# Prove that the function tends to the delta function

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.

• If $f$ is $L^1$ and $\int f = 1$ and $g$ is continuous bounded then replace $g$ by $g-g(0)$ you get $\int nf(nx)g(x)dx=\int f(y)g(y/n)dy=O(\sup_{|y|< n^{1/2}}|g(y/n)|)+O(\int_{|y|>n^{1/2}} f(y)dy)$. Can you finish from there ? Commented Mar 2, 2020 at 5:40

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 $$|\varphi(x)| for all $$x$$. Therefore, $$|\exp(-y^2) \, \varphi(2\sqrt{\epsilon}y)| so by the dominated convergence theorem, the above expression converges to $$\frac{1}{\sqrt{\pi}} \int \exp(-y^2) \, \varphi(0) \, dy = \varphi(0).$$

• I don't understand lines 2 and 3 starting with "= dominated convergence theorem". What is it equal to and how does the conclusion follow?
– user744734
Commented Mar 2, 2020 at 17:39
• @sloupo. It was a sloppy way of saying that by the dominated convergence theorem, the former expression tends to the latter. I have now expanded the description. Commented Mar 3, 2020 at 10:58
• Thanks. Makes sense now.
– user744734
Commented Mar 4, 2020 at 0:40
• I was trying to fill in some details while following your solution. In the statement of the Lebesgue dominated convergence theorem, we have that a sequence of functions $f_{n}$ must convergence point-wise to a function $f$ (both being measurable functions). And then the conclusion is that we can pull the limit when integrating the sequence of functions inside the integral giving us the integral of the limiting function. How is that hypothesis fulfilled in your solution above?
– user744734
Commented Mar 4, 2020 at 5:32
• @sloupo. First take some sequence $(\epsilon_n)$ such that $\epsilon_n \to 0.$ Then let $f_n(y) = \exp(-y^2) \, \varphi(2\sqrt{\epsilon_n}y)$ and $f(y) = \exp(-y^2)$. Then $f_n, f \in L^1$ and $f_n \to f$ pointwise. Commented Mar 4, 2020 at 15:48