Let $J\_1$ be the Bessel function of the first kind and let $H\_1(x) = \frac{J\_1(|x|)}{|x|}$ for $n = 1$. Define the operator $Tf(x) = (f \* H\_1)(x)$ from $L^2$ to $L^2$.


Since the $H\_1$-function is the Fourier transform of something it must be in $L^2$, so we have a Hilbert-Schmidt operator which is in this case self-adjoint and compact, so the spectral theorem applies which for example says all the eigenvalues form a countable set, are real and go to zero.


What I am now interested in is the largest eigenvalue of $T$. What theorem or method could I try to obtain this? (I don't need a full solution, just a hint would suffice).



Eigenvalues convolution-type operator
In Fourier space, the operator $T$ is given by
$$\hat{(T f)}(\xi)=c\sqrt{1-\xi^2}\hat f(\xi)$$
for some constant $c\ne0$ ($c$ dependson the normalization chosen for thr Fourier transform.) If $\lambda$ is an eigenvalue and $f$ is a corresponding eigenfunction, then for almost all $\xi$
$$\lambda \hat f(\xi)=c\sqrt{1-\xi^2}\hat f(\xi).$$
It follows that the only eigenvalue is $\lambda=0$, and the coresponding eigenfunctions are all $L^2$ functions whose Fourier transform vanish almost everywhere on $[-1,1]$.


