# One-dimensional Quaternion Fourier Transform and Some Applications

Eusebio Ariza García, Claudia Jiménez Heredia, Carlos Chipantiza

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Eusebio Ariza García, Claudia Jiménez Heredia, Carlos Chipantiza. One-dimensional Quaternion Fourier Transform and Some Applications. Bull. Comput. Appl. Math., 2021, 9 (2), pp.125-149. hal-03579189

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# One-dimensional Quaternion Fourier Transform and Some Applications

Eusebio Ariza García¹², Claudia Jiménez Heredia<fnref n="3" />, Carlos Chipantiza<fnref n="4" />

CompAMa Vol.9, No.2, pp.125-149, 2021 - Received April 12, 2021 - Accepted September 30, 2021

Special Issue II JEM 2020, Eds. Carmen Judith Vanegas and Saúl Buitrago Boret

## Abstract

The Fourier transform is a very useful tool in many areas. This transform has been defined in many contexts, as complex numbers, quaternions, and Clifford algebras. This paper deals with a one-dimensional Fourier Transform defined for quaternion-valued functions. Some of its fundamental properties are proved, particularly, the inversion formula, which will be use to prove another important inversion formula for a two-dimensional quaternion Fourier transform. Using this Fourier transform, we also solve a partial differential equation.

**Keywords:** Quaternions; one-dimensional Quaternion Fourier Transform; inversion formula.

## Resumen

La transformada de Fourier es una herramienta muy útil en muchas áreas. Esta transformada ha sido definida en muchos contextos: en los números complejos, en los cuaterniones, y en las álgebras de Clifford. Este artículo se trata de una

<footnote n="1">¹School of Mathematical and Computational Sciences, Yachay Tech, Urcuquí, Ecuador (eariza@yachaytec.edu.ec). (ORCID iD: https://orcid.org/0000-0001-7754-2666).</footnote>

<footnote n="2">²Corresponding author.</footnote>

<footnote n="3">³Instituto de Ciencias Básicas, Universidad Técnica de Manabí, Portoviejo, Ecuador (claudia.jimenez@utm.edu.ec). (ORCID iD: https://orcid.org/0000-0003-4658-0031).</footnote>

<footnote n="4">⁴School of Mathematical and Computational Sciences, Yachay Tech, Urcuquí, Ecuador (carlos.chipantiza@yachaytech.edu.ec). (ORCID iD: https://orcid.org/0000-0002-0832-1840).</footnote>
transformada de Fourier unidimensional definida para funciones con valores en los
cuaterniones. Algunas de sus propiedades fundamentales son probadas, particular-
mente, la fórmula de inversión, que será usada para probar otra fórmula de inversión
importante para una transformada de Fourier bi-dimensional en los cuaterniones.
Usando esta transformada de Fourier, resolvemos una ecuación diferencial parcial.

**Palabras claves:** Cuaterniones; transformada de Fourier unidimensional en los cuaterniones; fórmula de inversión.

# 1 Introduction

Since the introduction of the Classical Complex Fourier Transform in the
earliest 1800s, it has been an important tool because of its applications in
many areas such as mathematics, physics and engineering (see [1] and [2]).
Among the various applications of the Fourier Transform, one of the most
important is in signal processing and in the analysis of medical images. It is in
this area that the use of the classical Complex Fourier Transform was limited.
Specifically in the analysis of nuclear magnetic resonance images (NMR).
The signal waveforms/samples measured in NMR are complex [3] and, since
quaternion Fourier transform is useful to avoid the mixing of information
presented in the case of complex Fourier transform, this quaternion aspect of
this transform was essential for maintaining the separation between the two
dimensions.

This situation aroused the interest of some mathematicians to begin to
work on the Quaternionic Fourier Transform and the Fourier Transform in
Clifford Algebras [1]. Quaternions were first applied to Fourier Transform by
Ernst [4] and Delsuc [3]. This quaternion transforms were two-dimensional
(that is, they had two independent variables) [1].

In 1998 in a paper by Sangwine and Ell, new ideas emerged [5]. These
were, firstly, the choice of a general root $\mu$ of -1 (a unit quaternion with zero
scalar part) rather than a basis unit $i, j, or, k$ of the quaternion algebra, and
secondly, the choice of a single exponential rather than two (giving a choice
of ordering relative to the quaternionic signal function) [1].

Using this idea, we define a one-dimensional Quaternion Fourier Trans-
form that, unlike the transform defined in [6], uses a pure unit quaternion in
the exponent of the transform kernel. We present some examples and prove
its principal algebraic properties with some applications.
The paper is organized as follows. In section 2 we recall the definition and properties of the Complex Fourier Transform. We also recall the quater-nion algebra, its basic properties and an estimate which will be used in the last part of section 3. In section 3 we focus on to obtain and prove some important properties of the one-dimensional Quaternion Fourier Transform, e.g., convolution, differentiation, inversion formula, Parseval and Plancherel's Theorems. Then, in section 4 we exhibit an example of a partial differential equation that can be solved using the method of Fourier transform, with this one-dimensional transform. In section 5 we give an application to a 2-dimensional Quaternion Fourier Transform, proving its inversion formula. Finally, we give some concluding remarks.

# 2 Preliminaries

## 2.1 Fourier Transform

We first recall what the complex Fourier transform is and some of its main properties (see [7] for more details).

**Definition 1.** Let $f$ be a complex-valued and integrable function on $\mathbb{R}$, its Fourier transform is the function $\hat{f}$ on $\mathbb{R}$ defined by

$$ \mathcal{F}\{f(x)\}(\xi) = \hat{f}(\xi) = \int_{\mathbb{R}} e^{-i\xi x} f(x)dx. $$

The following is a short list of properties of this integral transform.

**Proposition 1. (Properties)**

Suppose $f \in L^1(\mathbb{R}; \mathbb{C})$.

(a) For any $a \in \mathbb{R}$,

$$ \mathcal{F}\{f(x - a)\}(\xi) = e^{-ia\xi} \hat{f}(\xi) \quad \text{and} \quad \mathcal{F}\{e^{iax} f(x)\}(\xi) = \hat{f}(\xi - a). $$

(b) If $\delta > 0$ and $f_\delta(x) = \delta^{-1}f(x/\delta)$, then

$$ [\hat{f}_\delta](\xi) = \hat{f}(\delta\xi) \quad \text{and} \quad \mathcal{F}\{f(\delta x)\}(\xi) = [\hat{f}]_\delta(\xi) = \delta^{-1}\hat{f}(\xi/\delta). $$
(c) If $f$ is continuous and piecewise smooth and $f' \in L^1(\mathbb{R}; \mathbb{C})$, then

$$[f'] (\xi) = i\xi \hat{f}(\xi).$$

On the other hand, if $xf(x)$ is integrable, then

$$\mathcal{F}\{xf(x)\}(\xi) = i[\hat{f}]'(\xi).$$

(d) If also $g \in L^1(\mathbb{R}; \mathbb{C})$, then

$$\mathcal{F}\{(f \ast g)(x)\}(\xi) = \widehat{(f \ast g)}(\xi) = \hat{f}(\xi) \hat{g}(\xi),$$

where

$$(f \ast g)(x) := \int_{\mathbb{R}} f(x - y)g(y)dy.$$

**Theorem 1. (Inversion Formula)**

If $\hat{f} \in L^1(\mathbb{R}; \mathbb{C})$, then $f$ is continuous and $f$ can be recovered by using the formula

$$f(x) = \frac{1}{2\pi} \int_{\mathbb{R}} e^{i\xi x} \hat{f}(\xi) d\xi, \quad x \in \mathbb{R}.$$

## 2.2 Quaternions [8]

Let $e_0 = 1, e_1, e_2, e_3$ be the basic elements in $\mathbb{R}^4$. An arbitrary element $q \in \mathbb{R}^4$ can be represented as $q = q_0e_0 + q_1e_1 + q_2e_2 + q_3e_3$, where $q_0, q_1, q_2, q_3 \in \mathbb{R}$.

