Let $f$ be an square-integrable function on a bounded domain U. Moreover $\eta(x) = C e^{-\frac {1}{1-x^2}} $ on $[-1,1] $ and 0 elsewhere. C is taken in such a way that the integral over eta is equal to one. Define $\eta_h(x)=\frac{\eta(x/h^n)}{h^n}$. Now my textbook (Evans) says that the convolution of f and $\eta_h$ can only be nonzero on elements in U with distance at least h to the boundary.
Could anybody explain why this is true?
Observe that $$ (f*\eta_h)(x) = \int_{B_h} f(x-y)\,\eta_h(y)\,\mathrm dy $$ Since $f$ is defined on $U$ only, one requires that $x-y\in U$. Since $y$ runs through $B_h$, this amounts to $|y|<h$. In particular, if $x\in U$ and $\operatorname{dist}(x,\partial U)>h$, this condition is verified.