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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.

The mentioned Exercise 6.9 is as follows, i.e.,

Let $T \in \mathcal L(E, F)$.

  1. Prove that the following properties are equivalent:
  • (A) $\dim N(T) < \infty$ and $R(T)$ is closed.
  • (B) There exist a constant $C>0$ and a finite-rank operator $P \in \mathcal L(E, E)$ such that $P=P^2$ and $|u|_E \le C (|Tu|_F + |Pu|_E)$ for all $u \in E$.
  • (C) There exist a Banach space $G$, an operator $Q \in \mathcal K(E, G)$, and a constant $C>0$ such that $|u|_E \le C (|Tu|_F + |Qu|_G)$ for all $u \in E$.
  1. Assume that $T$ satisfies (A). Prove that $(T+S)$ also satisfies (A) for every $S \in \mathcal K(E, F)$.
  2. Proce that the set of all operators $T \in \mathcal L(E, F)$ satisfying (A) is open in $\mathcal L(E, F)$.
  3. Let $F_0$ be a closed linear subspace of $F$ and let $S \in \mathcal K(F_0, F)$. Prove that $R(I+S)$ is closed. Here $I$ is the identity map.

There are possibly subtle mistakes that I could not recognize in below attempt of $\dim N(I-T) < \infty$. Could you have a check on it? Could you elaborate on how to prove that $R(I-T)$ is closed?

Because $Q(1) \neq 0$, there is a nonzero real number $\alpha$ such that $Q(I) = \alpha I$. We have $$ N(I-T) \subset N(Q(I)-Q(T)) = N(\alpha I-Q(T)) = N \left (I-\frac{Q(T)}{\alpha} \right). $$

Clearly, $I$ satisfies property (A) in Exercise 6.9.1 and $-\frac{Q(T)}{\alpha}$ is a compact operator. By Exercise 6.9.4, $I-\frac{Q(T)}{\alpha}$ also satisfies property (A) and thus $\dim N \left (I-\frac{Q(T)}{\alpha} \right) < \infty$. It follows from $N(I-T) \subset N \left (I-\frac{Q(T)}{\alpha} \right)$ that $\dim N(I-T) < \infty$.

Akira
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1 Answers1

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I have figured out the proof that $R(I-T)$ is closed as below.

There is a nonzero real number $\alpha$ such that $I= \alpha Q(I)$. Because $E_0$ is a closed subspace of the Banach space $E$, it is itself a Banach space. For $u \in E_0$, $$ \begin{align*} |u|_{E} &=|I u|_{E} \\ &= |\alpha Q(I) u|_{E} \\ &= |\alpha (Q(T) + \widetilde Q (T) (I-T)) u|_{E} \\ &\le |\alpha Q(T) u|_E + \| \alpha \widetilde Q (T)\| \cdot | (I-T) u|_{E} \\ &= \| \alpha \widetilde Q (T)\| \left ( \left | \frac{\alpha}{\| \alpha \widetilde Q (T)\|} Q(T) u \right |_E + | (I-T) u|_{E} \right )\\ \end{align*} $$

We have $Q(T)$ is a compact operator, so is $\frac{\alpha}{\| \alpha \widetilde Q (T)\|} Q(T)$. The claim then follows from the equivalence in Exercise 6.9.1.

Akira
  • 17,367