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  "body": "\n## Definition\n\nLet $X$ be a [[Banach space|06GZ956657]] with closed unit ball $B$.\nLet $\\mathcal{K}$ be a set of bounded linear operators $X \\to X$.\nThe set $\\mathcal{K}$ is *collectively compact* if\n$$\\set{L\\psi : L \\in \\mathcal{K}, \\ \\psi \\in B}$$\nis relatively compact in $X$.\nThis means that its closure is [[compact|0R6EWFKRCF]].\nA sequence $(K_n)$ of bounded linear operators on $X$ is a *collectively compact approximation* of a bounded linear operator $K$ on $X$ if:\n\n1. $K_n \\psi \\to K\\psi$ for every $\\psi \\in X$, and\n2. the set $\\set{K_n : n \\ge 1}$ is collectively compact.\n\n## Remarks\n\nEvery operator in a collectively compact set is [[compact|M2NEP1K5AE]], and the limit $K$ in the definition is compact as well.\nThe main result assumes that $K_n$ is a collectively compact approximation of $K$ and that $I - K$ is injective.\nThen for large $n$ the operators $I - K_n$ are invertible with uniformly bounded inverses.\nThe solutions of $\\varphi_n - K_n\\varphi_n = f$ then satisfy $\\norm{\\varphi_n - \\varphi} \\le C \\norm{K_n\\varphi - K\\varphi}$.\nThe proof uses that $\\norm{(K_n - K) K} \\to 0$ and $\\norm{(K_n - K) K_n} \\to 0$, although $\\norm{K_n - K}$ need not tend to $0$.\n\nExample: let $X = C[0, 1]$, let $k$ be continuous on $[0, 1]^2$, and let $(K_n\\psi)(x) = \\frac{1}{n}\\sum_{j=1}^{n} k(x, \\frac{j}{n}) \\psi(\\frac{j}{n})$.\nFor $\\psi \\in B$ one has $\\abs{K_n\\psi(x)} \\le \\max \\abs{k}$ and $\\abs{K_n\\psi(x) - K_n\\psi(x')} \\le \\max_t \\abs{k(x, t) - k(x', t)}$, so the family is uniformly bounded and equicontinuous, hence relatively compact by the [[Arzelà–Ascoli theorem|0381S8Z5P0]].\nRiemann sums converge, so $K_n\\psi \\to K\\psi$ uniformly with $(K\\psi)(x) = \\int_0^1 k(x, t) \\psi(t) \\, dt$.\nIn general $\\norm{K_n - K} \\not\\to 0$ here: for $k = 1$, functions $\\psi$ in $B$ that equal $-1$ at the nodes and have integral close to $1$ show $\\norm{K_n - K} = 2$ for every $n$.\n\nNon-example: on $X = \\ell^2$, let $K_n$ be the projection onto the first $n$ coordinates.\nThen $K_n \\psi \\to \\psi$ for every $\\psi$, but the set $\\set{K_n e_n} = \\set{e_n}$ has no convergent subsequence, so clause 2 fails; indeed the limit $I$ is not compact.\n",
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