I'm looking for a proof that the Pontryagin dual $G^\*$ of a topological group $G$ is a topological group. 


It's very easy to prove that $G^\*$ is a group, my troubles are in proving that the map $G^\* \times G^\* \to G^\* : (f,g) \mapsto fg^{-1}$ is continuous and so $G^\*$ is topological.


I read in "Rudin - Fourier Analysis on Groups" a proof that $G^\*$ is a Locally Compact Abelian group when $G$ is LCA, but it's too much for my purposes and the proof involves the Fourier transform and so the Haar measure, I think these tools are not necessary.


Thanks very much for any suggestions.



Proof that the Pontryagin dual of a topological group is a topological group
I don't think this is a research level question, but here is an argument.


The topology of $G^\*$ is given by uniform convergence on compact subsets of $G$. Let $K\subset G$ be compact, then we need to show that if $f\_n\to f$ and $g\_n\to g$ uniformly on $K$, then $f\_ng\_n^{-1}\to fg^{-1}$ uniformly on $K$. This is immediate from the pointwise bound
$$ |f\_ng\_n^{-1}-fg^{-1}| \leq |f\_n(g\_n^{-1}-g^{-1})|+|(f\_n-f)g^{-1}| = |g\_n-g| + |f\_n-f|. $$


