Lemma 29.4.4 (Resolvent Equation).label Let $A$ be a unital Banach algebra, $x \in A$, and $\lambda, \mu \in \complex \setminus \sigma_{A}(x)$, then
\[R_{x}(\lambda) - R_{x}(\mu) = (\mu - \lambda)R_{x}(\lambda) R_{x}(\mu)\]
Proof.
\begin{align*}[R_{x}(\lambda) - R_{x}(\mu)](\mu - x)&= (\lambda - x)^{-1}(\mu - x) - 1 \\ (\lambda - x)[R_{x}(\lambda) - R_{x}(\mu)](\mu - x)&= (\mu - x) - (\lambda - x) = \mu - \lambda \\ R_{x}(\lambda) - R_{x}(\mu)&= (\mu - \lambda)R_{x}(\lambda)R_{x}(\mu)\end{align*}
$\square$
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