to treat each refinement step separately. In each stripe of the form $(x_{i+1},x_i)\times(0,H)$, $i=0, \dots, \tau-1$, we use $N2^i$ times the construction of Lemma \[eqconstronerecttheo2\]. In the set $(0,x_\tau)\times (0,L)$ we use the same procedure with Lemma \[eqconstronerecttheo2bl\]. The result is continuous thanks to (\[eqvperiodiclines\]).

It remains to estimate the energy. This is given by the sum over the stripes of $N2^i$ times the energy in each rectangle plus the boundary energy, which on the boundary of each rectangle is controlled by $\eps\|Dv-\Id\|_{L^\infty}
(h_i+\ell_i)\le c \alpha \eps \ell_i$. We obtain $$E_2^\eps[u,\Omega'] \leq c\sum_{i=0}^{\tau-1} 2^iN \left( 
\alpha^2 \frac{h_i^5}{\ell_i^3} + \alpha \eps \ell_i\right) + c N2^\tau
\left(\alpha^2 h_\tau\ell_\tau + \alpha \eps \ell_\tau\right) \,.$$ By definition of $\tau$, we have $h_\tau\le\ell_\tau\le 2 h_\tau$, therefore, the boundary term is comparable to the $i=\tau$ term in the series and can be included in the series. Using the definitions of $h_i$ and $\ell_i$, the estimate becomes $$E_2^\eps[u,\Omega'] \leq c\sum_{i=0}^{\tau}\left(
  \frac{\alpha^2H^5}{2^{4i}N^4\theta^{3i}L^3(1-\theta)^3} +
  \alpha\varepsilon2^{i}N\theta^{i}(1-\theta)L \right)\,.$$ If $\theta\in
(2^{-4/3},2^{-1})$, both $\sum_i 1/(2^4\theta^3)^{i}$ and $\sum_i (2\theta)^i$ are converging geometric series. The precise choice of $\theta$ in this interval does not influence the scaling, but only the prefactor, which we did not compute explicitly. In principle, one could choose the value 