Proof that the Pontryagin dual of a topological group is a topological group

2012-02-27T10:58:40Z

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.

Answer by GH for Proof that the Pontryagin dual of a topological group is a topological group

2012-02-27T11:42:38Z

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 $$ |f_ng_n^{-1}-fg^{-1}| \leq |f_n(g_n^{-1}-g^{-1})|+|(f_n-f)g^{-1}| = |g_n-g| + |f_n-f|. $$

Proof that the Pontryagin dual of a topological group is a topological group - MathOverflow

Proof that the Pontryagin dual of a topological group is a topological group

Answer by GH for Proof that the Pontryagin dual of a topological group is a topological group