I am seeking to show that if $\phi :\mathbb{R}^n\rightarrow\mathbb{C}$ is a character of the topological group $\mathbb{R}^n$ then $\phi$ must have the form $\phi(x)=e^{ix\cdot\xi}$ for some $\xi$ in $\mathbb{R}^n$.
I have had a thought to define a function $F(x)= \int_{B(0,|x|)}\phi(t)dt$ where $B(0,|x|)$ is the open ball at zero of radius $|x|$.
Then \begin{eqnarray} F(x+z) &=& \displaystyle\int_{B(0, |x+z|)} \phi(t)dt \\&=& \int_{B(0, |x|)} \phi(t)dt + \int_{B(0, |x+z|)\setminus B(0, |x|)} \phi(t)dt\\&=&F(x) +\int_{B(0, |z|)}\phi(t+x)dt \\&=&F(x) +\int_{B(0, |z|)} \phi(x)\phi(t)dt \;\;\;(\phi \text{ is a homomorphism}) \\&=&F(x) + \phi(x)F(z) \end{eqnarray}
Since $\phi$ is a homomorphism, it must take the identity to the identity. so, $\phi(0) = 1$. Furthermore, since $\phi$ is continuous, there must be some $z$ such that $F(z) \not =0$.
From the above chain of equalities, we get the following:
$\phi(x) =\dfrac{F(x+z)-F(x)}{F(z)}$, implying that $\phi$ is differentiable.
Therefore for any $x,y \in \mathbb{R}^n$, $\dfrac{\partial\phi}{\partial x_i}(x+y) = \phi(x)\dfrac{\partial\phi}{\partial x_i}(y).$
Set $y=0$ and let $k$ be equal to $\dfrac{\partial\phi}{\partial x_i}(0)$. So, we then have $\dfrac{\partial\phi}{\partial x_i}(x) = k\phi(x)$
We then have a differential equation. I want to use integrating factors of some sort, but my knowledge of PDEs is not the best.
Does the following hold?
$\dfrac{\partial}{\partial x_i}(\phi(x) e^{-kx}) = (\dfrac{\partial\phi}{\partial x_i}(x)-k\phi(x))e^{-kx} = 0$ so that $\phi(x) e^{-kx}$ is constant. We can set $x=0$ to see that $\phi(x) e^{-kx} = 1$. Since we know that $|\phi(x)| = 1$, this then implies that $k = i \xi$ for some $\xi \in \mathbb{R}^n$.
Does this show that every character of $\mathbb{R}^n$ as a topological group must be of form $\phi(x)=e^{ix\cdot\xi}$?