Lower bound for the eigenvalue

Question

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).

Consider a compact integral operator $$ F_c(\psi)(x)=\int_{-1}^1\frac{\sin(c(x-y))}{\pi(x-y)}\psi(y)dy. $$

It is known that $\psi_{k,c}$ are the eigenfunctions of the operator $F_c$. So, $$ F_c=\lambda_k\psi_{k,c}, $$ where $\lambda_k$ are eigenvalues of $F_c$. (Note, it is known that $\lambda_k\leq \frac{c}{2}\left(\frac{ec}{4k}\right)^{2k}$.)

I would like to find a lower bound for $\lambda_k$.

Any references and ideas will be very helpful.

Thank you.

Answer 1

OK, here is the sketch of the bound I mentioned. The operator in question is just $PQP$ where $P$ is the space projection to the interval $[-1,1]$ and $Q$ is the Fourier projection to the interval $[-c,c]$. So, we just need a $k$-dimensional space $V$ of functions such that $\|QPf\|\ge A(k,c)\|f\|$ for every function $f$ in that space. Then $\lambda_k\ge A(k,c)^2$. Without thinking too much, let us take $V$ to be the space of functions that are constant on each of $k$ intervals of length $2/k$ partitioning $[-1,1]$. Assume also that $c/k\ll 1$ (otherwise the bound I mentioned is total nonsense because the norm of the operator is at most $1$). Then if the function $f$ takes value $a_j$ on the $j$th interval, we have to estimate from below $\int_{[-c,c]}|\widehat f|^2$ under the condition $\sum_j|a_j|^2=k$ (or something like that). However,
$$ \widehat f(y)=g(y)\sum_i a_ie^{4\pi ij/k} $$ where $g(y)$ is the Fourier transform of the characteristic function of an interval of length $2/k$, which, under the assumption $c/k\ll 1$ is essentially $1/k$ on the whole interval $[-c,c]$ in absolute value. Thus we are down to estimating a trigonometric polynomial of given degree on a small arc, more precisely, on $c/k$ of the full period, from below. In $L^2$, it is at least $(ac/k)^{k-1/2}$ of the full norm (choose between Berstein, Remez, and Turan for the reference name), whence $\lambda_k\ge (ac/k)^{2k-1}$ (Here I start enumeration with $1$ unlike in your post and my comment where it starts with $0$).

That's it.

Answer 2

I suppose you order your eigenvalues by $\lambda_{k} > \lambda_{k+1}$ and that it should be clear that they are positive. If yes, using the mini-max principle one should be able to get lower bounds on the eigenvalues. Of course, this would require having some idea what the correct test functions look like, but one should be able to obtain this from the Sturm--Liouville operator picture.

Of course, this is just what I would do having no idea on the background of the problem. There might be a smarter solution available.