Is integral transform an expansion over a self-ajoint operator's eigenfunctions? In my understanding, integral transform is actually computing expansion coefficient over some particular (over)complete orthonormal basis (e.g. $e^{i\omega x}$ and $e^{-\frac{x^2}{2}}H_n(x)$ for the Fourier and Hermite transform).
It seems that these basis functions can be regarded as the eigenfunctions of some self-adjoint operator on a Hilbert space (e.g. $\frac{d^2}{{dt}^2}$ and $\frac{d^2}{{dx}^2}-x^2$ under vanishing Dirichlet boundary condtion for the Fourier and Hermite transform).
According to the spectral theorem, a self-adjoint operator has (over)complete eigenfunction set under proper conditions.
So does integral transform amount to, at least under certain circumstances, expansion over a self-ajoint operator's eigenfunctions?
 A: It depends on what you mean.
For a concrete example, the Abel transform $A$ is defined by
$$
(Af)(x)
=
2\int_x^\infty[1-(y/r)^2]^{-1/2}f(r)dr
$$
and inverted by a similar-looking formula.
Or for a perhaps silly example, consider the integral operator $I$ defined by $(If)(x)=\int_0^xf(y)dy$.
These are both integral operators and they do appear in practice, but it doesn't seem to be very useful to think of them as as eigenfunction expansions.
However, if you have an invertible integral transform $T$ whose inverse is given as
$$
f(x)=\int_\mathbb R k(x,y)(Tf)(y)dy,
$$
you can always argue that $f(x)$ is expanded in terms of the functions $k(\cdot,y)(Tf)(y)$, where $y$ ranges over the real line (or any other measure space of your choice).
Then the question is whether you can engineer a linear operator whose eigenfunctions are $k(\cdot,y)$.
If these functions are suitably linearly independent, you can define an operator $M$ so that $[Mk(\cdot,y)](x)=yk(\cdot,y)$, so that you formally have an eigenfunction expansion.
Whether there is a useful or tangible operator is a whole other problem.
