Convolution of 3 functions Assume that I have an equation like this:
$f(t)=\int{g(\tau)h(2t-\tau)k(t-\tau)d\tau}$, in which g(t), h(t) and k(t) are three arbitrary functions.

It can be seen that it is similar to convolution function and convolution can be expressed by the Fourier transform. I am wondering, is it possible to write this equation based on the Fourier transform, as well?

 A: In the following derivation, we will denote the various Fourier Transforms as follows
$$F(s) = \mathcal{F}\left\{f(t)\right\}$$
$$G(s) = \mathcal{F}\left\{g(t)\right\}$$
$$H(s) = \mathcal{F}\left\{h(t)\right\}$$
$$K(s) = \mathcal{F}\left\{k(t)\right\}$$
Assuming all of the integrals below and Fourier Transforms exist, and generalizing $2t-\tau$ to $at-\tau$,
$$\begin{align*}F(s) &= \mathcal{F}\left\{f(t)\right\}\\
\\
&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}g(\tau)h(at-\tau)k(t-\tau)\mathrm{d}\tau\space e^{-2\pi i ts} \mathrm{d}t\\
\\
&=\int_{-\infty}^{\infty}g(\tau)\int_{-\infty}^{\infty}h(at-\tau)k(t-\tau)\space e^{-2\pi i ts} \mathrm{d}t\space \mathrm{d}\tau\\
\\
&=\int_{-\infty}^{\infty}g(\tau)\space\mathcal{F}\left\{h(at-\tau)k(t-\tau)\right\}\space \mathrm{d}\tau\\
\\
&=\int_{-\infty}^{\infty}g(\tau)\space\left[\mathcal{F}\left\{h(at-\tau)\right\}*\mathcal{F}\left\{k(t-\tau)\right\}\right]\space \mathrm{d}\tau\\
\\
&=\int_{-\infty}^{\infty}g(\tau)\space\left[e^{-2\pi i \tau\frac{s}{a}}\dfrac{1}{|a|}H\left(\dfrac{s}{a}\right)*e^{-2\pi i \tau s}K(s)\right]\space \mathrm{d}\tau\\
\\
&=\int_{-\infty}^{\infty}g(\tau)\int_{-\infty}^{\infty}e^{-2\pi i \tau\frac{\nu}{a}}\dfrac{1}{|a|}H\left(\dfrac{\nu}{a}\right)e^{-2\pi i \tau (s-\nu)}K(s-\nu)\space \mathrm{d}\nu\space \mathrm{d}\tau\\
\\
&=\int_{-\infty}^{\infty}\dfrac{1}{|a|}H\left(\dfrac{\nu}{a}\right)K(s-\nu)\int_{-\infty}^{\infty}g(\tau)e^{-2\pi i \tau\frac{\nu}{a}}e^{-2\pi i \tau (s-\nu)}\space \mathrm{d}\tau\space \mathrm{d}\nu\\
\\
&=\int_{-\infty}^{\infty}\dfrac{1}{|a|}H\left(\dfrac{\nu}{a}\right)K(s-\nu)\int_{-\infty}^{\infty}g(\tau)e^{-2\pi i \tau\left(s-\frac{a-1}{a}\nu\right)}\space \mathrm{d}\tau\space \mathrm{d}\nu\\
\\
&=\int_{-\infty}^{\infty}\dfrac{1}{|a|}H\left(\dfrac{\nu}{a}\right)K(s-\nu)G\left(s-\frac{a-1}{a}\nu\right)\space \mathrm{d}\nu\\
\\
\end{align*}$$
So for $a = 2$
$$ F(s) =\int_{-\infty}^{\infty}\dfrac{1}{2}H\left(\dfrac{\nu}{2}\right)K(s-\nu)G\left(s-\frac{\nu}{2}\right)\space \mathrm{d}\nu$$
You are still left with a "convolution-like" integral, so there is not much to be gained in the general case.  Although if any of $G(s), H(s), K(s)$ have a special form, such as a dirac delta function or a rectangle function, there may be some benefit.
