convolution, associativity of


Proposition.

Convolution is associative.

Proof.

Let f, g, and h be measurable functionsMathworldPlanetmath 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) =(f*g)(x)h(u-x)𝑑x
=[f(y)g(x-y)𝑑y]h(u-x)𝑑x
=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) =f(y)g(x-y)h(u-x)𝑑x𝑑y
=f(y)[g(x-y)h(u-x)𝑑x]𝑑y.

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

g(x-y)h(u-x)𝑑x =g((x+y)-y)h(u-(x+y))𝑑x
=g(x)h((u-y)-x)𝑑x
=(g*h)(u-y).

So we have shown that

((f*g)*h)(u)=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