convolution, associativity of

Proposition.

Convolution is associative.

Proof.

Let $f$, $g$, and $h$ be measurable functions on the reals, and suppose the convolutions $(f*g)*h$ and $f*(g*h)$ exist. We must show that $(f*g)*h=f*(g*h)$. By the definition of convolution,

$((f*g)*h)(u)$ $={\displaystyle {\int }_{ℝ}}(f*g)(x)h(u-x)𝑑x$

$={\displaystyle {\int }_{ℝ}}\left[{\displaystyle {\int }_{ℝ}}f(y)g(x-y)𝑑y\right]h(u-x)𝑑x$

$={\displaystyle {\int }_{ℝ}}{\displaystyle {\int }_{ℝ}}f(y)g(x-y)h(u-x)𝑑y𝑑x.$

By Fubini’s theorem we can switch the order of integration. Thus

$((f*g)*h)(u)$ $={\displaystyle {\int }_{ℝ}}{\displaystyle {\int }_{ℝ}}f(y)g(x-y)h(u-x)𝑑x𝑑y$

$={\displaystyle {\int }_{ℝ}}f(y)\left[{\displaystyle {\int }_{ℝ}}g(x-y)h(u-x)𝑑x\right]𝑑y.$

Now let us look at the inner integral. By translation invariance,

${\displaystyle {\int }_{ℝ}}g(x-y)h(u-x)𝑑x$ $={\displaystyle {\int }_{ℝ}}g((x+y)-y)h(u-(x+y))𝑑x$

$={\displaystyle {\int }_{ℝ}}g(x)h((u-y)-x)𝑑x$

$=(g*h)(u-y).$

So we have shown that

$$((f*g)*h)(u)={\int }_{ℝ}f(y)(g*h)(u-y)𝑑y,$$

which by definition is $(f*(g*h))(u)$. Hence convolution is associative. ∎

Title	convolution, associativity of
Canonical name	ConvolutionAssociativityOf
Date of creation	2013-03-22 16:56:36
Last modified on	2013-03-22 16:56:36
Owner	mps (409)
Last modified by	mps (409)
Numerical id	5
Author	mps (409)
Entry type	Derivation
Classification	msc 94A12
Classification	msc 44A35