Let $(G,+)$ be an Hausdorff topological abelian group. Assume that $G$ is complete, i.e. every Cauchy sequence in $G$ converges in $G$.
Let $\widehat G:=\operatorname{Hom}_{\text{cont}}(G, S^{1})$ be the dual group of $G$ and put on $\widehat G$ the compact-open topology.
Here my question:
Is also $\widehat{G}$ a complete group?
Thanks in advance
I don't know the answer in general, but I suspect the following answer is good enough for your purposes (since completeness just with respect to sequences is not very useful anyways in full generality). Namely, if $G$ is any compactly generated topological abelian group (not necessarily complete), then $\widehat{G}$ is complete. Most spaces you will run into are compactly generated unless you are specifically trying to find pathological examples; in particular, for instance, any metrizable space is compactly generated.
To prove $\widehat{G}$ is complete if $G$ is compactly generated, suppose $(f_n)$ is a Cauchy sequence in $\widehat{G}$. Then in particular $(f_n(x))$ is Cauchy for each $x\in G$, and so we may define a function $f$ as the pointwise limit of $f_n$. We then see that $f$ is a homomorphism since it is a pointwise limit of homomorphisms. Moreover, $(f_n)$ is uniformly Cauchy on each compact subset of $G$, so $f_n\to f$ uniformly on compact sets. Thus $f$ is continuous when restricted to each compact subset of $G$. Since $G$ is compactly generated, this means that $f$ is continuous. Thus $f\in\widehat{G}$, and $f_n\to f$ in $\widehat{G}$ since the convergence is uniform on compact sets.