Spectral theory - continuous spectrum imagine that I have some differential operator $D$ that is defined on an interval $[a,b]$.
Now, assume that we take the boundary conditions in such a way that this operator is self-adjoint. Then, I have a few questions about this:
1.) Is it true that this operator cannot have a residual spectrum due to its self-adjointness.
2.) Where is the difference between continuous spectrum and a classical solution?
Probably, I need to specify question (ii). The classical theory of ODEs tells me that for any eigenvalue $\lambda $ I will find two independet solutions that solve $D f = \lambda f$. The thing is of course, that these two solutions do not satisfy the boundary conditions of the operator. 
But what happens when I have a continuous spectrum? Will I find eigenvalues ( I heard that one gets a whole interval of eigenvalues, is this true?) ? Will I get solutions to my eigenvalue that satisfy the boundary conditions? I think I still did not understand what I get in the case of a continuous spectrum.
Unfortunately, I am not a mathematician, so I am not so good at teaching this to myself from books in funtional analysis, so I would very much appreciate any assistance.
 A: This is a continuation of what I posted earlier. What follows are two examples of how the theory is applied to the trigonometric functions where $V=0$. The equation is in the limit point case on $[0,\infty)$ because $e^{i\sqrt{\lambda}x}\in L^{2}[0,\infty)$ while $e^{-i\sqrt{\lambda}x}$ is not, where $\sqrt{\lambda}$ is the branch whose branch cut is along the positive $x$ axis. So there is only one condition at $x=0$ of the form $Af(0)+Bf'(0)=0$ which can be imposed in order to arrive at a selfadjoint problem. No condition is needed at $\infty$, and none can be imposed--that's the consequence of the limit point case.
The most general domain for $Lf = -\frac{d^{2}}{dx^{2}}f$ consists of all twice continuously differentiable functions $f$ on $(0,\infty)$ such that $f$, $f'$ are continuous at $0$, and $f,Lf \in L^{2}[0,\infty)$. Then $L$ becomes selfadjoint when imposing one condition at $x=0$ on the domain of the form $Af(0)+Bf'(0)=0$ for real constants $A$, $B$ with $A^{2}+B^{2}\ne 0$. I'll look at a couple of cases because one gives only continuous spectrum, while the other gives rise to continuous spectrum and an eigenvalue, too.
Example 1 (Continuous Spectrum Only): In this example, consider the equation
$$
              -f''-\lambda f = g,\\
               f(0)=0.
$$
Start by letting $\phi_{\lambda}$, $\psi_{\lambda}$ be the unique classical solutions of $Lf=\lambda f$ with
$$
       \begin{array}{cc}
              \phi_{\lambda}(0) = 0 &  \psi_{\lambda}(0)=1 \\
              \phi_{\lambda}'(0) = 1 & \psi_{\lambda}'(0)=0.
       \end{array}
$$
These solutions are
$$
            \phi_{\lambda}(x) = \frac{\sin{\lambda x}}{\sqrt{\lambda}},\;\;
            \psi_{\lambda}(x) = \cos(\sqrt{\lambda}x).
$$
The Wronskian $W(\phi_{\lambda},\psi_{\lambda})=\phi_{\lambda}\psi_{\lambda}'-\phi_{\lambda}'\psi_{\lambda}\equiv -1$. For the limit point case, the trick to this problem is choosing the unique $m(\lambda)$ such that
$$
      \psi_{\lambda}+m(\lambda)\phi_{\lambda} \in L^{2}[0,\infty),\;\;\; \lambda\notin\mathbb{R}.
$$
The above condition requires the coefficient of $e^{-i\sqrt{\lambda}x}$ to be 0. So
$$
    \frac{1}{2}-m(\lambda)\frac{1}{2i\sqrt{\lambda}}=0 \implies m(\lambda)=i\sqrt{\lambda}.
$$
Then $\eta_{\lambda}=\psi_{\lambda}+i\sqrt{\lambda}\phi_{\lambda} \in L^{2}[0,\infty)$, and $W(\phi_{\lambda},\eta_{\lambda})=W(\phi_{\lambda},\psi_{\lambda})\equiv -1$. The resolvent operator associated with $L$ is $f=(\lambda I-L)^{-1}g$ or, equivalently,
$$
              f''+\lambda f = g,\\
               f(0)= 0.    
$$
The solution of this equation is
$$
           f=(\lambda I-L)^{-1}g = -\eta_{\lambda}(x)\int_{0}^{x}g(t)\phi_{\lambda}(t)\,dt
  -\phi_{\lambda}(x)\int_{x}^{\infty}g(t)\eta_{\lambda}(t)\,dt.
