Proof that the spectrum of an element of a Banach Algebra is non-empty

Question

I don't see why the line indicated with ***** in the following proof is true in the proof that spectrum of an element of a Banach Algebra is non-empty (Arveson, p.27) :

For every $\lambda_0 \not\in \sigma(x)$, $(x - \lambda)^{-1}$ is defined for all $\lambda$ sufficiently close to $\lambda_0$ because $\sigma(x)$ is closed, and we claim that

$$ \lim\limits_{\lambda \rightarrow \lambda_0} \frac{1}{\lambda - \lambda_0}\big[ (x-\lambda)^{-1} - (x- \lambda_0)^{-1}\big] = (x-\lambda_0)^{-2}$$ in the norm topology of $A$. Indeed, we can write

\begin{equation}\begin{aligned}***** (x-\lambda)^{-1}- (x- \lambda_0)^{-1} &= (x-\lambda)^{-1}\big[(x-\lambda_0)-(x-\lambda)\big](x-\lambda_0)^{-1} ***** \\&= (\lambda - \lambda_0)(x-\lambda)^{-1}(x-\lambda_0)^{-1}. \end{aligned}\end{equation}

Divide by $\lambda - \lambda_0$ and use the fact that $(x- \lambda)^{-1} \rightarrow (x - \lambda_0)^{-1}$ as $\lambda \rightarrow \lambda_0$.

Answer 1

In the indicated line, we can just expand the right hand side

\begin{align} (x-\lambda)^{-1}[(x-\lambda_0) - (x-\lambda)](x-\lambda_0)^{-1} &= [(x-\lambda)^{-1}(x-\lambda_0) - (x-\lambda)^{-1}(x-\lambda)](x-\lambda_0)^{-1}\\ &=[(x-\lambda)^{-1}(x-\lambda_0) - I](x-\lambda_0)^{-1}\\ &= (x-\lambda)^{-1}(x-\lambda_0)(x-\lambda_0)^{-1} (x-\lambda_0)^{-1}\\ &= (x-\lambda)^{-1} - (x-\lambda_0)^{-1} \end{align}

to obtain the left hand side.

Maybe it is easier to see more abstractly:

\begin{align} A^{-1}[B-A]B^{-1} &= [A^{-1}B - A^{-1}A]B^{-1}\\ &= [A^{-1}B - I]B^{-1}\\ &= A^{-1} BB^{-1} - B^{-1}\\ &= A^{-1} - B^{-1}. \end{align}