Convolution composed with an invertible matrix Let $T$ be an invertible $n \times n$ matrix and let $(h \circ T)(x)$ mean $h(Tx)$.
Take functions $f,g$.
Does it hold that $(f*g) \circ T = |det(T)| (f \circ T) * (g\circ T)?$
I have had some thoughts about using the fact that $f * g = g*f,$ but I cannot see wholly how this will allow us to compose the matrix with both functions.
It should come down to showing $\int_{\mathbb{R}^n} f(y)g(Tx-y)dy =  |det(T)| \int_{\mathbb{R}^n}f(Ty)g(T(x-y))dy$ 

Let $h(x) = f\ast g(x)$. Then 
$$
h(x) = \int_{\mathbb{R}^n} f(y)g(x - y)dy.
$$
Now, since $T : \mathbb{R}^n \to \mathbb{R}^n$ is an invertible linear map, we can apply the change of variables formula with the change of variable $y \to Ty$ and the change of variables formula says that 
$$
\int_{\mathbb{R}^n} f(y)g(x-y)dy = |\det\,T|\int_{T^{-1}(\mathbb{R}^n)}f(Ty)g(x-Ty)dy = |\det\,T|\int_{\mathbb{R}^n}f(Ty)g(x-Ty)dy
$$
So,
$$
h(x) =  |\det\,T|\int_{\mathbb{R}^n}f(Ty)g(x-Ty)dy.
$$
Therefore
$$
(f\ast g)\circ T = h(Tx)=|\det \,T|\int_{\mathbb{R}^n}f(Ty)g(Tx-Ty)dy
$$
and you have your inequality. 
