Operator exponential and translation

Remarks

For $t \ge \lambda$, $(h_\lambda * f)(t) = \int_0^{t - \lambda} f(\tau) \, d\tau$, and for $t < \lambda$ it is $0$. So $h_\lambda * f$ is the integral from zero of the delayed function $f_\lambda(t) = H(t - \lambda) f(t - \lambda)$, that is, $h_\lambda f = h f_\lambda$. Multiplying by $s = 1/h$ gives clause 1. Clause 2 follows from clause 1, since a delay by $\mu$ followed by a delay by $\lambda$ is a delay by $\lambda + \mu$, and $\mathcal{M}$ has no zero divisors. For example, $e^{-\lambda s} h = h_\lambda$: delaying the unit step $h$ by $\lambda$ gives the jump of size one at time $\lambda$. The notation matches the Laplace transform, where a delay by $\lambda$ multiplies the transform by $e^{-\lambda s}$.

Going deeper

For a student

The result explains Heaviside's formal rule that $e^{-\lambda p}$ acts as a delay, where $p$ is differentiation. In Mikusiński's setting no power series in $s$ is needed: the operator is defined directly from the delayed step. It lets one solve equations with delays and treat piecewise defined inputs, such as a switch that turns on at time $\lambda$. The same rule for the Laplace transform is the time shift theorem.