Nonempty intersection between approximate point spectrum and residual spectrum

Question

On the Wikipedia page on "Spectrum (functional analysis)", it is mentioned that the approximate point spectrum and residual spectrum are not necessarily disjoint. Is there a straightforward example to illustrate this? I can't come up with one.

Answer 1

Assume $X$ is a Banach space and $A$ is a bounded linear operator on $X$. $\lambda$ is in the point spectrum iff $\mathcal{N}(A-\lambda I) \ne \{0\}$. $\lambda$ is in the continuous spectrum iff $\mathcal{N}(A-\lambda I)=\{0\}$ and $\overline{\mathcal{R}(A-\lambda I)}=X$. Everything else is the residual spectrum.

You want $\lambda$ to be in the residual spectrum and in approximate point spectrum. So, $\mathcal{N}(A-\lambda I)=\{0\}$ is required, $\overline{\mathcal{R}(A-\lambda I)}\ne X$ is required, and there must exist a sequence of unit vectors $\{ x_n \}$ such that $(A-\lambda I)x_n \rightarrow 0$. Let $X=\ell^2$ and define $$ A(x_1,x_2,x_3,\cdots) = (0,x_1,\frac{1}{2}x_2,\frac{1}{3}x_3,\cdots). $$ Clearly $\mathcal{N}(A)=\{0\}$, $(1,0,0,0,\cdots)\in\mathcal{R}(A)^{\perp}$, and $\{(1,0,0,\cdots),(0,1,0,\cdots),(0,0,1,\cdots),\cdots\}$ is a sequence of unit vectors whose images under $A$ converge in norm to $0$.