## The Invariant Subspace Problem and Lomonosov’s Theorem (Part 1 of 3).

One of the biggest open problems in functional analysis is the invariant subspace problem, which asks if every operator $T$ on a Banach space $X$ admits a proper non-zero closed subspace $M\subseteq X$ for which $T(M)\subseteq M$. Actually, as stated, this is quite a lie. The problem stated above was answered in the negative by Swedish mathematician Per Enflo in the late seventies, and also answered negatively by Charles Read (academic brother of Jamie Radcliffe!) in the mid-eighties. They proved the existence of a Banach space and an operator who admitted no non-trivial closed subspaces (Enflo constructed a wild counterexample, and Read constructed an operator on $\ell^1$). In fact, Read proved quite a bit more by constructing an operator that admits no nontrivial closed subset (I believe this counterexample is not on $\ell^1$, however). In no less than eighty pages, of course.

The counterexamples of Enflo and Read show that there are some infinite dimensional spaces in which everything can go (depending on your viewpoint) terribly wrong. After all, every operator on a finite dimensional complex vector space admits a nontrivial invariant subspace. To see why, take such an operator $T$ and find an eigenvalue $\lambda$ for $T$ and its corresponding eigenvector $v$ (which exists since its characteristic polynomial admits a root, and hence an eigenvalue, over $\mathbb{C}$). Then span$(v)$ is an invariant subspace for $T$. First, it’s clearly a closed subspace. Moreover, if $\alpha x\in \text{span}(v)$, then $T(\alpha x)=\alpha T(v)=\alpha \lambda v\in \text{span}(v)$. Herein lies a fundamental and tremendously important difference between infinite and finite dimensional vector spaces.

The invariant subspace problem, as its stated now, asks whether every operator on a/the separable infinite dimensional Hilbert space admits a nontrivial invariant subspace. Since there’s really only one separable, finite dimensional Hilbert space one can exclusively restrict their attention to $\ell^2$ if they’d like. Since Hilbert spaces admit a more rigid structure, it might be behaved well-enough to force operators to have a nontrivial invariant subspace. In passing, I’ll mention that recently (2009) two mathematicians constructed an infinite dimensional Banach space for which every operator admits a nontrivial invariant subspace. Again, I’ve lied to you again and the example is much cooler than what I’ve just stated. Before we understand why, we need to get to the meat of this post; Lomonosov‘s Theorem.

Lomonosov’s Theorem is hailed as one of the most beautiful theorems in Functional Analysis. First, for its remarkable result to be discussed below. Second, because of its simplicity. After trying to read Charles Read’s paper this summer, I’ve come to know first hand how inaccessible a lot of material in functional analysis is without the proper background. Lomonosov’s proof is beautiful in its own right, but relies on nothing more than a little bit of definitions, Topology, Convex Geometry, and a fixed point theorem. In order, I’ll present;

• A few definitions.
• The Schauder Fixed Point Theorem
• Mazur’s Theorem
• Lomonosov’s Lemma
• Lomonosov’s Theorem

As an aside I should mention that, in my opinion, I think Lomonosov’s Lemma should in fact be the theorem and Lomonosov’s Theorem should be a corollary. It seems the statute of limitations has run out on that though. Let’s do a few definitions and establish some notation. You should look up what a Banach space is. Also, all operators are linear. Lastly, all subspaces are, by definition, closed subspaces. A non-closed subspace is traditionally called a linear manifold. As expected, we’ll work exclusively in the metric topology induced by the norm.

Given an operator $T$ between two Banach spaces $T:X\to Y$, we define the operator norm, denoted $||T||_{op}$ as $\sup\{||Tx||: ||x||=1\}$. Of course the norm of the domain and image are measured via $X$‘s and $Y$‘s norm, respectively. If the operator norm of $T$ is finite, we say $T$ is bounded. This is, in fact, equivalent to $T$ being continuous.

We say that a bounded operator $T$ is a compact operator provided that $T$ maps bounded sets in $X$ to pre-compact sets in $Y$. That is, if $B\subseteq X$ is bounded, then the closure of $T(X),$ denoted $\overline{T(X)}$, is compact in $X$. To be less precise, the intuition is that $T$ puts bounded sets into compact sets. This intuition applies to just continuous maps (not necessarily linear), in which we say that a continuous map $f$ on some set $S$ is compact if $T(S)$ is contained in a compact set.

Given a Banach space $X$ we denote the algebra of bounded operators of $X$ by $B(X)$. A sub-algebra of $B(X)$ is, as you might have guessed, a subset of $B(X)$ that is also an algebra.

Given a Banach space $X$ and an operator $T\in B(X)$, we denote the collection of all invariant subspaces of $T$ by $Lat(T)$.  The notation is appropriate since $Lat(T)$ forms a lattice under the relation $M_1\vee M_2 =\overline{M_1+M_2}$ and $M_1\wedge M_2=M_1\cap M_2$ . Observe that the zero subspace $(0)$ and the whole space $X$ are always invariant for $T$ and so we say that $T$ admits a nontrivial invariant subspace if $Lat(T)\supsetneq \{(0), X\}$. If ${A}$ is a family of operators, then we define $Lat(A)=\bigcap_{T\in {A}}Lat(T)$.

Moreover, we define the commutant $T\in B(X)$, and denoted $\{T\}'$, to be the set of all operators that commute with $T$. I’ll leave it as an exercise that $T$ is a sub-algebra of $B(X).$  This fact will be important later. Even more special than an invariant subspace is what’s called a hyper-invariant subspace. We say that a closed subspace $M$ is hyper-invariant for $T$ provided $M\in Lat\{T\}'$, i.e. not only is $M$ is an invariant subspace for $T$, but its an invariant subspace for every operator that commutes with $T$ as well.

Lastly. A set $S$ of a Banach space $X$ is convex provided that $\lambda a+(1-\lambda)b\in S$ for all $a,b\in S$ and $\lambda\in [0,1]$. The intersection of convex sets is convex and so given any subset $S$ of $X$ we may define the convex hull of $S$ to be the intersection of all convex sets that contain $S$. Moreover, we may define the closed convex hull to be the intersection of all closed convex sets that contain $X$

Phew. Definitions are out of the way. I’m eager to publish this thing before everything gets deleted. So, I’ve made this purely motivational and definitional post the first of potentially three posts – all to come immediately.

I think they’re both accurate. Since $\{(0), X\}\subseteq Lat(T)$, then if $T$ admits a nontrivial invariant subspace, then there is some $M \in Lat(T)$ for which $M\ne (0), X$ and so $\{(0),X\}\subsetneq Lat(T)$ since its strictly proper. But this is the same as $\{(0),X\}\ne Lat(T)$ as well.