1
$\begingroup$

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!

$\endgroup$
5
  • 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

0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.