= Solution
For example, it suffices to assume $f,g\in L^1(\mathbb R)$. Indeed the absolute double integral is
$$
\int_{\mathbb R}\int_{\mathbb R}|f(t)g(x-t)|\,dt\,dx=\|f\|_1\|g\|_1<\infty.
$$
Thus <Fubini's theorem> permits interchange in the <Fourier transform> of the <convolution>. With $u=x-t$,
$$
\widehat{f*g}(\lambda)=\int\!\!\int f(t)g(x-t)e^{-i\lambda x}\,dt\,dx
=\left(\int f(t)e^{-i\lambda t}\,dt\right)\left(\int g(u)e^{-i\lambda u}\,du\right).
$$
Hence the <convolution theorem> in this normalization is
$$
\boxed{\widehat{f*g}(\lambda)=\hat f(\lambda)\hat g(\lambda).}
$$
There is no extra $2\pi$ in this formula; that factor belongs to the inverse transform and <Parseval identity> for the convention being used.
Back to article page
