#### Proof.

We will only consider the case *A*, since the case *B* is identical. Let $\eta >0$ be any number such that

$2\eta <\underset{x\in \overline{\mathrm{\Omega}}}{\mathrm{min}}(1-a(x)).$(1.5)

For small $\delta >0$, we have $a(x)<\frac{1}{2}$ if $dist(x,\partial {A}_{1})\le \delta $. Therefore, by (1.3), we deduce that ${u}_{\epsilon}\to 1$ uniformly on the compact subset of *A* that is described by $\{x\in \mathrm{\Omega}:dist(x,\partial {A}_{1})\le \frac{\delta}{2}\}$, as $\epsilon \to 0$. Consider the subset of Ω that is defined by ${A}_{2}={A}_{1}\cup \{x\in \mathrm{\Omega}:dist(x,\partial {A}_{1})<\frac{\delta}{2}\}$. We fix a small δ such that ${A}_{2}\supset {A}_{1}$ is smooth and ${\overline{A}}_{2}\subset \mathrm{\Omega}$. Since any global minimizer satisfies $0<{u}_{\epsilon}<1$ if ε is small, we have that

$1-{u}_{\epsilon}(x)\le \eta ,x\in \partial {A}_{2}.$(1.6)

We claim that $1-{u}_{\epsilon}(x)\le \eta $, $x\in {\overline{A}}_{2}$, which clearly implies the validity of the assertion of the theorem. Suppose that the claim is false. Then, for some sequence of small ε’s, there exists an ${x}_{\epsilon}\in {A}_{2}$ such that

$1-{u}_{\epsilon}({x}_{\epsilon})=\underset{x\in {\overline{A}}_{2}}{\mathrm{max}}(1-{u}_{\epsilon}(x))>\eta .$(1.7)

We will first exclude the possibility that

$1-{u}_{\epsilon}(x)\le 2\eta ,x\in {\overline{A}}_{2}.$(1.8)

To this end, we will argue by contradiction. Let

${\stackrel{~}{u}}_{\epsilon}(x)=\{\begin{array}{cc}\mathrm{max}\{{u}_{\epsilon}(x),2-2\eta -{u}_{\epsilon}(x)\},\hfill & x\in {A}_{2},\hfill \\ {u}_{\epsilon}(x),\hfill & x\in \mathrm{\Omega}\setminus {A}_{2}.\hfill \end{array}$

Since $\mathrm{max}\{{u}_{\epsilon},2-2\eta -{u}_{\epsilon}\}$ is the composition of a Lipschitz function with an ${H}^{1}({A}_{2})$ function, it follows from [8] that
${\stackrel{~}{u}}_{\epsilon}\in {H}^{1}({A}_{2})$. Furthermore, from (1.6) and the Lipschitz regularity of ${A}_{2}$ we obtain that ${\stackrel{~}{u}}_{\epsilon}\in {H}_{0}^{1}(\mathrm{\Omega})$, see again
[8]. Note that ${\stackrel{~}{u}}_{\epsilon}\in C(\overline{\mathrm{\Omega}})$. On the other hand, (1.8) implies that

$1-2\eta \le {u}_{\epsilon}(x)\le {\stackrel{~}{u}}_{\epsilon}(x)\le 1,x\in {\overline{A}}_{2}.$

In turn, recalling (1.2) and (1.5), this implies that

$F(x,{u}_{\epsilon}(x))\le F(x,{\stackrel{~}{u}}_{\epsilon}(x)),x\in {\overline{A}}_{2}.$(1.9)

To see this, observe that

$\text{for each}x\in \overline{\mathrm{\Omega}}\text{the function}F(x,t)\text{is increasing with respect to}t\in [1-2\eta ,1],$(1.10)

since ${F}_{t}(x,t)=t(t-a(x))(1-t)$. (Note that $t\mapsto F(t,x)$ changes monotonicity in $(0,1)$ only at $t=a(x)$). From (1.7), which implies that
${u}_{\epsilon}({x}_{\epsilon})<{\stackrel{~}{u}}_{\epsilon}({x}_{\epsilon})$, it follows that $F(x,{u}_{\epsilon}(x))<F(x,{\stackrel{~}{u}}_{\epsilon}(x))$ on an open subset of ${A}_{2}$ containing ${x}_{\epsilon}$. Hence,

${\int}_{\mathrm{\Omega}}F(x,{u}_{\epsilon}(x))\mathit{d}x<{\int}_{\mathrm{\Omega}}F(x,{\stackrel{~}{u}}_{\epsilon}(x))\mathit{d}x.$(1.11)

Moreover, it holds that

${\int}_{\mathrm{\Omega}}{|D{\stackrel{~}{u}}_{\epsilon}|}^{2}\mathit{d}x\le {\int}_{\mathrm{\Omega}}{|D{u}_{\epsilon}|}^{2}\mathit{d}x,$(1.12)

see [14, p. 93]. The above two relations yield that ${I}_{\epsilon}({\stackrel{~}{u}}_{\epsilon})<{I}_{\epsilon}({u}_{\epsilon})$, contradicting the fact that ${u}_{\epsilon}$ is a global minimizer of ${I}_{\epsilon}$ in ${H}_{0}^{1}(\mathrm{\Omega})$. Consequently, we have that

$0<1-{u}_{\epsilon}({x}_{\epsilon})<1-2\eta .$(1.13)

Now, let

${\widehat{u}}_{\epsilon}(x)=\{\begin{array}{cc}\mathrm{min}\{1,\mathrm{max}\{{u}_{\epsilon}(x),2-2\eta -{u}_{\epsilon}(x)\}\},\hfill & x\in {A}_{2},\hfill \\ {u}_{\epsilon}(x),\hfill & x\in \mathrm{\Omega}\setminus {A}_{2},\hfill \end{array}$

see also [11]. As before, it is easy to see that ${\widehat{u}}_{\epsilon}\in {H}_{0}^{1}(\mathrm{\Omega})$. Since $a(x)\le \frac{1}{2}$, $x\in {\overline{A}}_{2}$, it follows readily that

$F(x,t)<F(x,1)\mathit{\hspace{1em}}\text{for all}t\in (0,1),x\in {\overline{A}}_{2}.$

Hence, as before, making use of (1.10), (1.13), and the above relation, we get (1.11), (1.12) with ${\widehat{u}}_{\epsilon}$ in
place of ${\stackrel{~}{u}}_{\epsilon}$, which again contradict the minimality of ${u}_{\epsilon}$.
∎

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