Dirac delta as a limit of sequence of functions

Question

I have problem to proof that dirac delta function can be represented as following limits: $$ \lim_{\varepsilon\rightarrow 0}\frac{\varepsilon}{\pi(x^2+\varepsilon^2)}=\delta(x) $$ $$ \lim_{\varepsilon\rightarrow0}\frac{1}{\varepsilon\sqrt{\pi}}\text{exp}\Big(\frac{-x^2}{\varepsilon^2}\Big)=\delta(x) $$ I have intuition why this is true, but I don't know how to solve it analytically. I tried to put theese expretions under integral with some trial functions to check if: $$\int_{-\infty}^\infty \lim_{\varepsilon\rightarrow0}f_\varepsilon(x)\cdot g(x) dx= g(0),$$ where $f_\varepsilon$ is one of above sequence of functions and $g(x)$ is this trial function. This should hold by definition of Dirac delta: limit of some sequence of function with property that $ \int_{-\infty}^\infty \delta(x)\cdot g(x)dx = g(0)$. Checking this definition property should proof the convergence to Dirac delta, but I don't know how to compute such integrals (generality of trial function is quite problematic).

Answer 1

Let $g\in C_C^\infty$. Then, enforcing the substituion $x\mapsto \varepsilon x$ and appealing to the Dominated Convergence Theorem yields

$$\begin{align} \lim_{\varepsilon\to 0^+}\int_{-\infty}^{\infty}\frac{\varepsilon}{\pi(x^2+\varepsilon^2)}g(x)\,dx&\overbrace{=}^{x\mapsto \varepsilon x}\lim_{\varepsilon\to 0^+}\int_{-\infty}^\infty \frac1{\pi(x^2+1)}g(\varepsilon x)\,dx\\\\ &\overbrace{=}^{\text{DCT}}\int_{-\infty}^\infty \lim_{\varepsilon\to 0^+}\left(\frac1{\pi(x^2+1)}g(\varepsilon x)\right)\,dx\\\\ &=g(0) \end{align}$$

Similarly, we have

$$\begin{align} \lim_{\varepsilon\to 0^+}\int_{-\infty}^{\infty}\frac1{\varepsilon\sqrt\pi}e^{-x^2/\varepsilon^2}g(x)\,dx&\overbrace{=}^{x\mapsto \varepsilon x}\lim_{\varepsilon\to 0^+}\int_{-\infty}^\infty \frac1{\sqrt \pi}e^{-x^2}g(\varepsilon x)\,dx\\\\ &\overbrace{=}^{\text{DCT}}\int_{-\infty}^\infty \lim_{\varepsilon\to 0^+}\left(\frac1{\sqrt \pi}e^{-x^2}g(\varepsilon x)\right)\,dx\\\\ &=g(0) \end{align}$$

Therefore, in distribution we have

$$\lim_{\varepsilon\to 0^+} \frac{\varepsilon}{\pi(x^2+\varepsilon^2)}=\delta(x)$$

and

$$\lim_{\varepsilon\to 0^+} \frac1{\varepsilon \sqrt \pi}e^{-x^2/\varepsilon^2}=\delta(x)$$

as was to be shown!

Answer 2

Will it be sufficient to check that for $x=0$ both sequences have limits plus infinity, and for $x\neq 0$ they are convergent to $0$? Because I know how to do it, and I know that the Dirac delta behave that way, but I don't know how it is associated with the definition that $\int_{-\infty}^\infty\delta(x)\cdot g(x)\mathrm{d}x=g(0)$.

No, this is not sufficient. A proof would go along the following lines:

1. Note that for any $e>0$ we have that $\int_{-\infty}^\infty f_\varepsilon(x)\cdot g(x)\mathrm{d}x = \int_{-\infty}^{-e} f_\varepsilon(x)\cdot g(x)\mathrm{d}x+\int_{-e}^e f_\varepsilon(x)\cdot g(x)\mathrm{d}x+\int_e^\infty f_\varepsilon(x)\cdot g(x)\mathrm{d}x$
2. Note that for sufficiently small values of $\varepsilon$ (esp. small with respect to $e$) we have that $\int_{-\infty}^{-e} f_\varepsilon(x)\cdot g(x)\mathrm{d}x+\int_e^\infty f_\varepsilon(x)\cdot g(x)\mathrm{d}x$ is small (for which one needs integration criteria on $g$ and bounds on $f_\varepsilon$ away from $0$).
3. Note that by continuity of $g$ we can take $\min_{x\in[-e,e]}g(x)$ and $\mathrm{max}_{x\in[-e,e]}g(x)$ arbitrarily close to $g(0)$ by choosing $e$ sufficiently small.
4. Note that for sufficiently small $\varepsilon$ we have that $\int_{-e}^e f_\varepsilon(x)\mathrm{d}x$ is arbitrarily close to $1$

For a formal proof similar to this, take a look at peek-a-boo's link in the comments.

In this fashion, we can obtain: $$\lim_{\varepsilon\to0}\int_{-\infty}^\infty f_\varepsilon(x)\cdot g(x)\mathrm{d}x = \lim_{\varepsilon\to0}\int_{(-\infty,-e)\cup(e,\infty)} f_\varepsilon(x)\cdot g(x)\mathrm{d}x + \lim_{\varepsilon\to0}\int_{-e}^e f_\varepsilon(x)\cdot g(x)\mathrm{d}x$$$$\leq 0+\left(\mathrm{max}_{x\in[-e,e]}g(x)\right)\lim_{\varepsilon\to0}\int_{-e}^e f_\varepsilon(x)\mathrm{d}x = \mathrm{max}_{x\in[-e,e]}g(x)$$

And similarly we have $\lim_{\varepsilon\to0}\int_{-\infty}^\infty f_\varepsilon(x)\cdot g(x)\mathrm{d}x\geq\min_{x\in[-e,e]}g(x)$.

We have thus found $$\min_{x\in[-e,e]}g(x)\leq\lim_{\varepsilon\to 0}\int_{-\infty}^\infty f_\varepsilon(x)\cdot g(x)\mathrm{d} x\leq\mathrm{max}_{x\in[-e,e]} g(x)$$ for any $e$. This can only hold for any $e$ (by continuity of $g$) if this is equal to $g(0)$.

The tricky part here is proving 2, for which we need some assumptions on the behaviour of $g$. Part 4. is also not that straightforward to do, but the comment by md2perpe will help you for that.