Self-adjoint operator and eigenbasis Let us assume that we have a self-adjoint operator $A: D(A) \subset L^2 \rightarrow L^2$ and we know that $A$ has a purely discrete spectrum and the eigenvalues of $A$ are simple. Does that mean that there is an eigenbasis $(b_n)$ such that $(b_n)$ is an ONB of $L^2$?
Could anybody explain, if this holds, where this actually follows from?
 A: There is a potential problem with this. You can construct a selfadjoint operator $A$ on a Hilbert space $X$ with $\sigma(A)=[0,1]$ such that every point in the spectrum is in the point spectrum. The problems is this: once you remove the discrete space, you can still be left with an operator with only continuous spectrum. The definition of point spectrum does not preclude the possibility of continuous spectrum being mixed in. Once a $\lambda$ is tainted by calling is an element of the point spectrum, there are no further qualifiers for it. So, hiding beneath of this can be continuous spectrum, even though--technically speaking--$\sigma(A)=\sigma_{p}(A)$.
For example, Let $H=L^{2}_{\mu}[0,1]$ where $\mu$ is counting measure, and let $K=L^{2}_{m}[0,1]$ where $m$ is Lebesgue measure. Define $A : H \times K \rightarrow H\times K$ by
$$
                  A\langle f,g \rangle = \langle xf(x), xg(x)\rangle.
$$
For every $\lambda\in[0,1]$, the function $\delta_{\lambda}(x)\in H$ which is $0$ except at $\lambda$, where it is $1$ is an eigenvector with eigenvalue $\lambda$. Therefore,
$$
                    A\langle \delta_{\lambda},0\rangle = \lambda \langle \delta_{\lambda},0\rangle
$$
That means every $\lambda\in[0,1]$ is in the point spectrum of $A$. The eigenvectors
$\{ \langle \delta_{\lambda},0\rangle \}_{\lambda\in[0,1]}$ form an orthonormal basis of the subspace $H\times\{0\}\subset H\times K$. And the restriction of $A$ to orthogonal complement of this space has only continuous spectrum. So $A$ does not have an orthonormal basis of eigenvectors, even though it has only 'discrete' spectrum.
