Let $f$ be a smooth function whose laplacian equals 1 everywhere in $\mathbb{R}^{p}$ and $p>1$. Let $B(r)$ be the ball of radius $r>0$ and centered at the origin in $\mathbb{R}^{p}$, and let $ \chi_{r} $ designate the characteristic function of $B(r)$ (it equals 1 on the ball and 0 elsewhere). Finally set $g=\chi_{r}f$ and for $n=1,2,...$, let $\psi_{n}$ be the $ C^{\infty}-$regularization of $ g $, i.e., $$ \psi_{n}(\zeta)=\int_{\mathbb{R}^{p}}g(y)\phi_{n}(\zeta-y)d\lambda_{p}(y) $$ with $ \phi_{n}(t)=\phi(1-n^{2}\mid x\mid^{2}) $, and $\phi(t)$ is equal to $C_{p}e^{-1/t^{}}$ for $|t|>0$ and equals $0$ for $|t|\leq 0$, and $ C_{p} $ is a constant chosen so that $$ \sigma_{p}\int_{0}^{1}t^{p-1}\phi(1-t^{2} )dt=1.$$ My questions are:
1) If $R>r>0$ and for $n$ big enough, is each $\psi_{n}$ a smooth function with compact support in $B(R)$; i.e., $\psi_{n}\in C_{c}^{\infty}(B(R))$?
2) Is it true that $\lim_{n\to\infty}\Delta \psi_{n}=\chi_{r}?$($\Delta$ is the laplacian).
My answers to both questions is yes. I am right?