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My understanding is that in finite dimensions, every linear space $V$ is isomorphic to its dual $V^\ast$. In infinite dimensions, we have that any Hilbert space $\mathcal{H}$ is isomorphic (specifically, anti-isomorphic) to its dual $\mathcal{H}^\ast$ (Riesz Representation Theorem). Furthermore, every Hilbert space is also isomorphic to the square summable sequence space $\ell^2$.

I am wondering if there are examples of infinite dimensional linear spaces where the dual is equal to itself, and the space is not isomorphic to $\ell^2$.

Edit: We assume the underlying field to be $\mathbb{R}$ or $\mathbb{C}$.

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    $\begingroup$ A simple way to construct such examples is to consider a space $X$ which is isometrically isomorphic to its second dual $X^{**}$ (for example, reflexive spaces) but it is not a Hilbert space. Then $Y=X\oplus X^*$ is isometrically isomorphic to its dual $Y^*$ and it fails to be a Hilbert space. Additionally, the James space $J$ is an example of a nonreflexive space isometrically isomorphic to its second dual, so the space $Y=J\oplus J^*$ is self dual but nonreflexive. $\endgroup$ – tree detective Feb 28 '18 at 17:55
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There's some confusion here:

  • In the context of Hilbert spaces, $\mathcal{H}^*$ is not the full (algebraic) dual of $\mathcal H$. It's the topological dual, that is, the space of all continuous linear forms.
  • It is not true that every infini-dimensional Hilbert space is isomorphic to the space $\ell^2$ of square summable sequences. Only those which are separable.

If we are talking only about the algebraic dual, then no infinite-dimention vector space $V$ is isomorphic to $V^*$.

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  • $\begingroup$ Thanks for highlighting this! So the "topological dual" is the space of all continuous linear functionals that map $\mathcal{H} \to \mathbb{F}$, while the "algebraic dual" is a superset: just the space of linear functionals $\mathcal{H} \to \mathbb{F}$? $\endgroup$ – Anton Xue Feb 28 '18 at 3:10
  • $\begingroup$ @AntonXue That's exactly right! $\endgroup$ – José Carlos Santos Feb 28 '18 at 3:18
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If $V$ is a normal infinite dimensional vector space, no additional structures then $$\dim V^*=2^{\dim V}$$ so they can never be isomorphic.

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  • $\begingroup$ From context I think it means the continuous dual (which of course means one needs to specify a topological vector space, not just a vector space.) $\endgroup$ – Ian Feb 28 '18 at 2:59
  • $\begingroup$ @Ian yeah I dont know what exactly he means, $\endgroup$ – Rene Schipperus Feb 28 '18 at 3:02
  • $\begingroup$ Thanks for the response! This is going a bit over my head, but what if we assume the underlying field $\mathbb{F}$ is either $\mathbb{R}$ or $\mathbb{C}$? $\endgroup$ – Anton Xue Feb 28 '18 at 3:02
  • $\begingroup$ @AntonXue This answer holds for that case, if $*$ means the algebraic dual. $\endgroup$ – Ian Feb 28 '18 at 3:02
  • $\begingroup$ Its not the field that is in question. In my answer I take $V^*$ to be all linear functions $f:V\to k$, with no other restrictions. $\endgroup$ – Rene Schipperus Feb 28 '18 at 3:03

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