$$
Choosing $\eta_{\lambda}$ the way we did is important in order to guarantee that the integral on the right exists for $g \in L^{2}[0,\infty)$. As I mentioned before, finding the spectral measure of the interval $[\alpha,\beta]$--provided neither $\alpha$ nor $\beta$ are eigenvalues--is obtained by finding the generalized residue of the resolvent associated with the singularities in $[\alpha,\beta]\subset \mathbb{R}$:
$$
       E[\alpha,\beta]g=\lim_{\epsilon\downarrow 0}\frac{1}{2\pi i}\int_{\alpha}^{\beta}((u-i\epsilon)I-A)^{-1}g-((u+i\epsilon)I-A)^{-1}g \,du
$$
All of the singularities of the resolvent are found in the Titchmarsh m-function $m$. To see this, assume for the moment that $f$ is $0$ outside some interval $[0,R]$. Then the resolvent expression can be split into entire functions of $\lambda$ and one singular component. This is because $\phi_{\lambda}(x)$, $\psi_{\lambda}(x)$ are holomorphic in $\lambda$ for fixed $x$, and uniformly differentiable in $\lambda$ for all $x$ in such an interval $[0,R]$. The resolvent splits into
$$
    (\lambda I-L)^{-1}g = -\psi_{\lambda}\int_{0}^{x}g\phi_{\lambda}\,dt
            - \phi_{\lambda}\int_{x}^{\infty}g\psi_{\lambda}\,dt+
            -m(\lambda)\phi_{\lambda}\int_{0}^{\infty}g\phi_{\lambda}\,dt
$$
So, integrating the resolvent in $\lambda$ gives $0$ for all but the last term, and the last term is only non-zero because of the branch cut for $m(\lambda)$. So the spectrum of $L$ is $[0,\infty)$ and it's all continuous spectrum. Integrating over everything gives back $g$. That is $g=\lim_{\alpha\downarrow -\infty,\beta\uparrow\infty}E[\alpha,\beta]g$, and the convergence is guaranteed in $L^{2}[0,\infty)$. This gives the Fourier transform and its inverse:
$$
\begin{align}
       g(x) & =\frac{1}{2\pi}\int_{0}^{\infty}\left(\int_{0}^{\infty}g(t)\phi_{u}(t)\,dt\right)\phi_{u}(x)\sqrt{u}\,du \\
            & = \frac{1}{2\pi}\int_{0}^{\infty}\left(\int_{0}^{\infty}f(t)\frac{\sin(\sqrt{u}t)}{\sqrt{u}}\,dt\right)\frac{\sin(\sqrt{u}x)}{\sqrt{u}}\sqrt{u}\,du.
\end{align}
$$
To be sure this is a non-standard form, but becomes recognizable after substituting $u=v^{2}$ so that $du=2v\,dv = 2\sqrt{u}\,dv$. Now the above becomes just about as classical as it gets:
$$
          g(x)=\frac{1}{\pi}\int_{0}^{\infty}\left(\int_{0}^{\infty}g(t)\sin(vt)\,dt\right)\sin(vx)\,dv.
$$
Example 2 (Mixed Point and Continuous Spectrum): This example is a good way to see that trigonometric expansions can be fairly complex. Here I'm using the same operator as before, but with a different endpoint condition
$$
                      -f''-\lambda f = g,\\
                       f(0)+f'(0) = 0.
$$
So the general unrestricted $L$ is as before, but this domain is further restricted to functions for which $f(0)+f'(0)=0$. In keeping with the previous ideas, choose solutions $\phi_{\lambda}$ and $\psi_{\lambda}$ such that
$$
            \begin{array}{cc}
  \phi_{\lambda}(0)+\phi_{\lambda}'(0)= 0, & \psi_{\lambda}(0)+\psi_{\lambda}'(0) = 2, \\
  \phi_{\lambda}(0)-\phi_{\lambda}'(0)= 2, & \psi_{\lambda}(0)-\psi_{\lambda}'(0) = 0.
            \end{array}
$$
You can check that the solutions are
$$
      \phi_{\lambda}(x) = \cos(\sqrt{\lambda}x)-\frac{\sin(\sqrt{\lambda}x)}{\sqrt{\lambda}},\;\;\;
      \psi_{\lambda}(x) = \cos(\sqrt{\lambda}x)+\frac{\sin(\sqrt{\lambda}x)}{\sqrt{\lambda}}.
