u_\epsilon \partial_i (\epsilon N^\epsilon_l \lambda_l+\bm{\lambda}\cdot \bm{x}) \overset{*}{\rightharpoonup} \bm{p}_0 \cdot \bm{\lambda} .$$ On the other side, notice $A^\epsilon$ and $\hat{A}$ are both symmetric, from the definition of correctors we have $\partial_j{\left( A^\epsilon_{ij} \partial_i (\epsilon N^\epsilon_l \lambda_l + \bm{\lambda}\cdot \bm{x}) \right)}=0$, and $$A^\epsilon_{ij}\partial_j u_\epsilon \partial_i (\epsilon N^\epsilon_l \lambda_l+\bm{\lambda}\cdot \bm{x}) = A^\epsilon_{ij} \partial_j(\epsilon N^\epsilon_l \lambda_l + \bm{\lambda}\cdot \bm{x})\partial_i u_\epsilon \overset{*}{\rightharpoonup} \hat{A}\bm{\lambda} \cdot \nabla u_0 = \hat{A} \nabla u_0 \cdot \bm{\lambda} .$$ It asserts that $\bm{p}_0=\hat{A}\nabla u_0$.

Recall that $\gamma_j$ is a compact operator from $V$ to $V_j$, then up to a subsequence $\{\gamma_ju_\epsilon \}$ converges strongly in $V_j$. Because of $u_\epsilon \rightharpoonup u_0$, $\gamma_ju_\epsilon \rightarrow \gamma_ju_0$ in $L^2(\Gamma_C)$ holds. Utilize the fact that $j^0(x; h)$ is upper semicontinuous with $x$ (see [@Denkowski2003] Prop. 5.6.6), we arrive at $j^0(\gamma_ju_0; \gamma_j v) \geq \limsup_\epsilon j^0(\gamma_ju_\epsilon; \gamma_j v)$, take the sup limit in the left hand of \[eq:contact problem hemiform\], we have $\forall v \in V$: $$\int_{\Omega} \hat{A} \nabla u_0 \cdot \nabla v +j^0(\gamma_ju_0; \gamma_j v) \geq {\left\langle \tilde{f},v \right\rangle},$$ and this finishes the proof.

$\epsilon^{1/2}$ estimation {#sec:estimation}
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Following \[lem:3.1\], \[lem:3.2\] and \[lem:3.4\] are quoted from [@Shen2016]. In those the standard smoothing operator ${\mathbf{S}}_\epsilon u = \phi_\epsilon \ast u$ is defined as usual (see [@Grisvard2011] 