Let T be a distribution on a domain $\Omega$ in $\mathbb{R}^n$ and $f:\Omega\mapsto\Omega'$(another domain in $\mathbb{R}^n$) be a smooth map which is also proper on $supp(T)$ (i.e. $f^{-1}(K)\bigcap supp(T)$ is compact for every compact set $K\subset\Omega'$). Then the push-forward $f_{*}(T)$ is a distribution on $\Omega'$ given by $f_{*}(T)(\phi):=T(\phi\circ f)$ for all $\phi\in\mathcal{D}(\Omega')$
In the above, even though $supp(\phi)$ is compact, $supp(\phi\circ f)$ may not be. Then why $T(\phi\circ f)$ is well defined$?$ (I guess $f$ being proper has been assumed to address this issue but I don't see how it is working here).
Push forward of a distribution
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functional-analysis
measure-theory
distribution-theory
1 Answers
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Let $\chi_\phi \in C_c^\infty(\Omega)$ be identical $1$ on a relative compact neighbourhood of $f^{-1}(supp(\phi)\cap supp(T))$, then $\phi\circ(\chi_\phi\cdot f)$ hast compact support and the value of $T(\phi\circ(\chi_\phi\cdot f)$ does not depend on the choice of $\chi_\phi$.
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0Why $\phi\circ(\chi_{\phi} .f)$ has compact support? – 2017-02-26
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0$\chi_\phi$ has compact support and so the product $\chi_\phi\cdot f$. – 2017-02-26
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0Yes, you are right. The rest follows from the fact that $supp(\phi)\subseteq supp(\chi_{\phi}.f$) due to the linearity of $\phi$. I think even $\chi_{V}. (\phi\circ f)$ will work where $V$ is any compact nbd. of $f^{-1}(supp(\phi)\cap supp(T))$ and $\chi_{V}$ is similar to $\chi_{\phi}$. – 2017-02-26