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

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?

• Which Banach space is $c$? Commented May 11 at 17:55
• @SeverinSchraven $c=\{(x(n)) n\in \mathbb{N} | x(n) \in \mathbb{C} \text{and the sequence converges}\}$ with sup|x(n)| - norm Commented May 11 at 18:18
• The operator is compact. It seems that $\sigma(A)=\{0\}.$ Commented May 11 at 18:49
• The range of $A$ is not dense, as it is contained in the subspace consisting of all sequences convergent to $0.$ Commented May 12 at 1:56

## 1 Answer

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 $$(\lambda I-A)x=y.$$