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  "body": "\n## Statement\n\nLet $a \u003c b$ and let $K \\in L^2([a, b] \\times [a, b])$ be a [[square-integrable kernel|WWMWX7AD4V]].\nLet $K^*(x, y) = \\overline{K(y, x)}$ be the adjoint kernel, and let $\\mu \\in \\C$.\nConsider, in $L^2[a, b]$, the [[Fredholm equation of the second kind|DEYYNVCV9J]] and its [[adjoint equation|XEZXGBT42J]]\n$$\\varphi(x) - \\mu \\int_a^b K(x, y) \\varphi(y) \\, dy = f(x), \\qquad \\psi(x) - \\overline{\\mu} \\int_a^b K^*(x, y) \\psi(y) \\, dy = g(x).$$\nThen:\n\n1. Either the first equation has exactly one solution $\\varphi \\in L^2[a, b]$ for every $f \\in L^2[a, b]$, or the [[homogeneous equation|W6CYS54QXS]] with $f = 0$ has a nonzero solution.\n2. The homogeneous equation ($f = 0$) and the homogeneous adjoint equation ($g = 0$) have the same finite number of linearly independent solutions.\n3. Assume the homogeneous equation has a nonzero solution. Then the first equation is solvable if and only if $\\int_a^b f \\overline{\\psi} = 0$ for every solution $\\psi$ of the homogeneous adjoint equation.\n4. The set of [[characteristic values|DC7HMHVREP]] of $K$ has no accumulation point in $\\C$. These are the $\\mu$ for which the homogeneous equation has a nonzero solution.\n\n## Remarks\n\nLet $T$ be the integral operator with kernel $K$; it is a [[compact operator|M2NEP1K5AE]] on $L^2[a, b]$, and its adjoint is the integral operator with kernel $K^*$.\nFor $\\mu = 0$ the statements are trivial, and for $\\mu \\ne 0$ they are statements about $I - \\mu T$ with $\\mu T$ compact.\nOne proof approximates $K$ in $L^2$ by a degenerate kernel $K_0$ with $\\abs{\\mu} \\norm{K - K_0} \u003c 1$, with operator $T_0$.\nThen $I - \\mu (T - T_0)$ is invertible by a Neumann series, and the equation reduces to a finite linear system.\nThe [[Fredholm alternative|4HH8C9V8JQ]] for matrices applies to that system.\n\nExample: on $[0, 1]$ with $K(x, y) = xy$ and $\\mu = 3$, the homogeneous equation has the solution $\\varphi(x) = x$, and $K^* = K$.\nThe equation $\\varphi(x) - 3 \\int_0^1 x y \\varphi(y) \\, dy = f(x)$ is solvable exactly when $\\int_0^1 f(x) \\, x \\, dx = 0$; for instance $f = 1$ fails this, and indeed integrating $\\varphi - 3x \\int_0^1 y \\varphi = 1$ against $x$ gives the contradiction $0 = 1/2$.\n\nThe square-integrability of $K$ matters because it makes $T$ compact.\nFor the identity operator on $L^2[a, b]$, which is bounded but not compact and not given by a square-integrable kernel, the equation $\\varphi - \\varphi = 0$ has infinitely many independent solutions, against part 2.\n\n## Explanations\n\n### L1 human\n\nAn integral equation of the second kind behaves like a square system of linear equations.\nEither it always has exactly one solution, or its homogeneous version has nonzero solutions.\nIn the second case, a right-hand side is allowed exactly when it is perpendicular to all solutions of the adjoint homogeneous equation, just as for matrices.\n\n### L2 human\n\nThese are Fredholm's results, obtained first with his determinants for continuous kernels and later extended to square-integrable kernels.\nIn the language of operators, they say that $I - \\mu T$ is a [[Fredholm operator|8HWGKWD6ZM]] of index $0$ when $T$ is compact.\nThey also say that its image is the orthogonal complement of the kernel of its adjoint.\nPart 4 is the statement that the nonzero spectrum of a compact operator is discrete, as in the [[spectral theory of compact operators|JEPDM5B5FC]].\nThey are the reason why boundary value problems that reduce to integral equations of the second kind have a solvability theory as clean as linear algebra.\n",
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