Proof that the Pontryagin dual of a topological group is a topological group - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T20:14:15Z http://mathoverflow.net/feeds/question/89651 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/89651/proof-that-the-pontryagin-dual-of-a-topological-group-is-a-topological-group Proof that the Pontryagin dual of a topological group is a topological group michael-grade83 2012-02-27T10:58:40Z 2012-02-27T11:42:38Z <p>I'm looking for a proof that the Pontryagin dual $G^*$ of a topological group $G$ is a topological group. </p> <p>It's very easy to prove that $G^*$ is a group, my troubles are in proving that the map <code>$G^* \times G^* \to G^* : (f,g) \mapsto fg^{-1}$</code> is continuous and so <code>$G^*$</code> is topological.</p> <p>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.</p> <p>Thanks very much for any suggestions.</p> http://mathoverflow.net/questions/89651/proof-that-the-pontryagin-dual-of-a-topological-group-is-a-topological-group/89653#89653 Answer by GH for Proof that the Pontryagin dual of a topological group is a topological group GH 2012-02-27T11:42:38Z 2012-02-27T11:42:38Z <p>I don't think this is a research level question, but here is an argument.</p> <p>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 $$|f_ng_n^{-1}-fg^{-1}| \leq |f_n(g_n^{-1}-g^{-1})|+|(f_n-f)g^{-1}| = |g_n-g| + |f_n-f|.$$</p>