I have some doubts on the following problem :
Let us consider $T : \ell^1(\mathbb N) \to \ell^1(\mathbb N) $by $(x_1,x_2….. ) \to (x_2, x_3 ……..) $.
I want to find the eigen values and spectrum of T and also of $T' : \ell^\infty (\mathbb N)\to \ell^\infty(\mathbb N)$
let us consider $\lambda $ to be the eigen value , then $Tx=\lambda x$ for a $x \in \ell^1$ then we get $(x_2,x_3,……)=(\lambda x_1, \lambda x_2 ……..)$ which holds equality if $x_1=x_2=…..=0$ , which means there is no eigen value for $T$ .
How do i find the spectrum of $T$ and $T'$ ?
Thank you for your help.
Best Answer
The "iff" part is wrong. Consider: $(x_i)$ with $x_i=\lambda^i$. For what $\lambda$ is this in $\ell^1$? In $\ell^\infty$?
You can write out explicitly for $|\lambda|>1$ the inverse of $\lambda I -T$ by writing:
$$S_\lambda = (\lambda I - T)^{-1} = \lambda^{-1}(I-\lambda^{-1}T)^{-1} = \lambda^{-1}\sum_{k=0}^\infty \lambda^{-k} T^k$$
Writing $x=(x_i)$ and $(y_i)=S_\lambda x$, we get:
$$y_i = \sum_{k=0}^\infty \lambda^{-(k+1)} x_{i+k}$$
You need to show that if $x\in\ell^1$ (resp. $\ell^\infty$), then this series for $y_i$ coverges for all $i$, and $(y_i)\in\ell^1$ (resp. $\ell^\infty$.)