Hartshorne Exercise IV.1.5.
For an effective divisor $D$ on a curve $X$ of genus $g$, show that $\dim |D|\le\deg D$. Furthermore, equality holds if and only if $D=0$ or $g=0$.
We have $h^0(X,\OO_X(D))-1=h^1(X,\OO_X(D))+\deg D-g$, so it suffices to show that $h^1(X,\OO_X(D))\le g$. By Serre Duality,
$$h^1(X,\OO_X(D))=h^0(X,\OO_X(K_X-D)),$$
and since
$$H^0(X,\OO_X(K_X-D))\ra H^0(X,\OO_X(K_X))$$
is an injection, it follows that
$$h^0(X,\OO_X(K_X-D))\le h^0(X,\OO_X(K_X))=g,$$
as desired.
It is clear that if $D=0$ then $\dim |D|=\deg D=0$ (by Serre duality) and if $g=0$ then $X\cong \PP{k}{1}$ so $h^1$ of any effective divisor is 0 (again using SD), hence $\dim |D|=\deg D$.
Now assume $\dim |D|=\deg D$.
Clearly $\deg D=\dim |D|\iff h^0(X,\OO_X(K_X-D))=g\iff H^0(X,\OO_X(K_X-D))\cong H^0(X,\OO_X(K_X))$ by RR and SD. Note that $D\sim 0$ automatically implies $D=0$ since $D$ is effective. Indeed by RR and SD,
$$ g=h^0(X,\OO_X(K_X-D))=h^0(X,\OO_X(D))+g-1=g+\deg D.$$
Therefore, $\deg D=0$, and it follows that $D\sim 0$ because $D\ge 0$.
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