# Can integral transforms be viewed as change of basis formulas?

Forgive any lack of rigor.

If you have a countable orthonormal basis $B$ for a Hilbert space $H$ , then any function $f \in H$ can be expressed as $$f(t) = \sum\limits_{g \, \in \, B} \langle f, \, g \rangle \, g(t)$$ where $\langle f, g \rangle$ might be something like $$\int_{t_1}^{t_2}f(t) \, g(t) \, \mathrm{d}t$$

So essentialy $(a_1, a_2, \dots, a_{N-1})$ is the coordinate vector of $f$ for basis $B$.

Now my question is: are integral transforms the continuous version of this? Can the kernel function of an integral transform be viewed as a sort of basis? For example suppose you do have a basis $B' = \{ K_u(t) : u \in (u_1, u_2) \}$, and a function $f'$. Can you use a similar expression for calculating the coordinate vector of $f'$ relative to $B'$ as with $f$? If so what conditions does $B'$ have to satisfy?

Here's an example of the finite case using sines and cosines for the basis - http://en.wikipedia.org/wiki/Fourier_series#Hilbert_space_interpretation

Thanks for helping.

• Here's a similar question math.stackexchange.com/questions/123551/… Commented Aug 14, 2014 at 0:11
• In brief, "yes". The kernel-function of an integral transform (generally known as "Schwartz kernel") is not so much an analogue of a "basis", but an analogue of a "matrix". Commented Aug 14, 2014 at 0:48
• Not quite. If the resulting integral transform is unitary, then it's not a bad analogy. If the resulting integral transform is, say, compact (e.g. a Hilbert-Schmidt integral transform), then it's a terrible analogy as it will have nontrivial kernel. Commented Sep 16, 2014 at 17:08