Convolution

Remarks

The integral converges absolutely for almost every $x$, and $f * h$ lies in $L^1(G)$ with $\norm{f * h}_1 \le \norm{f}_1 \norm{h}_1$. Convolution is associative, and it is commutative when $G$ is abelian.

Example: $G = (\R, +)$ with Lebesgue measure, $f = h = \mathbf{1}_{[0, 1]}$. Then $(f * h)(x) = \int_\R \mathbf{1}_{[0,1]}(y) \mathbf{1}_{[0,1]}(x - y) \, dy$ is the length of $[0, 1] \cap [x - 1, x]$. This equals $x$ for $0 \le x \le 1$, $2 - x$ for $1 \le x \le 2$, and $0$ otherwise: a triangle function.

Non-example: for $f = h = 1$ on $\R$, the defining integral $\int_\R 1 \, dy$ diverges at every $x$, so the convolution is not defined; the hypothesis $f, h \in L^1$ fails.

On $(\Z, +)$ with counting measure, $(f * h)(n) = \sum_{m \in \Z} f(m) h(n - m)$, the Cauchy product of coefficient sequences. The convolution of probability distributions is the special case of measures, which describes sums of independent random variables.