Let $X$ be a closed subscheme of $P=\mathbb{P}^N_k$ of dimension $n$. Theorem III.7.6(b) of Hartshorne states that:
Suppose that for any $\mathcal{F}$ locally free on $X$, we have $H^i(X,\mathcal{F}(-q))=0$ for $i<n$ and $q>>0$. Then $X$ is Cohen-Macaulay and equidimensional.
I have an issue with the beginning of the proof, where he deduces from a previous argument that $\mathcal{Ext}^i(\mathcal{O}_X,\omega_P)=0$ for $i>N_n$. Specifically, this previous argument is the following. From the duality on $P$ (Theorem 7.1), we know that $Ext_P^i(\mathcal{F},\omega_P(q))=H^{N-i}(X,\mathcal{F}(-q))=0$. On the other hand, by Proposition 6.9, we have $Ext_P^i(\mathcal{F},\omega_P(q))=\Gamma(P,\mathcal{Ext}_P^i(\mathcal{F},\omega_P(q)))$ for $q>>0$.
My question is then: how does Hartshorne go from $\Gamma(P,\mathcal{Ext}_P^i(\mathcal{F},\omega_P(q)))=0$ to $\mathcal{Ext}_P^i(\mathcal{F},\omega_P)=0$ ?
My line of reasoning is that $\mathcal{Ext}_P^i(\mathcal{F},\omega_P)$ is coherent by Proposition 6.5, so we can apply Theorem II.5.17 of Serre to get $\mathcal{Ext}_P^i(\mathcal{F},\omega_P(q))=\mathcal{Ext}_P^i(\mathcal{F},\omega_P)\otimes\mathcal{O}(q)=0$ since it is generated by global section, from which we get the result.
However it seems a bit much to stuff into the "previous argument", and I feel I may have missed something simpler. Does anyone have any insight?