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From the convolution theorem, we know that the multiplication in the frequency domain is equivalent to convolution in the time domain, and vice-versa. I am wondering if there is some kind of extension to the following "frequency-time"/"time-frequency" scenario (these are in quotes, since I've never seen these terms, and never encountered these kinds of functions).

Let's use the ordinary frequency Fourier transform unitaries $\hat{f}(\xi)=\int_{\mathbb{R}}f(t)e^{-2\pi t\xi}dt$ and $f(t)=\int_{\mathbb{R}}\hat{f}(\xi)e^{2\pi\xi t}d\xi$. Let's call two-dimensional function $g(\xi_1,t_2)$ a "frequency-time" domain function. We can transform this function to "time-frequency" domain by applying the inverse Fourier transform and the Fourier transform to the respective domain. While in general we cannot claim that the order of transforms application does not matter (since that changes the order of integration), let's assume that flipping this order produces the same result. Thus, we have the following:

$$\hat{g}(t_1,\xi_2)=\int\int g(\xi_1,t_2) e^{2\pi\xi_1 t_1}e^{-2\pi t_2\xi_2}d\xi_1dt_2 =\int\int g(\xi_1,t_2) e^{-2\pi t_2\xi_2}e^{2\pi\xi_1 t_1}dt_2d\xi_1$$.

Now, let $\{h_1(t),\hat{h}_1(\xi)\}$ and $\{h_2(t),\hat{h}_2(\xi)\}$ be Fourier transform pairs. I am wondering when the following is true for the function pair $\{g(t_1,\xi_2),\hat{g}(\xi_1,t_2)\}$ defined above:

$$\int\int g(\xi_1,t_2)\hat{h}_1(\xi_1)h_2(t_2) e^{2\pi\xi_1 t_1}e^{-2\pi t_2\xi_2}d\xi_1dt_2=\int\int \hat{g}(\tau,f)h_1(t_1-\tau)\hat{h}_2(\xi_2-f)d\tau df.$$

In other words, is there an extension of the convolution theorem to my case? This holds trivially when $g(\xi_1,t_2)\equiv \hat{g}_1(\xi_1)g_2(t_2)$, however, I am wondering what is known when $g(\xi_1,t_2)$ cannot be decomposed into a product like this. In particular, I am interested in $g(\xi_1,t_2)=ce^{-a(\xi_1^2+t_2^2)+b\xi_1t_2}$ for $a,b,c>0$ and similar forms.

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Lets assume enough about our functions for operations to commute etc. In less symbols, your identity is (where $\hat{g}$ confusingly is the result of 2 transforms)

$$(\mathcal{F}_2\mathcal{F}_1^{-1}g(\xi_1,t_2)\widehat{h_1}(\xi_1)h_2(t_2))(t_1,\xi_2) \overset{?}{=} ([\hat{g} \ast_1 h_1 ]\ast_2 \widehat{h_2})(t_1,\xi_2)$$

Since $\mathcal{F}^{-1}[f(\xi)](x) = \mathcal{F}[f(-\xi)](x)$ (the distinction between frequency and time is superficial) without loss we may replace $\mathcal{F}^{-1}$ with $\mathcal{F}$,

$$\mathcal{F}_2\mathcal{F}_1 g(t_1,t_2)h_1(t_1)h_2(t_2)(\xi_1,\xi_2) \overset{?}{=} [\mathcal{F}_2\mathcal{F}_1g \ast_1 \mathcal{F}_1h_1 ]\ast_2 \mathcal{F}_2h_2(t_1,\xi_2)])(\xi_1,\xi_2)$$

To show this we apply $\mathcal{F}_i$s in sequence and apply the known Convolution Theorem. Let $g_{t_2} = g(·,t_2)$.

For each $t_2$ and for each $\xi_1$, we know by Convolution Theorem, $$\mathcal{F}_1 [g_{t_2}(t)h_1(t)](\xi_1) =[\mathcal{F}_1 g_{t_2} \ast_1 \mathcal{F}_1 h_1](\xi_1) $$ This is a function of $t_2$, for each $\xi_1$: call it $G_{\xi_1}(t_2)$.

Similarly, we also know $$\mathcal{F}_2(G_{\xi_1} h_2)(\xi_2) = \mathcal{F}_2G_{\xi_1} \ast_2 \mathcal{F}_2h_2(\xi_2) $$ holds true for each $\xi_1$ and $\xi_2$. Unpacking the definitions gives your identity.

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Your question is very similar to the convolution theorem for the so-called sympletic Fourier transform, which is your transform, combined with a flip of variables. So it maps functions of time-frequency again on functions of time-frequency! The convolution theorem for the symplectic Fourier transform is well established in the literature on sympletic geometry.

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