Prove that if $\phi \in C\left(\mathbb{R}^{n}\right)$ is absolutely integrable with $\int \phi(x) d x=1$, and $\phi_{\epsilon}(x)=\epsilon^{-n} \phi(x / \epsilon)$ for $\epsilon>0$, then for every Schwartz function $f \in \mathcal{S}\left(\mathbb{R}^{n}\right)$ the convolution

$$\phi_{\epsilon} * f(x) \rightarrow f(x)$$

uniformly in $x$ as $\epsilon \downarrow 0$.

Show that the function $N_{\epsilon} \in C^{\infty}\left(\mathbb{R}^{3}\right)$ given by

$$N_{\epsilon}(x)=\frac{1}{4 \pi \sqrt{|x|^{2}+\epsilon^{2}}}$$

for $\epsilon>0$ satisfies

$$\lim _{\epsilon \rightarrow 0} \int_{\mathbb{R}^{3}}-\Delta N_{\epsilon}(x) f(x) d x=f(0)$$

for $f \in \mathcal{S}\left(\mathbb{R}^{n}\right)$. Hence prove that the tempered distribution determined by the function $N(x)=(4 \pi|x|)^{-1}$ is a fundamental solution of the operator $-\Delta .$

[You may use the fact that $\int_{0}^{\infty} r^{2} /\left(1+r^{2}\right)^{5 / 2} d r=1 / 3 .$ ]