If S is the shift operator on $l^2$ which means $Se_n=e_{n+1}$ and $|\lambda|>1$, then $$||(\lambda-S)^{-1}||=\frac{1}{|\lambda|-1}$$ I need to prove this statement.
We can easily get that $||(\lambda-S)^{-1}||<=\frac{1}{|\lambda|-1}$ but I can't find a sequence $\{x_n\}$ to obtain the equality.
We want to find a sequence $x=(x_{n})\in\ell_{2}$ such that $$\|(\lambda-S)x\|_{2}=\|\lambda x_{1}e_{1}+\sum^{\infty}_{n=2}(\lambda x_{n}-x_{n-1})e_{n}\|_{2}=|\lambda|\|x\|-\|x\|.$$ Idea, pick $x\in\ell_{2}$ such that $x_{n}=0$ for all $n$ even and $\|x\|=(|\lambda|-1)^{-1}$. Then $$\|(\lambda-S)x\|_{2}=\|\sum^{\infty}_{n=1}\lambda x_{2n-1}e_{2n+1}-\sum_{n=1}^{\infty}x_{2n-1}e_{2n}\|_{2}=|\lambda|\|x\|-\|x\|=1,$$ and thus $\|(\lambda-S)^{-1}(\lambda-S)x\|=(|\lambda|-1)^{-1}$.