1 Introduction

The integral Fourier transform, as well as Fourier series, are widely used in various fields of calculus, computational mathematics, mathematical physics, etc. Certain applications of this transform are described in a number of fundamental monographs (for example, see [25, 26]). Numerical estimates of the Fourier transform are presented in [30].

The classical linear canonical transform (LCT) is considered as a generalization of the Fourier transform (FT), and was first proposed in the 1970 s by Collins [9] and Moshinsky and Quesne [21]. It is an effective processing tool for chirp signal analysis, such as the parameter estimation, sampling progress for non bandlimited signals with nonlinear Fourier atoms [20], and the LCT filtering [27].

In this paper we will give our results in a more general context, that of quaternion linear canonical transform (QLCT). There are many studies in the literature that are concerned with the QLCT (see, for example, [5, 7, 16,17,18, 29]). They established some important properties of the QLCT, such as the uncertainty principle, the inversion formula and the study of generalized swept-frequency filters.

Recently several results of estimation have been proved in several different versions and for several different types of transforms (for example for the Fourier transform [3, 13], for the Bessel transform, [10], for the Dunkl transform [11], for the Laguerre Hypergroup transform [22]). In [1] the authors estimated the integral \(\displaystyle \int _{|x|\ge R}|{\mathcal {F}}\{f\}(x)|^2dx \) in certain classes of functions in \(L^2({\mathbb {R}}^n)\) where \({\mathcal {F}}\{f\}\) is the Fourier transform (FT) of f. Since the QLCT is a generalization of the FT, so for this reason we want in this paper, to estimate the integral \(\displaystyle \int _{|\omega |\ge N^2}|{\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega )|^2d^2\omega \), where \({\mathcal {L}}_{A_{1},A_{2}}\{f\}\) stands for the QLCT transform of f and \(N \ge 1\).

In order to describe our results, we first need to introduce some facts about harmonic analysis related to the QLCT. We cite here, as briefly as possible, some properties. For more details we refer to [2, 6, 8, 12, 14, 16,17,18, 28, 29].

The quaternion algebra \({\mathcal {H}}\) was first invented by W. R. Hamilton in 1843 for extending complex numbers to a 4D algebra [24]. A quaternion \(q \in {\mathcal {H}}\) can be written in this form

$$\begin{aligned} q=q_{0}+{\underline{q}}=q_{0}+iq_{1}+jq_{2}+kq_{3} \end{aligned}$$

where i, j, k satisfy Hamilton’s multiplication rules

$$\begin{aligned} i^2= & {} j^2=k^2=ijk=-1,~~ij=-ji=k.\\ jk= & {} -kj=i,~~ki=-ik=j. \end{aligned}$$

Using Hamilton’s multiplication rules, the multiplication of two quaternions \(p=p_{0}+{\underline{p}}\) and \(q=q_{0}+{\underline{q}}\) can be expressed as

$$\begin{aligned} pq=p_{0}q_{0}+p_{0}{\underline{q}}+q_{0}{\underline{p}} +{\underline{p}}{\underline{q}}. \end{aligned}$$

We define the conjugation of \(q\in {\mathcal {H}}\) by \({\overline{q}}=q_{0}-iq_{1}-jq_{2}-kq_{3}.\) Clearly, \(q{\overline{q}}=q_{0}^{2}+q_{1}^{2}+q_{2}^{2}+q_{3}^{2}.\) So the modulus of a quaternion q is defined by

$$\begin{aligned} |q|=\sqrt{q{\overline{q}}}=\sqrt{q_{0}^{2}+q_{1}^{2}+q_{2}^{2}+q_{3}^{2}}. \end{aligned}$$

In this paper, we study the quaternion-valued signal \(f: {\mathbb {R}}^{2} \rightarrow {\mathcal {H}}\) that can be expressed as

$$\begin{aligned} f(x)=f_{0}(x)+if_{1}(x)+jf_{2}(x)+kf_{3}(x) \end{aligned}$$

where \(x=x_{1}e_{1}+x_{2}e_{2}\in {\mathbb {R}}^{2}\) and \(f_{0},~f_{1},~f_{2}~\text {and}~f_{3}\) are real-valued functions. For \(1\le r<\infty ,\) the quaternion modulus \(L^{r}({\mathbb {R}}^{2},{\mathcal {H}})\) is defined as

$$\begin{aligned} L^{r}=L^{r}({\mathbb {R}}^{2},{\mathcal {H}})=\{f/f:{\mathbb {R}}^{2}\rightarrow {\mathcal {H}},~\Vert f\Vert _{L^{r}({\mathbb {R}}^{2},{\mathcal {H}})}^r =\int _{{\mathbb {R}}^{2}}|f(x)|^{r}dx<\infty ~ \}. \end{aligned}$$

Let \(f\in L^{r}({\mathbb {R}}^{2},{\mathcal {H}}).\) The quaternion Fourier transform (QFT) of f is defined by

$$\begin{aligned} {\mathcal {F}}(f)(\omega )=\frac{1}{2\pi }\int _{{\mathbb {R}}^{2}} e^{-ix_{1}\omega _{1}}f(x)e^{-jx_{2}\omega _{2}}dx. \end{aligned}$$

The inner product of f, \(g\in L^{2}({\mathbb {R}}^{2},{\mathcal {H}})\) is defined by

$$\begin{aligned} \langle f,g\rangle =\int _{{\mathbb {R}}^{2}}f(x)\overline{g(x)}dx. \end{aligned}$$

Clearly, \(\Vert f\Vert _{2}^2=\langle f,f\rangle \). For all \(\theta \in {\mathbb {R}}\) we have

$$\begin{aligned} |e^{i\theta }|=|e^{j\theta }|=1. \end{aligned}$$
(1)

Now, we define a norm of \({\mathcal {F}}(f)\) as

$$\begin{aligned} |{\mathcal {F}}(f)(\omega )|_{Q}=\left( |{\mathcal {F}}(f_0)(\omega )|^{2} +|{\mathcal {F}}(f_{1})(\omega )|^{2}+|{\mathcal {F}}(f_{2})(\omega )|^{2} +|{\mathcal {F}}(f_3)(\omega )|^{2}\right) ^{1/2}. \end{aligned}$$

Furthermore, we obtain the \(L^{r}({\mathbb {R}}^{2},{\mathcal {H}})\)-norm

$$\begin{aligned} \Vert {\mathcal {F}}(f)\Vert _{Q,r}=\left( \int _{{\mathbb {R}}^{2}}|{\mathcal {F}} (f)(\omega )|_{Q}^{r}d^2\omega \right) ^{1/r}. \end{aligned}$$

