In the following definition of convolution involving mollification . When I make change of variable $x-y=z$, I have $$\int_{U}\eta_{\epsilon}(x-y)f(y)dy=\int_{?}\eta_{\epsilon}(z)f(x-z)dz=\int_{?}\eta_{\epsilon}(y)f(x-y)dy,$$ why does the integration change from $U$ to $B(0,\epsilon)$?
Hope to receive a comprehensive detailed explanation from anyone. (Preferably in a geometric way...I come across long time ago, but I really couldn't find those scribbles...hope someone could help drawing some pictures...two or three circles perhaps, in my memory)
By definition, $\eta$ is supported in the unit ball $\overline{B(0,1)}$, i.e. $\eta(x) = 0$ for $x \in \mathbb{R}^N \setminus \overline{B(0,1)}$. Again by definition, $$\eta_{\epsilon}(x) = \epsilon^{-N}\eta(x\epsilon^{-1}).\tag 1$$
Now I claim that $\eta_{\epsilon}(x) = 0$ for $x \in \mathbb{R}^N \setminus \overline{B(0,\epsilon)}$. Indeed, fix such a $x$, notice that $x\epsilon^{-1} \in \mathbb{R}^N \setminus \overline{B(0,1)}$ and use $(1)$ to prove the claim.
This implies that $\eta_{\epsilon}(x - y) = 0$ for $y \in \mathbb{R}^N \setminus \overline{B(x,\epsilon)}$. Moreover, if $x \in U_{\epsilon}$ then $B(x,\epsilon) \subset U$. Then we have
$$ \int_U\eta_{\epsilon}(x - y)f(y)\,dy = \int_{B(x,\epsilon)}\eta_{\epsilon}(x - y)f(y)\,dy. $$
Now it should be much easier to understand how the domain of integration changes with the change of coordinates $z = x - y$.