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The definition of a Hilbert space is a complete inner product space. I am struggling to imagine what an incomplete inner product space would look like, especially since inner products by definition must be maps on vector spaces whose scalar fields are either $\mathbb R$ or $\mathbb C$. Since $\mathbb R$ and $\mathbb C$ are complete and inner products are (sesqui)linear, it seems that any inner product must be complete. Is the definition of Hilbert space redundant?

In short, what's an example of an inner product space that is not complete?

It seems one of the following must be the case:

  1. All inner product spaces are complete, and the definition of Hilbert space is redundant.

  2. Inner products can be defined on more general fields than $\mathbb R$ and $\mathbb C$, for example $\mathbb Q$ and other subfields of $\mathbb R$. The definition of Hilbert space simply rules out these incomplete subfields from consideration.

  3. There exist genuine inner product spaces over $\mathbb R$ and/or $\mathbb C$ that are not complete, which would be the most interesting and unexpected situation.

WillG
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2 Answers2

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Take the space of countable sequences $(x_1,x_2,\dots)$ that are non-zero for only finitely many entries.

You can see that square-summable sequences would be their limit points.

Roy D.
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  • Thanks, this example is simple and shows that my "3" is true. Although perhaps others are true, as the other answer indicates. – WillG Jan 10 '20 at 15:51
  • Yep! And I'd also sort of add a bit of intuition on this. The notions that define an inner product space really only deal with finite relations, e.g. the space has to be closed under finite sums.

    When looking for examples of inner product spaces that are not Hilbert spaces, you'll have to look for situations where the infinite limits break something that was closed under these finite arrangements, if that makes sense. This example does that.

     – Roy D. Jan 10 '20 at 15:53
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Well, 2's true for starters. An inner product space can be defined on many fields, including $\Bbb Q$, by thinking of the field as a $1$-dimensional inner product space over itself, viz. $\langle u|v\rangle:=u^\ast v$ (and for $\Bbb Q$ and $\Bbb R$, we don't even need the $^\ast$): we just need such multiplication to satisfy $v^\ast v\ge0$. Well, $\Bbb Q$'s not a Hilbert space because it's an incomplete metric space. And $\Bbb R$ and the $p$-adics are different ways to metric-complete it, giving us Hilbert spaces.

J.G.
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