For \(f\in L_{}^{1}({\mathbb {R}}^{2},{\mathcal {H}})\), we have

$$\begin{aligned} \Vert {\mathcal {F}}(f)\Vert _{Q,\infty }\le \Vert f\Vert _{1}. \end{aligned}$$
(2)

(QFT Plancherel) If \(f\in L^{1}({\mathbb {R}}^{2},{\mathcal {H}})\cap L^{2}({\mathbb {R}}^{2},{\mathcal {H}})\), then

$$\begin{aligned} \int _{{\mathbb {R}}^{2}}|f(x)|^{2}dx=\int _{{\mathbb {R}}^{2}}|{\mathcal {F}} (f)(\omega )|^{2}d\omega . \end{aligned}$$
(3)

Moreover,

$$\begin{aligned} \int _{{\mathbb {R}}^{2}}|f(x)|^{2}dx=\int _{{\mathbb {R}}^{2}}|{\mathcal {F}} (f)(\omega )|_{Q}^{2}d\omega , \end{aligned}$$

then we can rewrite the QFT Plancherel as follows

$$\begin{aligned} \Vert {\mathcal {F}}(f)\Vert _{Q,2}=\Vert f\Vert _{2}. \end{aligned}$$
(4)

Indeed, we have

$$\begin{aligned} \begin{aligned} \int _{{\mathbb {R}}^2}\left| {\mathcal {F}}\{f\}(\omega )\right| _Q^2 d \omega&= \int _{{\mathbb {R}}^2}\left( \left| {\mathcal {F}}\left\{ f_0\right\} (\omega ) \right| ^2+\left| {\mathcal {F}}\left\{ f_1\right\} (\omega )\right| ^2\right. \\&\quad \left. +\left| {\mathcal {F}}\left\{ f_2\right\} (\omega )\right| ^2 +\left| {\mathcal {F}}\left\{ f_3\right\} (\omega )\right| ^2\right) d \omega \\&=\left( \int _{{\mathbb {R}}^2}\left| {\mathcal {F}}_{\textrm{Q}} \left\{ f_0\right\} (\omega )\right| ^2 d \omega \right. +\int _{{\mathbb {R}}^2}\left| {\mathcal {F}}_{\textrm{Q}} \left\{ f_1\right\} (\omega )\right| ^2 d \omega \\&\quad \left. +\int _{{\mathbb {R}}^2} \left| {\mathcal {F}}_{\textrm{Q}}\left\{ f_2\right\} (\omega )\right| ^2 d \omega +\int _{{\mathbb {R}}^2}\left| {\mathcal {F}} \left\{ f_3\right\} (\omega )\right| ^2 d \omega \right) \\ \end{aligned} \end{aligned}$$

Applying 3 into the right-hand side of the above identity gives

$$\begin{aligned} \begin{aligned} \int _{{\mathbb {R}}^2}\left| {\mathcal {F}}\{f\}(\omega )\right| _Q^2 d \omega&= \int _{{\mathbb {R}}^2}\left| f_0(x)\right| ^2 d x+\int _{{\mathbb {R}}^2} \left| f_1(x)\right| ^2 d x \\&\quad +\int _{{\mathbb {R}}^2}\left| f_2(x)\right| ^2 d x +\int _{{\mathbb {R}}^2}\left| f_3(x)\right| ^2 d x \end{aligned} \end{aligned}$$

Since \(f_i(x), i=0,1,2,3\), is real-valued, the above equation can be written in the form

$$\begin{aligned} \begin{aligned} \int _{{\mathbb {R}}^2}\left| {\mathcal {F}}\{f\}(\omega )\right| _Q^2 d \omega&=\int _{{\mathbb {R}}^2}\left( f_0^2(x)+f_1^2(x)+f_2^2(x)+f_3^2(x)\right) d x. \end{aligned} \end{aligned}$$

Suppose that \({\mathcal {F}}(f)\in L^{1}({\mathbb {R}}^{2},{\mathcal {H}})\) and \({\mathcal {F}}(\frac{\partial ^n f}{\partial x_{1}^{n}})\in L^{1}({\mathbb {R}}^{2},{\mathcal {H}})\). Then

$$\begin{aligned} {\mathcal {F}}( \frac{\partial ^n f}{\partial x_{1}^{n}}i^{-n})(w) =w_{1}^{n}{\mathcal {F}}(f)(w), \quad \forall n\in {\mathbb {N}}. \end{aligned}$$
(5)

Moreover, if \({\mathcal {F}}(f)\in L^{1}({\mathbb {R}}^{2},{\mathcal {H}})\) and \({\mathcal {F}}(\frac{\partial ^m f}{\partial x_{2}^{m}})\in L^{1}({\mathbb {R}}^{2},{\mathcal {H}})\), then

$$\begin{aligned} {\mathcal {F}}( \frac{\partial ^m f}{\partial x_{2}^{m}})(w) ={\mathcal {F}}(f)(w)(jw_{2})^{m}, \ \ \forall m\in {\mathbb {N}}. \end{aligned}$$
(6)

Let \(A_{s}=\begin{pmatrix} a_s &{} b_s \\ c_s &{} d_s \end{pmatrix}\in {\mathbb {R}}^{2\times 2}\) be a real matrix parameter such that \(\det (A_s)=1\), for \(s=1,2.\) The two-sided (sandwich) QLCT of \(f\in L^{1}({\mathbb {R}}^{2},{\mathcal {H}})\) is defined by

$$\begin{aligned} {\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega )=\int _{{\mathbb {R}}^2} K_{A_{1}}^{i}(x_{1},\omega _{1})f(x)K_{A_{2}}^{j}(x_{2},\omega _{2})dx, \end{aligned}$$

where the kernel functions of the QLCT above are given by

$$\begin{aligned} K_{A_{1}}^{i}(x_{1},\omega _{1})= & {} {\left\{ \begin{array}{ll} \frac{1}{\sqrt{2\pi b_{1}}}e^{(i/2)((a_{1}/b_{1})x^{2}_{1} -(2/b_{1})x_{1}\omega _{1}+(d_{1}/b_{1})\omega ^{2}_{1}-(\pi /2))}&{}\quad \text {for}~~ b_{1}\ne 0,\\ \sqrt{d_{1}}e^{i(c_{1}d_{1}/2)\omega _{1}^{2}} &{} \quad \text {for}~~ b_{1}=0. \end{array}\right. }\end{aligned}$$
(7)
$$\begin{aligned} K_{A_{2}}^{j}(x_{2},\omega _{2})= & {} {\left\{ \begin{array}{ll} \frac{1}{\sqrt{2\pi b_{1}}}e^{(j/2)((a_{2}/b_{2})x^{2}_{2} -(2/b_{2})x_{2}\omega _{2}+(d_{2}/b_{2})\omega ^{2}_{2}-(\pi /2))}&{}\quad \text {for}~~ b_{2}\ne 0,\\ \sqrt{d_{2}}e^{j(c_{2}d_{2}/2)\omega _{2}^{2}} &{}\quad \text {for}~~ b_{2}=0. \end{array}\right. } \end{aligned}$$
(8)

