= Solution
For a <divisor> $D=\sum_p n_pp$ on the <smooth projective curve>,
$$
L(D)=\{f\in k(X)^\times:\operatorname{div}(f)+D\geq0\}\cup\{0\}.
$$
This is a $k$-vector space of <rational functions> with the prescribed pole bounds. A nonzero rational function on a projective curve has divisor of degree zero. Thus if $\deg D<0$ there can be no nonzero element of $L(D)$.
For any point $p$ with coefficient $n_p$ in $D$, use a <local parameter> $t$ at $p$. The linear map sending $f\in L(D)$ to the residue-field value of $t^{n_p}f$ has kernel $L(D-p)$ and target $k$, since $k$ is algebraically closed. Therefore $\dim L(D)/L(D-p)\leq1$. If $d=\deg D\geq0$, successively subtract a point $d+1$ times. The final divisor has degree $-1$ and no nonzero sections, giving
$$
\boxed{\dim L(D)\leq d+1\quad(d\geq0),\qquad\dim L(D)=0\quad(d<0).}
$$
Equivalently the universally valid bound is $\dim L(D)\leq\max(0,\deg D+1)$. The PDF states the unqualified bound $\dim L(D)\leq\deg D+1$; literally this is false when $\deg D\leq-2$, because its right side is negative. For example $D=-2p$ has $L(D)=0$.
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