TITLE: Dual Group of $\Bbb{R}$
QUESTION [5 upvotes]: The Dual Group of $\mathbb{R}$ is isomorphic to $\Bbb{R}$ itself in the following way:
The map
$$\Bbb{R} \to \hat{\Bbb{R}}, \quad y \mapsto \exp(ixy) $$
is an isomorphism. Further it is stated in the literature that this map is also an homeomorphism. See for exmaple Conway, A course in functional analysis Theorem 9.11. I am trying to prove this result. While proving the continuity is straightforward, i am stuck proving that the inverse is continous. 
One should note that $ \Bbb{R}$ is equipped with the standard topology and $\hat{\Bbb{R}}$ with the topology induced by compact convergence. That means a function converges iff it converges unfirom on every compact subset. 
I tried some ways, none of wich was promising. I am grateful for any hint on how to do this. 
Edit 2: 
I have found a proof based on non trivial results as seen below. However i am still interested if this can be proven more direct without the help of said results. 
Edit 1:
The proof that the map is bijective is often done in the following way: 
One shows that $\gamma \in \hat{\Bbb{R}}$ fullfills this differential equation:
$$ \left\{ \begin{array}{cl}
 \gamma'(x) &= \gamma'(0) \gamma(x) \\
 \gamma(0) &= 1\\
 \end{array}\right. $$
with the condition $| \gamma(x)| = 1 $ for all $x \in \Bbb{R}$.
Then it is clear that $\gamma(x)  = \exp(x\gamma'(0))$ and we now that $y = -  i\gamma'(0)$ in $\gamma(x) = \exp(ixy)$. If one could show that the map 
$\hat{\mathbb{R}}  \to \Bbb{R}, \; \gamma \mapsto \gamma'(0) $ is continous this would prove the continouity of the inverse. 
Comments and ideas considering this approach are greatly appreciated.

REPLY [2 votes]: This comes a little late, but here is a direct proof that the inverse is continuous:
Let's write the continuous group isomorphism as $$\Phi: \mathbb{R} \to \widehat{\mathbb{R}}, \ y \mapsto \chi_y \ \text{ with } \ \chi_y(x) := e^{ixy}.$$
Now let $\psi \in \widehat{\mathbb{R}}$ be a character of $\mathbb{R}$ and let $y_0 \in \mathbb{R}$ with $\psi = \Phi(y_0)$. We need to show that $\Phi^{-1}$ is continuous at $\psi$, i.e. that for every neighborhood $N \subseteq \mathbb{R}$ of $y_0 = \Phi^{-1}(\psi)$ the set $$\Phi(N) = (\Phi^{-1})^{-1}(N)$$
is a neighborhood of $\psi = \Phi(y_0)$. It suffices to show this for sufficiently small neighborhoods $N$ of $y_0$, e.g. intervals $(y_0 - \varepsilon, y_0 + \varepsilon)$ with $\varepsilon\in (0, \frac{\pi}{2})$, since every neighborhood of $y_0$ contains such a neighborhood.
The map $$f: \left[- \frac{\pi}{2}, \frac{\pi}{2} \right] \to \mathbb{C}, \ x \mapsto e^{ix}$$
is a continuous injection and thus a continuous bijection onto $L := f([- \frac{\pi}{2}, \frac{\pi}{2}])$.
Since $L \subseteq \mathbb{C}$ is Hausdorff and $[- \frac{\pi}{2}, \frac{\pi}{2}]$ is compact, this map also is closed and hence a homeomorphism. The continuity of the inverse implies that we find a $\delta \in (0,1)$ such that $$\forall x \in \left[-\frac{\pi}{2}, \frac{\pi}{2} \right]: |e^{ix}-1| < \delta \implies |x-0| < \varepsilon.$$
Furthermore we set $K := [0,1]$. Now let $$\chi \in N(\psi; K, \delta) := \{\gamma \in \widehat{\mathbb{R}}: \max_{x \in K} |\gamma(x) - \psi(x)| < \delta\}$$
lie in a neighborhood of $\psi$, set $y := \Phi^{-1}(\chi)$ and let $x \in K$. Then we have $$|e^{ix(y-y_0)} - 1| = |e^{-ixy_0}| \cdot |e^{ixy} - e^{ixy_0}| = |(\Phi(y))(x) - (\Phi(y_0))(x)| = |\chi(x) - \psi(x)| < \delta < 1,$$
so $\cos(x(y-y_0)) = \operatorname{Re}(e^{ix(y-y_0)}) > 0$. Hence there is a $k_x \in \mathbb{Z}$ such that $$x(y - y_0) - 2\pi k_x \in \left[- \frac{\pi}{2}, \frac{\pi}{2} \right], \ \text{ i.e. } \frac{x(y-y_0)}{2 \pi} - k_x \in \left[- \frac{1}{4}, \frac{1}{4}\right]  \subseteq \left(- \frac{1}{2}, \frac{1}{2} \right)$$
and the triangle inequality shows that $k_x$ is the only integer with this property: If we have $k, l \in \mathbb{Z}$ and $a \in \mathbb{R}$ with $|a- k| < \frac{1}{2}$ and $|a-l| < \frac{1}{2}$, then $$|k-l| = |k-a+a-l| \leq |a-k| + |a- l| < 1$$
and therefore $k = l$. We obtain the map $$k: K = [0, 1] \to \mathbb{Z}, \ x \mapsto k_x.$$
Again let $x \in K$ and choose $r >0$ such that $r \cdot |y-y_0| < \frac{\pi}{2}$. For any $z \in K$ with $|z-x| < r$ we then have
\begin{align*}
\left| \frac{z(y-y_0)}{2 \pi} - k_x \right| &= \left|\frac{z(y-y_0)}{2 \pi} - \frac{x(y-y_0)}{2 \pi} +\frac{x(y-y_0)}{2 \pi} - k_x \right| \\
&\leq \frac{|z-x|}{2 \pi}|y-y_0| + \left| \frac{x(y-y_0)}{2 \pi} - k_x \right| \\
&\leq \frac{1}{2\pi} r \cdot |y-y_0| + \left| \frac{x(y-y_0)}{2 \pi} - k_x \right| < \frac{1}{4} + \frac{1}{4} = \frac{1}{2}
\end{align*}
and the uniqueness shown above yields $k_x = k_z$. Therefore the map $k$ is locally constant, so the connectedness of $K = [0,1]$ implies that $k$ is constant. In particular, $$k_1 = k_0 = 0.$$
Going back to the definition of $k_x$, we see that $$y-y_0 = 1 \cdot (y-y_0) - 2\pi k_1 \in \left[- \frac{\pi}{2}, \frac{\pi}{2} \right]$$
and $|e^{i(y-y_0)} - 1| = |\chi(1) - \psi(1)| < \delta$ finally implies $|y-y_0| < \varepsilon$, i.e. $\chi \in \Phi((y_0 - \varepsilon, y_0 + \varepsilon))$. Therefore $$N(\psi; K, \delta) \subseteq \Phi((y_0 - \varepsilon, y_0 + \varepsilon)),$$
so $\Phi((y_0 - \varepsilon, y_0 + \varepsilon))$ is a neighborhood of $\psi$ and we conclude that $\Phi^{-1}$ is continuous at $\psi$.