I'm looking for a rigorous proof of the statement: $\delta(x) = \lim_{\epsilon->0} \frac{\sin(x/\epsilon)}{\pi x}$ (see (37)).

For any non-zero value of x, LHS of the above is by definition zero. But, for any non-zero value of x, the limit in RHS simply does not exist. So, how is this statement proved?

  • $\begingroup$ With the new "interpretation" of the limit (sifting property at the origin) as suggested by @Ian, it is possible to prove the statement. However, in the classical definition of the limit per se, I believe the statement is not correct simply because the limit does not exist so not sure why it is quoted like that in text-books. To me, that statement is misleading without the additional important qualification of the "interpretation" of the limit as underlined by Ian. $\endgroup$ Jan 2, 2015 at 14:26
  • $\begingroup$ Related Phys.SE question: physics.stackexchange.com/q/240191/2451 $\endgroup$
    – Qmechanic
    Sep 19, 2019 at 2:08

2 Answers 2


The Dirac delta is to be defined as a distribution: a linear functional acting on the space of smooth compactly supported functions. So this limit is to be understood as:

$$\lim_{\varepsilon \to 0^+} \int_{-\infty}^\infty \frac{\sin \left ( \frac{x}{\varepsilon} \right )}{\pi x} f(x) dx = f(0)$$

whenever $f$ is smooth and has compact support. (Actually, the Dirac delta may be extended to continuous compactly supported functions, but this is the starting point.)

That said, the argument here is similar to the usual situation of showing that a certain family of functions is an approximate identity. First you should rewrite in a more convenient form:

$$g_\varepsilon(x) \equiv \frac{\sin \left ( \frac{x}{\varepsilon} \right )}{\pi x} = \frac{1}{\varepsilon} \frac{\sin \left ( \frac{x}{\varepsilon} \right )}{\pi \frac{x}{\varepsilon}}.$$

Since $\int_{-\infty}^\infty \frac{\sin(x)}{x} dx = \pi$, we see that $g_\varepsilon$ has integral $1$. The next condition to be checked is that for any compact set $K$ and any $\delta > 0$:

$$\lim_{\varepsilon \to 0^+} \int_{K \setminus (-\delta,\delta)} g_\varepsilon(x) dx = 0.$$

If you haven't seen this before, the intuition here is that $g_\varepsilon$ has the same mass for any $\varepsilon$ but that this mass is becoming concentrated in an arbitrarily small interval around $0$.

Most commonly, we choose $g_\varepsilon$ to be absolutely integrable on the whole real line, and in this case we could look at $\mathbb{R} \setminus (-\delta,\delta)$ instead. In this situation, we don't have absolute integrability. But if we only work with test functions with compact support then we can make this generalization to get things to work out, by taking $K$ to be the support of $f$ in the hypothesis above.

A related proof is by Fourier transforms. Here's a sketch of this proof:

The sinc function (with appropriate scaling) is the Fourier transform of the indicator function of an interval centered at $0$. The delta function is the Fourier transform of the constant function $1$ (again with appropriate scaling). When $\varepsilon$ is small (when the mass has gotten concentrated), $f_\varepsilon$ is the Fourier transform of an indicator function of a large interval centered at zero. (The small/large phenomenon here is more generally the uncertainty principle in Fourier analysis.) Since the indicator functions are converging to the constant function $1$, their Fourier transforms are converging to the Fourier transform of $1$, i.e. $\delta$.

This proof with Fourier transforms is harder to formalize. For instance, we need to show that the Fourier transform is continuous in some appropriate sense. But it may be more intuitive.

  • 2
    $\begingroup$ What theorem do you use to proof your first equation. You seem to.imply it is enough to check unit mass, and vanishing outside.of delta region. Can you proof how this implies the original statement? $\endgroup$
    – lalala
    Aug 6, 2018 at 5:46
  • $\begingroup$ @lalala I don't think this theorem has any standard name, it is just "the conditions for a sequence to be an approximate identity". $\endgroup$
    – Ian
    Apr 28, 2020 at 14:19
  • $\begingroup$ I believe now it is not true. You could add the derivative of the delta function to the deltabfunction and both of your conditions would still be true but the conclusion not $\endgroup$
    – lalala
    Apr 29, 2020 at 5:29
  • $\begingroup$ @lalala As I mentioned, there are additional technicalities involved when your function isn't absolutely integrable. The "standard theorem" has this absolute integrability assumption. If that assumption isn't there then you need to do some more problem-specific massaging. $\endgroup$
    – Ian
    Apr 29, 2020 at 13:24
  • 2
    $\begingroup$ Na. Unfortunately not. Question is asking for a rigerous proof (which interests me at well). Then your third formula uses a compactly supported function f, but f is not used only supp(f), so you mean for every compact set (or you forgot to write f in the integral?). If you use a theorem (which is actually the beef of this question), can you either reference it properly or write it down in detail (and then the original question is basically how this theorem is proven). I have no beef with the intuition part, but the whole answer is definitely not rigorous. $\endgroup$
    – lalala
    May 7, 2020 at 20:06

In addition to the answer by Ian, one very important lemma you need to prove the second part, i.e $$\lim_{\varepsilon \to 0^+} \int_{K \setminus (-\delta,\delta)} g_\varepsilon(x) dx = 0.$$ is the Riemann-Lesbegue lemma which shows how an increasingly oscillatory function under integration approaches zero measure. You can find it mentioned here :- https://onlinelibrary.wiley.com/doi/pdf/10.1002/9780470723876.app3(Page 498)


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