What I know so far is

  1. In finite dimensions the dual space is continuous, has the same dimension as the space, and the "dual basis" is in fact a basis.
  2. the Riesz representation theorem proves the existence of a conjugate-linear isometric bijection between an IPS and its continuous dual.
  3. a linear bijection between vector spaces is an isomorphism, but does this extend to a conjugate-linear bijection ? (presumably so for a real IPS, but what about a complex IPS).
  4. the "dual basis" of linear functionals is linearly independent in the dual space and (I think) continuous. so that $dim(V') \ge dim(V)$
  5. I haven't seen any statement anywhere affirming my question, so I suspect that the answer is no.

So, is an infinite dimensional IPS isomorphic to its continuous dual, and is the "dual basis" a basis for the continuous dual ? If not, is there a simple counterexample ?

Update: I see some reference suggesting that this may be true in a Hilbert space, but not in general (http://www.solitaryroad.com/c855.html)

  • $\begingroup$ What do you mean by "isomorphic"? Isomorphic as topological vector spaces? Just as vector spaces? As inner product spaces? $\endgroup$ – Eric Wofsey Mar 25 '16 at 10:12
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    $\begingroup$ @EricWofsey my understanding of the terminology in this context is that isomorphism relates to linearity and isometry relates to the metric and associated topological properties. So the essence of my question is whether the IPS and its continuous dual have the same dimension. But, I'm open to correction and clarification. $\endgroup$ – Tom Collinge Mar 25 '16 at 10:23
  • $\begingroup$ Usually when talking about normed vector spaces, "isomorphism" means a topological isomorphism, not just an algebraic isomorphism of vector spaces. There seems to be a lot of confusion about what kind of objects and maps you're talking about in your five statements. $\endgroup$ – Eric Wofsey Mar 25 '16 at 10:25
  • $\begingroup$ @EricWofsey If you could find time to add an answer that unconfuses me I'd appreciate it. $\endgroup$ – Tom Collinge Mar 25 '16 at 10:30
  • $\begingroup$ In particular: (2) is only true for complete inner product spaces. For (3), the answer is trivially no, since an isomorphism is by definition linear. For (4), you need to define what you mean by "basis": any vector space has an algebraic basis, and an infinite-dimensional inner product space may or may not have an orthonormal "basis" in an analytic sense. If you have an algebraic basis for an inner product space, the dual basis functionals need not even be continuous. It is unclear what you mean by "dimension". $\endgroup$ – Eric Wofsey Mar 25 '16 at 10:32

There are various confusions in the statements you've made that I've discussed briefly in the comments. Let me now directly answer the question in your title, for any possible meaning of "isomorphic". Consider the real inner product space $\mathbb{R}^\infty$ consisting of all sequences of real numbers which have only finitely many nonzero terms. The inner product is just the usual dot product (which is always just a finite sum). This vector space has countable (algebraic) dimension, since the set of sequences for which one entry is $1$ and all other entries are $0$ forms an algebraic basis.

The continuous dual of $\mathbb{R}^\infty$ is the usual (real) Hilbert space $\ell^2$ of all square-summable sequences. This space has uncountable algebraic dimension (see here, for instance), so it cannot even be algebraically isomorphic to $\mathbb{R}^\infty$.

What is true is that a complete inner product space (i.e., Hilbert space) is isomorphic to its continuous dual, for any possible sense of "isomorphic". For real Hilbert spaces, this follows immediately from the Riesz representation theorem, which gives a canonical isometric isomorphism between a real Hilbert space and its continuous dual.

For a complex Hilbert space, the Riesz representation theorem only gives a conjugate-linear isometry, and to get an actual linear isomorphism you have to do a bit more work (and the isomorphism will not be canonical). For instance, you can choose an orthonormal basis $B$ of your Hilbert space, and then the image of $B$ under the conjugate-linear isometry given by the Riesz representation theorem will be an orthonormal basis for the continuous dual. Since a Hilbert space is determined up to isometric isomorphism by the cardinality of an orthonormal basis, this implies your Hilbert space is isometrically isomorphic to its continuous dual.

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