[proofplan] Part (1) is a direct degree argument: a nonzero $f \in \mathcal{L}(D)$ would force $0 = \deg \operatorname{div}(f) \geq -\deg D$ via [Principal Divisors Have Degree Zero](/theorems/2177), contradicting $\deg D < 0$. Parts (2) and (3) reduce to each other: part (3) is the core local statement, and part (2) follows from part (3) by induction on $\deg D$, peeling off one point at a time until the divisor becomes negative and Part (1) terminates the induction. The proof of part (3) constructs an explicit $k$-linear evaluation map $\alpha: \mathcal{L}(D) \to k$ whose kernel is exactly $\mathcal{L}(D - [p])$, so the rank-nullity theorem gives the bound. [/proofplan] [step:Prove that $\mathcal{L}(D) = 0$ when $\deg D < 0$] Suppose for contradiction that $\mathcal{L}(D) \neq 0$, and pick $f \in \mathcal{L}(D)$ with $f \neq 0$. By the definition of $\mathcal{L}(D)$, the divisor $\operatorname{div}(f) + D$ is effective: \begin{align*} \operatorname{div}(f) + D \geq 0. \end{align*} Taking degrees and using additivity of the degree homomorphism $\deg: \operatorname{Div}(C) \to \mathbb{Z}$: \begin{align*} \deg \operatorname{div}(f) + \deg D \geq 0. \end{align*} Since $f \in K(C)^\times$, the divisor $\operatorname{div}(f)$ is principal. By [Principal Divisors Have Degree Zero](/theorems/2177) (which applies because $C$ is a smooth projective irreducible curve over an algebraically closed field and $f$ is a nonzero rational function), $\deg \operatorname{div}(f) = 0$. Hence $\deg D \geq 0$, contradicting the hypothesis $\deg D < 0$. Therefore $\mathcal{L}(D) = 0$, proving part (1). [/step] [step:Construct the evaluation map $\alpha: \mathcal{L}(D) \to k$ encoding the leading coefficient at $p$] We prove part (3). Fix a closed point $p \in C$ and let \begin{align*} n := \operatorname{ord}_p(D) \in \mathbb{Z} \end{align*} denote the coefficient of $[p]$ in $D$. Choose a uniformiser $t_p \in \mathfrak{m}_p \subset \mathcal{O}_{C, p}$ at $p$ — that is, a generator of the maximal ideal $\mathfrak{m}_p$ of the discrete valuation ring $\mathcal{O}_{C, p}$, characterised by $\operatorname{ord}_p(t_p) = 1$. Such $t_p$ exists by [Element Outside $\mathfrak{m}_p^2$ Generates $\mathfrak{m}_p$](/theorems/2169) applied to any element of $\mathfrak{m}_p \setminus \mathfrak{m}_p^2$. Define the candidate evaluation map \begin{align*} \alpha: \mathcal{L}(D) &\to k \\ f &\mapsto \begin{cases} (t_p^n f)(p) & \text{if } f \neq 0, \\ 0 & \text{if } f = 0, \end{cases} \end{align*} where $(t_p^n f)(p)$ denotes the residue of $t_p^n f \in \mathcal{O}_{C, p}$ in the residue field $\mathcal{O}_{C, p}/\mathfrak{m}_p \cong k$ (the isomorphism is the composition of the natural quotient with the identification of the residue field with $k$, valid because $C$ is a curve over the algebraically closed field $k$ and $p$ is a closed point). We verify that $\alpha$ is well-defined, i.e. that $t_p^n f \in \mathcal{O}_{C, p}$ for every $f \in \mathcal{L}(D)$. For $f = 0$ this is trivial. For $f \neq 0$, the condition $\operatorname{div}(f) + D \geq 0$ implies in particular at $p$: \begin{align*} \operatorname{ord}_p(f) + \operatorname{ord}_p(D) \geq 0, \end{align*} i.e. $\operatorname{ord}_p(f) \geq -n$. By additivity of the [Order Homomorphism](/theorems/2170): \begin{align*} \operatorname{ord}_p(t_p^n f) = n \cdot \operatorname{ord}_p(t_p) + \operatorname{ord}_p(f) = n + \operatorname{ord}_p(f) \geq 0, \end{align*} so $t_p^n f \in \mathcal{O}_{C, p}$ as required, and $(t_p^n f)(p) \in k$ is well-defined. [guided] The map $\alpha$ extracts the "leading coefficient" of $f$ at $p$ relative to the pole-order budget $n = \operatorname{ord}_p(D)$ that $D$ allots there. Why does this make sense? A function $f \in \mathcal{L}(D)$ is allowed to have a pole at $p$ of order at most $n$ (or, if $n < 0$, must vanish at $p$ to order at least $-n$). Multiplying by $t_p^n$ shifts the order by $+n$, converting any "exactly order $-n$" pole into a "exactly order $0$" — i.e., a regular function with nonzero value at $p$. Functions with milder pole behaviour at $p$ (smaller $|\operatorname{ord}_p(f)|$ in the right direction) become regular functions vanishing at $p$. \textbf{Why does $t_p^n f \in \mathcal{O}_{C, p}$?} The condition $\operatorname{div}(f) + D \geq 0$ unpacks point-by-point: at $p$, $\operatorname{ord}_p(f) + n \geq 0$, i.e. $\operatorname{ord}_p(f) \geq -n$. By the [Order Homomorphism](/theorems/2170), $\operatorname{ord}_p$ is a homomorphism $K(C)^\times \to \mathbb{Z}$, so $\operatorname{ord}_p(t_p^n f) = n + \operatorname{ord}_p(f) \geq 0$. The non-negative-order condition on $\mathcal{O}_{C, p}$ (which holds because $\mathcal{O}_{C, p}$ is the discrete valuation ring of $\operatorname{ord}_p$) means $t_p^n f$ is regular at $p$. \textbf{Why is the residue field $k$?} Because $C$ is a curve over the algebraically closed field $k$ and $p$ is a closed point, the residue field $\mathcal{O}_{C, p}/\mathfrak{m}_p$ is a finite extension of $k$; since $k$ is algebraically closed, this extension equals $k$. So $(t_p^n f)(p) := (t_p^n f) + \mathfrak{m}_p \in \mathcal{O}_{C, p}/\mathfrak{m}_p = k$ is genuinely an element of $k$. \textbf{$k$-linearity.} For $f, g \in \mathcal{L}(D)$ and $c \in k$, the elements $t_p^n f, t_p^n g \in \mathcal{O}_{C, p}$ are regular at $p$, and evaluation $\mathcal{O}_{C, p} \to k$ at $p$ is a ring homomorphism (by definition the quotient by $\mathfrak{m}_p$); in particular it is $k$-linear. So $\alpha(cf + g) = (t_p^n (cf + g))(p) = c (t_p^n f)(p) + (t_p^n g)(p) = c\alpha(f) + \alpha(g)$. [/guided] [/step] [step:Identify the kernel of $\alpha$ as $\mathcal{L}(D - [p])$] We compute $\ker(\alpha) = \mathcal{L}(D - [p])$. For $f \in \mathcal{L}(D)$ nonzero, $\alpha(f) = 0$ if and only if $(t_p^n f)(p) = 0$, i.e. $t_p^n f \in \mathfrak{m}_p$, i.e. $\operatorname{ord}_p(t_p^n f) \geq 1$. By the [Order Homomorphism](/theorems/2170), this is equivalent to \begin{align*} n + \operatorname{ord}_p(f) \geq 1 \quad\iff\quad \operatorname{ord}_p(f) \geq -n + 1 = -(n - 1). \end{align*} The divisor $D - [p]$ has coefficient $n - 1$ at $p$ and the same coefficient as $D$ at every other point. Hence the condition $\operatorname{div}(f) + (D - [p]) \geq 0$ is equivalent to: \begin{align*} \operatorname{ord}_q(f) + \operatorname{ord}_q(D) \geq 0 \quad \text{for all } q \neq p, \qquad \operatorname{ord}_p(f) + (n - 1) \geq 0. \end{align*} The first family of conditions (for $q \neq p$) is automatic from $f \in \mathcal{L}(D)$; the condition at $p$ is exactly $\operatorname{ord}_p(f) \geq -(n - 1)$. So \begin{align*} \ker(\alpha) = \{ f \in \mathcal{L}(D) : \operatorname{ord}_p(f) \geq -(n-1) \} = \mathcal{L}(D - [p]). \end{align*} (The inclusion $\mathcal{L}(D - [p]) \subseteq \mathcal{L}(D)$ holds because lowering the coefficient at $p$ from $n$ to $n - 1$ tightens the constraint, and $f$ already satisfies all other constraints.) The case $f = 0 \in \ker(\alpha) \cap \mathcal{L}(D - [p])$ is consistent. [/step] [step:Apply rank-nullity to conclude part (3)] Apply the rank-nullity theorem to the $k$-linear map $\alpha: \mathcal{L}(D) \to k$: \begin{align*} \dim_k \mathcal{L}(D) = \dim_k \ker(\alpha) + \dim_k \operatorname{Range}(\alpha) \leq \dim_k \mathcal{L}(D - [p]) + \dim_k k = \dim_k \mathcal{L}(D - [p]) + 1. \end{align*} The bound $\dim_k \operatorname{Range}(\alpha) \leq \dim_k k = 1$ is because $\operatorname{Range}(\alpha) \subseteq k$. This proves part (3). [guided] The rank-nullity theorem states: for a linear map $T: V \to W$ between vector spaces with $V$ finite-dimensional, $\dim V = \dim \ker T + \dim \operatorname{Range} T$. We need $\mathcal{L}(D)$ to be finite-dimensional for this to apply directly — but the bound we are proving will itself imply finite-dimensionality, by induction on $\deg D$ in part (2). To avoid circularity here, we argue dimension-freely: if $\mathcal{L}(D)$ is infinite-dimensional, the bound $\dim \mathcal{L}(D) \leq 1 + \dim \mathcal{L}(D - [p])$ would force $\mathcal{L}(D - [p])$ to also be infinite-dimensional, and the conclusion $\dim \mathcal{L}(D) \leq \dim \mathcal{L}(D - [p]) + 1$ holds vacuously in $\mathbb{Z} \cup \{\infty\}$ as the natural extension. (Finite-dimensionality of $\mathcal{L}(D)$ for all $D$ is in fact a separate theorem, but is logically posterior to the present bound.) The cleaner statement is: for any subspace $W \leq V$ and any $k$-linear map $\alpha: V \to k$ whose kernel is $W$, we have either (i) $\alpha = 0$ and $V = W$, or (ii) $\alpha$ is surjective with kernel $W$, in which case $V/W \cong k$ as $k$-vector spaces, hence $\dim V = \dim W + 1$. Either way $\dim V \leq \dim W + 1$. Applied here with $V = \mathcal{L}(D)$, $W = \mathcal{L}(D - [p])$, $\alpha$ as constructed, the bound follows. [/guided] [/step] [step:Deduce part (2) by induction on $\deg D$ via iterated point removal] Assume $\deg D \geq 0$. We prove $\dim_k \mathcal{L}(D) \leq 1 + \deg D$ by induction on the integer $\deg D \geq 0$. \textbf{Base case ($\deg D = 0$).} If $\deg D = 0$, we must show $\dim_k \mathcal{L}(D) \leq 1$. Suppose $f, g \in \mathcal{L}(D)$ are both nonzero. Then $f/g \in K(C)^\times$ and $\operatorname{div}(f/g) = \operatorname{div}(f) - \operatorname{div}(g)$ by additivity of $\operatorname{ord}_q$ for each $q$. We have $\operatorname{div}(f) + D \geq 0$ and $\operatorname{div}(g) + D \geq 0$, so: \begin{align*} \operatorname{div}(f/g) + (D - D) = \operatorname{div}(f) - \operatorname{div}(g) \quad \text{and} \quad \operatorname{div}(g/f) = -\operatorname{div}(f/g). \end{align*} Now $\operatorname{div}(f/g)$ is principal, hence has degree $0$ by [Principal Divisors Have Degree Zero](/theorems/2177). Moreover, by part (1) applied to the divisor $D' := D + \operatorname{div}(g)$ (which satisfies $\operatorname{div}(g) + D \geq 0$ so $D' \geq 0$), we have $f \in \mathcal{L}(D)$ iff $f/g \in \mathcal{L}(D')$ (multiplication by $g^{-1}$ is a $k$-linear isomorphism $\mathcal{L}(D) \to \mathcal{L}(D')$). So WLOG we may assume $g$ is a constant (replace $g$ by a constant of $\mathcal{L}(D')$ — but at this stage we have not yet shown such a constant exists), and we proceed differently: pick any closed point $p \in C$. Apply part (3): $\dim_k \mathcal{L}(D) \leq 1 + \dim_k \mathcal{L}(D - [p])$. The divisor $D - [p]$ has degree $\deg D - 1 = -1 < 0$, so by part (1), $\mathcal{L}(D - [p]) = 0$. Hence $\dim_k \mathcal{L}(D) \leq 1 + 0 = 1$, as required. \textbf{Inductive step.