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Natural way to find an example of banach spaces is to look at closed subspaces of Banach spaces. Initially, It was really hard to find examples of closed subspaces of $L^2(R)$. Then, my professor gave me this example.

For $ f \in L^1(R) \cap L^2(R)$ such that $\hat f(0)=0.$ Then closure( in $L^2(R)$) of translational invariant linear subspace containing $f$ is the closed subspace of $L^2(R)$.

Then the question arises, what are all closed subspaces of $L^2(R)$ ?

Can we characterize all closed subspaces of $L^2(R)$ ?

I would like to know more examples of closed subspaces of $L^2(R)$, if you are familiar with them.

P.S. There is nothing special about $L^2(R)$, question is also applicable to $L^p(R)$. But $L^2(R)$ has nice properties ( Fourier transformation isometry).

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  • $\begingroup$ As Charles Matthews remarks below, you really need to impose some extra conditions on your subspaces before it is reasonable to ask for a characterization. The clue is that your question is phrased in terms of $L^2(R)$ and not, say, $\ell^2$ of the free group on two generators; as abstract Hilbert spaces they are "the same", but in many contexts one is interested in extra structure that comes from the "particular" Hilbert space you start with. $\endgroup$
    – Yemon Choi
    Commented Mar 27, 2011 at 9:28
  • $\begingroup$ A related example in a slightly different setting: consider $\ell_2({\mathbb Z}_+)$ where ${\mathbb Z}_+$ denotes the set of non-negative integers, and let $S$ be the backwards shift operator. Then one can characterize the closed subspaces of $\ell_2({\mathbb Z}_+)$ which are $S$-invariant; this is, essentially, Beurling's theorem, the mere statement of which already requires complex analysis and some single-variable function theory. $\endgroup$
    – Yemon Choi
    Commented Mar 27, 2011 at 9:34

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What kind of characterization are you thinking of? There are plenty of closed linear subspaces in a Hilbert space $H$, and they really have no special extra structure coming from closedness and linearity. Closed linear subpaces have a Hilbert dimension and co-dimension, and this is the only invariant to distinguish two of them, in the sense that, of course, if $W$ and $V$ have the same dimension and co-dimension, there is a unitary operator $U$ on $H$ mapping $V$ onto $W$. The set of all linear closed subspaces of $H$ has the structure of an analytic manifold (with connected components labeled by the above pair of cardinal numbers).

There are several characterization for closed linear subspaces, e.g. $V\subset H$ is a linear closed subspaces iff $V= V^{\perp \perp}$; and you may also like this: $V$ is a closed additive subgroup of $H$, arc-wise connected by $\alpha$-Hölder paths, for some $\alpha > 1/2$.

In the case of $L^2(\mathbb{R})$, a question richer in consequences woud be: characterize the closed linear subspaces that are stable for additional operations, like translations, dilatations, positive part. A good reference for these questions is Rudin's Real and Complex analysis.

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I wouldn't say there is "nothing special" about $ L^2(R)$, which is a Hilbert space. Abstractly the structure of its closed subspaces (as orthocomplemented lattice, say) is the same as for any other separable Hilbert space of infinite dimension: there are a large number of them, and this structure is well-known in the subject of quantum logic. There is actually no great mystery about finding closed subspaces, since you can take any subset S and look at the vectors orthogonal to it.

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