Let $E$ be a real Banach space. Let $\mathcal L(E)$ be the space of bounded linear operators on $E$ and $\mathcal K(E)$ its subspace consisting of compact operators. Let $I:E \to E$ be the identity map. For $T \in \mathcal L(E)$,
- we denote by $N(T)$ its kernel and by $R(T)$ its range.
- we denote by $\rho(T)$ its resolvent set, by $\sigma(T)$ its spectrum, and by $EV(T)$ its set of eigenvalues. Then $EV(T) \subset \sigma(T) = \mathbb R \setminus \rho(T)$.
- for $\lambda \in EV(T)$, the set $N(T-\lambda I)$ is called the eigenspace corresponding to $\lambda$.
I'm trying to solve an exercise in Brezis' Functional Analysis, i.e.,
Let $E$ be a real Banach space, and let $T \in \mathcal{L}(E)$. Given a polynomial $Q(t)=\sum_{k=0}^p a_k t^k$ with $a_k \in \mathbb{R}$, let $Q(T)=\sum_{k=0}^p a_k T^k$.
- Prove that $Q(E V(T)) \subset E V(Q(T))$.
- Prove that $Q(\sigma(T)) \subset \sigma(Q(T))$.
- Construct an example in $E=\mathbb{R}^2$ for which the above inclusions are strict.
In what follows we assume that $E$ is a Hilbert space (identified with its dual space $H^*$) and that $T^* = T$.
- Assume here that the polynomial $Q$ has no real root, i.e., $Q(t) \neq 0$ for all $t \in \mathbb{R}$. Prove that $Q(T)$ is bijective.
- Deduce that for every polynomial $Q$, we have
- (i) $Q(E V(T)) = E V(Q(T))$,
- (ii) $Q(\sigma(T)) = \sigma(Q(T))$.
There are possibly subtle mistakes that I could not recognize in below attempt of (5). Could you please have a check on it?
Let $\lambda \in \mathbb R$. We write $$ Q(t) - \lambda = P(t) \prod_{k=1}^m (t-t_k), $$ where $P$ has no real root and $t_1, t_2, \ldots, t_m$ are real roots of $Q(t) - \lambda$.
Let $\lambda \in E V(Q(T))$, i.e., $Q(T) - \lambda I$ is not injective. Then $P(T) \prod_{k=1}^m (T-t_k I)$ is not injective. By (4), $P(T)$ is bijective. Then $\prod_{k=1}^m (T-t_k I)$ is not injective. Then $T-t_n I$ is not injective for some $n \in \{1, 2, \ldots, m\}$. By (1), it suffices to prove $\lambda \in Q(E V(T))$, i.e., there is $\mu \in EV(T)$ such that $\lambda = Q(\mu)$ and $T-\mu I$ is not injective. Clearly, $Q(t_k)=\lambda$ for all $k=1, \ldots,m$. The claim then follows by picking $\mu = t_n$.
With similar reasoning, we can prove that $Q(\sigma(T)) = \sigma(Q(T))$.
