Invariant Subspaces:

Definition 6.1.1   For $T \in B(X)$, $X$ a Banach spaces, we call $M \leq X$ an invariant subspace for $T$ is for all $x \in M$, $Tx \in M$.

Because subsapces in $X$ do not have orthogonal complements (only Hilbert spaces have there), there is no way to define reducing subspaces. A subspace may have many complements.

For $T \in B(X)$, define Lat $T$ to be the invariant subspaces of $T$. Notice, $0, X \in$   Lat $T$. There are called trivial. For $M,N \in$   Lat $T$, define $M \wedge N$ ($M$ meet $N$) to be $M \cap N$. Note $M \cap N \in$   Lat $T$. Define $M \vee N = \overline{\\ text{span}\{ M \cup N \}}$ ($M$ join $N$). Thus, $M \vee N \in$   Lat $T$.

Similarly, we can define $\vee$ and $\wedge$ for arbitrary collections of elements of Lat $T$. That is Lat $T$ is a complete lattice with 0 as the smallest element and $X$ the biggest.

Theorem 6.1.2   Reed (1984), Enflow/Beauzamy (1985)
There is a non-reflexive Banach space $X$ and $T \in B(X)$ such that Lat $T = \{ 0, X \}$

Later, Reed showed that $X$ can be taken to be $\ell^1$.

Open Question: If $X$ is reflexive, is it possible to have $T \in B(X)$ where Lat $T = \{ 0, X \}$. In particular, this is open for $X$ a separable Hilbert space.

In general, it is very hard to find Lat $T$ (in general).

Examples:

1. $V:L^2[0,1] \rightarrow L^2[0,1]$ where $V$ is the Volterra operator $(Vf)(x) = \int_0^x f dm$.
   Let $M_\alpha \leq L^2[0,1]$ be $f \in L^2 : f\vert _{[0,\alpha]} \equiv 0$. Clearly, $M_\alpha \in$   Lat $V$. In fact,
      Lat $$V = \{ M_\alpha : \alpha \in [0,1] \}$$
   Notice Lat $V$ is a (totally ordered) chain.
   
2. Let $S:\ell^2 \rightarrow \ell^2$ be the shift operator
   $$S(x_1,x_2, \ldots) = (0, x_1, x_2, \ldots )$$
   Lat $S$ is large and is parameterized using analytic functions (Beurling's Theorem in Rudin). E.g., subspace of elements of the form $(0,0,0,x_4,x_5, \ldots)$.
   
3. Let $T = \begin{bmatrix}0 & 1 \ 1 & 1 \end{bmatrix}$ acting on $\mathbb{R}^2$. Then Lat $T$ is trivial, since $T$ is a rotation by $90$ degrees.
4. Every matrix/operator on $B(\mathbb{C}^n$ has nontrivial invariant subspaces. This is because there exists $\alpha \neq 0$ such that $(T - \alpha I)$ is not invertible (This is in the book).
5. (Not in the book). Suppose $T \in B(H)$, $\Vert T\Vert = 1$. If
   $$\{ z \in \mathbb{C}: T - zI$$    is not invertible $$\} \supset \mathbb{T}$$
   then $T$ has a nontrivial invariant subspace where
   $$\mathbb{T}= \{ z \in \mathbb{C}: \vert z\vert = 1 \}.$$

Theorem 6.1.3   Lomonosov, 1973:
Let $T \in B(X)$, $X$ is a Banach space over $\mathbb{C}$, and $T$ is not a multiple of the identity. If there is a nonzero compact operator $K$ such that $TK = KT$, then $T$ has a nontrivial hyper-invariant subspace. That is, there is $M \leq X$ such that $\forall A \in B(X)$ where $AT = TA$, $AM \subset M$.

We need two preliminary results first:

Theorem 6.1.4   Mazur's Theorem:
Let $X$ be a Banach space and $K \subset X$ be compact. Then $\overline{\text{co }}(K)$ is compact

Proof. We need to show $\overline{\text{co }}(K)$ is totally bounded. Let $\epsilon > 0$. Since $K$ is compact, it has an $\frac\epsilon 4$-net $x_1, \ldots, x_n$.

Let $C =$   co $\{ x_1, \ldots, x_n \}$ (which is a finite-dimensional set). This is easily seen to be compact since it is bounded and finite-dimensional (use Heine-Borel).

Let $y_1, \ldots, y_m$ be an $\frac\epsilon 4$-net for $C$. Let $w \in \overline{\text{co }}(K)$. Thus there is $z \in$   co $(K)$ such that $\Vert z - w\Vert < \frac\epsilon 4$.
Thus, there is $\alpha_1, \ldots , \alpha_p$ such that

$$z = \sum_{p=1}^\ell \alpha_p k_p,$$    where $$k_p \in K, \alpha_i \neq 0, \sum_{p=1}^n \alpha_p = 1$$

For each $p$ there is $x_\lambda(p)$ such that

$$\Vert x_p - x_\sigma(p)\Vert < \frac\epsilon 4$$

Let

$$a = \sum_{p=1}^\ell \alpha_p x_\sigma(p) \in C$$

Then,

$$\Vert z - a\Vert \leq \sum_{p=1}^\ell \vert\alpha_p\vert \Vert k_p - x_\sigma(p)\Vert \leq \frac\epsilon 4$$

As $a \in C$, there exists $y_i$ such that $\Vert y_i - a\Vert < \frac\epsilon 4$. Thus,

$$\Vert w - y_i \Vert < \epsilon$$

So the $y_i$ are an $\epsilon$-net for $\overline{\text{co }}(K)$. $\qedsymbol$

Monday, April 3, 2006:

Lemma 6.1.5   Lomonosor's Lemma
Let $\mathcal{A}$ be a subalgebra of $B(H)$ and assume $I \in \mathcal{A}$, Lat $\mathcal{A}= \{ 0, \mathcal{A}\}$. Let $K$ be a nonzero compact operator on $X$. There is $A \in \mathcal{A}$ such that $\ker( AK - I) \neq 0$. That is, there is $y \in X$ such that $AKy = y$.

