# Dirac delta as a limit of sequence of functions

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).

• What are the assumptions on $g$? (Continuous? Differentiable? square-integrable?) What integration do you use? (Riemann-integration?) I think your best bet would be using the definition of integral. Feb 20 at 12:20
• Well I'm using dirac delta in quantum mechanics course in my university so I suppose it is reasonable to assume g is square-integrable and differentiable. Also I suppose to have more generality it will be better to use lebesque integration (not even sure if dirac delta is riemann integrable). Feb 20 at 12:44
• I think the only important property of $g$ here is it's continuous at $0$. The Dirac delta is not Riemann integrable or Lebesgue integrable since it's not a function at all, just a "generalized function". $\lim_{\epsilon\to 0} f_\epsilon(x)$ does not exist at $x=0$ and is $0$ everywhere else. But rather than the integral of the limit, you can prove the limit of the integral has the wanted property, and that's the key point of the Dirac delta. Feb 20 at 13:20
• @aschepler with e.g. $g(x)=e^{x^4}$ none of the integrals would converge for any value of $\varepsilon>0$, so some integration conditions on $g$ are necessary. Feb 20 at 13:49
• I want to point out that your definition is wrong. You’re supposed to show $\lim\limits_{\epsilon\to 0^+}\int_{\Bbb{R}^n}f_{\epsilon}g=g(0)$. It is very important the limit is outside.. The fancy terminology is that $\lim\limits_{\epsilon\to 0^+}f_{\epsilon}=\delta_0$ in the weak* topology on the space of distributions, this is NOT a pointwise limit. Anyway, I’ve provided a general statement and proof here. Feb 20 at 22:12

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!

• It should be $g(\epsilon x)$ for all these equations, no? Feb 21 at 8:11
• @peek-a-boo Ugh! Of course. Thank you for the heads up. Feb 21 at 15:15
• Thanks for the solution, but I'm not sure how DCT works in this case. As I read the theorem says about convergence of sequence to measurable function. So how it allows you to put this limit inside integral? Feb 22 at 9:45
• @XaveryXavier THIS POST should answer your question. Feb 22 at 14:20
• @MarkViola just curious (clearly your answer is better, you got my +1), where did I use DCT? It is not needed for steps 1,3 or 4 and how step 2 is solved depends on the criteria on $g$, so I guess you mean I do so in step 2? Or do you mean in the part after the 4 steps? Feb 23 at 12:26

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.

• by the way, I’ve given a full proof in my answer to Dirac's $\delta$ distribution smooth approximation. Feb 20 at 22:09
• Thank you, that's a nice read! Note that in this case, we have that $\lim_{\varepsilon\to0}\int_{-\infty}^\infty f_\varepsilon(x)\cdot g(x)\mathrm{d}x=0$ for an even broader class of functions than $g\in L^1(\mathbb{R})$. e.g. for the $f_\varepsilon(x)=\frac1{\varepsilon\sqrt\pi}e^{-(\frac x\varepsilon)^2}$ case you can even take $g$ to be polynomials of arbitrary degree (And polynomials do sometimes appear in these kind of situations in my experience), and you can still make this proof work. Feb 20 at 23:18
• Just curious. Why go through the trouble of splitting the integral when clearly the substitution $x\mapsto \varepsilon x$ at the start along with application of the DCT works? See my posted solution herein. Feb 21 at 4:56