## The Invariant Subspace Problem and Lomonosov’s Theorem (Part 2 of 3).

With those pesky definitions out of the way. I will merely state these two lemmas (which are nontrivial theorems in their own right), but the proofs both long approximations that won’t contribute to this post.

Lemma 1 (The Schauder Fixed Point Theorem)

Let $E$ be a closed, bounded, convex subset of a normed space $X$. If $f: E\to X$ is a compact map such that $f(E)\subseteq E$, then there is an $x\in E$ such that $f(x)=x$.

The intuition is the following. Compact sets are topologically small. They attain supremums, you can extract finite subcovers from arbitrary covers, they’re punks. For similar reasons to every other fixed point theorem (Banach, Brouwer, Kakutani, etc.) the fact that you’re mapping a set into a small subset (in a topological sense) gives you a fixed point. This is a dramatic over simplification of the proofs of these nontrivial theorems, but the intuition is fairly accurate.

Lemma 2 (Mazur’s Theorem)

If $K$ is a compact subset of a Banach space $X$, then the closed convex hull $\overline{co}{(K)}$ is compact.

The intuition for this is similar. Compact sets are small, and since the convex hull fits them very tightly, they can’t get much bigger (in a topological sense). The proof is easy to follow and can be found in Conway (I believe he’s the only one to refer to this as Mazur’s Theorem anyhow).

Now standing on the shoulders of these theorems, we can proceed with a lemma (read: theorem) that I find particularly beautiful.

Theorem (Lomonosov’s Lemma)

If $A$ is a subalgebra of $B(X)$ such that $I\in B(X)$ [the identity] and $Lat(A)=\{(0), X\}$, and if $K$ is a nonzero compact operator on $X$, then there exists a $A'\in A$ such that $\ker(A'K-I)\supsetneq \{(0)\}$.

Before I proceed I’ll offer some intuition. The fact that this algebra has no nontrivial invariant subspace is an indication of how diverse this family of operators is. If they were similar in some sense (whatever that would mean), then they would share invariant subspaces when composed with a very uninteresting compact operator. When I say uninteresting, I mean that compact operators are as close to finite-dimensional operators as you can get. In fact, there is a very precise sense for which this is true. The way this proof will proceed is as follows, we’ll define a closed bounded convex set $S$ and construct a continuous map $\psi$ on $S$ for which $S$ is invariant under $\psi$. We’ll use Mazur’s theorem to show that $\psi$ is a compact map. Then we’ll use Schauder’s Fixed Point Theorem to get a fixed point, which will allow us to construct an operator in our algebra who, when composed with $K$, contains the span of our fixed point in its kernel. Onwards!

Let $K$ be a nonzero compact operator and assume that $||K||_{op}=1$. Find $x_0\in X$ with $||K(x_0)||>1$ and let $S=\{x\in {X}: ||x-x_0||\le 1\}$ be the unit ball about $x_0$. It is easy to see that $0\not\in S$, for if $0\in S$ then $||x_0||\le 1$ and so $11$ and so $K(S)$ is bounded away from zero, i.e. $0\not\in \overline{K(S)}$. Since compact operators map bounded sets to pre-compact sets, it follows that $\overline{K(S)}$ is compact. Since ${A}$ is an algebra, it follows that for all nonzero $x\in X$ that $\overline{\{Tx: T\in {A}\}}$ is an invariant subspace for ${A}$. Moreover, since $x$ is nonzero and $I\in {A}$, this guarantees that $\overline{{\{Tx: T\in {A}\}}}\supset \{Ix\}=\{x\}\supsetneq (0)$. Thus $\overline{\{Tx: T\in {A}\}}\in Lat({A})$, whence it follows that $\overline{\{Tx: T\in {A}\}}={X}$. So $\overline{{\{Tx: T\in {A}\}}}$ is dense in ${X}$

Since $\overline{\{Tx: T\in {A}\}}$ is dense in ${X}$ for each $x\in {X}$, in particular for every $y\in \overline{K(S)}$ we may find a $T\in {A}$ with $||Ty-x_0||<1$, i.e.

$\overline{K(S)}\subseteq \bigcup_{T\in {A}}\{y: ||Ty-x_0||0$. Now, define

$b_j(y)=\frac{a_j(y)}{\sum_{i=1}^{n}a_i(y)}$

Then $b_j(y)$ is continuous, being a quotient of two continuous functions (with the denominator non-zero).

