1 Introduction

The cross-Wigner distribution has been studied in a number of papers [5, 9, 10, 18, 19] and has been proved the continuity properties on modulation spaces, given in [11, 13]. More precisely, continuity property of cross-Wigner distribution was obtained for \(M^{1,p}\) modulation spaces with polynomial weights in [9], for modulation spaces both without weights and with polynomial weights (which satisfy the equality (4.10) in [19]) in [18, 19], respectively. Cordero and Nicola thereafter proved in [10] that cross-Wigner distribution is continuous on all modulation spaces \(M^{p,q},\) \(p,q \in {[1,\infty ]}\), with polynomial weights.

The \(\tau \)-Wigner transform, \(\tau \in {[0,1]}\), was first introduced in [6]. There was proved basic properties and \(L^p\)-boundedness of this transform. We refer to [4, 7, 8, 15] for more details on the \(\tau \)-Wigner transform.

Multilinear localization operators in the context of modulation spaces were studied by many authors [2, 3, 14, 16, 17]. In a very recent papers [16, 17], Teofanov derived the bilinear and the multilinear Wigner transform and proved the continuity properties of these transforms for modified version of modulation spaces.

In this paper, in the second section, we first define the multilinear \(\varvec{\tau }\)-Wigner transform, \(\varvec{\tau }\in {[0,1]^d}\). We give a simple relation between short-time Fourier transform and multilinear \(\varvec{\tau } \)-Wigner transform. Further, we prove that multilinear \(\varvec{\tau }\)-Wigner transform is bounded on products of Lebesgue spaces. In the third section, we list some properties for multilinear \(\varvec{\tau }\)-Wigner transform. From these results we then prove the boundedness properties of multilinear \(\varvec{\tau }\)-Wigner transform on modulation spaces.

We have compiled some basic facts as in follows:

We denote \({\mathcal {S}}({\mathbb {R}}^d)\) as the space of complex-valued continuous functions on \({\mathbb {R}}^d\) rapidly decreasing at infinity. Let f be a complex valued measurable function on \({\mathbb {R}}^d\). The operators \(T_xf(t)=f(t-x)\) and \(M_wf(t)=e^{2{\pi }iw\cdot t}f(t)\) are called translation and modulation operators for \(x,w\in {{\mathbb {R}}^d}\), respectively. The compositions

$$\begin{aligned} T_xM_wf(t)=e^{2{\pi }iw\cdot (t-x)}f(t-x)\qquad \text {or}\qquad M_wT_xf(t)=e^{2{\pi }iw\cdot t}f(t-x) \end{aligned}$$

are called time frequency shifts. We write \((L^p({\mathbb {R}}^d),\parallel {.}\parallel _p)\) for the Lebesgue spaces with \(1\le p\le \infty \).

For \(f\in {L^1({\mathbb {R}}^d)}\) the Fourier transform \(\widehat{f} \) (or \({\mathcal {F}}f\) ) is defined as

$$\begin{aligned} \widehat{f}(t)=\int _{{\mathbb {R}}^d}f(x)e^{-2{\pi }ix\cdot t}dx, \end{aligned}$$

where \(x\cdot t=\sum _{i=1}^{d}x_i\cdot t_i\) is the usual scalar product on \({\mathbb {R}}^d\).

Fix a function \(g\ne 0\) (called the window function). The short-time Fourier transform (STFT) of a function f with respect to g is given by

$$\begin{aligned} V_gf(x,w)=\int _{{\mathbb {R}}^d}f(t)\overline{g(t-x)}e^{-2{\pi }it\cdot w}dt, \end{aligned}$$

for all \(x,w\in {{\mathbb {R}}^d}\). It is known that if \(f,g\in {L^2({\mathbb {R}}^d)}\) then \(V_gf\in {L^2({{\mathbb {R}}^d}\times {{\mathbb {R}}^d})}\) and \(V_gf\) is uniformly continuous (see [13]).

The cross-Wigner distribution of \(f,g\in {L^2({\mathbb {R}}^d)}\) is defined to be

$$\begin{aligned} W(f,g)(x,w)=\int _{{\mathbb {R}}^d}f(x+\frac{t}{2})\overline{g(x-\frac{t}{2})}e^{-2{\pi }it\cdot w}dt. \end{aligned}$$

If \(f=g\), then \(W(f,f)=Wf\) is called the Wigner distribution of \(f\in {L^2({\mathbb {R}}^d)}\).

For \({\tau }\in [0,1]\) and \(f,g\in {{\mathcal {S}}({\mathbb {R}}^d)}\), the \(\tau \)-Wigner transform is defined as

$$\begin{aligned} W_{\tau }(f,g)(x,w)=\int _{{\mathbb {R}}^d}f(x+{\tau }t){\overline{g(x-(1-\tau )t)}e^{-2{\pi }it\cdot w}dt}. \end{aligned}$$

If \(\tau =\frac{1}{2}\), then the \(\tau \)-Wigner transform is the cross-Wigner distribution. Moreover, for \(\tau =0\), \(W_0\) is the Rihaczek transform,

$$\begin{aligned} W_0(f,g)(x,w)=R(f,g)(x,w)=e^{-2{\pi }ix\cdot w}f(x){\overline{\widehat{g}(w)}}, \end{aligned}$$

and the conjugate Rihaczek transform is

$$\begin{aligned} W_1(f,g)(x,w)={\overline{R(g,f)}(x,w)=e^{2{\pi }ix\cdot w}{\overline{g(x)}}{\widehat{f}(w)}}, \end{aligned}$$

if \(\tau =1\). For more details we refer the reader to [6].

The Wigner transform was extended to the multilinear case in [17] as follows:

$$\begin{aligned} W(\mathbf {f},\mathbf {g})(x,w)=\int _{{\mathbb {R}}^{nd}}\prod _{j=1}^{n} \left( f_j\left( x_j+\frac{t_j}{2}\right) \overline{g_j\left( x_j-\frac{t_j}{2}\right) }\right) \, e^{-2\pi {i}{\mathcal {I}}{w\cdot t}}\, dt, \end{aligned}$$

where \(f_j,g_j\) are in the Gelfand–Shilov type space of analytic functions, \(x_j,w_j, t_j\in {{\mathbb {R}}^d}\), \(j=1,\ldots ,n\), \( x=(x_1, x_2,\ldots , x_n), w=(w_1,w_2,\ldots ,w_n), t=(t_1,t_2,\ldots ,t_n)\) and \({\mathcal {I}}{w\cdot t}=\sum \nolimits _{j=1}^{n}{w_j\cdot t_j}\). Here \(\mathbf {f}\) denotes both the vector \(\mathbf {f}=(f_1,f_2,\ldots ,f_n)\) and the tensor product \(\mathbf {f}=f_1\otimes {f_2}\otimes \cdots \otimes {f_n}\). So it is written

$$\begin{aligned} \mathbf {f}(t)=f_1(t_1)f_2(t_2)\cdots f_n(t_n)={\otimes }_{j=1}^nf_j(t_j), \end{aligned}$$

where \(t=(t_1,t_2,\ldots ,t_n), t_j\in {{\mathbb {R}}^d}, j=1,\ldots ,n\). We will use the same notations in our text as well.

A weight function w on \({\mathbb {R}}^{d}\) is a non-negative, continuous and locally integrable function. The weight v is called submultiplicative if \(v\left( x+y\right) \le v\left( x\right) v\left( y\right) \) for all \(x,y\in {\mathbb {R}}^{d}.\) Let v be a submultiplicative function on \({\mathbb {R}}^{d}.\) A weight function w on \({\mathbb {R}}^{d}\) is v-moderate if \(w\left( x+y\right) \le Cv\left( x\right) w\left( y\right) \) for all \(x,y\in {\mathbb {R}}^{d}.\) Further, w is a weight of polynomial growth if

$$\begin{aligned} w\left( x\right) \le Cv_{s}\left( x\right) =C{\left\langle x \right\rangle }^s \end{aligned}$$

where \({\left\langle x \right\rangle }^s=\left( 1+\left| x\right| ^{2}\right) ^{\frac{s}{2}},\) for some \(C>0,\) \(s\ge 0\) and \(x\in {\mathbb {R}}^{d}.\)

Fix a non-zero window \(g\in {\mathcal {S}}\left( {\mathbb {R}}^{d}\right) \) and \(1\le p,q\le \infty \). Let m be a weight function of polynomial growth and \(v_{s}\)-moderate on \({\mathbb {R}}^{2d}.\) Then the modulation space \(M_{m}^{p,q}\left( {\mathbb {R}}^{d}\right) \) consists of all tempered distributions \(f\in {\mathcal {S}}^{\prime }\left( {\mathbb {R}}^{d}\right) \) such that the short-time Fourier transform \(V_{g}f\) is in the weighted mixed-norm space \(L_{m}^{p,q}\left( {\mathbb {R}}^{2d}\right) \), given in [1]. The norm on \(M_{m}^{p,q}\left( {\mathbb {R}} ^{d}\right) \) is \(\left\| f\right\| _{M_{m}^{p,q}}=\left\| V_{g}f\right\| _{L_{m}^{p,q}}.\) If \(p=q,\) then we write \(M_{m}^{p}\left( {\mathbb {R}} ^{d}\right) \) instead of \(M_{m}^{p,p}\left( {\mathbb {R}}^{d}\right) \) and if \(m=1\), we have the standard modulation space \(M^{p,q}\left( {\mathbb {R}}^{d}\right) \) (see [11, 13]).

In [17], Teofanov gave a modified version of modulation spaces by restricting the weights to the form \({m(x,w)={\left\langle x \right\rangle }^t{\left\langle w \right\rangle }^s},\) \(s,t\in {\mathbb {R}}.\) These spaces were denoted by \({\mathcal {M}}_{s,t}^{p,q}\left( {\mathbb {R}}^{nd}\right) \) and defined as the set of \( \mathbf {f}=(f_1,\ldots ,f_n), f_j\in {\mathcal {S'}({\mathbb {R}}^d)},j=1,\ldots ,n,\) such that

$$\begin{aligned} \Vert \mathbf {f}\Vert _{{\mathcal {M}}_{s,t}^{p,q}} =\left( \int _{{\mathbb {R}}^{nd}}\left( \int _{{\mathbb {R}}^{nd}} |V_{\mathbf {g}}\mathbf {f}(x,w){\left\langle x \right\rangle }^t{\left\langle w \right\rangle }^s|^p dx \right) ^{q/p}dw \right) ^{1/q}<\infty , \end{aligned}$$

for \(1\le p,q\le \infty \), where \(\mathbf {g}=(g_1,\ldots ,g_n), g_j\in {{\mathcal {S}}({\mathbb {R}}^d)}, j=1,\ldots ,n\) and multilinear case of short-time Fourier transform is

$$\begin{aligned} V_{\mathbf {g}}\mathbf {f}(x,w)=\int _{{\mathbb {R}}^{nd}}\mathbf {f}(t) \prod _{j=1}^{n} \overline{M_{w_j}T_{x_j}g_j(t_j)}dt. \end{aligned}$$
(1.1)

In the third section of this paper, we will use this modified version of modulation spaces.

2 Main results

We begin by characterizing the definition of multilinear \(\varvec{\tau }\)-Wigner transform.

Definition 1

Let \(f_j,g_j\in {{\mathcal {S}}({\mathbb {R}}^d)},j=1,\ldots ,n, \mathbf {f}=(f_1,\ldots ,f_n),\mathbf {g}=(g_1,\ldots ,g_n)\). For \(\varvec{\tau }=(\tau _1,\tau _2,\ldots ,\tau _d)\in {[0,1]^d}\), the multilinear \(\varvec{\tau }\)-Wigner transform \(W_{\varvec{\tau }}(\mathbf {g},\mathbf {f})\) is defined by

$$\begin{aligned} W_{\varvec{\tau }}(\mathbf {g},\mathbf {f})(x,w)=\int _{{\mathbb {R}}^{nd}}\prod _{j=1}^{n} \Big (g_j(x_j+\varvec{\tau }{t_j}) \overline{f_j(x_j-(\mathbf {1}-\varvec{\tau })t_j)}\Big )\, e^{-2\pi {i}{\mathcal {I}}{w\cdot t}}\, dt, \end{aligned}$$
(2.1)

where \(x_j,w_j, t_j\in {{\mathbb {R}}^d}, x=(x_1,\ldots , x_n), w=(w_1,\ldots ,w_n), t=(t_1,\ldots ,t_n)\), \(\mathbf {1}:=(1,\ldots ,1)\), \(\varvec{\tau } t_j:=(\tau _1t_{j1},\ldots ,\tau _dt_{jd})\) and \({\mathcal {I}}{w\cdot t}=\sum \nolimits _{j=1}^{n}{w_j\cdot t_j}\).

