How does $(T-\lambda)R_T(\lambda_0)=B_{\lambda}$ come?

Question

Symbols used $B(X)$ denotes the set of all bounded operators on $X$.

$C(X)$ denotes the set of all closed operators.

$R_T(\lambda)=(T-\lambda)^{-1}$

Resolvent set={$\lambda\in \mathbb C:(T-\lambda)^{-1} \in B(X)$}

In the proof of Theorem 1.7.3. I understood till $B_{\lambda}\in B(X)$(Using the Lemma 1.7.2)

Doubt:- 1.How does $(T-\lambda)R_T(\lambda_0)=B_{\lambda}$ come?

My attempts:-

1. $(T-\lambda)R_T(\lambda_0)=(T-\lambda)(T-\lambda_0)^{-1}$. I just apply the Value of $R_T(\lambda_0)$. I am not able to reduce further.

2. Another try was, I tried backward

$B_{\lambda}=I-R_T(\lambda_0)(\lambda-\lambda_0)=T-T+I-R_T(\lambda_0)(\lambda-\lambda_0)$. Still not able to reduce.

3.$T-\lambda)R_T(\lambda_0)=T-\lambda_0 R_T(\lambda_0)+\lambda_0 R_T(\lambda_0)-\lambda R_T(\lambda_0)=T-\lambda_0 R_T(\lambda_0) -(\lambda-\lambda_0) R_T(\lambda_0).$ Is $T-\lambda_0 R_T(\lambda_0)=I$?

I am not getting the expression. I go through the complete proof. I am having doubt here only. I also not sure where the author used the closed operator property of $T$.

Answer 1

The equality $(T-\lambda)R_T(\lambda_0) = B_\lambda$ follows by multiplying both sides by $T-\lambda_0$. In particular, \begin{align*} (T-\lambda)R_T(\lambda_0)\cdot (T - \lambda_0) &= (T-\lambda)(T-\lambda_0)^{-1} \cdot (T-\lambda_0)\\ &= T- \lambda \end{align*} On the other hand, \begin{align*} B_\lambda \cdot (T-\lambda_0) &= (I - (T-\lambda_0)^{-1}(\lambda - \lambda_0))\cdot(T-\lambda_0)\\ &= T - \lambda_0 - (\lambda - \lambda_0)\\ &= T-\lambda. \end{align*}