0$, then Claim~\ref{claim} holds.
\end{lemm}
\begin{proof}
Remark that $(\alpha_i -1)^+ = \alpha_i -1$ for every $i$. We are going to show that $t_2 \leq r$. First, let's compute $t_2$:
\begin{equation*}
t_2 = \sum_{i=1}^n (\alpha_i -1)^+ + B + 1 = \sum_{i=1}^n \alpha_i - n + B + 1 = an + bm - h +m + n -nm - 1.
\end{equation*}
So we have that
\begin{align*}
r-t_2 &= \left((a+1)(b+1) - h -1\right) - \left(an + bm - h +m + n -nm - 1\right) \\
& = (m-a-1)(n-b-1).
\end{align*}
By hypothesis $a\leq m-1$ and $b\leq n-1$, hence $r-t_2 \geq 0$.
\end{proof}
From now on, we add the hypothesis $A+2-m \leq 0$. To prove the claim, we will proceed as follows: first we introduce a specific integer sequence, which we show to be the ``worst-case scenario''. Afterwards, we prove the claim for this particular sequence.
For each $p \in \{1, \ldots, n\}$ and $q \in \{A+2-m, \ldots, A\}$, we define the sequence $\beta^{(p,q)} = \big(\beta_i^{(p,q)} \big)=\big(\beta_i \big)$ as follows:
\begin{equation}
\tag{$\triangle$} \label{sequence}
\beta_1 = \dots = \beta_p = A+1, \quad \beta_{p+1} = q, \quad \beta_{p+2}= \dots =\beta_{n} = A+2-m.
\end{equation}
We want that the sequence $\beta^{(p,q)}$ satisfies the same conditions $(*)$ as the sequence $\alpha$. In particular, we need that
\[
\sum_{i=1}^n \beta_i = an + bm - h +m + 2n -nm -B -2.
\]
This equation allows us to compute $p$ and $q$. Indeed, since
\begin{align*}
\sum_{i=1}^n \beta_i &= (A+1)p + q + (A+2-m)(n-p-1) \\
&= p(m-1) + \big(q - (A+2-m)\big) + n (A+2-m),
\end{align*}
it follows that the desire equation is equivalent to
\begin{equation}
\tag{$\Diamond$} \label{pandq}
p(m-1) + \big(q - (A+2-m)\big) = an + bm - h + m -An -B -2.
\end{equation}
Since $0\leq q -(A+2-m) \leq m-2$, the parameters $p$ and $q$ are uniquely determined by Euclidean division of the right hand side of the equation~\eqref{pandq} by $m-1$. Moreover, since clearly
\[
\sum_{i=1}^n\beta_{i}^{(1,A+2-m)}\leq \sum_{i=1}^n\alpha_i \leq \sum_{i=1}^n\beta_{i}^{(n,A)},
\]
we have indeed $p\in \{1,2,\dots,n\}$.
\begin{lemm}\label{worstcase}
Assume that $A+2-m\leq 0$ and
let $\alpha = (\alpha)_{i=1, \ldots, n}$ be a sequence of integer numbers that satisfies the conditions $(*)$. If $p,q$ are the integers such that the sequence $\beta^{(p,q)}$ satisfies $\sum_{i=1}^n \beta_i = \sum_{i=1}^n \alpha_i$, then
\[
\sum_{i=1}^n \alpha_i^+ \leq \sum_{i=1}^n \beta_i^+ \quad \text{and} \quad \sum_{i=1}^n (\alpha_i-1)^+ \leq \sum_{i=1}^n (\beta_i-1)^+.
\]
\end{lemm}
\begin{proof}
If $\sum_{i=1}^n \alpha_i^+ > \sum_{i=1}^n \beta_i^+$, then $\sharp\{i\,|\,\alpha_i>0\}\geq p+1$ since $\beta_1,\ldots,\beta_p$ reach the maximal value $A+1>0$. This implies that
\[
\sum_{i=1}^n \alpha_i\geq \sum_{i=1}^n \alpha_i^+ + (n-p-1)(A+2-m)>\sum_{i=1}^n \beta_i^+ + (n-p-1)(A+2-m)\geq \sum_{i=1}^n \beta_i,
\]
a contradiction. Similarly, we can see that
\[
\sum_{i=1}^n (\alpha_i-1)^+ \leq \sum_{i=1}^n (\beta_i-1)^+.\qedhere
\]
\end{proof}
Now we are able to prove Claim~\ref{claim} which will complete the proof of Theorem~\ref{thm_Kmn}.
\begin{proof}
Because of Lemma~\ref{worstcase}, the sequence $\beta^{(p,q)}$ defined by~\eqref{sequence} maximizes $\sum_{i=1}^n \beta_i^+$ and $\sum_{i=1}^n (\beta_i -1)^+$. Therefore it is enough to check our claim for this kind of sequence. Recall that we may assume that $A+2-m \leq 0$ by Lemma~\ref{lem_easycase}. We distinguish the following four cases:
\begin{enumerate}[label=(\alph*), ref=\alph*]
\item $q>0$ and $B0$ and $B \geq p$; \label{case2}
\item $q \leq 0$ and $B

0$. If $D''+(e-1,f)$ is non-special on $C'$, then
\[
h^0 \big(\LL_{C'} (D''+(e-1,f))\big) = d'-n-f-(m-2)(n-1)+1,
\]
so
\[
h^0 \big(\LL_C (D')\big) = d'-n-(m-2)(n-1)+2 = d'-(m-1)(n-1)+1,
\]
and $D'$ is non-special on $C$.
Otherwise, by induction we have
\[
d'-n-f = \deg \big(D''+(e-1,f)\big) \geq \min_{(a,b,h) \in I_{r-f-1}} a(m-1) + bn - h.
\]
In other words, there exist constants $a,b$, and $h$ such that
\[
d' \geq a(m-1) + bn - h+n+f
\]
and
\[
(a+1)(b+1) - h = r-f.
\]
Note that $a \geq f$, hence $h+a-f \geq 0$. Let $a'=a$, $b'=b+1$, and $h'=h+a-f$. Then
\[
d' \geq a'm+b'n-h'
\]
and
\[
(a'+1)(b'+1) - h' = (a+1)(b+1)-h + (f+1) = r+1.
\]
Hence
\[
d' \geq \min_{(a,b,h) \in I_r} am + bn - h.
\]
Finally, notice that in both cases our choice for $a'$ and $b'$ do not decrease, so the inequalities $a \geq d_2, b \geq d_1$ follow.
\end{proof}
\bibliographystyle{amsplain-ac}
\bibliography{ALCO_Cools_55}
\end{document}