For \(\varvec{\tau }=\mathbf {0}\), multilinear \(\varvec{\tau }\)-Wigner transform becomes multilinear Rihaczek transform as follows:

$$\begin{aligned}&W_{\mathbf {0}}(\mathbf {g},\mathbf {f})(x,w)=\int _{{\mathbb {R}}^{nd}}\prod _{j=1}^{n} \Big (g_j(x_j) \overline{f_j(x_j-t_j)}\Big )\, e^{-2\pi {i}{\mathcal {I}}{w\cdot t}}\,dt \nonumber \\&\quad =e^{-2\pi {i}{\mathcal {I}}{x\cdot w}}\prod _{j=1}^{n} g_j(x_j)\,\overline{\int _{{\mathbb {R}}^{nd}}f_1(u_1)e^{-2\pi {i}u_1\cdot w_1} \cdots f_n(u_n)e^{-2\pi {i}u_n \cdot w_n}\, du_1 \cdots du_n}\nonumber \\&\quad =e^{-2\pi {i}{\mathcal {I}}{x\cdot w}}\prod _{j=1}^{n} g_j(x_j)\overline{\hat{f_j}(w_j)}={\mathcal {R}}(\mathbf {g},\mathbf {f})(x,w) \end{aligned}$$
(2.2)

by changing of variables \(x_j-t_j=u_j\,, j=1,\ldots ,n.\)

If \(\varvec{\tau }=\mathbf {1}\), then we obtain the multilinear conjugate Rihaczek transform from the multilinear \(\varvec{\tau }\)-Wigner transform by changing variable \(x_j+t_j=u_j,\, j=1,\ldots ,n\), as follows:

$$\begin{aligned}&W_{\mathbf {1}}(\mathbf {g},\mathbf {f})(x,w)=\int _{{\mathbb {R}}^{nd}}\prod _{j=1}^{n} \Big (g_j(x_j+t_j) \overline{f_j(x_j)}\Big )\, e^{-2\pi {i}{\mathcal {I}}{w \cdot t}}\, dt \nonumber \\&\quad = e^{2\pi {i}{\mathcal {I}}{x\cdot w}}\prod _{j=1}^n \overline{f_j(x_j)}\Big (\int _{{\mathbb {R}}^{nd}}e^{-2\pi {i}{u_1\cdot w_1}}g_1(u_1) \nonumber \\&\quad \quad \cdots e^{-2\pi {i}{u_n \cdot w_n}}g_n(u_n)du_1 \cdots du_n\Big )\nonumber \\&\quad = e^{2\pi {i}{\mathcal {I}}{x\cdot w}}\prod _{j=1}^n \overline{f_j(x_j)}\hat{g_j}(w_j)={\mathcal {R}}^{*}(\mathbf {g},\mathbf {f})(x,w). \end{aligned}$$
(2.3)

If \(\varvec{\tau }=\varvec{\frac{1}{2}}=(\frac{1}{2},\ldots ,\frac{1}{2})\), multilinear \(\varvec{\tau }\)-Wigner transform coincides with multilinear Wigner transform in [17].

Now, let us define ratio of vectors \(u,v \in {{\mathbb {R}}^d}\) as follows:

$$\begin{aligned} \frac{u}{v}:=\left( \frac{u_1}{v_1},\ldots ,\frac{u_d}{v_d}\right) . \end{aligned}$$

For \(\varvec{\tau }\in (0,1)^d\), let \(A_{\varvec{\tau }}\) be the mapping defined by

$$\begin{aligned} A_{\varvec{\tau }}:\mathbf {h}(t)\rightarrow \tilde{\mathbf {h}} \left( \frac{\mathbf {1}-\varvec{\tau }}{\varvec{\tau }}t\right) =\prod _{j=1}^{n} \tilde{h_j}{\left( \frac{\mathbf {1}-\varvec{\tau }}{\varvec{\tau }}t_j\right) } \end{aligned}$$

where \(t=(t_1,t_2,\ldots ,t_n), t_j\in {{\mathbb {R}}^d}, j=1,\ldots ,n\), \({\big (\frac{\mathbf {1}-\varvec{\tau }}{\varvec{\tau }}t_j\big )}=\big (\frac{1-\tau _1}{\tau _1}t_{j1},\ldots ,\frac{1-\tau _d}{\tau _d}t_{jd}\big )\) and \(\tilde{\mathbf {h}}(t)=\mathbf {h}(-t)\). If \(\mathbf {h}=(h_1, h_2,\ldots , h_n)\in {L^p({\mathbb {R}}^d)\times {L^p({\mathbb {R}}^d)}\times \cdots \times {L^p({\mathbb {R}}^d)}}, p\in {[1,\infty ]}\), then we have for \(\varvec{\tau }\in (0,1)^d\)

$$\begin{aligned} |A_{\varvec{\tau }}\mathbf {h}(t)|=\,&\left| \tilde{\mathbf {h}}\left( \frac{\mathbf {1}-\varvec{\tau }}{\varvec{\tau }}t\right) \right| =\left| \tilde{h_1}\left( \frac{\mathbf {1}-\varvec{\tau }}{\varvec{\tau }}t_1\right) \right| \cdots \left| \tilde{h_n}\left( \frac{\mathbf {1}-\varvec{\tau }}{\varvec{\tau }}t_n\right) \right| \end{aligned}$$

and so making the change of variables \(\frac{1-\tau _k}{\tau _k}t_{jk}=y_{jk}\), \(k=1,\ldots ,d\),

$$\begin{aligned}&\Vert A_{\varvec{\tau }}\mathbf {h}\Vert ^p_{L^p({\mathbb {R}}^{nd})}\\&\quad =\int _{{\mathbb {R}}^{nd}}|A_{\varvec{\tau }}\mathbf {h}(t)|^p dt=\int _{{\mathbb {R}}^{nd}}\left| \tilde{h_1}\left( \frac{\mathbf {1}-\varvec{\tau }}{\varvec{\tau }}t_1\right) \right| ^p\cdots \left| \tilde{h_n}\left( \frac{\mathbf {1}-\varvec{\tau }}{\varvec{\tau }}t_n\right) \right| ^pdt\\&\quad =\int _{{\mathbb {R}}^{d}}\left| \tilde{h_1}\left( \frac{1-\tau _1}{\tau _1}t_{11},...,\frac{1-\tau _d}{\tau _d}t_{1d}\right) \right| ^pdt_1 \\&\quad \quad \cdots \int _{{\mathbb {R}}^{d}}\left| \tilde{h_n}\left( \frac{1-\tau _1}{\tau _1}t_{n1},...,\frac{1-\tau _d}{\tau _d}t_{nd}\right) \right| ^pdt_n\\&\quad =\prod _{k=1}^{d}\Big |\frac{\tau _k}{1-\tau _k}\Big |\int _{{\mathbb {R}}^{d}}\Big |\tilde{h_1}(y_1)\Big |^pdy_1 \cdots \prod _{k=1}^{d}\Big |\frac{\tau _k}{1-\tau _k}\Big | \int _{{\mathbb {R}}^{d}}\Big |\tilde{h_n}(y_n)\Big |^pdy_n\\&\quad =\left( \prod _{k=1}^{d}\left| \frac{\tau _k}{1-\tau _k}\right| \right) ^n\Vert h_1\Vert ^p_p\Vert h_2\Vert ^p_p\cdots \Vert h_n\Vert ^p_p. \end{aligned}$$

From here, we get

$$\begin{aligned} \Vert A_{\varvec{\tau }}\mathbf {h}\Vert _{L^p({\mathbb {R}}^{nd})}=\left( \prod _{k=1}^{d}\left| \frac{\tau _k}{1-\tau _k}\right| \right) ^{\frac{n}{p}}\prod _{j=1}^{n}\Vert h_j\Vert _{L^p({\mathbb {R}}^d)}. \end{aligned}$$
(2.4)

Also, we need to define the multilinear \(\varvec{\tau }\)-short-time Fourier transform which will be used later as follows:

$$\begin{aligned} V^{\varvec{\tau }}_{\mathbf {f}}\mathbf {g}(x,w)=V_{\mathbf {f}}\,\mathbf {g}{\left( \frac{\mathbf {1}}{\mathbf {1}-\varvec{\tau }}x,\frac{\mathbf {1}}{\varvec{\tau }}w\right) }=\prod _{j=1}^{n}V_{f_j}{g_j}{\left( \frac{\mathbf {1}}{\mathbf {1}-\varvec{\tau }}x_j,\frac{\mathbf {1}}{\varvec{\tau }}w_j\right) } \end{aligned}$$
(2.5)

where \(x=(x_1,x_2,\ldots ,x_n), w=(w_1,w_2,\ldots ,w_n), x_j, w_j\in {{\mathbb {R}}^d}, j=1,\ldots ,n\), \({\big (\frac{\mathbf {1}}{\mathbf {1}-\varvec{\tau }}x_j\big )}=\big (\frac{1}{1-\tau _1}x_{j1},\ldots ,\frac{1}{1-\tau _d}x_{jd}\big )\) and \({\big (\frac{\mathbf {1}}{\varvec{\tau }}w_j\big )}=\big (\frac{1}{\tau _1}w_{j1},\ldots ,\frac{1}{\tau _d}w_{jd}\big ).\) Then, we have for \(\varvec{\tau }\in {(0,1)^d}\)

$$\begin{aligned} \Vert V^{\varvec{\tau }}_{\mathbf {f}}\mathbf {g}\Vert _{L^p({\mathbb {R}}^{2nd})}=\left( \prod _{k=1}^{d}|\tau _k||1-\tau _k|\right) ^\frac{n}{p}\Vert V_{\mathbf {f}}\,\mathbf {g}\Vert _{L^p({\mathbb {R}}^{2nd})}. \end{aligned}$$
(2.6)

Indeed, by a change of variables \(\frac{1}{1-\tau _k}x_{jk}=u_{jk}, \frac{1}{\tau _k}w_{jk}=v_{jk}, j=1,\ldots ,n\) and \(k=1,\ldots ,d,\)

$$\begin{aligned} \Vert V^{\varvec{\tau }}_{\mathbf {f}}\mathbf {g}\Vert _{L^p({\mathbb {R}}^{2nd})}^p=\,&\int _{{\mathbb {R}}^{2nd}}|V^{\varvec{\tau }}_{\mathbf {f}}\mathbf {g}(x,w)|^pdxdw\\ =\,&\int _{{\mathbb {R}}^{2nd}}\left| \prod _{j=1}^{n}V_{f_j}{g_j}{\left( \frac{\mathbf {1}}{\mathbf {1}-\varvec{\tau }}x_j,\frac{\mathbf {1}}{\varvec{\tau }}w_j\right) }\right| ^pdxdw\\ =\,&\left( \prod _{k=1}^{d}|\tau _k||1-\tau _k|\right) ^n\int _{{\mathbb {R}}^{2nd}}\left| \prod _{j=1}^{n}V_{f_j}{g_j}{\left( u_j,v_j\right) }\right| ^pdudv\\ =\,&\left( \prod _{k=1}^{d}|\tau _k||1-\tau _k|\right) ^n\Vert V_{\mathbf {f}}\mathbf {g}\Vert ^p_{L^p({\mathbb {R}}^{2nd})} \end{aligned}$$

and so, we get the equality (2.6).

Our first lemma shows that multilinear \(\varvec{\tau }\)-Wigner transform can be rewritten as a multilinear short time Fourier transform. The proof is similar to the proof of Lemma 6.2 in [6]. But for the sake of clarity, we give the proof for the multilinear case.

Lemma 1

For \(\varvec{\tau }\in {(0,1)^d}\), we have

$$\begin{aligned} W_{\varvec{\tau }}(\mathbf {g},\mathbf {f} )(x,w)=\left( \prod _{k=1}^{d}\frac{1}{|\tau _k|}\right) ^ne^{2\pi i {\mathcal {I}}{\frac{\mathbf {1}}{\varvec{\tau }}w \cdot x}}V_{A_{\varvec{\tau }}\mathbf {f}}\mathbf {g}\left( \frac{\mathbf {1}}{\mathbf {1}-\varvec{\tau }}x,\frac{\mathbf {1}}{\varvec{\tau }}w\right) . \end{aligned}$$

Proof

Taking into account (2.1) and making the change of variables \(x_{jk}+\tau _k{t_{jk}}=q_{jk}, j=1,\ldots ,n\), \(k=1,\ldots ,d,\) we obtain

$$\begin{aligned}&W_{\varvec{\tau }}(\mathbf {g},\mathbf {f} )(x,w)\\&\quad = \int _{{\mathbb {R}}^{nd}}\prod _{j=1}^{n} \left( g_j(x_j+\varvec{\tau } t_j) \overline{f_j(x_j-(\mathbf {1}-\varvec{\tau })t_j)}\right) e^{-2\pi i{\mathcal {I}}{w\cdot t}}dt\\&\quad =\left( \prod _{k=1}^{d}\frac{1}{|\tau _k|}\right) ^n\int _{{\mathbb {R}}^{nd}}\prod _{j=1}^{n} \left( g_j(q_j)\overline{f_j(x_j-\frac{\mathbf {1}-\varvec{\tau }}{\varvec{\tau }}(q_j-x_j))}\right) e^{-2\pi i{\mathcal {I}}(\frac{\mathbf {1}}{\varvec{\tau }}w)\cdot (q-x)}dq\\&\quad =\left( \prod _{k=1}^{d}\frac{1}{|\tau _k|}\right) ^ne^{2\pi i{\mathcal {I}}({\frac{\mathbf {1}}{\varvec{\tau }}w)\cdot x}}\int _{{\mathbb {R}}^{nd}}\prod _{j=1}^{n} \left( g_j(q_j)\overline{\widetilde{f_j}(\frac{\mathbf {1}-\varvec{\tau }}{\varvec{\tau }}q_j-\frac{\mathbf {1}}{\varvec{\tau }}x_j)}\right) e^{-2\pi i{\mathcal {I}}(\frac{\mathbf {1}}{\varvec{\tau }}w)\cdot q}dq\\&\quad =\left( \prod _{k=1}^{d}\frac{1}{|\tau _k|}\right) ^ne^{2\pi i{\mathcal {I}}({\frac{\mathbf {1}}{\varvec{\tau }}w)\cdot x}}\int _{{\mathbb {R}}^{nd}}\prod _{j=1}^{n} \left( g_j(q_j)\overline{A_{\varvec{\tau }}{f_j}(q_j-\frac{\mathbf {1}}{\mathbf {1}-\varvec{\tau }}x_j)}\right) e^{-2\pi i{\mathcal {I}}(\frac{\mathbf {1}}{\varvec{\tau }}w)\cdot q}dq\\&\quad =\left( \prod _{k=1}^{d}\frac{1}{|\tau _k|}\right) ^ne^{2\pi i{\mathcal {I}}({\frac{\mathbf {1}}{\varvec{\tau }}w)\cdot x}}V_{A_{\varvec{\tau }}\mathbf {f}} \mathbf {g}\left( \frac{\mathbf {1}}{\mathbf {1}-\varvec{\tau }}x,\frac{\mathbf {1}}{\varvec{\tau }}w\right) . \end{aligned}$$

\(\square \)

The boundedness property of short-time Fourier transform on Lebesgue spaces was examined in [5]. In the next proposition, we will examine the boundedness of the multilinear case of it.

