# Dirac's $\delta$ distribution smooth approximation

Is there a family of functions $$(\delta_{\varepsilon})_{\varepsilon>0}\subset C^{\infty}_{c}(\mathbb{R})$$ with compact support in $$[-\epsilon,\epsilon]$$ such that:

$$\lim\limits_{\varepsilon\to 0} \int_{\mathbb{R}} \delta_{\varepsilon}(x)f(x)=f(0),\ \forall\ f\in C_{c}(\Omega)$$

This should give smooth approximations of Dirac's delta distribution in the vague topology (dual of the space $$C_c(\mathbb{R})$$).

I found in an article that we can even choose $$\delta_{\varepsilon}(x)=\dfrac{1}{\varepsilon}\zeta\left (\dfrac{x}{\varepsilon}\right )$$ with $$\zeta\in C^{\infty}(\mathbb{R})$$ with compact support in $$[-1,1]$$. But there is given no $$\zeta$$ as an example.

I found here Dirac Delta limiting representation a discontinuous approximation. Is there any smooth one? I didn't found one yet.

• You can use derivatives of a sequence of smooth approximations to the step function (whose derivative is the distribution you want). math.stackexchange.com/questions/1264681/… Jul 27, 2021 at 16:37
• Start with $\exp(-1/(1 - x^2))$ on $(-1, 1)$. Jul 27, 2021 at 16:42
• @Eric Towers I know the basics. But asking for $f$ to be $C$ instead $C^{\infty}$ is difficult to handle. We can't use density arguments here since $\delta_\epsilon$ will approach $\infty$ near $0$. Do you have a proof? Jul 27, 2021 at 17:07
• en.wikipedia.org/wiki/Bump_function Jul 27, 2021 at 18:50
• Is your qualm with the article's suggestion that you interpret it as saying that there exists some working choice of the function $\zeta$, but it doesn't provide that choice? The claim is actually that it works for any such $\zeta$ (with the additional hypothesis that one has scaled it to normalize $\int_\mathbb{R} \zeta$ to $1$)-- pick any you like, such as the function suggested by a rural reader. Jul 28, 2021 at 0:08

Let $$\zeta\in L^1(\Bbb{R}^n)$$, and define $$c:=\int_{\Bbb{R}^n}\zeta(x)\,dx$$, and for each $$t>0$$, let $$\zeta_t(x)=\frac{1}{t^n}\zeta\left(\frac{x}{t}\right)$$. Then, for any bounded Lebesgue measurable function $$f:\Bbb{R}^n\to\Bbb{C}$$ which is continuous at the origin, we have $$\lim\limits_{t\to0^+}\int_{\Bbb{R}^n}\zeta_t(x)f(x)\,dx = cf(0)$$.

(we assume $$f$$ is bounded and Lebesgue measurable so that $$\zeta_tf\in L^1$$, and thus the integral on the LHS is well defined for each $$t>0$$)

Edit:

Thanks to @MarkViola’s comment, there’s a much shorter proof. We have, \begin{align} \left|\int_{\Bbb{R}^n}\zeta_t(x)f(x)\,dx-cf(0)\right|&\leq\int_{\Bbb{R}^n}|\zeta_t(x)||f(x)-f(0)|\,dx =\int_{\Bbb{R}^n}|\zeta(y)||f(ty)-f(0)|\,dy. \end{align} Since $$f$$ is continuous at the origin, the integrand approaches $$0$$ pointwise everywhere as $$t\to 0^+$$, and the integrand is dominated by $$2\|f\|_{\infty}|\zeta|\in L^1(\Bbb{R}^n)$$, it follows by Lebesgue’s dominated convergence theorem that the RHS approaches $$0$$, so $$\lim\limits_{t\to 0^+}\int_{\Bbb{R}^n}\zeta_t(x)f(x)\,dx=cf(0)$$.

Original long winded proof.

The proof is pretty straight forward. Say $$M>0$$ is a bound for $$f$$, and let $$\epsilon>0$$ be arbitrary. By continuity of $$f$$ at the origin, there is a $$\delta>0$$ such that for all $$x\in\Bbb{R}^n$$ with $$\|x\|\leq \delta$$, we have $$|f(x)-f(0)|\leq \epsilon$$. Now, $$\int_{\Bbb{R}^n} \zeta=c$$ implies each $$\int_{\Bbb{R}^n}\zeta_t=c$$, and thus for each $$t>0$$, \begin{align} \left|\int_{\Bbb{R}^n}\zeta_t(x)f(x)\,dx- cf(0)\right| &= \left|\int_{\Bbb{R}^n}\zeta_t(x)[f(x)-f(0)]\,dx\right|\\ &\leq \int_{\|x\|\leq \delta}|\zeta_t(x)|\cdot|f(x)-f(0)|\,dx + \int_{\|x\|> \delta}|\zeta_t(x)|\cdot|f(x)-f(0)|\,dx\\ &\leq \epsilon\int_{\|x\|\leq \delta}|\zeta_t(x)|\,dx + 2M\int_{\|x\|>\delta}|\zeta_t(x)|\,dx\\ &=\epsilon\int_{\|y\|\leq \frac{\delta}{t}}|\zeta(y)|\,dy + 2M\int_{\|y\|>\frac{\delta}{t}}|\zeta(y)|\,dy\\ &\leq \epsilon \cdot \|\zeta\|_{L^1}+2M\int_{\|y\|>\frac{\delta}{t}}|\zeta(y)|\,dy, \end{align} where in the second last line, I simply made the change of variables $$y=tx$$. Observe that as $$t\to 0^+$$, in the second term we're integrating over smaller and smaller sets. By the dominated convergence theorem, the limit is $$0$$. Thus, \begin{align} \limsup_{t\to 0^+}\left|\int_{\Bbb{R}^n}\zeta_t(x)f(x)\,dx- cf(0)\right| &\leq \epsilon\|\zeta\|_{L^1}+0 = \epsilon \|\zeta\|_{L^1}. \end{align} Finally, since $$\epsilon>0$$ was arbitrary, it follows that the LHS is in fact equal to $$0$$, so that $$\lim\limits_{t\to 0^+}\int_{\Bbb{R}^n}\zeta_t(x)f(x)\,dx$$ exists and equals $$cf(0)$$.

