norm of the resolvent inequality

Question

Let $x$ be an element of a Banach Algebra. Let $\lambda \in \rho(x)$, where $\rho(x)$ is the resolvent of $x$.

Let ${d(\lambda, \sigma(x))}$ be the distance between $\lambda$ and $\sigma(x)$.

Show that $\| (\lambda-x)^{-1} \| \geq \frac{1}{d(\lambda, \sigma(x))}$.

Information I know:

$\rho(x)$ is an open set.

$\sigma(x)$ is compact.

I've seen many places using this inequality, but I cannot find a proof of it. And I'm not sure how to prove this by breaking down the definitions I know.

Thanks in advance!

Answer 1

Let $\mu \in \sigma (x)$. If $|\lambda -\mu| <\frac 1 {\|(\lambda -x)^{-1}\|}$ then $e+(\lambda -x)^{-1} (\mu-\lambda)$ is invertible and the identity $(\mu -x) =(\lambda -x)[e+(\lambda -x)^{-1} (\mu-\lambda)]$ shows that $\mu-x$ is invertible. This contradicts the fact that $\mu \in \sigma (x)$. We have proved that $|\lambda -\mu| \geq \frac 1 {\|(\lambda -x)^{-1}\|}$ whenever $\mu \in \sigma (x)$. Taking infimum over all such $\mu$ we get the desired inequality.

Answer 2

In general, if $a$ is invertible and $y$ is not, then $$\|a-y\|\ge\frac{1}{\|a^{-1}\|}$$ This is because the set of invertible elements form an open set, and $a$ has a ball of radius at least $\|a^{-1}\|^{-1}$ containing invertible elements.

Applying this to $y=\mu-x$ and $a=\lambda-x$, where $\mu\in\sigma(x)$, $\lambda\in\rho(x)$, gives $$|\mu-\lambda|\ge\frac{1}{\|(\lambda-x)^{-1}\|}$$ The required formula is the limit as $\mu$ minimizes the distance $d(\lambda,\sigma(x))$