We will call the part $q_0e_0 =: \text{Sc } q$ the **scalar part** of $q$ and $\underline{q} = q - q_0e_0$ the **vector part** of $q$ (we also write $\text{Vec } q$ instead of $\underline{q}$). Now we define a product in $\mathbb{R}^4$ which fulfills the following conditions:

$$i) e_1^2 = e_2^2 = e_3^2 = -1,$$

$$ii) e_1e_2 = e_3, e_2e_3 = e_1, e_3e_1 = e_2,$$

$$iii) e_i e_j + e_j e_i = 0 \ (i, j = 1, 2, 3; \ i \neq j).$$

For $q, p \in \mathbb{R}^4$ we write for short

$$(\underline{q}, \underline{p}) := q_1p_1 + q_2p_2 + q_3p_3$$
and

$$ \underline{q} \times \underline{p} := (q_2p_3 - q_3p_2)e_1 + (q_3p_1 - q_1p_3)e_2 + (q_1p_2 - q_2p_1)e_3. $$

By using of the algebraic rules (i), (ii) (iii) the **quaternionic product xy** is given by

$$ qp = q_0p_0 - (\underline{q}, \underline{p}) + q_0\underline{p} + p_0\underline{q} + \underline{q} \times \underline{p}. $$

The pair $(\mathbb{R}^4, \cdot)$ is called **the algebra of real quaternions** and is denoted by $\mathbb{H}$, in honor to Sir William Rowan Hamilton.

If $q = \underline{q}$ then $q$ is called a **pure quaternion**. The subset of all pure quaternions is denoted by $\text{Vec } \mathbb{H}$, while the subset of all scalars is denoted by $\text{Sc } \mathbb{H}$. The quaternion $\bar{q} = q_0 - \underline{q}$ is called the **conjugate** to $q$. The mapping $q \mapsto \bar{q}$ is called **conjugation**.

The **norm** or **absolute value** of a quaternion $q$ is defined as

$$ |q| := \sqrt{\bar{q}\bar{q}}. $$

If $|q| = 1$, the quaternion $q$ is said to be a **unit quaternion**. The set of pure unit quaternions is given by

$$ S_{\mathbb{H}}^3 := \{q = e_1q_1 + e_2q_2 + e_3q_3 \mid q_1^2 + q_2^2 + q_3^2 = 1\}. $$

**Lemma 1.** *qp is an $\mathbb{R}$-bilinear and associative product, but it is not commutative.*

For $q, p \in \mathbb{H}$ and $a \in \mathbb{R}$ one has the relations

$$ (i) \quad \overline{q+p} = \overline{q} + \overline{p}, $$

$$ (ii) \quad \overline{aq} = a\overline{q}, $$

$$ (iii) \quad \overline{q} = q, $$

(iv) $\overline{qp} = \overline{p}\overline{q}$, which means that the conjugation is an anti-homomorphism.

**Proposition 2.** *For all $q, p \in \mathbb{H}$ the identity $|qp| = |q||p|$ holds. If $q \neq 0$, then $q^{-1}$ exists and we have*

$$ q^{-1} = \overline{q}|q|^{-2} \qquad (1) $$

and

$$ |q|^{-1} = |q^{-1}|. $$
**Proposition 3.** An arbitrary quaternion $q \in \mathbb{H}$, $\underline{q} \neq 0$ permits the binary representation

$$q = |q|(\cos \phi + \omega(q) \sin \phi),$$

where $\phi = \arccot(q_0/|\underline{q}|)$, $\cos \phi = \frac{q_0}{|\underline{q}|}$, $\sin \phi = \frac{|\underline{q}|}{|\underline{q}|}$ and $\omega(q) = \underline{q}/|\underline{q}| \in S_{\mathbb{H}}^3$.

**Corollary 1. (MOIVRE's formula)**

Let $q \in \mathbb{H}$, $\underline{q} \neq 0$, $n \in \mathbb{N}$, then the following formula is valid:

$$(\cos \phi + \omega(q) \sin \phi)^n = \cos n\phi + \omega(q) \sin n\phi.$$

**Remark 1.** Each element of $S_{\mathbb{H}}^3$ creates a distinct copy of the complex numbers. In fact, given $\mu \in S_{\mathbb{H}}^3$, it's clear that $\mu^2 = -1$. So, for each $\mu \in S_{\mathbb{H}}^3$, we can associate a complex sub-field of $\mathbb{H}$ denoted by

$$C_{\mu} = \{\alpha + \beta\mu; |\alpha, \beta \in \mathbb{R}, \mu \in S_{\mathbb{H}}^3\}.$$

**Definition 2.** Let $\omega \in \text{Vec} \mathbb{H}$, $\omega^2 = -1$. The exponential function $e^{\omega\theta}$, for $\theta \in \mathbb{R}$, is defined by the formal expansion

$$e^{\omega\theta} := 1 + \omega\theta - \frac{1}{2!}\theta^2 - \frac{1}{3!}\omega\theta^3 + \dots,$$

where we have uniform convergence of the expansion

$$e^{\theta} = 1 + \theta + \frac{1}{2!}\theta^2 + \dots, \theta \in \mathbb{R}.$$

**Proposition 4.** Let $\omega \in \text{Vec} \mathbb{H}$, with $\omega^2 = -1$. Then

$$(i) \ e^{\omega\theta} = \cos \theta + \omega \sin \theta.$$

$$(ii) \ |e^{\omega\theta}| = 1.$$

**Proposition 5.** Let $p, q \in \text{Vec} \mathbb{H}$. Then

$$(i) \ pq = qp \implies e^p e^q = e^{p+q}.$$

$$(ii) \text{ From } e^p e^q = e^{p+q} \text{ it does \underline{not} follow that } pq = qp.$$

Let's verify (ii). Taking $p := 3\pi e_1$ and $q := 4\pi e_2$, we obtain

$$e^{3\pi e_1} = -1, \ e^{4\pi e_2} = 1 \text{ and } e^{3\pi e_1 + 4\pi e_2} = e^{(\frac{3}{5}e_1 + \frac{4}{5}e_2)5\pi}.$$

