Define $f(x)=e^{\frac{1}{||x||^2 -1}}$ if $\|x\|<1$ and $f(x)=0$ otherwise, and set $I=\int_{\mathbb{R}^n} f dm$.
Define $\phi=f/I$ and $\phi_\epsilon(x)={\epsilon}^{-n} \phi(x/\epsilon)$ for $\epsilon>0$.
It is written in wikipedia that $\phi_\epsilon(x) \rightarrow \delta(x)$ in the space of Schwarz distribution, but how do I prove this?
(By the way, I know that $\phi\in C_c^\infty(\mathbb{R}^n)$ and $\int_{\mathbb{R}^n} \phi dm= 1$. )
Note that, for $g\in L^2$, \begin{eqnarray} &&\int_{\mathbb{R}^n} g (x)\phi_\epsilon (x)dm\\ &=&\frac{1} {\epsilon^nI}\int_{\mathbb{R}^n} g (x)f (\frac {x}{\epsilon})dm\\ &=&\frac{1} {I}\int_{\mathbb{R}^n} g (\epsilon x)f (x)dm\\ &=&\frac{1} {I}\int_{\|x\|<1} g (\epsilon x)f (x)dm\\ &=&\frac{1} {I}\int_{\|x\|<1} g (0)f (x)dm+\frac{1} {I}\int_{\|x\|<1} [g(\epsilon x)-g (0)]f (x)dm\\ &=&g (0)+\frac{1} {I}\int_{\|x\|<1} [g(\epsilon x)-g (0)]f (x)dm. \end{eqnarray} It is easy to show that $$ \lim_{\epsilon\to0}\int_{\|x\|<1} [g(\epsilon x)-g (0)]f (x)dm=0. $$ So $$ \lim_{\epsilon\to0}\int_{\mathbb{R}^n} g (x)\phi_\epsilon (x)dm=g(0). $$ Thus $$ \phi_\epsilon (x)\to\delta(x) $$ in distribution.