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  "body": "\n## Definition\n\nLet $X$ be a set and let $f \\colon X \\to X$ be a map such that each $x \\in X$ has finitely many preimages.\nLet $g \\colon X \\to \\C$ be a function, called the weight.\nThe *transfer operator* $\\mathcal{L}_g$ acts on functions $\\varphi \\colon X \\to \\C$ by\n$$(\\mathcal{L}_g \\varphi)(x) = \\sum_{y \\in f^{-1}(x)} g(y) \\, \\varphi(y),$$\nwhere an empty sum is $0$.\n\n## Remarks\n\nIn practice $X$ carries more structure, and $\\mathcal{L}_g$ is studied on a suitable Banach space of functions, such as Hölder or bounded variation functions.\n\nExample: let $f(x) = 2x \\bmod 1$ on $[0,1)$ and $g = 1/2$.\nThen $(\\mathcal{L}_g \\varphi)(x) = \\frac{1}{2}\\big(\\varphi(x/2) + \\varphi((x+1)/2)\\big)$, and $\\mathcal{L}_g 1 = 1$.\nThe substitutions $y = x/2$ and $y = (x+1)/2$ give, for integrable $\\varphi$ and bounded $\\psi$,\n$$\\int_0^1 (\\mathcal{L}_g \\varphi) \\, \\psi \\, dx = \\int_0^1 \\varphi \\, (\\psi \\circ f) \\, dx.$$\nSo $\\mathcal{L}_g$ is the adjoint of the [[Koopman operator|37TRG5DBQC]] $\\psi \\mapsto \\psi \\circ f$ for [[Lebesgue measure|2Z2QVWZ262]], and taking $\\psi$ an indicator function shows that $\\varphi \\, dx$ is $f$-invariant exactly when $\\mathcal{L}_g \\varphi = \\varphi$.\n\nNon-example of this duality: with the weight $g = 1$ for the same map, $\\mathcal{L}_1 1 = 2$, and $\\int \\mathcal{L}_1 \\varphi \\, dx = 2 \\int \\varphi \\, dx$, so $\\mathcal{L}_1$ is not the adjoint of a Koopman operator for Lebesgue measure.\n\nFor a piecewise smooth expanding map of an interval, the weight $g = 1/\\abs{f'}$ gives the Perron–Frobenius operator, the adjoint of the Koopman operator for Lebesgue measure.\n",
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