This is the natural continuation to this Is the point spectrum always countable?
Premise: Let $T$ be a linear densely defined symmetric/self-adjoint operator in a complex separable Hilbert space $\frak{H}$ (typical example in Quantum Mechanics).
Definition: Let $T_l :=T-lI$. The set of points $B(T)$ such that $T_{l}^{-1}$ exists and is an unbounded linear transformation with domain $\frak{R}_l$ everywhere dense in $\frak{H}$ is called the continous spectrum of $T$.
Is the following true?
Statement: The continous spectrum of $T$ is either uncountable of the empty set.
P.S. The definition I took is from M.H. Stone ""Linear Transformations in Hilbert Space and their Applications to Analysis", AMS, 1932, page 129.
Best Answer
The continuous spectrum can be finite, even when $T$ is bounded.
Let $\{e_n\}$ be an orthonormal basis of a separable Hilbert space $H$. Define a bounded operator $T:H\to H$ by $$ Te_n=\tfrac1n\,e_n. $$ So $T$ is the selfadjoint operator that with respect to the aforementioned basis is diagonal with diagonal $\{1,\tfrac12,\tfrac13,\ldots\}$. The spectrum of $T$ is $$ \sigma(T)=\{0\}\cup\{\tfrac1n:\ n\in\mathbb N\}. $$ Each $\tfrac1n$ is an eigenvalue. And the continuous spectrum of $T$ is $\{0\}$, since $T$ is injective with dense range.
Tweaking of the above example allows you to produce examples where the continuous spectrum is any finite (or countable, bounded, and discrete) subset of $\mathbb C$. Concretely, write $T_0$ for the operator above. Given $\{q_n\}\subset\mathbb C$ bounded, let $$ T=\bigoplus_n (q_n I+ T_0). $$ Then $\sigma(T)=\overline{\bigcup_n\sigma(q_n I+T_0)}=\overline{\bigcup_n\{q_n\}}$.