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.
There are possibly subtle mistakes that I could not recognize in below attempt of (4). Could you please have a check on it?
WLOG, we assume $Q (t) >0$ for all $t \in \mathbb R$. Then there are polynomials $Q_1, \ldots, Q_m$ and a constant $\alpha>0$ such that $Q(t) = \sum_{k=1}^m Q_k^2(t) + \alpha$. We have $(Q_k(T))^* = Q_k(T^*)$, so $Q_k (T)$ is self-adjoint for $k=1, \ldots,m$. Let $P(t) := \sum_{k=1}^m Q_k^2(t)$.
First, we prove that $P(T)$ is positive, i.e., $\langle P(T)u, u \rangle \ge 0$ for all $u \in H$. Indeed, $$ \begin{align*} \langle P(T)u, u \rangle &= \sum_{k=1}^m \langle Q_k^2 u, u \rangle \\ &=\sum_{k=1}^m \langle Q_k u, Q_k u \rangle \text{ because } Q_k \text{ is self-adjoint}\\ &= \sum_{k=1}^m |Q_ku|^2 \ge 0. \end{align*} $$
Second, we verify that $P(T)+\alpha I$ is injective. Let $u\in H$ such $(P(T)+\alpha I)u=0$. Then $P(T)u = -\alpha u$. Then $u=0$ because $$ 0 \le \langle P(T)u, u \rangle = -\alpha \langle u, u \rangle = -\alpha|u|^2. $$
Finally, we prove that $P(T)+\alpha I$ is surjective. It's clear that $P(T)$ and thus $P(T)+\alpha I$ are self-adjoint. For $u\in H$, we have $$ \begin{align*} |(P(T)+\alpha I)u|^2 &= |P(T)u|^2 +2\alpha \langle P(T)u, u \rangle + \alpha^2|u|^2 \\ &\ge \alpha^2|u|^2 + 2\alpha \langle P(T)u, u \rangle \\ &\ge \alpha^2|u|^2 \text{ because } P(T) \text{ is positive}. \end{align*} $$
The claim then follows from below Theorem 2.21. in the same book, i.e.,
Let $E, F$ be real Banach spaces. For a linear map $T$, we denote by $N(T)$ its kernel and by $R(T)$ its range. Let $A: D(A) \subset E \to F$ be an unbounded linear operator that is densely defined and has a closed graph.
Theorem 2.20. The following properties are equivalent:
- (a) $A$ is surjective, i.e., $R(A)=F$,
- (b) there is a constant $C$ such that $$ |v| \leq C |A^* v| \quad \forall v \in D (A^*), $$
- (c) $N (A^*) = \{0\}$ and $R(A^*)$ is closed.
Theorem 2.21. The following properties are equivalent:
- (a) $A^*$ is surjective, i.e., $R(A^*)=E^*$,
- (b) there is a constant $C$ such that $$ |u| \leq C |A u| \quad \forall u \in D (A), $$
- (c) $N (A) = \{0\}$ and $R(A)$ is closed.
