We first consider $\tilde{\omega}$. As $z\rightarrow\pm\infty$ the behavior of $\tilde{\omega}$ is governed by (3.349), so for $T\gg 1$ we know $\tilde{\omega}$ is positive over the region where $z\in[-k_{-}^{-1}(T-1),-k_{+}^{-1}(T-1)]\setminus[-C,C]$ for some number $C>0$ independent of $T$. By the expansion of $\psi$ in a neighborhood of $P$ given in Proposition 3.24, for $T$ sufficiently large, $\tilde{\omega}$ is also positive when $z\in[-C,C]$. Hence $\tilde{\omega}$ is positive over the region where $z\in[-k_{-}^{-1}(T-1),-k_{+}^{-1}(T-1)].$ Since this contains $Q_{T}$ we see in particular $\tilde{\omega}$ is positive over $Q_{T}\setminus P$.

To deal with $h$ we need to analyze $q(z)$. When $|z|\geq 1$, we have

(4.18)

$$q(z)=q_{0}(z)=T^{2-n}(T+k_{\pm}z)^{n-1}-(n-1)(T+k_{\pm}z),$$

where the choice of $+$ or $-$ depends on whether $z>0$ or $z<0$. By (3.349) we then get

(4.19)

$$h=T^{2-n}(T+k_{\pm}z)^{n-1}+\epsilon(z).$$

So we can find $C>0$ such that $h$ is positive when $z\in[T_{-},T_{+}]\setminus[-C,C]$. On the other hand, on $[-C,C]$ we know by definition

(4.20)

$$q(z)=(2-n)T+T^{-1}B(z).$$

Hence by the expansion in Proposition 3.28 we obtain (4.17). This implies that for $T\gg 1$, $h$ is also positive when $z\in[-C,C]$. ∎