Classification of operators

I have a collection of questions about the limit point/circle concept and self-adjointness that are kind of connected, so I would like to ask them in a row.

Apparently, an operator that is limit point/ limit circle has either deficiency indices one or two.

This means for me that in Sturm-Liouville theory we do not want to study self-adjoint operators that would always have deficiency indices zero, but only these minimal operators, but I don’t quite see why?

1.) I found the following definition: A minimal operator is a densely-defined, closed and symmetric operator (i.e. $T \subset T^{**} \subset T^*$). Thus, this operator cannot be unique, right? Especially, a self-adjoint extension (if one exists) of this operator would also be a minimal operator by this definition?

Does this mean that the maximal operator is just the adjoint of the minimal operator (this would mean that the maximal operator is also closed, densely defined and an extension of the minimal operator)?

2.) Apparently for 2nd order Sturm-Liouville operators, it is said that there is this l.c. /l.p. dichotomy (so an operator is either l.p. or l.c.), but the self-adjoint extension would be neither l.c. or l.p. as the deficiency indices are clearly zero. This somehow contradicts the dichotomy in that way. I think there I am probably wrong about the definitions, so maybe anybody could help me with that?

3.) If we take a domain for a Sturm-Liouville operator in the l.c. case for example, then we have to impose boundary conditions sometimes in order to exclude unwanted solutions ( as we do for Legendre’s differential equation). Now, if we study its self-adjoint extension ( which has a larger domain(!)), then I don’t see how we could a priori know that this extension has not reintroduced the solutions that we tried to exclude or can this never happen?

4.) If we know that a Sturm-Liouville operator on a finite interval is l.p. and we know that the spectrum is purely discrete, does this always mean that there is an orthonormal basis of eigenvectors that span $L^2$? ( I mean I know that it is true for l.c. operators in general and we know in the l.c. case a priori that the spectrum is discrete, but can we conclude the existence of an ONB in the l.p. case, if we know that all elements in the spectrum must be discrete)?

Solutions Collecting From Web of "Classification of operators"

You cannot understand this problem until you understand endpoint conditions and their relation to the deficiency spaces. So, please bear with me.

Start with a regular Sturm-Liouville problem. Assume the following basic form to keep the discussion simple:
$$
Lf = \left[-\frac{d^{2}}{dx^{2}} + q\right]f,\;\;\; a \le x \le b.
$$
Assume $q \in L^{1}[a,b]$ so that everything can be defined and analyzed in a classical way. Define the minimal operator $L_{0}$ to be $L$ on functions in $\mathcal{C}_{0}^{\infty}(a,b)$, which are the infinitely differentiable functions that are compactly supported in $(a,b)$. It is trivial to check that
$$
(Lf,g)=(f,Lg),\;\;\;f,g \in \mathcal{D}(L_{0}).
$$
Of course, we’ll use the complex inner-product so that deficiency indices can be studied, and we’ll assume that $q$ is real. The operator $L$ is closable because it is densely-defined and symmetric.

It turns out the closure $L_{\min}$ of $L$ has domain $\mathcal{D}(L_{\min})$ consisting of all twice absolutely continuous functions $f \in L^{2}$ with $Lf \in L^{2}$ and $f(a)=f'(a)=f(b)=f'(b)=0$. The adjoint $L_{\max}=L_{\min}^{\star}$ of $L_{\min}$ has domain $\mathcal{D}(L_{\max})$ equal to all twice absolutely continuous functions $f \in L^{2}$ for which $Lf \in L^{2}$, without any endpoint conditions. One can verify that
$$
(L_{\min}f,g)=(f,L_{\max}g),\;\;\; f\in\mathcal{D}(L_{\min}),\;g\in\mathcal{D}(L_{\max}).
$$
The fact that $f \in\mathcal{D}(L_{\max})$–which requires only that $f,Lf\in L^{2}$–is enough to guarantee the existence of the following limits
$$
f(a+0),f'(a+0),f(b-0),f'(b-0).
$$
In fact, these limits are continuous linear functionals on the graph of $L_{\max}$. That is, there is a constant $M$ such that
$$
|f(a+0)| \le M(\|f\|+\|L_{\max}f\|),\;\;\; f\in\mathcal{D}(L_{\max}),
$$
and the same is true for all of the other 3 limits.

