Problem involving the Spectral Mapping theorem.

Question

Consider the following problem:

Let $T$ be a bounded operator in a Banach space $X$. Use the Spectral Mapping theorem to show that $|\lambda^n|\le\|T^n\|$ for all $\lambda\in\sigma(T).$

Here's what I have considered so far:

If $p$ is a polynomial, then, by the Spectral Mapping theorem,

$$\sigma(p(T))=\{p(z):z\in\sigma(T)\}$$

$$\implies\sigma(p(T^n))=\{p(x):x\in\sigma(T^n)\}$$

where $x=z^n$ for $z\in\sigma(T)$.

I am really not sure how to proceed with this. I had thought to try and use the spectral radius of $T$, lets call it $z_1$ to bound from above all other elements of the spectrum, but am not sure how exactly to tie it with $\|T^n\|$.

Answer 1

It's an easy thing to show that $\sigma(T) \subseteq \{ \lambda : |\lambda| \le \|T\| \}$ because the following inverse series converges in operator norm for $\|T\| < |\lambda|$ \begin{align} (T-\lambda I)^{-1} &= \frac{1}{\lambda}(\frac{1}{\lambda}T-I)^{-1} \\ & = -\frac{1}{\lambda}(I-\frac{1}{\lambda}T)^{-1} \\ & = -\frac{1}{\lambda}\sum_{n=0}^{\infty}\frac{1}{\lambda^{n}}T^{n}. \end{align} Therefore, $T^{n}-\lambda I$ is invertible if $\|T^{n}\| < |\lambda|$. That gives you $\sigma(T^{n})\subseteq \{ \lambda : |\lambda| \le \|T^{n}\| \}$. The spectral mapping theorem also gives $\sigma(T^n) = \{ \lambda^{n} : \lambda\in\sigma(T)\}$. Putting these pieces together: $$ \{ \lambda^{n} : \lambda \in \sigma(T) \} \subseteq \{ \lambda : |\lambda| \le \|T^n\| \} \\ \implies \sigma(T) \subseteq \{ \lambda : |\lambda| \le \|T^{n}\|^{1/n} \}. $$ And this is true for every $n=1,2,3,\cdots$.