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  "body": "\n## Statement\n\nLet $a \u003c b$ and let $K \\colon [a, b] \\times [a, b] \\to \\R$ be a [[continuous kernel|ZM5G144PZG]] with $K(x, y) \u003e 0$ for all $x, y \\in [a, b]$.\nLet $T_K$ be the integral operator $(T_K \\varphi)(x) = \\int_a^b K(x, y) \\varphi(y) \\, dy$, and let $r$ be its [[spectral radius|9V7YQA85PC]].\nThen:\n\n1. $r \u003e 0$, and $r$ is an eigenvalue of $T_K$ with an [[eigenfunction|4NSACSWRJZ]] $u$ that is continuous and satisfies $u(x) \u003e 0$ for all $x \\in [a, b]$;\n2. the eigenspace of $r$ is spanned by $u$;\n3. every other eigenvalue $\\mu$ of $T_K$, real or complex, satisfies $\\abs{\\mu} \u003c r$.\n\n## Remarks\n\nThe operator $T_K$ is compact on $C[a, b]$ by the [[compactness of integral operators|MF77N00SK8]], and it maps the cone of nonnegative functions into itself.\nSince $K$ has a positive minimum on the square, $T_K$ sends every nonnegative function that is not identically $0$ to a strictly positive function.\nSo it maps the cone minus $0$ into the interior of the cone.\nThe strong form of the [[Krein–Rutman theorem|VADXDK3TKM]] then gives all three parts.\n\nExample: for $K(x, y) = 1 + xy$ on $[0, 1]$, $T_K$ has rank $2$ and maps into the span of $1$ and $x$.\nIts matrix in this basis is $\\begin{pmatrix} 1 \u0026 1/2 \\\\ 1/2 \u0026 1/3 \\end{pmatrix}$, with eigenvalues $\\frac{4 \\pm \\sqrt{13}}{6}$.\nThe larger one, $r = \\frac{4 + \\sqrt{13}}{6}$, has the eigenfunction $u(x) = 1 + 2(r - 1) x = 1 + \\frac{\\sqrt{13} - 2}{3} x$, which is positive on $[0, 1]$.\nThe other eigenvalues, $\\frac{4 - \\sqrt{13}}{6}$ and $0$, have modulus less than $r$.\n\nStrict positivity matters.\nTake $K(x, y) = 1$ when $y \\le x$ and $0$ otherwise, which is nonnegative but neither continuous nor positive.\nIts operator is the Volterra operator, whose spectral radius is $0$ and which has no eigenvalue at all.\n\n## Explanations\n\n### L1 human\n\nIf every point influences every other point positively, the integral operator has a dominant mode.\nThis mode is a positive function, it is unique up to scaling, and it grows faster under repeated application than every other mode.\nThis is the integral version of the [[Perron–Frobenius theorem|62NS7PV1M9]] for matrices with positive entries.\n\n### L2 human\n\nThe theorem explains the long-time behaviour of models with a positive integral operator, such as population models structured by size.\nAfter many steps, the state lines up with the positive eigenfunction $u$ and grows by the factor $r$ per step.\nIt is a special case of the Krein–Rutman theorem for strongly positive compact operators on ordered Banach spaces.\n",
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