If $X$ is a separable Hilbert Space and $T : X \to X$ selfadjoint and bounded, then the point spectrum $$ \sigma_p(T) $$ is only countable as explained here.
I have the following three questions:
- Would it be possible that $\sigma_p(T) = \emptyset$ ?
- Assume that $||T|| \in \sigma_p(T) $. Is $||T||$ then an isolated eigenvalue?
- Does $\sigma_p(T) \neq \emptyset$ imply that $T$ has any isolated eigenvalues? If so, can one show that it also has infinitly many isolated eigenvalues?
Edit: I think I found an answer too the last two questions. The operator $Tx=x$ is self-adjoint and has no isolated eigenvalues.
It is possible for a selfadjoint operator $T$ that $\sigma_p(T)=\emptyset$. I guess that an example of such an operator is $$ T:L^2(0,1)\to L^2(0,1);\qquad (Tf)(x)=xf(x). $$
Let $X=\ell^2$ and $(e_j)$ $(j\geq 0)$ be the standard orthonormal basis. For each positive integer $n$ let $\lambda_n=1-\tfrac{1}{n}$ and $\lambda_0=1$. Define $$Te_j=\lambda_j e_j\qquad (j\geq 0)$$ ($T$ is linearly and continuously extended to the whole space). Then $\| T\|=1\in \sigma_{p}(T)$ and its is not isolated since every $\lambda_n$ is in the point spectrum.
Again let $X=\ell^2$ and $(e_j)$ $(j\geq 0)$ be the standard orthonormal basis. Let $(\mu_j)_{j\geq 0}$ be a sequence of all rational numbers on the interval $[0,1]$. Define $$Te_j=\mu_j e_j\qquad (j\geq 0)$$ ($T$ is linearly and continuously extended to the whole space). Then each $\mu_j$ is an eigenvalue of $T$ and $\sigma(T)=[0,1]$, hence no eigenvalue is isolated although $\sigma_p(T)\ne \emptyset.$