Since $(\frac{3}{5}e_1 + \frac{4}{5}e_2)^2 = -1$, the latter identity is minus one. So,

$$e^{3\pi e_1} e^{4\pi e_2} = e^{3\pi e_1 + 4\pi e_2}.$$

However,

$$qp = 4\pi e_2 \cdot 3\pi e_1 = -3\pi e_1 \cdot 4\pi e_2 \neq 3\pi \cdot e_1 4\pi e_2 = pq.$$
## 2.3 An important estimate

It's clear from the definition of the norm of a real quaternion $q = q_0 + q_1 e_1 + q_2 e_2 + q_3 e_3$, that

$$|q|^2 = \sum_{i=0}^{3} |q_i|^2,$$

so,

$$|q_i| \le |q|.$$

On the other hand, given an integrable function $f: \mathbb{R} \to \mathbb{H}$, since

$$\int_{\mathbb{R}} f(x) dx = \sum_{i=0}^{3} \int_{\mathbb{R}} f_i(x) dxe_i,$$

we see that

$$\begin{aligned} \left|\int_{\mathbb{R}} f(x) dx\right|^2 &= \sum_{i=0}^{3} \left|\int_{\mathbb{R}} f_i(x) dx\right|^2 \\ &\le \sum_{i=0}^{3} \left(\int_{\mathbb{R}} |f_i(x)| dx\right)^2 \\ &\le \sum_{i=0}^{3} \left(\int_{\mathbb{R}} |f(x)| dx\right)^2 \\ &= \left(2 \int_{\mathbb{R}} |f(x)| dx\right)^2. \end{aligned}$$

That is,

$$\left|\int_{\mathbb{R}} f(x) dx\right| \le 2 \int_{\mathbb{R}} |f(x)| dx.$$

A similar argument yields

$$\left|\int_{-\infty}^{a} f(x) dx\right| \le 2 \int_{-\infty}^{a} |f(x)| dx, \text{ with } a \le 0 \quad (2)$$

and

$$\left|\int_{b}^{\infty} f(x) dx\right| \le 2 \int_{b}^{\infty} |f(x)| dx, \text{ with } b \ge 0. \quad (3)$$
The same estimate is true for integrals of the form $\int_a^0$ and $\int_0^b$.

# 3 One dimensional Quaternion Fourier Transform

**Definition 3.** Let $f \in L^1(\mathbb{R}; \mathbb{H})$. We define the One-dimensional (left) Quaternion Fourier Transform of $f$ (1D-QFT for short), by

$$ \mathcal{F}\{f(x)\}(t) = \hat{f}(t) = \int_{\mathbb{R}} e^{-\mu tx} f(x) dx, \quad (4) $$

where $\mu \in \mathbb{H}$ is a pure unit quaternion.

**Example 1.** For a positive real number $a$, consider $f(x) = e^{-ax^2}$. Then,

$$ \mathcal{F}\{e^{-ax^2}\}(t) = \int_{\mathbb{R}} e^{-\mu tx} e^{-ax^2} dx = \sqrt{\frac{\pi}{a}} e^{-\frac{t^2}{4a}}. $$

In particular, if $a = \frac{1}{2}$, we have

$$ \mathcal{F}\{e^{-\frac{x^2}{2}}\}(t) = \sqrt{2\pi}e^{-\frac{t^2}{2}}. $$

**Example 2.** Let $f(x) = e^{-a|x|}$, for $a > 0$. Then,

$$ \mathcal{F}\{e^{-a|x|}\}(t) = \int_{\mathbb{R}} e^{-\mu tx} e^{-a|x|} dx = \frac{2a}{a^2 + t^2}. $$

## 3.1 Algebraic Properties

In this subsection we prove some algebraic properties of the 1D-QFT.

**Proposition 6.** $(\mathbb{R}-\text{linearity})$

The 1D-QFT is $\mathbb{R}$-linear. That is, given $a, b \in \mathbb{R}$ and $f, g \in L^1(\mathbb{R}; \mathbb{H})$,

$$ \mathcal{F}\{(af + bg)(x)\}(t) = af(t) + bg(t). \quad (5) $$

**Remark 2.** It should be noted that (left) $\mathbb{C}$-linearity and (left) $\mathbb{H}$-linearity are false in general, but (right) $\mathbb{H}$-linearity holds. In fact, if $a, b \in \mathbb{H}$, then

$$ \begin{aligned} \mathcal{F}\{(fa + gb)(x)\}(t) &= \int_{\mathbb{R}} e^{-\mu tx}(f(x)a + g(x)b)dx \\ &= \int_{\mathbb{R}} (e^{-\mu tx}f(x))adx + \int_{\mathbb{R}} (e^{-\mu tx}g(x))bdx \\ &= \hat{f}(t)a + \hat{g}(t)b. \end{aligned} $$
Now, we see the following result.

**Proposition 7. (Dilation or Scaling Property)**

Let a be a positive real number and $f \in L^1(\mathbb{R}; \mathbb{H})$. Then,

$$ \frac{1}{a} \mathcal{F} \left\{ f \left( \frac{x}{a} \right) \right\} (t) = \hat{f}(at) \qquad (6) $$

or, analogously,

$$ a\mathcal{F}\{f(ax)\}(t) = \hat{f}\left(\frac{t}{a}\right). \qquad (7) $$

*Proof.*

$$ \hat{f}(at) = \int_{\mathbb{R}} e^{-\mu(at)x} f(x) dx = \frac{1}{a} \int_{\mathbb{R}} e^{-\mu(ty)} f\left(\frac{y}{a}\right) dy = \frac{1}{a} \mathcal{F}\left\{f\left(\frac{x}{a}\right)\right\}(t). $$

<figure><img src="image_2.png" /></figure>

**Proposition 8. (Shift Property)**

Let $x_0 \in \mathbb{R}$ and $f \in L^1(\mathbb{R}; \mathbb{H})$. Then,

$$ \mathcal{F}\{f(x - x_0)\}(t) = e^{-\mu tx_0} \hat{f}(t). \qquad (8) $$

*Proof.* Since $e^{\alpha+\beta} = e^{\alpha}e^{\beta}$ if $\alpha\beta = \beta\alpha$, where $\alpha, \beta \in \text{Vec } \mathbb{H}$, we have that

$$ \mathcal{F}\{f(x - x_0)\}(t) = \int_{\mathbb{R}} e^{-\mu tx} f(x - x_0) dx $$

<figure><img src="image_3.png" /></figure>

**Proposition 9. (Modulation Property)**

Let $t_0 \in \mathbb{R}$ and $f \in L^1(\mathbb{R}; \mathbb{H})$. Then,

$$ \mathcal{F}\{e^{\mu t_0 x} f(x)\}(t) = \hat{f}(t - t_0). \qquad (9) $$

*Proof.*

$$ \mathcal{F}\{e^{\mu t_0 x} f(x)\}(t) = \int_{\mathbb{R}} e^{-\mu t x} e^{\mu t_0 x} f(x) dx = \int_{\mathbb{R}} e^{-\mu(t-t_0)x} f(x) dx = \hat{f}(t-t_0). $$

<figure><img src="image_4.png" /></figure>
## 3.2 Right 1-dimensional Quaternion Fourier Transform

Some properties are more easy to present when the left 1D-QFT is related with the right 1-dimensional Quaternion Fourier Transform defined as follows.