Let the kernel function \(K_{A}\) be defined by (7) or (8). Then

  • \(K_{A}(-x,\omega )=K_{A}(x,-\omega ).\)

  • \(K_{A}(-x,-\omega )=K_{A}(x,\omega ).\)

  • \(\overline{K_{A}(x,\omega )}=K_{A}^{-1}(\omega ,x).\)

From the definition of the QLCT, we can easily see that when \(b_{1}b_{2} = 0\) and \(b_{1}=b_{2}=0\), the QLCT of a signal is essentially a quaternion chirp multiplication. Therefore, in this work, we always assume \(b_{1}b_{2} \ne 0\).

(Inversion formula) The (Two-sided) inverse quaternion linear canonical transform of \(g \in L^{1}({\mathbb {R}}^{2},{\mathcal {H}})\)

$$\begin{aligned} {\mathcal {L}}_{A_{1},A_{2}}^{-1}\{g\}(\omega )=\int _{{\mathbb {R}}^2} K_{A_{1}^{-1}}^{i}(x_{1},\omega _{1})g(x)K_{A_{2}^{-1}}^{j}(x_{2},\omega _{2})dx. \end{aligned}$$
(9)

(Hausdorff–Young inequality) If \(1\le r< 2\) and letting \(r'\) be such that \(1/r+1/r'=1\) then for all \(f\in L^r({\mathbb {R}}^{2},{\mathcal {H}})\) it holds that

$$\begin{aligned} \Vert {\mathcal {L}}_{A_{1},A_{2}}\{f\}\Vert _{Q,r'}\le \frac{|b_{1}b_{2}|^{-1/2+1/r'}}{2\pi }\Vert f\Vert _{r}. \end{aligned}$$
(10)

(Plancherel theorem of QLCTs) Let \(f\in L^{2}({\mathbb {R}}^{2},{\mathcal {H}})\). Then

$$\begin{aligned} \Vert {\mathcal {L}}_{A_{1},A_{2}}\{f\}\Vert _{Q,2}=\Vert f\Vert _{2}. \end{aligned}$$

(Shift property) For a quaternion function \(f\in L^{1}({\mathbb {R}}^{2},{\mathcal {H}})\), we denote by \(\tau _kf(x)\) the shifted (translated) function defined by \(\tau _kf(x) = f(x-k)\), where \(k=k_{1}e_{1}+k_{1}e_{1}\in {\mathbb {R}}^2\). Then we obtain

$$\begin{aligned} {\mathcal {L}}_{A_{1},A_{2}}\{\tau _kf\}(\omega )= & {} e^{-ia_{1}c_{1}k_{1}^{2}/2 +ic_{1}k_{1}\omega _{1}}\nonumber \\{} & {} \times {\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega _{1}-a_{1}k_{1},\omega _{2} -a_{2}k_{2})e^{-ja_{2}c_{2}k_{2}^{2}/2+jc_{2}k_{2}\omega _{2}}. \end{aligned}$$
(11)

(Modulation property). We define a modulation operator \({\mathbb {M}}_{\omega _0}f\) by

$$\begin{aligned} {\mathbb {M}}_{\omega _0}f(x)=e^{ix_{1}u_0}f(x)e^{jx_{2}v_0} \end{aligned}$$

with \(\omega _0=u_0e_{1}+v_0e_{1}\). So

$$\begin{aligned} {\mathcal {L}}_{A_{1},A_{2}}\{{\mathbb {M}}_{\omega _0}f\}(\omega )= & {} {\mathcal {L}}_{A_{1},A_{2}}\{e^{ix_{1}u_0}f(x)e^{jx_{2}v_0}\}(\omega )\nonumber \\= & {} e^{-ib_{1}d_{1}u_{0}^{2}/2+id_{1}u_{0}\omega _{1}} {\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega _{1}-u_{0}b_{1}, \omega _{2}-v_{0}b_{2})e^{-jb_{2}d_{2}v_{0}^{2}/2+jd_{2}v_{0}\omega _{2}}.\nonumber \\ \end{aligned}$$
(12)

(Time-frequency shift). Let a quaternion function \(f\in L^{1}({\mathbb {R}}^{2},{\mathcal {H}})\). Then we obtain that

$$\begin{aligned}{} & {} {\mathcal {L}}_{A_{1},A_{2}}\{{\mathbb {M}}_{\omega _0}\tau _kf\}(\omega )\nonumber \\{} & {} \quad ={\mathcal {L}}_{A_{1},A_{2}}\{e^{ix_{1}u_0}f(x-k)e^{jx_{2}v_0}\}(\omega ) \nonumber \\{} & {} \quad =e^{-i(a_{1}c_{1}k_{1}^{2}+b_{1}d_{1}u_{0}^{2})/2+i(c_{1}k_{1} +d_{1}u_{0})\omega _{1}-ib_{1}c_{1}k_{1}u_0}\nonumber \\{} & {} \qquad {\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega _{1}-a_{1}k_{1} -u_{0}b_{1},\omega _{2}-a_{2}k_{2}-v_{0}b_{2})\nonumber \\{} & {} \qquad e^{-j(a_{2}c_{2}k_{2}^{2}+b_{2}d_{2}v_{0}^{2})/2+j(c_{2}k_{2} +d_{2}v_{0})\omega _{2}-jb_{2}c_{2}k_{2}v_0}. \end{aligned}$$
(13)

