a constant. Furthermore, we can choose $\beta_m$ sufficiently large ($\beta_m\to\infty$ as $m\to\infty$) such that for any compact, proper subset $K\subset N_{u_0}(\Omega')$, $$\label{e415}
>    K\subset N_{u_m}(\Omega')$$ provided $m$ is sufficently large, where $N_u$ is the normal mapping introduced in Section 2.

The proof of the claim is contained in [@TW10] for the case $\alpha=\frac{1}{n+2}$, see also [@Z1] for the case $\alpha=0$. The idea is similar to the proof of Lemma \[l302\]. But instead of considering the second boundary value problem with inhomogeneous term , we consider a weighted one $$\label{e416}
        f_{m,j}=\left\{\begin{array}{ll}
            f+\beta_m\chi_{D_m} & \mbox{in }\Omega' \\
            H_j'(u-u_0) & \mbox{in }B_R\setminus\Omega'
        \end{array}\right.$$ where $H_j(t)=H(4^jt)$ given by , $B_R$ is a large ball enclosing $\Omega'$. By Lemma \[l303\], there is a solution $u_{m,j}$ satisfying $$\label{e417}
        |u_{m,j}-u_0|\leq 4^{-j},\quad x\in B_R\setminus\Omega'.$$ By the convexity, $u_{m,j}$ sub-converges to a convex function $u_m$ as $j\to\infty$ and $u_m=u_0$ in $B_R\setminus\Omega'$. Note that $u_m\in\mathcal{S}[u_0,\Omega']$ when restricted in $\Omega'$, therefore, $u_m$ converges to a convex function $u_\infty$ in $\mathcal{S}[u_0,\Omega']$ as $m\to\infty$. Similarly, one can show that $u_\infty$ is the maximizer of $J_\alpha$ over the set $\mathcal{S}[u_0,\Omega']$. By the uniqueness of maximizer, we have $u_\infty=u_0$ and obtain the claim. See [@TW10; @Z1] for more details.

Now, suppose that $\ell$ is a 