# Automorphism induced automorphism of Lp spaces

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 $$f\mapsto \Phi^{-1}\circ f\circ\Phi?$$

• 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 $L^p(\mu)$. – Shalop May 9 at 9:36
• I will accept this answer if you put it :) – AIM_BLB May 9 at 10:12