For a function f on \(L^{1}({\mathbb {R}}^{2},{\mathcal {H}})\) and for any \(h_{1}, h_{2}\in {\mathbb {R}}\), we define the operator \(\Delta _{h_{1},h_{2}}\) by

$$\begin{aligned} \Delta _{h_{1},h_{2}}f(x)= & {} e^{i\frac{a_{1}h_{1}}{b_{1}}x_{1}} f(x_{1}+h_{1}, x_{2}+h_{2})e^{j\frac{a_{2}h_{2}}{b_{2}}x_{2}} -e^{i\frac{a_{1}h_{1}}{b_{1}}x_{1}}f(x_{1}+h_{1}, x_{2})\\{} & {} -f(x_{1}, x_{2}+h_{2})e^{j\frac{a_{2}h_{2}}{b_{2}}x_{2}}+f(x_{1},x_{2}).\nonumber \end{aligned}$$
(14)

Definition 1.1

Let \(f(x)=f(x_{1},x_{2})\) belongs to \(L^{2}({\mathbb {R}}^{2},{\mathcal {H}})\). We say that f is in the Lipschitz space \(\text {Lip}_{A_{1},A_{2}}(\alpha _{1}, \alpha _{2} )\) if

$$\begin{aligned} \Vert \Delta _{h_{1},h_{2}}f(x)\Vert _{2}=O(h_{1}^{\alpha _{1}}h_{2}^{\alpha _{2}}), \end{aligned}$$
(15)

as \(h_{1},h_{2}\) tend to zero, \(0 <\alpha _{1}, \alpha _{2}\le 1\).

In \(L^{2}({\mathbb {R}}^{2},{\mathcal {H}})\), consider the operator

$$\begin{aligned} F_{h}^{A_{1}, A_{2}}f(x_{1}, x_{2})= & {} \dfrac{1}{4h^2}\int _{-h}^{h}\int _{-h}^{h} e^{i\frac{a_{1}\xi }{b_{1}}x_{1}+i\frac{a_{1}\xi ^{2}}{2b_{1}}} f(x_{1}+\xi ,x_{2}+\eta )\nonumber \\{} & {} \times e^{j\frac{a_{2}\eta }{b_{2}}x_{2} +j\frac{a_{2}\eta ^{2}}{2b_{2}}}d\xi d\eta , \ \ h>0. \end{aligned}$$
(16)

Observe that if \(a_{1}=a_{2}=0\), then

$$\begin{aligned} F_{h}^{A_{1}, A_{2}}f(x_{1}, x_{2})=\dfrac{1}{4h^2} \int _{-h}^{h}\int _{-h}^{h} f(x_{1}+\xi ,x_{2}+\eta )d\xi d\eta . \end{aligned}$$

This is analogous to the Steklov operator.

Let the function \(f\in L^{2}({\mathbb {R}}^{2},{\mathcal {H}})\). The finite differences of the order m (\(m\in {1,2,3,\ldots }\)) are defined as follows:

$$\begin{aligned} \Delta _{h}^{m}f(x_{1},x_{2})=(I-F_{h}^{A_{1}, A_{2}})^mf(x_{1},x_{2}), \end{aligned}$$

here I is the unit operator, and the mth order generalized continuity modulus of the function f is defined by the formula

$$\begin{aligned} w_{m}(f,\delta )_{2}=\sup _{0<h\le \delta }\Vert \Delta _{h}^{m}f\Vert _{2}, \end{aligned}$$

where \(\delta > 0\).

For a function f on \(L^{2}({\mathbb {R}}^{2},{\mathcal {H}})\), we define the function \(g_f\) by

$$\begin{aligned} g_{f}(x)=e^{-i(a_{1}/2b_{1})x_{1}^{2}}f(x) e^{-j(a_{2}/2b_{2})x_{2}^{2}}. \end{aligned}$$

From the definition of \(g_f\), we easily obtain

$$\begin{aligned} {\mathcal {F}}\lbrace g_f\rbrace ^2(w)= \sqrt{b_{1}b_{2}}e^{-i(d_{1}/2b_{1})w_{1}^{2}+i\pi /4} {\mathcal {L}}_{A_{1},A_{2}}\lbrace f\rbrace (b_1w_1,b_2w_2) e^{-j(d_{2}/2b_{2})w_{2}^{2}+j\pi /4}. \nonumber \\ \end{aligned}$$
(17)

Consider in \(L^{2}({\mathbb {R}}^{2},{\mathcal {H}})\) the operator

$$\begin{aligned} D_{A_{1},A_{2}}f(x)=e^{-i(a_{1}/2b_{1})x_{1}^{2}}Dg_{f}(x) e^{-j(a_{2}/2b_{2})x_{2}^{2}}, \end{aligned}$$
(18)

where \(D=\dfrac{\partial ^2}{\partial x_{1}^{2}}+\dfrac{\partial ^2}{\partial x_{2}^{2}},\) \(D_{A_{1},A_{2}}^0f=f\), \(D_{A_{1},A_{2}}^rf=D_{A_{1},A_{2}}(D_{A_{1},A_{2}}^{r-1}f)\), \(r=1,2,\ldots \).In view of formulas (5), (6) and (18), we have

$$\begin{aligned} {\mathcal {L}}_{A_{1},A_{2}}\lbrace D_{A_{1},A_{2}}f \rbrace ^2(w) =-\left( \left( \dfrac{w_{1}}{b_{1}}\right) ^2+\left( \dfrac{w_{2}}{b_{2}}\right) ^2\right) {\mathcal {L}}_{A_{1},A_{2}}\lbrace f \rbrace ^2(w), \end{aligned}$$

and hence

$$\begin{aligned} {\mathcal {L}}_{A_{1},A_{2}}\lbrace D_{A_{1},A_{2}}^{r}f \rbrace ^2(w) =(-1)^r\left( \left( \dfrac{w_{1}}{b_{1}}\right) ^2+\left( \dfrac{w_{2}}{b_{2}}\right) ^2 \right) ^r{\mathcal {L}}_{A_{1},A_{2}}\lbrace f \rbrace ^2(w). \end{aligned}$$
(19)

Denote by \(W_{2,\phi }^{2,k}({\mathbb {R}})\) the class of functions \(f\in L^{2}({\mathbb {R}}^{2},{\mathcal {H}})\) having the generalized derivatives \(\dfrac{\partial f}{\partial x_1}\), \(\dfrac{\partial ^2 f}{\partial x_1\partial x_2}\),...in the sense of Levi (see [19, 23]) in \(L^{2}({\mathbb {R}}^{2},{\mathcal {H}})\) are estimated by

$$\begin{aligned} \omega _{k}(D^{r}f;\delta )=O(\phi (\delta ^k)). \end{aligned}$$

where \(\phi (t)\) is a continuous steadily increasing function on \([0,+\infty )\) and \(\phi (0)=0\).

