Prove that the function
$$\frac{1}{2 \sqrt{\pi \epsilon}}exp(\frac{-x^{2}}{4 \epsilon})$$
tends to $\delta(x)$ when $\epsilon \rightarrow +0$.
As I understand it, in order to show $f_{\epsilon}(x) \rightarrow \delta(x)$ as $\epsilon \rightarrow +0$ we must show that for any continuous function $\phi(x)$ we have $\lim_{\epsilon \rightarrow 0} \int f_{\epsilon}\phi(x) dx = \phi(0)$
i.e. that for any $\eta > 0$ there is a $\delta_{0} > 0$ such that $|x – 0| = |x| < \delta_{0}$ implies $| \int f_{\epsilon}\phi(x) dx – \phi(0) | < \eta$.
We have
$$
\begin{align}
| \int f_{\epsilon}\phi(x) dx – \phi(0)| &= | \int f_{\epsilon}\phi(x) – \phi(0) dx | \\
& = | \int \frac{1}{2 \sqrt{\pi \epsilon}}exp(\frac{-x^{2}}{4 \epsilon})\phi(x) – \phi(0) dx | \\
& = \frac{1}{2 \sqrt{\pi \epsilon}} | \int exp(\frac{-x^{2}}{4 \epsilon})\phi(x) – \phi(0) dx | \\
& \leq \frac{1}{2 \sqrt{\pi \epsilon}} \int |exp(\frac{-x^{2}}{4 \epsilon}) | |\phi(x) – \phi(0)| dx \\
& < \frac{1}{2 \sqrt{\pi \epsilon}} \eta \int |exp(\frac{-x^{2}}{4 \epsilon}) | dx \\
\end{align}
$$
By continuity of $\phi$; since this means that for each $\eta > 0$ there is a $\delta_{0} > 0$ such that $|x – x_{0}| < \delta_{0}$ implies $|\phi(x) – \phi({0}) | < \eta$. In particular for $x_{0} = 0$.
Can't proceed further than this.
Best Answer
First we do a variable change by setting $x = 2\sqrt{\epsilon}y$: $$ \int \frac{1}{2 \sqrt{\pi \epsilon}} \exp(\frac{-x^{2}}{4 \epsilon}) \, \varphi(x) \, dx = \frac{1}{\sqrt{\pi}} \int \exp(-y^2) \, \varphi(2\sqrt{\epsilon}y) \, dy . $$
Then, since $\varphi$ is continuous and has compact support, there exists $M>0$ such that $|\varphi(x)|<M$ for all $x$. Therefore, $|\exp(-y^2) \, \varphi(2\sqrt{\epsilon}y)|<M \exp(-y^2) \in L^1$ so by the dominated convergence theorem, the above expression converges to $$ \frac{1}{\sqrt{\pi}} \int \exp(-y^2) \, \varphi(0) \, dy = \varphi(0). $$