# If $T:D(T)\longrightarrow H$ is an unbounded operator in a Hilbert Space, is it true that $\sigma_r(T) = \emptyset$?

I`ve been studiyng functional analysis for a while now, and I'm sure that every self adjoint operator in a Hilbert Space has empty residual spectrum. I am also sure that every bounded self-adjoint operator on a Hilbert Space has empty residual spectrum, but is it true that if I have an unbounded linear operator $T:\mathcal{D}(T)\longrightarrow \mathscr{H}$, defined in a dense subspace $\mathcal{D}(T)\subset\mathscr{H}$, where $\mathscr{H}$ is a Hilbert Space, then I have $\sigma_r(T) = \emptyset$?

I always have some trouble finding references of functional analysis that talk specifically about Hilbert Spaces, so I would also apreciate some recommendations.

• Is your unbounded operator necessarily self-adjoint (or symmetric)? – Omnomnomnom Mar 25 '18 at 23:21
• What I know of works for self-adjoint operators or essentialy self-adjoint operators. But I want to find out if the $\sigma_r(T)$ is empty without this assumption. – Tales Rick Perche Mar 25 '18 at 23:47

This is not true in general. For instance, consider the "scaled shift operator" defined on (some dense subspace in) $\ell^2$ by $$T: (x_1,x_2,x_3,x_4,\dots)\mapsto (0,1x_1,2x_2,3x_3,4x_4,\dots)$$ Verify that $0$ is in the residual spectrum of $T$.
The residual spectrum is the left-over spectrum after you take away point spectrum and continuous spectrum. So $\lambda\in \sigma_r(T)$ for an operator $T \in \mathcal{B}(H)$ on a Hilbert space $H$ iff $\mathcal{N}(T-\lambda I)=\{0\}$ and $\overline{\mathcal{R}(T-\lambda I)}\ne H$ which implies that $\mathcal{N}(T^*-\overline{\lambda}I)\ne \{0\}$.
The right shift operator $S$ on $\ell^{2}(\mathbb{N})$ is a good example of an operator with $0\in\sigma_r(S)$. $S$ is defined as $$S : (x_0,x_1,x_2,\cdots)\mapsto (0,x_0,x_1,x_2,\cdots).$$ The range of $S$ is closed, and $(1,0,0,0,\cdots)$ is orthogonal to $\mathcal{R}(S)$. In fact, every $\lambda$ in the open unit disk is in the residual spectrum o $S$. Because of this, $$\{0\} \ne \mathcal{R}(S-\lambda I)^{\perp} = \mathcal{N}(S^*-\overline{\lambda}I).$$ So everything in the open unit disk is an eigenvalue of $S^*$.