\frac{|z-\zeta_1(j)|}{{{\rm{diam}}}(L_j)},   \\
    |\tilde H_{j,2}(z)| &\le& \frac{|z-\zeta_2(j)|}{{{\rm{diam}}}(L_j)},
    \end{aligned}$$ for $z\in L_j$. Setting $H_j=\tilde H_{j,1}\tilde H_{j,2}$, and then $G=\prod_0^4H_j$ thus provides the desired function.\
Fixing $\beta>0$, if $\{z_k\}\in\gamma_j$ are points satisfying $$\label{sepbeta}|z_k-z_\ell|\ge\beta(1-|z_k|),\qquad k\neq\ell,$$ then $\sum_k(1-|z_k|)\delta_{z_k}$, where $\delta_{z_k}$ in this case denotes the unit point mass at $z_k$, is a Carleson measure with norm depending only on $\beta$, and so $\{z_k\}$ is an interpolating sequence with constant of interpolation depending only on $\beta$ by Carleson’s interpolation theorem for $H^\infty(\mathbb D)$ [@Carint]. Due to the hyperbolic separation of the arcs $\gamma_j$ in $\mathbb D$, this remains true for a sequence $S=\bigcup_1^4 S_j$ with each $S_j$ a sequence in $\gamma_j$ satisfying this condition. Then if $G(z)$ is the function provided by Lemma \[outerfunction\], given $\{w_j\}\in\ell^\infty$ we can find an interpolating function $f\in H^\infty(\mathbb D)$ with $f(z_j)=\frac{w_j}{G(z_j)}$. The function $g=fG\in H^\infty(\mathbb D)$ then satisfies $$g(z_j)=w_j,$$ $$\Vert g\Vert_{L^\infty(\mathbb D)}\le A'c_0,$$ $$|g(z)|\le\frac{A'd(z,\{\zeta_1,\zeta_2\})}{n^3{{\rm{diam}}}(L_j)},\qquad z\in L_j,$$ for $L_j$ and $\zeta_1(j),\zeta_2(j)$ defined as in the lemma, where $A'$ bounds the largest constant of interpolation for a maximal sequence on $\bigcup_1^4\gamma_j$ satisfying (\[sepbeta\]). Fixing $\beta$ for the remainder of the proof, let $A=c_0A'$.

\[lem2\] Let $S=\{z_n\}$ be a maximal sequence on $\bigcup_1^4\gamma_j$ satisfying $$|z_k-z_\ell|\ge\frac{1-|z_j|}{8A^2},\qquad k\neq\ell.$$ Then there are functions $h_j\in H^\infty(\mathbb D)$ 