In Evans' PDE, §6.2, in Theorem 4 he proves a Fredholm alternative based dichotomy for second order uniformly elliptic pde.
The general idea: For $\mu\geq\gamma\geq0$ with $\gamma\geq0$ from Thm. 2, for all $f\in L^2(U)$ there is a unique weak solution $u\in H_0^1(U)$ of $Lu+\mu u=f$ (Thm. 3). Thus, we get an operator $L_\mu^{-1}:L^2(U)\longrightarrow L^2(U),f\mapsto u$.
In the proof of Thm. 4 he now states, that, given $f\in L^2(U)$, $u\in H_0^1(U)$ is a weak solution of $Lu=f$ iff \begin{equation} \text{(16)}\quad\quad u=L_\mu^{-1}(\mu u+f) \end{equation} (p. 322) and then he uses the fredholm alternative.
Now (16) troubles me, as this in an implicit equation. Shouldn't we rather say: $u\in H_0^1(U)$ is a weak solution of $Lu=f$ iff $u$ is a fixed point of \begin{equation} \Phi_\mu:L^2(U)\longrightarrow L^2(U), v\mapsto L_\mu^{-1}(\mu v+f) \end{equation} ? And now the question arises whether such a fixed point exists and whether it is unique. Thinking of the Fixed-point theorem this is to ask whether \begin{equation} \|\Phi_\mu(g-h)\|_{L^2(U)}\leq\alpha\|g-h\|_{L^2(U)} \end{equation} for all $g,h\in A$ where $A$ is a certain closed subset of $L^2(U)$ and $0<\alpha<1$. As far as I can see, from Thm. 6 we get at least $\|\Phi_\mu\|_{L^2(U)}\leq|\mu|C$ where $C$ blows up when $-\mu$ approaches an eigenvalue of $L$.
This must not be a problem, as we still can choose a smaller $A$ (which is natural as we assume the fixed point to be in $H_0^1(U)$) and as we can pick an abitrary $\mu\geq\gamma$ (but gamma could be big).
At least that's something to think about, so does anyone see a way out?