Here comes the genius of Lomonosov. Define $\psi: S\to {X}$ by

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

As $b_j, K, T_j$ are continuous for each $j$, it follows that $\psi$ is continuous. I claim that $\psi(S)\subseteq S$. Now suppose $x\in S$. We want to show that $||\psi(x)-x_0||\le 1$. But this is clear for the following reasons. If $T_j\circ K(x)\not\in S$, then by definition it follows that $||T_j(K(x))-x_0||\ge 1$, but then $a_j(K(x))=0$ and consequently $b_j(K(x))=0$. So $b_j(K(x))$ kills every problematic term! Thus the only nonzero terms of $\psi$ are those $T_j(K(x))$ for which $T_j(K(x))\in S$. Moreover, by definition we see that $\sum_{j=1}^{n}b_j(K(x))=\frac{\sum_{j=1}^{n}a_j(K(x))}{\sum_{j=1}^{n}a_j(K(x))}=1$.

Thus $\psi(x)$ is simply a convex combination of elements in $S$ and since $S$ is clearly convex, it follows that $\psi(x)\in S$. Thus $\psi(S)\subseteq S$.

I claim that each $T_jK$ is compact. Indeed, if $B$ is any bounded set, then since $K$ is compact it follows that $\overline{K(B)}$ is compact. Since $T_j$ is continuous, it maps compact sets to compact sets and thus $T_j(\overline{K(B)})$ is compact. But $\overline{T_jK(B)}\subseteq T_j(\overline{K(B)})$ and so $\overline{T_jK(B)}$ is compact, being a closed subset of a compact set. Thus $T_jK$ maps bounded sets to pre-compact sets, and so $T_jK$ is compact. Since a finite union of pre-compact sets is pre-compact, and $S$ is clearly bounded, it follows that $\bigcup_{j=1}^{n}T_jK(S)$ has compact closure. Moreover, by Mazur’s theorem, $\overline{co}(\bigcup_{j=1}{n}T_j(K(S)))$ is compact, being the closed convex hull of a compact set. But this convex set contains $\psi(S)$, since $\psi(S)$ is a convex combination of elements in $T_jK(S)$. Thus $\overline{\psi(S)}$ is compact, being a closed subset of a compact set. But then $\psi$ is a compact map on $S$. By the Schauder Fixed-Point Theorem, there is some vector $x_1\in S$ with $\psi(x_1)=x_1$. Now let $\beta_j=b_j(K(x_1))$ and let $A'=\sum_{j=1}^{n}\beta_jT_j$. Then $A'\in {A}$ being a linear combination of elements form the algebra ${A}$. Moreover, $A'K(x_1)=\psi(x_1)=x_1$. Since $x_1\in S$ and $0\not\in S$, it follows that $x_1\neq 0$. Suppose $\alpha\in \mathbb{F}$. Then $(A'K-I)(\alpha x_1)=\alpha (A'K-I)(x_1)=\alpha(A'K(x_1)-x_1)=\alpha(x_1-x_1)=0$. So $0\subsetneq span(x_1)\subseteq \ker(A'K-1)$, as desired.

Brilliant. Coming up in the third post is a corollary to Lomonosov’s Lemma, known as Lomonosov’s Theorem.

This Fall I will be a first year mathematics PhD student at UCLA. I enjoy doing analysis - particularly of the functional variety.
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### 2 Responses to The Invariant Subspace Problem and Lomonosov’s Theorem (Part 2 of 3).

1. Z Norwood says:

I don’t believe that ‘a union of pre-compact sets is pre-compact’, but I think you (luckily) only need the fact for finite unions.

There are a couple of mysterious characters ($11$?) in the paragraphs following ‘Onwards!’.

A union of pre-compact sets is certainly not pre-compact. Take $\mathbb{R}$ with its Euclidean topology. Then $(n,n+1)$ is pre-compact for each $n\in \mathbb{Z}$, yet the infinite union $\bigcup_{n\in \mathbb{Z}} (n,n+1)$ is not pre-compact, since its closure is $\mathbb{R}$. You are right, I meant finite unions of pre-compact sets is pre-compact.