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.

  • $\begingroup$ 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/… $\endgroup$ Commented Jul 27, 2021 at 16:37
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    $\begingroup$ Start with $\exp(-1/(1 - x^2))$ on $(-1, 1)$. $\endgroup$ Commented Jul 27, 2021 at 16:42
  • $\begingroup$ @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? $\endgroup$
    – Bogdan
    Commented Jul 27, 2021 at 17:07
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    $\begingroup$ en.wikipedia.org/wiki/Bump_function $\endgroup$
    – Hyperplane
    Commented Jul 27, 2021 at 18:50
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    $\begingroup$ 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. $\endgroup$
    – jawheele
    Commented Jul 28, 2021 at 0:08

2 Answers 2


After reading the comments in hyperplane's answer, here's a very standard theorem:

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 $t>0$)


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".

  • $\begingroup$ 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? $\endgroup$
    – Mark Viola
    Commented Feb 21, 2023 at 4:54
  • $\begingroup$ @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 $\endgroup$
    – peek-a-boo
    Commented Feb 21, 2023 at 5:41
  • $\begingroup$ I was just curious since I've seen many of your excellent posts on distributions. $\endgroup$
    – Mark Viola
    Commented Feb 21, 2023 at 15:21
  • $\begingroup$ (+1) for the solution, before and after the edit ;-) $\endgroup$
    – Mark Viola
    Commented Feb 23, 2023 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 $\epsilon$.
  • $\begingroup$ 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 $C_c(\mathbb{R})$. $\endgroup$
    – Bogdan
    Commented Jul 28, 2021 at 6:10
  • $\begingroup$ Something similar was asked here, without any response too: math.stackexchange.com/questions/3481835/… $\endgroup$
    – Bogdan
    Commented Jul 28, 2021 at 6:32
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    $\begingroup$ @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. $\endgroup$
    – Hyperplane
    Commented Jul 28, 2021 at 8:24

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