Fredholm index for 1-d Schroedinger operator if I look at a Schroedinger-operator $-\frac{d^2}{dx^2} +V$ on a compact intervall $[a,b] \subset \mathbb{R}$ and I take boundary conditions that this operator is self-adjoint (for example periodic ones). $V$ shall be a continuous function on $[a,b]$. Is it always true ( no matter what the potential explicitely is) that the Fredholm index is zero?
 A: Classical Solutions: First, assume $V\in L^{1}[a,b]$, and show the existence of classical solutions of
$$
          -f''+Vf = \lambda f,\;\;\; f(a)=A,\;f'(a)=B.
$$
This can be done by considering the equivalent integral equation
$$
           f(x)=A+B(x-a)+\int_{a}^{x}\int_{a}^{t}(V(u)-\lambda)f(u)\,du\,dv.
$$
This is a fixed point problem for $C[a,b]$ associated with the function
$$
             Kf = A+B(x-a)+\int_{a}^{x}\int_{a}^{t}(V(u)-\lambda)f(u)\,du\,dv.
$$
By restricting to a smaller interval $[a,b']$ if necessary, you get a contractive map
$$
\begin{align}
       \|Kf-Kg\|_{C[a,b']} 
           & \le \int_{a}^{x}\int_{a}^{t}|V(u)-\lambda|\,du\,dv\cdot\|f-g\|_{C[a,b']} \\
           & \le \left[\|V(u)-\lambda\|_{L^{1}[a,b]}(b'-a)\right]\|f-g\|_{C[a,b']}
\end{align}
$$
This leads to integral operators that are contractive on each interval $[a_{k}',b_{k}']$ of a finite partition of $[a,b]$. You can stitch solutions together by taking values of $f$, $f'$ at the right-endpoint of a partition interval to use as $A$, $B$ on the next interval. This gives a solution of the original ODE on the full interval $[a,b]$. The solution on each interval is a sum of iterated integrals, and it is not hard to show that $\lambda \mapsto f_{\lambda}(x)$ is an entire function of $\lambda$ for fixed $x$. The solution $f_{\lambda}$ is a solution of the integral equation, and that gives you classical differentiability: $f_{\lambda}$ is continuously differentiable and $f_{\lambda}'$ is absolutely continuous on $[a,b]$ with $f_{\lambda}''=(V-\lambda)f$ a.e.. These solutions are unique.
Green Function: Choose the unique solutions $\phi_{\lambda}$ and $\psi_{\lambda}$ of $-f''+Vf=\lambda f$ which satisfy
$$
             \phi_{\lambda}(a)=1,\; \phi_{\lambda}'(a)=0,\\ \psi_{\lambda}(a)=0,\;\psi_{\lambda}'(a)=1.
$$
Any solution of $f$ of $-f''+Vf =\lambda$ can be written as $f=f(a)\phi_{\lambda}+f'(a)\psi_{\lambda}$ because of uniqueness of such solutions.
The Wronskian $W(\phi_{\lambda},\psi_{\lambda})=\phi_{\lambda}\psi_{\lambda}'-\phi_{\lambda}'\psi_{\lambda}$ is constant on $[a,b]$ because its derivative is $0$. And the Wronskian at $a$ is $1$. So the Wronskian is identically $1$. Using this, you can verify that the general solution of the inhomogenous equation $-f''+(V-\lambda)f = g$ has the form
$$
          f(x) = \psi_{\lambda}(x)\int_{a}^{x}\phi_{\lambda}(t)g(t)\,dt
                -\phi_{\lambda}(x)\int_{a}^{x}\psi_{\lambda}(t)g(t)\,dt
               +A\psi_{\lambda}(x)+B\phi_{\lambda}(x).
$$
It is easy to verify that $f$ is twice absolutely continuous with $f'' \in L^{1}[a,b]$.
Let $\mathscr{I}_{\lambda}g$ denote the integral terms. The function $h=\mathscr{I}_{\lambda}g$ is the unique solution of
$$
                  -h''+(V-\lambda)h = g,\\
                   h(a)=h'(a) =0.
$$
The constants $A$ and $B$ are determined from any required endpoint conditions required for a given solution.
You are interested in solutions $f$ of $-f''+(V-\lambda)f=g$ where $f(a)=f(b)$, $f'(a)=f'(b)$, and that's the next topic. For convenience, define $\Delta h = h(b)-h(a)$ and $\Delta' h=h'(b)-h'(a)$. There is a periodic solution $f$ iff $\Delta f=\Delta' f = 0$ for some such $f$. That comes down to being able to solve a $2\times 2$ system for constants $A$ and $B$ such that
$$
        \left[\begin{array}{c}\Delta f \\ \Delta'f\end{array}\right]=
    \left[\begin{array}{c} 0\\0\end{array}\right]
     = \left[\begin{array}{c}\Delta\mathscr{I}_{\lambda}g \\ \Delta'\mathscr{L}_{\lambda}g\end{array}\right] + 
       \left[\begin{array}{cc}
             \Delta\phi_{\lambda} & \Delta\psi_{\lambda} \\
             \Delta'\phi_{\lambda} & \Delta'\psi_{\lambda}
            \end{array}\right]\left[\begin{array}{c}A \\ B\end{array}\right].
$$
This equation can definitely be solved if the determinant $D(\lambda)$ of the $2\times 2$ matrix on the right is non-zero. Notice that $D(\lambda)$ is an entire function of $\lambda$. So $D$ has only a discrete set of zeros that cannot accumulate anywhere in the finite plane. Using Sturm-Liouville arguments, the zeros of $D$ must also lie on the real line. Given $g \in L^{1}[a,b]$, for all $\lambda$ for which $D(\lambda) \ne 0$ there is a unique classical solution $f_{\lambda}$ of
$$
               -f_{\lambda}''+(V-\lambda)f_{\lambda} = g\\
                  f(a)=f(b),\;\; f'(a)=f'(b).
$$
So, for all such $\lambda$, the Fredholm index is $0$ on $L^{1}[a,b]$. If $D(\lambda)=0$, then there are constants $A$ and $B$ which are not both $0$ such that $f=A\phi_{\lambda}+B\psi_{\lambda}$ satisfies $\Delta f=\Delta' f=0$. That is, if $D(\lambda)=0$, then there is a non-zero periodic eigenfunction with eigenvalue $\lambda$. In this case the nullity index is either $1$ or $2$. In the case that $D(\lambda)=0$, you can show that there is a unique solution of
$$
                       -f''+(V-\lambda)f = g,\\
                       f(a)=f(b),\;f'(a)=f'(b)
$$
iff $g$ satisfies $\int_{a}^{b}g(t)h(t)\,dt =0$
for every solution $h$ of the periodic eigenfunction problem
$$
                             -h''+(V-\lambda)h =0,\\
                             h(a)=h(b),\; h'(a)=h(b).
$$
I'll give you a hint to prove this last part: This ties in directly with the Green function solution where evaluations at $b$ involve such integrals.
Because of this, the nullity and deficiency indices are the same, regardless of whether or not $D(\lambda)=0$. Restricting to $L^{2}$ gives the same result. But the result is generally true on any $L^{p}$, $1 \le p < \infty$, if $V \in L^{1}[a,b]$.
