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Let $M$ be a compact space equipped with a Borel probability measure $\mu$, let $L^2(M,\mu)$ be the corresponding $L^2$-space, and let $f:M\to M$ be a homeomorphism.

Question: If $\varphi\in L^2(M,\mu)$, is it true that $\varphi\circ f\in L^2(M,\mu)$? If it is the case, do we have an estimation for the norm $\|\varphi\circ f\|_{L^2(M,\mu)}$?

(I am not interested in cases where the measure $\mu$ is smooth with respect to some volume element.)

john_math
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    Based on your parenthetical, Lebesgue measure on $[0,1]$ is not allowed? Otherwise, on $[0,1]$ with Lebesgue measure, how about $\varphi(x)=\frac{1}{\sqrt[4] x}$, $f(x) = x^2$. – Jonas Meyer Feb 03 '15 at 17:22
  • My answer given here (http://math.stackexchange.com/questions/1118835/is-the-composition-function-again-in-l2a-b/1118878#1118878) also applies to your situation and shows that this is the case iff you have $\mu(f^{-1}(E))\leq C \mu(E)$ for all measurable $E$. – PhoemueX Feb 03 '15 at 17:35
  • Thank you very much. It answers my question. – john_math Feb 03 '15 at 21:32

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