It is well known that the orthonormal basis for $L_{2}[-\pi, \pi]$ is $\Omega = \{ e^{-jmt} \}_{m \in \mathbb{Z}}$. We extend this to $L_{2}(\mathbb{R})$ via the Fourier Transform, which is an integral transform of the form:

$$ f(\omega) = \displaystyle\int\limits_{-\infty}^{\infty} f(t) K(\omega, t) dt = \displaystyle\int\limits_{-\infty}^{\infty} f(t) e^{-j \omega t} dt $$

I.e. where $K(\omega, t) = e^{-j\omega t}$ We know this kernel is unitary, so the extension to $L_{2}(\mathbb{R})$ is simply:

$$ f(t) = \displaystyle\int\limits_{-\infty}^{\infty} f(\omega) K^{\ast}(\omega, t) d\omega = \displaystyle\int\limits_{-\infty}^{\infty} f(\omega) e^{j \omega t} d\omega $$


Now suppose I apply some unitary transform $U$ to $\Omega$, giving us the new orthonormal basis for $L_{2}[-\pi, \pi]$ denoted by $\Omega_{U} = \{ Ue^{-jmt} \}_{m \in \mathbb{Z}}$. I want to extend this to $L_{2}(\mathbb{R})$ in the same way. So this would mean:

$$ f(\omega) = \displaystyle\int\limits_{-\infty}^{\infty} f(t) K_{U}(\omega, t) dt = \displaystyle\int\limits_{-\infty}^{\infty} f(t) \big(Ue^{-j \omega t}\big) dt $$

Which means the extension to $L_{2}(\mathbb{R})$ takes the form:

$$ f(t) = \displaystyle\int\limits_{-\infty}^{\infty} f(\omega) K_{U}^{\ast}(\omega, t) d\omega = \displaystyle\int\limits_{-\infty}^{\infty} f(\omega) \big(Ue^{-j \omega t}\big)^{\ast} d\omega = \displaystyle\int\limits_{-\infty}^{\infty} f(\omega) \big(U^{-1}e^{j \omega t}\big) d\omega$$

In other words:

$K_{U}(\omega, t) = Ue^{-j\omega t}$ and $K_{U}^{\ast}(\omega, t) = U^{-1}e^{j\omega t}$


1.) Is this the correct way to build an integral transform from an orthonormal basis? If not, where did I go wrong?

2.) If $\Omega$ was some arbitrary orthonormal basis $\{e_{m}\}_{m \in \mathbb{Z}}$ for some space $L_{2}[a,b]$, would this still be the way to go to extend the basis into an integral transform for all of $\mathbb{R}$?

3.) If $\Omega$ is no longer orthogonal, can I still do these two extensions (one with $U$ and one without)?

This is of course assuming we can actually compute the integrals.


  • $\begingroup$ The individual exponentials are not in $L^2$. What meaning does unitary transformation have when you are applying it to objects that are not in the space? $\endgroup$ – DisintegratingByParts Aug 3 '18 at 13:20

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