# 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,

 $\displaystyle((f*g)*h)(u)$ $\displaystyle=\int_{\mathbb{R}}(f*g)(x)h(u-x)\,dx$ $\displaystyle=\int_{\mathbb{R}}\left[\int_{\mathbb{R}}f(y)g(x-y)\,dy\right]h(u% -x)\,dx$ $\displaystyle=\int_{\mathbb{R}}\int_{\mathbb{R}}f(y)g(x-y)h(u-x)\,dy\,dx.$

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

 $\displaystyle((f*g)*h)(u)$ $\displaystyle=\int_{\mathbb{R}}\int_{\mathbb{R}}f(y)g(x-y)h(u-x)\,dx\,dy$ $\displaystyle=\int_{\mathbb{R}}f(y)\left[\int_{\mathbb{R}}g(x-y)h(u-x)\,dx% \right]\,dy.$

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

 $\displaystyle\int_{\mathbb{R}}g(x-y)h(u-x)\,dx$ $\displaystyle=\int_{\mathbb{R}}g((x+y)-y)h(u-(x+y))\,dx$ $\displaystyle=\int_{\mathbb{R}}g(x)h((u-y)-x)\,dx$ $\displaystyle=(g*h)(u-y).$

So we have shown that

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

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

Title convolution, associativity of ConvolutionAssociativityOf 2013-03-22 16:56:36 2013-03-22 16:56:36 mps (409) mps (409) 5 mps (409) Derivation msc 94A12 msc 44A35