2
$\begingroup$

I would like to know if the Prolate Spheroidal Wavefunctions (PSWFs, defined below) are in $L^1(\mathbb{R})$. I know that they are square integrable, but cannot decide about absolute integrability.

The Prolate Spheroidal Wave Functions are eigenfunctions of the following integral equation: $$\int_{-T}^T\varphi_n(x) \text{sinc}(t-x) dx = \lambda_n \varphi_n(t)$$

where $\text{sinc}(t) = \sin(\pi t)/ \pi t$. Alternatively (as discovered by Slepian et al.) they are also the eigenfunctions of the following differential operator: $$(1-t^2)\frac{d^2\varphi_n}{dt^2}-2t\frac{d\varphi_n}{dt} -(2 \pi T \Omega)^2t^2 \varphi_n = \mu_n \varphi_n$$

The Prolates are bandlimited to $[-\Omega/2, \Omega/2]$ and maximally time-concentrated on the interval $[-T, T]$ (see the series of papers by David Slepian, Landau, and Pollack). As such they are entire functions in the complex variable $t$.

$\endgroup$
3
  • $\begingroup$ By just staring at the integral equation, it seems we obtain that $\varphi\simeq 1/t$. $\endgroup$ Commented May 31, 2018 at 1:17
  • $\begingroup$ @ChristianRemling Yep. Just define the energy in the usual way, conclude that it decays at most like $t^{-2}$ and recall that for bandlimited functions the integrability of $\varphi$ implies that of $\varphi'$. $\endgroup$
    – fedja
    Commented May 31, 2018 at 12:26
  • $\begingroup$ @fedja could you kindly elaborate a little. Do we conclude from this that Prolates are not absolutely integrable? $\endgroup$
    – Iconoclast
    Commented May 31, 2018 at 14:31

1 Answer 1

2
$\begingroup$

They are not in $L^1$. The principal term of the asymptotics is $$\frac{e^{\pm iTx}}{x}.$$ This asymptotics is written for example here:

Richard-Jung, F.; Ramis, J.-P.; Thomann, J.; Fauvet, F. New characterizations for the eigenvalues of the prolate spheroidal wave equation. (English summary) Stud. Appl. Math. 138 (2017), no. 1, 3–42.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .