$\displaystyle=\overline{IJ}\otimes f_{0}g_{0}$
$\displaystyle=\overline{J}\otimes g_{0}\cdot\overline{I}\otimes f_{0}$
$\displaystyle=\eta(g)*\eta(f).$

Then it follows by linearity that $\eta(f*g)=\eta(g)*\eta(f)$ for any
$f,g\in\mathcal{R}^{l}(U)$. Similarly one also has $\eta(f*g)=\eta(g)*\eta(f)$
for any $f,g\in\mathcal{R}^{r}(U)$. Since $\eta^{2}=1$,
$\eta:\mathcal{R}^{l}(U)\rightarrow\mathcal{R}^{r}(U)$ is an anti-isomorphism
with inverse $\eta:\mathcal{R}^{r}(U)\rightarrow\mathcal{R}^{l}(U)$, which is
also an anti-isomorphism.

(2) and (3) can be verified directly by definition.

(4) As $\phi$ and $\text{ext}\otimes 1$ are both isomorphisms of associative
algebras, their composition $\psi$ is also an isomorphism of associative
algebras. Let $g\otimes J\in H(U_{I})\bigotimes_{\mathbb{R}}\mathbb{H}$,
$\widetilde{g}=\text{ext}(g)\in\mathcal{R}^{0}(U)$ and $f=\psi(g\otimes
J)\in\mathcal{R}^{l}(U)$. Then for $q\in U$, $f(q)=\psi(g\otimes
J)(q)=\phi(\widetilde{g}\otimes J)(q)=\widetilde{g}(q)J$. So for $q\in U_{I}$,

$f(q)=\widetilde{g}(q)J=g(q)J=(g\otimes J)(q).$

Thus $f|_{U_{I}}=g\otimes J$ and $f$ is the unique regular extension of
$g\otimes J$.

∎

##  4 Left and right quaternionic Laplace transforms

Let $g(t)$ be a complex-valued function, defined for all real numbers $t\geq
0$. Then its classical Laplace transform $L\\{g\\}$ is defined by

$L\\{g\\}(s)=\int_{0}^{\infty}e^{-ts}g(t)dt,$

for those complex numbers $s\in\mathbb{C}$ such that the integral convergents.
It will be referred as the complex Laplace transform in the paper. For the
main properties of complex Laplace transform, please refer to the classical
book [DB].

Let $f(t)$ be a quaternion-valued function defined for all real numbers $t\geq
0$. Note that in last section $f(t)$ is a quaternion-valued function of a
quaternionic variable, but in this section $f(t)$ is a quaternion-valued
function of a real variable!

Define

$L^{l}\\{f\\}(s)=\int_{0}^{\infty}e^{-ts}f(t)dt,$

for those quaternions $s\in\mathbb{H}$ such that the integral convergents, to
be the left quaternionic Laplace transform of $f$; and

$L^{r}\\{f\\}(s)=\int_{0}^{\infty}f(t)e^{-ts}dt,$

for those quaternions $s\in\mathbb{H}$ such that the integral convergents, to
be the right quaternionic Laplace transform of $f$.

Note that if $f$ is real-valued, then $L^{r}\\{f\\}=L^{l}\\{f\\}$.

Let $f(t)$ be a quaternion-valued function defined for all real numbers $t\geq
0$. Assume $a\geq 0$. The function $f(t)$ is said to be of exponential order
$a$ on $t\geq 0$ if there exist two positive constants $K$ and $T$, such that
for all $t>T$

$|f(t)|\leq Ke^{at}.$

Once we say that $f(t)$ is of exponential order $a$ then we always assume that
$a\geq 0$.

######  Theorem 4.1.

Let $f(t)$ be a quaternion-valued function defined for all real numbers $t\geq
0$, which is of exponential order $a$ (with $a\geq 0$), and is continuous or
piecewise continuous in every finite interval $(0,T).$ Then

(1) The left Laplace transform $L^{l}\\{f\\}(s)$ of $f(t)$ convergent
absolutely for all $s\in\mathbb{H}$ provided $\operatorname{Re}(s)>a$.

(2) The left Laplace transform
$L^{l}\\{f\\}(s)=\int_{0}^{\infty}e^{-st}f(t)dt$ is uniformly convergent with
respect to $s\in\mathbb{H}$ provided $\operatorname{Re}(s)\geq a_{1}$ where
$a_{1}>a$.

