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  "body": "\n## Statement\n\nLet $H_1$, $H_2$ and $H$ be [[Hilbert spaces|HM2HHE3J51]], and let $A \\colon H_1 \\to H$ and $B \\colon H_2 \\to H$ be bounded linear operators.\nThe following are equivalent.\n\n1. The range of $A$ is contained in the range of $B$.\n2. There is $\\lambda \\ge 0$ with $AA^* \\le \\lambda^2 BB^*$, that is, $\\norm{A^* x} \\le \\lambda \\norm{B^* x}$ for all $x \\in H$.\n3. There is a bounded linear operator $C \\colon H_1 \\to H_2$ with $A = BC$.\n\nHere $A^*$ and $B^*$ are the [[Hilbert space adjoints|6HSJBJ8TTQ]], and $\\le$ is the order given by [[positive operators|77020CY5ZW]].\nMoreover, if these hold, there is a unique $C$ with $A = BC$ whose range lies in the closure of the range of $B^*$.\nIt satisfies $\\ker C = \\ker A$ and $\\norm{C}^2 = \\inf \\set{\\mu \\ge 0 : AA^* \\le \\mu BB^*}$.\n\n## Remarks\n\nProof idea: (3) implies (1) at once, and (3) implies (2) because $AA^* = BCC^*B^* \\le \\norm{C}^2 BB^*$.\nFor (2) implies (3), the map $B^* x \\mapsto A^* x$ is well defined and bounded by $\\lambda$ on the range of $B^*$.\nExtend it to the closure, set it to $0$ on the orthogonal complement, and let $C$ be the adjoint of the result.\nFor (1) implies (3), define $Cx$ as the unique $y \\perp \\ker B$ with $By = Ax$, and use the closed graph theorem to see that $C$ is bounded.\n\nExample: on $\\ell^2$ let $B$ be the diagonal operator with entries $1/n$ and $A$ the diagonal operator with entries $1/n^2$.\nThen $A = BC$ with $C$ diagonal with entries $1/n$, and indeed $AA^* \\le BB^*$ since $1/n^4 \\le 1/n^2$.\n\n## Explanations\n\n### L1 human\n\nThe lemma relates three ways in which one operator can be smaller than another.\nThe range of $A$ fits inside the range of $B$ exactly when $A$ factors through $B$, and exactly when $AA^*$ is dominated by a multiple of $BB^*$.\nIt turns a question about ranges, which are often not closed, into an inequality between positive operators.\n\n### L2 human\n\nDouglas' lemma is a standard tool for factorization and for range inclusion questions.\nFor example, with $B = (AA^*)^{1/2}$, the [[square root|3JY5HNFRMF]] of $AA^*$, condition 2 holds in both directions, so $A$ and $(AA^*)^{1/2}$ have the same range.\nIt is used in dilation theory, in the theory of de Branges spaces, and in the study of operator equations $A = BX$.\n",
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