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.

  • $\begingroup$ Here's a similar question math.stackexchange.com/questions/123551/… $\endgroup$
    – sudowoodo
    Commented Aug 14, 2014 at 0:11
  • 2
    $\begingroup$ 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". $\endgroup$ Commented Aug 14, 2014 at 0:48
  • $\begingroup$ 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. $\endgroup$ Commented Sep 16, 2014 at 17:08

2 Answers 2


No, it is not really related to the concept of basis. The strongest analogy I know to a basis is via the singular value decomposition for compact operators.


I would (belatedly) say yes. In particular, regarding Cameron's comment, I'd note that there's strictly speaking nothing preventing you from using a transformation whose Jacobian has a nontrivial kernel -- that's basically just a projection operator. It won't be invertible, but while that may or may not rigorously align with what mathematicians call a "basis" or "basis transformation" it certainly has a sensible and consistent meaning.


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