Isolated spectral value

Question

My question is about self-adjoint operators (general unbounded) in a separable Hilbert space. Must an isolated spectral value of such operator belong to the point spectrum? Does the answer for the above question change for non separable Hilbert spaces?

Answer 1

The answer is yes. And it follows from the spectral theorem. For example, you can consider it in the multiplication operator form. For instance if $T$ is a bounded self-adjoint operator on a Hilbert space $\mathcal H$ then it is unitarily equivalent to multiplication by the identity. More precisely, there exists a unitary operator $U:{\mathcal H}\rightarrow \oplus_{n=1}^N L_2({\mathbb R},d\mu_n)$ for some measures $\mu_n$, such that

$$ (M\phi)_n(x)=x\phi_n(x),\qquad M=UTU^{-1}. $$

Now, if $\lambda$ is an isolated value of the spectrum, then proving that it is an eigenvalue is equivalent to proving that $\mu_k(\{\lambda\})>0$ for some $k$ (with eigenvector given by $\phi_n(\lambda)=1$ and zero everywhere else $\phi_m$ identically zero for $m\neq k$).

But if every $\mu_n(\lambda)=0$, then $M-\lambda I$ will have an inverse given by multiplication by $1/(x-\lambda)$, which is well defined since