The Fourier transform of Dirac delta is often naively calculated by considering Delta function as a function that makes sense within an integral and by using its fundamental property:

$$ \int_{-\infty}^{+\infty} f(x) \delta(x-x_0)dx = f(x_0) $$

Then the Fourier transform of Dirac delta function can be evaluated to be $\hat{\delta(p)} = \frac{1}{\sqrt{2\pi}}$ simply by applying such property to the definition of Fourier transform.

However, here I am seeking for a formal proof using theory of distributions. In particolar, we know that the succession $D_n(x) := nD(nx)$ approximates Dirac delta $\delta(x)$, with $D(x) = \frac{1}{\sqrt{\pi}} e^{-x^2}$. My objective is to derive the same result above discussed using this approach.

The Fourier transform of a Gaussian-like function is known from theory, therefore it is straightforward that:

$$ \mathcal{F} [ D(x) ] = \mathcal{F} \left[ \frac{e^{-x^2}}{\sqrt{\pi}} \right] = \frac{e^{-\frac{p^2}{4}}}{\sqrt{2\pi}} $$

But this doesn't itself give me directly the solution, as there's still the term $e^{-\frac{p^2}{4}}$; how should I proceed to prove my thesis formally?


1 Answer 1


In the language of distributions, the Dirac delta distribution is the map $\delta$ from the space of test functions (smooth compactly supported functions) to, say $\mathbb{R}$ with the "operation" $(\delta, f) = f(0)$ for every test function $f$.

To figure out the Fourier transform of a distribution, you need to determine the Fourier transform of a test function $f$.

$$\widehat{f}(\xi) = \frac{1}{\sqrt{2\pi}}\int_\mathbb{R} e^{-ix \xi}f(x) \, dx$$

By definition, the Fourier transform of a distribution $\varphi$ is defined by $(\widehat{\varphi},f)=(\varphi,\widehat{f})$ for every test function $f$.

EDIT: As commenters below pointed out, I should say Schwartz function instead of test function and tempered distribution instead of distribution..


$$(\widehat{\delta},f) = (\delta,\widehat{f}) = \widehat{f}(0) = \frac{1}{\sqrt{2\pi}}\int_\mathbb{R} e^{-i x \cdot 0} f(x) \, dx =\frac{1}{\sqrt{2\pi}}\int_\mathbb{R} f(x) \, dx = (\frac{1}{\sqrt{2\pi}},f)$$

where the last equality is because the "constant" distribution 1 is regular, i.e., can be represented in integral form. Therefore, as a distribution, $\widehat{\delta} = (2\pi)^{-1/2}$.

  • $\begingroup$ Your answer is indeed complete and formal, which is pretty satisfactory itself. What confuses me is that my question arises from a problem in which I am asked to derive $\hat{\delta}$ specifically using the succession written above; however it seems to me in conflict with the necessity to use a generic test function as you did. Could you elaborate on this? $\endgroup$
    – John
    Feb 1, 2018 at 16:34
  • 3
    $\begingroup$ RIght, except you should say "tempered distribution" and "schwarz function" instead of "distribution" and "test function". In general a distribution doesn't have a Fourier transform. (The definition doesn't work because if $f$ is a test function it does not follow that $\hat f$ is a test function.) $\endgroup$ Feb 1, 2018 at 16:35
  • $\begingroup$ @John sorry, I misread what you were asking. Try looking at $(\widehat{D_n},f) = (D_n,\widehat{f})$, and apply Fubini's theorem. $\endgroup$ Feb 1, 2018 at 17:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.