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That is, what are some good examples of vector spaces which are inner product spaces but in which not every Cauchy sequence converges?

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up vote 10 down vote accepted

Any non-closed subspace of a Hilbert space will do. For example, the linear span of a Hilbert basis of an infinite-dimensional Hilbert space.

For an explicit example, take $H = \ell^2(\mathbb{R})$, the vector space of square summable sequences of real numbers with pointwise operations and inner product $$\Bigl\langle \{a_n\},\{b_n\}\Bigr\rangle = \sum_{k=1}^{\infty} a_kb_k.$$ Let $\mathbf{e}_i$ be the sequence that has a $1$ in the $i$th coordinate and $0$s elsewhere. Then $\{\mathbf{e}_i\mid i\in\mathbb{N}\}$ is a Hilbert basis for $H$, and its linear span (the vector space of all almost null sequences) is an inner product space that is not complete.

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Lots of natural subspaces of Hilbert spaces aren't Hilbert spaces (which is the same thing as saying they're not closed). Two examples that come to mind are the subspace of sequences in $\ell^2(\mathbb{Z})$ at most finitely many terms of which are nonzero (the compactly supported sequences) and the continuous functions $C([0, 1])$ under the $L^2$-norm. (Note that you don't have to quotient out by anything here because it's already true that a continuous function with zero $L^2$-norm is identically zero.)

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Can you give an example on a nonconvergent Cacuhy sequence of continuous functions on $C([0,1])$? – Gadi A May 18 '11 at 7:39
@Gadi A: The sequence of continuous functions given by $u_n (x) = max( (2 x)^n, 1)$ is a Cauchy sequence in $L^2$, but converges in $L^2$ norm to the non-continuous function that is 0 on (0, 1/2), and 1 on (1/2, 1). – Michael Ulm May 18 '11 at 7:51
Or use f(x)=$x^n$, which converges to f=0 for x in [0,1), and f(1)=1 – gary May 19 '11 at 6:37
@gary: no. In $L^2$ it just converges to the (equivalence class of the) zero function. – Qiaochu Yuan May 19 '11 at 6:47
I did not read that the OP was using the $L^2$ norm, my bad. – gary May 19 '11 at 13:21

A slightly different example than already mentioned would be a (sub-) vector space over the rationals. You can take any real or complex Hilbert space, for example, and consider the linear subspace over the rational numbers.

Edit: Sorry for the confusion: It would seem that "inner product space" is usually defined to be over the real or complex numbers, although the definition also works if one uses the rational numbers instead (or any other normed field). So, striktly speaking, the restriction of the ground field to the rational numbers is not an example of an inner product space that is incomplete.

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This won't be a (real or complex) inner product space. I've never heard of anyone considering inner product spaces over $\mathbb{Q}$. – Qiaochu Yuan May 18 '11 at 8:19
Tim: strictly speaking, there is no meaning to "the linear subspace over Q" of a Hilbert space since there is no one "linear subspace over Q". The construction depends on a choice of basis. – KCd May 18 '11 at 8:51
Qiaochu: Perhaps the term "inner product space" is usually applied over R or C, but vector spaces over other fields with nondegenerate bilinear forms on them (perhaps pos. definite) definitely get used when you study quadratic forms over different fields. – KCd May 18 '11 at 8:52
@Qiaochu: Also for me, when I hear "incomplete" my mind turns to the rationals. It seems $\mathbb{Q}^n$ satisfies the axioms of an inner product space with the inner product inherited from $\mathbb{R}^n$, though Wikipedia demands that the field be $\mathbb{R}$ or $\mathbb{C}$. – Ross Millikan May 18 '11 at 13:44
@KCd: sure, but in my experience the particular term "inner product space" refers to one over $\mathbb{R}$ or $\mathbb{C}$ (maybe over $\mathbb{H}$). – Qiaochu Yuan May 18 '11 at 13:46

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