$$
The Wronskian is constant, $W(\phi_{\lambda},\psi_{\lambda})=\phi_{\lambda}\psi_{\lambda}'-\phi_{\lambda}'\psi_{\lambda}\equiv 2$. So the resolvent will have a factor of $1/2$ that didn't exist previously because we must divide by the Wronskian in forming the Green function solution. The function $m$ is again chosen so that $\psi_{\lambda}+m(\lambda)\phi_{\lambda} \in L^{2}[0,\infty)$ for $\lambda\notin\mathbb{R}$, which means that $m$ must be chosen so that the $e^{-i\sqrt{\lambda}t}$ term vanishes. So,
$$
      \left(\frac{1}{2}-\frac{1}{2i\sqrt{\lambda}}\right)+m(\lambda)\left(\frac{1}{2}+\frac{1}{2i\sqrt{\lambda}}\right) = 0\\
      m(\lambda) = \frac{i\sqrt{\lambda}-1}{i\sqrt{\lambda}+1} =
             \frac{\sqrt{\lambda}+i}{\sqrt{\lambda}-i}
        =\frac{(\sqrt{\lambda}+i)^{2}}{\lambda+1}
        =\frac{\lambda+2i\sqrt{\lambda}+1}{\lambda+1}.
$$
Now there are two non-holomorphic pieces: a pole at $\lambda=-1$ and the $\sqrt{\lambda}$ term on $[0,\infty)$. So there are two kinds of spectrum: point spectrum at $\lambda=-1$ and continuous spectrum on $[0,\infty)$. This time what we're integrating is
$$
    \frac{1}{2}m(\lambda)\phi_{\lambda}(x)\int_{0}^{\infty}g(t)\phi_{\lambda}(t)\,dt
$$
The $1/2$ comes because of having to divide by the Wronskian. The negative of the residue at $\lambda=-1$ is what we what (the integration is clockwise):
$$
         \phi_{-1}(x)\int_{0}^{\infty}g(t)\phi_{-1}(t)\,dt \\
$$
And,
$$
     \phi_{-1}(x)=\frac{e^{i^{2}x}+e^{-i^{2}x}}{2}-\frac{e^{i^{2}x}-e^{-i^{2}x}}{2i^{2}}=e^{i^{2}x}=e^{-x}
$$
Notice this comes out normalized already so that its $L^{2}$ norm is $1$. The expansion for this case is
$$
       g=e^{-x}\int_{0}^{\infty}g(t)e^{-t}\,dt+\frac{1}{\pi}\int_{0}^{\infty}\left(\int_{0}^{\infty}g(t)\phi_{u}(t)\,dt\right)\phi_{u}(x)\frac{\sqrt{u}}{u+1}\,du,
$$
where $\phi_{u}(x)=\cos(\sqrt{u}x)-\frac{\sin{\sqrt{u}x}}{\sqrt{u}}$. Warning: Check all the constants for yourself; I'm always getting those things wrong.
A: The classical operator
$$
                  L = -\frac{d^{2}}{dx^{2}}+V,\;\;\; a \le x \le b,
$$
is different if $V$ is very singular. If $V \in L^{1}[a,b]$, then things are nice because there are 2 linearly-independet classical solutions of $Lf = \lambda f$ for every $\lambda$. That is, such solutions are continuous on $[a,b]$, their first derivatives are continuous on $[a,b]$, and their first derivatives are absolutely continuous with $-f''+Vf=\lambda f$. In that case, everything is nice. Whether you impose separated endpoint conditions
$$
              Af(a)+Bf'(a) =0, \;\;\; Cf(b)+Df'(b)=0,
$$
or periodic conditions
$$
                f(a)=f(b),\;\;\; f'(a)=f'(b),
$$
there is a discrete set of eigenvalues $\lambda_{1} < \lambda_{2} < \lambda_{3} < \cdots$ for which eigenfunctions exist satisfying these conditions, and there is a complete orthonormal system of eigenfunctions in $L^{2}[a,b]$. The eigenfunction expansions have the same pointwise convergence properties as the trigonometric series where $V=0$.
If $V$ is not integrable on $[a,b]$, then things change, and the situation is different on the half-line, too. The typical way to deal with a strong singularity of $V$ is to put the singularity at $b$, and to assume that $V$ is integrable on $[a,b-\epsilon)$ for all $\epsilon > 0$. If the potential is well-behaved in the finite plane, but unbounded at $\infty$, then you might consider the problem on $[a,b=\infty)$, for example.
The interesting part of the analysis in the singular case of $[a,b)$ is that there always exists one classical eigenfunction for eigenvalue $\lambda \notin\mathbb{R}$ that is in $L^{2}$ near the singular endpoint. But it can still happen that both eigenfunctions for $\lambda\notin\mathbb{R}$ are in $L^{2}[a,b)$. (These are classified as limit point or limit circle cases, and have to with how these theorems were originally proved by Weyl ~1905 using purely classical methods.) In the limit point case, one need only impose a single classical condition at $a$, and assume that all functions in the domain of the operator are in $L^{2}[a,b)$ in order to obtain a selfadjoint operator. In the limit circle case where both classical eigenfunctions for $\lambda \notin\mathbb{R}$ are in $L^{2}[a,b)$, then an additional condition must be imposed near $b$; such a condition often has to do with asymptotic behavior of functions near $b$, and is rarely obvious. In any event, after imposing the required conditions, the operator with domain of functions satisfying these conditions becomes selfadjoint.
