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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. Assume that $T^2 = I$. Prove that $\sigma(T) \subset \{\pm 1\}$ and determine $(T-\lambda I)^{-1}$ for $\lambda \neq \pm 1$.
  2. More generally, assume that there is an integer $n\ge 2$ such that $T^n=I$. Prove that $\sigma(T) \subset \{\pm 1\}$ and determine $(T-\lambda I)^{-1}$ for $\lambda \neq \pm 1$.
  3. Assume that there is an integer $n\ge 2$ such that $T^n=0$. Prove that $\sigma(T) = \{0\}$ and determine $(T-\lambda I)^{-1}$ for $\lambda \neq 0$.
  4. Assume that there is an integer $n\ge 2$ such that $\|T^n\| < 1$. Prove that $I-T$ is bijective and give an expression for $(I-T)^{-1}$ in terms of $\left(I-T^n\right)^{-1}$ and the iterates of $T$.

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

First, we prove that $T^n-I$ is bijective. Let $v \in E$. We define a map $K:E \to E, u \mapsto T^nu-v$. Then $$ |K(u_1) - K (u_2) | = |T^n(u_1-u_2)| \le \|T^n\| |u_1-u_2|. $$

Then $K$ is a strict contraction because $\|T^n\|<1$. By Banach fixed-point theorem, there is a unique $u \in E$ such that $K(u)=u$ or equivalently $(T^n-I)u=v$. Hence $T^n-I$ is bijective.

Consider the polynomial $p(t) := t^n - 1$ for $t \in \mathbb R$. Then $$ p(t)= (t-1) \sum_{k=0}^{n-1} t^k. $$

We have $p(T)= T^n-I$ is bijective. On the other hand, we have $$ (T-I) \sum_{k=0}^{n-1} T^k \overset{(1)}{=} p(T) \overset{(2)}{=} \left ( \sum_{k=0}^{n-1} T^k \right ) (T- I) \tag{4}. $$

It follows from (1) that $(T- I)$ is surjective and $\sum_{k=0}^{n-1} T^k$ is injective. It follows from (2) that $\sum_{k=0}^{n-1} T^k$ is surjective and $(T-\lambda I)$ is injective. Then both $(T- I)$ and $\sum_{k=0}^{n-1} T^k$ are bijective. It follows from (1) that $$ (I-T) \sum_{k=0}^{n-1} T^k = I - T^n. $$

Then $$ (I-T) \left( \sum_{k=0}^{n-1} T^k \right ) (I - T^n)^{-1} = I. $$

Then $$ (I-T)^{-1} = \left( \sum_{k=0}^{n-1} T^k \right ) (I - T^n)^{-1}. $$

Akira
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