Operator classification Imagine that we have a second-order Sturm-Liouville problem on an interval $(a,b)$. 
What is the relationship between the structure of solutions and the dimension of $\ker(T^* \pm i)$, does anybody know this?
 A: Suppose $A$ is a densely-defined symmetric linear operator on a complex Hilbert space $H$. Symmetry means
$$
                (Af,g)=(f,Ag),\;\;\; f,g \in \mathcal{D}(A).
$$
The operator $A$ is closable to a closed linear operator $A^{c}$ which is also symmetric. So, without loss of generality, assume $A$ is closed. That is, the graph
$$
     \mathcal{G}(A)=\{ \langle x, Ax\rangle \in H\times H : x \in \mathcal{D}(A) \}
$$
is a closed subspace of $H\times H$ ($\langle x,Ax\rangle$ is an ordered pair, and should not be confused with an inner product.) The symmetry of $A$ is equivalent to saying that $\mathcal{G}(A)\subseteq\mathcal{G}(A^{\star})$. The graph of $A^{\star}$ is always closed. Therefore, there is a subspace $\mathcal{N}\subset H\times H$ such that
$$
                   \mathcal{G}(A)\oplus\mathcal{N}=\mathcal{G}(A^{\star}).
$$
This decomposition is orthogonal in the inner product of $H\times H$ given by
$$
            (\langle x,y\rangle,\langle x',y'\rangle)_{X\times X}=(x,x')_{X}+(y,y')_{X}.
$$
To determine $\mathcal{N}$ suppose $\langle y,z\rangle\in\mathcal{N}\subset\mathcal{G}(A^{\star})$. Then $y\in\mathcal{D}(A^{\star})$ and $A^{\star}y=z$. The orthogonality of $\mathcal{N}$ and $\mathcal{G}(A)$ then gives
$$
           (x,y)_{X}+(Ax,A^{\star}y)=0,\;\;\; x\in\mathcal{D}(A).
$$
The fact that the above holds for all $x\in\mathcal{D}(A)$ is equivalent to the condition that $A^{\star}y\in\mathcal{D}(A^{\star})$ and
$$
                     (x,y+A^{\star}A^{\star}y)=0,\;\;\; x\in\mathcal{D}(A).
$$
Because $\mathcal{D}(A)$ is dense, then the above is equivalent to
$$
                  y \in \mathcal{N}(A^{\star 2}+I).
$$
Every such $y$ can be uniquely written as $y=y_{-}+y_{+}$, where
$$
     A^{\star}y_{-}=-iy_{-},\;\; A^{\star}y_{+}=iy_{+}.
$$
This decomposition is $y_{-}=\frac{1}{2}(iI-A^{\star})y$, $y_{+}=\frac{1}{2}(iI+A^{\star})y$. This leads immediately to the expanded orthogonal decomposition
$$
    \mathcal{G}(A)\oplus\mathcal{D}_{-}\oplus\mathcal{D}_{+}=\mathcal{G}(A^{\star}),
$$
where
$$
    \mathcal{D}_{-}=\{ \langle y_{-},-iy_{-}\rangle : y_{-}\in\mathcal{N}(A^{\star}+iI)\},\\
    \mathcal{D}_{+}=\{ \langle y_{+},+iy_{+}\rangle : y_{+}\in\mathcal{N}(A^{\star}-iI)\}.
$$
When looking at a Sturm-Liouville operator $A$ on $[a,b)$ with a regular endpoint at $a$ and a regular or singular endpoint at $b$, one usually applies this theory by defining $A$ on $\mathcal{C}_{0}^{\infty}(a,b)$ where the symmetry $(Af,g)=(f,Ag)$ holds for all $f,g\in\mathcal{C}_{0}^{\infty}(a,b)$ because the integration-by-parts evaluation terms vanish. Then $f \in \mathcal{N}(A^{\star}-iI)$ iff $f$ is a classical solution of $Af=if$ which is also in $L^{2}[a,b)$. So the deficiency indices are the same as the dimensions of the classical eigenspaces with eigenvalue $\pm i$ which are also square-integrable. That, of course, has to do with the Weyl limit point and limit circle cases. 
Note: for real Sturm-Liouville problems, $\mathcal{D}_{+}$ and $\mathcal{D}_{-}$ are isomorphic because complex conjugation maps one to the other.
