I want to show, that for some functions $ f,g \in L^1(\mathbb{R^n})$ hold: $$f*g=g*f$$
My proof so far:
$ f*g= \int_{\mathbb{R^n}} f(x-y)f(y) dy$ Substitution: $\phi(y) =x-y$ This leads to:
$$\int_{\mathbb{R^n}} f(\phi(y)) g(y) | Det(D \phi(y)| = \int_{\mathbb{R^n}} f(\phi(y)) g(y) (-1)^n dy$$ How can I go on from there?
The Jacobian is $1$, not $(-1)^{n}$. (You forgot the absolute value). So $\int f(x-y)g(y)dy=\int f(y)g(x-y)dy$ and $(f*g)(x)=(g*f)(x)$.