Vanishing of global sections with very negative twists. Let $k$ be an algebraically closed field and let $\mathcal{F}$ be a coherent sheaf on the projective space $\mathbb{P}^n_k$. I would like to know when is it true that
$$ H^0(\mathbb{P}^n_k, \mathcal{F}(m)) = 0 \qquad \text{ for } m<<0  $$
For example, if $\mathcal{F}=\mathcal{O}_p$ is the structure sheaf corresponding to a closed point $p\in \mathbb{P}^n_k$ we have that
$$ H^0(\mathbb{P}^n_k,\mathcal{O}_p(m))= H^0(p,\mathcal{O}_p)=k \qquad \text{ for all } m\in \mathbb{Z}$$
In particular, is it true that the vanishing above holds if and only if $\mathcal{F}$ has no closed associated points?

Motivation:
The motivation for my problem is the following: for every coherent sheaf $\mathcal{F}$ on $\mathbb{P}^n_k$ we can build the associated group as
$$ R(\mathcal{F}) = \bigoplus_{m\in \mathbb{Z}}H^0(\mathbb{P}_k^n, \mathcal{F}(m)) $$
and this has a natural structure of a graded $S$-module, where 
$$ S=R(\mathcal{O})=\bigoplus_{n\geq 0} H^0(\mathbb{P}_k^n,\mathcal{O}(n)) $$
is the homogeneous coordinate ring of $\mathbb{P}^n_k$.
I would like to know when is $R(\mathcal{F})$ finitely generated over $S$. I know that every truncated module 
$$ \bigoplus_{m\geq m_0}H^0(\mathbb{P}^n_k,\mathcal{F}(m)) $$
is finitely generated, so that the problem is equivalent to the vanishing in the question above. In particular, in Eisenbud's book "The Geometry of Syzygies" it says (p. 67) that the problem in finite generation comes about when $\mathcal{F}$ has an associated closed point (so that the answer to the question above should be "yes"), but there is no proof of this.
 A: I have thought of a solution and I am posting it here in case somebody is interested.

Claim: let $\mathcal{F}$ be a coherent sheaf on $\mathbb{P}^N_k$, with $k$ algebraically closed and $N\geq 0$. Then $H^0(\mathcal{F}(n))=0$ for $n\ll 0$ if and only if $\mathcal{F}$ has no associated closed points.
Proof of Claim: $(\implies)$ First, we prove that if $\mathcal{F}$ has an associated closed point, then $H^0(\mathcal{F}(n))\ne 0$ for any $n\in \mathbb{Z}$. Since the associated points of $\mathcal{F}$ are the same as the associated points of $\mathcal{F}(n)$, we can suppose $n=0$. Let $S=S[X_0,\dots,X_N]$ be the coordinate ring of $\mathbb{P}^N$ and let $M$ be a graded $S$-module such that $\mathcal{F}=\widetilde{M}$. Then if $p\in \mathbb{P}^N$ is a closed point associated to $\mathcal{F}$ and $\mathfrak{p}\subseteq S$ is the corresponding homogeneous ideal, then $\mathfrak{p}$ is an associated prime of $M$, meaning that there is an embedding
$$ 0 \longrightarrow S/\mathfrak{p}(n) \longrightarrow M $$
for a certain $n\in \mathbb{Z}$. After sheafification we get a morphism of sheaves
$$ 0 \longrightarrow \kappa(p)(n) \longrightarrow \mathcal{F} $$
and taking global sections it follows that $H^0(\mathcal{F})\supseteq H^0(\kappa(p)(n)) \ne 0$.

$(\impliedby)$ For the converse, suppose that we have a coherent sheaf $\mathcal{F}$ on $\mathbb{P}^N$ such that $H^0(\mathcal{F}(n))\ne 0$ for infinitely many $n<0$, then we want to prove that $\mathcal{F}$ has an associated closed point. Let $M=\sum_{n\in \mathbb{Z}} H^0(\mathcal{F}(n))$: then this is a graded $S$-module such that $M_n$ is of finite dinmension over $k$ for every $n$, and such that, if $m\in M_n$ and $\ell\cdot m =0$ for each $\ell \in S_1$, then $m=0$. Indeed, that is equivalent to saying that the global section $m\in H^0(\mathcal{F}(n))$ vanishes on $\{ \ell \ne 0 \}$ for every $\ell\in S_1$, and since these open subsets cover $\mathbb{P}^n$, it must be that $m=0$. Moreover, the associated primes of $M$ give associated points of $\mathcal{F}$, so that we are done if we can prove the following. 
Lemma: Let $S=k[X_0,\dots,X_N]$ and let $M=\sum _{n\in \mathbb{Z}}M_n$ be a graded $S$-module such that:


*

*every $M_n$ is of finite dimension as a vector space over $k$. 

