I need help understanding the proof of Lemma $3$ in the paper Commutators of Operators by Paul R. Halmos.

Lemma $3$. Every Hermitian operator on an infinite-dimensional Hilbert space leaves invariant at least one large subspace with a large orthogonal complement.

A subspace $H$ of a Hilbert space is large if $H$ contains infinitely many orthogonal copies of its orthogonal complement, or, in other words, if $\dim H \ge \aleph_0 \dim (H^\perp)$.

Proof. The underlying Hilbert space, if it is not already separable, can be expressed as a direct sum of separable, infinite-dimensional subspaces invariant under the given operator. There is, therefore, no loss of generality in restricting attention to separable Hilbert spaces.

Q1. I understand that $\mathcal H$ can be written as an uncountable direct sum of separable subspaces, but why does it suffice to consider the separable case? Suppose $A \in \mathcal B(H)$ is Hermitian and $\mathcal H = \bigoplus_{n=1}^\infty \mathcal H_n$ where $\{\mathcal H_n\}_{n=1}^\infty$ is a family of $A$-invariant separable subspaces of $\mathcal H$. So, $A_i:= A\vert_{\mathcal H_i}: \mathcal H_i \to \mathcal H_i$ is also Hermitian for every $i\ge 1$. Assuming we have worked out the separable case; for every $i\ge 1$ there exists a large $A_i$-invariant subspace $M_i \le H_i$ such that $H_i - M_i$ (orthogonal complement in $H_i$) is also large. Consider $M = \bigoplus_{n=1}^\infty M_i$. It is clear that $M$ is $A$-invariant. How do I show that $M$ and $M^\perp$ are also large?

If $A$ is Hermitian and $E$ is the spectral measure of $A$, and if, for every Borel subset $M$ of the real line, $E(M) = \mathbf{0}$ or $\mathbf{I}$, then $A$ is a scalar multiple of $\mathbf{I}$. It follows easily that if, for every $M$, the dimension of the range of $E(M)$ is finite or co-finite, then $A$ differs from a scalar multiple of $\mathbf{I}$ by a finite-dimensional operator. In the contrary case both $E(M)$ and $\mathbf{I} - E(M)$ have infinite-dimensional ranges for some $M$. In either case, the conclusion of the lemma is obvious.

Q2. Could someone please explain the details of the above proof? It seems very cryptic and not at all straightforward.

Thank you!

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  • 1
    $\begingroup$ If $\mathcal H$ is not separable, it can be decomposed into an uncountable direct sum of invariant separable spaces. Now the problem reduces to the following. Given a set of an uncountable cardinality. It can be decomposed into two sets of the same cardinality as the entire set. $\endgroup$ Feb 25, 2022 at 4:15
  • 1
    $\begingroup$ A countable sum of separable spaces is separable. $\endgroup$ Feb 25, 2022 at 4:25

1 Answer 1


Take nonzero $x_0\in H$. Then $$ H_0=\overline{\operatorname{span}}\{A^n x_0:\ n\in\mathbb N\cup\{0\}\} $$ is a separable invariant subspace for $A$. If $H_0\subsetneq H$, take $x_1\in H_0^\perp$ and repeat. Using Zorn's Lemma (details at the end) one can construct a maximal family $\{x_j\}$ of such elements, and so $$H=\bigoplus_{j\in J}H_j,$$ with $H_j$ separable and invariant. If $J$ is infinite, split it into two disjoint equally infinite subsets $J_0$ and $J_1$, and now $\bigoplus_{J_0}H_j$ is a large invariant subspace with large orthogonal complement.

So now it remains to treat the case where $J$ is finite. We may as well assume $H$ is separable infinite dimensional. If there exists $M$ such that $P=E(M)$ and $I-P$ are infinite-dimensional, then $PH$ is invariant. Because it's infinite-dimensional with infinite-dimensional orthogonal complement in a separable space (so both have equal dimension), both are large.

We are now left with the case where either $E(M)$ or $I-E(M)$ is finite for all Borel sets $M$. Suppose there exist infinitely many pairwise disjoint Borel subsets with $E(M_k)$ finite. Then we can consider the invariant subspaces $H_k=E(M_k)H$ and again play the game of dividing them into two infinite families to get a large invariant subspace with large orthogonal complement. This leaves us with the case where any family of pairwise disjoint Borel subsets with $E(M_k)$ finite, is finite. A maximal family of such sets will be finite, then, and by taking the union we get $M$ such that $E(M)$ is finite and $E(N)$ is infinite for all $N\subset M^c$. If $E(M^c)\neq0$ and $M^c$ has at least two points we can get $E(N_1),E(N_2)$ infinite for $N_1\cup N_2=M^c$ and disjoint, and again we get $E(N_1)H$ invariant, large, with large orthogonal complement. This leaves us with the case where $E(M)$ is finite and $M^c$ has at most one point. Writing $P=E(M)$ we have $AP$ finite rank and $$A=AP+\lambda(I-P)=\lambda I+(A-\lambda I)P$$ as Halmos says. Since $P$ is finite rank, $I-P$ is infinite. Writing $I-P=Q_1+Q_2$ pairwise orthogonal and infinite, the subspaces $Q_1H$ and $Q_2H$ are invariant, countably infinite dimensional, and orthogonal. So $Q_1H$ is invariant, large, with large orthogonal complement.

Use of Zorn's Lemma. Let $$ \mathcal F=\{(x_j):\ \langle A^nx_j,A^mx_k\rangle=0\ \text{ for all } k\ne j,\ \text{ and all } n,m\}, $$ ordered by inclusion. This is trivially nonempty since $A$ is selfadjoint and so the orthogonal complement of an invariant subspace is invariant. Given a chain $\{(x_j)_{j\in J_\alpha}: \alpha\}$, where $J_\alpha\subset J_\beta$ if $\alpha\leq\beta$, let $J=\bigcup_\alpha J_\alpha$. Then $(x_j)_{j\in J_\alpha}\subset (x_j)_{j\in J}\in\mathcal F$ for all $\alpha$, so it is an upper bound. By Zorn's Lemma there exists a maximal element $(x_j)_{j\in J}$. With $H_j=\overline{\operatorname{span}}\{A^nx_j:\ n\}$, let $\tilde H=\bigoplus_jH_j$. Then $\tilde H=H$, because otherwise there exists $y\in \tilde H^\perp$, and since $\tilde H^\perp$ is invariant, this contradicts the maximality of $(x_j)$. So $$ H=\bigoplus_jH_j. $$

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    Feb 27, 2022 at 12:03

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