Operator $f \mapsto u(f)$ solution of non-homogeneous Laplace equation is compact and self-adjoint Let $u : L^2_0(D) \to L^2_0(D): = \lbrace f \in L^2 : \int_D f = 0 \rbrace $ be the linear operator which associates $f$ to $u(f)$ the solution of
$$ 
\begin{cases}
\Delta u = f & \text{in } D \\
\dfrac{\partial u}{\partial \nu} = 0 & \text{on } \partial D
\end{cases}
$$
with $D \subseteq \mathbb R^d$ a bounded smooth domain.
I am trying to show that $u$ is compact and self adjoint, i.e.


*

*if $ f_n$ is bounded in $L^2$ then $u(f_n)$ has a convergent subsequence,

*$\int_D u(f) g = \int_D f u(g)$.


Any hint?

For 1, I tried reasoning by taking a weakly converging subsequence $f_n \rightharpoonup f $. Then by Green's identity I can show that $\Delta u(f_n) \rightharpoonup \Delta u(f)$, but I don't know how to conclude from here.
For 2, it is immediate to show that  $\int_D g \Delta u(f)  = \int_D f \Delta u(g)$, but how to conclude from this?
 A: Answer for (1): Using the Poincare-Wirtinger inequality, there is $C>0$ so that (write $u  = u(f)$)
$$\tag{1} \| u\|_2 = \| u - u_D\|_2 \le C\| \nabla u\|_2,$$
where $u_D := \int_D u $ is assumed to be zero here. Then we have 
$$\begin{split}
\|\nabla u\|_2^2 &= \int_D |\nabla u|^2 dx \\
&= \sum_{i=1}^n \int_D u_i u_i dx \\ 
&= -\sum_{i=1}^n \int_D u u_{ii} dx \ \ \ (\text{Boundary condition used)}\\
& = - \int_D uf dx \\ 
&\le \| u\|_2 \|f\|_2 \\
&\le C \| \nabla u\|_2 \|f\|_2 \\
&\le  \frac 12 \|\nabla u\|_2^2 + C \|f\|_2^2 \\
\Rightarrow  \|\nabla u\|_2 &\le C\|f\|_2.
\end{split}$$
(Note that we used $2ab\le \epsilon a^2 +  \frac{1}{\epsilon} b^2$ with some small $\epsilon$ and $C$ denotes different constants) With $(1)$ again we have 
$$\|u\|_{W^{1,2}} \le C\|f\|_2$$
Thus $u : L^2_0(D) \to L_0^2(D)$ can be written as 
$$ u : L_0^2 (D) \to W^{1,2}(D) \cap L_0^2(D) \overset{\iota}{\to} L_0^2(D).$$
Since $\iota$ is compact by the Rellich compactness theorem, $u : L_0^2(D) \to L_0^2(D)$ is compact, being a composition of compact operator and bounded operator. 
