Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For a given real number $c>0$ define functions $\left(\psi_k^c(\cdot)\right)_{k\ge0}$, as an eigenfunctions of the Sturm-Liouville operators $L_c$ defined $$ L_c(\psi)=(1-x^2)\frac{d^2\psi}{dx^2}-2x \frac{d\psi}{dx}-c^2x^2\psi $$ ( in fact, $\psi_k^c$ is called the prolate spheroidal wave function).

Let $f \in L_1(R)$ and $\hat{f}(\xi)=\int_Rf(t)e^{-it\xi}dt$ denotes the Fourier transform of $f$.

Let for $T=[-a,a]$,$\Omega=[-b,b]$

$$ \int_{|x|\ge T}|f(x)|^2dx\le \epsilon^2 $$ and $$ \int_{|\xi|\ge \Omega}|\hat{f}(\xi)|^2d\xi\le \epsilon^2 $$

By sutable changes of variables $f\longrightarrow \Omega^{-1/2}f\left(\frac {x}{\Omega}\right)$, $T\longrightarrow \Omega T$ one can get $T=\Omega=[-c,c]$.

Given the above two inequalities one can get $$ f(x)\approx \sum_{|k|<2\Omega T}\langle f, \psi_k^c\rangle\psi_k^c. \tag{1} $$

Help me please with the following two questions:

1). What the error in (1)?

2). How does $\epsilon$ come out?

Thank you.

share|improve this question
You say "let $T = [-a,a]$" and later have $|x|\geq T$. Do you mean $x\geq a$? –  Thomas Andrews May 16 '12 at 17:26
@ Thomas Andrews: We can say that the integrals are iver complements $T^c$ and $\Omega^c$. So, $|x|\geq a$ and $\xi\ge b$. –  Alex K. May 16 '12 at 18:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.