Definition 1: For every divisor $D=\sum_{P\in C}n_PP$ over a curve $C$, we define the vectorial space:
$L(D)\doteqdot\{f\in k(C);\text{ord}_P(f)\ge -n_P,\forall P\in C\}$
Furthermore, $L(D)$ is a vectorial space over $k$ and we denote $l(D)\doteqdot\dim L(D)$.
Following these definitions why $L(0)=k$?
Remark1: Fulton proves this using a corollary and a proposition, I would like to know if there would be a more direct (maybe easier?) proof of this fact, i.e., using only these definitions.
Thanks in advance
Using your definition, $L(0)$ is the set of elements $f\in k(C)$ such that $ord_P(f)\ge 0$ for each point $P$. Of course every constant function is like this. Then, you need to see that only constant functions are possible. Your element $f\in k(C)$ corresponds to a regular morphism $f\colon C\to \mathbb{A}^1$, so you need to show that only constant morphisms are possible.
This is true for any irreducible projective algebraic variety $C\subset \mathbb{P}^n$. Indeed, the image by a morphism of a projective variety is closed (see Do there exist regular functions on the projective space over a field? ), and so is here a constant as the image is in $\mathbb{A}^1$.