Proposition 1

Let \(q\ge 2, q'\le {p_j}\le {q}, j=1,\ldots ,n\), where \(\frac{1}{q}+\frac{1}{q'}=1\). Then

$$\begin{aligned} V:\Big (L^{{p'_1}}({\mathbb {R}}^d)\times \cdots \times L^{{p'_n}}({\mathbb {R}}^d)\Big )\times \Big (L^{p_1}({\mathbb {R}}^d)\times \cdots \times L^{p_n}({\mathbb {R}}^d)\Big )\rightarrow L^q({\mathbb {R}}^{2nd}) \end{aligned}$$

is continuous and we have

$$\begin{aligned} \Vert V_{\mathbf {f}}\mathbf {g}\Vert _q\le \prod _{j=1}^{n}\Vert g_j\Vert _{{p'_j}} \Vert f_j\Vert _{p_j} \end{aligned}$$
(2.7)

where \(\mathbf {g}=(g_1,\ldots ,g_n)\in {L^{{p'_1}}({\mathbb {R}}^d)\times \cdots \times L^{{p'_n}}({\mathbb {R}}^d)}\) and \(\mathbf {f}=(f_1,\ldots ,f_n)\in {L^{p_1}({\mathbb {R}}^d)\times \cdots \times L^{p_n}({\mathbb {R}}^d)}\).

Proof

By using the formula (1.1) and by applying the Parseval’s formula, we have

$$\begin{aligned} \Vert V_{\mathbf {f}}\mathbf {g}\Vert _2^2=\,&\int _{{\mathbb {R}}^{nd}}\int _{{\mathbb {R}}^{nd}} |V_{\mathbf {f}}\mathbf {g}(x,w)|^2 dx dw\\ =\,&\int _{{\mathbb {R}}^{nd}}\int _{{\mathbb {R}}^{nd}}\Big |\Big <\prod _{j=1}^{n}g_j,\prod _{j=1}^{n} f_j(.-x_j)e^{2\pi iw_j}\Big >\Big |^2 dx dw\\ =\,&\int _{{\mathbb {R}}^{nd}}\int _{{\mathbb {R}}^{nd}}\Big |\prod _{j=1}^{n}g_j(u_j)\Big |^2\Big |\prod _{j=1}^{n} f_j(v_j)\Big |^2 du dv\\ =\,&\prod _{j=1}^{n}\Vert g_j\Vert ^2_2 \prod _{j=1}^{n}\Vert f_j\Vert ^2_2 \end{aligned}$$

and so, we get

$$\begin{aligned} \Vert V_{\mathbf {f}}\mathbf {g}\Vert _2= \Vert f_1\Vert _2\Vert f_2\Vert _2\cdots \Vert f_n\Vert _2\Vert g_1\Vert _2\Vert g_2\Vert _2\cdots \Vert g_n\Vert _2. \end{aligned}$$
(2.8)

Moreover, for \(q=\infty \), we write

$$\begin{aligned} \Vert V_{\mathbf {f}}\mathbf {g}\Vert _{\infty }=\,&\sup _{x,w\in {{\mathbb {R}}^{nd}}}\Big |\int _{{\mathbb {R}}^{nd}}e^{-2\pi i{\mathcal {I}}{t\cdot w}}\mathbf {g}(t)\overline{\mathbf {f}(t-x)}dt\Big |\\ \le&\sup _{x,w\in {{\mathbb {R}}^{nd}}}\int _{{\mathbb {R}}^{nd}}|\mathbf {g}(t)||\mathbf {f}(t-x)|dt. \end{aligned}$$

Using the Hölder inequality for \(r=(r_1,\ldots ,r_n)\in {[1,\infty ]^n}\), we have

$$\begin{aligned}&\int _{{\mathbb {R}}^{nd}}|\mathbf {g}(t)||\mathbf {f}(t-x)|dt\\&\quad =\int _{{\mathbb {R}}^{d}}|{g_1}(t_1)||{f_1}(t_1-x_1)dt_1\cdots \int _{{\mathbb {R}}^{d}}|{g_n}(t_n)||{f_n}(t_n-x_n)|dt_n\\&\quad \le \ \Vert g_1\Vert _{r'_1}\Vert f_1\Vert _{r_1}\cdots \Vert g_n\Vert _{r'_n}\Vert f_n\Vert _{r_n} \end{aligned}$$

where \(r'=({r'_1},\ldots ,{r'_n})\) and \(\frac{1}{r_j}+\frac{1}{r'_j}=1\), \(j=1,\ldots ,n\). Hence, we obtain

$$\begin{aligned} \Vert V_{\mathbf {f}}\mathbf {g}\Vert _{\infty }\le&\prod _{j=1}^{n}\Vert {g_j}\Vert _{r'_j}\Vert {f_j}\Vert _{r_j} \end{aligned}$$
(2.9)

Thus, we have obtained in (2.8) and (2.9) the claimed boundedness property in the case \(q=2\) and \(q=\infty \). Hence we get the inequality (2.7) by multilinear interpolation (see [12]) for general q. \(\square \)

Proposition 2

Let \(1\leqslant p\leqslant \infty \) and \(f\in L^1({\mathbb {R}}^d)\). The short-time Fourier transform \(V_fg\) has the following boundedness property:

$$\begin{aligned} \Vert V_fg\big \Vert _{L^{p,\infty }({\mathbb {R}}^{2d})}\leqslant \Vert f\big \Vert _{L^1({\mathbb {R}}^d)}\Vert g\big \Vert _{L^p({\mathbb {R}}^d)}\,. \end{aligned}$$

Proof

The case \(p=\infty \) is known already. Let first \(f\in C^\infty _{0}({\mathbb {R}}^d)\). For \(1<p<\infty \), by applying the Hölder inequality we get the following

$$\begin{aligned} \Vert V_fg\big \Vert _{L^{p,\infty }({\mathbb {R}}^{2d})}^p\!\!&=\sup _{\omega \in {\mathbb {R}}^d} \int \limits _{{\mathbb {R}}^d}\left| \int \limits _{{\mathbb {R}}^d}\overline{f(t-x)}g(t)e^{-2\pi it\cdot \omega }dt\right| ^p\!\!\!dx \nonumber \\&\quad =\sup _{\omega \in {\mathbb {R}}^d}\int \limits _{{\mathbb {R}}^d}\left| \int \limits _{{\mathbb {R}}^d} \overline{f(t-x)}^\frac{1}{p'}\overline{f(t-x)}^\frac{1}{p}g(t)e^{-2\pi it\cdot \omega }dt\right| ^p\!\!\!dx\nonumber \\&\quad \leqslant \int \limits _{{\mathbb {R}}^d}\left( \int \limits _{{\mathbb {R}}^d} |f(t-x)|dt\right) ^\frac{p}{p'}\int \limits _{{\mathbb {R}}^d} |f(t-x)||g(t)|^p\,dt dx\nonumber \\&\quad =\Vert f\big \Vert _{L^1({\mathbb {R}}^d)}^\frac{p}{p'}\int \limits _{{\mathbb {R}}^d} |g(t)|^p\int \limits _{{\mathbb {R}}^d}|f(t-x)|dx dt \nonumber \\&\quad =\Vert f\big \Vert _{L^1({\mathbb {R}}^d)}^{\frac{p}{p'}+1}\int \limits _{{\mathbb {R}}^d}|g(t)|^pdt =\Vert f\big \Vert _{L^1({\mathbb {R}}^d)}^p\Vert g\big \Vert _{L^p({\mathbb {R}}^d)}^p\,, \end{aligned}$$
(2.10)

since \(\displaystyle \frac{p}{p'}+1=p\) and \(\displaystyle \frac{1}{p}+\displaystyle \frac{1}{p'}=1\). Also since \(f\in C^\infty _{0}({\mathbb {R}}^d)\) is dense in \(L^1({\mathbb {R}}^d)\), from (2.10) follows that this inequality can be extended to all \(f\in L^1({\mathbb {R}}^d)\) by taking the sequence \(f_n \rightarrow f\) in \(L^1({\mathbb {R}}^d)\) norm.

For \(p=1\) the proof proceeds as follows:

$$\begin{aligned} \Vert V_fg\big \Vert _{L^{1,\infty }({\mathbb {R}}^{2d})}\!\!&=\sup _{\omega \in {\mathbb {R}}^d} \int \limits _{{\mathbb {R}}^d}\left| \int \limits _{{\mathbb {R}}^d}\overline{f(t-x)}g(t)e^{-2\pi it\cdot \omega }dt\right| \, \!\!\!\, dx\nonumber \\&\leqslant \int \limits _{{\mathbb {R}}^d}\int \limits _{{\mathbb {R}}^d} |f(t-x)||g(t)|dt\,dx\nonumber \\&=\int \limits _{{\mathbb {R}}^d}|f(y)|dy\int \limits _{{\mathbb {R}}^d}|g(t)|dt =\Vert f\big \Vert _{L^1({\mathbb {R}}^d)}\Vert g\big \Vert _{L^1({\mathbb {R}}^d)}. \end{aligned}$$

The proof is complete. \(\square \)

Proposition 3

Let \(1\le p \le \infty \) and \(f_j \in {L^1({\mathbb {R}}^d)},\, j=1,\ldots ,n, \, \mathbf {f}=(f_1,\ldots ,f_n)\). The short-time Fourier transform \(V_{\mathbf {f}}\mathbf {g}\) has the following property:

$$\begin{aligned} \Vert V_{\mathbf {f}}\mathbf {g}\big \Vert _{L^{p,\infty }({\mathbb {R}}^{2nd})}\leqslant \prod _{j=1}^n \Vert f_j\big \Vert _{L^1({\mathbb {R}}^d)}\Vert g_j\big \Vert _{L^p({\mathbb {R}}^d)} \end{aligned}$$

Proof

Let \(1<p<\infty \).If we apply the proof technique of Proposition 2 to the multilinear case, we get for \(x_j,w_j\in {{\mathbb {R}}^d}, x=(x_1,\ldots ,x_n), w=(w_1,\ldots ,w_n)\)

$$\begin{aligned}&\Vert V_{\mathbf {f}}\mathbf {g}\big \Vert ^p_{L^{p,\infty }({\mathbb {R}}^{2nd})}=\sup \limits _{w\in {{\mathbb {R}}^{nd}}}\int _{{\mathbb {R}}^{nd}}|V_{\mathbf {f}}\mathbf {g}(x,w)|^p dx \\&\quad \le \sup \limits _{w\in {{\mathbb {R}}^{nd}}}\int _{{\mathbb {R}}^{nd}}\left( \int _{{\mathbb {R}}^{nd}}|\mathbf {g}(t)|\,|\mathbf {f}(t-x)|dt\right) ^p dx\\&\quad = \sup \limits _{w\in {{\mathbb {R}}^{nd}}}\int _{{\mathbb {R}}^{nd}}\left( \int _{{\mathbb {R}}^{d}}|g_1(t_1)|\,|f_1(t_1-x_1)|^{1/p}|f_1(t_1-x_1)|^{1/p'}dt_1\cdots \right. \\&\left. \qquad \int _{{\mathbb {R}}^{d}}|g_n(t_n)|\,|f_n(t_n-x_n)|^{1/p}|f_n(t_n-x_n)|^{1/p'}dt_n\right) ^p dx\\&\quad \le \int _{{\mathbb {R}}^{nd}}\left( \Vert f_1\Vert ^{1/p'}_{L^1 ({\mathbb {R}}^d)}\left( \int _{{\mathbb {R}}^{d}}|f_1(t_1-x_1)|\,|g_1(t_1)|^{p} dt_1\right) ^{1/p} \right. \\&\left. \quad \quad \cdots \Vert f_n\Vert ^{1/p'}_{L^1 ({\mathbb {R}}^d)}\left( \int _{{\mathbb {R}}^{d}}|f_n(t_n-x_n)|\,|g_n(t_n)|^{p} dt_n\right) ^{1/p}\right) ^p dx\\&\quad =\prod _{j=1}^n \Vert f_j\Vert ^{p/p'}_{L^1 ({\mathbb {R}}^d)} \int _{{\mathbb {R}}^{nd}}(|f^\sim _1|*|g_1|^p)(x_1)\cdots (|f^\sim _n|*|g_n|^p)(x_n) dx\\&\quad =\prod _{j=1}^n \Vert f_j\Vert ^{p/p'}_{L^1 ({\mathbb {R}}^d)} \Big (\Vert |f^\sim _1|*|g_1|^p\Vert _{L^{1}({\mathbb {R}}^d)}\cdots \Vert |f^\sim _n|*|g_n|^p\Vert _{L^{1}({\mathbb {R}}^d)}\Big )\\&\quad \le \prod _{j=1}^n \Vert f_j\Vert ^{p/p'}_{L^1 ({\mathbb {R}}^d)} \Big (\Vert f_1\Vert _{L^1 ({\mathbb {R}}^d)}\Vert |g_1|^p\Vert _{L^1 ({\mathbb {R}}^d)}\cdots \Vert f_n\Vert _{L^1 ({\mathbb {R}}^d)}\Vert |g_n|^p\Vert _{L^1 ({\mathbb {R}}^d)}\Big )\\&\quad =\prod _{j=1}^n \Vert f_j\Vert ^{p}_{L^1 ({\mathbb {R}}^d)}\Vert g_j\Vert _{L^p ({\mathbb {R}}^d)}^p, \end{aligned}$$

where \(\frac{p}{p'}+1=p.\)

