Proof of Lemma 5

It is easy to check, by substitution, that conditions (2.10)-(2.12) are satisfied. It is easy to see $0<p<1$, since $m_3^G<2m_2^G-1$. Also, $m_2^X\geq\frac{N+2}{N+1}$ implies $w>0$. Thus, it suffices to prove that condition (2.9) is satisfied.

We first consider the first inequality of condition (2.9). The assumption on $r^G$ in the lemma gives

$$\frac{3}{2}-r^G \leq \frac{3}{2} - \frac{(N+1)m_2^G+(N+4)}{2(N+2)} = \frac{N+1}{N+2}\cdot\frac{2-m_2^G}{2}.$$

Therefore, since $\frac{3}{2}>r^G$, it follows that

$$m_2^X = 2w = \frac{2-m_2^G}{2}\cdot\frac{1}{\frac{3}{2}-r^G} \geq \frac{N+2}{N+1}.$$

We next consider the second inequality of condition (2.9). We begin by bounding the range of $m_2^G$ for $G$ considered in the lemma. Condition $G\in \widehat{\cal L}_N$ implies $m_2^G \geq \frac{N+2}{N+1}$. Also, if $m_2^G>2$, then by the assumption on $r^G$ in lemma,

$$r^G \geq \frac{(N+1)m_2^G+(N+4)}{2(N+2)} > \frac{2(N+1)+(N+4)}{2(N+2)} = \frac{3}{2}.$$

This contradicts $r^G<\frac{3}{2}$. Thus, $m_2^G\leq 2$. So far, we derived the range of $m_2^G$ as $\frac{N+2}{N+1} < m_2^G \leq 2$.

We prove $m_2^X<\frac{N+1}{N}$ in two cases: (i) $\frac{N+1}{N}\leq m_2^G\leq 2$ and (ii) $\frac{N+2}{N+1}\leq m_2^G<\frac{N+1}{N}$. (i) When $\frac{N+1}{N}\leq m_2^G\leq 2$,

$$m_2^X = \frac{2-m_2^G}{2\left(\frac{3}{2}-r^G\right)}. < \f... ...}}{2\left(\frac{3}{2}-\frac{N+2}{N+1}\right)} = \frac{N+1}{N}.$$

The inequality follows from $m_2^G<\frac{N+1}{N}$ and $r^G \leq \frac{N+2}{N+1}$. (ii) When $\frac{N+2}{N+1}\leq m_2^G<\frac{N+1}{N}$,

$$m_2^X = \frac{2-m_2^G}{2\left(\frac{3}{2}-r^G\right)}. < \f... ...{3}{2}-\frac{2m_2^G-1}{m_2^G}\right)} = m_2^G < \frac{N+1}{N}.$$

The inequality follows from

$$r^G=\frac{m_3^G}{m_2^G}<\frac{2m_2^G-1}{m_2^G},$$

which follows from $G\in \widehat{\cal L}_N$.     width 1ex height 1ex depth 0pt