D)$ of norm one with the following properties.

1.  There is a constant $c_2$, not depending on $n$, such that $|G(z)|\ge c_2$ for $z\in\gamma_j^\ast$.

2.  If $\tilde\alpha<\frac\pi2$ and $L_j$ is the lens domain in $\mathbb D$ bounded by $E_j$ and the circular arc $\gamma_j$ meeting the endpoints $\zeta_1(j),\zeta_2(j)$ of $E_j$ in angle $\tilde\alpha$ then $$|G(z)|\le\frac1{n^3}\frac{d(z,\{\zeta_1,\zeta_2\})}{{{\rm{diam}}}(L_j)}.$$

[**Proof:** ]{} We will obtain $G$ by multiplying a number of outer functions. For the first factor, we define $$H_0(z)=\exp\left\{-\int_{\partial\mathbb D}\frac{e^{i\theta}+z}{e^{i\theta}-z}6\log n\,
        \chi_{\bigcup E_j^\ast}(\theta)\frac{d\theta}{2\pi}\right\},$$ noting that $$\begin{aligned}
    |H_0(z)| &=& \exp\left\{-\int_{\partial\mathbb D}P_z(\theta)6\log n\,\chi_{\bigcup E_j^\ast}(\theta)\frac{d\theta}{2\pi}\right\}    \\
        &=& \exp\left\{-6\log n\,\omega\bigg(z,\bigcup E_j^\ast,\mathbb D\bigg)\right\},
    \end{aligned}$$ where $P_z$ is the Poisson kernel. Due to our choice of angle $\theta_n$, we have $$\omega\bigg(z,\bigcup E_j^\ast,\mathbb D\bigg)\le4\omega(z,E_k,\mathbb D)\le\frac{4c_1}{\log n},\qquad z\in\gamma_k^\ast,$$ so that $|H_0(z)|\ge e^{-24c_1}$ on each $\gamma_j^\ast$. Also, on $L_j$ we can easily compute that $\omega(z,E_j^\ast,\mathbb D)\ge\frac12$, whereby $|H_0(z)|\le\frac1{n^3}$ on $L_j$.

For the remaining factors we first note that on the imaginary axis the functions $$u_\pm(z) := \int_{-\frac\pi2}^\frac\pi2\log^-|e^{\pm i\frac\pi6}-e^{i\theta}|P_{iy}(\theta)\frac{d\theta}{2\pi}$$ are bounded below by $-\log2$, and so after precomposing these functions by appropriate Möbius transformations and exponentiating, we obtain functions $\tilde H_{j,1}$, $\tilde H_{j,2}$ of norm one such that $|\tilde H_{j,1}(z)|,|\tilde H_{j,2}(z)|\ge\frac12$ on each $\gamma_k$, $1\le k\le 4$, and such that $$\begin{aligned}
    |\tilde H_{j,1}(z)| &\le& 