I've got a question about something. We consider $\mathbb{T}=\{ z \in \mathbb{C} \; | \; |z|=1 \}$, and :
$car(\mathbb{R}) = \{ f : \mathbb{R} \rightarrow \mathbb{T} \; | \; f$ continuous and $\forall x,y \in \mathbb{R} \; f(x+y)=f(x)f(y) \}$
We have already show that $(1)$ $\forall \epsilon > 0 \; f(x)=g(x)$ on $]-\epsilon;\epsilon[$, then $f=g$, and also : if $f : ]-1;1[ \rightarrow ]-1;1[$ is continuous and $f(x+y)=f(x)+f(y)$ for $x,y \in ]-1;1[$ such as $x+y \in ]-1;1[$, then $f(x)=cx$ for $x \in ]-1;1[$, where $c = 4f(\frac{1}{4})$.
Now, I want to show that : $\phi : \mathbb{R} \rightarrow car(\mathbb{R})$ defined by $u \rightarrow \hat{u}$ where $\hat{u}(x) = e^{iux}$ is a bijection.
This is what I did :
Let's $f \in car(\mathbb{R})$ which verify the condition wanted. Then, $f(x) = e^{iu(x)}$ for a certain function $u$. Now, I consider $arg : \mathbb{R} \rightarrow \mathbb{R}$ defined by :
$arg(x) = u(x) \; $(mod)$\; (2\pi)$ and $arg(x) \in [0;2\pi[$ for all $x$. Then, from $f(x+y)=f(x)f(y)$, we can show that : $arg(x+y)=arg(x)+arg(y)$, and $arg$ is also continuous. Then, I wanted to consider the restriction of $arg$ on $]-1;1[$, and by the result $(1)$, we have for all $x \in ]-1;1[$ : $arg(x) = cx$, $c = 4arg(\frac{1}{4})$. Then, $u = c$ and $\hat{u} = f$ on $]-1;1[$, and then, again by $(1)$, $\hat{u}=f$. The only problem is that the hypothesis of $(1)$ implies that we should have $arg(]-1;1[) \subseteq ]-1;1[$, which is not necessarily true.
But in my proof of $(1)$, I don't use anywhere that $f(]-1;1[) \subseteq ]-1;1[$, so I don't know if it's my proof which is wrong or if it's not a condition which is needed.
Someone to help ? :)