For $M$ compact Riemann surface, divisors $D_{1}$ and $D_{2}$ on $M$, $D_{2}\geq 0$ show that: $$\dim L(D_{1}+D_{2})\leq \dim L(D_{1})+\deg D_{2}.$$
One method is using Riemann-Roch theorem, but I do not want to use it. And my idea is that using the inductive method by $D_{2}$. It is obvious as $D_{2}=0$, but how to do next step?
Here is a inequality: if $D \geq 0$, then $$ \dim L(D)\leq \deg (D)+1$$
P.S. $L(D):=\{f: (f)+D \geq 0\}, \deg D=\sum_{p\in M} D(p),$ which $(D)=\sum_{p\in M} D(p)p$.
Thanks for any advanced suggestions.
Using induction on $\deg D$, this reduces to the case where $\deg D=1$, that is $D=P$, a single point. Then $L(D)\subseteq L(D+P)$, and I claim that $\dim(L(D+P)/L(D))\le1$, which will suffice. Suppose $P$ occurs with coefficient $m$ in $D$, and let $w$ be a local uniformiser at $P$; that is $w$ a function defined in a neighbourhood of $P$ with a simple zero there. Each element of $L(D+P)$ has a Laurent expansion $$f=a_{-m-1}w^{-m-1}+a_{-m}w^{-m}+\cdots$$ at $P$. Then $f\in L(D)$ iff $a_{-m-1}=0$ so $L(D) $ has codimension at most $1$ in $L(D+P)$.