**Definition 4.** If $f \in L^1(\mathbb{R}; \mathbb{H})$, its right 1D-QFT is given by

$$ \mathcal{F}_r \{f(x)\} (t) = \int_{\mathbb{R}} f(x)e^{-\mu tx} dx, $$

where $\mu$ is as before.

**Remark 3.** It's clear that if $f$ commutes with $e^{-\mu\theta}$, where $\theta \in \mathbb{R}$, then

$$ \mathcal{F}_r \{f(x)\} (t) = \mathcal{F} \{f(x)\} (t), $$

i.e., in this case the left 1D-QFT, given by Definition 3, agrees with the right 1D-QFT.

**Proposition 10. (Conjugation Property)**

Let $f \in L^1(\mathbb{R}; \mathbb{H})$. Then,

$$ \mathcal{F}_r \{\overline{f(-x)}\} (t) = -\overline{\hat{f}(t)}. \qquad (10) $$

*Proof.*

$$ \begin{aligned} \mathcal{F}_r \{\overline{f(-x)}\} (t) &= \int_{\mathbb{R}} \overline{f(-x)} e^{-\mu tx} dx \\ &= \int_{\mathbb{R}} \overline{f(-x)} \overline{e^{\mu tx}} dx \\ &= \int_{\mathbb{R}} \overline{e^{\mu tx} f(-x)} dx \\ &= -\overline{\int_{\mathbb{R}} e^{-\mu tx} f(x) dx} \\ &= -\overline{\hat{f}(t)}. \end{aligned} $$

□
**Remark 4.** The last property can be rewritten as

$$ \mathcal{F}\{\overline{f(-x)}\}(t) = -\overline{\mathcal{F}_r\{f(x)\}}(t). $$

Other properties that can be shown are

**Proposition 11. (Composition Property)**

Let $f, g \in L^1(\mathbb{R}; \mathbb{H})$. Then,

$$ \int_{\mathbb{R}} g(t)e^{\mu t x} \mathcal{F}\{f(x)\}(t)dt = \int_{\mathbb{R}} \mathcal{F}_r\{g(t)\} (y-x)f(y)dy. \quad (11) $$

*Proof.*

$$ \begin{aligned} \int_{\mathbb{R}} g(t)e^{\mu t x} \mathcal{F}\{f(x)\}(t)dt &= \int_{\mathbb{R}} g(t)e^{\mu t x} \left[ \int_{\mathbb{R}} e^{-\mu ty} f(y)dy \right] dt \\ &= \int_{\mathbb{R}} \left( \int_{\mathbb{R}} g(t)e^{-\mu t(y-x)} dt \right) f(y)dy \\ &= \int_{\mathbb{R}} \mathcal{F}_r\{g(t)\} (y-x)f(y)dy. \end{aligned} $$

□

**Remark 5.** If $g$ commutes with $e^{\mu\theta}$, for example, when $g = g_0 + \mu g_1$, where $g_0, g_1 : \mathbb{R} \to \mathbb{R}$, then

$$ \int_{\mathbb{R}} g(t)e^{\mu t x} \mathcal{F}\{f(x)\}(t)d(t) = \int_{\mathbb{R}} \mathcal{F}\{g(t)\}(y-x)f(y)dy. $$

**Proposition 12.** If $f, g \in L^1(\mathbb{R}; \mathbb{H})$, then

$$ \int_{\mathbb{R}} g(\xi) \mathcal{F}\{f(x)\}(\xi) d\xi = \int_{\mathbb{R}} \mathcal{F}_r\{g(\xi)\}(x)f(x)dx. \quad (12) $$

*Proof.*

$$ \begin{aligned} \int_{\mathbb{R}} g(\xi) \mathcal{F}\{f(x)\}(\xi) d\xi &= \int_{\mathbb{R}} g(\xi) \left( \int_{\mathbb{R}} e^{-\mu \xi x} f(x) dx \right) d\xi \\ &= \int_{\mathbb{R}} \left( \int_{\mathbb{R}} g(\xi) e^{-\mu \xi x} d\xi \right) f(x) dx. \end{aligned} $$

□
### 3.3 Differentiation

Some applications of the 1D-QFT are related with differential equations. That's why in this subsection we prove some properties of this transform which will help us in this direction.

**Proposition 13.** Let $f, f' \in L^1(\mathbb{R}; \mathbb{H})$ and $f(x) \to 0$ as $|x| \to \infty$, then

$$ \mathcal{F}\{f'(x)\}(t) = \mu t \hat{f}(t). \quad (13) $$

*Proof.*

$$
\begin{align*}
\mathcal{F}\{f'(x)\}(t) &= \int_{\mathbb{R}} e^{-\mu t x} f'(x) dx \\
&= \int_{-\infty}^{0} e^{-\mu t x} f'(x) dx + \int_{0}^{\infty} e^{-\mu t x} f'(x) dx \\
&= \lim_{b \to -\infty} \left(e^{-\mu t x} f(x)\right)|_{b}^{0} + \mu t \int_{-\infty}^{0} e^{-\mu t x} f(x) dx \\
&\quad + \lim_{b \to \infty} \left(e^{-\mu t x} f(x)\right)|_{0}^{b} + \mu t \int_{0}^{\infty} e^{-\mu t x} f(x) dx \\
&= \mu t \int_{\mathbb{R}} e^{-\mu t x} f(x) dx.
\end{align*}
$$

□

As a consequence of this result we see that

**Corollary 2.** If $f, f', \dots, f^{(n)} \in L^1(\mathbb{R}; \mathbb{H})$ and $f^{(k)}(x) \to 0$ as $|x| \to \infty$, for $k = 1, \dots, n-1$, then

$$ \mathcal{F}\{f^{(n)}(x)\}(t) = (\mu t)^n \hat{f}(t). $$

**Remark 6.** If $u$ is a function of two or more independent variables, similar operational results are valid in this case. As an example, for $u(x, y)$ and $\mathcal{F}_x\{u(x, y)\}(t) = U(t, y)$, we have that

$$ \mathcal{F}_x\left\{\frac{\partial}{\partial x}u(x, y)\right\}(t) = \mu t U(t, y), \quad \mathcal{F}_x\left\{\frac{\partial^2}{\partial x^2}u(x, y)\right\}(t) = -t^2 U(t, y), $$
$$ \mathcal{F}_x \left\{ \frac{\partial}{\partial y} u \right\} (t) = \frac{\partial}{\partial y} U(t, y), \text{ and so on.} $$

**Proposition 14.** Let $xf(x) \in L^1(\mathbb{R}; \mathbb{H})$, then

$$ \mathcal{F}\{xf(x)\}(t) = \mu \frac{d}{dt} \hat{f}(t). \qquad (14) $$

*Proof.*

$$ \mathcal{F}\{xf(x)\}(t) = \int_{\mathbb{R}} e^{-\mu tx}xf(x)dx = \mu \int_{\mathbb{R}} \frac{d}{dt}(e^{-\mu tx}f(x))dx = \mu \frac{d}{dt}\hat{f}(t). $$

<figure><img src="image_5.png" /></figure>

From the last proposition we also have that

**Corollary 3.** If $x^n f(x) \in L^1(\mathbb{R}; \mathbb{H})$, then

$$ \mathcal{F}\{x^n f(x)\}(t) = \mu^n \frac{d^n}{dt^n} \hat{f}(t). $$

## 3.4 Convolution

It's well known that one useful tool related with the Fourier transform is the convolution. In this context we introduce the concept of convolution and give some of its properties.

