I need to find a region $D \subset \mathbb{R}^2$ such that the following equailty holds for any pair of integrable functions $f, g$:
$\iint_D f(x+y) g(x-y)dx dy = \int_0^1f(u) du \cdot \int_0^1 g(v) dv$
My first thought was to define $T(x, y) = (x+y, x-y)$. But that has $|\det(DT)| = 2$, so the equality doesn't hold.
Is that the correct path? Is there any other $T$ that I am not seeing? Can I do something with the region generated by $T^-1$?
The substitution is -- no question about that --:
$$u=x+y, \ v=x-y.$$
Then
$$x=\frac{u+v}2, \ y=\frac{u-v}2.$$ So, the Jacobian determinant is
$$\begin{vmatrix} \frac{\partial x}{\partial u}&\frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u}&\frac{\partial y}{\partial v} \end{vmatrix}= \begin{vmatrix} \frac12&\ \ \ \frac12\\ \frac12&-\frac12\end{vmatrix}=-\frac12.$$
That is, the absolute value of the Jacobian is $\frac12$.
So, the the original requirement cannot be met.
As far as the domain of integration:
That is,
$$D=\{(x,y):-x \le y\le -x+1 \}\cap\{(x,y):x-1\le y\le x\}.$$
Over this domain
$$0\le x+y=u \le 1 \text{ and } 0\le x-y=v \le 1.$$
Finally, for any integrable $f$ and $g$
$$\int_Df(x+y)g(x-y)\ dxdy=\frac12\int_0^1f(u)\ du\cdot\int_0^1g(v) \ dv.$$