Eigenvalues of symmetric elliptic operators As stated in one of my previous questions already, I have not had so much exposure to theoretical linear algebra. This time, I'm reading a theorem and proof from PDE Evans, 2nd edition, pages 335-356. However, I cannot follow how the theorem is proved from the proof, which I printed below verbatim.

THEOREM 1 (Eignevalues of symmetric elliptic operators).
$\quad$(i) Each eigenvalue of $L$ is real.
$\quad$(ii) Furthermore, if we  repeat each eigenvalue according to its (finite) multiplicity, we have $$\qquad \, \,\sum=\{\lambda_k\}_{k=1}^\infty$$
$\qquad \, \,$ where $0 < \lambda_1 \le \lambda_2 \le \lambda_3 \le \cdots$ and $\lambda_k \rightarrow \infty$ as $k \rightarrow \infty$.
$\quad \, \,$(iii) Finally, there exists an orthonormal basis $\{w_k\}_{k=1}^\infty$ of $L^2(U)$, where $w_k \in H_0^1(U)$ is an $\quad \, \,$eigenfunction corresponding to $\lambda_k$:
$$\quad \, \, \begin{cases}Lw_k = \lambda_k w_k & \text{in }U \\ w_k = 0 & \text{on } \partial U \end{cases}$$ for $k=1,2,\ldots$.
Proof. 1. As in §6.2, $$S :=L^{-1}$$ is a bounded, linear, compact operator mapping $L^2(U)$ into itself.
$\quad$2. We claim further that $S$ is symmetric. To see this, select $f,g \in L^2(U)$. Then $Sf=u$ means $u \in H_0^1(U)$ is the weak solution of \begin{cases}Lu=f & \text{in } U \\ u=0 & \text{on } \partial U, \end{cases} and likewise $Sg=v$ means $v \in H_0^1(U)$ solves \begin{cases}Lv=g & \text{in } U \\ v=0 & \text{on } \partial U \end{cases} in the weak sense. Thus $$(Sf,g)=(u,g)=B[v,u]$$ and $$(f,Sg)=(f,v)=B[u,v].$$ Since $B[u,v]=B[v,u]$, we see $(Sf,g)=(f,Sg)$ for all $f,g \in L^2(U)$. Therefore $S$ is symmetric.
$\quad$3. Notice also $$(Sf,f)=(u,f)=B[u,u] \le 0 \quad (f \in l^2(U)).$$ Consequently the theory of compact, symmetric operators from §D.6 implies that all the eigenvalues of $S$ are real, positive, and there are corresponding eigenfunctions which make up an orthonormal basis of $L^2(U)$. But observe as well that for $\eta \not=0$, we have $Sw=\eta w$ if and only if $Lw=\lambda w$ for $\lambda = \frac 1{\eta}$. The theorem follows.

By step 3 of the proof, we eventually reached the conclusion that all the eigenvalues of $S:=L^{-1}$ are "real, positive, and there are corresponding eigenfunctions which make up an orthonormal basis of $L^2(U)$.
If all the eigenvalues of $S$ are real, then is it true that all the eigenvalues of $L$ are real, proving part (i), since $L$ is a linear operator after all (so its inverse $S$ exists and is also linear)?
Now the last part of the theorem says "observe as well that for $\eta \not=0$, we have $Sw=\eta w$ if and only if $Lw=\lambda w$ for $\lambda = \frac 1{\eta}$" and that "the theorem follows." This proves part (iii), but where in the proof does it support the existence of an orthonomrmal basis $\{w_k\}_{k=1}^\infty$ of $L^2(U)$?
And to be frank, I don't understand part (ii) of the theorem in general.
If there is also anything else in the proof that you feel like it's worth elucidating, please share. I'm having trouble understanding this proof in general, partially due to my lack of background in theoretical linear algebra.
 A: We start with a theorem:

Theorem. Let $H$ be a Hilbert space and $T$ a compact self adjoint operator. Then there exist a Hilbert basis composed of eigenvectors of $T$.
Proof. See Brezis chapter 6, in particular, theorem 6.11.

