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For every finite Schwartz function $f$ and every $t \in \mathbb{R}$, define the Dirac delta as

$$\int_{-\infty}^\infty dt f(t) \delta(t-t') := f(t').$$

How do I proof with this definition that $$\int_{-\infty}^\infty d\omega e^{i\omega(t-t')} =2\pi \delta(t-t'). $$ When I use complex analysis, this gives me zero for all $t, t'$ since there are no poles and the contour, which clearly is wrong.

I would like to know this way of a proof to understand what the following integral is in distribution sense: $$\int_{-\infty}^\infty d\omega e^{i\omega(t-t')} |\omega|^{a}, $$ with real $a > 0$.

Jendrik Stelzner
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Audrique
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  • See THIS ANSWER in which I developed the Fourier Transform for $|x|^\alpha$ for all real $\alpha$. – Mark Viola Dec 09 '22 at 13:55
  • Thanks! This is really helpful! – Audrique Dec 10 '22 at 13:07
  • You're welcome. My pleasure. – Mark Viola Dec 10 '22 at 18:47
  • Hi,@MarkViola, I have been going throught the answer you suggested above, is there some more reading material on the $D_2$ distribution that you use in that answer. And I'm also a bit stuck on why the second contribution in equation 8 disappears. I was not sure if this was the place to ask but I didn't know of any other way. (Amazing answer btw.) – Audrique Dec 11 '22 at 16:03
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    For the distribution $D_2$, it is a generalization o the $\lambda_r$ functional that Terry Tao defines in Exercise 13 on THIS PAGE. I'm not sure what is meant by the "second contribution in equation 8" of THIS ANSWER – Mark Viola Dec 11 '22 at 18:36
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    And feel free to up vote any answer that you find useful. ;-) – Mark Viola Dec 11 '22 at 18:41
  • Thanks! I meant how did the second equality in (8) hold. But I think I found now how it worked, combining the two exponentials gives $e^{in\pi} = 1$ and using $\sin(sign(k)a)=sign(k)\sin(a)$. – Audrique Dec 11 '22 at 22:56

2 Answers2

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Given $t'$ for each $A$ let $$g_A(t)=\int_{-A}^A e^{i\omega(t-t')}d\omega = 2 \frac{\sin(A(t-t'))}{t-t'}$$ Then $\lim_{A\to \infty} g_A$ converges to $2\pi \delta(.-t')$ in the sense of tempered distributions.

To see this let $$h(t)=\int_{-\infty}^t \frac{2\sin(u)}{u}du$$ For any $\phi \in S(\Bbb{R})$ $$\int_{-\infty}^\infty \phi(t) g_A(t)dt = -\int_{-\infty}^\infty \phi'(t+t') h(At)dt$$ $h(At)$ converges to $2\pi 1_{t >0}$ in $L^1_{loc}$ and uniformly away from $(-\epsilon,\epsilon)$ so $-\int_{-\infty}^\infty \phi'(t) h(At)dt$ converges to

$$-\int_{t'}^\infty 2\pi\phi'(t)= \phi(t')$$

Note that this is the proof of the Fourier inversion theorem at least for the Schwartz functions.

reuns
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Let $f(t)=e^{-i\omega t'}$ and $\phi\in \mathbb{S}$. Then we have

$$\begin{align} \langle \mathscr{F}^{-1}\{f\},\phi\rangle&=\langle f,\mathscr{F}^{-1}\{\phi\}\rangle\\\\ &=\int_{-\infty}^\infty e^{-i\omega t'} \frac1{2\pi}\int_{-\infty}^\infty \phi(t)e^{i\omega t}\,dt\,d\omega\\\\ &=\phi(t') \end{align}$$

where the last equality is a consequence of the Fourier Inversion Theorem (See Appendix). Therefore, in distribution we have

$$\mathscr{F}^{-1}\{f\}(t)=\delta(t-t')$$

APPENDIX: PROOF OF THE INVERSION THEOREM FOR SCHWARTZ FUNCTIONS

In THIS ANSWER, THIS ONE and in the Alternative Development section of THIS ANSWER, I proved that $\frac{\sin(Lt)}{\pi t}$ is a nascent Dirac Delta. Using this result, we see that

$$\begin{align} \int_{-\infty}^\infty e^{-i\omega t'} \frac1{2\pi}\int_{-\infty}^\infty \phi(t)e^{i\omega t}\,dt\,d\omega&=\frac1{2\pi}\int_{-\infty}^\infty \int_{-\infty}^\infty \phi(t+t')e^{i\omega t}\,dt\,d\omega\\\\ &=\lim_{L\to \infty}\frac1{2\pi} \int_{-\infty}^\infty \phi(t+t') \int_{-L}^L e^{i\omega t}\,d\omega\,dt\\\\ &=\lim_{L\to \infty} \int_{-\infty}^\infty \phi(t+t') \frac{2\sin(Lt)}{\pi t}\,dt\\\\ &= \phi(t') \end{align}$$

Mark Viola
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