Example Fredholm alternative Let $T \in \mathcal{K}(E)$, that is, $T$ is a compact operator in the Banach space $E$. Consider $N(T)=\{x \in E: Tx=0\}$ and $R(T)=\{y \in E:y=Tx 
 ~(\exists x \in E)\}$.
In Brézis book we have the following result:
Theorem 6.6 (Fredholm Alternative) Let $T \in \mathcal{K}(E).$ Then
a) $N(I-T)$ is finite dimensional,
b) $R(I-T)$ is closed, and more precisely $R(I-T)=N(I-T^*)^\perp$,
c) $N(I-T)=\{0\} \Leftrightarrow R(I-T)=E$,
d) $\dim N(I-T)=\dim N(I-T^*)$.
After the theorem the author says:
Remark 4: The Fredholm alternative deals with the solvability of the equation
$u − T u = f$. It says that
$\bullet $ either for every $f \in E$ the equation $u − T u = f$ has a unique solution,
$\bullet $ or the homogeneous equation $u−T u = 0$ admits $n$ linearly independent solutions,
and in this case, the inhomogeneous equation $u−T u = f$ is solvable if and only
if $f$ satisfies $n$ orthogonality conditions, i.e., $N(I-T^*)^\perp$.
We know that $T:C([a,b]) \rightarrow C([a,b])$ given by
$$Tf(t)=\lambda\int_{a}^{b}k(t,s)f(s)ds$$
is a compact operator, where $k:[a,b] \times [a,b] \rightarrow \mathbb{R}$ is continuous ($|k(t,s)|<c$). Moreover, for all $g \in C([a,b])$ there exists a unique $f \in C([a,b])$ such that
$$f(t)-\lambda \int_{a}^{b} k(t,s)f(s)ds=g(t)$$
provided that $|\lambda|<\frac{1}{c}(b-a)$.
Hence, $T$ is in the first case of the Fredholm alternative.
My question: What would be a example of a compact operator in a infinite dimensional Banach space satisfying the second case of the Fredholm alternative and how to solve this equation by using the $n$ orthogonality conditions?
 A: One example is to prove the solvability (in the weak sense) of the following partial differential equation:
$$
-\Delta u = f \text{ on } \Omega, \quad \frac{\partial u}{\partial n} = 0 \text{ on } \partial \Omega,
$$
where $\Omega \subset \mathbb R^d$ is a domain and $f$ a given function.
One can easily see that if $u$ is a solution then $u+c$, $c\in \mathbb R$, is also a solution. On the other hand, if $u$ is a solution, then by integrating the equation we obtain the compatibility condition $\int_\Omega 0$.
The setup is in the sense of weak solutions in the space $H^1(\Omega)$.
There the equation
$$
-\Delta w  + w= f, \quad \frac{\partial w}{\partial n} = 0
$$
is uniquely solvable with solution $w = (I-\Delta)^{-1}f$ by Lax-Milgram theorem.
Now $u$ is a solution of the original problem if and only if
$$
(I - (I-\Delta)^{-1})u = (I-\Delta)^{-1} f.
$$
Here, $T:=(I-\Delta)^{-1}$ is compact on $L^2(\Omega)$.
The orthogonality condition to get solvability is $\int_\Omega f =0$, which is not only necessary (as one can easily figure out) but also sufficient by Fredholm.
