When is a subspace of a Hilbert space large? In this paper, Halmos defines what it means for a subspace of a Hilbert space to be large.

Definition. A subspace $H$ of a Hilbert space $K$ is large if $H$ contains infinitely many orthogonal copies of its orthogonal complement, or, in other words, if $\dim H \ge \aleph_0 \dim (K\setminus H)$. Thus, for example, a subspace of a separable Hilbert space is large if and only if it is infinite-dimensional.


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*What does the author mean by "infinitely many orthogonal copies of its orthogonal complement"? Does it refer to infinitely many subspaces inside $H$, which are all orthogonal, and also isomorphic to $H^\perp$? A concrete example would help me appreciate this more!


*If $H$ is not closed, $K\setminus H$ may not be a vector space, so how can we talk about its dimension?


*How would you prove that a subspace of a separable Hilbert space is large if and only if it is infinite-dimensional?
This is the first time I have seen this definition, so it would be great if you could point me to any other related references. Thank you!
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*"Copy of X" here means "space that is isomorphic to X".  So the condition is indeed that $H$ should contain infinitely many subspaces, each of which is isomorphic to the orthogonal complement $H^\perp$, and which are pairwise orthogonal to each other.
A concrete example might be $K = L^2(\mathbb{R})$, $H = L^2(0,\infty)$.  Then $H^\perp = L^2(-\infty,0)$.  Let $H_i = L^2(i, i+1)$, which you can show is isomorphic to $L^2(-\infty, 0)$, and the $H_i$ are all pairwise orthogonal.  So $H$ is an example of a large subspace.


*The notation $K \setminus H$ here doesn't mean set difference, but orthogonal complement.  So $K \setminus H$ is just the orthogonal complement of $H$ in $K$, which you could also denote $H^\perp$ when the ambient space $K$ is understood.  It is certainly a closed subspace of $K$, and it makes sense to speak of its dimension.
The $K \setminus H$ notation is fairly common, and although it is technically ambiguous with set difference, in practice it is usually clear from context which one is meant.


*Let $H$ be a subspace of the separable Hilbert space $K$.  It is clear that a large subspace must be infinite-dimensional (it contains infinitely many nontrivial orthogonal subspaces).
For the converse, suppose $H$ is infinite-dimensional, and fix an orthonormal basis $\{e_1, e_2, \dots\}$ for $H$.  Let $H^\perp$ be its orthogonal complement.
Suppose first that $H^\perp$ has finite dimension $n$.  For each $i > 0$ let $H_i$ be the subspace of $H$ spanned by $\{e_{ni+1}, \dots, e_{ni+n}\}$.    Each $H_i$ is isomorphic to $H^\perp$ because they have the same dimension $n$, and the $H_i$ are pairwise orthogonal because they are spanned by disjoint subsets of an orthogonal set.  Thus $H$ is large.
Otherwise, suppose $H^\perp$ is infinite-dimensional; since it is separable, it has a countable orthonormal basis $\{f_1, f_2, \dots\}$.  Let  $\{ k_{1,1}, k_{1,2}, \dots\}$, $\{k_{2,1}, k_{2,2}, \dots\}$ be infinitely many disjoint sequences in $\mathbb{N}$; for instance you could enumerate the primes as $p_n$ and let $k_{n,i} = p_n^i$.  Let $H_i$ be the closed linear span of $\{e_{k_{i,1}}, e_{k_{i,2}}, \dots\}$.  Each $H_i$ is isomorphic to $H^\perp$, because $H^\perp, H_i$ are both infinite-dimensional separable Hilbert spaces, or because they both have orthonormal bases of the same cardinality $\aleph_0$.  And the $H_i$ are again pairwise orthogonal because they are the closed linear spans of disjoint orthogonal sets.  Thus $H$ is also large in this case.
In essence, it's really just the fact that a countably infinite set can be written as a countable union of countable sets (or of finite sets), i.e. that $\aleph_0 = \aleph_0^2$, applied to dimension instead of cardinality.
The term "large subspace" seems to have been introduced by Halmos and I haven't seen it used by anyone else; the concept is not one that I think comes up especially often.  I wouldn't particularly expect there to be a lot of reference material on the topic.  It does appear in Halmos's Hilbert Space Problem Book on page 131, where he says that "the idea has appeared before, even if the term has not".
