Prove of some properties about unitary operators

Question

Let $X$ be a Hilbert space and $T \in L(X)$ be an unitary operator. Show

1. $\sigma(T)\subset\{\lambda \in \mathbb C:|\lambda|=1\}$
2. for $\lambda \in \mathbb C$ with $|\lambda|\neq1$ holds: $||(T-\lambda)^{-1}||_{L(X)}\le||\lambda|-1|^{-1}$

Answer 1

(1) If $T$ is a unitary, then $\|T\| = 1$, so $\sigma(T) \subset \{ z \in \mathbb{C}: | z | \le 1\}$. Also, if $\lambda \in \sigma(T)$, then $\overline{\lambda} \in \sigma(T^{\ast}) = \sigma(T^{-1})$. Since $$ (T-\alpha)^{-1} = -\alpha^{-1}(T^{-1} - \alpha^{-1})T^{-1} $$ and so $\overline{\lambda}^{-1} \in \sigma(T)$, whence $\sigma(T) \subset \{ z \in \mathbb{C}: | z | = 1\}$.

(2) If $|\lambda| > 1$, then $\|T/\lambda\| < 1$, so $$ \left( 1- \frac{T}{\lambda}\right)^{-1} = \sum_{n = 0}^{\infty} \left(\frac{T}{\lambda}\right)^n $$ Now use this to compute $\|(T-\lambda)^{-1}\|$.

If $|\lambda| < 1$, do the same with $(1-T^{-1}\lambda)^{-1}$ instead.