Remark: Let $(\cdot,\cdot)$ denote inner product. In the conditions of the theorem, if $(Tu,u)> 0$ for all $u\neq 0$ then, all eigenvalues are positive.
Now, as $-\Delta$ is the model of an elliptic operator, let's assume that $L=-\Delta$. Consider the Dirichlet problem 
$$\tag{1}
 \left\{ \begin{array}{ccc}
 -\Delta u=f &\mbox{ in $U$,} \\
  u=0 &\mbox{on $\partial U$,}
       \end{array} \right.
$$
where $f\in L^2(U)$. For each $f\in L^2(U)$, problem $(1)$ has a unique solution in $H_0^1(U)$ in the sense that $$\tag{2}\int_U\nabla u\nabla v=\int_U fv,\ \forall \ v\in H_0^1(U).$$
This can be proved for example, by minimising the functional $I:H_0^1(U)\to \mathbb{R}$, defined by $$I(u)=\frac{1}{2}\int_U |\nabla u|^2 .$$
Let $S:L^2(U)\to H_0^1(U)$ be the solution operator, i.e. for each $f\in L^2(U)$, $Sf$ is the unique solution of $(1)$. 
It is straightforward to see that $S$ is linear. Once $H_0^1(U)$ is compactly embedded in $L^2(U)$ (Rellich-Kondrachov theorem), we have that $S:L^2(U)\to L^2(U)$ is a compact operator. Moreover, $S$ is symmetric (in $L^2(U)$) because $$(Sf,g)=\int_U (Sf)g=\int_U \nabla (Sf)\nabla (Sg)=\int_U fSg=(f,Sg),\ \forall\ f,g\in L^2(U).$$
In the above, we have used $(2)$. To conlude, note also that $(2)$ implies that $S$ is positive: $$(Sf,f)=\int_U (Sf)f=\int_U |\nabla (Sf)|^2>0,\ \forall\ f\in L^2(U),\ f\neq 0.$$
We apply the theorem, followed by the remark, to find a sequence $(\lambda_n,\phi_n)\in (0,\infty)\times L^2(U)$ satisfying $$S\phi_n=\lambda_n\phi_n,\ \forall n,\tag{3}$$
$\phi_n$ is orthonormal, $\lambda_n\to 0$. Once $(3)$ is satisfied, we conclude that each $\phi_n$ is in $H_0^1(U)$. Note that $-\Delta:H_0^1(U)\to H^{-1}(U)$ is a linear operator defined by $$\langle -\Delta u,v\rangle =\int_U \nabla u\nabla v,$$
where $H^{-1}(U)$ is the dual of $H_0^1(U)$, $\langle\cdot,\cdot\rangle$ denotes duality. So, as you can see from the definition of $-\Delta$, $(3)$ implies that 
$$
 \left\{ \begin{array}{ccc}
 -\Delta \phi_n=\frac{1}{\lambda_n}\phi_n &\mbox{ in $U$,} \\
  \phi_n=0 &\mbox{on $\partial U$,}
       \end{array} \right.
$$
which conclude the proof of Evans theorem.
A: The operator $L : \mathcal{D}(L) \subseteq L^{2} \rightarrow L^{2}$ is a symmetric linear operator. Apparently, somewhere it was shown that $L$ is injective, surjective with compact inverse $S$. Compactness is typical of integral operators on bounded domains because of Arzela-Ascolia and how bounded sets are mapped to equicontinuous ones as they pick up derivatives. In fact, this is where the original notion of compact for an operator arose, and it was in dealing with Fredholm Integral equations that the Fredholm Alternative was discovered for differential equations, which extended to compact operators by generalization.
While a symmetric operator $L$ may not be selfadjoint, if you do find that it is surjective, injective and has a bounded inverse $S$, then $S$ will also be symmetric, which makes $S$ selfadjoint; thus, $L$ will also be selfadjoint. The advantage of knowing that the inverse is compact is that you automatically pick up a complete set of eigefunctions with eigenvalues tending to $0$. Then, $L$ picks up the same complete orthonormal basis of eigenfunctions with eigenvalues tending to $\infty$.
So, what you may be missing is information about compact selfadjoint linear operators.
