$\epsilon>0$, there is $m_{i,\epsilon}\in\mathcal X_i$ satisfying that $$\|m_i-m_{i,\epsilon}\|_{\mathcal X_i}<\epsilon,$$ $b_i(x)+\lambda_i\widehat m_{i, \epsilon}(x)$ is $C^N$, $b_i(x)+\lambda_i\widehat m_{i, \epsilon}(x)$ attains its maximum at some $x_0\in {\rm Int}(D_i)$, and the derivatives of $b_i+\lambda_i\widehat m_{i, \epsilon}(x)$ up to order $N-1$ at $x_0$ are zero.

It follows from [@RaSh Lemma 4.1]. For the self-completeness, we provide a proof in the following.

We prove the case $i=1$ or $2$. The case $i=3$ can be proved similarly. Without loss of generality, we assume $\lambda_i=1$, and recall that $\widehat h_i(x)=-b_i(x)+\widehat m_i(x)$ (see ).

First, let $\tilde x_0\in \bar D_i$ be such that $$\widehat h_i(\tilde x_0)=\max_{x\in\bar D_i}  \widehat h_i(x).$$ For any $\epsilon>0$, there is $\tilde x_\epsilon \in {\rm Int}(D_i)$ such that $$\label{eq1}
\widehat h_i(\tilde x_0)-\widehat h_i(\tilde x_\epsilon)<\frac{\epsilon}{3}.$$ Let $\tilde\sigma>0$ be such that $$B(\tilde x_\epsilon,\tilde\sigma)\Subset D_i,$$ where $B(\tilde x_\epsilon,\tilde\sigma)$ denotes the open ball with center $\tilde x_\epsilon$ and radius $\tilde\sigma$.

Note that there is $\xi_i(\cdot)\in C(\bar D_i)$ such that $0\leq \xi_i(x)\leq 1$, $ \xi_i(\tilde x_\epsilon)=1$, and ${\rm supp}( \xi_i)\subset B(\tilde x_\epsilon,\tilde\sigma)$. Let $$\overline m_{i, \epsilon}(t, x)=m_i(t, x)+\frac{\epsilon}{3} \xi_i(x),$$ and $$\label{eq0}
 \overline h_{i,\epsilon}(x)=-b_i(x)+\widehat m_{i}(x)+\frac{\epsilon}{3} \xi_i(x).$$ Then $\overline m_{i, \epsilon}(t, \cdot)$ and $\overline h_{i,\epsilon}(\cdot)$ is continuous on $\bar D_i$, \[overline-ai\] m\_[i, ]{}(t, )-m\_i(t, ) and $\overline h_{i,\epsilon}(\cdot)$ attains its maximum in ${\rm 