Solver Equation

Question

If $A : D(A) \subset X \rightarrow X$ is a closed operator, then \begin{eqnarray*} R(\lambda : A) - R(\mu : A) = (\mu - \lambda) R(\lambda : A)R(\mu : A), \ \ \forall \mu , \lambda \in \rho(A). \end{eqnarray*}

Answer 1

For $\lambda \in \rho(A)$, \begin{eqnarray} AR(\lambda : A) & = & A(\lambda I - A)^{-1} \nonumber \\ & = & (A - \lambda I + \lambda I )(\lambda I - A)^{-1} \nonumber \\ & = & \lambda (\lambda I - A)^{-1} - (\lambda I - A)(\lambda I - A)^{-1} \nonumber \\ & = & \lambda (\lambda I - A)^{-1} - I \nonumber \\ & = & \lambda R(\lambda : A) - I. \end{eqnarray}

From the above equality, it follows that for $\lambda , \mu \in \rho(A)$ the following identities are valid \begin{eqnarray*} R(\lambda : A) = R(\lambda : A)[ \mu R(\mu : A) - AR(\mu : A)] \end{eqnarray*} and \begin{eqnarray*} R(\mu : A) = R(\mu : A)[ \lambda R(\lambda : A) - AR(\lambda : A)] \end{eqnarray*} Subtracting the equations, we get \begin{eqnarray*} R(\lambda : A) - R(\lambda : A) & = & R(\lambda : A)[ \mu R(\mu : A) - AR(\mu : A)] \\ & - & R(\mu : A)[ \lambda R(\lambda : A) - AR(\lambda : A)] \\ & = & \mu R(\lambda : A)R(\mu : A) - AR(\lambda : A)R(\mu : A) \\ & - & \lambda R(\mu : A)R(\lambda : A) + AR(\lambda : A)R(\mu : A) \\ & = & (\mu - \lambda ) R(\lambda : A)R(\mu : A), \end{eqnarray*}

Answer 2

Start with $(I) = R(\lambda : A) - R(\mu : A) = (\lambda I - A)^{-1} - (\mu I - A)^{-1}$

You want to show $(I) = (\mu - \lambda)(\lambda I - A)^{-1}(\mu I - A)^{-1}$.

So multiply $(I)$ by $(\mu I - A)(\lambda I - A)$ and show the answer is $(\mu - \lambda)$.