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.
There are possibly subtle mistakes that I could not recognize in below attempt of (1, 2). Could you please have a check on it and give an example for (3)?
We have $Q(t) - Q(s)=(t-s) P(t, s) = P(t, s) (t-s)$ where $P(t, s)$ is a polynomial in two indeterminates.
1.
Let $\lambda \in Q(E V(T))$, i.e., there is $\mu \in EV(T)$ such that $\lambda = Q(\mu)$. We have $T - \mu I$ is not injective. We have $$ \begin{align*} Q(T) - \lambda I &= Q(T) - Q(\mu) I \\ &= Q(T) - Q(\mu I) \\ &= P(T, \mu I)(T-\mu I). \end{align*} $$
Because $T-\mu I$ is not injective, $Q(T) - \lambda I$ is not injective. Hence $\lambda \in EV(Q(T))$.
2.
Let $\lambda \in Q(\sigma(T))$, i.e., there is $\mu \in \sigma(T)$ such that $\lambda = Q(\mu)$. We have $T - \mu I$ is not bijective. We have $$ \begin{align*} Q(T) - \lambda I &= Q(T) - Q(\mu) I \\ &= Q(T) - Q(\mu I) \\ &\overset{(*)}{=} P(T, \mu I)(T-\mu I) \\ &\overset{(**)}{=} (T-\mu I) P(T, \mu I). \end{align*} $$
It follows from $(*)$ and $(**)$ that $Q(T) - \lambda I$ is not bijective. Hence $\lambda \in \sigma(Q(T))$.
