finite. Roughly speaking, each square annulus $A^n_J$ contains one critical point, and $g(z)$ is of the same size as $\omega(Q^n_J)$ (up to a constant factor) for $z\in A^n_J$ when there are $0<a<b<\frac12$ such that $a\le\lambda_n\le b$ for every $n$. Thus by Harnack’s inequality the convergence of $\sum g(z_j)$ is equivalent to the convergence of $\sum_{n,J}\omega(Q^n_J)$. But $\sum_{n,J}\omega(Q^n_J)=\sum_n \omega(K)=\infty$, so $\{z_j\}$ cannot be an interpolating sequence.

If one takes slightly more care in comparing harmonic measure to Green’s function, this can be extended to show that $\{z^n_J\}_{n,J}$ is not an interpolating sequence when $\delta_n\le\frac{c_0}{\log n}$ for large $n$, where $c_0$ is some absolute constant.

In the other direction, it has been shown by Jones (unpublished) that the critical points of Green’s function form an interpolating sequence in the case that $K$ has positive area. (This occurs iff $\sum\delta_n<\infty$.) This can be done via the techniques of [@jones-marshall] using harmonic measure estimates (see Theorems 4.1 and 4.5 of that paper), or, alternatively, by solving a $\bar\partial$-problem.

In the present work, we prove the following result.

\[corthmdim2\] If $K=K(\{\lambda_n\})$ is a square Cantor set with $\delta_n=o\left(\frac1{\log\log n}\right)$ and $\{f_\mu\}^M_{\mu=1}\in H^\infty(\Omega)$ satisfy $$0<\eta\le\max_{1\le\mu\le M}|f_\mu(z)|\le1,$$ then there are $\{g_\mu\}_{\mu=1}^M\in H^\infty(\Omega)$ such that $$\sum_{\mu=1}^Mf_\mu(z)g_\mu(z)=1,\qquad z\in\Omega.$$\

This 