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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.

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  • $\begingroup$ Whenever something is not clear in Rudin, my first reflex is to check Conway. He offers a different proof using Schauder's fixed point theorem, but refers the reader to an alternative proof not using said theorem. This proof seems at first glance similar to Rudin's proof, though care is taken to talk about the closure of what Rudin calls $\Gamma(y)$. This leads me to seriously consider the possibility that Rudin made a mistake. But maybe we're missing something... Anyway, it's way past bed time over here, so I'll think about it some other day! $\endgroup$ Jan 29, 2016 at 0:35

1 Answer 1

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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 $\Gamma$?

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  • $\begingroup$ So Rudin was definitely wrong with "it is easy to see …". Whether he was wrong with "$\Gamma(y)$ is closed" remains open. However, the proof works if one considers $H(y) = \overline{\Gamma(y)}$ instead of $\Gamma(y)$ itself, with practically no changes. $\endgroup$ Feb 2, 2016 at 0:20
  • $\begingroup$ Oh yes, I didn't even think to check the rest of the proof (in an exit on first error kind of way). :-) Interestingly, I just discovered that the word closed has been removed in the book I own, which is the fourteenth reprint (2014) in its cheapo Indian edition. (At the time this was the only version I could still buy.) So the error was found before! $\endgroup$ Feb 26, 2016 at 14:14

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