If $V:\mathbb R\to \mathbb R$ is an $L$ periodic function in $\operatorname L^{\infty}$ we can always find two independent solutions for $$\psi''(x)+V(x)\psi(x)=E\psi(x)$$
$\psi^{\pm}(x)=e^{\pm ipx}\phi^{\pm}(x)$, where $p\in\mathbb C$ is the quasi-momentum and $\phi^{\pm}:=e^{-i(\pm)px}\psi^{\pm}(x)$ is an $L$-periodic function. In particular we'll have $\psi^{\pm}(x+L)=\mu^{\pm}\psi^{\pm}(x)$, where $\mu^+$ and $\mu^-$ are the associated Floquet multipliers which correspond to the eigenvalues of the Wronskian matrix evalued in $x=L$
I'm not understanding this: If $\Im p\ne 0\implies $ the associated value of energy $E$ of the equation is not in the (continuous) spectrum of $\mathcal H:H^2(\mathbb R)\subset \operatorname L^2\to\operatorname L^2$.
The proof of this property is based on the fact that if we are taking indpendent solutions of the form $\psi^{\pm}(x)=e^{\pm ipx}\phi^{\pm}(x)$, then these functions are exponentially increasing/decreasing and we can define the Green function of $\mathcal H-E$ in such a way that it's a linear continuous operator $G_E:\operatorname L^2\to H^2$.
The Green function is defined as
$$G_E(x,y):=\begin{cases}\dfrac{\psi^+(x)\psi^-(y)}{[\psi^+,\psi^-]}&&x\ge y\\\dfrac{\psi^-(x)\psi^+(y)}{[\psi^-,\psi^+]}&&y\ge x \end{cases}$$
where $[f,g](x):=f(x)g'(x)-g(x)f'(x)$ which is constantly equal to 1 in our case.
I don't understand if it's using the fact of the exponential monotonicity of the function for $x\to\infty$ or the fact that the functions are independent. The conclusion it's clear because if $G\equiv \mathcal (H-E)^{-1}$ and $G_E$ is bounded than $E\notin\sigma _c(\mathcal H)$.
Thank you!