**Definition 5.** Let $f$ and $g$ be two integrable functions. The convolution of $f$ and $g$, denoted by $(f * g)(x)$, is defined by

$$ (f * g)(x) = \int_{\mathbb{R}} f(x - \xi) g(\xi) d\xi. $$

**Remark 7.** In view of the non-commutativity of quaternions, in general

$$ (f * g)(x) \neq (g * f)(x). $$

When $f$ and $g$ commute, we also have commutativity for the convolution. This is true, in particular, if either $f$ or $g$ are real-valued functions.
**Proposition 15.** If $a, b \in \mathbb{R}$, the convolution defined above has the following properties.

$$af * h = f * ah = a(f * h),$$

$$(af + bg) * h = a(f * h) + b(g * h),$$

$$f * (ag + bh) = a(f * g) + b(f * h),$$

and

$$f * (g * h) = (f * g) * h.$$

*Proof.* It's direct from the definition.

<figure><img src="image_6.png" /></figure>

**Proposition 16. (Derivative of a convolution)**

Let $f$ be a differentiable function and suppose that $f * g$ and $f' * g$ are well-defined. Then

$$(f * g)' = f' * g. \qquad (15)$$

*Proof.*

$$(f * g)'(x) = \frac{d}{dx} \int_{\mathbb{R}} f(x - \xi) g(\xi) d\xi = \int_{\mathbb{R}} f'(x - \xi) g(\xi) d\xi = (f' * g)(x).$$

<figure><img src="image_7.png" /></figure>

**Remark 8.** Since, in general, $(f * g)(x) \neq (g * f)(x)$, we have that

$$(f' * g)(x) \neq (f * g')(x).$$

**Theorem 2. (Convolution Theorem)**

Let $f, g \in L^1(\mathbb{R}; \mathbb{H})$. Write

$$g(\xi) = g_0(\xi) + g_1(\xi)e_1 + g_2(\xi)e_2 + g(\xi)e_3,$$

with $g_0, g_1, g_2, g_3 \in L^1(\mathbb{R}; \mathbb{R})$, then

$$\mathcal{F}\{(f * g)(x)\}(t) = \hat{g}_0(t)\hat{f}(t) + \hat{g}_1(t)\hat{f}(t)e_1 + \hat{g}_2(t)\hat{f}(t)e_2 + \hat{g}_3(t)\hat{f}(t)e_3. \quad (16)$$
Proof.

$$
\begin{align*}
\mathcal{F}\{(f * g)(x)\}(t) &= \int_{\mathbb{R}} e^{-\mu t x} (f * g)(x) dx \\
&= \int_{\mathbb{R}} \int_{\mathbb{R}} e^{-\mu t x} f(x - \xi) g(\xi) d\xi dx \\
&= \int_{\mathbb{R}} \left( \int_{\mathbb{R}} e^{-\mu t x} f(x - \xi) dx \right) g(\xi) d\xi \\
&= \int_{\mathbb{R}} \mathcal{F}\{f(x - \xi)\}(t) g(\xi) d\xi \\
&= \int_{\mathbb{R}} e^{-\mu t \xi} \hat{f}(t) g(\xi) d\xi \\
&= \int_{\mathbb{R}} e^{-\mu t \xi} \hat{f}(t) (g_0(\xi) + g_1(\xi)e_1 + g_2(\xi)e_2 + g(\xi)e_3) d\xi \\
&= \hat{g}_0(t) \hat{f}(t) + \hat{g}_1(t) \hat{f}(t)e_1 + \hat{g}_2(t) \hat{f}(t)e_2 + \hat{g}_3(t) \hat{f}(t)e_3,
\end{align*}
$$

which proves (16).

<figure><img src="image_8.png" /></figure>

**Remark 9.** Note that

(a) If $f$ and $g$ are such that $\hat{f}$ commutes with $e^{\mu\theta}$, $\theta \in \mathbb{R}$, then

$$
\mathcal{F}\{(f * g)(x)\}(t) = \hat{f}(t)\hat{g}(t). \qquad (17)
$$

(b) If $\hat{f}$ commutes with $g$, then

$$
\mathcal{F}\{(f * g)(x)\}(t) = \hat{g}(t)\hat{f}(t). \qquad (18)
$$

**3.5 Inversion Formula**

In this section we will prove the inversion formula for the 1D–QFT. In order to do that, we first prove the following theorem.
**Theorem 3.** Let $g \in L^2(\mathbb{R}; \mathbb{R})$ and $\int_{\mathbb{R}} g(y)dy = 1$. Let

$$ \alpha := \int_{-\infty}^{0} g(y)dy \quad \text{and} \quad \beta := \int_{0}^{\infty} g(y)dy, $$

so that $\alpha + \beta = 1$ and $\alpha = \beta = \frac{1}{2}$ if $g$ is even. Let $f: \mathbb{R} \to \mathbb{H}$ be a piecewise continuous function on $\mathbb{R}$ and suppose either that $f$ is bounded or that $g$ vanishes outside a finite interval so that $f * g$ is well-defined for all $x$. If $g_{\varepsilon}(x) = \frac{1}{\varepsilon}g(\frac{x}{\varepsilon})$, $f(x^{+}) := \lim_{\varepsilon \to 0^{+}} f(x+\varepsilon)$ and $f(x^{-}) := \lim_{\varepsilon \to 0^{+}} f(x-\varepsilon)$, then

$$ \lim_{\varepsilon \to 0} (g_{\varepsilon} * f)(x) = \alpha f(x^{+}) + \beta f(x^{-}) \quad (19) $$

for all $x$. If $f$ is continuous at $x$, we have

$$ \lim_{\varepsilon \to 0} (g_{\varepsilon} * f)(x) = f(x). \quad (20) $$

*Proof.* First, note that $(g_{\varepsilon} * f)(x) = (f * g_{\varepsilon})(x)$ and

$$ \begin{aligned} (f * g_{\varepsilon})(x) - (\alpha f(x^{+}) + \beta f(x^{-})) &= \int_{-\infty}^{0} [f(y-x) - f(x^{+})]g_{\varepsilon}(y)dy \\ &\quad + \int_{0}^{\infty} [f(y-x) - f(x^{-})]g_{\varepsilon}(y)dy. \end{aligned} \quad (21) $$

So, in order to prove (19), we need to check that the previous expression goes to zero as $\varepsilon \to 0$.

