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  "body": "\n## Statement\n\nLet $\\mathcal{M}$ be the [[Mikusiński field of operators|82YAXZMMQ1]], the field of fractions of the convolution ring of locally integrable functions on $[0, \\infty)$.\nLet $s$ be the inverse of the constant function $1$ in $\\mathcal{M}$.\nFor $\\lambda \\ge 0$, let $h_\\lambda(t) = H(t - \\lambda)$ for $t \\ge 0$, where $H$ is the [[Heaviside step function|09C4JP7Y0M]], and define the translation operator $e^{-\\lambda s} = s \\, h_\\lambda$.\nThen:\n\n1. For every locally integrable $f$ on $[0, \\infty)$, the product $e^{-\\lambda s} f$ is the function $t \\mapsto H(t - \\lambda) f(t - \\lambda)$.\n2. $e^{-0 s} = 1$, and $e^{-\\lambda s} e^{-\\mu s} = e^{-(\\lambda + \\mu) s}$ for all $\\lambda, \\mu \\ge 0$.\n\n## Remarks\n\nFor $t \\ge \\lambda$, $(h_\\lambda * f)(t) = \\int_0^{t - \\lambda} f(\\tau) \\, d\\tau$, and for $t \u003c \\lambda$ it is $0$.\nSo $h_\\lambda * f$ is the [[integral from zero|BDBDJSP6DF]] of the delayed function $f_\\lambda(t) = H(t - \\lambda) f(t - \\lambda)$, that is, $h_\\lambda f = h f_\\lambda$.\nMultiplying by $s = 1/h$ gives clause 1.\nClause 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.\nFor example, $e^{-\\lambda s} h = h_\\lambda$: delaying the unit step $h$ by $\\lambda$ gives the jump of size one at time $\\lambda$.\nThe notation matches the Laplace transform, where a delay by $\\lambda$ multiplies the transform by $e^{-\\lambda s}$.\n\n## Explanations\n\n### L1 human\n\nShifting a signal later in time by $\\lambda$ is, in the field of operators, multiplication by one fixed operator.\nThat operator behaves like an exponential: two shifts compose by adding their lengths.\nThis is why it is written $e^{-\\lambda s}$.\n\n### L2 human\n\nThe result explains Heaviside's formal rule that $e^{-\\lambda p}$ acts as a delay, where $p$ is differentiation.\nIn Mikusiński's setting no power series in $s$ is needed: the operator is defined directly from the delayed step.\nIt lets one solve equations with delays and treat piecewise defined inputs, such as a switch that turns on at time $\\lambda$.\nThe same rule for the Laplace transform is the time shift theorem.\n",
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