} Assume the bound holds for all divisors of degree $\leq d - 1$, where $d \geq 1$. Let $D$ have $\deg D = d$. Pick any closed point $p \in C$ — such a $p$ exists because $C$ is a smooth projective curve over an algebraically closed field, so $C$ has infinitely many closed points (in particular, at least one). By part (3): \begin{align*} \dim_k \mathcal{L}(D) \leq 1 + \dim_k \mathcal{L}(D - [p]). \end{align*} The divisor $D - [p]$ has $\deg(D - [p]) = d - 1 \geq 0$, so the inductive hypothesis applies: \begin{align*} \dim_k \mathcal{L}(D - [p]) \leq 1 + \deg(D - [p]) = 1 + (d - 1) = d. \end{align*} Combining: \begin{align*} \dim_k \mathcal{L}(D) \leq 1 + d = 1 + \deg D. \end{align*} This completes the induction and proves part (2). [guided] The strategy is to chip away at $D$ one closed point at a time, decreasing the degree by $1$ at each step while losing at most one dimension. After $\deg D + 1$ chips, the divisor has degree $-1 < 0$, so part (1) terminates the recursion with $\mathcal{L}(\cdot) = 0$. Working backwards, $\mathcal{L}(D)$ has dimension at most $1 + \deg D$. \textbf{Why does the base case need part (3)?} In the base case $\deg D = 0$, we cannot directly conclude $\dim \mathcal{L}(D) \leq 1$ from just the definition: $\mathcal{L}(D)$ contains all rational functions with $\operatorname{div}(f) + D \geq 0$, which is a nontrivial condition. The trick is to apply part (3) once: removing any closed point $p$ takes $D$ to a divisor $D - [p]$ of degree $-1$, where part (1) gives $\mathcal{L}(D - [p]) = 0$. Then part (3) yields $\dim \mathcal{L}(D) \leq 1$. \textbf{Why is the inductive step legitimate?} The key observation is that part (3) gives a strict-monotonicity statement: the dimension drops by at most $1$ when we remove a single point from $D$. So $\dim \mathcal{L}(D) \leq \dim \mathcal{L}(D - [p]) + 1$, and iterating $\deg D$ times we reach a divisor of degree $0$, where the base case applies. The total bound is $\dim \mathcal{L}(D) \leq \deg D + \dim \mathcal{L}(D_0) \leq \deg D + 1$ where $D_0$ has degree $0$. \textbf{Why does $C$ have a closed point?} Smooth projective curves over algebraically closed fields are nonempty (otherwise they would have no points in the closed embedding into $\mathbb{P}^N_k$, hence be the empty variety, which is not a curve). The set of closed points is in bijection with the set of $k$-points by Hilbert's Nullstellensatz applied at each affine chart, and $\mathbb{P}^N_k$ has infinitely many $k$-points for $N \geq 1$. [/guided] [/step] [step:Combine parts (1)-(3) to obtain the full theorem] Parts (1), (2), and (3) together give the full statement: for any divisor $D$ on $C$, \begin{align*} \deg D < 0 &\implies \mathcal{L}(D) = 0, \\ \deg D \geq 0 &\implies \dim_k \mathcal{L}(D) \leq 1 + \deg D, \\ \text{for any } p \in C: \quad &\dim_k \mathcal{L}(D) \leq 1 + \dim_k \mathcal{L}(D - [p]). \end{align*} This completes the proof. [/step]