If a self-adjoint operator has a mixed spectrum, is it necessarily bounded from below?

Question

Let $A:\mathcal{D}(A)\subsetneq\mathscr{H} \rightarrow \mathscr{H}$ be an unbounded self-adjoint operator in a complex separable Hilbert space. Let $\sigma_{\text{pp}}(A)\neq \emptyset$ and $\sigma_{\text{c}}(A)\neq\emptyset$. Does it follow that $\sigma (A)= \sigma_{\text{pp}}(A)\cup \sigma_{\text{c}}(A)$ has a lower bound? How can one show it?

Answer 1

First note that if $A$ satisfies this condition, then so does $-A$. If your assertion were true, this would imply that $A$ and $-A$ are lower bounded, in other words, $A$ would have to be bounded. But that is of course not true.

For an explicit counterexample take $\mathscr H=L^2(\mathbb R,\lambda|_{[0,1]}+\sum_{k\in\mathbb Z}\delta_k)$, where $\lambda$ is the Lebesgue measure, and $A$ the operator of multiplication by the identity function. Then $\sigma_c(A)=[0,1]$, $\sigma_{pp}(A)=\mathbb Z\setminus \{0,1\}$, yet $A$ is clearly not bounded or semi-bounded.