In all that follows all operators are taken to be densely defined on a Hilbert space $H$.
Some textbooks state that an operator $A$ on $H$ has pure point spectrum if $H$ admits a complete ONS (Hilbert space basis) of eigenvectors of $A$. Naively the term "pure point spectrum" suggests a relation to the point spectrum $\sigma_p (A)$ of $A$. I searched in lots of books (e.g. Brezis, analyse fonctionelle; Reed/Simon, etc) and could not find anything. At last I found something along the lines that if $A$ is self-adjoint and has pure point spectrum then $\sigma_p (A)$ is dense in $\sigma(A)$. Could someone point me to a reference where this and related connections (is there a converse perhaps?) are proven?
Remark: the mathematical physics tags is totally appropriate, this is obviously of importance in mathematical physics.
Best Answer
If $A=A^{\star}$ is a densely-defined selfadjoint linear operator on a complex Hilbert space $H$, and if there is a complete orthonormal basis of $H$ consisting of eigenvectors of $A$, then it is true that the point spectrum $\sigma_{p}(A)$ of $A$ is dense in $\sigma(A)$. The converse is not true.
To show this, let $H$ be a Complex Hilbert space and suppose $A:\mathcal{D}(A)\subseteq H\rightarrow H$ is a densely-defined selfadjoint linear operator. Suppose $H$ has an orthonormal basis $\{ e_{\alpha} \}_{\alpha \in \Lambda}$ of eigenvectors of $A$ with corresponding eigenvalues $\{\lambda_{\alpha}\}_{\alpha\in\Lambda}$, which must be real. If $x\in\mathcal{D}(A)$, then $$ Ax = \sum_{\alpha\in\Lambda}(Ax,e_{\alpha})e_{\alpha}=\sum_{\alpha\in\Lambda}(x,Ae_{\alpha})e_{\alpha} = \sum_{\alpha\in\Lambda}\lambda_{\alpha}(x,e_{\alpha})e_{\alpha}. $$ If $\mu \in \sigma_{p}(A)$, then $Ax=\mu x$ for some $0\ne x \in \mathcal{D}(A)$, which gives $$ 0=(\mu x- Ax)=\sum_{\alpha\in\Lambda}(\mu-\lambda_{\alpha})(x,e_{\alpha})e_{\alpha}. $$ Because $x \ne 0$, then $(x,e_{\alpha})\ne 0$ for some $\alpha\in\Lambda$, which forces $\mu=\lambda_{\alpha}$. So $\sigma_{p}(A)\subseteq\{ \lambda_{\alpha} : \alpha \in \Lambda\}$, while the opposite inclusion is obvious. Hence, $\sigma_{p}(A)=\{\lambda_{\alpha} : \alpha\in\Lambda\}$.
The spectrum $\sigma(A)$ is closed; so $\sigma_{p}(A)^{c}\subseteq \sigma(A)$ (here 'c' denotes topological closure.) To see that $\sigma_{p}(A)^{c} =\sigma(A)$, we assume $\lambda \notin\sigma_{p}(A)^{c}$ and show that $\lambda\notin\sigma(A)$. To show this, note that if $\lambda\notin\sigma_{p}(A)^{c}$, then there exists $\delta > 0$ such that $$ |\lambda_{\alpha}-\lambda| \ge \delta > 0,\;\;\; \alpha\in\Lambda. $$ Therefore, if $x \in \mathcal{D}(A)$, $$ \|(A-\lambda I)x\|^{2}=\sum_{\alpha}|\lambda_{\alpha}-\lambda|^{2}|(x,e_{\alpha})|^{2} \ge \delta^{2}\|x\|^{2}. $$ Because $A=A^{\star}$, the above is enough to prove that $\lambda\notin\sigma(A)$, which proves that $\sigma_{p}(A)^{c}=\sigma(A)$.
To see that the converse is false, let $H=L^{2}[0,1]\times L^{2}[0,1]$ and let $\{ e_{n}\}_{n=1}^{\infty}$ be a complete orthonormal subset of $L^{2}[0,1]$. Define $$ A(f,g) = (tf(t),\sum_{n=1}^{\infty}q_{n}(g,e_{n})e_{n}), $$ where $\{ q_{n}\}$ is an enumeration of the rational numbers in $[0,1]$. Then $$ A(0,e_{n})=q_{n}(0,e_{n}), $$ which implies that $\{ q_{n}\}\subseteq \sigma_{p}(A)$. And $A(f,g)=\lambda (f,g)$ implies $f=0$ and $g=e_{n}$ for some $n$. So, even though $\sigma_{p}(A)^{c}=[0,1]=\sigma(A)$, $H$ cannot have a basis of eigenvectors for $A$.