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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...?

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Something here sounds fishy. If $X$ is an incomplete inner product space and $H$ is its completion then an orthonormal basis for $H$ which consists of elements of $X$ is in particular an orthonormal basis for $X$, but some incomplete inner product spaces (which are necessarily not separable) do not have an orthonormal basis. –  Mark Aug 26 '10 at 9:53
    
Ah, well that would give a counter-example for sure! Do you have a reference? –  Matthew Daws Aug 26 '10 at 9:57
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Ah, Google comes to the rescue: secure.wikimedia.org/wikipedia/en/wiki/… –  Matthew Daws Aug 26 '10 at 10:03
    
Mark: if you write that up into an answer, I'll accept it (as it was news to me that non-separable (incomplete) inner-product spaces might fail to have an o.n. basis. –  Matthew Daws Aug 26 '10 at 10:07
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Sorry, last commment. If you access, a better reference is jstor.org/stable/2318908 –  Matthew Daws Aug 26 '10 at 10:12

4 Answers 4

up vote 2 down vote accepted

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.]

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As Mark hasn't typed his comment into an answer, I'm accepting this. Thanks all. –  Matthew Daws Aug 28 '10 at 12:24

Sci.math, March 8, 2000 LINK

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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.

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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.

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