The proof is similar for the cases \(p=1\) and \(p=\infty \). \(\square \)

Theorem 1

Let \(q\ge 2\) and \(q'\le p_j \le q, j=1,\ldots ,n\) where \(\frac{1}{q}+\frac{1}{q'}=1\). Then

  1. (i)

    \(W_{\varvec{\tau }}:\Big (L^{{p'_1}}({\mathbb {R}}^d)\times \cdots \times L^{{p'_n}}({\mathbb {R}}^d)\Big )\times \Big (L^{p_1}({\mathbb {R}}^d)\times \cdots \times L^{p_n}({\mathbb {R}}^d)\Big )\rightarrow L^q({\mathbb {R}}^{2nd}) \) is continuous for \(\varvec{\tau } \in {(0,1)^d}\) and we have

    $$\begin{aligned} \Vert W_{\varvec{\tau }}(\mathbf {g},\mathbf {f})\Vert _q \le \prod _{k=1}^d\left( \frac{|1-\tau _k|^{n\left( \frac{1}{q}-\sum \limits _{j=1}^n \frac{1}{p_j}\right) }}{|\tau _k|^{n\left( 1-\frac{1}{q}-\sum \limits _{j=1}^n \frac{1}{p_j}\right) }}\right) \prod _{j=1}^n \Vert g_j\Vert _{p'_j}\Vert f_j\Vert _{p_j} \end{aligned}$$
  2. (ii)

    For \(\varvec{\tau }=\mathbf {0}\), \(W_{\mathbf {0}}:\Big (L^{{q}}({\mathbb {R}}^d)\times \cdots \times L^{{q}}({\mathbb {R}}^d)\Big )\times \Big (L^{q'}({\mathbb {R}}^d)\times \cdots \times L^{q'}({\mathbb {R}}^d)\Big )\rightarrow L^q({\mathbb {R}}^{2nd}) \) is continuous and we get the inequality

    $$\begin{aligned} \Vert W_{\mathbf {0}}(\mathbf {g},\mathbf {f})\Vert _q \le \prod _{j=1}^n \Vert g_j\Vert _{q_j}\Vert f_j\Vert _{q'_j} \end{aligned}$$
  3. (iii)

    For \(\varvec{\tau }=\mathbf {1}\), \(W_{\mathbf {1}}:\Big (L^{{q'}}({\mathbb {R}}^d)\times \cdots \times L^{{q'}}({\mathbb {R}}^d)\Big )\times \Big (L^{q}({\mathbb {R}}^d)\times \cdots \times L^{q}({\mathbb {R}}^d)\Big )\rightarrow L^q({\mathbb {R}}^{2nd}) \) is continuous, in particular

    $$\begin{aligned} \Vert W_{\mathbf {1}}(\mathbf {g},\mathbf {f})\Vert _q \le \prod _{j=1}^n \Vert g_j\Vert _{q'_j}\Vert f_j\Vert _{q_j} \end{aligned}$$

Proof

  1. (i)

    From Lemma 1, Proposition 1 and Eq. (2.6), we have

    $$\begin{aligned} \Vert W_{\varvec{\tau }}(\mathbf {g},\mathbf {f})\Vert ^q_q=\,&\left\| \left( \prod _{k=1}^{d}\frac{1}{|\tau _k|^n}\right) e^{2\pi i {\mathcal {I}}(\frac{\mathbf {1}}{\varvec{\tau }}w)\cdot x} V^{\varvec{\tau }}_{{A_{\varvec{\tau }}}\mathbf {f}} \mathbf {g}\right\| ^q_q\\ =\,&\left( \prod _{k=1}^{d}\frac{1}{|\tau _k|^{nq}}|1-\tau _k|^{n}|\tau _k|^{n}\right) \left\| V_{{A_{\varvec{\tau }}}\mathbf {f}} \mathbf {g}\right\| ^q_q\\ \le&\left( \prod _{k=1}^{d}\frac{|1-\tau _k|^{n}}{|\tau _k|^{n(q-1)}}\right) \left( \prod _{j=1}^n \Vert g_j\Vert _{p'_j}\Vert A_{\varvec{\tau }} f_j\Vert _{p_j}\right) ^q \end{aligned}$$

    And so, we get by Eq. (2.4)

    $$\begin{aligned} \Vert W_{\varvec{\tau }}(\mathbf {g},\mathbf {f})\Vert _q\le&\left( \prod _{k=1}^{d}\frac{|1-\tau _k|^{\frac{n}{q}}}{|\tau _k|^{n(1-\frac{1}{q})}}\right) \left( \prod _{j=1}^n \Vert g_j\Vert _{p'_j}\left( \prod _{k=1}^{d}\frac{|\tau _k|}{|1-\tau _k|}\right) ^{\frac{n}{p_j}}\Vert f_j\Vert _{p_j}\right) \\ =\,&\left( \prod _{k=1}^{d}\frac{|1-\tau _k|^{\frac{n}{q}}}{|\tau _k|^{n(1-\frac{1}{q})}}\right) \left( \prod _{k=1}^{d}\frac{|\tau _k|}{|1-\tau _k|}\right) ^{n\left( \sum \limits _{j=1}^n \frac{1}{p_j}\right) }\prod _{j=1}^n \Vert g_j\Vert _{p'_j}\Vert f_j\Vert _{p_j}\\ =\,&\prod _{k=1}^d\left( \frac{|1-\tau _k|^{n\left( \frac{1}{q}-\sum \limits _{j=1}^n \frac{1}{p_j}\right) }}{|\tau _k|^{n\left( 1-\frac{1}{q}-\sum \limits _{j=1}^n \frac{1}{p_j}\right) }}\right) \prod _{j=1}^n \Vert g_j\Vert _{p'_j}\Vert f_j\Vert _{p_j}. \end{aligned}$$
  2. (ii)

    Now, we will show that the multilinear Rihaczek transform is continuous. Using the equality (2.2) and the Hausdorff-Young’s inequality, we obtain

    $$\begin{aligned} \Vert W_{\mathbf {0}}(\mathbf {g},\mathbf {f})\Vert _{L^q}^{q} =\,&\int _{{\mathbb {R}}^{2nd}}|{\mathcal {R}}(\mathbf {g},\mathbf {f})(x,w)|^q dxdw\\ =\,&\int _{{\mathbb {R}}^{2nd}}\Big | g_1(x_1)\overline{\hat{f_1}(w_1)}\Big |^q \cdots \Big | g_n(x_n)\overline{\hat{f_n}(w_n)}\Big |^q dx_1dw_1 \cdots dx_ndw_n\\ =\,&\Vert g_1\overline{\hat{f_1}}\Vert ^q_{L^q}\Vert g_2\overline{\hat{f_2}}\Vert ^q_{L^q}\cdots \Vert g_n\overline{\hat{f_n}}\Vert ^q_{L^q}\\ =\,&\Vert |g_1\Vert ^q_{L^q({{\mathbb {R}}^d_x})}\Vert |\hat{f_1}\Vert ^q_{L^q({{\mathbb {R}}^d_w})}\cdots \Vert |g_n\Vert ^q_{L^q({{\mathbb {R}}^d_x})}\Vert |\hat{f_n}\Vert ^q_{L^q({{\mathbb {R}}^d_w})}\\ \le&\Vert |g_1\Vert ^q_{L^q({{\mathbb {R}}^d_x})}\Vert |f_1\Vert ^q_{L^{q'}({{\mathbb {R}}^d_t})}\cdots \Vert |g_n\Vert ^q_{L^q({{\mathbb {R}}^d_x})}\Vert |f_n\Vert ^q_{L^{q'}({{\mathbb {R}}^d_t})}. \end{aligned}$$

    Hence we get

    $$\begin{aligned} \Vert W_{\mathbf {0}}(\mathbf {g},\mathbf {f})\Vert _{L_q} \le \prod _{j=1}^n \Vert g_j\Vert _{L^q}\Vert f_j\Vert _{L^{q'}}. \end{aligned}$$
  3. (iii)

    To obtain the continuity property of the (2.3) multilinear conjugate Rihaczek transform, we will follow the same way in (ii):

    $$\begin{aligned} \Vert W_{\mathbf {1}}(\mathbf {g},\mathbf {f})\Vert ^q_{L_q}=\,&\int _{{\mathbb {R}}^{2nd}}|{\mathcal {R}}^{*}(\mathbf {g},\mathbf {f})(x,w)|^q dxdw=\,\Vert \overline{f_1}\hat{g_1}\Vert ^q_{L_q}\cdots \Vert \overline{f_n}\hat{g_n}\Vert ^q_{L_q}\\ =\,&\Vert |f_1\Vert ^q_{L^q({{\mathbb {R}}^d_x})}\Vert |\hat{g_1}\Vert ^q_{L^q({{\mathbb {R}}^d_w})}\cdots \Vert |f_n\Vert ^q_{L^q({{\mathbb {R}}^d_x})}\Vert |\hat{g_n}\Vert ^q_{L^q({{\mathbb {R}}^d_w})}\\ \le&\prod _{j=1}^n\Vert |f_j\Vert ^q_{L^q({{\mathbb {R}}^d_x})}\Vert |{g_j}\Vert ^q_{L^{q'}({{\mathbb {R}}^d_t})} \end{aligned}$$

    and so we have

    $$\begin{aligned} \Vert W_{\mathbf {1}}(\mathbf {g},\mathbf {f})\Vert _{L_q} \le \prod _{j=1}^n \Vert f_j\Vert _{L^q}\Vert g_j\Vert _{L^{q'}}. \end{aligned}$$

\(\square \)

3 Boundedness of multilinear \(\varvec{\tau }\)-Wigner transform on modulation spaces

We purpose in this section to study of boundedness of multilinear \(\varvec{\tau }\)-Wigner transform when acting on modified version of modulation spaces. To do this, we need the following features of the multilinear \(\varvec{\tau }\)-Wigner transform.

Proposition 4

Let \(f_j, g_j \in {{\mathcal {S}}({\mathbb {R}}^d)}\), \(\mathbf {f}=(f_1,\ldots ,f_n)\), \(\mathbf {g}=(g_1,\ldots ,g_n)\) and \(u_j,v_j,\eta _j,\gamma _j\in {\mathbb {R}}^d\), \(j=1,\ldots ,n\), where \(u=(u_1,\ldots ,u_n)\in {\mathbb {R}}^{nd}\) and similarly \(v,\eta ,\gamma \in {\mathbb {R}}^{nd}\).

