Support of Convolution and Smoothing 

I just want to know how it follows that $v^{\epsilon} \in C^{\infty}(\bar{V})$? I could see how $v^{\epsilon} \in C^{\infty}(V)$ by using the translations, but I'm having difficulty seeing how it extends to $\bar{V}$, since it says that $u_{\epsilon}(x) := u(x^{\epsilon}) \text{ for } x \text{ in } V$, we are mollifying on $V$. I also don't think we can use the same type of extension as in the previous question I asked, since previously we extended a function that had already become zero to zero on the rest of the domain. Do you have any idea of how this follows? Thanks for the assistance. 

$\DeclareMathOperator \supp {supp}$For functions on $\mathbb R^n$ we have in general $\supp (f*g) \subset \supp f+\supp g$, where $A+B = \{a + b \mid a \in A, b \in B\}$. To see this note that $x$ must be in this set for $f(x-y)g(y)$ to be non-zero.
Thus we have $\supp f^\epsilon \subset \supp f + B(0,\epsilon) = \{x \mid d(x,\supp f)<\epsilon\}$, so $\bigcap_{\epsilon > 0} \supp f^\epsilon \subset \supp f$ (which I guess is what you were trying to get at with your "limit').
Intuitively we expect $\supp f \subset \supp f^\epsilon$, and this is true if $\eta(0)>0, \eta\ge 0, f \ge 0$. (I'm assuming $\eta$ is at least continuous here; usually we take it to be smooth.) In general, however, I do not think this always holds.
The only time we would have $\supp f ^\epsilon \subset \supp f$ is when $\supp f$ is the whole space: there will always be some "leakage" of $f^\epsilon$ over the boundary of $\supp f$.