2 Some new estimates for quaternion linear canonical transform

In order to prove the main result, we shall need some preliminary results.

Lemma 2.1

If f belongs to \(L^{2}({\mathbb {R}}^{2},{\mathcal {H}})\), then

$$\begin{aligned} \Vert \Delta _{h}^{m}f\Vert _{2}\le 2^m\Vert f\Vert _{2}. \end{aligned}$$
(20)

Proof

Using the inequality (16), we have \(\Vert F_{h}^{A_{1}, A_{2}}(f)\Vert _{2}\le \Vert f\Vert _{2}\). Then \(\Vert \Delta _{h}^{1}f\Vert _{2}\le 2\Vert f\Vert _{2}\). Thus the result follows easily by using the recurrence for m. \(\square \)

Lemma 2.2

If quaternion function \(f\in L^{2}({\mathbb {R}}^{2},{\mathcal {H}})\), then

$$\begin{aligned} {\mathcal {L}}_{A_{1},A_{2}}\lbrace F_{h}^{A_{1}, A_{2}}f\rbrace ^2(w) =\dfrac{\sin (w_{1}h/b_{1})}{w_{1}h/b_{1}}\dfrac{\sin (w_{2}h/b_{2})}{w_{2}h/b_{2}}{\mathcal {L}}_{A_{1},A_{2}}\lbrace f\rbrace ^2(w). \end{aligned}$$
(21)

Proof

Let \(f\in L^{2}({\mathbb {R}}^{2},{\mathcal {H}})\). Taking into account the formula (1), we have

$$\begin{aligned} J= & {} {\mathcal {L}}_{A_{1},A_{2}}\lbrace F_{h}^{A_{1}, A_{2}}f\rbrace ^2(w)\\= & {} \frac{1}{4h^2}\int _{-h}^{h}\int _{-h}^{h} e^{i\frac{a_{1}\xi ^{2}}{2b_{1}}} {\mathcal {L}}_{A_{1}, A_{2}}\lbrace e^{i\frac{a_{1}\xi }{b_{1}}x_{1}} f(x_{1} +\xi ,x_{2}+\eta )e^{j\frac{a_{2}\eta }{b_{2}}x_{2}} \rbrace ^2(w)e^{j\frac{a_{2}\eta ^{2}}{2b_{2}}}d\xi d\eta \\= & {} \dfrac{1}{4h^2}\int _{-h}^{h}\int _{-h}^{h} e^{i\frac{a_{1}\xi ^{2}}{2b_{1}}} e^{i\frac{\omega _{1}h_{1}}{b_{1}}-i\frac{a_{1}\xi ^{2}}{2b_{1}}} {\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega ) e^{j\frac{\omega _{2}h_{2}}{b_{2}}-j\frac{a_{2}\eta ^{2}}{2b_{2}}} e^{ j\frac{a_{2}\eta ^{2}}{2b_{2}}}d\xi d\eta \\= & {} \dfrac{1}{4h^2}\int _{-h}^{h}\int _{-h}^{h} e^{i\frac{\omega _{1}\xi }{b_{1}}}{\mathcal {L}}_{A_{1},A_{2}} \{f\}(\omega ) e^{j\frac{\omega _{2}\eta }{b_{2}}}d\xi d\eta \\= & {} \left( \dfrac{1}{2h}\int _{-h}^{h} e^{i\frac{\omega _{1}\xi }{b_{1}}}d\xi \right) {\mathcal {L}}_{A_{1}, A_{2}}\{f\}(\omega ) \left( \dfrac{1}{2h}\int _{-h}^{h} e^{j\frac{\omega _{2}\eta }{b_{2}}} d\eta \right) . \end{aligned}$$

It is easily seen that

$$\begin{aligned} \dfrac{1}{2h}\int _{-h}^{h} e^{i\frac{\omega _{1}\xi }{b_{1}}}d\xi =\dfrac{\sin (w_{1}h/b_{1})}{w_{1}h/b_{1}} \end{aligned}$$

and

$$\begin{aligned} \dfrac{1}{2h}\int _{-h}^{h} e^{j\frac{\omega _{2}\eta }{b_{2}}} d\eta =\dfrac{\sin (w_{2}h/b_{2})}{w_{2}h/b_{2}}. \end{aligned}$$

So that the transform of \( F_{h}^{A_{1}, A_{2}}f(x)\) is given as

$$\begin{aligned} \dfrac{\sin (w_{1}h/b_{1})}{w_{1}h/b_{1}}\dfrac{\sin (w_{2}h/b_{2})}{w_{2}h/b_{2}}{\mathcal {L}}_{A_{1},A_{2}}\lbrace f\rbrace ^2(w). \end{aligned}$$

This completes the proof. \(\square \)

Corollary 2.3

For any function f in \(L^{2}({\mathbb {R}}^{2},{\mathcal {H}})\), we have

$$\begin{aligned} {\mathcal {L}}_{A_{1},A_{2}}\lbrace \Delta _{h}^{m}f\rbrace ^2(w)=\left( 1- \dfrac{\sin (w_{1}h/b_{1})}{w_{1}h/b_{1}} \dfrac{\sin (w_{2}h/b_{2})}{w_{2}h/b_{2}}\right) ^m {\mathcal {L}}_{A_{1},A_{2}}\lbrace f\rbrace ^2(w). \end{aligned}$$
(22)

In the next, in order to describe our results we will use the following notation:

  • \(\displaystyle G=\{(w_1,w_2):(\frac{w_{1}}{b_1})^2+(\frac{w_{2}}{b_2})^2\ge N^2\}\).

  • \(\displaystyle |w|=(\frac{w_{1}}{b_1})^2+(\frac{w_{2}}{b_2})^2.\)

  • \(\displaystyle \varphi _{h}(\frac{w_{1}}{b_1},\frac{w_{2}}{b_2}) =\frac{\sin (w_{1}h/b_1)}{w_{1}h/b_1}\frac{\sin (w_{2}h/b_2)}{w_{2}h/b_2}.\)

  • \(\displaystyle I_G=\left( \int _{G}|{\mathcal {L}}_{A_{1},A_{2}} \{f\}(\omega )|^2d^2\omega \right) .\)

In the following result, we estimate the integral

$$\begin{aligned} \int _{|\omega |\ge N^2}|{\mathcal {L}}_{A_{1},A_{2}} \{f\}(\omega )|^2d^2\omega \end{aligned}$$

in certain classes of functions in \(L^2({\mathbb {R}}^2,{\mathcal {H}})\).

