Proving that the Fourier transform of the identity is the Dirac delta using test functions.

Question

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 $a > 0$.

Answer 1

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.

Answer 2

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

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

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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

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