a similar result comparing to \[lem:3.2\] by assuming ${\left\lVert f \right\rVert}_{L^\infty(\Omega)} < \infty$, but in here the periodic property plays key role and leads to relax on regularity assumption for $f$, which provides us more generality in the estimation.

Recall the domain we consider has Lipschitz boundary, then the extension operator $\mathbf{E}: H^1(\Omega)\mapsto H^1({\mathbb{R}}^d)$ exists and is bounded. Let $w_\epsilon = u_\epsilon - u_0-\epsilon N^\epsilon_l {\mathbf{S}}_\epsilon(\partial_l {\mkern 1.5mu\overline{\mkern-1.5muu\mkern-1.5mu}\mkern 1.5mu}_0)$, ${\mkern 1.5mu\overline{\mkern-1.5muu\mkern-1.5mu}\mkern 1.5mu}_0 = \mathbf{E}u_0 \in H^1({\mathbb{R}}^d)$ is the extension of $u_0$, our goal in this section is to prove following key lemma:

\[lem:key\] Let \[ass:A\] be satisfied, and $u_\epsilon, u_0$ the solution of \[eq:contact problem hemiform\] \[eq:contact problem homohemiform\] respectively, and suppose that $u_0 \in H^2(\Omega)$, then $\forall v \in V$: $$\int_{\Omega} A^\epsilon \nabla w_\epsilon \cdot \nabla v \leq j^0(\gamma_ju_0; \gamma_j v)+j^0(\gamma_j u_\epsilon; -\gamma_j v) +  C {\left\lVert u_0 \right\rVert}_{2,\Omega}
    {\left( \epsilon^{1/2}{\left\lVert \nabla v \right\rVert}_{0,\Omega_{2\epsilon}}+\epsilon{\left\lVert \nabla v \right\rVert}_{0,\Omega} \right)},$$ here constant $C$ dependents on $\Omega, \kappa_1, \kappa_2$ and $w_\epsilon = u_\epsilon - u_0-\epsilon N^\epsilon_l {\mathbf{S}}_\epsilon(\partial_l {\mkern 1.5mu\overline{\mkern-1.5muu\mkern-1.5mu}\mkern 1.5mu}_0)$.

It is harmless to assume $v \in C^\infty (\Omega) \cap V$. Since $u_0 \in H^2(\Omega)$, we have ${\left\lVert {\mkern 1.5mu\overline{\mkern-1.5muu\mkern-1.5mu}\mkern 1.5mu}_0 \right\rVert}_{2, {\mathbb{R}}^d} \leq C {\left\lVert u \right\rVert}_{2,\Omega}$. In following proof, constant 