Part of Lomonosov's Invariant Subspace Theorem

Question

Let $X$ be a complex Banach space of infinite dimension, let $T\in\mathcal{B}(X)\backslash\{0\}$ be compact.

Define $$\Gamma := \{S\in\mathcal{B}(X)\,|\,S\circ T=T\circ S\}$$and define, for each $y\in X$, $$\Gamma(y):=\{S(y)\,|\,S\in\Gamma\}$$

I am trying to prove that $\Gamma(y)\in Closed(X)$ for each $y\in X$. I have already proven that $\Gamma \in Closed(\mathcal{B}(X))$, which was easy, but I don't see how from that it follows that $\Gamma(y)$ is closed.

This is part of theorem $\boxed{10.35}$ ("Lomonosolv's Invariant Subspace Theorem") in Rudin's Functional Analysis book.

What I have tried:

1. I tried to take a convergent sequence in $\Gamma(y)$ and show it converges to a point necessarily within $\Gamma(y)$. That entails taking a sequence of points in $\Gamma$, $\{S_n\}_{n\in\mathbb{N}}$ such that the limit exits in $X$: $$\lim_{n\to\infty} S_n(y)$$Now if it would be possible to show that $\{S_n\}_{n\in\mathbb{N}}$ converges to some $S\in\Gamma$ then we would be finished. However, I'm not sure how to use the data to show that, because in order for $\{S_n\}_{n\in\mathbb{N}}$ to converge you need to know something about, let's say, $||S_{n_1}-S_{n_2}|| $ whereas you only know something about $||S_{n_1}(y)-S_{n_2}(y)||$ and you then only have $||S_{n_1}(y)-S_{n_2}(y)||\leq||S_{n_1}-S_{n_2}||||y|| $ by linearity.

2. I tried to define a mapping $\Psi_y:\mathcal{B}(X)\to X$ by $S\mapsto S(y)$. Then $\Psi$ is linear and continuous. The goal would be to prove $\Psi_y$ is a closed mapping, but I am not sure how to do that.

Answer 1

That $\Gamma(y)$ should be closed doesn't follow from the fact that $\Gamma$ is a closed subalgebra.

Example. Let $\mathbb N^+ = \mathbb N \setminus \{0\}$ be the set of positive integers. Consider $X = \ell^2(\mathbb N^+)$ and let $R : X \to X$ be the following weighted right shift: $$ (Rx)(n) = \begin{cases} \quad 0, & \quad\text{if $n = 1$}; \\[2ex] \dfrac{1}{n-1}\cdot x(n-1), & \quad\text{if $n \geq 2$}. \end{cases} $$ Now let $\mathcal A \subseteq B(X)$ be the closed subalgebra generated by $\text{id}$ and $R$. In other words, $\mathcal A$ is the closed linear span of $\{\text{id},R,R^2,R^3,\ldots\}$. Then $\mathcal A$ is a closed subalgebra of $B(X)$ containing the identity $\text{id}$.

For $y\in X$, let $\mathcal A(y) \subseteq X$ be defined as in Rudin's proof: $$ \mathcal A(y) = \{Sy : S\in \mathcal A\}. $$ It is easy to see that $\mathcal A(y)$ is a subspace. We present an example where $\mathcal A(y)$ is not closed.

Let $\{e_n\}_{n=1}^\infty$ denote the standard orthonormal basis for $X$: $$ e_n(m) = \begin{cases} 1,&\quad\text{if $m = n$};\\[1ex] 0, & \quad\text{if $m \neq n$}. \end{cases} $$ We show that $A(e_2)$ is not closed. For all $n\in\mathbb N$ we have $$ R^n e_2 = \frac{1}{(n+1)!}\cdot e_{n+2}.\tag{1} $$ Therefore we have $\text{span}(e_2,e_3,e_4,\ldots) \subseteq \mathcal A(e_2)$. Define $v \in \ell^2$ by setting $$ v(n) = \begin{cases} \quad 0,&\quad\text{if $n = 1$}; \\[2ex] \dfrac{1}{n - 1},&\quad\text{if $n\geq 2$}. \end{cases} $$ Clearly we have $v \in \overline{\text{span}}(e_2,e_3,e_4,\ldots)$. We show that $v \notin \mathcal A(e_2)$ holds. Intuitively, this is because $||R^n e_2||$ is much smaller than $||R^n||$ when $n$ goes to infinity. More precisely, it is readily verified that we have $||R^n|| = \frac{1}{n!}$ and $||R^n e_2|| = \frac{1}{(n+1)!}$, hence $$ \frac{||R^n||}{||R^n e_2||} = n+1,\qquad\text{for all $n\in\mathbb N$}. $$ The straightforward term-by-term approximation of $v$ does not work: we have $$ v \: = \: \sum_{n=2}^\infty \frac{1}{n-1}\cdot e_n \: = \: \sum_{k=0}^\infty k!\cdot R^k e_2,\tag{2} $$ but the series $$ \sum_{k=0}^\infty k!\cdot R^k $$ does not converge, since every term has norm $1$ (and we know that the summands of a convergent series must converge to $0$). Therefore we see that the approximation (2) does not correspond with an element $S\in\mathcal A$ such that $Se_2 = v$ holds.

