I am asked to prove the identities of $(12)$ and $(13)$, which are given on page 438 of the textbook PDE Evans, 2nd edition as follows:
THEOREM 3 (Properties of resolvent operators).
(i) If $\lambda,\mu \in \rho(A)$, we have $$R_\lambda – R_\mu=(\mu-\lambda)R_\lambda R_\mu \quad \text{(resolvent identity)} \tag{12}$$ and $$R_\lambda R_\mu = R_\mu R_\lambda \tag{13}.$$
If it helps, here are the relevant definitions on the previous page. Also, $A$ is a closed linear operator on the real Banach space $X$, with domain $D(A)$.
DEFINITIONS. (i) We say a real number $\lambda$ belongs to $\rho(A)$, the resolvent set of $A$, provided the operator $$\lambda I – A : D(A) \to X$$ is one-to-one and onto.
(ii) If $\lambda \in \rho(A)$, the resolvent operator $R_\lambda : X \to X$ is defined by $$R_\lambda u := (\lambda I – A)^{-1} u.$$
What can I do to prove the first identity $$R_\lambda – R_\mu=(\mu-\lambda)R_\lambda R_\mu$$ at least? I got stuck after writing the following: $$R_\lambda – R_\mu = (\lambda I – A)^{-1} – (\mu I – A)^{-1}$$ and $$(\mu – \lambda) R_\lambda R_\mu = (\mu – \lambda) (\lambda I – A)^{-1} (\mu I – A)^{-1}$$
Perhaps the second identity requires similar justification, so I can try to do that on my own after getting help with the first one.
Best Answer
By definition,
$$R_{\lambda} (\lambda-A)= (\lambda-A) R_{\lambda} = \text{id}.$$
Consequently,
$$\begin{align*} R_{\lambda}-R_{\mu} &= R_{\lambda} (\mu-A) R_{\mu} - R_{\lambda}(\lambda-A) R_{\mu} \\ &= R_{\lambda} ((\mu-A)-(\lambda-A)) R_{\mu}. \end{align*}$$
This proves (12). In order to show (13) use (12). (Hint: What happens if we switch $\lambda \leftrightarrow \mu$ in (12)?)