Fourier Inversion Theorem

Question

I have come across a $\textbf{non-formal}$ proof of Fourier's Inversion theorem and I wanted to know if reversing the order of the proof one may still use it and gain formality. The proof is:

$$\int_{-\infty}^{+\infty}\hat{f}(\omega)e^{2\pi i \omega t}d\omega = \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}f(y)e^{-2\pi i \omega y}dye^{2\pi i \omega t} d\omega = \\ =\int_{-\infty}^{+\infty}f(y)\int_{-\infty}^{+\infty}e^{-2\pi i \omega(y-t)}d\omega dy=\int_{-\infty}^{+\infty}f(y)\delta(y-t)dy = f(t)$$ were $\delta(\cdot)$ is the Dirac delta function. This proof is not formal, since we are applying the Fubini theorem to a function $\hat{f}(\omega)e^{2\pi i \omega t}$, which is not in $\mathcal{L}^{1}(\mathbb{R}\times\mathbb{R})$. But if we go through it starting from $f(t)$, we can justify using the Fubini theorem using the Young convolution inequality; then we have $f,\delta \in \mathcal{L}^{1}(\mathbb{R})$ and therefore $f\ast\delta \in \mathcal{L}^{1}(\mathbb{R}\times\mathbb{R})$, making it a formal proof.

Is this a correct assumption or am I missing something?

Answer 1

Your proof can be altered slightly to make it rigorous for Schwartz functions. Let $f\in \mathbb{S}$. Then, we have

$$\begin{align} \mathscr{F^{-1}}\{\mathscr{F}\{f\}\}(t)&=\int_{-\infty}^\infty e^{2\pi i \omega t}\int_{-\infty}^\infty f(y)e^{-2\pi i \omega y}\,dy\,d\omega\\\\ &=\lim_{L\to\infty}\int_{-L}^L \int_{-\infty}^\infty f(y) e^{2\pi i \omega(t-y)}\,dy\,d\omega\tag1\\\\ &=\lim_{L\to\infty} \int_{-\infty}^\infty f(y) \int_{-L}^L e^{2\pi i \omega(t-y)}\,d\omega\,dy\tag2\\\\ &=\lim_{L\to\infty} \int_{-\infty}^\infty f(y) \frac{\sin(2\pi L(t-y))}{\pi (t-y)}\,dy\tag3\\\\ &=f(t)\tag4 \end{align}$$

NOTES:

In going from $(1)$ to $(2)$, we applied the Fubin-Tonelli theorem, which is legitimate since $f$ is a Schwartz function.

In going from $(2)$ to $(3)$ we carried out the integral over $\omega$.

In going from $(3)$ to $(4)$ we made use of THIS ANSWER, which showed that $\frac{\sin(kL)}{\pi k}$ is a nascent Dirac Delta.