Orthonormal base of eigenfunctions Let $A\colon H\to H$ be a compact symmetric operator with dense range in a Hilbert space. Show that the eigenfunctions form an orthonormal basis of $L^2([-L,L])$
Hint: First consider the case of a point in the range. Consider the finite orthogonal projection onto the first n eigenfunctions. Use symmetry to infer convergence of the representation as $n\to\infty$. Then use that the range is dense.
I have some problems to work with the given hint but here is what I did try so far: 

Consider the case of a point in the range

Let $A(x)=y$ in the range of A.

Consider the finite orthogonal projection onto the first n eigenfunctions.

I think this means that we are assuming to have $n$ eigenvalues $\lambda_i, i=1,\ldots,n$ with $n$ eigenfunctions $v_i, i=1,\ldots,n$ and are looking for the subspace $V_n$ spanned by $v_i$.
Then, we consider the orthogonal Projection $P_{V_n}\colon H\to V_n$ 
$$
P_{V_n}(x)=\sum_{i=1}^n \langle x,v_i\rangle v_i, x\in H.~~(*)
$$

Use symmetry to infer convergence of the representation as $n\to\infty$.

No idea. 

Then use that the range is dense.

No idea
 A: This is the Hilbert-Schmidt Theorem. The proof uses Fredholm Alternative and results pertaining to the spectrum of a compact operator on a Hilbert space. First by the compactness of $A$ you get that it has at most countably many distint eigenvalues, and by the Fredholm Alternative all eigenspaces (associated to nonzero eigenvalues) are finite-dimensional. Symmetry of $A$ provides that they are orthogonal too.
By taking the direct sum of distinct eigenspaces one obtains a subspace of $L^2$, which turns out to be dense (If $\{\lambda_n\}_n$ is the sequence of eigenvalues of $A$, then $\left(\oplus_{n} \ker(A-\lambda_n I)\right)^\perp\subseteq \ker(A)$). Finally one takes bases (consisting of finitely many vectors) of each distinct eigenspace. The union of these bases is guaranteed to be an orthogonal basis of $L^2$.
A: If $A$ is a bounded selfadjoint operator, then $\|A\|=\sup_{\|x\|=1}|(Ax,x)|$. Now assume that $A$ is compact. Then the eigenvalues have finite multiplicity, and they tend to $0$ as a sequence, whether or not they are listed accordingly to multiplicity. Let $\lambda_1,\lambda_2,\lambda_3,\cdots$ be the distinct eigenvalues listed so that
$$
     |\lambda_1| \ge |\lambda_2| \ge |\lambda_3| \ge \cdots \ge |\lambda_n|\ge \cdots > 0.
$$
Then $|\lambda_n|\rightarrow 0$ as $n\rightarrow\infty$, and 
$|\lambda_n| \ne 0$ for all $n$ because the range $\mathcal{R}(A)$ is dense, and because $A=A^*$ gives
$$
            \mathcal{N}(A)=\mathcal{R}(A^*)^{\perp}=\mathcal{R}(A)^{\perp} = \{0\}.
$$
Let $P_k$ be the orthogonal projection onto the eigenspace with eigenvalue $\lambda_k \ne 0$. Then
$$
       AP_k = \lambda_k P_k = (\lambda_k P_k)^* = (AP_k)^* = P_k A.
$$
The operator
$$
        A_n = A - \sum_{k=1}^{n}\lambda_k P_k = A(I-\sum_{k=1}^{n}P_k) = (I-\sum_{k=1}^{n}P_k)A
$$
is selfadjoint.
To see that the non-zero eigenvalues of $A_n$ are $\lambda_{n+1},\lambda_{n+2},\cdots$, suppose that $A_nx=\lambda x$ for some $\lambda\ne 0$ and $x\ne 0$. Then $x=\frac{1}{\lambda}(I-\sum_{k=1}^{n}P_k)Ax$ implies that $(I-\sum_{k=1}^{n}P_k)x=x$ and, hence,
$$
     Ax = A(I-\sum_{k=1}^{n}P_k)x = A_n x = \lambda x.
$$
The eigenvalue $\lambda$ cannot be $\lambda_1,\lambda_2,\cdots,\lambda_n$ because $P_k x=0$ for $k=1,2,\cdots,n$. Therefore, the non-zero eigenvalues of $A_n$ are $\lambda_{n+1},\lambda_{n+2},\lambda_{n+3},\cdots$. The operator $A_n$ selfadjoint, and $A_n$ is compact because $A$ is compact. Hence,
$$
              \|A_n\|=|\lambda_{n+1}|,
$$
because $\lambda_{n+1}$ is the largest magnitude eigenvalue of $A_n$. Therefore $\|A_n\|\rightarrow 0$ as $n\rightarrow\infty$, which gives
$$
              \left\|A-\sum_{k=1}^{n}\lambda_k P_k\right\|\rightarrow 0 \mbox{ as } n\rightarrow \infty.
$$
Because the range of $A$ is dense, and because $Ax=\sum_{k=1}^{\infty}\lambda_kP_k$ converges to $Ax$, then it follows that the linear subspace spanned by the eigenvectors of $A$ is dense in $H$, which implies that the set of eigenvectors of $A$ is a complete orthornomal basis of $H$.
