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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 $N(I-T) = \{0\} \iff R(I-T) = E$.
  3. Prove that $\dim N(I-T) = \dim N(I-T^*)$.

There are possibly subtle mistakes that I could not recognize in below attempt of (3). Could you please have a check on it? Is there a shorter proof by Fredholm alternative?

We have $Q(T^*) =Q(T)^*$ is compact by Schauder's theorem. Let $d := \dim N(I-T)$ and $d^* := \dim N(I-T^*)$. Then $d, d^*<\infty$ by (1). Then $N(I-T)$ has a complement (closed) subspace $L$ in $E$. Then $E = N(I-T) \oplus L$. Then there is a (surjective bounded linear) projection $P:E \to N(I-T)$. By Corollary 2.18 in the same book, $\overline{R(I-T)} = N(I-T^*)^\perp$. Then $R(I-T) = N(I-T^*)^\perp$ by (1). By Examples 2 at page 38,

  • $\operatorname{codim} R(I-T) = \dim N(I-T^*)$.
  • $R(I-T)$ has a complement (closed) subspace $F$ in $E$. Then $E = F \oplus R(I-T)$ and $\dim F = d^*$.

First, we assume that $d \le d^*$. Then there is a continuous linear injection $\Lambda:N(I-T) \to F$. We will prove that $\Lambda$ is surjective. Because $E = F \oplus R(I-T)$, it suffices to prove that $\Lambda P+(I-T) = I-(T-\Lambda P)$ is surjective. Because $\Lambda P$ is a finite-rank operator and $Q(T)$ is a compact operator, we get $Q(T-\Lambda P)$ is a compact operator. Then $N(I-(T-\Lambda P)) = \{0\} \iff R(I-(T-\Lambda P)) =E$ by (2). Then it suffices to prove that $I-(T-\Lambda P)$ is injective. Let $u\in E$ such that $v:=(I-T)u = \Lambda P u$. Then $v \in R(I-T) \cap F$. Then $v=0$ and $u \in N(I-T)$. Because $\Lambda$ is injective, we get $Pu=0$ and thus $u\in L$. Then $u \in N(I-T) \cap L$ and thus $u=0$. Hence $d = d^*$.

Second, we assume that $d^* \le d$. Then there is a continuous linear injection $\Lambda:F \to N(I-T)$. With the same argument as above, we get $d^* = d$. This completes the proof.

Akira
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