(3) The map $f\mapsto L^{l}\\{f\\}$, where $f$ is of exponential order $a,$ is
right $\mathbb{H}$-linear.

We will denote $F=L^{l}\\{f\\}$ in the rest of the theorem.

(4)If $f(t)$ is real-valued, then $F(s)$ is intrinsic regular on
$\operatorname{Re}(s)>a$ and $F{{}^{\prime}}(s)=L^{l}\\{-tf(t)\\}$. The
restriction of $F$ to each $\mathbb{C}_{I}$ is just the complex Laplace
transform of $f$.

(5)In general, $F(s)$ is left regular on $\operatorname{Re}(s)>a$, and
$\partial F(s)=L^{l}\\{-tf(t)\\}$.

###### Proof.

(1) and (2) can be proved as in the complex case.

(3) follows by definition.

(4) Let $U=\\{s\in\mathbb{H}|\operatorname{Re}(s)>a\\}.$ For any
$I\in\mathbb{S}$, $F({U_{I}})\subseteq U_{I}$ and $F|_{U_{I}}$ is just the
usual complex Laplace transform of $f$, which is holomorphic on $U_{I}$. So
$F$ is intrinsic regular. For each $I\in\mathbb{S}$ and $s\in U_{I}$ with
$\operatorname{Re}(s)>a$, one has that
$F{{}^{\prime}}(s)=\int_{0}^{\infty}e^{-st}(-tf(t))dt$ , thus for any $s\in
U$, $F{{}^{\prime}}(s)=L^{l}\\{-tf(t)\\}$.

(5) In general, $f=f_{0}+f_{1}i+f_{2}j+f_{3}k$ where $f_{m}$ are real valued
functions. Then by (3),
$L^{l}\\{f\\}=L^{l}\\{f_{0}\\}+L^{l}\\{f_{1}\\}i+L^{l}\\{f_{2}\\}j+L^{l}\\{f_{3}\\}k$,
which is left regular as each $L^{l}\\{f_{m}\\}$ is intrinsic regular. ∎

######  Remark 4.2.

The results for right Laplace transform are completely symmetric and are
omitted.

For a quaternion-valued function $f$ defined for all real numbers $t\geq 0$,
$\overline{f}$ is the function defined by $\overline{f}(t)=\overline{f(t)}$.
Then $f$ is real-valued if and only if $f=\overline{f}$.

######  Proposition 4.3.

Let $f(t)$ be a quaternion-valued function defined for all real numbers $t\geq
0$. Then $\eta(L^{l}\\{f\\})=L^{r}{\\{\overline{f}\\}}$ and
$\eta(L^{r}\\{f\\})=L^{l}{\\{\overline{f}\\}}$.

###### Proof.

Write $f(t)=\sum_{m=0}^{3}f_{m}(t)\cdot J_{m}$, where
$J_{0}=1,J_{1}=i,J_{2}=j,J_{3}=k$ and $f_{m}(t)$ are all real-valued
functions. Then $\overline{f}(t)=\sum_{m=0}^{3}\overline{J_{m}}\cdot
f_{m}(t)$.

One has

$\displaystyle L^{l}\\{f\\}(s)$
$\displaystyle=\int_{0}^{\infty}e^{-ts}f(t)dt=\int_{0}^{\infty}e^{-ts}(\sum_{m=0}^{3}f_{m}(t)\cdot
J_{m})dt$
$\displaystyle=\sum_{m=0}^{3}(\int_{0}^{\infty}e^{-ts}f_{m}(t)dt)\cdot
J_{m}=\sum_{m=0}^{3}L^{l}\\{f_{m}\\}(s)\cdot J_{m}$

$\displaystyle L^{r}\\{\overline{f}\\}(s)$
$\displaystyle=\int_{0}^{\infty}\overline{f}(t)e^{-ts}dt=\int_{0}^{\infty}(\sum_{m=0}^{3}\overline{J_{m}}\cdot
f_{m}(t))e^{-ts}dt$
$\displaystyle=\sum_{m=0}^{3}\overline{J_{m}}\cdot(\int_{0}^{\infty}e^{-ts}f_{m}(t)dt)=\sum_{m=0}^{3}\overline{J_{m}}\cdot
L^{l}\\{f_{m}\\}(s)$

So $\eta(L^{l}\\{f\\})=L^{r}\\{\overline{f}\\}$. The result
$\eta(L^{r}\\{f\\})=L^{l}\\{\overline{f}\\}$ can be similarly proved.

