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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?

3 Answers 3

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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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    It's a Hilbert basis, so the completion is the entire space $\ell^2(\mathbb{R})$. Linear spans just don't do it in Hilbert spaces. You take the closure of the linear span.2017-08-11
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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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    I did not read that the OP was using the $L^2$ norm, my bad.2011-05-19
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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, strictly 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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    Can't we just use any $L^p$ space for p not 2? All these spaces have an inner-product defined on them, but only in $L^2$ the norm is generated by the inner-product.2011-05-19