  1. (i)

    If \(\varvec{\tau }\in {(0,1)^d}\), then we have

    $$\begin{aligned}&W_{\varvec{\tau }}\Big (T_uM_{\eta }\mathbf {g},T_vM_{\gamma }\mathbf {f}\Big )(x,w)\\&\quad =e^{2\pi i{\sum \nolimits _{j=1}^n}x_j\cdot (\eta _j-\gamma _j)}e^{2\pi i{\sum \nolimits _{j=1}^n}w_j\cdot (v_j-u_j)}e^{2\pi i{\sum \nolimits _{j=1}^n}(\gamma _j-\eta _j)\cdot (\varvec{\tau } v_j+(\mathbf {1}-\varvec{\tau })u_j)}.\\&\quad \qquad W_{\varvec{\tau }}(\mathbf {g},\mathbf {f})(x-\varvec{\tau } v-(\mathbf {1}-\varvec{\tau })u,w-\varvec{\tau } \eta -(\mathbf {1}-\varvec{\tau })\gamma ). \end{aligned}$$

    In particular, we have

    $$\begin{aligned} W_{\varvec{\tau }}(T_uM_\eta \mathbf {f})(x,w)=W_{\varvec{\tau }}\mathbf {f}(x-u,w-\eta ). \end{aligned}$$
    (3.1)
  2. (ii)

    If \(\varvec{\tau }=\mathbf {0}\), then

    $$\begin{aligned} W_{\mathbf {0}}\Big (T_uM_{\eta }\mathbf {g},T_vM_{\gamma }\mathbf {f}\Big )(x,w)=\,&{\mathcal {R}}\Big (T_uM_{\eta }\mathbf {g},T_vM_{\gamma }\mathbf {f}\Big )(x,w)\\ =\,&e^{2\pi i{\sum \nolimits _{j=1}^n}(x_j-u_j)\cdot (\eta _j-\gamma _j)}e^{2\pi i{\sum \nolimits _{j=1}^n}w_j\cdot (v_j-u_j)}\\&W_{\mathbf {0}}(\mathbf {g},\mathbf {f})\big (x-u,w-\gamma \big ). \end{aligned}$$
  3. (iii)

    If \(\varvec{\tau }=\mathbf {1}\), then we get

    $$\begin{aligned} W_{\mathbf {1}}\Big (T_uM_{\eta }\mathbf {g},T_vM_{\gamma }\mathbf {f}\Big )(x,w)=\,&{\mathcal {R}}^*\Big (T_uM_{\eta }\mathbf {g},T_vM_{\gamma }\mathbf {f}\Big )(x,w)\\ =\,&e^{-2\pi i{\sum \nolimits _{j=1}^n}(x_j-v_j)\cdot (\gamma _j-\eta _j)}e^{-2\pi i{\sum \nolimits _{j=1}^n}w_j\cdot (u_j-v_j)}\\&W_{\mathbf {1}}(\mathbf {g},\mathbf {f})\big (x-v,w-\eta \big ). \end{aligned}$$

Proof

  1. (i)

    For \(\varvec{\tau }\in {(0,1)^d},f_j, g_j \in {{\mathcal {S}}({\mathbb {R}}^d)}\) and \(u_j,v_j,\eta _j,\gamma _j\in {\mathbb {R}}^d\), \(j=1,\ldots ,n\), we have

    $$\begin{aligned}&W_{\varvec{\tau }}\Big (T_uM_{\eta }\mathbf {g},T_vM_{\gamma }\mathbf {f}\Big )(x,w)\\&=\int _{{\mathbb {R}}^{nd}}\prod _{j=1}^n \Big (T_{u_j}M_{\eta _j}{g_j}\,(x_j+\varvec{\tau } t_j)\,\overline{T_{v_j}M_{\gamma _j}{f_j}(x_j-(\mathbf {1}-\varvec{\tau })t_j)}\Big )e^{-2\pi i{\mathcal {I}}w\cdot t}dt\\&=\int _{{\mathbb {R}}^{nd}}\prod _{j=1}^n \Big (g_j(x_j-u_j+\varvec{\tau } t_j)\overline{{f_j}(x_j-v_j-(\mathbf {1}-\varvec{\tau })t_j)}\,e^{2\pi i\eta _j\cdot (x_j-u_j+\varvec{\tau } t_j)}\\&e^{-2\pi i\gamma _j\cdot (x_j-v_j-(\mathbf {1}-\varvec{\tau })t_j)}\Big )e^{-2\pi i{\mathcal {I}}w\cdot t}dt. \end{aligned}$$

    By making the substitution \(z_{jk}=x_{jk}-u_{jk}+\tau _k t_{jk}, \, j=1,\ldots ,n\), \(k=1,\ldots ,d,\) we get

    $$\begin{aligned}&W_{\varvec{\tau }}\Big (T_uM_{\eta }\mathbf {g},T_vM_{\gamma }\mathbf {f}\Big )(x,w)\\&\quad =\left( \prod _{k=1}^{d}\frac{1}{|\tau _k|^n}\right) \int _{{\mathbb {R}}^{nd}}\prod _{j=1}^n \\&\quad \quad \left[ (g_j(z_j)\overline{{f_j}\Big (x_j-v_j-\frac{(\mathbf {1}-\varvec{\tau })}{\varvec{\tau }}(z_j-x_j+u_j)\Big )}\,e^{2\pi i\eta _j\cdot z_j} \right. \\&\left. \quad \quad \times e^{-2\pi i\gamma _j\cdot (x_j-v_j-\frac{(\mathbf {1}-\varvec{\tau })}{\varvec{\tau }}(z_j-x_j+u_j))}\right] e^{-2\pi i{\mathcal {I}}(\frac{\mathbf {1}}{\varvec{\tau }}w)\cdot (z-x+u)}dz\\&\quad =\left( \prod _{k=1}^{d}\frac{1}{|\tau _k|^n}\right) e^{-2\pi i\sum \nolimits _{j=1}^n \gamma _j\cdot (\frac{\mathbf {1}}{\varvec{\tau }}x_j-v_j-\frac{\mathbf {1}-\varvec{\tau }}{\varvec{\tau }}u_j)} e^{2\pi i \sum \nolimits _{j=1}^n (\frac{\mathbf {1}}{\varvec{\tau }}w_j)\cdot (x_j-u_j)}\\&\quad \quad \times \int _{{\mathbb {R}}^{nd}}\prod _{j=1}^n \left[ g_j(z_j)\overline{A_{\varvec{\tau }}{f_j}\Big (z_j-\Big (\frac{\mathbf {1}}{\mathbf {1}-\varvec{\tau }}x_j-\frac{\varvec{\tau }}{\mathbf {1}-\varvec{\tau }}v_j-u_j\Big )\Big )} \right. \\&\left. \quad \quad \times e^{-2\pi iz_j\cdot (\frac{\mathbf {1}}{\varvec{\tau }}w_j-\eta _j-\frac{\mathbf {1}-\varvec{\tau }}{\varvec{\tau }}\gamma _j) }\right] dz\\&\quad =\left( \prod _{k=1}^{d}\frac{1}{|\tau _k|^n}\right) e^{-2\pi i\sum \nolimits _{j=1}^n \gamma _j\cdot (\frac{\mathbf {1}}{\varvec{\tau }}x_j-v_j-\frac{\mathbf {1}-\varvec{\tau }}{\varvec{\tau }}u_j)} e^{2\pi i \sum \nolimits _{j=1}^n (\frac{\mathbf {1}}{\varvec{\tau }}w_j)\cdot (x_j-u_j)}\\&\quad \quad \times \prod _{j=1}^n V_{A_{\varvec{\tau }}f_j}g_j\left( \frac{\mathbf {1}}{\mathbf {1}-\varvec{\tau }}(x_j-\varvec{\tau } v_j-(\mathbf {1}-\varvec{\tau })u_j),\frac{\mathbf {1}}{\varvec{\tau }}(w_j-\varvec{\tau }\eta _j-(\mathbf {1}-\varvec{\tau })\gamma _j)\right) . \end{aligned}$$

    Now if Lemma 1 is applied, we obtain

    $$\begin{aligned}&W_{\varvec{\tau }}\Big (T_uM_{\eta }\mathbf {f},T_vM_{\gamma }\mathbf {g}\Big )(x,w)\\&\quad =\left( \prod _{k=1}^{d}\frac{1}{|\tau _k|^n}\right) e^{-2\pi i\sum \nolimits _{j=1}^n \gamma _j\cdot (\frac{\mathbf {1}}{\varvec{\tau }}x_j-v_j-\frac{\mathbf {1}-\varvec{\tau }}{\varvec{\tau }}u_j)} e^{2\pi i \sum \nolimits _{j=1}^n (\frac{\mathbf {1}}{\varvec{\tau }}w_j)\cdot (x_j-u_j)}\\&\quad \quad \times \left( \prod _{k=1}^{d}{|\tau _k|^n}\right) e^{-2\pi i\sum \nolimits _{j=1}^n \frac{\mathbf {1}}{\varvec{\tau }}(x_j-\varvec{\tau } v_j-(\mathbf {1}-\varvec{\tau })u_j)\cdot (w_j-\varvec{\tau }\eta _j-(\mathbf {1}-\varvec{\tau })\gamma _j)} \\&\quad \quad \times \prod _{j=1}^n \Big (W_{\varvec{\tau }}(g_j,f_j)(x_j-\varvec{\tau } v_j-(\mathbf {1}-\varvec{\tau })u_j,w_j-\varvec{\tau }\eta _j-(\mathbf {1}-\varvec{\tau })\gamma _j)\Big )\\&\quad = e^{2\pi i\sum \nolimits _{j=1}^n x_j\cdot (\eta _j-\gamma _j)} e^{2\pi i \sum \nolimits _{j=1}^n w_j\cdot (v_j-u_j)} e^{2\pi i \sum \nolimits _{j=1}^n (\gamma _j-\eta _j)\cdot (\varvec{\tau } v_j+(\mathbf {1}-\varvec{\tau })u_j)} \\&\quad \quad \times W_{\varvec{\tau }}(\mathbf {g},\mathbf {f})(x-\varvec{\tau } v-(\mathbf {1}-\varvec{\tau })u,w-\varvec{\tau } \eta -(\mathbf {1}-\varvec{\tau })\gamma ). \end{aligned}$$

    From here, we also get (3.1) as follows

    $$\begin{aligned} W_{\varvec{\tau }}(T_uM_\eta \mathbf {f})(x,w)=\,&W_{\varvec{\tau }}(T_uM_\eta \mathbf {f},T_uM_\eta \mathbf {f})(x,w) \\ =\,&W_{\varvec{\tau }}(\mathbf {f},\mathbf {f})(x-\varvec{\tau } u-(\mathbf {1}-\varvec{\tau })u,w-\varvec{\tau } \eta -(\mathbf {1}-\varvec{\tau })\eta )\\ =\,&W_{\varvec{\tau }}\mathbf {f}(x-u,w-\eta ). \end{aligned}$$
  2. (ii)

    Let \(\varvec{\tau }=\mathbf {0}\). For \(u_j,v_j,\eta _j,\gamma _j\in {\mathbb {R}}^d,\, j=1,\ldots ,n\), let us apply the equality \(\big (T_{v_j}M_{\gamma _j}f_j\big )^\wedge =e^{-2\pi i {v_j}\cdot {\gamma _j}}T_{\gamma _j}M_{-v_j}\hat{f_j}\) to (2.2), we conclude that

    $$\begin{aligned}&W_{\mathbf {0}}\Big (T_uM_{\eta }\mathbf {g},T_vM_{\gamma }\mathbf {f}\Big )(x,w)\\&\quad = e^{-2\pi i{\sum \nolimits _{j=1}^n}x_j\cdot w_j}\prod _{j=1}^n(T_{u_j}M_{\eta _j}g_j)(x_j)\overline{\big (T_{v_j}M_{\gamma _j}f_j\big )^\wedge (w_j)}\\&\quad =e^{-2\pi i{\sum \nolimits _{j=1}^n}x_j\cdot w_j}e^{2\pi i{\sum \nolimits _{j=1}^n}\eta _j\cdot (x_j-u_j)}e^{2\pi i{\sum \nolimits _{j=1}^n}v_j\cdot w_j}\prod _{j=1}^ng_j(x_j-u_j)\overline{\hat{f_j}(w_j-\gamma _j)}\\&\quad =e^{2\pi i{\sum \nolimits _{j=1}^n}(-x_j\cdot \gamma _j-u_j\cdot w_j+u_j\cdot \gamma _j)}e^{2\pi i{\sum \nolimits _{j=1}^n}\eta _j\cdot (x_j-u_j)}e^{2\pi i{\sum \nolimits _{j=1}^n}v_j\cdot w_j}\\&\quad \quad \times e^{-2\pi i{\sum \nolimits _{j=1}^n}(x_j-u_j)\cdot (w_j-\gamma _j)}\prod _{j=1}^ng_j(x_j-u_j)\overline{\hat{f_j}(w_j-\gamma _j)}\\&\quad =e^{2\pi i{\sum \nolimits _{j=1}^n}(x_j-u_j)\cdot (\eta _j-\gamma _j)}e^{2\pi i{\sum \nolimits _{j=1}^n}w_j\cdot (v_j-u_j)} W_{\mathbf {0}}(\mathbf {g},\mathbf {f})\big (x-u,w-\gamma \big ). \end{aligned}$$
  3. (iii)

    For \(\varvec{\tau }=\mathbf {1}\), by using the same arguments in (ii), we obtain by (2.3)

    $$\begin{aligned}&W_{\mathbf {1}}\Big (T_uM_{\eta }\mathbf {g},T_vM_{\gamma }\mathbf {f}\Big )(x,w)\\&\quad = e^{2\pi i{\sum \nolimits _{j=1}^n}x_j\cdot w_j}\prod _{j=1}^n\overline{\big (T_{v_j}M_{\gamma _j}f_j\big )(x_j)}(T_{u_j}M_{\eta _j}g_j)^\wedge (w_j)\\&\quad =e^{2\pi i{\sum \nolimits _{j=1}^n}x_j\cdot w_j}e^{-2\pi i{\sum \nolimits _{j=1}^n}\gamma _j\cdot (x_j-v_j)}e^{-2\pi i{\sum \nolimits _{j=1}^n}u_j\cdot w_j}\prod _{j=1}^n\overline{f_j(x_j-v_j)}{\hat{g_j}(w_j-\eta _j)}\\&\quad =e^{-2\pi i{\sum \nolimits _{j=1}^n}(-x_j\cdot \eta _j-v_j\cdot w_j+v_j\cdot \eta _j)}e^{-2\pi i{\sum \nolimits _{j=1}^n}\gamma _j\cdot (x_j-v_j)}e^{-2\pi i{\sum \nolimits _{j=1}^n}u_j\cdot w_j}\\&\quad \quad \times e^{2\pi i{\sum \nolimits _{j=1}^n}(x_j-v_j)\cdot (w_j-\eta _j)}\prod _{j=1}^n\overline{f_j(x_j-v_j)}{\hat{g_j}(w_j-\eta _j)}\\&\quad =e^{-2\pi i{\sum \nolimits _{j=1}^n}(x_j-v_j)\cdot (\gamma _j-\eta _j)}e^{-2\pi i{\sum \nolimits _{j=1}^n}w_j\cdot (u_j-v_j)}W_{\mathbf {1}}(\mathbf {g},\mathbf {f})\big (x-v,w-\eta \big ). \end{aligned}$$

\(\square \)

Now, we are able to find the short-time Fourier transform of the multilinear \(\varvec{\tau }\)-Wigner transform. In the remainder of this section, unless otherwise stated, we assume that \(z=(z_1,z_2)\in {{\mathbb {R}}^{2nd}}, \xi =(\xi _1,\xi _2)\in {{\mathbb {R}}^{2nd}}\) and \(z_{1_{j}},z_{2_{j}},\xi _{1_{j}},\xi _{2_{j}}\in {{\mathbb {R}}^{d}}\) for \(j=1,\ldots ,n\).

