Brezis' exercise 6.15.6: $\sigma(U) = f (\sigma(T))$

Question

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., $\sigma(T) := \mathbb R \setminus \rho(T)$.

1. Let $\lambda \in \mathbb R$ such that $\|T\| < |\lambda|$. Prove that $$ \|I + \lambda (T-\lambda I)^{-1}\| \le \frac{\|T\|}{|\lambda|- \|T\|}. $$
2. Let $\lambda \in \rho(T)$. Check that $(T-\lambda I)^{-1} T = T (T-\lambda I)^{-1}$ and prove that $$ d(\lambda, \sigma(T)) := \inf_{h \in \sigma(T)} |\lambda-h| \ge \frac{1}{\| (T-\lambda I)^{-1} \|}. $$
3. Assume that $0 \in \rho(T)$. Prove that $\sigma(T^{-1}) = \frac{1}{\sigma(T)}$.

In what follows, we assume that $1 \in \rho(T)$; let $$ U := (T+I)(T-I)^{-1} = (T-I)^{-1}(T+I). $$

4. Check that $1 \in \rho(U)$ and give a simple expression for $(U-I)^{-1}$ in terms of $T$.
5. Prove that $T=(U+I)(U-I)^{-1}$.
6. Consider the function $f(t) = \frac{t+1}{t-1}$ for $t \in \mathbb R \setminus \{1\}$. Prove that $$ \sigma(U) = f (\sigma(T)). $$

There are possibly subtle mistakes that I could not recognize in below attempt of (4, 5, 6). Could you please have a check on it? I'm also happy to see other approaches.

---

We have $$ \begin{align} U-I &= (T+I)(T-I)^{-1}- (T-I)(T-I)^{-1} \\ &= 2I(T-I)^{-1} \\ &= 2(T-I)^{-1}. \end{align} $$

Then $U-I$ is bijective and thus $1 \in \rho(U)$. Clearly, $(U-I)^{-1} = \frac{T-I}{2}$.

5.

We have $$ \begin{align} T &= I+2 (U-I)^{-1} \\ &= (U-I)(U-I)^{-1}+2I (U-I)^{-1} \\ &= (U-I+2I)(U-I)^{-1} \\ &= (U+I) (U-I)^{-1}. \end{align} $$

6.

Let $\lambda \in \mathbb R \setminus \{1\}$. It suffices to prove that $\lambda \in \sigma(T) \iff f(\lambda) \in \sigma(U)$. It suffices to prove that $T-\lambda I$ is bijective IFF $U - f(\lambda) I$ is bijective. We have $$ \begin{align} T-\lambda I &= (U+I) (U-I)^{-1} - \lambda (U-I) (U-I)^{-1} \\ &= ((1-\lambda)U+(1+\lambda)I) (U-I)^{-1} \\ &= (1-\lambda) (U-f(\lambda) I) (U-I)^{-1}. \end{align} $$

The claim then follows.