Theorem 2.4

For functions \(f\in L^{2}({\mathbb {R}}^{2},{\mathcal {H}})\) in the class \(W_{2,\phi }^{2,k}\),

$$\begin{aligned} \sup _{f\in W_{2,\phi }^{2,k}}\left( \int _{|\omega |\ge N^2}|{\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega )|^2d^2\omega \right) ^{\frac{1}{2}}=O\left( N^{-2r}\phi ^2((\frac{\pi }{4N})^k)\right) \end{aligned}$$

where \(r=1,\ldots \); \(k=1,2,\ldots \); and \(\phi (t)\) is any nonnegative function defined on the interval \([0,\infty )\).

Proof

Let \(f\in W_{2,\phi }^{2,k}\). Thanks to Hölder’s inequality, we obtain

$$\begin{aligned} J^{'}= & {} \int _{G}\left( 1- \varphi _{h}\left( \frac{w_{1}}{b_1},\frac{w_{2}}{b_2}\right) \right) | {\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega )|^2d^2\omega \nonumber \\= & {} \int _{G} \left( 1- \varphi _{h}\left( \frac{w_{1}}{b_1},\frac{w_{2}}{b_2}\right) \right) |{\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega )|^{\frac{1}{k}} |{\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega )|^{2-\frac{1}{k}}d^2\omega \nonumber \\\le & {} \left( \int _{G}\left( \left| 1- \varphi _{h}\left( \frac{w_{1}}{b_1}, \frac{w_{2}}{b_2}\right) \right| \right) ^{2k}|{\mathcal {L}}_{A_{1},A_{2}} \{f\}(\omega )|^2d^2\omega \right) ^{\frac{1}{2k}} \left( \int _{G}|{\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega )|^2d^2\omega \right) ^{1-\frac{1}{2k}}\nonumber \\= & {} \left( \int _{G}|w|^{2r}\frac{\left( \left| 1- \varphi _{h} \left( \frac{w_{1}}{b_1},\frac{w_{2}}{b_2}\right) \right| \right) ^{2k}}{|w|^{2r}}|{\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega )|^2d^2\omega \right) ^{\frac{1}{2k}} \left( \int _{G}|{\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega )|^2d^2\omega \right) ^{1-\frac{1}{2k}}\nonumber \\\le & {} N^{-\frac{2r}{k}}\left( \int _{G}|w|^{2r}\left( \left| 1- \varphi _{h}\left( \frac{w_{1}}{b_1},\frac{w_{2}}{b_2}\right) \right| \right) ^{2k}| {\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega )|^2d^2\omega \right) ^{\frac{1}{2k}} I_{G}^{1-\frac{1}{2k}} \end{aligned}$$
(23)

where \(\displaystyle I_G=\left( \int _{G}|{\mathcal {L}}_{A_{1},A_{2}} \{f\}(\omega )|^2d^2\omega \right) \).

In view of Holder inequality, we have that

$$\begin{aligned} \int _{G}|w|^{2r}\left( \left| 1- \varphi _{h}\left( \frac{w_{1}}{b_1},\frac{w_{2}}{b_2}\right) \right| \right) ^{2k}|{\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega )|^2d^2\omega \le \Vert \Delta _{h}^kD^rf(x)\Vert _{2}^{2}. \end{aligned}$$
(24)

Using the inequalities (23) and (24),

$$\begin{aligned} \int _{G}\left( 1- \varphi _{h}\left( \frac{w_{1}}{b_1},\frac{w_{2}}{b_2}\right) \right) |{\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega )|^2d^2\omega\le & {} N^{-\frac{2r}{k}}\Vert \Delta _{h}^kD^rf(x)\Vert _{2}^{\frac{1}{k}}I_{G}^{1-\frac{1}{2k}} \end{aligned}$$

Now, let us estimate the integral

$$\begin{aligned} \int _G \varphi _{h}\left( \frac{w_{1}}{b_1},\frac{w_{2}}{b_2}\right) | {\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega ))|^2d^2\omega . \end{aligned}$$

It is easy to see that, for this purpose, it is sufficient to consider the domain of integration

$$\begin{aligned} E=\left\{ (w_1,w_2):\frac{w_{1}}{b_1}\ge 0,\frac{w_{2}}{b_2}\ge 0,|w|\ge N^2\right\} \end{aligned}$$

Divide the domain E into the two subdomains

$$\begin{aligned} E_1=\left\{ (w_1,w_2):\frac{w_{1}}{b_1}\ge \frac{w_{2}}{b_2},~|w|\ge N^2\right\} ~\text {and}~ E_2=\left\{ (w_1,w_2):\frac{w_{1}}{b_1}< \frac{w_{2}}{b_2},~|w|\ge N^2\right\} . \end{aligned}$$

Then,

$$\begin{aligned} I= & {} \int _{E}\varphi _{h}\left( \frac{w_{1}}{b_1},\frac{w_{2}}{b_2}\right) |{\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega )|^2d^2\omega \\= & {} \int _{E_1}\varphi _{h}\left( \frac{w_{1}}{b_1},\frac{w_{2}}{b_2}\right) |{\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega )|^2d^2\omega +\int _{E_2} \varphi _{h}\left( \frac{w_{1}}{b_1},\frac{w_{2}}{b_2}\right) |{\mathcal {L}}_{A_{1}, A_{2}}\{f\}(\omega )|^2d^2\omega \end{aligned}$$

Since \(\displaystyle |\sin x| \le |x|\), \(\displaystyle \frac{w_{1}}{b_1}\ge \frac{N}{\sqrt{2}}((w_1,w_2)\in E_1)\) and \(\displaystyle \frac{w_{2}}{b_2}\ge \frac{N}{\sqrt{2}}((w_1,w_2)\in E_2)\) it is clear that

$$\begin{aligned} |I|= & {} \int _{E_1}\left| \frac{\sin (w_{1}h/b_1)}{w_{1}h/b_1}\right| |{\mathcal {L}}_{A_{1}, A_{2}}\{f\}(\omega )|^2d^2\omega + \int _{E_2}\left| \frac{\sin (w_{2}h/b_2)}{w_{2}h/b_2}\right| |{\mathcal {L}}_{A_{1}, A_{2}}\{f\}(\omega )|^2d^2\omega \\\le & {} \frac{\sqrt{2}}{Nh}\left( \int _{E_1}|{\mathcal {L}}_{A_{1}, A_{2}}\{f\}(\omega )|^2d^2\omega + \int _{E_2}|{\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega )|^2d^2\omega \right) \\\le & {} \frac{\sqrt{2}}{Nh}\int _{E}|{\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega )|^2d^2\omega \\\le & {} \frac{\sqrt{2}}{Nh}\int _{G}|{\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega )|^2d^2\omega \end{aligned}$$

