Is continuous spectrum closed?

Question

I’m learning spectrum theory. I know the definition of point or continuous spectrum, knowing that spectrum of a bounded (or unbounded but self-adjoint) operator consisting only these two type. I have seen several operator with a closed continuous spectrum, but I don’t think it has to be closed when the point spectrum isn’t empty. Can someone show me a counter-example? And, does a continuous spectral point have to be a limit point of continuous spectral points? Thank you so much !

Answer 1

No, the continuous spectrum need not be closed. Let $f$ be defined on $[-1,1]$ by $$ f(x) = \left\{\matrix{ 0, & \text{if } -1\leq x\leq0,\cr x, & \text{if } 0\leq x\leq1,\hfill }\right. $$ and let $T$ be the corresponding multiplication operator on $L^2[-1,1]$, namely $$ T(\xi)|_x = f(x)\xi(x). $$ Then the spectrum of $T$ is $[0,1]$, the point spectrum is $\{0\}$, and the continuous spectrum is $(0,1]$.

The answer to the second question is also no. If $T$ is the diagonal operator on $\ell^2$, with diagonal entries $\{1/n: n\geq1\}$, then the point spectrum is $\{1/n: n\geq1\}$ and the continuous spectrum is $\{0\}$, and clearly 0 is not the limit of continuous spectral points (different from itself).