Lower bound for the norm of the resolvent I need to prove next statement (I want to do it for general case)
$\|R_A(z)\| = \lVert \frac{1}{A-zI} \rVert \ge \text{dist}(z,\sigma(A))^{-1}$
I think it could be like this
let $a\in \sigma(A) z \notin \sigma (A)$
then $(A-a)^{-1}$ is unbounded.
$(A-z-(a-z))=(A-z)(I-(a-z)(A-z)^{-1})
||(a-z)(A-z)^{-1}||&gt;1$
then we get statement.
but my proof doesn't cover eigenvalue.
See Corollary 6.9(1), p242 in Invitation To Operator Theory (here).
See Corollary 6.9(1), p242 in Invitation To Operator Theory (here).