Question about the Dirac Delta

Question

I started reading about the Dirac Delta Distribution. In particular the basic idea was presented in the Introduction of the book I am reading. It states:

Define $\rho_\epsilon(x):=\frac{A}{\epsilon}\exp(-\frac{\epsilon^2}{\epsilon^2-x^2})$ for $|x|<\epsilon$ and

$0$ for $|x| \geq \epsilon$

where $A^{-1}:=\int_{-1}^{1}\exp(-\frac{1}{1-x^2})$

The integral $\int_{\mathbb{R}} \rho_{\epsilon}(x)dx$=1

If one modifies the Integral above with a continuous function $\phi$, we get the Integral

$\int_{\mathbb{R}}\rho_{\epsilon}(x) \phi(x) dx$ which is approximate $\phi(0)$.

My Question is: Why does $\int_{\mathbb{R}}\rho_{\epsilon}(x) \phi(x) dx$ approximate $\phi(0)$.

My Calculations: I did some thinking and, using the Mean Value Theorem one would get

$\int_{-\infty}^{\infty}\rho_{\epsilon}(x) \phi(x) dx=\phi(c)\underbrace{\int_{-\infty}^{\infty}\rho_{\epsilon}(x)dx}_{=1}=\phi(c)$

This is how far I got. But I don't know why $c=0$ in particular.

Answer 1

If $\phi$ is continuous, it has a minimum and a maximum value on $[-\epsilon,\epsilon]$, say $m,M$ respectively. Since $\rho_\epsilon$ is only “supported” on $[-\epsilon,\epsilon]$: $$\int_{\Bbb R}\rho_\epsilon\phi=\int_{-\epsilon}^{\epsilon}\rho_\epsilon\phi$$And we can bound that between $(m,M)\int_{-\epsilon}^{\epsilon}\rho_\epsilon=(m,M)$ as lower and upper bounds respectively. Since $\phi$ is continuous, as $\epsilon\to0^+$ the values $m,M$ are squeezed to $\phi(0)$, so that is what the integral converges to.

The choice of $\rho$ is quite arbitrary, by the way; in fact many different constructions are used to the same effect.

Answer 2

For a given $\epsilon>0$ it's not necessary that $c_\epsilon=0$ (I added subscript $\epsilon$ to $c$ since there is a dependency). But we must have $c_\epsilon \in \operatorname{supp} (\rho_\epsilon \phi) \subseteq [-\epsilon, \epsilon].$ So as $\epsilon\to 0$ we have $c_\epsilon \to 0.$