approximate spectrum ${\sigma}_{ap}$ of a bounded linear transformation

Question

if $\lambda\in{\sigma}_{ap}(T)$, where T is bounded linear transformation, how do i show that $|\lambda|≤\|T|\|$ ?

if $\lambda\in{\sigma}_{ap}(T)$, then for every $\varepsilon >0$ there exists $x\neq0$ such that $\|Tx−λx\|≤\varepsilon \|x\|$. I imagine that next it should be developed in some way to use the boundibility of T to conclude what is required, but I can't think of how to relate them, could someone give me a clue, or illustrate the path I should take?

Thanks in advance.

Answer 1

For any $\varepsilon> 0$ there is $x\ne 0$ such that \begin{align*} \varepsilon\|x\|\ge\|Tx-\lambda x\|\ge\big|\|Tx\|-|\lambda|\|x\|\big|\ge |\lambda|\|x\| - \|Tx\|. \end{align*} Hence, $$ |\lambda|\|x\|\le\|Tx\| + \varepsilon\|x\|\le(\|T\| + \varepsilon)\|x\|. $$ Dividing by $\|x\|\neq0$ gives $|\lambda|\le\|T\|+\varepsilon$. Letting $\varepsilon\to 0$ yields the desired result.