I'm studying harmonic analysis, and I'm trying to understand the fact that $$\phi: \mathbb{R}\to \widehat{\mathbb{R}}: s \mapsto (t \mapsto \exp(2\pi i st))$$ is an isomorphism of topological groups (here we consider $\mathbb{R}$ as a topological group for the usual topology and the usual addition $+$).

So far, I have managed to show that $\phi$ is a group isomorphism (surjectivity was the harder part). It remains to show that $\phi$ is a homeomorphism, i.e. that both $\phi$ and $\phi^{-1}$ are continuous.

To show continuity of $\phi$, we need to show that if $\{s_\lambda\}$ is a net that converges to $0$ in $\mathbb{R}$, then $$\exp(2\pi i s_\lambda t) \to 1$$ uniformly in $t\in K$ where $K\subseteq \mathbb{R}$ is compact. I think I can prove this using the fact that continuous functions are uniformly continuous on compact subsets.

Conversely, I also struggle to show that if $\exp(2\pi i s_\lambda t) \to 1$ uniformly on compact subsets, then $s_\lambda \to 0$. I guess we need some form of complex logarithm for this?

Thanks in advance for any hints/suggestions!

  • 2
    $\begingroup$ Define $\hat{\Bbb R}$ $\endgroup$
    – FShrike
    May 16 at 16:38
  • $\begingroup$ @FShrike Sure, it is the set of continuous group homomorphisms $\mathbb{R}\to \mathbb{T}$ endowed with the topology of compact convergence. This is tagged with harmonic analysis though, and it is pretty standard notation for the dual group. $\endgroup$
    – Andromeda
    May 16 at 16:59
  • $\begingroup$ I don't study the subject, is why I ask :) By "compact convergence" do you mean the compact-open topology? $\endgroup$
    – FShrike
    May 16 at 17:07
  • $\begingroup$ @FShrike Uhm, a of functions $f_\lambda$ converges to $f$ in the topology of compact convergence if the convergence is uniform on every compact subset of the domain. It turns out that this topology coincides with the compact-open topology in this case. $\endgroup$
    – Andromeda
    May 16 at 17:23
  • 2
    $\begingroup$ You should edit your question to include the information you gave in the comments. $\endgroup$
    – Paul Frost
    May 17 at 7:21

1 Answer 1


Assume that $\phi(s_\lambda) \to 1$ uniformly on compact subsets. Assume to the contrary that $s_\lambda \not\to 0$. Then there is a subnet $\{s_\mu\}$ and $\epsilon > 0$ such that $|s_\mu| \ge \epsilon$ for all $\mu$. By passing to a further subnet, we may assume that $s_\mu$ has the same sign for all $\mu$. By possibly replacing $\{s_\lambda\}$ by $\{-s_\lambda\}$, we may assume that $s_\mu \ge 0$ for all $\mu$, so that $s_\mu \ge \epsilon$ for all $\mu$.

By assumption, there is an index $\mu$ such that $$\sup_{t \in [0,\epsilon^{-1}]}|\exp(2\pi is_\mu t)-1| < 1.$$ However, $s_\mu [0, \epsilon^{-1}] = [0, s_\mu \epsilon^{-1}] \supseteq [0,1]$ so that $$2=\sup_{v \in [0,1]} |\exp(2\pi iv)-1| \le \sup_{t \in [0,\epsilon^{-1}]}| \exp(2\pi i s_\mu t)-1| < 1$$ a contradiction. Hence, $s_\lambda \to 0$, and this is enough to show that the group homomorphism $\phi^{-1}$ is continuous.


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