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Let $(\mathbb{R}^d,\mathbb{B}(\mathbb{R}^d),\mu)$ where $\mu$ is a $\sigma$-finite Radon measure. If $\Phi:\mathbb{R}^d\rightarrow \mathbb{R}^d$ is a homeomorphism, then does $\Phi$ induce a homeomorphism of the Bochner-Lebesgue space $L^p_{\mu}(\mathbb{R}^d;\mathbb{R}^d)$ onto itself by $$ f\mapsto \Phi^{-1}\circ f\circ\Phi? $$

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    $\begingroup$ No. Take $d=1$, let $\mu(dx) = e^{-x^2}dx$, let $f(x)=e^x$, and let $\Phi(x) = x^3$. Notice that $f \in L^p(\mu)$ for all $1\le p < \infty$. However $(\Phi^{-1}f\;\Phi)(x) = e^{\frac13 x^3} $ which is never in any $L^p(\mu)$. $\endgroup$ – Shalop May 9 at 9:36
  • $\begingroup$ I will accept this answer if you put it :) $\endgroup$ – AIM_BLB May 9 at 10:12

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