I am trying to solve this:
If $D$ is a divisor over a algebraic curve X of genus $g$ such that $\deg(D)=2g-2$ and $\dim L(D)=g$, then $D$ is a canonical divisor
Using Riemann-Roch theorem, being $K$ an canonical divisor over $X$, I have:
$$ \begin{array}{l} \dim L(D)-\dim L(K-D)=\deg(D)+1-g\\ \dim(K-D)=1 \end{array} $$
By Serre's duality, $$ \dim L(K-D)=\dim L^{(1)}(-D)=1 $$
This means that there exists only one 1-form $\omega \in L^{(1)}(-D)$ such that $ord_p(\omega)\geq D(p)$, for all $p\in X$.
I don't know how to finish this. Can someone help me? Thanks in advance.
You have successfully shown that $h^1(O_X(D))=1,$ i.e., that $h^0(\mathcal O_X(K-D)) = 1.$ Now try Riemann-Roch again: $$h^0(\mathcal O_X(K-D))-h^1(\mathcal O_X(K-D))=\deg(K-D)+1-g,$$ and using that $h^1(\mathcal O_X(K-D))=h^0(\mathcal O_X(D))$ we get that $\deg(K-D)=0.$
Now apply Hartshorne Lemma IV.1.2, which gives $K-D\sim 0,$ i.e., $K\sim D.$