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  "body": "\n## Definition\n\nLet $\\mathcal{C}$ be the set of continuous functions $[0, \\infty) \\to \\C$, with pointwise addition and the [[convolution on the half-line|FCKX8KFVGN]] $(f * g)(t) = \\int_0^t f(\\tau) g(t - \\tau) \\, d\\tau$ as multiplication.\nThis is a commutative ring without unit and [[without zero divisors|E9QNNPQD2P]].\nThe *Mikusiński field of operators* $\\mathcal{M}$ is its [[field of fractions|0F2PTGX1A2]]:\n\n1. its elements are the classes $f/g$ of pairs $(f, g) \\in \\mathcal{C} \\times (\\mathcal{C} \\setminus \\set{0})$, where $(f, g)$ and $(f_1, g_1)$ are identified when $f * g_1 = f_1 * g$;\n2. addition is $f/g + f_1/g_1 = (f * g_1 + f_1 * g)/(g * g_1)$;\n3. multiplication is $(f/g)(f_1/g_1) = (f * f_1)/(g * g_1)$.\n\nThe elements of $\\mathcal{M}$ are called *operators*.\n\n## Remarks\n\nThe construction works without a unit in $\\mathcal{C}$.\nThe identification is an equivalence relation because $\\mathcal{C}$ has no zero divisors, and $\\mathcal{M}$ is a field with unit $g/g$ for any $g \\ne 0$.\nEach $f \\in \\mathcal{C}$ is identified with $(f * g)/g$, which does not depend on $g$, and this embeds $\\mathcal{C}$ as a subring.\nLet $l \\in \\mathcal{C}$ be the constant function $1$, so that $l * f$ is the integral $\\int_0^t f$.\nA complex number $\\alpha$ is identified with $(\\alpha l)/l$; then the product of $\\alpha$ and $f$ in $\\mathcal{M}$ is the ordinary multiple $\\alpha f$.\nThe operator $s = l/(l * l) = 1/l$ satisfies $s f = f' + f(0)$ for every continuously differentiable $f$, because $l * f' = f - f(0) l$.\n\nExample: the function $t$ equals $l * l$, so its image in $\\mathcal{M}$ is $(l * l * l)/l$, and $s \\cdot t = \\frac{l}{l * l} \\cdot \\frac{l * l * l}{l} = \\frac{l * l * l * l}{l * l * l} = l$.\nSo $s t = l$, the constant function $1$, in line with $s f = f' + f(0)$ for $f(t) = t$.\n\nNon-example: the pair $(l, 0)$ does not define an element of $\\mathcal{M}$, since clause 1 requires a nonzero denominator.\nAllowing 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.\n",
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