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I wanted to ask if anyone knew of a simple way to prove this.

The proof I know: Identify any such character with a continuous, periodic homomorphism $f: \mathbb R \rightarrow S^1 \subset \mathbb C^{\ast}$. Then $f$ has a Fourier expansion $$f(x) = \sum\limits_n c_n e^{2\pi i nx}$$

with

$$c_n = \int\limits_{\mathbb R/\mathbb Z} f(x)e^{-2\pi i nx}dx = \begin{cases} 1 & \textrm{if $f(x) = e^{2\pi i nx}$} \\ 0 &\textrm{else} \end{cases}$$

This shows that if $f$ is not any of the $e^{2\pi i nx}$, it is zero. $\blacksquare$

This proof could be considered circular, because the Fourier expansion of $f$ is essentially using the Peter-Weyl theorem, which says that the characters of a compact topological group form an orthonormal basis for the $L^2$ space. So we are essentially already assuming that $f$ is one of the $e^{2\pi inx}$.

How can this be proved without using Fourier series/circular reasoning?

D_S
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    Have a look here: https://math.stackexchange.com/questions/123588/varphi-in-operatornamehoms1-s1-are-of-the-form-zn – Con Jun 30 '19 at 14:17
  • Any homomorphism $\Bbb{R/Z} \to \Bbb{C}^$ is an homomorphism $\Bbb{R} \to \Bbb{C}^$. You are supposed to know they are of the form $f(x) = e^{ax}$. And $f(0) = f(1)$ implies $a = 2i \pi n$. – reuns Jun 30 '19 at 19:13

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