# Is a certain map $\mathbb{R}\to \widehat{\mathbb{R}}$ a homeomorphism?

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!

• Define $\hat{\Bbb R}$ May 16 at 16:38
• @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. May 16 at 16:59
• I don't study the subject, is why I ask :) By "compact convergence" do you mean the compact-open topology? May 16 at 17:07
• @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. May 16 at 17:23
• You should edit your question to include the information you gave in the comments. May 17 at 7:21

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.