I'm trying to solve an exercise in Brezis' Functional Analysis, i.e.,
Let $(E, |\cdot|)$ be a real Banach space. Let $T \in \mathcal L(E)$, i.e., $T:E \to E$ is a bounded linear operator. Let $I:E \to E$ be the identity map. Let $\rho(T)$ be the resolvent set of $T$. Let $\sigma(T)$ be the spectrum of $T$, i.e., $\sigma(T) := \mathbb R \setminus \rho(T)$.
- Let $\lambda \in \mathbb R$ such that $\|T\| < |\lambda|$. Prove that $$ \|I + \lambda (T-\lambda I)^{-1}\| \le \frac{\|T\|}{|\lambda|- \|T\|}. $$
- Let $\lambda \in \rho(T)$. Check that $(T-\lambda I)^{-1} T = T (T-\lambda I)^{-1}$ and prove that $$ d(\lambda, \sigma(T)) := \inf_{h \in \sigma(T)} |\lambda-h| \ge \frac{1}{\| (T-\lambda I)^{-1} \|}. $$
There are possibly subtle mistakes that I could not recognize in below attempt. Could you please have a check on it? I'm also happy to see other approaches.
We need a result from the same book, i.e.,
Exercise 6.14 Assume that $\|T\| < 1$.
- Prove that $(I-T)$ is bijective and that $$ \|(I-T)^{-1}\| \le \frac{1}{1- \|T\|}. $$
- Let $S_n := I+T+\cdots+T^{n-1}$. Prove that $$ \|S_n-(I-T)^{-1}\| \le \frac{\|T\|^n}{1- \|T\|}. $$
Clearly, $\lambda \neq 0$. Let $K:= \frac{T}{\lambda}$. Then $\|K\| < 1$. By exercise 6.14.2, we get $$ \|I-(I-K)^{-1}\| \le \frac{\|K\|}{1- \|K\|}. $$
The claim then follows.
2.
Let $A := (T-\lambda I)^{-1} T$ and $B:=T (T-\lambda I)^{-1}$. Then $(T-\lambda I) A= T$. Because $(T-\lambda I)$ is bijective, it suffices to prove that $(T-\lambda I) B= T$. First, we have $$ \begin{align*} B - \lambda I(T-\lambda I)^{-1} &= T (T-\lambda I)^{-1} - \lambda I (T-\lambda I)^{-1} \\ &= (T-\lambda I) (T-\lambda I)^{-1}= I. \end{align*} $$
Then $B= I + \lambda I(T-\lambda I)^{-1}$ and thus $$ \begin{align*} (T-\lambda I) B &= (T-\lambda I) + (T-\lambda I) \lambda I(T-\lambda I)^{-1} \\ &= (T-\lambda I) + \lambda I= T. \end{align*} $$
By Proposition 6.7 (in the same book), $\sigma(T)$ is compact and $\sigma(T) \subset [-\|T\|, \|T\|]$. Then $\rho(T)$ is open. Let $r:= \frac{1}{\| (T-\lambda I)^{-1} \|}$. As in the proof of Proposition 6.7, $B(\lambda, r) :=\{h \in \mathbb R : |h-\lambda| < r\} \subset \rho(T)$. The claim then follows.
