Why this dimension is $0$ using Riemann-Roch theorem?

Question

If $C$ is a curve of genus $g$, I'm trying to prove the dimension of the divisor $(2g-1)P$ associated to this curve is $g$.

I'm using the Riemann-Roch theorem which says:

Let $W$ be a canonical divisor on $C$. Then for any divisor $D$: $$l(D)=deg(D)+1-g+l(W-D)$$

Using this theorem I get $l((2g-1)P)=g+l(W-(2g-1)P)$. So why $l(W-(2g-1)P)=0$?

Remark: I'm studying Fulton's Algebraic Curves book.

Thanks in advance

Answer 1

Remember that the canonical divisor has degree $2g - 2$, and that if $\deg D < 0$ then $l(D) = 0$. The reason is that it's impossible to find a non-zero rational function $f$ satisfying $D + \operatorname{div}(f) \geq 0$ since $\operatorname{div}(f)$ always has degree $0$.

[Looking at the book, this is given as Proposition 8.2.3 (2).]