Convolution power

Remarks

By associativity of convolution, $f^{*k}$ is the convolution of $k$ copies of $f$ in any bracketing, and $f^{*j} * f^{*k} = f^{*(j+k)}$. Each $f^{*k}$ lies in $L^1(\R^d)$ with $\norm{f^{*k}}_1 \le \norm{f}_1^k$. For finite Borel measures $\mu$ the same recursion applies, and one sets $\mu^{*0} = \delta_0$, the Dirac measure at $0$, which is the unit for convolution.

Example: for $\lambda > 0$ let $f(x) = \lambda e^{-\lambda x}$ for $x \ge 0$ and $f(x) = 0$ for $x < 0$, the exponential density. Then $f^{*k}(x) = \lambda^k x^{k-1} e^{-\lambda x} / (k-1)!$ for $x \ge 0$. The induction step is $$\int_0^x \frac{\lambda^k y^{k-1} e^{-\lambda y}}{(k-1)!} \, \lambda e^{-\lambda (x - y)} \, dy = \frac{\lambda^{k+1} e^{-\lambda x}}{(k-1)!} \int_0^x y^{k-1} \, dy = \frac{\lambda^{k+1} x^k e^{-\lambda x}}{k!}.$$ This is the Erlang density of the sum of $k$ independent exponential waiting times.

Non-example: the constant $f = 1$ on $\R$ is not in $L^1(\R)$, and indeed $f * f$ is defined nowhere, since $\int_\R 1 \, dy = \infty$.