Is the Sturm-Liouville operator self-adjoint? Let $L$ be the Sturm-Liouville operator with domain
$$  D(L)=\Bigg\{\psi\in C^\infty([a,b])\cap L^2([a,b]) : 
\begin{cases}
\alpha\,\psi(a)+\beta\, \psi'(a)&=0\\ \gamma\,\psi(b)+\delta \,\psi'(b)&=0
\end{cases}
\Bigg\}\,.$$
and action given by
$$
L\psi=-(p\psi')'+q\psi\,,
$$
where $p,q\in C^\infty([a,b])$ and $p>0$ on $[a,b]$.
I'm wondering whether $L$ is self-adjoint (in the sense of abstract functional analysis). $L$ is an unbounded linear operator so it isn't enough to check that $L$ is symmetric.
I suspect that $L$ is essentially self-adjoint (i.e. its closure is self-adjoint) but I don't know how to prove it.
Can you give me a hint? Thanks in advance.
 A: $L$ is essentially self-adjoint in the regular case, as you noted. You can show this by exhibiting explicit bounded inverses $(L\pm iI)^{-1}$ on the domain $\mathcal{D}(L)$ consisting of twice absolutely continuous functions $f$ on $[a,b]$ that satisfy the required endpoint conditions at both $x=a$ and $x=b$. This is done using a Green function solution for the resolvent operator $R(\lambda)=(L-\lambda I)^{-1}$ of the form
$$
      R(\lambda)f=\frac{\varphi_{\lambda}(x)}{\omega(\lambda)}\int_a^xf(x')\psi_{\lambda}(x')dx+\frac{\psi_{\lambda}(x)}{\omega(\lambda)}\int_x^bf(x')\varphi_{\lambda}(x')dx'
$$
where $\psi_{\lambda},\varphi_{\lambda}$ are non-trivial classical solutions of $(L-\lambda I)f=0$ which satisfy
$$
        \alpha\varphi_{\lambda}(a)+\beta\varphi_{\lambda}'(a)=0 \\
        \gamma \psi_{\lambda}(b)+\delta\psi_{\lambda}'(b)=0.
$$
A simple normalization that eliminates some complexity can be specified by requiring
$$
     \varphi_{\lambda}(a)=\beta,\;\;\varphi_{\lambda}'(a)=-\alpha \\
     \psi_{\lambda}(b)=\delta,\;\;\psi_{\lambda}'(b)=-\gamma.
$$
Then the Wronskian $\omega(\lambda)$ of these solutions is a function that does not depend on $x$:
$$
      \omega(\lambda)=(p\varphi_{\lambda}')\psi_{\lambda}-\varphi_{\lambda}(p\psi_{\lambda}')
$$
The Wronskian either vanishes at no $x\in [a,b]$ or it vanishes identically, because it depends only on $\lambda$. The Wronskian vanishes iff $\{ \varphi_{\lambda},\psi_{\lambda} \}$ is a dependent set of functions of $x$, which is precisely when both functions satisfy $(L-\lambda I)h=0$ as well as the specified conditions at $x=a$ and $x=b$ (in other words, $\lambda$ is an eigenvalue of $L$.) This cannot happen for non-real $\lambda$, and it must happen for an infinite sequence of real values of $\lambda$ that has its only cluster point at $\infty$ (otherwise, the entire function $\omega(\lambda)$ would vanish identically, making every $\lambda$ an eigenvalue of $L$, which is impossible.)