So, the idea of the proof is just to note that $$\int \zeta_t = c$$, and that we can break up the region of integration into two pieces: one where $$|f(x)-f(0)|$$ is small, and another which becomes small as $$t\to 0^+$$ due to $$\zeta_t$$ getting "more concentrated".

• Just curious. Why go through the trouble of splitting the integral when clearly the substitution $y=tx$ at the start along with the DCT works? Feb 21 at 4:54
• @MarkViola good point. I think at the time of writing, I was thinking of more general hypotheses which required splitting up into the different regions, but now I can’t remember what precisely (something like $f\in L^1$ and proving the identity at Lebesgue points or something). And indeed, bounded $f$ with continuity at the origin is super straightforward with DCT Feb 21 at 5:41
• I was just curious since I've seen many of your excellent posts on distributions. Feb 21 at 15:21
• (+1) for the solution, before and after the edit ;-) Feb 23 at 14:08

Note that $$\lim\limits_{\epsilon\to 0} \frac{1}{\epsilon} \zeta(\frac{x} {\epsilon})$$ works for a way bigger function class than $$\mathcal C^\infty(\mathbb R)$$.

In fact, it is enough that $$\zeta\in L^1_\text{loc}(\mathbb R)$$ and $$\int_{\mathbb R}\zeta(x) {\rm d}x =1$$.

In particular, any probability density function on the real line would work.

• $$p(x) =\tfrac{1}{2}\mathbf{1}_{[-1, +1]}(x)$$, the pdf of the uniform distribution on $$[-1, +1]$$ gives $$\lim\limits_{\epsilon\to 0} \frac{1}{\epsilon} p(\frac{x} {\epsilon})=\delta(x)$$
• $$p(x) = \mathbf{1}_{[7, 8]}(x)$$, the pdf of the uniform distribution on $$[7, 8]$$ gives $$\lim\limits_{\epsilon\to 0} \frac{1}{\epsilon} p(\frac{x} {\epsilon})=\delta(x)$$
• You can take any bump-function (don't forget to normalize) such as $${\displaystyle \Psi (x)={\begin{cases}\exp \left(-{\frac {1}{1-x^{2}}}\right),&x\in (-1,1)\\0,&{\mbox{otherwise}}\end{cases}}}$$ like rural reader suggested. See this thread for some further examples
• The function doesn't even need to be positive / a pdf, something like the sinc or Airy function will work as well - you should try plotting $$\frac{1}{\epsilon} \text{Ai}(\frac{x} {\epsilon})$$ for small $$\epsilon$$.
• You really misunderstood the question. The point is how can you prove that $\lim\limits_{\varepsilon\to 0} \int_{\mathbb{R}}\dfrac{1}{\epsilon} \Psi\left (\dfrac{x}{\epsilon}\right )f(x)\ dx=f(0)$ for any $f\in C_c(\mathbb{R})$ where $\Psi$ is the bump function you wrote above. In the space of distribution this relation should be proven by considering only $f\in C^{\infty}_c(\mathbb{R})$. My question was how can we pass from $C^{\infty}(\mathbb{R})$ to $C_c(\mathbb{R})$. Jul 28, 2021 at 6:10
• Something similar was asked here, without any response too: math.stackexchange.com/questions/3481835/… Jul 28, 2021 at 6:32
• @Bogdan Well your posed question is "Is there any smooth one?" and not, given normalized, smooth function $ζ$ compactly supported on $[-1,+1]$ how do I prove $\lim_{\epsilon\to 0} \frac{1}{ε}ζ(\frac{x}{ε})= δ(x)$. You can't expect people to answer things you didn't ask for. In any case, the proof is simply (1) show that the primitive (anitderivative) of $\frac{1}{ε}ζ(\frac{x}{ε})$ converges to the Heaviside step function as $ε \to 0$ (e.g. math.stackexchange.com/q/1083721), and (2) proof that then $f$ must be equal to the dirac delta distribution. Jul 28, 2021 at 8:24