Brezis' exercise 6.10.2: how does the author obtain $\left\|Q(T) u_n-Q(T) u_m\right\| \geq|Q(1)| / 2$?

Question

Let $E,F$ be real Banach spaces. Let $\mathcal L(E, F)$ be the space of bounded linear operators from $E$ to $F$, and $\mathcal K(E, F)$ its subspace consisting of compact operators. Let $\mathcal L(E) := \mathcal L(E, E)$ and $\mathcal K(E) :=\mathcal K(E, E)$. For a linear map $T$, we denote by $R(T)$ its range and by $N(T)$ its kernel.

There is an exercise in Brezis' Functional Analysis, i.e.,

Let $Q(t) = \sum_{k=1}^p a_k t^k$ be a polynomial such that $Q(1) \neq 0$. Let $T \in \mathcal L(E)$ such that $Q(T) \in \mathcal K(E)$. Let $I:E \to E$ be the identity map.

1. Prove that $\dim N(I-T) < \infty$, and that $R(I-T)$ is closed. More generally, prove that $(I-T) (E_0)$ is closed for every closed subspace $E_0$ of $E$. [Hint: Write $Q(1) - Q(t)=\widetilde Q (t) (1-t)$ for some polynomial $\widetilde Q$ and apply Exercise 6.9.
2. Prove that $N(I-T) = \{0\} \iff R(I-T) = E$.

The solution of (2) by the author is

Proof of the implication $N(I-T)=\{0\} \Rightarrow R(I-T)=E$. Suppose by contradiction that $R(I-T)=E_1 \neq E$. Set $E_n=(I-T)^n E$. Then $\left(E_n\right)$ is a decreasing sequence of closed subspaces. Choose $u_n \in E_n$ such that $\left\|u_n\right\|=1$ and $\operatorname{dist}\left(u_n, E_{n+1}\right) \geq 1 / 2$. Write $$ \begin{align} &Q(T) u_n-Q(T) u_m \\ ={} &Q(T) u_n-Q(1) u_n+Q(1) u_n-Q(1) u_m+Q(1) u_m-Q(T) u_m \tag{1} \end{align} $$ Thus, for $m>n$, we have $$ \left\|Q(T) u_n-Q(T) u_m\right\| \geq|Q(1)| / 2, \tag{2} $$ and this is impossible. For the converse, follow the argument described in the proof of Theorem 6.6.

Could you explain how the author goes from (1) to (2)? Thank you so much for your elaboration!

Answer 1

Choosing the polynomial $\tilde{Q}$ with $Q(1) - Q(t) = \tilde{Q}(t) (1 - t)$ (as in point "1."), for any $n$ we have \begin{align*} Q(T) u_n - Q(1) u_n & = -\tilde{Q}(T) (I - T) u_n & \text{because $Q(t) - Q(1) = -\tilde{Q}(t) (1-t)$} \\ & \in -\tilde{Q}(T) E_{n+1} & \text{because $u_n \in E_n$ and $(I-T) E_n =: E_{n+1}$} \\ & \subseteq E_{n+1} & \text{because $-\tilde{Q}(T)$ commutes with $I-T$.} \end{align*} And hence for any $m$ we also have $(Q(1) - Q(T)) u_m \in E_{m+1}$, and if $m > n$ we can deduce $E_{m+1} \subseteq E_{n+1}$ and hence $(Q(1) - Q(T)) u_m \in E_{n+1}$.

Similarly if $m > n$ then $u_m \in E_m \subseteq E_{n+1}$.

To summarize: if $m > n$ then the expression labeled (1) has the form $a_{n,m} + Q(1) u_n$, where $a_{n,m} \in E_{n+1}$. Now we use how $u_n$ is chosen. Can you finish?