Question about the Bessel operator

Question

For $\nu>-1$ denote by $\{\lambda_{k,\nu}\}_{k\in\mathbb{N}}$ the succesive positive zeros of the Bessel function of the first kind $J_{\nu}$. The Bessel operator is given by \begin{equation*} L_\nu=-\frac{d^2}{dx^2}+\frac{\nu^2-1/4}{x^2}. \end{equation*} It is known that functions \begin{equation*} \psi_k^\nu(x)=d_{k,\nu}(\lambda_{k,\nu}x)^{1/2}J_\nu(\lambda_{k,\nu}x), \quad d_{k,\nu}=\frac{\sqrt{2}}{\lambda_{k,\nu}^{1/2}|J_{\nu+1}(\lambda_{k,\nu})|} \end{equation*} are eigenfunctions of $L_\nu$ with $$ L_\nu\psi_k^\nu=\lambda_{k,\nu}^2 \psi_k^\nu. $$ Moreover, the system $\{\psi_k^\nu\}_{k=1}^\infty$ is orthonormal and complete in $L^2((0,1), dx)$.

The self-adjoint extension of $L_{\nu}$ associated with the Fourier-Bessel system is defined in $L^2((0,1), dx)$ as $$ \widetilde{L}_{\nu}f = \sum_{k=1}^{\infty} \lambda_{k,\nu}^2 \langle f, \psi_k^{\nu} \rangle \psi_k^{\nu}, $$ on the domain $$ \textrm{Dom } \widetilde{L}_{\nu} = \bigg\{ f \in L^2((0,1), dx) : \sum_{k=1}^{\infty} \big| \lambda_{k,\nu}^{2} \big\langle f, \psi_k^{\nu}\big\rangle \big|^2 < \infty \bigg\}. $$

I would like to show that if $\nu, \alpha>-1$ are such that $\nu\neq\alpha$, then $$ \textrm{Dom } \widetilde{L}_{\nu}\neq\textrm{Dom } \widetilde{L}_{\alpha}. $$ I have made some little progress on the above question, but I am not able to settle it fully. I can construct a smooth function $f$ such that $L_\nu(f)\in L^2$ and $L_\alpha(f)\notin L^2$. Morally, that should imply $f\in\textrm{Dom }\widetilde{L}_{\nu}$ and $f\notin\textrm{Dom } \widetilde{L}_{\alpha}$, but I do not see how to make it rigorous with the above definition of the domain.

I would very much appreciate any hints.

Answer 1

It is probably best to do this by hand, by looking at the domains described in the usual way as those $f$ with $Lf\in L^2$ that satisfy certain boundary conditions at $x=0,1$ (in fact, there will be no boundary condition at $x=0$ when $\nu^2\ge 1$).

But we can also avoid dealing with boundary conditions explicitly, as follows: if we had $D(L)=D(L')$, then $D(L)\subseteq D(M)$, with $M$ denoting the (self-adjoint) operator of multiplication by $1/x^2$ on its natural domain.

But from the behavior of the Bessel functions near $x=0$, we also know that the eigenfunctions of $L$ satisfy $\psi(x)\simeq x^{\nu+1/2}$, so $\psi\in D(M)$ only if $\nu> 1$.

If indeed $\nu\ge 1$, then we transition to limit point case at $x=0$, and no boundary condition is imposed here. It seems possible or even plausible that $D(L)\subseteq D(M)$ now. If this is correct, then $M$ is $L$-bounded, and the bound would drop to values $<1$ if we multiply $M$ by a small constant. Moreover, it cannot get larger if we make $\nu$ larger. That would mean that $D(L)$ is actually constant in the range $\nu>1$.

Answer 2

This is not an answer to the question but the main argument to show that the domain is the intersecion between the domain of the second derivative and of the multiplication operator when $\nu>1$. Write $b=\nu^2- \frac 14$ and for $u \in C_c^\infty (0,\infty)$ integrate by parts $Lu=f$ against $x^{-2}u$. One gets \begin{equation} \label{1} \int_0^\infty \left (\frac{u'^2}{x^2} +b \frac{u^2}{x^4}-2 \frac{u u'}{x^3}\right )=\int_0^\infty \frac{fu}{x^2} \leq \|f\|_2 \|x^{-2} u\|_2 \end{equation} and the aim is to estimate the $L^2$ norm of $x^{-2} u$ in terms of $f$ under $b > \frac 34$. By Cauchy -Schwartz we have $\|x^{-3}u u'\|^2_1 \leq \|x^{-1} u'\|_2 \|x^{-2}u\|_2$ and by Hardy's inequality (see below) $\|x^{-2} u\|_2 \leq \frac 23 \|x^{-1} u'\|_2$. With $A^2=\|x^{-1} u'\|_2^2, B^2=\|x^{-2}u\|_2^2$ the RHS of the displayed equation is bigger than $A^2+bB^2-2AB$. With the constraint $B \leq \frac 23 A$ the quadratic expression above is non-negative for $b \geq \frac 34$, hence dominates $\epsilon B^2$ for $b >\frac 34$ and allows to estimate, by Cauchy-Schwartz again, $B$ in terms of $\|f\|_2$.

Concering the Hardy inequality, I used $\frac u x=\int_0^1 u'(sx)ds$ and take norms with respect to the measure $x^{-2} dx$. By Minkowski inequality (for integrals) $\|x^{-1} u\|_{L^2(x^{-2} dx)} \leq \int_0^1 \|u'(sx)\|_{L^2(x^{-2} dx)}ds$ and the result follows by simple computations.