Prove the resolvent equation

Question

Let $\lambda,\mu\in\rho(x)$. Prove the resolvent equation $$v(\mu)-v(\lambda)=(\mu-\lambda)v(\mu)v(\lambda)$$ where $v(\lambda)=(x-\lambda e)^{-1}$.

It is clear that I have to show $\mu v(\lambda)=e=\lambda v(\mu)$ since $$(\mu-\lambda)v(\mu)v(\lambda) = \mu v(\mu)v(\lambda)-\lambda v(\mu)v(\lambda).$$

I am not sure how to show this. I have been messing around with the property $$v(\mu)=(e-(\mu-\lambda)v(\lambda))^{-1}v(\lambda)$$ but I still can not figure it out. Any suggestion would be really appreciated!

Answer 1

The equality you're trying to show, $\mu v(\lambda)=e=\lambda v(\mu)$, is not true. If $\mu v(\lambda)=e$, then assuming $\mu\neq0$ we have $$x-\lambda e=\mu e,$$ in which case $x=(\lambda+\mu)e$, which is most certainly not true in general.

One simple way to show the resolvent equation in question is as follows:

Since $(x-\lambda e)(x-\mu e)=(x-\mu e)(x-\lambda e)$, we have $$(x-\lambda e)(x-\mu e)\left[(x-\mu e)^{-1}-(x-\lambda e)^{-1}\right] =(x-\lambda e)-(x-\mu e)=(\mu-\lambda)e.$$ Left multiplying by $(x-\mu e)^{-1}(x-\lambda e)^{-1}$ then gives us the result.

Answer 2

Observe that for $u$ and $v$ be two invertible elements $u^{-1}-v^{-1}=u^{-1}(v-u)v^{-1}$. Note that for $\lambda, \mu \in \rho (x)$, $\nu (\mu)$ and $\nu (\lambda)$ are two invertible elements in the Banach algebra. So choose $u=\nu (\lambda)$ and $v=\nu (\mu)$. Then we will get our desired result.