Theorem: Let $L_{\min},L_{\max}$ be as stated. Let $\Phi$ be a linear functional on $\mathcal{D}(L_{\max})$ which is continuous with respect to the graph norm of $L_{\max}$. Then $\Phi$ vanishes on $\mathcal{D}(L_{\min})$ iff there exists $g \in \mathcal{N}(L_{\max}+iI)\oplus\mathcal{N}(L_{\max}-iI)$ such that
$$
\Phi(f) = (f,g)+(L_{\max}f,L_{\max}g).
$$
The representing $g$ is unique. Any such functional $\Phi$ has an alternative unique representation as a linear combination
$$
\Phi(f) = \alpha f(a+0)+\alpha’f'(a+0)+\beta f(b-0)+\beta’f'(b-0).
$$

The details of the proof are not difficult, but they cloud the issue, which is this: Continuous endpoint functionals and elements of the deficiency spaces are the same. Intuitively, you can see that any functional defined on $\mathcal{D}(L_{\max})$ which vanishes on $\mathcal{D}(L_{\min})$ must have something to do with the endpoint behavior only, because you can change the function inside a compact subset of $(a,b)$ and not affect the value of the functional applied to the function. This intuition works even when dealing with PDEs on finite domains with smooth boundaries. Such functionals have to do with boundary behavior of the functions in $\mathcal{D}(L_{\max})$. And this boundary space is essentially isomorphic to the deficiency spaces $\mathcal{N}(L_{\max}\pm iI)$. For ODEs this puts you in a finite-dimensional space because you’re dealing with two values that you can specify at each regular endpoint. The deficiency spaces are much more complex for PDEs, but the ideas are the same.

This last theorem tells you what is different for a singular problem: you may or may not have endpoint functionals at a singular endpoint $b$. The classical Weyl Alternative Theorem is cast in the setting of one regular endpoint at $a$ and a singular endpoint at $b$. So, here we consider $L$ on $[a,b)$ where $b$ may be finite or infinite. For simplicity, assume $q \in L^{1}[a,c]$ for every $a \le c < b$. This does not imply that $q$ is in $L^{1}[a,b)$ because $\int_{a}^{c}|q|\,dx$ is allowed to blow up as $c\uparrow b$. And, the endpoint $b$ can be infinite, which is also very different than the regular case.

What changes in the singular case is that you may or may not have endpoint conditions at $b$; sometimes there are no non-zero limiting endpoint functionals at $b$. If there is one such functional, it will be a limit of some expression as $x\rightarrow b$, and that functional will still have a Riesz Representation in the graph norm as described above, because the functional will vanish on the graph of the minimal operator. The boundary functional comes out of looking at integration by parts to obtain
$$
\left.(L_{\max}f,g)-(f,L_{\max}g)= \{\cdots\}\right|_{a}^{b}
$$
The limits on the right will exist at $a$ and at $b$ because the integrals on the left are absolutely convergent due to the assumption that $L_{\max}f, f \in L^{2}$ for all $f\in\mathcal{D}(L_{\max})$. There are no non-zero endpoint functionals at $b$ iff the evaluation terms at $b$ on the right are always $0$. That’s the limit point case where no endpoint conditions exist or are required to get a selfadjoint operator. In that case, you can specify one condition at the endpoint $a$ and the operator becomes selfadjoint. This happens iff $\dim[ \mathcal{N}(L_{\max}+iI)\oplus\mathcal{N}(L_{\max}-iI)]=2$, which means one square integrable classical solution of $Lf=if$ and one of $Lf=-if$. They have to come in pairs that like because the original operator commutes with complex conjugation. So the only other case is where $\dim[ \mathcal{N}(L_{\max}+iI)\oplus\mathcal{N}(L_{\max}-iI)]=4$, which is the limit circle case. In that case, you have to have two linearly independent endpoint functionals at $b$. To get a selfadjoint operator in this case requires at condition at $a$, a condition at $b$, or some strange set of 2 periodic conditions mixing endpoint functionals at $a$ and $b$, which are rarely physical for singular problems.

