Mikusiński field of operators

Remarks

The construction works without a unit in $\mathcal{C}$. The identification is an equivalence relation because $\mathcal{C}$ has no zero divisors, and $\mathcal{M}$ is a field with unit $g/g$ for any $g \ne 0$. Each $f \in \mathcal{C}$ is identified with $(f * g)/g$, which does not depend on $g$, and this embeds $\mathcal{C}$ as a subring. Let $l \in \mathcal{C}$ be the constant function $1$, so that $l * f$ is the integral $\int_0^t f$. A complex number $\alpha$ is identified with $(\alpha l)/l$; then the product of $\alpha$ and $f$ in $\mathcal{M}$ is the ordinary multiple $\alpha f$. The operator $s = l/(l * l) = 1/l$ satisfies $s f = f' + f(0)$ for every continuously differentiable $f$, because $l * f' = f - f(0) l$.

Example: the function $t$ equals $l * l$, so its image in $\mathcal{M}$ is $(l * l * l)/l$, and $s \cdot t = \frac{l}{l * l} \cdot \frac{l * l * l}{l} = \frac{l * l * l * l}{l * l * l} = l$. So $s t = l$, the constant function $1$, in line with $s f = f' + f(0)$ for $f(t) = t$.

Non-example: the pair $(l, 0)$ does not define an element of $\mathcal{M}$, since clause 1 requires a nonzero denominator. Allowing the denominator $0$ would make every pair $(f, 0)$ equivalent to every $(f_1, 0)$, since $f * 0 = f_1 * 0 = 0$, and the division rules would fail.