Consider the second integral in (21). Let $\delta > 0$ and choose $c$ small enough so that $|f(x-y) - f(x^-)| < \delta$ if $0 < y < c$. Breaking up the integral as
$\int_0^c + \int_c^\infty$, we have, first, that

$$ 
\begin{aligned}
\left| \int_0^c [f(x-y) - f(x^-)] g_\varepsilon(y) dy \right| &\le 2 \int_0^c |f(x-y) - f(x^-)| |g_\varepsilon(y)| dy \\
&\le 2\delta \int_0^c |g_\varepsilon(y)| dy \\
&= 2\delta \int_0^c \left| \frac{1}{\varepsilon} g\left(\frac{y}{\varepsilon}\right) \right| dy \\
&= 2\delta \int_0^{c/\varepsilon} |g(y)| dy \\
&\le 2\delta \int_0^\infty |g(y)| dy.
\end{aligned}
$$

Clearly, for a suitable chosen $\delta$, the last expression can be made small enough.

On the other hand, in order to estimate the integral from $c$ to $\infty$ we consider the two following cases:

**Case 1:** $f$ is bounded, say $|f| \le M$, for some positive $M$.

In this case,

$$ 
\begin{aligned}
\left| \int_c^\infty [f(x-y) - f(x^-)] g_\varepsilon(y) dy \right| &\le 2 \int_c^\infty |f(x-y) - f(x^-)| |g_\varepsilon(y)| dy \\
&\le 4M \int_c^\infty |g_\varepsilon(y)| dy \\
&\le 4M \int_{c/\varepsilon}^\infty |g(y)| dy,
\end{aligned}
$$

which goes to zero as $\varepsilon \to 0$.

**Case 2:** $g$ vanishes outside a finite interval, say $g(x) = 0$ for $|x| > R$.

In this case, $g_\varepsilon(x) = 0$ for $|x| > R\varepsilon$ and, in particular, $g_\varepsilon(x) = 0$ for $x > c$ if $c > R\varepsilon$, i.e., if $\varepsilon < \frac{c}{R}$.
So, we have that

$$ \int_c^\infty [f(x-y) - f(x^-)]g_\varepsilon(y) \to 0 $$

as $\varepsilon \to 0$.

A similar argument is valid when one considers the first integral in (21). Summarizing, we have that

$$ \lim_{\varepsilon \to 0} (g_\varepsilon * f)(x) = \alpha f(x^+) + \beta f(x^-). $$

If $f$ is continuous at $x$ we have that

$$ \lim_{\varepsilon \to 0} (g_\varepsilon * f)(x) = f(x). $$

<figure><img src="image_9.png" /></figure>

**Theorem 4. (Fourier Inversion Formula for the 1D-QFT)**

Let $f$ be an integrable and piecewise continuous on $\mathbb{R}$, with values in $\mathbb{H}$, defined at its points of discontinuity as to satisfy $f(x) = \frac{1}{2}[f(x^-) + f(x^+)]$ for all $x$. Then

$$ f(x) = \lim_{\varepsilon \to 0} \frac{1}{2\pi} \int e^{\mu tx} e^{-\varepsilon^2 t^2/2} \hat{f}(t)dt, \quad x \in \mathbb{R}. \qquad (22) $$

Moreover, if $\hat{f} \in L^1(\mathbb{R}; \mathbb{H})$, then $f$ is continuous and

$$ f(x) = \frac{1}{2\pi} \int e^{\mu tx} \hat{f}(t)dt, \quad x \in \mathbb{R}. \qquad (23) $$

*Proof.* Let $\varepsilon > 0$ and consider the function $e^{-\varepsilon^2 t^2/2}$, which decreases rapidly as $t \to \pm\infty$. Let $f \in L^1(\mathbb{R}; \mathbb{H})$ and consider the integral

$$ \frac{1}{2\pi} \int_{\mathbb{R}} e^{\mu tx} e^{-\varepsilon^2 \frac{t^2}{2}} \hat{f}(t) dt = \frac{1}{2\pi} \int_{\mathbb{R}} \int_{\mathbb{R}} e^{\mu tx} e^{-\varepsilon^2 \frac{t^2}{2}} e^{-\mu ty} f(y) dy dt. $$

This double integral is absolutely convergent, so we can interchange the order of integration. That is, we can write

$$ \frac{1}{2\pi} \int_{\mathbb{R}} e^{\mu tx} e^{-\varepsilon^2 \frac{t^2}{2}} \hat{f}(t) dt = \frac{1}{2\pi} \int_{\mathbb{R}} \left( \int_{\mathbb{R}} e^{\mu t(x-y)} e^{-\varepsilon^2 \frac{t^2}{2}} dt \right) f(y) dy. $$
Note that, putting $a = \varepsilon^2/2$, we have that

$$ \int_{\mathbb{R}} e^{\mu t(x-y)} e^{-\varepsilon^2 \frac{t^2}{2}} dt = \int_{\mathbb{R}} e^{-\mu t(y-x)} e^{-at^2} dt = \mathcal{F}\{e^{-at^2}\}(y-x) = \sqrt{\frac{\pi}{a}} e^{-\frac{(y-x)^2}{4a}}. $$

That is,

$$ \int_{\mathbb{R}} e^{\mu t(x-r)} e^{-\varepsilon^2 t^2/2} dt = \frac{\sqrt{2\pi}}{\varepsilon} e^{-\frac{(x-y)^2}{2\varepsilon^2}}. $$

Thus,

$$ \frac{1}{2\pi} \int_{\mathbb{R}} e^{\mu t x} e^{-\varepsilon^2 \frac{t^2}{2}} \hat{f}(t) dt = \frac{1}{\varepsilon\sqrt{2\pi}} \int_{\mathbb{R}} e^{-\frac{(x-y)^2}{2\varepsilon^2}} f(y) dy = (\phi_{\varepsilon} * f)(x), $$

$$ \text{where } \phi(x) = \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} \text{ and } \phi_{\varepsilon}(x) = \frac{1}{\varepsilon} \phi\left(\frac{x}{\varepsilon}\right) = \frac{1}{\varepsilon\sqrt{2\pi}} e^{-\frac{x^2}{2\varepsilon^2}}. $$

Now, since $f$ is piecewise continuous, in view of Theorem 3, we see that

$$ \lim_{\varepsilon \to 0} (\phi_{\varepsilon} * f)(x) = \frac{1}{2} [f(x^{-}) + f(x^{+})] $$

for all $x$. Finally, since $|e^{\mu t x}| = 1$ and $|e^{-\varepsilon^2 \frac{t^2}{2}}| \le 1$ for small enough $\varepsilon$,

$$ |e^{\mu t x} e^{-\varepsilon^2 \frac{t^2}{2}} \hat{f}(t)| = |e^{\mu t x}| |e^{-\varepsilon^2 \frac{t^2}{2}}| |\hat{f}(t)| \le |\hat{f}(t)|. $$

Thus, if $\hat{f} \in L^1(\mathbb{R}; \mathbb{H})$, the dominated convergence theorem implies that

$$ \begin{aligned} \frac{1}{2\pi} \int_{\mathbb{R}} e^{\mu t x} \hat{f}(t) dt &= \frac{1}{2\pi} \lim_{\varepsilon \to 0} \left( e^{\mu t x} e^{-\varepsilon^2 \frac{t^2}{2}} \hat{f}(t) \right) dt \\ &= \lim_{\varepsilon \to 0} \frac{1}{2\pi} \int_{\mathbb{R}} e^{\mu t x} e^{-\varepsilon^2 \frac{t^2}{2}} \hat{f}(t) dt \\ &= \frac{1}{2} [f(x^{-}) + f(x^{+})]. \end{aligned} $$

Since

$$ \int_{\mathbb{R}} e^{\mu t x} \hat{f}(t) dt = \mathcal{F}\{\hat{f}(t)\}(-x) $$

and the Fourier transform of an integrable function is continuous, we have that $f$ is continuous and, thus,

$$ f(x) = \frac{1}{2} [f(x^{-}) + f(x^{+})], $$

which proves (20). □

<figure><img src="image_10.png" /></figure>
## 3.6 Plancherel and Parseval's Theorems

Now, we want to establish the Plancherel and Parseval's theorems in this context. We begin with the definition of inner product and norm in $L^2(\mathbb{R}; \mathbb{H})$.

