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  "body": "\n## Statement\n\nLet $X$ be a complex [[Banach space|0BDTTM24DP]], and let $(T(t))_{t \\ge 0}$ be a [[$C_0$-semigroup|EVF180QJXY]] on $X$ with [[generator|60HSWZBTK9]] $A$.\nLet $M \\ge 1$ and $\\omega \\in \\R$ be constants with $\\norm{T(t)} \\le M e^{\\omega t}$ for all $t \\ge 0$.\nThen every complex $\\lambda$ with $\\operatorname{Re} \\lambda \u003e \\omega$ lies in the [[resolvent set|X0243CZD9D]] of $A$, and for every $x \\in X$,\n$$(\\lambda I - A)^{-1} x = \\int_0^\\infty e^{-\\lambda t} \\, T(t) x \\, dt = \\lim_{r \\to \\infty} \\int_0^r e^{-\\lambda t} \\, T(t) x \\, dt.$$\nMoreover $\\norm{(\\lambda I - A)^{-1}} \\le M / (\\operatorname{Re} \\lambda - \\omega)$.\n\n## Remarks\n\nConstants $M$ and $\\omega$ as above exist for every $C_0$-semigroup, by the uniform boundedness principle.\nLet $R(\\lambda) x$ denote the integral; it converges absolutely by the bound on $\\norm{T(t)}$, which also gives the norm estimate.\nA direct computation with the semigroup law shows $\\frac{1}{h}(T(h) - I) R(\\lambda) x \\to \\lambda R(\\lambda) x - x$ as $h \\to 0^+$, so $R(\\lambda) x \\in D(A)$ and $(\\lambda I - A) R(\\lambda) x = x$.\nFor $x \\in D(A)$, $A$ commutes with the integral, so $R(\\lambda)(\\lambda I - A) x = x$.\nExample: for a square matrix $A$ and $T(t) = e^{t A}$, the formula reads $(\\lambda I - A)^{-1} = \\int_0^\\infty e^{-\\lambda t} e^{t A} \\, dt$ when $\\operatorname{Re} \\lambda$ exceeds the real parts of all eigenvalues of $A$.\n\n## Explanations\n\n### L1 human\n\nFor a number $a$, $\\int_0^\\infty e^{-\\lambda t} e^{a t} \\, dt = 1/(\\lambda - a)$.\nThe theorem says that the same identity holds for operators: the [[Laplace transform|0V49JEZW06]] of the semigroup $e^{t A}$ is the inverse of $\\lambda - A$.\n\n### L2 human\n\nThe formula is the bridge between semigroups and their generators.\nIt gives the necessity part of the [[Hille–Yosida theorem|5YR4GJKPCV]], and the norm bound $M/(\\operatorname{Re} \\lambda - \\omega)$ is exactly the resolvent estimate in that theorem.\nConversely, one recovers $T(t)$ from the resolvent by a complex inversion integral or by $T(t) x = \\lim_{n \\to \\infty} \\left(\\frac{n}{t} R(\\frac{n}{t})\\right)^n x$, an operator form of [[Post's inversion formula|D6QKPBS2EN]].\n",
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