∎

######  Example 4.4.

Let $b\in\mathbb{H}$ and $f(t)=e^{bt}$. Let $F=L^{l}\\{f\\}$ and
$G=L^{r}\\{f\\}$.

For $s\in\mathbb{R}$ and $\operatorname{Re}(s)>\operatorname{Re}(b)$,
$F(s)=G(s)=\int_{0}^{\infty}e^{-ts}e^{bt}dt=(s-b)^{-1}$.

As $F$ is left regular, it is the left regular reciprocal of $s-b$. So
$F(s)=(s^{2}-2\operatorname{Re}(b)s+|b|^{2})^{-1}(s-\overline{b})$, where
$\operatorname{Re}(s)>\operatorname{Re}(b)$.

As $G(s)$ is right regular, it is the right regular reciprocal of $s-b$. Then
$G(s)=(s-\overline{b})(s^{2}-2\operatorname{Re}(b)s+|b|^{2})^{-1}$, where
$\operatorname{Re}(s)>\operatorname{Re}(b)$.

In the rest of the paper we will consider only left Laplace transforms, and
$f(t)$ will always be a quaternion-valued function defined for all real
numbers $t\geq 0$.

######  Proposition 4.5.

If ${L}\\{f(t)\\}=F(s),$ then

$\mathcal{L}^{l}\left\\{e^{-at}f(t)\right\\}=F(s+a),$

where $a$ is a real constant.

###### Proof.

We have, by definition,

$\mathcal{L}^{l}\left\\{e^{-at}f(t)\right\\}=\int_{0}^{\infty}e^{-(s+a)t}f(t)dt=F(s+a).$

∎

Note that one does not have
$\mathcal{L}^{l}\left\\{f(t)e^{-at}\right\\}=F(s+a)$ because of the
noncommutativity of $\mathbb{H}$.

Recall the Heaviside function

$H(x)=\left\\{\begin{array}[]{ll}0,&x<0;\\\ 1,&x\geq 0.\end{array}\right.$

######  Proposition 4.6.

If $\mathcal{L}^{l}\\{f(t)\\}=F(s),$ then for $a>0$ one has:

$\mathcal{L}^{l}\\{f(t-a)H(t-a)\\}=e^{-as}F(s)=e^{-as}\mathcal{L}^{l}\\{f(t)\\}.$

The proof is as in the complex case and is omitted.

######  Proposition 4.7.

(Laplace Transforms of Derivatives). Let $f(t)$ be a quaternion-valued
function defined for all real numbers $t\geq 0$, which is of exponential order
$a$. Let $U=\\{s\in\mathbb{H}|\operatorname{Re}(s)>a\\}$. If
$\mathcal{L}^{l}\\{f(t)\\}=F(s),$ then
$\mathcal{L}^{l}\left\\{f^{\prime}(t)\right\\}=s\mathcal{L}^{l}\\{f(t)\\}-f(0+)=sF(s)-f(0+)$
for those $s\in U$; More generally one has,

$\mathcal{L}^{l}\left\\{f^{(n)}(t)\right\\}=s^{n}F(s)-s^{n-1}f(0+)-s^{n-2}f^{\prime}(0+)-\cdots-
sf^{(n-2)}(0+)-f^{(n-1)}(0+)$

for those $s\in U$, where $f^{(r)}(0+)=lim_{t\rightarrow 0+}f^{(r)}(t)$,
$r=0,1,\cdots,n-1$.

###### Proof.

Integrate by parts, one has

$\begin{array}[]{l}\mathcal{L}^{l}\left\\{f^{\prime}(t)\right\\}=\int_{0}^{\infty}e^{-st}f^{\prime}(t)dt\\\
=\left[e^{-st}f(t)\right]_{0}^{\infty}-\int_{0}^{\infty}[d(e^{-st})\cdot
f(t)]\\\
=\left[e^{-st}f(t)\right]_{0}^{\infty}+s\int_{0}^{\infty}e^{-st}f(t)dt\\\
=-f(0+)+sF(s),\end{array}$

which holds for those $s\in\mathbb{H}$ with $\operatorname{Re}(s)>a$. The
formula for $\mathcal{L}^{l}\left\\{f^{(n)}(t)\right\\}$ can be proved
similarly.∎

Let $f(t)$ and $g(t)$ be quaternion-valued functions defined for all real
numbers $t\geq 0$. Let $f\circ g$ be the convolution of $f(t)$ and $g(t)$,
which is defined by the integral

$(f\circ g)(t)=\int_{0}^{t}f(t-\tau)g(\tau)d\tau.$

Note that $(f\circ g)(t)$ may not equal to $(g\circ f)(t)$.

