TITLE: Pontriagin reflexivity of the character group
QUESTION [6 upvotes]: 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).

REPLY [2 votes]: 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 $f$.