Proposition 5

  1. (i)

    If \(f_j, g_j \in {{\mathcal {S}}({\mathbb {R}}^d)}\) and \(\varvec{\tau }\in {[0,1]^d}\) then \(W_{\varvec{\tau }}(\mathbf {g},\mathbf {f})\in {{\mathcal {S}}({\mathbb {R}}^{2nd})}\).

  2. (ii)

    Let \(\varphi _j \in {{\mathcal {S}}({\mathbb {R}}^d)}\) and set \(\phi =W_{\varvec{\tau }}(\varvec{\varphi },\varvec{\varphi })=W_{\varvec{\tau }}(\varvec{\varphi })\in {{\mathcal {S}}({\mathbb {R}}^{2nd})}\). For \(\varvec{\tau }\in {(0,1)^d}\), we have

    $$\begin{aligned}&V_{W_{\varvec{\tau }}(\varvec{\varphi },\varvec{\varphi })}W_{\varvec{\tau }}(\mathbf {g},\mathbf {f})(z,\xi )\\&\quad =e^{-4\pi i \sum \nolimits _{j=1}^n (\varvec{\tau }z_{2_{j}})\cdot \xi _{2_{j}}}\\&\quad \quad V_{\varvec{\varphi }}\mathbf {g}\,(z_1-\varvec{\tau } \xi _2,z_2+(\mathbf {1}-\varvec{\tau }) \xi _1)\overline{V_{\varvec{\varphi }}\mathbf {f}\,(z_1+\varvec{\tau } \xi _2,z_2-(\mathbf {1}-\varvec{\tau }) \xi _1)}. \end{aligned}$$
  3. (iii)

    For \(\varvec{\tau }=\mathbf {0}\), \(W_{\mathbf {0}}(\varvec{\varphi },\varvec{\varphi })=W_{\mathbf {0}}(\varvec{\varphi })={\mathcal {R}}(\varphi )\in {{\mathcal {S}}({\mathbb {R}}^{2nd})}\) and \(W_{\mathbf {0}}(\mathbf {g},\mathbf {f})={\mathcal {R}}(\mathbf {g},\mathbf {f})\), we have

    $$\begin{aligned} V_{W_{\mathbf {0}}(\varvec{\varphi })}W_{\mathbf {0}}(\mathbf {g},\mathbf {f})(z,\xi )=e^{-2\pi i \sum \nolimits _{j=1}^n z_{2_j}\xi _{2_j}} V_{\varvec{\varphi }}\mathbf {g}\,(z_1,z_2+ \xi _1)\overline{V_{\varvec{\varphi }}\mathbf {f}\,(z_1+ \xi _2,z_2)}. \end{aligned}$$
  4. (iv)

    For \(\varvec{\tau }=\mathbf {1}\), \(\phi =W_{\mathbf {1}}(\varvec{\varphi },\varvec{\varphi })=W_{\mathbf {1}}(\varvec{\varphi })={\mathcal {R}}^*(\varvec{\varphi })\in {{\mathcal {S}}({\mathbb {R}}^{2nd})}\), we have

    $$\begin{aligned} V_{W_{\mathbf {1}}(\varvec{\varphi })}W_{\mathbf {1}}(\mathbf {g},\mathbf {f})(z,\xi )=e^{-2\pi i \sum \nolimits _{j=1}^n z_{2_j}\xi _{2_j}} V_{\varvec{\varphi }}\mathbf {g}\,(z_1-\xi _2,z_2)\overline{V_{\varvec{\varphi }}\mathbf {f}\,(z_1,z_2-\xi _1)}. \end{aligned}$$

Proof

  1. (i)

    By using the Theorem 11.2.5 in [13] and Lemma 1, we obtain \(W_{\varvec{\tau }}(\mathbf {g},\mathbf {f})\in {{\mathcal {S}}({\mathbb {R}}^{2nd})}\) for \(\varvec{\tau }\in (0,1)^d\).

    If \(\varvec{\tau }=\mathbf {0}\), let \(\mathbf {g}\otimes \mathbf {f}\) be the tensor product

    $$\begin{aligned} (\mathbf {g}\otimes \mathbf {f})(x,t)=\mathbf {g}(x) \mathbf {f}(t)=\,&g_1(x_1)\cdots g_n(x_n)f_1(t_1)\cdots f_n(t_n)\\ =&\prod _{j=1}^n g_j(x_j)f_j(t_j), \end{aligned}$$

    where \(x=(x_1,\ldots , x_n), t=(t_1,\ldots ,t_n), x_j, t_j\in {{\mathbb {R}}^d}, j=1,\ldots ,n,\) and set \({\mathcal {T}}_a F(x,t)=F(x,x-t)\). Then we write

    $$\begin{aligned}&W_{\mathbf {0}}(\mathbf {g},\mathbf {f})(x,w)\\&\quad =\int _{{\mathbb {R}}^{nd}}\prod _{j=1}^n \Big ((g_j(x_j)\overline{{f_j}(x_j-t_j)}\Big )e^{-2\pi i{\mathcal {I}}w\cdot t}dt\\&\quad =\int _{{\mathbb {R}}^{nd}}({g}_1\otimes \overline{f_1})(x_1,x_1-t_1)e^{-2\pi i w_1\cdot t_1}\\&\quad \quad \cdots ({g}_n\otimes \overline{f_n})(x_n,x_n-t_n)e^{-2\pi i w_n\cdot t_n}dt_1 \cdots dt_n\\&\quad =\int _{{\mathbb {R}}^{nd}}{\mathcal {T}}_a({g}_1\otimes \overline{f_1})(x_1,t_1)e^{-2\pi i w_1\cdot t_1} \\&\quad \quad \cdots {\mathcal {T}}_a({g}_n\otimes \overline{f_n})(x_n,t_n)e^{-2\pi i w_n\cdot t_n}dt_1 \cdots dt_n\\&\quad ={\mathcal {F}}_2{\mathcal {T}}_a({g}_1\otimes \overline{f_1})(x_1,w_1) \cdots {\mathcal {F}}_2{\mathcal {T}}_a({g}_n\otimes \overline{f_n})(x_n,w_n)\\&\quad ={\mathcal {F}}_2{\mathcal {T}}_a({\mathbf {g}}\otimes \varvec{\overline{f}})(x,w) \end{aligned}$$

    where \({\mathcal {F}}_2\) is the Fourier transform with respect to \(t_j, \, j=1,\ldots ,n\). Since \({g}_j\otimes \overline{f_j}\in {{\mathcal {S}}({\mathbb {R}}^{2d})}\) and \({{\mathcal {S}}({\mathbb {R}}^{2d})}\) is invariant under the transformation and the Fourier transform, then we have \(W_{\mathbf {0}}(\mathbf {g},\mathbf {f})\in {{\mathcal {S}}({\mathbb {R}}^{2nd})}\).

    For \(\varvec{\tau }=\mathbf {1}\), if we set \({\mathcal {T}}_b F(x,t)=F(x+t,x)\) we obtain

    $$\begin{aligned} W_{\mathbf {1}}(\mathbf {g},\mathbf {f}) (x,w)={\mathcal {R}}^*(\mathbf {g},\mathbf {f})=\overline{{\mathcal {R}}(\mathbf {g},\mathbf {f})} (x,w)={\mathcal {F}}_2 {\mathcal {T}}_b ({\mathbf {g}}\otimes \varvec{\overline{f}})(x,w) \end{aligned}$$

    and \(W_{\mathbf {1}}(\mathbf {g},\mathbf {f})\in {{\mathcal {S}}({\mathbb {R}}^{2nd})}\).

  2. (v)

    Let \(\varphi _j, g_j,f_j \in {{\mathcal {S}}({\mathbb {R}}^d)}, \, j=1,\ldots ,n, \, \varvec{\tau }\in {(0,1)^d}\). By using Lemma 1 and the equality (3.1), we write

    $$\begin{aligned}&V_{W_{\varvec{\tau }}(\varvec{\varphi },\varvec{\varphi })}W_{\varvec{\tau }}(\mathbf {g},\mathbf {f})\big ((z_1,z_2)(\xi _1,\xi _2)\big ) \nonumber \\&\quad =\int _{{\mathbb {R}}^{2nd}}W_{\varvec{\tau }}(\mathbf {g},\mathbf {f})(x,w) \overline{M_{(\xi _1,\xi _2)}T_{(z_1,z_2)}W_{\varvec{\tau }}(\varvec{\varphi },\varvec{\varphi })(x,w)}dxdw \nonumber \\&\quad =\int _{{\mathbb {R}}^{2nd}}W_{\varvec{\tau }}(\mathbf {g},\mathbf {f})(x,w)e^{-2\pi i \sum \nolimits _{j=1}^n (x_j\cdot \xi _{1_j}+w_j\cdot \xi _{2_j})} \overline{W_{\varvec{\tau }}(\varvec{\varphi },\varvec{\varphi })(x-z_1,w-z_2)}dxdw\nonumber \\&\quad =\left( \prod _{k=1}^{d}\frac{1}{|\tau _k|^n}\right) \int _{{\mathbb {R}}^{2nd}}e^{2\pi i \sum \nolimits _{j=1}^n (\frac{\mathbf {1}}{\varvec{\tau }}x_j)\cdot w_j}V_{A_{\varvec{\tau }}\mathbf {f}}\,\mathbf {g}\left( \frac{\mathbf {1}}{\mathbf {1}-\varvec{\tau }}x,\frac{\mathbf {1}}{\varvec{\tau }}w\right) \nonumber \\&\quad \quad \times e^{-2\pi i \sum \nolimits _{j=1}^n (x_j\cdot \xi _{1_j}+w_j\cdot \xi _{2_j})} \overline{W_{\varvec{\tau }}(T_{z_1}M_{z_2}\varvec{\varphi })(x,w)}dxdw \nonumber \\&\quad =\left( \prod _{k=1}^{d}\frac{1}{|\tau _k|^{2n}}\right) \int _{{\mathbb {R}}^{2nd}}V_{A_{\varvec{\tau }}\mathbf {f}}\,\mathbf {g}\left( \frac{\mathbf {1}}{\mathbf {1}-\varvec{\tau }}x,\frac{\mathbf {1}}{\varvec{\tau }}w\right) e^{-2\pi i \sum \nolimits _{j=1}^n (x_j\cdot \xi _{1_j}+w_j\cdot \xi _{2_j})} \nonumber \\&\quad \quad \times \overline{V_{A_{\varvec{\tau }}(T_{z_1}M_{z_2}\varvec{\varphi })}(T_{z_1}M_{z_2}\varvec{\varphi })\left( \frac{\mathbf {1}}{\mathbf {1}-\varvec{\tau }}x,\frac{\mathbf {1}}{\varvec{\tau }}w\right) }dxdw \nonumber \\&\quad =\left( \prod _{k=1}^{d}\frac{|1-\tau _k|^n}{|\tau _k|^n}\right) \int _{{\mathbb {R}}^{2nd}}V_{A_{\varvec{\tau }}\mathbf {f}}\,\mathbf {g}(x,w)e^{-2\pi i \sum \nolimits _{j=1}^n ((\mathbf {1}-\varvec{\tau })x_j\cdot \xi _{1_j}+(\varvec{\tau } w_j)\cdot \xi _{2_j})}\nonumber \\&\quad \quad \times \overline{V_{A_{\varvec{\tau }}(T_{z_1}M_{z_2}\varvec{\varphi })}(T_{z_1}M_{z_2}\varvec{\varphi })(x,w)}dxdw \end{aligned}$$
    (3.2)

