If the degree of a divisor on a Riemann surface is $\deg(D) \geq 2g$, then $L(D-(p)) \subsetneq L(D)$ for any point $p$

Question

Let $S$ be a compact connected Riemann surface, $D$ a divisor on $S$, and $p\in S$ a point. I want to show that if $\deg(D) \geq 2g$ (where $g$ is the genus of $S$), then we have a strict inclusion between Riemann-Roch spaces:

$$L(D-(p)) \subsetneq L(D)$$

My thoughts so far: I have to show the existence of a function $f\in L(D) \setminus L(D-(p))$, i.e. such that

$$(f) \geq -D$$ $$(f) \ngeq -D+(p)$$

In particular, at the point $p$, we would have

$$(f)_p \geq -D_p$$ $$(f)_p \ngeq -D_p+1$$

where $D_p:=D(p)$ for a divisor $D:S\rightarrow \mathbb{Z}$. We would then have $(f)_p=-D_p$, but I’m not sure if this is helpful.

I’ve then shown that the dimension of $L(D)$ can only be at most the dimension of $L(D-[p])$ plus one: $l(D)=l(D-[p])$ or $l(D)=l(D-[p])+1$ (where $l(D):=\dim(L(D))$), but I’m not sure if this is helpful either.

I don’t know how to use the degree condition. Any help would be appreciated.

Answer 1

Your problem is equivalent to Corollary 3 on page 109 of Algebraic Curves by William Fulton

Corollary 3. If $\deg(D) ≥ 2g$, then $l(D −P) = l(D)−1$ for all $P ∈ X$.

which follows from Corollary 2 applied to $D$ and $D-P$:

Corollary 2. If $\deg(D) \geq 2 g-1$, then $l(D)=\operatorname{deg}(D)+1-g$.

which follows from Riemann-Roch Theorem and $\S8.2$ Proposition 3(2): $L(D) = 0$ if $\deg(D) < 0$.