CONVOLUTION THEOREM

In mathematics, the 'convolution theorem' states that under suitable
conditions the Fourier transform of a convolution is the point-wise product of Fourier transforms. In other words, convolution in one domain (e.g., time domain) equals point-wise multiplication in the other domain (e.g., frequency domain). Versions of the convolution theorem are true for various Fourier-related transforms.
Let $f$ and $g$ be two functions with convolution $f$
★ g . (Note that the asterisk denotes convolution in this context, and not multiplication. The symbol $otimes$ is sometimes used instead.)
Let $mathcal\{F\}$ denote the Fourier transform operator, so $mathcal\{F\}\{f\}$ and $mathcal\{F\}\{g\}$ are the Fourier transforms of $f$ and $g$, respectively.
Then
: $mathcal\{F\}\{f$
★ g} = mathcal{F}{f} cdot mathcal{F}{g}
where · denotes point-wise multiplication. It also works "the other way round":
: $mathcal\{F\}\{f\; cdot\; g\}=\; mathcal\{F\}\{f\}$
★ mathcal{F}{g}
By applying the inverse Fourier transform $mathcal\{F\}^\{-1\}$, we can write:
: $f$
★ g= mathcal{F}^{-1}ig{mathcal{F}{f}cdotmathcal{F}{g}ig}
Note that the relationships above are only valid for the form of the Fourier transform shown in the Proof section below. The transform may be normalised in other ways, in which case constant scaling factors (typically $2pi$ or $sqrt\{2pi\}$) will appear in the relationships above.
This theorem also holds for the Laplace transform, the two-sided Laplace transform and, when suitably modified, for the Mellin transform and Hartley transform (see Mellin inversion theorem). It can be extended to the Fourier transform of abstract harmonic analysis defined over locally compact abelian groups.
This formulation is especially useful for implementing a numerical convolution on a computer: The standard convolution algorithm has quadratic computational complexity. With the help of the convolution theorem and the fast Fourier transform, the complexity of the convolution can be reduced to O(''n'' log ''n''). This can be exploited to construct fast multiplication algorithms.

Contents

Proof

Proof

''The proof here is shown for a particular normalisation of the Fourier transform. As mentioned above, if the transform is normalised differently, then constant scaling factors will appear in the derivation.''
Let $F$ be the Fourier transform of $f$ and $G$ be the Fourier transform of $g$:
:$F($
u) = mathcal{F}{f} = int_{mathbb{R}} f(x) e^{-2 pi i xcdot
u} ,dx
:$G($
u) = mathcal{F}{g} = int_{mathbb{R}}g(x) e^{-2 pi i xcdot
u} ,dx .
Let $h$ be the convolution of $f$ and $g$
:$h(z)\; =\; intlimits\_\{mathbb\{R\}\}\; f(x)\; g(z-x),\; mathrm\{d\}\; x.$
Now notice that
:$intint\; |f(z)g(x-z)|,dx,dz=int\; |f(z)|\; int\; |g(z-x)|,dx,dz\; =\; int\; |f(z)|,|g|\_1,dz=|f|\_1\; |g|\_1.$
Hence by Fubini's theorem we have that $hin\; L^1(mathbb\{R\}^n)$ so its Fourier transform is defined. Let $H$ be the Fourier transform of $h$:
:$$
egin{align}
H(
u) = mathcal{F}{h} &= int_{mathbb{R}} h(z) e^{-2 pi i zcdot
u}, dz \
&= int_{mathbb{R}} int_{mathbb{R}^n} f(x) g(z-x), dx, e^{-2 pi i zcdot
u}, dz.
end{align}

Observe that $|f(x)g(z-x)e^\{-2pi\; i\; zcdot$
u}|=|f(x)g(z-x)| and hence by the argument above we may apply Fubini's theorem again:
:$H($
u) = int_{mathbb{R}} f(x)left(int_{mathbb{R}^n} g(z-x)e^{-2 pi i zcdot
u},dz
ight),dx.
Substitute $y=z-x$; then $dy\; =\; dz$, so:
:$H($
u) = int_{mathbb{R}} f(x) left( int_{mathbb{R}} g(y) e^{-2 pi i (y+x)cdot
u},dy
ight) ,dx
:::$=int\_\{mathbb\{R\}\}\; f(x)e^\{-2pi\; i\; xcdot$
u} left( int_{mathbb{R}} g(y) e^{-2 pi i ycdot
u},dy
ight) ,dx
:::$=int\_\{mathbb\{R\}\}\; f(x)e^\{-2pi\; i\; xcdot$
u},dx int_{mathbb{R}} g(y) e^{-2 pi i ycdot
u},dy.
These two integrals are the definitions of $F($
u) and $G($
u), so:
:$H($
u) = F(
u) cdot G(
u),
which was to be demonstrated.