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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?

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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..

Therefore

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

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

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  • $\begingroup$ 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 '18 at 16:34
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    $\begingroup$ 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.) $\endgroup$ – David C. Ullrich Feb 1 '18 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. $\endgroup$ – welshman500 Feb 1 '18 at 17:10

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