Estimate $\Vert (\lambda - x)^{-1}\Vert$

Question

Let $A$ be a unital Banach algebra and $\omega: A \to \mathbb{C}$ a continuous functional. Let $\lambda \in \mathbb{C}, x \in A$. I'm reading a proof that claims that (note that we identify $\lambda :=\lambda 1_A$) $$\Vert (\lambda-x)^{-1}\Vert \leq(|\lambda|-\Vert x \Vert)^{-1}$$ if $|\lambda|$ is sufficiently large (note that part of the assumption is that $\lambda-x$ is invertible)

Why is this true? I tried to do

$$1 = \Vert 1 \Vert = \Vert (\lambda-x)^{-1} (\lambda -x) \Vert \leq \Vert (\lambda-x)^{-1}\Vert \Vert \lambda -x \Vert$$

but this gives an estimate in the wrong direction.

Answer 1

Suppose $|\lambda|\gt \|x\|$. Then $\|\lambda^{-1}x\|\lt 1$ and hence $1-\lambda^{-1}x$ is invertible. It is easy to see that $$\|(1-\lambda^{-1}x)^{-1}\|\leq \frac{1}{1-\|\lambda^{-1}x\|}=\frac{1}{1-|\lambda|^{-1}\|x\|}.$$ So, $$\Vert (\lambda-x)^{-1}\Vert=\Vert (\lambda(1-\lambda^{-1} x))^{-1}\Vert=\frac1{|\lambda|}\|(1-\lambda^{-1} x)^{-1}\| \leq\frac1{|\lambda|-\Vert x \Vert}.$$

Answer 2

If $|\lambda|>\|x\|$, then $$ (\lambda-x)^{-1}=\lambda^{-1}(1-\lambda^{-1}x)^{-1}=\lambda^{-1}\sum_{n=0}^\infty \lambda^{-n}x^n. $$ Thus $$ \|(\lambda-x)^{-1}\|\leq |\lambda|^{-1}\sum_{n=0}^\infty |\lambda|^{-n}\|x\|^n=(|\lambda|-\|x\|)^{-1}. $$

Answer 3

Let $r(x)$ denote the spectral radius of $x$. If $| \lambda| >r(x)$ , then $\lambda -x$ is invertible and (Neumann series !)

$$(\lambda-x)^{-1}= \sum_{n=0}^{\infty}\frac{x^n}{\lambda^{n+1}}.$$

Since the series converges absolutely, we get

$$||(\lambda-x)^{-1}|| \le \sum_{n=0}^{\infty}\frac{||x||^n}{|\lambda|^{n+1}}.$$

The rest is your turn !