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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.

I'm trying to solve 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 $N(I-T) = \{0\} \iff R(I-T) = E$.

There are possibly subtle mistakes that I could not recognize in below attempt of (2). Could you please have a check on it? Is there a shorter proof by Fredholm alternative?

Because $Q(1) \neq 0$, there is a real number $\alpha \neq 0$ such that $Q(I) = \alpha I$. There is a polynomial $\widetilde Q$ such that $Q(t)-Q(1) =\widetilde Q (1-t)$ for $t \in \mathbb R$, so $$ \begin{align} Q(T) &= Q(I) + \widetilde Q (I-T) \\ &= \alpha I + \widetilde Q (I-T). \end{align} $$

  • $\implies$

Assume that $N(I-T) = \{0\}$. Let $$ E_{n} := (I-T)^n (E) \quad \forall n \ge 1 $$

By (1), $E_n$ is closed for $n \ge 1$. Assume the contrary that $E_1 \subsetneq E$. Because $I-T$ is injective, we have $E_{n+1} \subsetneq E_n$ for $n \ge 1$. By Riesz's lemma, there is a sequence $(u_n)$ such that $u_n \in E_n, |u_n|_E=1$ and $d(u_{n}, E_{n+1}) := \inf_{v \in E_{n+1}} |u_n-v|_E \ge 1/2$ for $n \ge 1$. For $m, n \ge 1$, $$ \begin{align*} \frac{Q(T)}{\alpha} (u_n-u_m) &= u_n-u_m - \frac{\widetilde Q (I-T)}{\alpha} u_n + \frac{\widetilde Q (I-T)}{\alpha} u_m. \end{align*} $$

Notice that $\widetilde Q (0)=0$, so $-\frac{\widetilde Q (I-T)}{\alpha} u_n \in E_{n+1}$ and $\frac{\widetilde Q (I-T)}{\alpha} u_m \in E_{m+1}$. For $m >n \ge 1$, we have $$ E_{m+1} \subseteq E_m \subseteq E_{n+1} \subset E_n, $$ and thus $$ -u_m - \frac{\widetilde Q (I-T)}{\alpha} u_n + \frac{\widetilde Q (I-T)}{\alpha} u_m \in E_{n+1}. $$

Then $$ \left |\frac{Q(T)}{\alpha} (u_n-u_m) \right | \ge d(u_n, E_{n+1}) \ge \frac{1}{2}. $$

So $(\frac{Q(T)}{\alpha} u_n)_n$ does not have any convergent sequence. This contradicts the fact that $Q(T)$ is compact. Hence $E_1 = E$.

  • $\impliedby$

Assume that $R(I-T)=E$. Then $I-T$ is surjective. By Corollary 2.18 in the same book, $N((I-T)^*)=N(I-T^*)=\{0\}$. We have $Q(T^*) =Q(T)^*$ is compact by Schauder's theorem. Then we apply direction "$\implies$" on $T^*$ and get $R((I-T)^*) = R(I-T^*) = E^*$, i.e., $(I-T)^*$ is surjective. By Corollary 2.18 again, $N(I-T) = \{0\}$. This completes the proof.

Akira
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    Looks OK. In particular it makes sense that much of the "work" would be in one direction ($I-T$ injective implies $I-T$ surjective), with the rest being deducible from that via general facts about adjoints and nullspaces. Re potential simplification, the result is already nontrivial if $T$ is compact (i.e. the case $Q(t) = t$), where it (and/or Schauder's theorem) is sometimes used to prove the Fredholm alternative. And here the point is that $T$ may not be compact (the usual hypothesis of the alternative). So while there might be shorter proofs, I'd bet they rest on similar ideas. – leslie townes Jun 26 '23 at 22:13
  • @leslietownes Thank you so much for your verification! – Akira Jun 26 '23 at 22:15

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