I am however not certain that every element of $\mathcal A$ can be written as the limit of a series of the form $\sum_{n=0}^\infty \alpha_n R^n$ with $\alpha_n\in\mathbb C$ (c.f. this answer), so I shall prove explicitly that other approximations fail as well. Suppose, for the sake of contradiction, that there is some $S \in \mathcal A$ such that $Se_2 = v$ holds. Let $\varepsilon > 0$ be given and choose some $T = \sum_{n=0}^k \alpha_nR^n \in \text{span}(\text{id},R,R^2,R^3,\ldots)$ such that $||S - T|| < \varepsilon$ holds. Now we have $$||Te_2 - v|| \: = \: ||(T - S)e_2|| \: \leq \: ||T - S||\cdot ||e_2|| \: < \: \varepsilon, $$ so in particular $\big|(Te_2)(n) - \frac{1}{n-1}\big| < \varepsilon$ holds for all $n \geq 2$. Define $M \subseteq \mathbb N_{\geq 2}$ by $$ M := \left\{n\in\mathbb N \: : \: n\geq 2\ \text{and}\ \varepsilon \leq \frac{1}{2}\cdot\frac{1}{n-1}\right\}. $$ Then for all $m\in M$ we have $\big|(Te_2)(m)\big| > \frac{1}{2}\cdot \frac{1}{m-1}$. It follows from (1) that we have $$ (Te_2)(m) \: = \: \begin{cases} \dfrac{\alpha_{m-2}}{(m+1)!},&\quad\text{if $2\leq m\leq k + 2$};\\[1em] \quad 0,&\quad\text{otherwise}. \end{cases} $$ Hence for $m\in M$ we have $ |\alpha_{m-2}| > \tfrac{1}{2}\cdot m!$. We find $$ ||Te_1||^2 \: = \: \left|\left|\sum_{n=0}^k \alpha_n R^ne_1\right|\right|^2 \: = \: \left|\left|\sum_{n=0}^k \frac{\alpha_n}{n!}\cdot e_{n+1}\right|\right|^2 \: = \: \sum_{n=0}^k \left(\frac{\alpha_n}{n!}\right)^2 \: \geq \: \sum_{m\in M} \frac{1}{4} \: = \: \frac{|M|}{4}. $$ It follows that $||T|| \geq \frac{1}{2}\cdot \sqrt{\:|M|\:}$ holds. But now we have $||S|| \geq ||T|| - \varepsilon \geq \frac{1}{2}\cdot \sqrt{\:|M|\:} - \varepsilon$. Making $\varepsilon$ smaller and smaller, this lower bound for $||S||$ grows larger and larger, showing that $||S||$ is arbitrarily large. This is a contradiction, so we may conclude that no such $S$ exists. Indeed we see that $\mathcal A(e_2)$ is not closed.

However, in Rudin's proof, $\Gamma$ is not just any closed subalgebra; it is the commutant of $T$. This tends to have much more structure than an arbitrary closed subalgebra. (For instance if $X$ is a Hilbert space and $T$ is self-adjoint, then $\Gamma$ is a Von Neumann algebra. But even if $X$ is not a Hilbert space, we might be able to choose an involution $* : B(X) \to B(X)$ for which $T$ is self-adjoint. In that case the commutant $\Gamma$ is a self-adjoint subalgebra.) In particular, I don't think the above example $\mathcal A$ occurs as the commutant of a compact operator $T$.

Maybe the theorem can still be salvaged by using other properties of $\Gamma$?