Sturm Liouville in Hilbert space: Question on spectrum of its inverse operator Consider the Sturm-Liouville operator
$\;L(u) = -(pu')' + qu\;$
where, $\;p \in C^1[a,b]\;$ and $\;q \in C[a,b]\;$ with $\;p(t) \neq 0\;$ for $\; t \in [a,b]\;$ are complex valued functions, with boundary conditions:
$\;\begin{align*}\alpha u(a) + \beta u'(a) = 0 \\ \gamma u(b) + \delta u'(b) = 0\end{align*}\;$
where $\;α,β,γ,δ \in \mathbb C\;$.
We know that if $\;λ=0\;$ isn't an eigenvalue of $\;L\;$ then $\;L\;$ is $"1-1"$ operator and its inverse operator is self-adjoint and compact. Moreover the relation between eigenvalues of $\;L\;$ and $\;L^{-1}\;$ is: 
$\;λ\;$ eigenvalue of $\;L\;$ iff $\;\frac{1}{λ}\;$ eigenvalue of $\;L^{-1}\;$.
My questions:


*

*Since $\;L^{-1}\;$ is compact , it follows $\;0 \in σ(L^{-1})\;$ but how is that possible? We 've proved everything assuming $\;0 \notin σ(L)\;$

*If $\;0 \in σ(L^{-1})\;$ then $\;L^{-1}\;$ has no inverse at all, am I right? Because in that case, $\;ker(L^{-1})\; \neq \{0\} \;$


I feel really confused about the spectrum of $\;L^{-1}\;$... I would appreciate any help through this. Thanks in advance!
 A: $L$ is a densely-defined unbounded linear operator. $L$ has an inverse iff $\mathcal{N}(L)=\{0\}$, in which case $L^{-1}$ is a bounded linear operator defined on all of $L^2[a,b]$ that is also compact. However, even if such $L^{-1}$ exists, this operator $L^{-1}$ will always have $0\in \sigma(L^{-1})$ because $(L^{-1})^{-1}=L$ is unbounded and only densely-defined.
Assuming $\alpha,\beta,\gamma,\delta$ are real and satisfy $\alpha^2+\beta^2\ne 0$, $\gamma^2+\delta^2\ne 0$, the spectrum of $L$ will consists of a discrete sequence of real eigenvalues
$$
                    \lambda_1 < \lambda_2 < \lambda_3  < \cdots.
$$
Assuming $0$ is not an eigenvalue, then $L^{-1}$ has a resolvent that can be expressed in terms of the resolvent of $L$:
\begin{align}
          (L^{-1}-\lambda I)^{-1} & =\left((I-\lambda L)L^{-1}\right)^{-1} \\
        & = \left(\lambda\left(\frac{1}{\lambda}I-L\right)L^{-1}\right)^{-1} \\
        & = -\frac{1}{\lambda}L\left(L-\frac{1}{\lambda}I\right)^{-1} \\
        & = -\frac{1}{\lambda}\left(L-\frac{1}{\lambda}I+\frac{1}{\lambda}I\right)\left(L-\frac{1}{\lambda}I\right)^{-1} \\
        & = -\frac{1}{\lambda}I-\frac{1}{\lambda^2}\left(L-\frac{1}{\lambda}I\right)^{-1}.
\end{align}
$\lambda \ne 0$ is in the resolvent set of $L^{-1}$ iff $1/\lambda$ is in the resolvent set of $L$.
