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?
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.