Proof. WLOG, $\Vert K\Vert = 1$. Fix $x_0 \in X$ such that $\Vert K x_0\Vert > 1$. Let

$$S = \{ x \in X : \Vert x - x_0\Vert \leq 1 \} = \overline{B_1(x_0)}$$

Notice: $0 \notin S$, $0 \notin \overline{K(S)}$ as $\overline{K(S)} \subset \overline{B_1(K x_0)}$. Consider

$$\{ Tx : T \in \mathcal{A}\}$$

for some fixed $x in X, x \neq 0$. This is in Lat $\mathcal{A}$ and it is not 0 (as $I \in \mathcal{A}$). Thus,

$$\overline{\{ Tx : T \in \mathcal{A}\}} = X$$

Thus, for any $y \in \overline{K(S)}$, there is $T \in \mathcal{A}$, such that $\Vert Ty - x_0 \Vert < 1$. Equivalently,

$$\overline{K(S)} \subset \bigcup_{T \in \mathcal{A}} \{ y : \Vert Ty - x_0\Vert < 1 \}$$

where the right hand side is open. By the compactness of $K$, $\overline{K(S)}$ is compact. So, there is $T_1, \ldots, T_n$ such that

$$\overline{K(S)} \subset \bigcup_{j=1}^n \{ y : \Vert T_j y - x_0 \Vert < 1 \}$$

For $y \in \overline{K(S)}$, $j \in \{ 1, \ldots, n \}$, let $a_j(y) = \max \{ 0, 1 - \Vert T_j y - x_0\Vert \}$. Then,

$$\sum_{j=1}^n a_j(y) > 0$$

Let

$$b_j(y) = \frac{a_j(y)}{\sum_{k=1}^n a_k(y)}$$

Then,

$$\sum_{j=1}^n b_j(y) = 1$$

for all $y$.
Define $\psi : S \rightarrow X$

$$\psi(x) = \sum_{j=1}^n b_j(K x ) T_j Kx.$$

It is easy to see that $a_j$ (hence, $b_j$ and $\psi$ are continuous). If $x \in S$, $Kx \in K(S)$. If $b_j(Kx) > 0$, then $a_j(Kx) > 0$ and $\Vert T_j Kx - x_0 \Vert < 1$. That is, $T_j Kx \in S$ whenever $b_j(Kx) > 0$. Thus, as $S$ is convex, $\psi(S) \subset S$. ($\psi(x)$ is a convex combination of elements of $S$).
Notice $T_j K$ is compact and $\textrm{ran }\psi$ is contained in

$$\overline{\text{co }}\{ T_J K(\text{ball }X) : j = 1, \ldots, n \}$$

By Mazur, this set is compact. By Schauder's Fixed Point Theorem, there is $y \in S$ such that $\psi(y) = y$.

Let $\beta_j = b_j(Ky)$. Let $A = \sum \beta_j T_j \in \mathcal{A}$. Then,

$$A Ky = \sum_{j=1}^n \beta_j t_j K y = \sum_{j=1}^n b_j(Ky) T_j(Ky) = \psi(y) = y$$

$\qedsymbol$

Proof. Proof of Lomonosor's Theorem:
Let

$$\mathcal{A}= \{ A \in B(X) : AT = TA \}$$

So $K \in A$. It suffices to prove that Lat $A \neq \{ 0, X \}$. Assume Lat $A = \{ 0, X \}$. By Lomonosor's Lemma, there is $A \in \mathcal{A}$ such that $\ker( AK - I) \neq 0$. Let $N = \ker (AK - I)$. Then,

$$N \in$$   Lat $$(AK)$$

$$AK\vert _N = I_N$$

As $AK$ is compact, $AK\vert _N$ is compact. The only compact identity operators are the ones on finite-dimensional spaces. Thus, $\dim N < \infty$.

As $AK \in \mathcal{A}$, so

$$Tx = T(AKx) = AK(Tx).$$

Hence, $T N \subset N$. Now, $T\vert _N$ has an eigenvalue, $\lambda$ So, $\ker(T-\lambda) \neq 0$. Now $\ker(T - \lambda) \neq X$, as $T$ is not a multiple of the identity.

Claim: $\ker(T - \lambda) \in$   Lat $A$.

Proof. If $\lambda \neq 0$,

$$Ax = A \frac1\lambda (\lambda x) = \frac1\lambda A\lambda x = \frac1\lambda A(Tx) = \frac1\lambda T(Ax)$$

$$T(\frac{Ax}{\lambda})$$

Thus,

$$T(Ax) = \lambda Ax$$

Thus, $Ax \in \ker T - \lambda$. This is true for all $x \in \ker$, $A \in \mathcal{A}$, proving the claim. $\qedsymbol$

This gives a contradiction, so Lat $\mathcal{A}\neq \{ 0, X \}$. $\qedsymbol$