Suppose $X$ is a proper normal noehterian scheme, I want to show dim of $L=L(D)=\{f|\operatorname{div}(f)+D>0\}$ is finite. I am not sure if my proof is right?
Write $D=\sum n_iP^i-m_jQ^j$ where $m,n$ s are positive integers and $P,Q$ s are prime divisors.
We choose an element $g$ in $L$ which has the smallest order at $P_i$, each element of $L$ can be written as $\operatorname{div}(g)+\operatorname{div}(h)$ where $h$ has smaller pole at $P_i$, after at most $\sum n_i$ steps we arrive at a global regular function which is constant, thus $l(D)\leq \sum n_i+1$.
I don't know if I missed something above? because Mumford wrote very long to prove the finite dimensionality of $l(D)$, (in his Algebraic Geometry I Complex Projective Varieties P102-103).
As Matt's answer below, there's something wrong. I am thinking where I had made the mistake. Any comments are welcome.
Let $X$ be $\mathbb P^2$, and let $D$ be a smooth irreducible curve of degree $d$. Then there is just one $P^i$, no $Q^j$'s, and $n_i = 1$. Your argument would then suggest that $l(D) \leq 2$. But in fact $l(D) = (d+2)(d+1)/2.$
So your argument is wrong. Maybe if you think about how your argument goes for these concrete examples of plane curves, you can find the mistake and come to a better understanding of the situation.