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 be a closed, bounded, convex subset of a normed space . If is a compact map such that , then there is an such that .

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 is a compact subset of a Banach space , then the closed convex hull 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 is a subalgebra of such that [the identity] and , and if is a nonzero compact operator on , then there exists a such that .

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 and construct a continuous map on for which is invariant under . We’ll use Mazur’s theorem to show that 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 , contains the span of our fixed point in its kernel. Onwards!

Let be a nonzero compact operator and assume that . Find with and let be the unit ball about . It is easy to see that , for if then and so and so is bounded away from zero, i.e. . Since compact operators map bounded sets to pre-compact sets, it follows that is compact. Since is an algebra, it follows that for all nonzero that is an invariant subspace for . Moreover, since is nonzero and , this guarantees that . Thus , whence it follows that . So is dense in

Since is dense in for each , in particular for every we may find a with , i.e.

. Now, define

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

Here comes the genius of Lomonosov. Define by

As are continuous for each , it follows that is continuous. I claim that . Now suppose . We want to show that . But this is clear for the following reasons. If , then by definition it follows that , but then and consequently . So kills every problematic term! Thus the only nonzero terms of are those for which . Moreover, by definition we see that .

Thus is simply a convex combination of elements in and since is clearly convex, it follows that . Thus .

I claim that each is compact. Indeed, if is any bounded set, then since is compact it follows that is compact. Since is continuous, it maps compact sets to compact sets and thus is compact. But and so is compact, being a closed subset of a compact set. Thus maps bounded sets to pre-compact sets, and so is compact. Since a finite union of pre-compact sets is pre-compact, and is clearly bounded, it follows that has compact closure. Moreover, by Mazur’s theorem, is compact, being the closed convex hull of a compact set. But this convex set contains , since is a convex combination of elements in . Thus is compact, being a closed subset of a compact set. But then is a compact map on . By the Schauder Fixed-Point Theorem, there is some vector with . Now let and let . Then being a linear combination of elements form the algebra . Moreover, . Since and , it follows that . Suppose . Then . So , as desired.

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

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 (?) in the paragraphs following ‘Onwards!’.

A union of pre-compact sets is certainly not pre-compact. Take with its Euclidean topology. Then is pre-compact for each , yet the infinite union is not pre-compact, since its closure is . You are right, I meant finite unions of pre-compact sets is pre-compact.