Approaching to PDE theory I was introduced to initial value problems for the linear one-dimensional transport equation like $$\begin{cases} u_t + cu_x = 0, &&x\in \mathbb{R},\,t>0 &&& (1) \\ \\ u(0,x) = g(x), && x\in \mathbb{R} &&& (2) \end{cases}$$ where $u=u(t,x)$ and $g:\mathbb{R} \to \mathbb{R}$ is a generic (possibly $C^1$) function.
I do not really understand what the conditions $x \in \mathbb{R}, t>0$ mean: Does the equality $u_t + cu_x = 0$ just hold in $(0,+\infty)\times \mathbb{R}$ and not in the whole $[0,+\infty)\times \mathbb{R}$?
By method of characteristics, I know that if $u=u(t,x)$ satisfies $u_t + cu_x = 0$ on $ (0,+\infty)\times \mathbb{R}$ then $u(t,x)=f(x-ct)$ on $ (0,+\infty)\times \mathbb{R}$ for an arbitrary $f:\mathbb{R} \to \mathbb{R}$. How do I have to use de IC $u(0,x)=g(x)$ for all $x \in \mathbb{R}$ to conclude that $u=g(x−ct)$ in the whole $[0,+\infty)\times \mathbb{R}$?
Many authors treat the problem like $(1)$ holds in the whole $[0,+\infty)\times \mathbb{R}$ and so they just plug in the initial condition to find that $u(t,x)=g(x-ct)$ for all $x\in \mathbb{R}$, $t\geq0$.
Any hint would be really appreciated!
As is explained by @user247327, equation (1) is only needed to hold on the interior of the area.
After using the method of characteristics to get $u(x, t) = f(x - ct)$ for some $f:\mathbb{R} → \mathbb{R}$, the next step is to prove that $f \in C(\mathbb{R})$. Since $u \in C(\mathbb{R} × [0, +∞))$, taking an arbitrary $t_0 \in [0, +∞)$, then $u(·, t_0) \in C(\mathbb{R})$. Note that $u(x, t_0) = f(x - ct_0)$, thus $f(x) = u(x + ct_0, t_0) \in C(\mathbb{R})$.
Now, because $u \in C(\mathbb{R} × [0, +∞))$ and $u|_{t = 0} = g$, then$$ g(x) = u(x, 0) = \lim_{t → 0^+} u(x, t) = \lim_{t → 0^+} f(x - ct) = f(x), \quad \forall x \in \mathbb{R} $$ i.e. $f = g$.