######  Proposition 4.8.

(Convolution Theorem). Assume that $f(t)$ and $g(t)$ are of exponential order
$a$ and $b$ respectively. Assume $\mathcal{L}^{l}\\{f(t)\\}=F(s)$ and
$\mathcal{L}^{l}\\{g(t)\\}=G(s).$ Let $c=max\\{a,b\\}$ and
$U=\\{s\in\mathbb{H}|\operatorname{Re}(s)>c\\}$. Then
$\mathcal{L}^{l}\\{f\circ g\\}\in\mathcal{R}^{l}(U)$ and
$\mathcal{L}^{l}\\{f\circ g\\}(s)=(F*G)(s)$ for $s\in U$. In particular, if
$s\in\mathbb{R}$ and $s>c$,

$\mathcal{L}^{l}\\{f\circ
g\\}(s)=\mathcal{L}^{l}\\{f(t)\\}\mathcal{L}^{l}\\{g(t)\\}=F(s)G(s).$

###### Proof.

It is directly verified that for any $\varepsilon>0$, $(f\circ g)(t)$ is of
exponential order $c+\varepsilon$. Then $\mathcal{L}^{l}\\{f\circ g\\}$ is
defined on $U=\\{s\in\mathbb{H}|\operatorname{Re}(s)>c\\}$, and it is left
regular on $U$.

By definition, for $s\in\mathbb{R}\cap U$,

$\displaystyle F(s)G(s)$
$\displaystyle=\int_{0}^{\infty}e^{-s\sigma}f(\sigma)d\sigma\int_{0}^{\infty}e^{-s\mu}g(\mu)d\mu$
$\displaystyle=\int_{0}^{\infty}\int_{0}^{\infty}e^{-s(\sigma+\mu)}f(\sigma)g(\mu)d\sigma
d\mu.$

We make the change of variables $\mu=\tau,\sigma=t-\mu=t-\tau$. Consequently,
for $s\in\mathbb{R}\cap U$, it becomes

$\displaystyle F(s)G(s)$
$\displaystyle=\int_{0}^{\infty}e^{-st}dt\int_{0}^{t}f(t-\tau)g(\tau)d\tau$
$\displaystyle=\mathcal{L}^{l}\left\\{\int_{0}^{t}f(t-\tau)g(\tau)d\tau\right\\}$
$\displaystyle=\mathcal{L}^{l}\\{f(t)\circ g(t)\\}.$

Thus, $\mathcal{L}^{l}\\{f\circ g\\}$ is the left regular function on $U$,
such that for $s\in\mathbb{R}$ and $s>c$, $\mathcal{L}^{l}\\{f\circ
g\\}(s)=F(s)G(s).$ As $F(s)$ and $G(s)$ are in $\mathcal{R}^{l}(U)$,
$\mathcal{L}^{l}\\{f\circ g\\}(s)=(F*G)(s)$ for $s\in U$ by Proposition 3.4
(1). ∎

######  Proposition 4.9.

(Derivatives of the Laplace Transform). Assume that $f(t)$ is of exponential
order $a$ and $U=\\{s\in\mathbb{H}|\operatorname{Re}(s)>a\\}$. If
$\mathcal{L}^{l}\\{f(t)\\}=F(s),$ then for $n\in\mathbb{N}$ and $s\in U$,

$\displaystyle\mathcal{L}^{l}\left\\{t^{n}f(t)\right\\}$
$\displaystyle=(-1)^{n}F^{(n)}(s),$

where $F^{(n)}(s)$ is the $n$-th left slice derivative of $F(s)$.

This can be proved as in the complex case.

######  Proposition 4.10.

(The Laplace Transform of an Integral). Assume that $f(t)$ is of exponential
order $a$ and $U=\\{s\in\mathbb{H}|\operatorname{Re}(s)>a\\}$. If
$\mathcal{L}^{l}\\{f(t)\\}=F(s),$ then for $s\in U$,

$\mathcal{L}^{l}\left\\{\int_{0}^{t}f(\tau)d\tau\right\\}=s^{-1}{F(s)}.$

This can also be proved as in the complex case using Proposition 4.7, see
THEOREM 3.6.4 of [DB].

So we have already seen that most of the properties for the complex Laplace
transform can be generalized to the quaternionic Laplace transform.

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