    Now, we need the following calculation to continue:

    $$\begin{aligned}&e^{-2\pi i \sum \nolimits _{j=1}^n ((\mathbf {1}-\varvec{\tau })x_j\cdot \xi _{1_j}+(\varvec{\tau } w_j)\cdot \xi _{2_j})}V_{A_{\varvec{\tau }}\mathbf {f}}\,\mathbf {g}(x,w)\nonumber \\&\quad = e^{-2\pi i \sum \nolimits _{j=1}^n ((\mathbf {1}-\varvec{\tau })x_j\cdot \xi _{1_j}+(\varvec{\tau } w_j)\cdot \xi _{2_j})}\left<\prod _{j=1}^n g_j,\prod _{j=1}^n M_{w_j}T_{x_j}A_{\varvec{\tau }}f_j \right> \nonumber \\&\quad =\left<\prod _{j=1}^n g_j,\prod _{j=1}^n e^{2\pi i ((\mathbf {1}-\varvec{\tau })x_j\cdot \xi _{1_j}+(\varvec{\tau } w_j)\cdot \xi _{2_j})}M_{w_j}T_{x_j}A_{\varvec{\tau }}f_j \right> \nonumber \\&\quad =\left<\prod _{j=1}^n g_j,\prod _{j=1}^n M_{(\mathbf {1}-\varvec{\tau })\xi _{1_j}}T_{-\varvec{\tau } \xi _{2_j}}M_{w_j} T_{x_j} T_{\varvec{\tau } \xi _{2_j}}M_{-(\mathbf {1}-\varvec{\tau })\xi _{1_j}} A_{\varvec{\tau }}f_j \right> \nonumber \\&\quad =\left<\prod _{j=1}^n T_{\varvec{\tau } \xi _{2_j}}M_{-(\mathbf {1}-\varvec{\tau })\xi _{1_j}}g_j,\prod _{j=1}^n M_{w_j} T_{x_j} T_{\varvec{\tau } \xi _{2_j}}M_{-(\mathbf {1}-\varvec{\tau })\xi _{1_j}} A_{\varvec{\tau }}f_j \right> \nonumber \\&\quad =V_{\prod \limits _{j=1}^n T_{\varvec{\tau } \xi _{2_j}}M_{-(\mathbf {1}-\varvec{\tau })\xi _{1_j}}A_{\varvec{\tau }}f_j }\left( \prod \limits _{j=1}^n T_{\varvec{\tau } \xi _{2_j}}M_{-(\mathbf {1}-\varvec{\tau })\xi _{1_j}}g_j \right) (x_j,w_j) \end{aligned}$$
    (3.3)

    Replacing (3.3) in (3.2) and using the orthogonality relations (Theorem 3.2.1 in [13]), we obtain

    (3.4)

    The first factor on the right side of (3.4) is

    $$\begin{aligned}&\prod \limits _{j=1}^n \Big< T_{\varvec{\tau } \xi _{2_j}}M_{-(\mathbf {1}-\varvec{\tau })\xi _{1_j}}g_j, T_{z_{1_j}}M_{z_{2_j}}\varphi _j\Big> \nonumber \\&\quad = \prod \limits _{j=1}^n \Big < g_j, M_{(\mathbf {1}-\varvec{\tau })\xi _{1_j}}T_{z_{1_j}-\varvec{\tau } \xi _{2_j}}M_{z_{2_j}}\varphi _j\Big > \nonumber \\&\quad =\prod \limits _{j=1}^n e^{2\pi i z_{2_j}\cdot ({z_{1_j}-\varvec{\tau } \xi _{2_j}})} V_{\varphi _j}g_j (z_{1_j}-\varvec{\tau } \xi _{2_j},z_{2_j}+(\mathbf {1}-\varvec{\tau })\xi _{1_j})\nonumber \\&\quad =e^{2\pi i \sum \nolimits _{j=1}^n z_{2_j}\cdot ({z_{1_j}-\varvec{\tau } \xi _{2_j}})} V_{\varvec{\varphi }}\,\mathbf {g}\, (z_{1}-\varvec{\tau } \xi _{2},z_{2}-(\mathbf {1}-\varvec{\tau })\xi _{1}) \end{aligned}$$
    (3.5)

    and the second factor on the right side of (3.4) is

    $$\begin{aligned}&\prod \limits _{j=1}^n \Big<T_{\varvec{\tau } \xi _{2_j}}M_{-(\mathbf {1}-\varvec{\tau })\xi _{1_j}}A_{\varvec{\tau }}f_j,A_{\varvec{\tau }}(T_{z_{1_j}}M_{z_{2_j}}\varphi _j)\Big>\nonumber \\&\quad =\prod \limits _{j=1}^n \Big <A_{\varvec{\tau }}(T_{-\varvec{\tau } \xi _{2_j}}M_{(\mathbf {1}-\varvec{\tau })\xi _{1_j}}f_j),A_{\varvec{\tau }}(T_{z_{1_j}}M_{z_{2_j}}\varphi _j)\Big >\nonumber \\&\quad =\prod \limits _{j=1}^n \int _{{\mathbb {R}}^{d}} (T_{-\varvec{\tau } \xi _{2_j}}M_{\left( \mathbf {1}-\varvec{\tau }\right) \xi _{1_j}}f_j)^{\sim }\left( \frac{\mathbf {1}-\varvec{\tau }}{\varvec{\tau }}x_j\right) \overline{(T_{z_{1_j}}M_{z_{2_j}}\varphi _j)^\sim \left( \frac{\mathbf {1}-\varvec{\tau }}{\varvec{\tau }}x_j\right) }dx_j\nonumber \\&\quad =\left( \prod _{k=1}^{d}{\Big |\frac{\tau _k}{1-\tau _k}\Big |^n}\right) \prod \limits _{j=1}^n e^{2\pi i z_{2_j}\cdot ({z_{1_j}+\varvec{\tau } \xi _{2_j}})}\nonumber \\&\quad \quad \times \int _{{\mathbb {R}}^{d}} f_j(x_j)\overline{\varphi _j(x_j-({z_{1_j}+\varvec{\tau } \xi _{2_j}}))} e^{-2\pi i x_j\cdot (z_{2_j}-(\mathbf {1}-\varvec{\tau })\xi _{1_j})}dx_j\nonumber \\&\quad =\left( \prod _{k=1}^{d}{\Big |\frac{\tau _k}{1-\tau _k}\Big |^n}\right) e^{2\pi i \sum \nolimits _{j=1}^n z_{2_j}\cdot ({z_{1_j}+\varvec{\tau } \xi _{2_j}})} V_{\varvec{\varphi }}\,\mathbf {f}\, (z_{1}+\varvec{\tau } \xi _{2},z_{2}-(\mathbf {1}-\varvec{\tau })\xi _{1}) \end{aligned}$$
    (3.6)

    Hence, by (3.4), (3.5) and (3.6) we obtain

    $$\begin{aligned}&V_{W_{\varvec{\tau }}(\varvec{\varphi },\varvec{\varphi })}W_{\varvec{\tau }}(\mathbf {g},\mathbf {f})\big ((z_1,z_2)(\xi _1,\xi _2)\big )\\&\quad =\left( \prod _{k=1}^{d}\Big |\frac{1-\tau _k}{\tau _k}\Big |^{n}\right) e^{2\pi i \sum \nolimits _{j=1}^n z_{2_j}\cdot ({z_{1_j}-\varvec{\tau } \xi _{2_j}})} V_{\varvec{\varphi }}\,\mathbf {g}\, (z_{1}-\varvec{\tau } \xi _{2},z_{2}+(\mathbf {1}-\varvec{\tau })\xi _{1})\\&\quad \quad \times \left( \prod _{k=1}^{d}\Big |\frac{\tau _k}{1-\tau _k}\Big |^{n}\right) e^{-2\pi i \sum \nolimits _{j=1}^n z_{2_j}\cdot ({z_{1_j}+\varvec{\tau } \xi _{2_j}})} \overline{V_{\varvec{\varphi }}\,\mathbf {f}\, (z_{1}+\varvec{\tau } \xi _{2},z_{2}-(\mathbf {1}-\varvec{\tau })\xi _{1})}\\&\quad =e^{-4\pi i \sum \nolimits _{j=1}^n (\varvec{\tau }z_{2_j})\cdot \xi _{2_j}} V_{\varvec{\varphi }}\mathbf {g}\,(z_1-\varvec{\tau } \xi _2,z_2+(\mathbf {1}-\varvec{\tau }) \xi _1)\\&\quad \quad \times \overline{V_{\varvec{\varphi }}\mathbf {f}\,(z_1+\varvec{\tau } \xi _2,z_2-(\mathbf {1}-\varvec{\tau }) \xi _1)}. \end{aligned}$$
  3. (vi)

    If \(\varvec{\tau }=\mathbf {0}\), by using the equality \(V_gf(x,w)=e^{-2\pi i x\cdot w}V_{\hat{g}}{\hat{f}}(w,-x) \) we have

    $$\begin{aligned} V_{W_{\mathbf {0}}(\varvec{\varphi })}W_{\mathbf {0}}(\mathbf {g},\mathbf {f})(z,\xi )&=\int _{{\mathbb {R}}^{2nd}}W_{\mathbf {0}} (\mathbf {g},\mathbf {f})(x,w)\overline{M_\xi T_zW_{\mathbf {0}} {\varvec{\varphi }(x,w)}}dx dw\\&=\int _{{\mathbb {R}}^{2nd}}e^{2\pi i {\mathcal {I}} x\cdot w}\left( \prod \limits _{j=1}^n g_j(x_j) \overline{\hat{f_j}(w_j)}\right) e^{-2\pi i {\mathcal {I}} (x\cdot \xi _1+w\cdot \xi _2)}\\&\quad \times W_{\mathbf {0}} \varvec{\varphi }(x-z_1,w-z_2)dx dw\\&=\int _{{\mathbb {R}}^{2nd}}\left( \prod \limits _{j=1}^n g_j(x_j) \overline{\hat{f_j}(w_j)}\overline{\varphi _j(x_j-z_{{1}_j})}\hat{\varphi }_j(w_j-z_{{2}_j})\right) \\&\quad \times e^{-2\pi i {\mathcal {I}} (x\cdot w+x\cdot \xi _1+w\cdot \xi _2)}e^{2\pi i {\mathcal {I}} (x-z_1)\cdot (w-z_2)}dx dw\\&=\int _{{\mathbb {R}}^{nd}}\left( \int _{{\mathbb {R}}^{nd}}\left( \prod \limits _{j=1}^n g_j(x_j)\overline{\varphi _j(x_j-z_{{1}_j})} e^{-2\pi i (z_{2_{j}}+\xi _{1_{j}})\cdot x_j}\right) dx\right) \\&\quad \times e^{-2\pi i {\mathcal {I}} (w\cdot \xi _2+w\cdot z_1-z_1\cdot z_2)}\left( \prod \limits _{j=1}^n \overline{\hat{f_j}(w_j)}\hat{\varphi }_j(w_j-z_{{2}_j})\right) dw\\&=e^{2\pi i \sum \nolimits _{j=1}^n z_{1_j}\cdot z_{2_j}}V_{\varvec{\varphi }}\mathbf {g}(z_1,z_2+\xi _1)\\&\quad \times \overline{\int _{{\mathbb {R}}^{nd}}\left( \prod \limits _{j=1}^n \hat{f_j}(w_j)\overline{\hat{\varphi }_j(w_j-z_{{2}_j})}e^{2\pi i w_j\cdot (z_{{1}_j}+\xi _{{2}_j})}\right) dw }\\&=e^{2\pi i \sum \nolimits _{j=1}^n z_{1_j}\cdot z_{2_j}}V_{\varvec{\varphi }}\mathbf {g}(z_1,z_2+\xi _1)\overline{\prod \limits _{j=1}^n V_{\hat{\varphi _j}}\hat{f_j}(z_{{2}_j},-z_{{1}_j}-\xi _{{2}_j})}\\&=e^{-2\pi i \sum \nolimits _{j=1}^n z_{2_j}\cdot \xi _{2_j}}V_{\varvec{\varphi }}\mathbf {g}(z_1,z_2+\xi _1)\overline{ V_{\varvec{\varphi }}\mathbf {f}(z_1+\xi _2,z_2)}. \end{aligned}$$
  4. (vii)

    Similar to (iii), we have the following calculation:

    $$\begin{aligned}&V_{W_{\mathbf {1}}\varvec{\varphi }}W_{\mathbf {1}}(\mathbf {g},\mathbf {f})(z,\xi )\\&\quad =\int _{{\mathbb {R}}^{2nd}} e^{-2\pi i {\mathcal {I}}x\cdot w }\left( \prod \limits _{j=1}^n \overline{{f_j}(x_j)}\hat{g_j} (w_j)\right) \overline{W_{\mathbf {1}}\varvec{\varphi } (x-z_1,w-z_2)}\\&\quad \quad \times e^{-2\pi i {\mathcal {I}}(x\cdot \xi _1+w\cdot \xi _2)}dx dw\\&\quad =\int _{{\mathbb {R}}^{2nd}} e^{2\pi i {\mathcal {I}}x\cdot w }\left( \prod \limits _{j=1}^n \overline{{f_j}(x_j)}\hat{g_j} (w_j)\right) e^{-2\pi i {\mathcal {I}}(x-z_1)\cdot (w-z_2)}\\&\quad \quad \times \left( \overline{\prod \limits _{j=1}^n \overline{\varphi _j(x_j-z_{1_{j}})}\hat{\varphi _j}(w_j-z_{2_{j}})}\right) e^{-2\pi i {\mathcal {I}}(x\cdot \xi _1+w\cdot \xi _2)}dx dw\\&\quad =e^{-2\pi i {\mathcal {I}} z_1\cdot z_2}\int _{{\mathbb {R}}^{nd}}\left( \overline{\int _{{\mathbb {R}}^{nd}}\left( \prod \limits _{j=1}^n {f_j(x_j)\overline{\varphi _j(x_j-z_{{1}_j)}}}e^{-2\pi i x_j \cdot (z_{{2}_j}+\xi _{{1}_j})}\right) dx }\right) \\&\quad \quad \times \prod \limits _{j=1}^n \hat{g_j}(w_j)\overline{\hat{\varphi }_j(w_j-z_{2_{j}})}e^{-2\pi i w_j\cdot (\xi _{{2}_j}-z_{{1}_j})}dw\\&\quad =e^{-2\pi i {\mathcal {I}} z_1\cdot z_2}\overline{V_{\varvec{\varphi }}\mathbf {f}(z_1,z_2-\xi _1)}\prod \limits _{j=1}^n V_{\hat{\varphi _j}}\hat{g_j}(z_{{2}_j},\xi _{{2}_j}-z_{{1}_j})\\&\quad =e^{-2\pi i {\mathcal {I}} z_2\cdot \xi _2}\overline{V_{\varvec{\varphi }}\mathbf {f}(z_1,z_2-\xi _1)}V_{\varvec{\varphi }}\mathbf {g}(z_1-\xi _2,z_2). \end{aligned}$$

