Differentiation operator p

Remarks

The two operators satisfy $$p \, p^{-1} f = f \quad (f \in C[0, \infty)), \qquad p^{-1} p f = f - f(0) \quad (f \in C^1[0, \infty)).$$ The first identity is the fundamental theorem of calculus: $p^{-1} f$ is the antiderivative of $f$ that vanishes at $0$. So $p^{-1}$ is a right inverse of $p$ but not a left inverse.

Example: for $f(t) = e^{at}$, $p f = a f$, and $p^{-1} f = (e^{at} - 1)/a$ for $a \ne 0$; then $p \, p^{-1} f = e^{at} = f$.

Non-example of a two-sided inverse: for the constant $f = 1$, $p^{-1} p f = p^{-1} 0 = 0 \ne 1$; the clause $p^{-1} p = \id$ fails exactly because $f(0) \ne 0$.

In the Mikusiński calculus, the integration operator is convolution with the constant $1$, and the corresponding element $s = 1/l$ of the field of operators satisfies $s f = f' + f(0)$, which repairs the defect above.