Theorem [Weyl Alternative]: Let $L$ be Sturm-Liouville operator on $[a,b)$ with regular endpoint at $a$. Then the square integrable solutions of $Lf=if$ form a one-dimensional or two-dimensional space, and the same is true of square integrable solutions of $Lf=-if$, and the dimensions of these spaces are equal. These dimensions are $1$ iff
$$
\lim_{x\rightarrow b}\{ fg’-f’g \} = 0
$$
for all twice absolutely continuous functions $f$, $g$ with square integrable $Lf$, $Lg$.

As a final remark, the Weyl alternative fails if you allow TWO singular endpoints. Then there may be no conditions at $a$ or at $b$, and the operator $L_{\max}=L_{\max}^{\star}$ is selfadjoint already. Symmetry occurs for $L_{\max}$ because the evaluation terms always vanish at $a$ and at $b$. This happens for the associated Legendre equations, for example.

Orthonormal Bases:
Suppose $A$ is a selfadjoint operator on a complex Hilbert space with $\sigma(A)$ consisting of isolated points. Then every point of $\sigma(A)$ is an isolated singularity of the resolvent operator $R(\lambda)=(A-\lambda I)^{-1}$. The residue $E_{\lambda}$ of $-R(\lambda)$ is the spectral projection onto the eigenspace associated with $\lambda$ (it is the negative of that because of using $(A-\lambda I)^{-1}$ instead of $(\lambda I-A)^{-1}$, but the first one is the usual defintion of the resolvent.) To see why this is, suppose that $[\lambda-r,\lambda+r]$ does not contain another point of the spectrum of $A$, and let $E$ be the spectral resolution of the identity for $A$. By Stone’s formula for the spectral measure:
$$
E[\lambda-r,\lambda+r]f=\lim_{\varepsilon\downarrow 0}\int_{\lambda-r}^{\lambda+r}\{R(u+i\varepsilon)-R(u-i\varepsilon)\}f\,du \\
= -\frac{1}{2\pi i}\oint_{C} R(\mu)f\,d\mu,
$$
where $C$ is the positively oriented rectangular contour connecting these four points:
$$
\lambda-r-i\varepsilon,\;\lambda+r-i\varepsilon,\;\lambda+r+i\varepsilon,\;\lambda-r+i\varepsilon .
$$
However, $E[\lambda-r,\lambda+r]=E\{\lambda\}$ because $[\lambda-r,\lambda+r]\cap\sigma(A)=\{\lambda\}$. Assuming all of these points are isolated as described, then the entire spectral measure is recovered by residues using Stone’s Theorem:
$$
Ax = \sum_{\lambda\in\sigma(A)}\lambda E\{\lambda\}x.
$$
Of course, one automatically has completeness because of the Spectral Theorem:
$$
Ix = E\sigma(A)x = \sum_{\lambda\in\sigma(A)}E\{\lambda\}x.
$$
Choose an orthonormal basis of $E\{\lambda\}X$ for each $\lambda\in\sigma(A)$, and you automatically have a complete orthonormal basis of the full space.