The existence of one eigenfunction in $L^{2}$ allows one to construct a resolvent $(\lambda I-L)^{-1}$ for $\lambda\notin\mathbb{R}$ using the classical Green function solution. Then the spectral measure associated with $A$ is constructed as a generalized residue on $(\alpha,\beta)$:
$$
         E(\alpha,\beta)f = \lim_{\epsilon\downarrow 0}\frac{1}{2\pi i}
           \int_{a}^{b}((u-i\epsilon)I-L)^{-1}f-((u+i\epsilon)I-L)^{-1}f\,du.
$$
The singularities of the resolvent $(\lambda I - L)^{-1}$ for the regular problem are poles of order 1 and the residues are the projections onto the eigenspaces. In the singular cases, continuous singularities (such as branch cuts) can arise, and you can end up with integral expressions instead of discrete ones, or in addition to discrete ones.
Eigenfunction Expansions and Functional Calculus: On the interval $[a,b)$ ($b$ finite or infinite) with regular endpoint at $a$, you end up with
$$
       f(x) = \int_{\sigma}\left(\int_{a}^{b}f(t)\phi_{\lambda}(t)\,dt\right)\phi_{\lambda}(x)\,d\mu(\lambda)
$$
In reasonable applications, the measure $\mu$ has two components: a continuous component and a discrete one. The singular continuous component is realistically absent. So, you can write this as
$$
     f(x) = \sum_{n}w_{n}\int_{a}^{b}f(t)\,\phi_{\lambda_{n}}(t)\,dt\,\phi_{\lambda_{n}}(x)+\int_{-\infty}^{\infty}\left(\int_{a}^{b}f(t)\phi_{\lambda}(t)\,dt\right)\phi_{\lambda}(x)\omega(\lambda)\,d\lambda
$$
where $\omega$ is a non-negative density function that can be $0$ over a large part of $\mathbb{R}$, and where $w_{n}$ are normalization weights. This is the general form of the expansion on $[a,b)$ when the singular continuous spectrum is absent (which is the case in everything you'll probably ever see.) Here $\phi_{\lambda}$ are the classical eigenfunctions for each real $\lambda$, and are chosen to satisfy a common condition at $a$. Furthermore, this expansion turns the operator $L$ into multiplication by $\lambda$, which is really just a statement of a form of diagonalization by the generalized eigenfunction expansion. Using the last form, the diagonalization is
$$
              Lf = \sum_{n}w_{n}\int_{a}^{b}f(t)\phi_{\lambda_{n}}(t)\,dt\,\lambda_{n}\phi_{\lambda_{n}}(x)
     + \int_{-\infty}^{\infty}\left(\int_{a}^{b}f(t)\phi_{\lambda}(t)\,dt\right)\lambda \phi_{\lambda}(x)\,\omega(\lambda)\,d\lambda
$$
And you get Parseval's equality:
$$
     \int_{a}^{b}|f(t)|^{2}\,dt = \sum_{n}w_{n}\left|\int_{a}^{b}f(t)\phi_{\lambda_{n}}(t)\,dt\right|^{2}
+\int_{a}^{b}\left|\int_{a}^{b}f(t)\phi_{\lambda}(t)\,dt\right|^{2}\omega(\lambda)\,d\lambda.
$$
Note: If you form a sum and integral over any part of the spectrum, and another over a disjoint part, then the two functions are orthogonal in $L^{2}[a,b)$.
Continuous Spectrum: The continuous spectrum exists wherever $\omega(\lambda)$ is positive, and you can see the reason for the original use of the term continuous spectrum. You have an integral sum of eigenfunctions over a continuous range of eigenvalues. Later, the definition evolved in order to study this is a more abstract setting. The continuous spectrum in Quantum corresponds to unbound states. The discrete spectrum $\{ \lambda_{n} \}$ corresponds to actual bound states with perfectly well-defined energies. In these cases $\phi_{\lambda}$ is not in $L^{2}$ for $\lambda$ in the continuous spectrum, even though integrals over the smallest intervals with respect to $\omega(\lambda)d\lambda$ are in $L^{2}$. Wave packets that you encounter in traditional trigonometric analysis are there in the most general context of continuous spectrum. You don't have pure states in the continuous spectrum, but you have approximate pure states--as close as you want.
