We recall Fourier algebra $ \mathcal{F}L^1(\mathbb R^2)= \{f:\mathbb R^2 \to \mathbb C: \hat{f}\in L^1(\mathbb R^2)\}.$ Given $f=f(x_1,x_2) \in \mathcal{F}L^1(\mathbb R^2),$ define $G(x_1,x_2)= f(x_1+x_2, x_1-x_2).$
My question: If $f\in \mathcal{F}L^1(\mathbb R^2),$ then can we say $G \in \mathcal{F}L^1(\mathbb R^2)$?
My attempt: Let $\xi= (\xi_1, \xi_2), x=(x_1,x_2) \in \mathbb R^2.$
$$\hat{G}(\xi)= \int_{\mathbb R^2} f(x_1+x_2, x_1-x_2) e^{-i \pi (x_1\xi_1 + x_2+\xi_2)} dx_1 dx_2$$
Put $x_1+x_2=u, x_1-x_2=v.$ Then $x_1= \frac{u+v}{2}, x_2=\frac{u-v}{2}.$ Using change of variable, we have $$ \hat{G}(\xi)= 2 \int_{\mathbb R^2} f(u,v) e^{-i\pi (\frac{u+v}{2}) \xi_1}e^{-i\pi (\frac{u-v}{2}) \xi_2}dudv$$
You have proved that $\hat {G} (\xi_1,\xi_2)=\hat {f} (\frac {\xi_1+\xi_2} 2,\frac {\xi_2-\xi_1} 2)$. You have to show that this function is integrable on $\mathbb R^{2}$. This is easy by another change of variable: $\zeta_1=\xi_1+\xi_2,\zeta_2=\xi_2-\xi_1$.