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For an Abelian topological group $G$ by $G^{\wedge}$ we denote the Pontryagin dual of $G$, i.e. the group of continuous homomorphisms $G\to\mathbb T:=\{z\in\mathbb C:|z|=1\}$. The group $G^{\wedge}$ is endowed with the topology of uniform convergence on compact subsets of $G$. A topological group $G$ is called Pontryagin reflexive if the canonical homomorphism $G\to (G^\wedge)^\wedge$ is a topological isomorphism.

Problem. Let $G$ be a metrizable Abelian topological group. Is the Pontriagin dual $G^{\wedge}$ of $G$ Pontryagin reflexive?

(This problem was posed 21.09.2017 by Lydia Aussenhofer on page 71 of Volume 1 of the Lviv Scottish Book).

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  • $\begingroup$ Is it clear that $G^{\wedge}$ is metrizable? $\endgroup$
    – YCor
    Commented Sep 25, 2017 at 22:40
  • $\begingroup$ @YCor, maybe I miss the point of your question, but $\hat G$ doesn't have to be metrisable to ask whether it is reflexive, right? $\endgroup$
    – LSpice
    Commented Sep 25, 2017 at 22:53
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    $\begingroup$ If G is metrizable then $G^\wedge$ is a hemicompact $k$--space. This implies that the canonical mapping $G^\wedge\to G^{\wedge\wedge\wedge}$ is continuous (even an embedding). Only the surjectivity of this mapping is a problem. $\endgroup$
    – LDA
    Commented Sep 26, 2017 at 9:06
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    $\begingroup$ I hope you know that in the case when $G$ is a (metrizable) locally convex space the answer is "yes", see e.g. link.springer.com/article/10.1023%2FA%3A1020929201133 or ams.org/mathscinet-getitem?mr=2037513 $\endgroup$
    – Sergei Akbarov
    Commented Sep 26, 2017 at 10:08
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    $\begingroup$ In my previous comment I forgot to indicate that $G$ must be complete, I am sorry. $\endgroup$
    – Sergei Akbarov
    Commented Sep 26, 2017 at 18:51

1 Answer 1

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Not an answer to this problem.
For a non-reflexive $G$ see example: Exercise (23.32) in Hewitt & Ross, Abstract Harmonic Analysis I (Springer 1963).

Consider a topological vector space $L^p(\mathbb R)$, $0 < p < 1$. They show that the (separable metrizable) topological group $G = L^p(\mathbb R)$ with operation of addition has no nonzero continuous characters. So $\widehat{G} = \{0\}$ and $\widehat{\widehat{G}} = \{0\}$, not isomorphic to $G$.

Facts used (Hewitt & Ross include outlines of proofs):

(Day, 1940) this topological vector space has no nonzero continuous linear functionals

(Hewitt & Zuckerman, 1950) every continuous character $\chi$ on $G$ has the form $\chi(x) = \exp(2\pi i f(x))$ for some continuous linear functional $f$.

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    $\begingroup$ This says that this $G$ isn't reflexive, but it doesn't seem to provide an example of a non-reflexive Pontryagin dual. I apologize if I'm missing something... $\endgroup$
    – Alon Amit
    Commented Sep 26, 2017 at 0:17
  • $\begingroup$ You are right. Not an answer to the original problem. $\endgroup$
    – Gerald Edgar
    Commented Sep 26, 2017 at 0:19

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