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) | =β«β(f*g)(x)h(u-x)πx | ||
=β«β[β«βf(y)g(x-y)πy]h(u-x)πx | |||
=β«ββ«βf(y)g(x-y)h(u-x)πyπx. |
By Fubiniβs theorem we can switch the order of integration. Thus
((f*g)*h)(u) | =β«ββ«βf(y)g(x-y)h(u-x)πxπy | ||
=β«βf(y)[β«βg(x-y)h(u-x)πx]πy. |
Now let us look at the inner integral. By translation invariance,
β«βg(x-y)h(u-x)πx | =β«βg((x+y)-y)h(u-(x+y))πx | ||
=β«βg(x)h((u-y)-x)πx | |||
=(g*h)(u-y). |
So we have shown that
((f*g)*h)(u)=β«βf(y)(g*h)(u-y)πy, |
which by definition is (f*(g*h))(u). Hence convolution is associative. β
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 |