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  "body": "\n## Statement\n\nLet $n \\ge 1$ and let $k \\in L^1(\\R_+; \\C^{n \\times n})$ be an integrable matrix-valued kernel on $\\R_+ = [0, \\infty)$.\nWrite $(k * r)(t) = \\int_0^t k(t - s) \\, r(s) \\, ds$.\nLet $r \\in L^1_{\\mathrm{loc}}(\\R_+; \\C^{n \\times n})$ be the resolvent of $k$, that is, the unique locally integrable solution of\n$$r + k * r = k \\quad \\text{and} \\quad r + r * k = k.$$\nLet $\\hat k(z) = \\int_0^\\infty e^{-z t} k(t) \\, dt$ be the [[Laplace transform|0V49JEZW06]] of $k$, defined for $\\operatorname{Re} z \\ge 0$.\nThen $r \\in L^1(\\R_+; \\C^{n \\times n})$ if and only if\n$$\\det\\big(I + \\hat k(z)\\big) \\ne 0 \\quad \\text{for all } z \\in \\C \\text{ with } \\operatorname{Re} z \\ge 0.$$\n\n## Remarks\n\nThe resolvent gives the solution of the [[Volterra equation|DND8CBEVXT]] $x + k * x = f$ as $x = f - r * f$.\nSo the theorem decides when the equation is [[stable|8M012XRPZM]] in the sense that the resolvent is integrable.\n\nProof idea.\nIf $r$ is integrable, taking Laplace transforms gives $(I + \\hat k(z))(I - \\hat r(z)) = I$ for $\\operatorname{Re} z \\ge 0$, so $I + \\hat k(z)$ is invertible.\nThe converse is the deep part.\nIt is a Wiener-type theorem for the convolution algebra $L^1(\\R_+)$ with a unit adjoined.\nIn that algebra, an element whose transform never vanishes on the closed right half-plane, including at infinity, is invertible.\nThe point at infinity causes no trouble, because $\\hat k(z) \\to 0$ as $\\abs{z} \\to \\infty$ in the closed half-plane.\nCompare [[Wiener's lemma|NC768TWZ3D]] for absolutely convergent Fourier series.\n\nExample: let $n = 1$ and $k(t) = a e^{-t}$ with $a \\in \\C$.\nThen $\\hat k(z) = a / (z + 1)$, and $1 + \\hat k(z) = 0$ only for $z = -1 - a$.\nThe theorem says that $r$ is integrable if and only if $\\operatorname{Re} a \u003e -1$.\nIndeed $r(t) = a e^{-(1 + a) t}$, as one checks from $(k * r)(t) = a e^{-t} - a e^{-(1 + a) t}$.\n\n## Explanations\n\n### L1 human\n\nA Volterra equation of convolution type feeds the past of the unknown back into the present through a kernel $k$.\nIts solution is obtained from the data by a second kernel, the resolvent.\nThe theorem tells when the resolvent is integrable, which means that bounded or integrable data give bounded or integrable solutions.\nThe test is a single condition on the Laplace transform of $k$: the matrix $I + \\hat k(z)$ must stay invertible on the closed right half-plane.\n\n### L2 human\n\nThe theorem is due to Paley and Wiener and is the basic tool for the stability of linear [[Volterra equations of convolution type|SNFYTC1TTA]].\nIt is the integral equation analogue of the test for linear ordinary differential equations $x' = A x$, where integrability of $e^{tA}$ holds exactly when all eigenvalues of $A$ have negative real part.\nWith $r$ integrable, Young's inequality shows that the solution map $f \\mapsto f - r * f$ is bounded on $L^p(\\R_+)$ for $1 \\le p \\le \\infty$, see [[Lp space|SASP14MA1Z]].\nThrough linearization, the integrability of the [[resolvent kernel|ARTJV1XCRX]] is also the starting point for the stability of nonlinear Volterra equations.\n",
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