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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.

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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 $A$.

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The definition you give of pure point spectrum is not the usual one. Usually the pure point spectrum is the spectrum associated with the pure point part of the spectral measure (hence it is usually associated with self-adjoint or normal operators, for which the spectral theorem holds). Therefore it could correspond to finite or infinite multiplicity eigenvalues.

Usually (e.g. in Reed-Simon) is the discrete spectrum the one associated with isolated eigenvalues of finite multiplicity. Now if an operator $A$ is compact, or with compact resolvent, it has only discrete spectrum (apart from zero that can be an accumulation point of eigenvalues if $A$ is compact). If $A$ is also self-adjoint, then it has a complete orthonormal eigenbasis. It seems that your reference is defining the "operators with pure point spectrum" (I repeat, unlikely most of the others) as the self-adjoint operators either compact or with compact resolvent (or more in general with purely discrete spectrum).

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