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 $$ d(\lambda, \sigma(T)) := \inf_{h \in \sigma(T)} |\lambda-h| \ge \frac{1}{\| (T-\lambda I)^{-1} \|}. $$
- Assume that $0 \in \rho(T)$. Prove that $\sigma(T^{-1}) = \frac{1}{\sigma(T)}$.
There are possibly subtle mistakes that I could not recognize in below attempt of (3). Could you please have a check on it? I'm also happy to see other approaches.
Let $\lambda \neq 0$. It suffices to prove that $\lambda \in \sigma(T) \iff \frac{1}{\lambda} \in \sigma(T^{-1})$. It suffices to prove that $T-\lambda I$ is not bijective IFF $T^{-1} - \lambda^{-1} I$ is not bijective. It suffices to prove that $T-\lambda I$ is bijective IFF $T^{-1} - \lambda^{-1} I$ is bijective.
We have $0 \in \rho(T)$, so $T$ is bijective. Let $u, v \in E$. We have $$ \begin{align} (T-\lambda I)u=v &\iff Tu = v+\lambda u \\ &\iff u = T^{-1} v+ \lambda T^{-1} u \\ &\iff (T^{-1} - \lambda^{-1} I)u = -\lambda^{-1} T^{-1} v, \end{align} $$ so $(T^{-1} - \lambda^{-1} I)u = -\lambda^{-1} T^{-1} (T-\lambda I)u$. Then $T^{-1} - \lambda^{-1} I = -\lambda^{-1} T^{-1} (T-\lambda I)$. The claim then follows.