$\varphi$ by $\varphi+\left(|x|-R+\frac12\right)_+^2$, where $$\left(|x|-R+\frac12\right)_+=\max\left\{|x|-R+\frac12,0\right\},$$ we also assume that $\varphi$ is uniformly convex in $\{x\in\mathbb{R}^n\,:\,R-\frac12<|x|<R\}$. Consider the second boundary value problem with $$\label{e314}
        f_j(x,u)=\left\{\begin{array}{ll}
            f & \mbox{in }\Omega \\
            H'_j(u-\varphi) & \mbox{in }B\setminus\Omega,
        \end{array}\right.$$ where $H_j(t)=H(4^jt)$ and $H$ is defined by . By Lemma \[l303\] there is a solution $u_j$ satisfying $$\label{e315}
        |u_j-\varphi|\leq 4^{-j},\quad x\in B\setminus\Omega.$$ By the convexity, $u_j$ sub-converges to a convex function $\bar u$ in $B$ as $j\to\infty$. Note that $\bar u=\varphi$ in $B\setminus\Omega$. Hence, $\bar u\in\mathcal{S}[\varphi,\Omega]$ when restricted in $\Omega$. Using a similar argument as in [@TW10] and [@Z1], one can show that $\bar u$ is the maximizer of over the set . By the uniqueness of maximizer, we obtain $\bar u=u$. The main ingredients of the argument in [@TW10] are the upper semicontinuity and the concavity of the functional , which hold for all $\alpha\in[0,\frac{1}{n+2}]$, see Lemma \[l201\].

Strict convexity
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In this section, we prove the strict convexity of $u_0$ in dimension two. Let $\mathcal{G}_0$ be the graph of $u_0$. If $u_0$ is not strictly convex, then $\mathcal{G}_0$ contains a line segment. Let $\ell(x)$ be a tangent function of $u_0$ at the segment and denote by $$\label{e401}
   \mathcal{C}=\{x\in\Omega\,:\,u_0(x)=\ell(x)\}$$ the contact set. The set $\mathcal{C}\subset\mathbb{R}^2$ 