What is the spectrum of the translation operator $(Ax)_n=\frac{x_{n+2}}{n^2}$?

Question

Let $\displaystyle A: c \rightarrow c, (Ax)(n) = \frac{x(n+2)}{n^2}$. The task is to find point spectrum, continuous spectrum, and residual spectrum of $A$ and of adjoint operator $A^*$.

$c$ is Banach space, so $\sigma(A) = \sigma(A^*)$. It is easy to see that $\Vert A\Vert = 1 \Rightarrow \sigma(A^*)=\sigma(A) \subseteq\{\lambda:|\lambda| \le 1\}$.

Attempt to find point spectrum gives $Ax=\lambda x \Rightarrow x(2k+2) = ((2k)!!)^2\lambda^kx(2)$ and we have similar formula for $x(2k+1)$. $(2k)!!$ grows fast, which implies $\lambda = 0$. Besides, $A(\{1,0,0,...\}) = 0 \Rightarrow \sigma_p(A) = \{0\}$.

$\displaystyle(A^*y)(k) = \frac{y(k-2)}{(k-2)^2} \text{ for } k\ge3; (A^*y)(1) = (A^*y)(2) =0$. Thus, $\text{Ker}A^*_\lambda = \{0\}\ \forall \lambda\in\mathbb{C} \Rightarrow \sigma_p(A^*) =\varnothing$. Moreover, closure of $\text{Im}A_\lambda$ is $^\perp(\text{Ker}A^*_\lambda) = ^\perp\{0\} = c$ and it gives $\sigma_r(A)=\varnothing$.

Bit I don't now how to find $\sigma_c(A), \sigma_c(A^*), \sigma_r(A^*)$. Didn't I make any mistakes?

Answer 1

The operator can be represented as $A=MS^2,$ where $(Mx)(k)=k^{-2}x(k)$ and $(Sx)(k)=x(k+1).$ The operator $M$ is the norm limit of the finite dimensional operators $M_n$ defined by $$(M_nx)(k)=\begin{cases}k^{-2}x_k & k\le n \\ 0 & k>n\end{cases}$$ Indeed it can be shown that $\|M_n-M\|=(n+1)^{-2}.$ Thus the operator $M$ is compact hence so is $A.$ The nonzero spectrum of a compact operator consists of eigenvalues, hence $\sigma(A)=\{0\},$ as $A$ does not admit nonzero eigenvalues. Consider the linear functional $\varphi(x)=\lim_nx_n.$ Then for $x\in c$ we have $(A^*\varphi)(x)=\varphi(Ax)=0,$ as the limit of $Ax$ is equal $0.$ Thus $0$ is an eigenvalue of $A^*.$

Remark The invertibility of $\lambda I-A,$ for $\lambda\neq 0,$ can be proved straightforward by solving explicitly the equation $(\lambda I-A)x=y.$