Consequently,

$$\begin{aligned} \int _{G}\varphi _{h}\left( \frac{w_{1}}{b_1},\frac{w_{2}}{b_2}\right) |{\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega )|^2d^2\omega\le & {} N^{-\frac{2r}{k}}\Vert \Delta _{h}^kD^rf(x)\Vert _{2}^{\frac{1}{k}}I_{G}^{1-\frac{1}{2k}} +\frac{4\sqrt{2}}{Nh} I_G. \end{aligned}$$

Setting \(\displaystyle h=\frac{\pi ^2}{N}\).

$$\begin{aligned} \left( 1-\frac{4\sqrt{2}}{Nh}\right) \int _{G}|{\mathcal {L}}_{A_{1}, A_{2}}\{f\}(\omega )|^2d^2\omega \le N^{-\frac{2r}{k}}\Vert \Delta _{h}^kD^r f(x)\Vert _{2}^{\frac{1}{k}}I_{G}^{1-\frac{1}{2k}}. \end{aligned}$$

Hence,

$$\begin{aligned} \int _{G}|{\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega )|^2d^2\omega =O\left( N^{-4r}\Vert \Delta _{h}^kD^rf(x)\Vert _{2}^{2}\right) \end{aligned}$$

and we have

$$\begin{aligned} \Vert \Delta _{h}^kD^rf(x)\Vert _{2}^{2}=O\left( N^{-2r}\phi \left[ \left( \frac{\pi ^2}{N}\right) ^k\right] \right) , \end{aligned}$$

which yields the desired result. \(\square \)

Theorem 2.5

Let \(\phi (t)=t^\alpha \) (\(\alpha >0\)). Then the next conditions are equivalent:

$$\begin{aligned} f\in W_{2,t^{\alpha }}^{2,k} \end{aligned}$$
(25)

and

$$\begin{aligned} \int _{|\omega |\ge N^2}|{\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega ) |^2d^2\omega =O( N^{-2r-2k\alpha }) \end{aligned}$$
(26)

where \(m=1,2\ldots .\); \(r=1,\ldots \); \(k=1,2,\ldots \); and \(0<\alpha <m\).

Proof

It follows from Theorem 2.4 that (25) entails (26).Suppose now that

$$\begin{aligned} \left( \int _{|\omega |\ge N^2}|{\mathcal {L}}_{A_{1},A_{2}}\{f\} (\omega )|^2d^2\omega \right) ^{\frac{1}{2}}=O( N^{-2r-k\alpha }). \end{aligned}$$

Thus, by Parseval’s identity, we have

$$\begin{aligned} \Vert \Delta _{h}^kD^rf(x)\Vert _{2}^{2}=\int _{{\mathbb {R}}^{2}}|\omega |^{2r} \left( \left| 1-\varphi _{h}\left( \frac{w_{1}}{b_1},\frac{w_{2}}{b_2}\right) \right| \right) ^{2k} |{\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega )|^2d^2\omega . \end{aligned}$$

Divide this integral into two, \(\underbrace{\int _{{\mathbb {R}}^{2}}}=\underbrace{\int _{|\omega |<N^2}}_{I_{1}}+\underbrace{\int _{|\omega |>N^2}}_{I_{2}},\) where \(N =[h^{-1}] \), and estimate each of them. Firstly, we estimate \(I_2\), since

$$\begin{aligned} \left| 1-\varphi _{h}\left( \frac{w_{1}}{b_1},\frac{w_{2}}{b_2}\right) \right| \le 2, \end{aligned}$$

it follows that

$$\begin{aligned} I_2&= \int _{|\omega |>N^2}|\omega |^{2r}\left( \left| 1- \varphi _{h} \left( \frac{w_{1}}{b_1},\frac{w_{2}}{b_2}\right) \right| \right) ^{2k} |{\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega )|^2d^2\omega \\&\le \int _{|\omega |>N^2}|\omega |^{2r}\left( 2\right) ^{2k} |{\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega )|^2d^2\omega .\\&=O\left( \int _{|\omega |>N^2}|\omega |^{2r}|{\mathcal {L}}_{A_{1},A_{2}} \{f\}(\omega )|^2d^2\omega \right) \\&=O\left( \sum _{l=0}^{\infty } \int _{(N+l)^2\le |\omega |<(N+l+1)^2} |\omega |^{2r}|{\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega )|^2d^2\omega \right) \\&=O\left( \sum _{l=0}^{\infty } (N+l+1)^{4r}\int _{N^2\le |\omega |} |{\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega )|^2d^2\omega \right) \\&\quad -\left( \sum _{l=0}^{\infty } (N+l+1)^{4r}\int _{|\omega |\ge (N+l +1)^2}|{\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega )|^2d^2\omega \right) \\&=O\left( N^{4r}\int _{N^2\le |\omega |}|{\mathcal {L}}_{A_{1}, A_{2}}\{f\}(\omega )|d^2\omega \right) \\&\quad +\left( \sum _{l=0}^{\infty } [(N+l+1)^{4r}-(N+l)^{4r}] \int _{|\omega |\ge (N+l+1)^2}|{\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega )|^2d^2\omega \right) \\&=O\left( N^{4r}\int _{N^2\le |w|}|{\mathcal {L}}_{A_{1}, A_{2}}\{f\}(\omega )|d^2\omega \right) \\&\quad +\left( \sum _{l=0}^{\infty } (N+l)^{4r-1}\int _{(N+l)^2 \le |\omega |}|{\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega )|^2d^2\omega \right) \\&=O\left( N^{4r}N^{-4r-2k\alpha }+\sum _{l=0}^{\infty } (N+l)^{4r-1}(N+l)^{-4r-2k\alpha }\right) \\&=O(N^{-2k\alpha })+O(N^{-2k\alpha })\\&=O(N^{-2k\alpha })\\&=O(h^{2k\alpha }), \end{aligned}$$

i.e.,

$$\begin{aligned} I_{2}=O(h^{2k\alpha }). \end{aligned}$$

Secondly, we estimate \(I_{1}\), since

$$\begin{aligned} 0\le 1- \dfrac{\sin (y)}{y}\le \dfrac{y^2}{6}, \ \ \mid \sin (y)\mid \le \mid y\mid , \ \ y\in {\mathbb {R}}, \end{aligned}$$

