Fourier transform of a time-variant convolution For a time-invariant convolution, given the convolution theorem, we know that $\mathcal{F}\{(h*x)(t)\}=\hat{h}(\omega).\hat{x}(\omega)$. My question is what if the convolution is time-variant.
Let $(h*x)(t)=\int h(t,\tau)x(t-\tau)d\tau$, here is my calculation so far:
Using the inverse Fourier transform:$$x(t)=\frac{1}{2\pi}\int \hat{x}(\omega)e^{i\omega t} d\omega$$
Therefore,
$$(h*x)(t)=\int h(t,\tau)x(t-\tau)d\tau = \frac{1}{2\pi}\int \int h(t,\tau) \hat{x}(\omega)e^{i\omega (t - \tau)} d\omega d\tau$$
$$=\frac{1}{2\pi}\int \hat{x}(\omega)(\int h(t,\tau) e^{-i\omega \tau} d\tau) e^{iwt} d \omega$$
$\int h(t,\tau) e^{-i\omega \tau} d\tau$ is the Fourier transform of $h(t,\tau)$ along the second dimension. Lets call it $\hat{h}(t,\omega)$. Therefore the above equality continues as:
$$(h*x)(t)=\frac{1}{2\pi} \int\hat{x}(\omega)\hat{h}(t,\omega)e^{i \omega t} d \omega = \frac{1}{2\pi} \int F e^{i \omega t} d \omega $$
where $F=\hat{x}(\omega)\hat{h}(t,\omega)$. Can we say, given the inverse Fourier transform, $F$ is the Fourier transform of $h*x$ ? $F$ being dependent on $t$, in addition to $\omega$, causes the confusion for me.
Thank you for your help.
 A: Please search the literature for linear time-variant (LTV) operators and/or pseudo-differential operators. Below you may find a few more keywords that are helpful for a literature search about the subject.
To keep the matters simple, let's assume $h\in L^2(\mathbb{R}^2)$ and $f\in L^2(\mathbb{R})$. Thus, $H:L^2(\mathbb{R})\to L^2(\mathbb{R})$ $$Hf(x) = \int h(x,t)f(x-t)\ dt $$
defines a bounded linear operator. It is customary to use the symbol $$\sigma(x,\xi) = \int h(x,t) e^{-2\pi it\xi}\ dt $$ which is called the Kohn-Nirenberg symbol of the linear time-variant operator $H$. Since $h\in L^2(\mathbb{R}^2)$, so is $\sigma$. $H$ has the representation $$Hf(x) = \int \hat{f}(\xi)\sigma(x,\xi)e^{2\pi i x\xi}\ d\xi$$
Similarly, let $$\eta(t,\gamma)= \int h(x,t) e^{-2\pi ix\gamma}\ dt $$ then $\eta\in L^2(\mathbb{R}^2)$ and $H$ has the representation
$$Hf(x) = \iint \eta(t,\gamma)e^{2\pi i x\gamma} f(x-t)\ dtd\gamma$$
$\eta$ is called the spreading function of $H$.

If $H$ and $T$ are two LTV operators with spreading functions $\eta_H$ and $\eta_T$ respectively, then
\begin{eqnarray}
THf(x) &=& \iint \eta_T(t,\gamma) e^{2\pi ix\gamma}Hf(x-t)\ dtd\gamma \\
&=& \iint \eta_T(t,\gamma) e^{2\pi ix\gamma} \iint \eta_H(\tau,\lambda) e^{2\pi i(x-t)\lambda}f(x-t-\tau)\ d\tau d\lambda\ dtd\gamma \\
&=& \iint \eta_T(t,\gamma) e^{2\pi ix\gamma} \iint \eta_H(\tau-t,\lambda-\gamma) e^{2\pi i(x-t)(\lambda-\gamma)}f(x-\tau)\ d\tau d\lambda\ dtd\gamma \\
&=& \iint \left(\iint\eta_T(t,\gamma) \eta_H(\tau-t,\lambda-\gamma) e^{-2\pi it(\lambda-\gamma)}\ dtd\gamma\right)e^{2\pi ix\lambda} f(x-\tau)\ d\tau d\lambda \\
\end{eqnarray}
where the quantity in the paranthesis in the last line above is denoted by $\eta_T\sharp\eta_H(\tau,\lambda)$. $\eta_T\sharp\eta_H\in L^2(\mathbb{R}^2)$ is the spreading function of the LTV operator $TH$. The operation $\sharp$ is called the skew-convolution, and it takes on the role of the convolution (in the time-invariant case) for the time-variant case.
