I've seen the standard proof that the Fourier transform is self-inverse (up to an overall factor determined by conventions), which is essentially equivalent to
$\int_{-\infty}^\infty e^{2 \pi i \omega x} dx = \delta(\omega)$.
That isn't too hard to see from definitions.
But intuitively, the Fourier transform of a function gives the amplitude of each component frequency in an obvious sense. Is it intuitively obvious that this operation is self-inverse or is there a deep reason why it must be so?
Best Answer
This won't be the most rigorous answer, but just a representation my intuition with regards to the Fourier transform.
In nonrelativistic quantum mechanics there are two important operators which act on $L^2(\mathbb{R})$.
The position operator,
$$ (\hat{x} \psi)(x) = x\psi(x),$$
and the momentum operator,
$$ (\hat{p} \psi)(x) = -i \frac{\partial \psi}{\partial x} (x).$$
Now the "eigenvectors" of $\hat{x}$ are $a_{x'}(x) = \delta(x-x')$ and the "eigenvectors" of $\hat{p}$ are $b_p(x) = e^{ipx}$.
Now if you study linear algebra at all you know that the choice of basis is arbitrary. Also you would know that the eigenvectors of a Hermitian operator always form a complete basis. Although our operators our Hermitian, their eigen-functions certainly don't live in $L^2(\mathbb{R})$. I'm going to completely ignore that fact and pretend that I can use them as basis.
For instance the expansion of $\psi(x)$ in terms of $a_x$'s will be,
$$ \psi(x) = \int \mathrm dx'\ c(x') a_{x'}(x),$$
we can use what we know about delta functions to conclude that the appropiate expansion coefficients, $c(x')$, are in fact $c(x')=\psi(x')$.
If we want to expand $\psi$ in the momentum basis (the $b_p$'s) then we write the superposition and try to determine the coefficients.
$$\boxed{ \psi(x) = \int \mathrm dp \ c(p) b_{p}(x)},$$
the coefficient function, $c(p)$, is a vector in $L^2(\mathbb{R})$ as well. In fact what we have writtend down here is an isomorphism between the x-basis and the p-basis. However the bahavior of the operators on functions expressed in this new bases is different.
$$ \hat{x} \psi(x) = \int \mathrm dp \ c(p) x e^{ipx} = \int \mathrm dp \ c(p) \left(-i\frac{\partial}{\partial p} \right)e^{ipx} = \int \mathrm dp \ i \frac{\partial c(p)}{\partial p} e^{ipx} $$
$$ \boxed{ \hat{x} = i \frac{\partial}{\partial p}} $$
$$ \hat{p} \psi(x) = \int \mathrm dp \ c(p) \left(-i \frac{\partial}{\partial x} \right) e^{ipx} = \int \mathrm dp \ c(p) p e^{ipx} = \int \mathrm dp \ i \frac{\partial c(p)}{\partial p} e^{ipx} $$
$$ \boxed{ \hat{p} = p} $$
This means that in the p-basis the eigenvalues of $\hat{x}$ are $a_x(p) = e^{-ipx}$. I can write $c(p)$ in terms of the x-basis then by the integral,
$$ c(p) = \int \mathrm dx \ \phi(x) e^{-ipx} $$
I've used $\phi(x)$ to represent the coefficients in this expansion, but in practice we actually know the vector in the x-basis which is isomorphic to $c$, that would be $\psi(x)$.
$$ \boxed{ c(p) = \int \mathrm dx \ \psi(x) e^{-ipx} }$$
One issue is the normalization of the transforms are off. This is because the don't have finite norms. For that you need to actually prove the Fourier inversion theorem.