Orthonormal basis for non-separable inner-product space

Question

Suppose X is an inner product space, with Hilbert space completion H (actually, I'm interested in the real scalar case, but I doubt there's any difference). If H is separable, then so is X, and I can find a (countable or finite) orthonormal basis of H inside X. Indeed, start with some countable subset Y of X which is dense in H. Then, by induction, we can move to a linearly independent subset of Y, and then apply Gram–Schmidt, again by induction. The point (to me, anyway) is that at any stage, we never take limits, and so we never leave X.

Now, what happens if H is not assumed separable? I've tried to use a Zorn's Lemma argument, but I keep end up wanting to take limits (or, rather, infinite sums) which gives me an orthonormal basis (in the generalised, non-countable, sense) in H, but I cannot ensure that it's in X. Am I just missing something obvious, or is there a slight technicality here...?

Answer 1

This is Problem 54 in Halmos' "A Hilbert Space Problem Book". However, I think this is a concrete counterexample. [Please let me know if not viewable.]

Answer 2

On the arXiv this (2010-09-09) morning:

1009.1441 [ps, pdf, other]

Title: Inner product space with no ortho-normal basis without choice.
Authors: Saharon Shelah
Primary Subject: math.LO

We prove in ZF that there is an inner product space, in fact, nicely definable with no orthonormal basis.

Answer 3

Sci.math, March 8, 2000 LINK

Answer 4

I don't think we can do that unless you can make sense out of uncountable sums. The Gram-Schmidt algorithm cannot transform a basis into an orthogonal one unless the original basis has no limit ordinal in its well-ordering. For example, take X as the space of square summable sequences. We can construct a Hamel basis by adding vectors to the set of standard basis vectors (1 at one position and 0 everywhere else). Obviously any non-zero vector in X cannot be orthogonal to every standard basis vector, so the Hamel basis cannot be made orthogonal. (In other words, if these standard basis vectors are considered the "first" vectors in our basis, the least upper bound of all standard basis vectors cannot be orthogonalized.) This shows that it may not be possible just to have an uncountable set of orthogonal vectors.

My argument is not comprehensive. It might be the case that some special choices of the first "countably many" vectors may lead to a valid construction of an uncountable set of orthogonal vectors.

However, there may be nice spaces that have the property you mentioned. I believe the following is an example:

Take X as a space of functions $f:R \to R$ such that $f^{-1}(0)$ is the complement of a countable set and $\sum_{f(x) \ne 0} f(x)^2$ is finite. X is pretty much like the space of square-summable sequences, but each sequence is indexed by a real number instead of a positive integer. We define standard basis vectors as functions that are 1 at only one point and 0 everywhere else. [I believe] these standard basis vectors form a complete orthogonal basis in your sense.

Answer 5

Theorem II.5 in Reed and Simon proves that any Hilbert space - separable or not - possesses an orthonormal basis. I don't see anywhere in the proof where it depends on the the space being complete, so, unless I'm missing something, it applies to any inner product space. It uses Zorn's lemma, so it's non-constructive. I think the disagreement here with other comments comes from a difference in the definition of "orthonormal basis": Reed and Simon merely define an orthonormal basis as a maximal orthonormal set. Perhaps that is insufficient to imply that the closure of the span is the whole space. For a Hilbert space it is (Theorem II.6 in Reed and Simon), but for a general inner product space, perhaps it is not.