*$M$ is saturated, i.e. if $m\in M_n$ is such that $\ell\cdot m =0$ for every $\ell \in S_1$, then $m=0$. 


Then, if $M_n\ne 0$ for infinitely many $n<0$, the module $M$ has the ideal of a closed point in $\mathbb{P}^N=Proj(S)$ as an associated prime.
Proof of Lemma:  First we observe that $M_n\ne 0$ for every $n\in \mathbb{Z}$. Indeed, it is enough to prove that if $M_n\ne 0$ then $M_{n+1}\ne 0$ as well. So, let $m\in M_n$ be a nonzero element: then by saturatedness there exists an element $\ell\in S_1$ such that $\ell\cdot m \ne 0$ and then this is a nonzero element in $M_{n+1}$. Now we prove the lemma by induction on $N$:
Basic step: we start with $N=0$: in this case, we know that since $M_n\ne 0$ for every $n$, the sheaf $\widetilde{M}$ on $\mathbb{P}^0$ is nonzero, so that it has an associated point, that must be the unique point of $\mathbb{P}^0$, that in particular is closed.
Induction step: let $N\geq 1$. Now, suppose that we can find an element $\ell \in S_1$ such that the submodule
$$ (0:_M \ell) = \{m\in M \,  |\, \ell\cdot m = 0\} $$
is nonzero in infinitely many negative degrees. Then we can regard this as a module over $S/\ell S$ and by induction hypothesis it follows that it must contain a prime associated to a closed point in $\mathbb{P}^{N-1}=Proj(S/\ell S)$. Such a prime is of the form $\mathfrak{p}/\ell S$ for $\mathfrak{p}\subseteq S$ and we have inclusions of $S$-modules
$$ S/\mathfrak{p}(n) = \frac{S/\ell S}{\mathfrak{p}/\ell S}(n) \subseteq (0:_M \ell) \subseteq M  $$
and this means that $\mathfrak{p}$ is associated to $M$ and is the prime that we were looking for. 
Now, if we cannot find any such $\ell$ this means that for every $\ell \in S_1$  the maps $M_n \overset{\cdot\ell}{\to} M_{n+1}$ are injective for all $n\ll 0$. In particular, it follows that $\dim_k M_n \leq \dim_k M_{n+1}$ for all $n\ll 0$ and since these dimensions are all positive and finite, they must stabilize at a certain point, meaning that $\dim_k M_n = \dim_k M_{n+1}$ for $n\ll 0$. In particular, we can find an $n_0 \in \mathbb{Z}$ such that all the maps $M_n \overset{\cdot X_0}{\to} M_{n+1}$ are isomorphisms if $n\leq n_0$. Suppose now that for every $\ell\in S_1$ the maps $M_n \overset{\cdot\ell}{\to} M_{n+1}$ are isomorphisms for $n\leq n_0$ and, for any such $n$ consider the multiplication map
$$ S_1 \otimes_k M_n \to M_{n+1} $$
this map is injective on both factor separatedly, so that, since $k$ is algebraically closed, by a theorem of Hopf (cfr ACGH-Moduli of Alg Curves I, p. 108) it follows that
$$ \dim_k M_{n+1} \geq \dim_k S_1 + \dim_k M_n -1 = \dim_k M_n + N > \dim_k M_n $$
but this is absurd because we know that both dimensions are the same. Hence, there must be an $\ell \in S_1$ such that  $\ell \cdot m = 0$ for a certain nonzero element $m\in M_{\overline{n}}$, with $\overline{n}\leq n_0$. But then we can prove that $(0:_M \ell)_n\ne 0$ for every $n\leq \overline{n}$, that would give a contradiction. To see this, recall that all the maps $M_{n}\overset{\cdot X_0}{\to} M_{n+1}$ are isomorphisms for $n\leq \overline{n}$ so that, for every $m\in M_n$ we can denote by $\frac{m}{X_0^r}$ the unique element in $M_{n-r}$ such that $X_0^r \cdot \frac{m}{X_0^r}=m$. Now observe that $X_0\cdot(\ell\cdot \frac{m}{X_0})=\ell \cdot m =0$ but since multiplication by $X-0$ is injective, it follows that $\ell \cdot \frac{m}{X_0}=0$ and then $(0:_{M}\ell)_{\overline{n}-1}\ne 0$. To show that $(0:_{M}\ell)_{\overline{n}-2}\ne 0$ we observe that $X_0\cdot(\ell\cdot \frac{m}{X_0^2})=\ell \cdot \frac{m}{X_0}=0$ and since multiplication by $X_0$ is injective it must be that $\frac{m}{X_0^2}=0$. Continuing like this one can prove that $(0:_M \ell)_n \ne 0$ for all $n\leq \overline{n}$ and this concludes the proof.
