Spectrum of isometry

Question

Let $X$ be a Banach space. Let $T\in \mathbb{B}(X)$. If $T$ is an isometry and not invertible, prove that $\sigma(T) = \overline{\mathbb{D}}$.

I can show that $\sigma(T) \subset \overline{\mathbb{D}}$. Since $T$ is not invertible, then $0 \in \sigma(T)$. Suppose $\sigma(T) \neq \overline{\mathbb{D}}$, then we can find $|\lambda|<1$ on the boundry of the spectrum, $\partial \sigma(T)$. Then $\lambda \in \sigma_{ap}(T)$. How can I go from here to a contradiction?

Answer 1

So you've gotten to the point that there must exists $\lambda \in \sigma_{ap}(T)$ with $\lvert \lambda \rvert < 1$. Then by definition of the approximate point spectrum there exists a sequence $(x_i) \subset X$ such that $\lVert x_i \rVert = 1$ for all $i$, and $Tx_i - \lambda x_i \to 0$ as $i \to \infty$. But $$ \lVert Tx_i - \lambda x_i \rVert \geq \lVert T x_i \rVert - \lvert \lambda \rvert \lVert x_i \rVert = \lVert x_i \rVert - \lvert \lambda \rvert \lVert x_i \rVert$$ since $T$ is an isometry. But then $\lVert x_i \rVert = 1$ and $\lvert \lambda \rvert < 1$ therefore $\lVert Tx_i - \lambda x_i \rVert \geq 1 - \lvert \lambda \rvert > 0$ thus $Tx_i - \lambda x_i \not \to 0$ as $i \to \infty$. Contradiction

Answer 2

Just completing the argument i.e., prove if $\lambda \in \partial \sigma(T)$ with $|\lambda|<1$ then $\lambda \in \sigma_{ap}(T)$ :

$T-\lambda I$ is not invertible implies there exists $v$ in the Banach space, $(T-\lambda I)v = 0 \implies Tv = \lambda v \implies ||Tv|| = |\lambda| ||v|| \implies |\lambda| = 1$ (since $||Tx ||= ||x||$ as $T$ is an isometry). So $\sigma_p(T) \subseteq \partial\mathbb{D}$.

Assume that $(T-\lambda I)x_n \rightarrow x \implies x_{n_k} \rightarrow x'$ for some $x'$ for any $|\lambda|<1$. Let $\lambda_r \in \sigma_r(T)$ and $|\lambda_r| < 1$, Then there exists a point $x$ in unit circle such that $B_r(x) \cap \text{Image}(T-\lambda_r I) = \emptyset$ for some $r>0$.

Let $\lambda$ be such that $|\lambda-\lambda_r| < \frac{r(1-|\lambda|)}{2}$ and $|\lambda|<1$. If $\lambda \notin \sigma_r(T) \cup \sigma_p(T)$ then for some $\{y_n\}$, $(T-\lambda I)y_n \rightarrow x \implies y_n \rightarrow y$ and $(T-\lambda_r I) y = x-(\lambda_r-\lambda)y$. Since $|\lambda-\lambda_r| < \frac{r(1-|\lambda|)}{2}$ and since we have shown that, $(T-\lambda_r I) y = x-(\lambda_r-\lambda)y \implies B_r(x) \cap \text{Image}(T-\lambda_r I) \neq \emptyset$ since $||(\lambda_r-\lambda)y|| < r$ leading to a contradiction. This is because $(T-\lambda I)y = x \implies ||Ty|| - |\lambda| ||y|| \leq ||x|| = 1 \implies ||y|| - |\lambda| ||y|| \leq ||x|| = 1 \implies ||y|| \leq \frac{1}{(1-|\lambda|)}$ Hence $\lambda \in \sigma_r(T) \cup \sigma_p(T)$ and since $|\lambda|<1$, we have that $\lambda \in \sigma_r(T)$. Hence for every $\lambda_r \in \sigma_r(T)$ with $|\lambda_r| < 1$, there is an $\gamma>0$ such that $B_{\gamma}(\lambda_r) \subseteq \sigma_r(T)$

Hence if $\lambda$ with $|\lambda|<1$ and $\lambda \in \partial \sigma(T)$ then $\lambda \in \sigma_{ap}(T)$ by the fact that $\sigma(T)$ is closed and $\lambda \notin \sigma_r(T)$. Hence we have proved the above two statements assuming $(T-\lambda I)x_n \rightarrow x \implies $ $x_{n_k} \rightarrow x'$ for some $x'$ for any $|\lambda|<1$. This assumption is true since $T$ is an isometry as $(1-|\lambda|) \times ||x_n-x_m|| \leq ||T(x_n-x_m)|| - ||\lambda(x_n-x_m)||\leq ||(T-\lambda I)(x_n-x_m)|| < \epsilon \implies$ $\{x_n\}$ is a cauchy sequence and hence $x_n \rightarrow x'$ for some $x'$ as the space is complete.

Bonus:

Further if there exists $\lambda \in \sigma(T) \setminus \sigma_p(T)$ with $|\lambda|<1$ and $\sigma(T)$ is a strict subset of $\mathbb{D}$ then since $\sigma(T)$ is closed, we can find a $\lambda$ with $\lambda \in \partial \sigma(T) \cap (\sigma(T) \setminus \sigma_p(T))$ with $|\lambda|<1$ then based in the above proof, we have $\lambda \in \sigma_{ap}(T)$. From this point we can follow the proof given in the answer by @bitesizebo

So the proof is complete if we show that there exists $\lambda \in \sigma(T) \setminus \sigma_{p}(T)$ such that $|\lambda|<1$. This is true since $0 \in \sigma(T) \setminus \sigma_{p}(T)$.