and

$$\begin{aligned} 1- \varphi _{h}\left( \frac{w_{1}}{b_1},\frac{w_{2}}{b_2}\right) =\left( 1- \dfrac{\sin (w_{1}h)}{w_{1}h}\right) \dfrac{\sin (w_{2}h)}{w_{2}h}+ \left( 1- \dfrac{\sin (w_{2}h)}{w_{2}h}\right) , \end{aligned}$$
$$\begin{aligned} I_1&= \int _{|\omega |<N^2}|\omega |^{2r}\left( \left| 1- \varphi _{h} \left( \frac{w_{1}}{b_1},\frac{w_{2}}{b_2}\right) \right| \right) ^{2k}|{\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega )|^2d^2\omega \\&\le \int _{|\omega |<N^2}|\omega |^{2r}\left( |\omega |\right) ^{2k} |{\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega )|^2d^2\omega \\&= O(h^{4k}) \int _{|\omega |<N^2}|\omega |^{2r+2k}|{\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega )|^2d^2\omega \\&= O(h^{4k})\sum _{n=0}^{N}\int _{n^2\le |\omega |\le (n+1)^2}|\omega |^{2r+2k} |{\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega )|^2d^2\omega \\&= O(h^{4k})\sum _{n=0}^{N}(n+1)^{4r+4k}\int _{n^2\le |\omega |\le (n+1)^2} |{\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega )|^2d^2\omega \\&= O(h^{4k})\sum _{n=0}^{N} (n+1)^{4r+4k}\left[ \int _{|\omega |\ge n^2} |{\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega )|^2d^2\omega \right] \\&\quad - \left[ \int _{|\omega |\ge (n+1)^2}|{\mathcal {L}}_{A_{1},A_{2}}\{f\} (\omega )|^2d^2\omega \right] \\&=O(h^{4k})\left[ \sum _{n=0}^{N} (n+1)^{4r+4k} \int _{|\omega |\ge n^2} |{\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega )|^2d^2\omega \right] \\&\quad - \left[ \sum _{n=0}^{N} (n+1)^{4r+4k} \int _{|\omega |\ge (n+1)^2} |{\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega )|^2d^2\omega \right] \\&=O(h^{4k})\left[ 1+\sum _{n=0}^{N} [(n+1)^{4r+4k} -n^{4r+4k}] \int _{|\omega |\ge n^2}|{\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega )|^2d^2\omega \right] \\&=O(h^{4k})\left[ 1+\sum _{n=0}^{N} (n+1)^{4r+4k-1} \int _{|\omega |\ge n^2}|{\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega )|^2d^2\omega \right] \\&=O(h^{4k})\left[ 1+\sum _{n=1}^{N} n^{4r+4k-1}n^{-4r-2k\alpha }\right] =O(h^{4k})\left[ 1+\sum _{n=1}^{N} n^{4k-2k\alpha -1} \right] \\&=O(h^{4k})O(N^{4k-2k\alpha })=O(h^{2k\alpha }); \end{aligned}$$

i.e.,

$$\begin{aligned} I_{1}=O(h^{2k\alpha }). \end{aligned}$$

Finally, combining the estimates for \(I_1\) and \(I_2\) gives

$$\begin{aligned} \Vert \Delta _{h}^kD^rf(x)\Vert _{2}=O(h^{k\alpha }), \end{aligned}$$

which means that \(f\in W_{2,t^{\alpha }}^{2,k}\). Hence the conditions (25) and (26) are equivalent. This proves the Theorem 2.5. \(\square \)

Remark

As in the article [15], the previous definition can be generalized as follows: For any two pure quaternions \(\alpha \) and \(\beta \) such that \(\alpha ^2 = \beta ^2 = -1\) used for re placing i and j in (7) and (8), and f in \(L^{1}({\mathbb {R}}^{2},{\mathcal {H}})\)

$$\begin{aligned} {\mathcal {L}}_{A_{1},A_{2}}^{\alpha ,\beta }\{f\}(\omega ) =\int _{{\mathbb {R}}^2}K_{A_{1}}^{\alpha }(x_{1},\omega _{1})f(x) K_{A_{2}}^{\beta }(x_{2},\omega _{2})dx. \end{aligned}$$

From linearity of \({\mathcal {L}}_{A_{1},A_{2}}^{\alpha ,\beta }\) we obtain the QLCT for the OPS split \(f=f_{+}+f_{-}\) where \(f_{ \pm }=\frac{1}{2}\left( f \pm \alpha f \beta \right) \)

$$\begin{aligned} {\mathcal {L}}_{A_{1},A_{2}}^{\alpha ,\beta }\{f\}(\omega )= & {} \int _{{\mathbb {R}}^2}K_{A_{1}}^{\alpha }(x_{1},\omega _{1})f(x) K_{A_{2}}^{\beta }(x_{2},\omega _{2})dx\\= & {} \int _{{\mathbb {R}}^2}K_{A_{1}}^{\alpha }(x_{1},\omega _{1}) f_{+}(x)K_{A_{2}}^{\beta }(x_{2},\omega _{2})dx\\{} & {} +\int _{{\mathbb {R}}^2} K_{A_{1}}^{\alpha }(x_{1},\omega _{1})f_{-}(x)K_{A_{2}}^{\beta }(x_{2},\omega _{2})dx \end{aligned}$$

so what was done above for the integral \(\displaystyle \int _{|\omega |\ge N^2}|{\mathcal {L}}_{A_{1},A_{2}}\{f\}(\omega )|^2d^2\omega \) can be done again for the two integrals \(\displaystyle \int _{|\omega |\ge N^2}|{\mathcal {L}}_{A_{1},A_{2}}^{\alpha ,\beta }\{f_{+}\}(\omega )|^2d^2\omega \) and \(\displaystyle \int _{|\omega |\ge N^2}|{\mathcal {L}}_{A_{1},A_{2}}^{\alpha ,\beta }\{f_{-}\}(\omega )|^2d^2\omega \) so that we can find a generalization of our results.

3 Conclusion

In this paper after having given two estimates for the quaternion linear canonical transform which generalize those of Abilov [1] we note so far the difficulty lies only in the fact that for quaternion fields we have no commutativity, whereas for the Fourier transform this does not pose a problem, but even for quaternion fields there is a question which arises, if we can have the same approximation for the right-sided Quaternion linear canonical transform [4], the answer is positive although there will be a slight difference concerning the calculations of certain steps.