I just want to know how it follows that $\displaystyle v^{\epsilon} \in C^{\infty}(\bar{V})$? I could see how $\displaystyle v^{\epsilon} \in C^{\infty}(V)$ by using the translations, but I'm having difficulty seeing how it extends to $\displaystyle \bar{V}$, since it says that $\displaystyle u_{\epsilon}(x) := u(x^{\epsilon}) \text{ for } x \text{ in } V$, we are mollifying on $\displaystyle V$. Do you have any idea of how this follows? Thanks for the assistance.