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  "body": "\n## Definition\n\nLet $n \\ge 1$, let $\\R_+ = [0, \\infty)$, and let $k \\in L^1_{\\mathrm{loc}}(\\R_+; \\C^{n \\times n})$ be a locally integrable matrix-valued kernel.\nWrite $(k * x)(t) = \\int_0^t k(t - s) \\, x(s) \\, ds$.\nConsider the linear [[Volterra equation|DND8CBEVXT]] of convolution type\n$$x(t) + (k * x)(t) = f(t), \\qquad t \\ge 0.$$\nIts resolvent is the unique $r \\in L^1_{\\mathrm{loc}}(\\R_+; \\C^{n \\times n})$ with $r + k * r = k$ and $r + r * k = k$.\nFor every $f \\in L^1_{\\mathrm{loc}}(\\R_+; \\C^n)$ the unique locally integrable solution is $x = f - r * f$.\nThe equation is called *stable*, or is said to have an *integrable resolvent*, if $r \\in L^1(\\R_+; \\C^{n \\times n})$.\n\n## Remarks\n\nIf the equation is stable and $1 \\le p \\le \\infty$, then $f \\in L^p(\\R_+)$ gives $x \\in L^p(\\R_+)$ with $\\norm{x}_p \\le (1 + \\norm{r}_1) \\norm{f}_p$, by Young's inequality in the [[Lp spaces|SASP14MA1Z]].\nIf moreover $f$ is bounded and $f(t) \\to f_\\infty$ as $t \\to \\infty$, then $x(t) \\to \\big(I - \\int_0^\\infty r(s) \\, ds\\big) f_\\infty$, by dominated convergence.\n\nExample: $n = 1$ and $k(t) = a e^{-t}$ with $\\operatorname{Re} a \u003e -1$.\nThe resolvent is $r(t) = a e^{-(1 + a) t}$.\nCheck: $(k * r)(t) = a^2 e^{-t} \\int_0^t e^{-a s} \\, ds = a e^{-t} - a e^{-(1 + a) t}$ for $a \\ne 0$, so $r + k * r = k$, and the case $a = 0$ is trivial.\nSince $\\operatorname{Re}(1 + a) \u003e 0$, $r$ is integrable and the equation is stable.\n\nNon-example: $n = 1$ and $k(t) = -1$.\nThen $r(t) = -e^t$, since $-e^t + \\int_0^t e^s \\, ds = -1$, and $r$ is not integrable.\nFor $f = 1$ the solution is $x(t) = 1 + (e^t - 1) = e^t$, which is unbounded although $f$ is bounded.\n\nFor nonlinear Volterra equations one speaks instead of stability of a particular solution, in the sense of Lyapunov.\n\n## Explanations\n\n### L1 human\n\nThe resolvent turns the data $f$ into the solution $x$.\nWhen the resolvent has finite total size, it cannot turn bounded data into unbounded solutions.\nThat is what stability means here.\n",
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