Dual Group of $\Bbb{R}$ 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. 
 A: Using two theorem from the book "Pontryagin Duality and the Structure of Locally Compact Abelian Groups" wirtten by Sidney A. Morris the proof is as follows:
First the two theorems from the book:

Theorem 1 (Open Mapping Theorem)
Let G be a locally compact Group that is $\sigma$-compact; that is, $G = \displaystyle \bigcup_{n=1}^{\infty} A_n$ where each $A_n$ is a compact. Let $f$ be any continous homomorphism of $G$ onto a locally compact group $H$. Then $f$ is an open mapping, that means the image of every open set is open.

This theorem will help showing that the map is open and therefore, as N.H. hinted, an homeomorphism. Before we can apply this theorem we need to know if $ \hat{\mathbb{R}}$ is a locally compact group.

Theorem 2
Let G be an LCA-group (locall compact abelian) then $\hat{G}$ the character space, endowed with the compact open topology , is an LCA-Group.

$ $

Claim: 
  The map
  $$\Bbb{R} \to \hat{\Bbb{R}}, \quad y \mapsto \exp(ixy) $$
  is an isomorphism and an homeomorphism. 

Proof: 
$(\Bbb{R},+)$ endowed with the standard topology surely is an LCA-group. It follows by Theorem 2 that $\hat{\Bbb{R}}$ endowed with the compact open topology is LCA. (Here the compact open topology is equivalent to the topology induced by compact convergence.) Further it clearly is 
$\mathbb{R} = \displaystyle \bigcup_{n=1}^\infty [-n,n]$ and therefore it follows that $(\mathbb{R},+)$ is $\sigma$-compact. 
The given Map is a continous homomorphism and  by applying  Theorem 1 we follow that the given map is open. Therefore it is an homeomorphism.  
A: 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 $\psi$.