**Definition 6.** Let $f, g \in L^2(\mathbb{R}; \mathbb{H})$, we define

$$ \langle f, g \rangle_{L^2(\mathbb{R}; \mathbb{H})} := \int_{\mathbb{R}} \overline{g(x)} f(x) dx \qquad (24) $$

and

$$ \|f\|_{L^2(\mathbb{R}; \mathbb{H})} = \left( \int_{\mathbb{R}} |f(x)|^2 dx \right)^{1/2} . \qquad (25) $$

**Remark 10.**

$$ \overline{\langle g, f \rangle}_{L^2(\mathbb{R}; \mathbb{H})} = \langle f, g \rangle_{L^2(\mathbb{R}; \mathbb{H})}. $$

Using the previous definition of inner product, we have the following theorem.

**Theorem 5. (Plancherel's Theorem)**

Let $f, g \in L^2(\mathbb{R}; \mathbb{H})$. Then,

$$ \langle f, g \rangle_{L^2(\mathbb{R}; \mathbb{H})} = \frac{1}{2\pi} \langle \hat{f}, \hat{g} \rangle_{L^2(\mathbb{R}; \mathbb{H})}. \qquad (26) $$

*Proof.*

$$ \begin{aligned} \langle f, g \rangle_{L^2(\mathbb{R}; \mathbb{H})} &= \int_{\mathbb{R}} \overline{g(x)} f(x) dx \\ &= \frac{1}{2\pi} \int_{\mathbb{R}} \overline{g(x)} \left( \int_{\mathbb{R}} e^{\mu tx} \hat{f}(t) dt \right) dx \\ &= \frac{1}{2\pi} \int_{\mathbb{R}} \left( \int_{\mathbb{R}} \overline{g(x)} e^{\mu tx} dx \right) \hat{f}(t) dt \\ &= \frac{1}{2\pi} \int_{\mathbb{R}} \left( \int_{\mathbb{R}} \overline{e^{-\mu tx} g(x)} dx \right) \hat{f}(t) dt \\ &= \frac{1}{2\pi} \int_{\mathbb{R}} \left( \int_{\mathbb{R}} \overline{e^{-\mu tx} g(x)} dx \right) \hat{f}(t) dt \\ &= \frac{1}{2\pi} \int_{\mathbb{R}} \overline{\hat{g}(t)} \hat{f}(t) dt. \end{aligned} $$
<figure><img src="image_11.png" /></figure>

As a corollary, we have the

**Corollary 4. (Parseval's Theorem)**

In the previous formula, if $f = g$, then

$$ \|f\|_{L^2(\mathbb{R};\mathbb{H})} = \frac{1}{\sqrt{2\pi}} \|\hat{f}\|_{L^2(\mathbb{R};\mathbb{H})}. $$

**4 Application to Partial Differential Equations**

In view of the definition and Remark 1 it is clear that, for functions $f: \mathbb{R} \to \mathbb{H}$ of the form $f = f_0 + \mu f_1$, where $f_0, f_1: \mathbb{R} \to \mathbb{R}$, the 1D-QFT defined in this paper works exactly in the same way as the complex Fourier transform. This illustrates the fact that our 1D-QFT is a generalization of the complex one. In order to see this, let's consider the initial value problem

$$ u_t = ku_{xx} \quad (-\infty < x < \infty), \quad u(x, 0) = f(x), $$

where $u(x, t)$ and $f(x)$ are supposed to vanish sufficiently rapidly as $x \to \pm\infty$, so that they are integrable over the whole line. Applying the 1D-QFT with respect to $x$ in the previous equation, we have that

$$ \frac{\partial \hat{u}}{\partial t}(\xi, t) = -k\xi^2\hat{u}(\xi, t), \quad \hat{u}(\xi, 0) = \hat{f}(\xi). $$

By fixing $\xi$, the last equation is an ordinary differential equation in the variable $t$ with initial condition $\hat{u}(\xi, 0) = \hat{f}(\xi)$ and its solution is given by

$$ \hat{u}(\xi, t) = \hat{f}(\xi)e^{-k\xi^2t}. $$

Since $e^{-k\xi^2t}$ is the 1D-QFT of

$$ K_t(x) = \frac{1}{\sqrt{4\pi k t}} e^{-x^2/4kt}, $$

(see Example 1 with $a = 1/2kt$) then

$$ \hat{f}(\xi)e^{-k\xi^2t} = \mathcal{F}\{(f * K_t)(x)\}(\xi). $$

Thus, by Theorem 2

$$ u(x, t) = (f * K_t)(x) = \frac{1}{\sqrt{4\pi k t}} \int_{\mathbb{R}} f(y)e^{-(x-y)^2/4kt} dy. $$
## 5 Application to the 2-D Quaternion Fourier Transform

Although we can apply the 1D-QFT to other topics, the original motivation to define it was the use of this transform to obtain an inversion formula for the 2-dimensional quaternion Fourier transform defined below.

**Definition 7.** Let $f \in L^1(\mathbb{R}^2; \mathbb{H})$. The 2-dimensional Quaternion Fourier Transform of $f$ (or 2D-QFT for short) is the function $\mathcal{F}_q\{f\} : \mathbb{R}^2 \to \mathbb{H}$ defined by

$$ \mathcal{F}_q\{f\}(\omega) = \hat{f}(\omega) = \int_{\mathbb{R}^2} e^{-\mu\omega \cdot x} f(x) d^2x, \quad (27) $$

where $x = x_1e_1 + x_2e_2$, $\omega = \omega_1e_1 + \omega_2e_2$, $d^2x = dx_1dx_2$, $\mu$ is as before and $e^{-\mu\omega \cdot x}$ is called the quaternion Fourier kernel. See [9].

**Theorem 6.** (Inversion formula for the 2D-QFT)

Suppose that $f \in L^2(\mathbb{R}^2; \mathbb{H})$ and $\mathcal{F}_q\{f\} \in L^1(\mathbb{R}^2; \mathbb{H})$. Then the 2D-QFT is invertible with inverse

$$ \mathcal{F}_q^{-1}[\mathcal{F}_q\{f\}](x) = f(x) = \frac{1}{(2\pi)^2} \int_{\mathbb{R}^2} e^{\mu\omega \cdot x} \mathcal{F}_q\{f(x)\}(\omega) d^2\omega, \quad (28) $$

where $d^2\omega = d\omega_1 d\omega_2$.

