Characterization of continuous homomorphisms on $(\mathbb{R}_+,+)$ Consider the semigroup $(\mathbb{R}_+,+)$. Is there a characterization of all continuous homomorphisms $\gamma \colon (\mathbb{R}_+,+) \to (\overline{\mathbb{D}},\cdot)$. I thought (in an analog way to the dual group of $\mathbb{R}$) of:
$$ \gamma_s(x) = \exp(-sx) $$
where $s \in \mathbb {C}$ with $\mathrm{Re}(s)  \ge 0$. However i am stuck proving that all of the continuous homomorphisms $\gamma \colon (\mathbb{R}_+,+)  \to (\overline{\mathbb{D}},\cdot)$ are of this form.
Where $\overline{\mathbb{D}} := \{ z \in \mathbb{C} | |z| \le 1 \}$ and $\mathbb{R}_+ = [0,\infty)$
Is my statement true or is there another characterization?
 A: 
Caim: Let $s \in \Bbb{C} $ mit $\mathrm{Re}(s) \ge 0$, then $\gamma_s(x) = e^{-sx}$ defines a continous homomorphism $\gamma_s \colon (\Bbb{R}_+,+) \to (\overline{\Bbb{D}},\cdot)$ and every continous homomorphism from $(\Bbb{R}_+,+)$ onto $(\overline{\Bbb{D}},\cdot)$ is of this form.

Proof:
Let $s \in \Bbb{C} $ with $\mathrm{Re}(s) \ge 0$, it follows  $|{\gamma_s(x)}| \le 1$ for all $x \in \Bbb{R}_+$ and $ \gamma_s(x_1+x_2) = \gamma_s(x_1)\gamma_s(x_2)$ for all $x_1, x_2 \in \Bbb{R}$. Also ist Therfore $\gamma_s$ is a cont. homomorphism.
Consider $\gamma \colon (\Bbb{R}_+,+) \to (\overline{\Bbb{D}},\cdot)$ a cont. homomorphism. It holds $\gamma(0) = \gamma(0)^2$ and therfore either $\gamma(0) = 1$ or $\gamma(0) = 0$. In the second case $\gamma$ would be identically zero, so it must be $\gamma(0)=1$. Choose $\delta > 0$, such that $|{\gamma(t)-1}|<1$ for all $t \in [0, \delta]$. Choose again $\tau = -\ln( \gamma(\delta))/\delta$ with argument $\ln(\gamma(\delta))$ in the intervall $[-\frac{\pi}{2},\frac{\pi}{2}]$.
It is $\gamma(\delta) = e^{-\tau \delta}$ and $\mathrm{Re}(\tau) \ge 0$. Due to the homomorphy of $\gamma$ it follows  $\gamma(\delta/2) \in \{e^{-\tau\delta/2}, -e^{-\tau \delta/2}\}$. Further due to $\delta/2 \in [0,\delta] $ it holds that $|{\gamma(\delta/2)-1}|<1 $ therfore one has  $\gamma(\delta/2) = e^{-\tau\delta/2}$ gelten. In an analog way it follows that $\gamma(\delta/2^k) = e^{-\tau\delta/2^k}$ for all $k \in \Bbb{N}$. We conclude $\gamma(\delta m/2^k) = e^{-\tau\delta m/2^k}$ for all $k,m \in \Bbb{N}$. For every positive real number $t$ exists a binary representation, especially ther is a sequence of the form $t_n = \sum_{k = 0}^{n} \frac{m_k}{2^k}$ with $t_n \to t $ for $n \to \infty$. We conclude
    \begin{align*}
  \gamma(\delta t) &= \gamma(\lim\limits_{n \to \infty} \delta t_n) = \lim\limits_{n \to \infty} \gamma(\delta t_n) = \lim\limits_{n \to \infty} \gamma(\delta \sum_{k = 0}^{n} \frac{m_k}{2^k} )= \lim\limits_{n \to \infty} \prod_{k=1}^{n} \gamma(\delta m_k/2^k) \\
  &= \lim\limits_{n \to \infty} \prod_{k=1}^{n} e^{-\tau\delta m_k/2^k} = \lim\limits_{n \to \infty} e^{-\tau \delta t_n} = e^{-\tau \delta  \lim\limits_{n \to \infty} t_n} = e^{-\tau \delta t}
 \end{align*}
    and finally  $\gamma(t) = e^{-\tau t}$ for all $t \in \Bbb{R}_+$.