\(\square \)

The continuity of multilinear \(\varvec{\tau }\)-Wigner transform is established by our next theorem. In this theorem, we will study on Teofanov’s modified version of modulation spaces \({{\mathcal {M}}_{s,t}^{p,q}({\mathbb {R}}^{nd})}\), where the weight function is in the form \({m(z,\xi )={\left\langle z \right\rangle }^t{\left\langle \xi \right\rangle }^s},\) \(s,t\in {\mathbb {R}}\). Our theorem are stated and proved on all modulation spaces without weights \({{\mathcal {M}}^{p,q}({\mathbb {R}}^{nd})}\), \(p,q \in {[1,\infty ]}\), for \(\varvec{\tau } \in {(0,1)^d}\). If \(\varvec{\tau }=\mathbf {0}\) and \(\varvec{\tau }=\mathbf {1}\), the boundedness property holds on all modulation spaces with polynomial weights \({{\mathcal {M}}_{s,0}^{p,q}({\mathbb {R}}^{nd})}\), \(p,q \in {[1,\infty ]}\), where the weight function is the function \(m_s(\xi )={\left\langle \xi \right\rangle }^s\) which is the special case of the function \({m(z,\xi )}\) for \(t=0\). Also we note that \(m_s\) is a \(m_{|s|}\)-moderate weight for \(s\in {\mathbb {R}}\).

Theorem 2

Let \(p_k,q_k,p,q \in {[1,\infty ]}\) satisfy the hypotheses of

$$\begin{aligned} p_k,q_k\le q, \quad k=1,2 \end{aligned}$$

and

$$\begin{aligned} \frac{1}{p_1}+ \frac{1}{p_2}\ge \frac{1}{p}+ \frac{1}{q}, \quad \frac{1}{q_1}+ \frac{1}{q_2}\ge \frac{1}{p}+ \frac{1}{q}. \end{aligned}$$
  1. (i)

    If \(\varvec{\tau } \in {(0,1)^d}\), \(\mathbf {g}\in {{\mathcal {M}}^{p_1,q_1}({\mathbb {R}}^{nd})}\) and \(\mathbf {f}\in {{\mathcal {M}}^{p_2,q_2}({\mathbb {R}}^{nd})}\), then the multilinear \(\varvec{\tau }\)-Wigner transform is continuous map from \({{\mathcal {M}}^{p_1,q_1}({\mathbb {R}}^{nd})}\times {{\mathcal {M}}^{p_2,q_2}({\mathbb {R}}^{nd})}\) to \({{\mathcal {M}}^{p,q}({\mathbb {R}}^{2nd})}\) and we have

    $$\begin{aligned} \Vert W_{\varvec{\tau }}(\mathbf {g},\mathbf {f})\Vert _{{\mathcal {M}}^{p,q}}\le C\Vert \mathbf {g}\Vert _{{\mathcal {M}}^{p_1,q_1}} \Vert \mathbf {f}\Vert _{{\mathcal {M}}^{p_2,q_2}}, \end{aligned}$$

    where \(C=\prod _{k=1}^{d}{\Big (4|\tau _k||1-\tau _k|\Big )^{-\frac{n}{q}}}\).

  2. (ii)

    Let \(\varvec{\tau }=\mathbf {0}\) and \(\varvec{\tau }=\mathbf {1}\). If \(\mathbf {g}\in {{\mathcal {M}}_{|s|,0}^{p_1,q_1}({\mathbb {R}}^{nd})}\) and \(\mathbf {f}\in {{\mathcal {M}}_{s,0}^{p_2,q_2}({\mathbb {R}}^{nd})}\), then the multilinear Rihaczek and conjugate Rihaczek transform are continuous maps from \({{\mathcal {M}}_{|s|,0}^{p_1,q_1}({\mathbb {R}}^{nd})}\times {{\mathcal {M}}_{s,0}^{p_2,q_2}({\mathbb {R}}^{nd})}\) to \({{\mathcal {M}}_{s,0}^{p,q}({\mathbb {R}}^{2nd})}\). Also we have the inequalities

    $$\begin{aligned} \Vert W_{\mathbf {0}}(\mathbf {g},\mathbf {f})\Vert _{{\mathcal {M}}_{s,0}^{p,q}}\lesssim \Vert \mathbf {g}\Vert _{{\mathcal {M}}_{|s|,0}^{p_1,q_1}} \Vert \mathbf {f}\Vert _{{\mathcal {M}}_{s,0}^{p_2,q_2}} \end{aligned}$$

    and

    $$\begin{aligned} \Vert W_{\mathbf {1}}(\mathbf {g},\mathbf {f})\Vert _{{\mathcal {M}}_{s,0}^{p,q}}\lesssim \Vert \mathbf {g}\Vert _{{\mathcal {M}}_{|s|,0}^{p_1,q_1}} \Vert \mathbf {f}\Vert _{{\mathcal {M}}_{s,0}^{p_2,q_2}}. \end{aligned}$$

Proof

  1. (i)

    We first assume that \(p,q<\infty \). Let \(\varphi _j\in {{\mathcal {S}}({\mathbb {R}}^d), j=1,\ldots ,n},\) and set \(\phi =W_{\varvec{\tau }}(\varvec{\varphi },\varvec{\varphi })\). We know \(\phi \in {{\mathcal {S}}({\mathbb {R}}^{2nd})}\) by Proposition 5 (i). For \(\varvec{\tau } \in {(0,1)^d}\), by writing \(\tilde{\xi }=(\varvec{\tau } \xi _2,-(\mathbf {1}-\varvec{\tau })\xi _1)\) in Proposition 5 (ii), we have

    $$\begin{aligned} |V_{\phi }W_{\varvec{\tau }}(\mathbf {g},\mathbf {f})(z,\xi )|=|V_{\varvec{\varphi }}\mathbf {f}(z+\tilde{\xi })||V_{\varvec{\varphi }}\mathbf {g}(z-\tilde{\xi })| \end{aligned}$$
    (3.7)

    By using (3.7) and chancing variable \(z\rightarrow z-\tilde{\xi }\) we get

    $$\begin{aligned}&\Vert W_{\varvec{\tau }}(\mathbf {g},\mathbf {f})\Vert _{{\mathcal {M}}^{p,q}}\\&\quad =\left( \int _{{\mathbb {R}}^{2nd}}\left( \int _{{\mathbb {R}}^{2nd}} |V_{\phi }W_{\varvec{\tau }}(\mathbf {g},\mathbf {f})(z,\xi )|^p dz \right) ^{q/p}d\xi \right) ^{1/q}\\&\quad =\left( \int _{{\mathbb {R}}^{2nd}}\left( \int _{{\mathbb {R}}^{2nd}} |V_{\varvec{\varphi }}\mathbf {f}(z)|^p|V_{\varvec{\varphi }}\mathbf {g}(z-2\tilde{\xi })|^p dz \right) ^{q/p}d\xi \right) ^{1/q}\\&\quad =\left( \int _{{\mathbb {R}}^{2nd}}\left( |V_{\varvec{\varphi }}\mathbf {f}|^p*|(V_{\varvec{\varphi }}\mathbf {g})^{\sim }|^p (2\tilde{\xi })\right) ^{q/p}d\xi \right) ^{1/q}\\&\quad =\left( \int _{{\mathbb {R}}^{2nd}}\left( \Big |\prod _{j=1}^{n}V_{\varphi _j}{f_j}\Big |^p*\Big |\prod _{j=1}^{n}(V_{\varphi _j}{g_j})^{\sim }\Big |^p (2\varvec{\tau }{\xi _{2_j}},-2(\mathbf {1}-\varvec{\tau })\xi _{1_j})\right) ^{q/p}d\xi \right) ^{1/q}\\&\quad =\left( \int _{{\mathbb {R}}^{2d}}\left( \Vert V_{\varphi _1}{f_1}|^p*|(V_{\varphi _1}{g_1})^{\sim }|^p (2\varvec{\tau }{\xi _{2_1}},-2(\mathbf {1}-\varvec{\tau })\xi _{1_1})\right) ^{q/p}d\xi _{1_1} d\xi _{2_1}\cdots \right. \\&\left. \quad \quad \int _{{\mathbb {R}}^{2d}}\left( \Vert V_{\varphi _n}{f_n}|^p*|(V_{\varphi _n}{g_n})^{\sim }|^p (2\varvec{\tau }{\xi _{2_n}},-2(\mathbf {1}-\varvec{\tau })\xi _{1_n})\right) ^{q/p}d\xi _{1_n} d\xi _{2_n}\right) ^{1/q}, \end{aligned}$$

    where \(-2(\mathbf {1}-\varvec{\tau })\xi _{1_j}=(-2(1-{\tau _1})\xi _{1_{j1}},\ldots ,-2(1-{\tau _d})\xi _{1_{jd}})\) and \(2\varvec{\tau }{\xi _{2_j}}=(2{\tau _1}{\xi _{2_{j1}}},\ldots ,2{\tau _d}{\xi _{2_{jd}}})\). After a change of variable \(-2(1-\tau _k) {\xi _{1_{jk}}}=v_k\) and \(2\tau _k {\xi _{2_{jk}}}=u_k\), \(j=1,\ldots ,n\), \(k=1,\ldots ,d\), in the last integral, we get

    $$\begin{aligned} \Vert W_{\varvec{\tau }}(\mathbf {g},\mathbf {f})\Vert _{{\mathcal {M}}^{p,q}}= \left( \prod _{k=1}^{d}{\Big (4|\tau _k||1-\tau _k|\Big )^{-\frac{n}{q}}}\right) \Vert |V_{\varvec{\varphi }}\mathbf {f}|^p*|(V_{\varvec{\varphi }}\mathbf {g})^{\sim }|^p\Vert ^{1/p}_{L^{q/p}}, \end{aligned}$$

    where \(u, v \in {{\mathbb {R}}^{nd}}.\)

    The rest of the proof runs by the same method as in ( [10], Theorem 3.1).

  2. (ii)

    Let \(\varvec{\tau }=\mathbf {0}\). Changing variable \(z_1\rightarrow z_1-{\xi _2}\) and writing \(\tilde{\xi }=(\xi _2,-\xi _1)\) at the Proposition 5 (iii), we write

    $$\begin{aligned} \Vert W_{\mathbf {0}}(\mathbf {g},\mathbf {f})\Vert _{{\mathcal {M}}_{s,0}^{p,q}}=\,&\Vert V_{W_{\mathbf {0}}(\varvec{\varphi })}W_{\mathbf {0}}(\mathbf {g},\mathbf {f})\Vert _{L_{s,0}^{p,q}}\\ =\,&\left( \int _{{\mathbb {R}}^{2nd}}\left( \int _{{\mathbb {R}}^{2nd}} |V_{W_{\mathbf {0}}(\varvec{\varphi })}W_{\mathbf {0}}(\mathbf {g},\mathbf {f})(z,\xi )|^p {\left\langle \xi \right\rangle }^{ps} dz \right) ^{q/p}d\xi \right) ^{1/q}\\ =\,&\left( \int _{{\mathbb {R}}^{2nd}}\left( \int _{{\mathbb {R}}^{2nd}} |V_{\varvec{\varphi }}\mathbf {f}(z)|^p|V_{\varvec{\varphi }}\mathbf {g}(z-\tilde{\xi })|^p {\left\langle \tilde{\xi } \right\rangle }^{ps} dz \right) ^{q/p}d\xi \right) ^{1/q}\\ =\,&\left( \int _{{\mathbb {R}}^{2nd}}\left( |V_{\varvec{\varphi }}\mathbf {f}|^p*|(V_{\varvec{\varphi }}\mathbf {g})^{\sim }|^p \right) ^{q/p}(\tilde{\xi }){\left\langle \tilde{\xi } \right\rangle }^{qs} d\xi \right) ^{1/q}\\ =\,&\Vert |V_{\varvec{\varphi }}\mathbf {f}|^p*|(V_{\varvec{\varphi }}\mathbf {g})^{\sim }|^p\Vert ^{1/p}_{L_{ps,0}^{q/p}}. \end{aligned}$$

    Hence we obtain

    $$\begin{aligned} \Vert W_{\mathbf {0}}(\mathbf {g},\mathbf {f})\Vert _{{\mathcal {M}}_{s,0}^{p,q}}\lesssim \Vert \mathbf {g}\Vert _{{\mathcal {M}}_{|s|,0}^{p_1,q_1}} \Vert \mathbf {f}\Vert _{{\mathcal {M}}_{s,0}^{p_2,q_2}} \end{aligned}$$

    by applying the same proof technique as in ( [10], Theorem 3.1).

    The proof for \(\varvec{\tau }=\mathbf {1}\) is similar.

\(\square \)