**Proof.**
We prove this by induction on $n$. If $n = 0$, then $\mathbf{P}^ n_ k = \mathop{\mathrm{Spec}}(k)$ and any coherent module is $0$-regular and any surjective map is surjective on global sections. Assume $n > 0$. Consider an exact sequence as in the lemma. Let $P' \in \mathbf{Q}[t]$ be the polynomial $P'(t) = P(t) - P(t - 1)$. Let $m'$ be the integer which works for $n - 1$, $r$, and $P'$. By Lemmas 33.35.8 and 33.33.4 we may replace $k$ by a field extension, hence we may assume $k$ is infinite. Apply Lemma 33.35.3 to the coherent sheaf $\mathcal{F}$. The Hilbert polynomial of $\mathcal{F}' = i^*\mathcal{F}$ is $P'$ (see proof of Lemma 33.35.14). Since $i^*$ is right exact we see that $\mathcal{F}'$ is a quotient of $\mathcal{O}_ H^{\oplus r} = i^*\mathcal{O}^{\oplus r}$. Thus the induction hypothesis applies to $\mathcal{F}'$ on $H \cong \mathbf{P}^{n - 1}_ k$ (Lemma 33.35.2). Note that the map $\mathcal{K}(-1) \to \mathcal{K}$ is injective as $\mathcal{K} \subset \mathcal{O}^{\oplus r}$ and has cokernel $i_*\mathcal{H}$ where $\mathcal{H} = i^*\mathcal{K}$. By the snake lemma (Homology, Lemma 12.5.17) we obtain a commutative diagram with exact columns and rows

\[ \xymatrix{ & 0 \ar[d] & 0 \ar[d] & 0 \ar[d] \\ 0 \ar[r] & \mathcal{K}(-1) \ar[r] \ar[d] & \mathcal{O}^{\oplus r}(-1) \ar[r] \ar[d] & \mathcal{F}(-1) \ar[d] \ar[r] & 0 \\ 0 \ar[r] & \mathcal{K} \ar[r] \ar[d] & \mathcal{O}^{\oplus r} \ar[r] \ar[d] & \mathcal{F} \ar[d] \ar[r] & 0\\ 0 \ar[r] & i_*\mathcal{H} \ar[r] \ar[d] & i_*\mathcal{O}_ H^{\oplus r} \ar[r] \ar[d] & i_*\mathcal{F}' \ar[r] \ar[d] & 0 \\ & 0 & 0 & 0 } \]

Thus the induction hypothesis applies to the exact sequence $0 \to \mathcal{H} \to \mathcal{O}_ H^{\oplus r} \to \mathcal{F}' \to 0$ on $H \cong \mathbf{P}^{n - 1}_ k$ (Lemma 33.35.2) and $\mathcal{H}$ is $m'$-regular. Recall that this implies that $\mathcal{H}$ is $d$-regular for all $d \geq m'$ (Lemma 33.35.10).

Let $i \geq 2$ and $d \geq m'$. It follows from the long exact cohomology sequence associated to the left column of the diagram above and the vanishing of $H^{i - 1}(H, \mathcal{H}(d))$ that the map

\[ H^ i(\mathbf{P}^ n_ k, \mathcal{K}(d - 1)) \longrightarrow H^ i(\mathbf{P}^ n_ k, \mathcal{K}(d)) \]

is injective. As these groups are zero for $d \gg 0$ (Cohomology of Schemes, Lemma 30.14.1) we conclude $H^ i(\mathbf{P}^ n_ k, \mathcal{K}(d))$ are zero for all $d \geq m'$ and $i \geq 2$.

We still have to control $H^1$. First we observe that all the maps

\[ H^1(\mathbf{P}^ n_ k, \mathcal{K}(m' - 1)) \to H^1(\mathbf{P}^ n_ k, \mathcal{K}(m')) \to H^1(\mathbf{P}^ n_ k, \mathcal{K}(m' + 1)) \to \ldots \]

are surjective by the vanishing of $H^1(H, \mathcal{H}(d))$ for $d \geq m'$. Suppose $d > m'$ is such that

\[ H^1(\mathbf{P}^ n_ k, \mathcal{K}(d - 1)) \longrightarrow H^1(\mathbf{P}^ n_ k, \mathcal{K}(d)) \]

is injective. Then $H^0(\mathbf{P}^ n_ k, \mathcal{K}(d)) \to H^0(H, \mathcal{H}(d))$ is surjective. Consider the commutative diagram

\[ \xymatrix{ H^0(\mathbf{P}^ n_ k, \mathcal{K}(d)) \otimes _ k H^0(\mathbf{P}^ n_ k, \mathcal{O}(1)) \ar[r] \ar[d] & H^0(\mathbf{P}^ n_ k, \mathcal{K}(d + 1)) \ar[d] \\ H^0(H, \mathcal{H}(d)) \otimes _ k H^0(H, \mathcal{O}_ H(1)) \ar[r] & H^0(H, \mathcal{H}(d + 1)) } \]

By Lemma 33.35.11 we see that the bottom horizontal arrow is surjective. Hence the right vertical arrow is surjective. We conclude that

\[ H^1(\mathbf{P}^ n_ k, \mathcal{K}(d)) \longrightarrow H^1(\mathbf{P}^ n_ k, \mathcal{K}(d + 1)) \]

is injective. By induction we see that

\[ H^1(\mathbf{P}^ n_ k, \mathcal{K}(d - 1)) \to H^1(\mathbf{P}^ n_ k, \mathcal{K}(d)) \to H^1(\mathbf{P}^ n_ k, \mathcal{K}(d + 1)) \to \ldots \]

are all injective and we conclude that $H^1(\mathbf{P}^ n_ k, \mathcal{K}(d - 1)) = 0$ because of the eventual vanishing of these groups. Thus the dimensions of the groups $H^1(\mathbf{P}^ n_ k, \mathcal{K}(d))$ for $d \geq m'$ are strictly decreasing until they become zero. It follows that the regularity of $\mathcal{K}$ is bounded by $m' + \dim _ k H^1(\mathbf{P}^ n_ k, \mathcal{K}(m'))$. On the other hand, by the vanishing of the higher cohomology groups we have

\[ \dim _ k H^1(\mathbf{P}^ n_ k, \mathcal{K}(m')) = - \chi (\mathbf{P}^ n_ k, \mathcal{K}(m')) + \dim _ k H^0(\mathbf{P}^ n_ k, \mathcal{K}(m')) \]

Note that the $H^0$ has dimension bounded by the dimension of $H^0(\mathbf{P}^ n_ k, \mathcal{O}^{\oplus r}(m'))$ which is at most $r{n + m' \choose n}$ if $m' > 0$ and zero if not. Finally, the term $\chi (\mathbf{P}^ n_ k, \mathcal{K}(m'))$ is equal to $r{n + m' \choose n} - P(m')$. This gives a bound of the desired type finishing the proof of the lemma.
$\square$

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