# Convolution theorem

Let $\mathfrak{F}$, $*$, and $\cdot$ denote the Fourier transform, convolution, and point-wise multiplication, respectively. The convolution theorem states $$\mathfrak{F}\{f*g\}=\mathfrak{F}\{f\}\cdot\mathfrak{F}\{g\} \label{eq:convolution}$$ and $$\mathfrak{F}\{f\cdot g\}=\mathfrak{F}\{f\}*\mathfrak{F}\{g\}. \label{eq:multiplication}$$

## 1   Proof using the shift theorem

The left-hand side of Eq. \eqref{eq:convolution} is $$\begin{split} \mathfrak{F}\{f*g\}(u) &=\int_{-\infty}^\infty f(x)*g(x)\,e^{-2i\pi ux}\,dx\\ &=\int_{-\infty}^\infty\int_{-\infty}^\infty f(y)g(x-y)\,dy\,e^{-2i\pi ux}\,dx\\ &=\int_{-\infty}^\infty\int_{-\infty}^\infty f(y)g(x-y)e^{-2i\pi ux}\,dy\,dx\\ &=\int_{-\infty}^\infty\int_{-\infty}^\infty f(y)g(x-y)e^{-2i\pi ux}\,dx\,dy\\ &=\int_{-\infty}^\infty f(y)\int_{-\infty}^\infty g(x-y)e^{-2i\pi ux}\,dx\,dy. \end{split} \label{eq:convolution_proof_shift}$$

Let $F=\mathfrak{F}\{f\}$ and $G=\mathfrak{F}\{g\}$. By the shift theorem, the inner integral $\int_{-\infty}^\infty g(x-y)e^{-2i\pi ux}\,dx=e^{-2i\pi uy}G(u)$ and Eq. \eqref{eq:convolution_proof_shift} reduces to $$\begin{split} \mathfrak{F}\{f*g\}(u) &=\int_{-\infty}^\infty f(y)e^{-2i\pi uy}G(u)\,dy\\ &=\int_{-\infty}^\infty f(y)e^{-2i\pi uy}\,dy\,G(u)\\ &=F(u)\cdot G(u)\\ &=\mathfrak{F}\{f\}(u)\cdot\mathfrak{F}\{g\}(u). \end{split} \label{eq:convolution_proof_2_shift}$$ where $F(u)=\mathfrak{F}\{f\}(u)$.

By taking the inverse Fourier transform of both sides of Eq. \eqref{eq:multiplication}, we obtain $$\begin{split} [f\cdot g](x) &=\mathfrak{F}^{-1}\{F*G\}(x)\\ &=\int_{-\infty}^\infty F(u)*G(u)\,e^{2i\pi ux}\,du\\ &=\int_{-\infty}^\infty\int_{-\infty}^\infty F(y)G(u-y)\,dy\,e^{2i\pi ux}\,du\\ &=\int_{-\infty}^\infty\int_{-\infty}^\infty F(y)G(u-y)\,e^{2i\pi ux}\,dy\,du\\ &=\int_{-\infty}^\infty\int_{-\infty}^\infty F(y)G(u-y)\,e^{2i\pi ux}\,du\,dy\\ &=\int_{-\infty}^\infty F(y)\int_{-\infty}^\infty G(u-y)\,e^{2i\pi ux}\,du\,dy. \end{split} \label{eq:multiplication_proof_shift}$$

By the shift theorem, the inner integral $\int_{-\infty}^\infty G(u-y)\,e^{2i\pi ux}\,du=e^{2i\pi yx}g(x)$ and Eq. \eqref{eq:multiplication_proof_shift} reduces to $$\begin{split} [f\cdot g](x) &=\int_{-\infty}^\infty F(y)e^{2i\pi yx}g(x)\,dy\\ &=\int_{-\infty}^\infty F(y)e^{2i\pi yx}\,dy\,g(x)\\ &=f(x)\cdot g(x). \end{split} \label{eq:multiplication_proof_2_shift}$$

## 2   Proof using Fubini’s theorem

The left-hand side of Eq. \eqref{eq:convolution} is $$\begin{split} \mathfrak{F}\{f*g\}(u) &=\int_{-\infty}^\infty f(x)*g(x)\,e^{-2i\pi ux}\,dx\\ &=\int_{-\infty}^\infty\int_{-\infty}^\infty f(y)g(x-y)\,dy\,e^{-2i\pi ux}\,dx.\\ \end{split} \label{eq:convolution_proof_fubini}$$

By letting $x’=x-y$ and using Fubini’s theorem, Eq. \eqref{eq:convolution_proof_fubini} can be written as $$\begin{split} \mathfrak{F}\{f*g\}(u) &=\int_{-\infty}^\infty\int_{-\infty}^\infty f(y)g(x’)\,dy\,e^{-2i\pi u(x’+y)}\,dx’\\ &=\int_{-\infty}^\infty f(y)e^{-2i\pi uy}\,dy\int_{-\infty}^\infty g(x’)e^{-2i\pi ux’}\,dx’\\ &=\mathfrak{F}\{f\}(u)\cdot\mathfrak{F}\{g\}(u). \end{split} \label{eq:convolution_proof_2_fubini}$$

Let $F=\mathfrak{F}\{f\}$ and $G=\mathfrak{F}\{g\}$. By taking the inverse Fourier transform of both sides of Eq. \eqref{eq:multiplication}, we obtain $$\begin{split} [f\cdot g](x) &=\mathfrak{F}^{-1}\{F*G\}(x)\\ &=\int_{-\infty}^\infty F(u)*G(u)\,e^{2i\pi ux}\,du\\ &=\int_{-\infty}^\infty\int_{-\infty}^\infty F(y)G(u-y)\,dy\,e^{2i\pi ux}\,du.\\ \end{split} \label{eq:multiplication_proof_fubini}$$

Similarly, by letting $u’=u-y$ and using Fubini’s theorem, Eq. \eqref{eq:multiplication_proof_fubini} can be written as $$\begin{split} [f\cdot g](x) &=\int_{-\infty}^\infty\int_{-\infty}^\infty F(y)G(u’)\,dy\,e^{2i\pi(u’+y)x}\,du’\\ &=\int_{-\infty}^\infty F(y)e^{2i\pi y x}\,dy\int_{-\infty}^\infty G(u’)e^{2i\pi u’x}\,du’\\ &=f(x)\cdot g(x). \end{split} \label{eq:multiplication_proof_2_fubini}$$