Question about Fourier series The Fourier series of a function $f: G \to  \mathbb C$ where $G$ is a group is the representation of $f$ in terms of characters $\chi_g \in \mathrm{Hom}(G, S^1)$ of $G$. 
I understand the case where $G$ is finite and discrete. Now I'm trying to generalise to periodic functions on $\mathbb R$ but I'm not so sure what I'm thinking is correct.
If $f: \mathbb R \to \mathbb C$ is a $[-R,R]$-periodic function, we set $G = \mathbb R / 2R\mathbb Z \cong [0, 2R)$. Then since $f$ is $2R$-periodic we can shift this by $R$ consider $G = [-R, R)$, makes no difference. 
In the discrete case $|G|=n$ the characters are $e^{\frac{2 \pi i k}{n}}$ for $0 \leq k \leq n$.
Now for some reason, the characters are $e^{i kx}$ for all $k \in \mathbb Z$. How do I see that these are the characters of $[-R,R)$?
 A: What you have is equivalent to the circle group $\mathbb R/\mathbb Z$, and its characters are indeed $e^{ikx}$ for $k\in \mathbb Z$. It is easy to verify that these are continuous group homomorphisms from $\mathbb R/\mathbb Z$ to itself (which is the definition of a character). To see that these are all characters of $\mathbb R/\mathbb Z$, suppose that $f:\mathbb R/\mathbb Z\to \mathbb R/\mathbb Z$ is a continuous homomorphism. Assume first $f$ is increasing. Let $k$ be the degree of $f$ and note that we can lift $f$ to a map $g:\mathbb R\to\mathbb R$ with $g(x)=f(x)+k\lfloor x\rfloor$. The map $g$ is in fact a continuous homomorphism, as if $\{x\}+\{y\}<1$ ($\{\cdot\}$ denotes fractional part) then 
$$g(x+y)=f(x+y)+k\lfloor x+y\rfloor= f(x)+f(y) + k\lfloor x\rfloor + k\lfloor y\rfloor=g(x)+g(y)$$
and otherwise 
$$g(x+y)=f(x+y)+k\lfloor x+y\rfloor=f(x)+f(y)-\lim\limits_{x\to 1^-}g(x) + k\lfloor x\rfloor + k\lfloor y\rfloor+k=g(x)+g(y)$$
as $\lim\limits_{x\to 1^-} g(x)=k$, which also proves continuity. Restricting our attention to $\mathbb Q$ we see that $n\cdot g(m/n)=g(m)=km$ so $g(m/n)=km/n$, thus $g|_{\mathbb Q}(x)=kx$. By continuity we have that $g(x)=kx$ on $\mathbb R$, and it follows that $f(x)=kx$. Viewing the circle as the group of unit complex numbers, this is $e^{ikx}$. Similarly, if $f$ is decreasing we get $e^{-ikx}$. Thus the characters of $\mathbb R/\mathbb Z$ are $e^{ikx}$ for $k\in\mathbb Z$.
