How to prove $\mathcal{l}(D+P) \leq \mathcal l{(D)} + 1$ Let $X$ be an irreducible curve, and define $\mathcal{L}(D)$ as usual for $D \in \mathrm{Div}(X)$. Define $l(D) = \mathrm{dim} \ \mathcal{L}(D)$. I'd like to show that for any divisor $D$ and point $P$, $\mathcal{l}(D+P)  \leq \mathcal l{(D)} + 1$.
Say $D = \sum n_i P_i$. I can prove this provided $P$ is not any of the $P_i$, by considering the map $\lambda : \mathcal{L}(D) \to k$, $f \mapsto f(P)$. This map has kernel $\mathcal{L}(D-P)$, and rank-nullity gives the result. 
But if $P$ is one of the $P_i$, say $P=P_j$ then I'm struggling. Any help would be appreciated.
Thanks

Here is an elementary formulation, without sheaves.    
Let $t\in Rat(X)$ be a uniformizing parameter at $P$ (that is, $t$ vanishes with order $1$ at $P$) and let $n_P\in \mathbb Z$ be the coefficient of $D=\sum n_QQ$ at $P$.
You then have en evaluation map $$\lambda: \mathcal L(D+P)\to k:f\mapsto (t^{n_P +1}\cdot f)(P)$$  and you can conclude with the reank-nullity theorem or in more sophisticated terminology with the exact sequence of $k$-vector spaces
$$  0\to   \mathcal L(D)\to \mathcal L(D+P)\stackrel {\lambda}{\to} k                           $$