Example: Consider the regular problem $Lf = -f”$ with domain $\mathcal{D}(L)$ consisting of all twice absolutely continuous functions on $[0,\pi]$ with endpoint conditions $f(0)=f(\pi)=0$. This is a selfadjoint operator on this domain. Define $\phi_{\lambda}$, $\psi_{\lambda}$ as the unique solutions of $Lf=\lambda f$ which satisfy these endpoint conditions:
$$
\phi_{\lambda}(0)=0,\;\;\phi_{\lambda}'(0) = 1,\\
\psi_{\lambda}(\pi)=0,\;\;\psi_{\lambda}'(0)=1.
$$
The solutions are
$$
\phi_{\lambda}=\frac{1}{\sqrt{\lambda}}\sin(\sqrt{\lambda} x), \\
\psi_{\lambda}=\frac{1}{\sqrt{\lambda}}\sin(\sqrt{\lambda}(x-\pi)).
$$
The Green function solution for the resolvent $R(\lambda)f=(L-\lambda I)^{-1}f$ is given by
$$
R(\lambda)f=\frac{1}{w(\lambda)}\left[\psi_{\lambda}(x)\int_{0}^{x}f(t)\phi_{\lambda}(t)\,dt+\phi_{\lambda}(x)\int_{x}^{\pi}f(t)\psi_{\lambda}(t)\,dt\right]
$$
where $w$ is the Wronskian:
$$
w(\lambda)=\psi_{\lambda}'(x)\phi_{\lambda}-\psi_{\lambda}(x)\phi_{\lambda}'(x)
\\ =\frac{1}{\sqrt{\lambda}}\{ \cos(\sqrt{\lambda}(x-\pi))\sin(\sqrt{\lambda}x)-\cos(\sqrt{\lambda}x)\sin(\sqrt{\lambda}(x-\pi))\}
\\ = \frac{1}{\sqrt{\lambda}}\sin(\sqrt{\lambda}\pi).
$$
So we have an explicit expression for the resolvent
$$
R(\lambda)f = \frac{\sqrt{\lambda}}{\sin(\sqrt{\lambda}\pi)}\left[\psi_{\lambda}(x)\int_{0}^{x}f(t)\phi_{\lambda}(t)\,dt+\phi_{\lambda}(x)\int_{x}^{\pi}f(t)\psi_{\lambda}(t)\,dt\right]
$$
I don’t show the exlicit form for $\phi_{\lambda}$, $\psi_{\lambda}$ because it confuses the fact that these functions are entire functions of $\lambda$, as guaranteed by classical existence theorems. The branch cuts disappear in these expressions, which is very odd, and very nice. So the only singularities of $R(\lambda)$ are at $\lambda = n^{2}$ for $n=1,2,3,\cdots$. These are the zeros of the Wronskin $w(\lambda)$. And $\lambda=0$ is a regular point of the Wronskian. So the spectrum of $L$ is $1^{2},2^{2},3^{2},4^{2},\cdots$. As expecte, $R(\lambda)f$ has a simple pole at each point of the spectrum, and the residue of $-R(\lambda)f$ must be the projection of $f$ onto the eigenspace associated with eigenvalue $\lambda$. The negative residue is computed with
$$
-\lim_{\lambda\rightarrow n^{2}} (\lambda-n^{2})R(\lambda)f =
-\lim_{\lambda\rightarrow n^{2}}(\sqrt{\lambda}-n)(\sqrt{\lambda}+n)R(\lambda)f \\
= -\left.\left(\frac{\sqrt{\lambda}(\sqrt{\lambda}+n)}{\pi\cos(\sqrt{\lambda}\pi)}
\left[\psi_{\lambda}(x)\int_{0}^{x}f(t)\phi_{\lambda}(t)\,dt+\phi_{\lambda}(x)\int_{x}^{\pi}f(t)\psi_{\lambda}(t)\,dt\right]\right)\right|_{\sqrt{\lambda}=n} \\
= -\frac{2n^{2}}{\pi (-1)^{n}}\left[\psi_{n^{2}}(x)\int_{0}^{x}f(t)\phi_{n^{2}}(t)\,dt+\phi_{n^{2}}(x)\int_{x}^{\pi}f(t)\psi_{n^{2}}(t)\,dt\right] \\
= \frac{2}{\pi} \sin(nx)\int_{0}^{\pi}f(t)\sin(nt)\,dt.
$$
Voilla! Normalized eigenfunctions and everything. The spectral projetion is
$$
E\{n^{2}\}f = \left[\frac{2}{\pi}\int_{0}^{\pi}f(t)\sin(nt)\,dt\right]\sin(nx).
$$
The sum of all of these is guaranteed by the Spectral Theorem to converge to $f$. That is, the following is guaranteed to converge in $L^{2}[0,\pi]$ for any $f \in L^{2}[0,\pi]$:
$$
f = \sum_{n=1}^{\infty}\left[\frac{2}{\pi}\int_{0}^{\pi}f(t)\sin(nt)\,dt\right]\sin(nx).
$$
Furthermore–and this is a big deal, too–$f \in L^{2}$ is twice absolutely continuous with $f(0)=f(\pi)=0$ and $f”\in L^{2}$ if and only if the following is convergent:
$$
\|Lf\|^{2} = \sum_{n=1}^{\infty}(n^{2})^{2}\left|\sqrt{\frac{2}{\pi}}\int_{0}^{\pi}f(t)\sin(nt)\,dt\right|^{2}
$$
And, in that case, the following converges in $L^{2}[0,\pi]$:
$$
Lf = \sum_{n=1}^{\infty}n^{2}\left[\frac{2}{\pi}\int_{0}^{\pi}f(t)\sin(nt)\,dt\right]\sin(nx).
$$