*Proof.* Denote the 1D-QFT with respect to $x_k$, for $k = 1, 2$, by $\mathcal{F}_{x_k}\{f(x)\}$, that is,

$$ \mathcal{F}_{x_1}\{f(x)\}(\omega_1, x_2) = \int_{\mathbb{R}} e^{-\mu\omega_1 x_1} f(x) dx_1 $$

and

$$ \mathcal{F}_{x_2}\{f(x)\}(x_1, \omega_2) = \int_{\mathbb{R}} e^{-\mu\omega_2 x_2} f(x) dx_2. $$

Denote also

$$ \mathcal{F}_{x_1,x_2}\{f(x)\}(\omega) = \mathcal{F}_{x_1}\{\mathcal{F}_{x_2}\{f(x)\}(x_1, \omega_2)\}(\omega_1) $$

and

$$ \mathcal{F}_{x_2,x_1}\{f(x)\}(\omega) = \mathcal{F}_{x_2}\{\mathcal{F}_{x_1}\{f(x)\}(\omega_1, x_2)\}(\omega_2). $$

Evidently

$$ \mathcal{F}_{x_1,x_2}\{f(x)\}(\omega) = \mathcal{F}_{x_2,x_1}\{f(x)\}(\omega). $$
Moreover, we have that

$$
\begin{align*}
\mathcal{F}_q\{f(\boldsymbol{x})\}(\boldsymbol{\omega}) &= \int_{\mathbb{R}^2} e^{-\mu\omega \cdot \boldsymbol{x}} f(\boldsymbol{x}) d^2\boldsymbol{x} \\
&= \int_{\mathbb{R}^2} e^{-\mu\omega_1 x_1} e^{-\mu\omega_2 x_2} f(\boldsymbol{x}) d^2\boldsymbol{x} \\
&= \int_{\mathbb{R}} e^{-\mu\omega_1 x_1} \left( \int_{\mathbb{R}} e^{-\mu\omega_2 x_2} f(\boldsymbol{x}) dx_2 \right) dx_1 \\
&= \int_{\mathbb{R}^2} e^{-\mu\omega_1 x_1} \mathcal{F}_{x_2}\{f(\boldsymbol{x})\}(x_1, \omega_2) dx_1 \\
&= \mathcal{F}_{x_1, x_2}\{f(\boldsymbol{x})\}(\boldsymbol{\omega}).
\end{align*}
$$

On the other hand, in view of the inversion formula for the 1D–QFT, we
can write

$$
\begin{align*}
f(\boldsymbol{x}) &= \frac{1}{2\pi} \int_{\mathbb{R}} e^{\mu\omega_1 x_1} \mathcal{F}_{x_1}\{f(\boldsymbol{x})\}(\omega_1, x_2) d\omega_1 \\
&= \frac{1}{(2\pi)^2} \int_{\mathbb{R}} e^{\mu\omega_1 x_1} \left( \int_{\mathbb{R}} e^{\mu\omega_2 x_2} \mathcal{F}_{x_2,x_1}\{f(\boldsymbol{x})\}(\boldsymbol{\omega}) d\omega_2 \right) d\omega_1 \\
&= \frac{1}{(2\pi)^2} \int_{\mathbb{R}^2} e^{\mu\omega_1 x_1} e^{\mu\omega_2 x_2} \mathcal{F}_{x_1,x_2}\{f(\boldsymbol{x})\}(\boldsymbol{\omega}) d^2\boldsymbol{\omega} \\
&= \frac{1}{(2\pi)^2} \int_{\mathbb{R}^2} e^{\mu\omega \cdot x} \mathcal{F}_q\{f(\boldsymbol{x})\}(\boldsymbol{\omega}) d^2\boldsymbol{\omega},
\end{align*}
$$

as stated.

<figure><img src="image_12.png" /></figure>

**6 Concluding Remarks**

We have introduced in this work a one dimensional Quaternion Fourier Trans-
form, given some of its more important properties. We applied this transform
to solve a partial differential equation and to obtain an inversion formula for
a two-dimensional Quaternion Fourier Transform. As a future work one
can develop this theory in some directions, in particular, one can work in to
prove an uncertainty principle for this transform. This 1D–QFT can be use to define a 1D–wavelet transform and to develop the corresponding theory of this new transform. It is also possible to consider multidimensional (3D or 4D) QFT and use the 1D–QFT in order to obtain an inversion formula and other properties for this multidimensional QFT. Finally, it must be noted that the theory introduce in this work can be easily developed for the right 1D–QFT defined in subsection 3.2.

References

[1] F. Brackx, E. Hitzer, and S. Sangwine. History of quaternion and Clifford-Fourier transforms and wavelets. In * Quaternion and Clifford Fourier transforms and wavelets, Trends in Mathematics (TIM) *, volume 27, pages XI-XXVII, Picassoplatz 4 4052 Basel Switzerland, 2013. Springer Basel AG. https://biblio.ugent.be/publication/4089431.

[2] L. Debnath and D. Bhatta. *Integral transforms and their applications*. International Series of Monographs on Chemistry. Chapman and Hall/CRC, 2014. ISBN 978-0429162633, doi: 10.1201/b17670.

[3] M.A. Delsuc. Spectral representation of 2D NMR spectra by hypercomplex numbers. * Journal of Magnetic Resonance (1969)*, 77(1):119-124, March 1988. doi: 10.1016/0022-2364(88)90036-4.

[4] R.R. Ernst, G. Bodenhausen, and A. Wokaun. *Principles of nuclear magnetic resonance in one and two dimensions*, volume 14 of *International Series of Monographs on Chemistry*. Clarendon press Oxford, 1987. ISBN 978-0198556473.

[5] S. Sangwine. The problem of defining the Fourier transform of a colour image. In *Proceedings 1998 International Conference on Image Processing. ICIP98 (Cat. No.98CB36269)*, volume 1, pages 171-175. Springer Basel AG, 1998. doi: 10.1109/ICIP.1998.723451.

[6] M. Bahri, S. Toaha, A. Rahim, and M.I. Azis. On one-dimensional quaternion Fourier transform. In *Journal of Physics: Conference Series*, volume 1341. IOP Publishing, 2019. ID 062004, https://iopscience.iop.org/article/10.1088/1742-6596/1341/6/062004/pdf.
[7] G.B. Folland. *Fourier analysis and its applications*, volume 4 of *Pure and Applied Undergraduate Texts*. American Mathematical Soc., 2009. ISBN 978-0821847909.

[8] K. Gürlebeck and W. Sprössig. * Quaternionic and Clifford calculus for physicists and engineers*. Chemistry of Organometallic Compounds. Whiley, 1997. ISBN 978-0-471-96200-7.

[9] M. Bahri, R. Ashino, and R. Vaillancourt. Continuous quaternion fourier and wavelet transforms. *International Journal of Wavelets, Multiresolution and Information Processing, Special issue: New theories and technological advances*, 12(4), 2014. ID 1460003, doi: 10.1142/S0219691314600030.