4.1. The concave region with $\underline{\boldsymbol{V}}''(\boldsymbol{x})<0$

Note that in this region, we have $\frac{\underline{V}'}{\underline{V}''}>0$ . Since $L^{a_1,a_2}\underline{V}(x)$ is a quadratic function in $a_1$ with a negative coefficient on the second-degree term, the minimizer of $L^{a_1,a_2}\underline{V}(x)$ in $a_1\in[0, A_1]$ is $\underline{a}^*(x,a_2)=A_1$ if its vertex point $\hat{a}_{1,a_2,\underline{V}}$ satisfies

\begin{align*}\hat{a}_{1,a_2,\underline{V}}(x)=-\theta_1\frac{\underline{V}'}{\underline{V}''}+\rho a_2\leq\frac{A_1}{2},\end{align*}

or equivalently, $a_2\in (\!-\!\infty,\tilde{a}_{2,\underline{V}}(x)]\cap[0,A_2]$ , where

(4.1) \begin{eqnarray}\tilde{a}_{2,\underline{V}}(x)=\frac{\theta_1}{\rho}\frac{\underline{V}'}{\underline{V}''}+\frac{A_1}{2\rho},\end{eqnarray}

which is positive in the concave region. Meanwhile, the minimizer in $a_1$ is $\underline{a}^*(x,a_2)=0$ if $\hat{a}_{1,a_2,\underline{V}}(x)=-\theta_1\frac{\underline{V}'}{\underline{V}''}+\rho a_2>\frac{A_1}{2}$ , that is, $a_2\in (\tilde{a}_{2,\underline{V}}(x),\infty)\cap[0,A_2]$ .

For convenience, we write the function $\tilde{a}_2=\tilde{a}_{2,\underline{V}}(x)$ . The minimizer of $L^{a_1,a_2}\underline{V}(x)$ in $a_1$ (given $a_2\in[0,A_2]$ ) can be expressed in the following form:

(4.2) \begin{equation}\underline{a}^*(x,a_2)=\begin{cases}0 & \tilde{a}_2 < A_2\ \ a_2\in(\tilde{a}_2, A_2],\\ A_1 & \tilde{a}_2 < A_2\ \ a_2\in[0, \tilde{a}_2],\\ A_1 & \tilde{a}_2\ge A_2\ \ a_2\in[0, A_2], \end{cases}\end{equation}

which takes the boundary values 0 and $A_1$ .

We now define two regions:

(4.3) \begin{equation}\begin{split}&\mathcal{R}_{1,-}=\{x\in (l,u)\;:\; \underline{V}'' < 0,\ \tilde{a}_2 < A_2\}=\{x\in (l,u)\;:\; \underline{V}'' < 0,\ {\underline{V}'}/{\underline{V}''} < B_1\},\\[5pt] &\mathcal{R}_{2,-}=\{x\in (l,u)\;:\; \underline{V}'' < 0,\ \tilde{a}_2\ge A_2\}=\{x\in (l,u)\;:\; \underline{V}'' < 0,\ {\underline{V}'}/{\underline{V}''}\ge B_1\},\end{split}\end{equation}

where

(4.4) \begin{eqnarray}B_1=(\rho A_2-A_1/2)/\theta_1.\end{eqnarray}

Next we give results to find the maximizer in $a_2$ of $L^{\underline{a}^*(x,a_2),a_2}\underline{V}$ in $\mathcal{R}_{1,-}$ and $\mathcal{R}_{2,-}$ . Note that the vertex point of $L^{A_1,a_2}\underline{V}$ (as a quadratic function of $a_2$ ) is $\hat{a}_{2,A_1,\underline{V}}=\theta_2\frac{\underline{V}'}{\underline{V}''}+\rho A_1 >0$ , and the vertex point of $L^{0,a_2}\underline{V}$ is $\hat{a}_{2,0,\underline{V}}=\theta_2\frac{\underline{V}'}{\underline{V}''}>0$ .

Considering the maximization in $a_2\in[0, \tilde{a}_2]$ of $L^{A_1,a_2}\underline{V}$ and in $a_2\in(\tilde{a}_2, A_2]$ of $L^{0,a_2}\underline{V}$ from (4.2) in ${R}_{1,-}$ with $\theta_1/\theta_2<\rho$ , we obtain the following lemma.

Lemma 4.1. For $x\in\mathcal{R}_{1,-}$ , if $\frac{\theta_1}{\theta_2}<\rho$ , we have

(4.5) \begin{equation}\underset{a_2\in[0,\tilde{a}_2]}{\sup}\underset{a_1\in[0,A_1]}{\mathrm{inf}}L^{a_1,a_2}\underline{V}=\begin{cases}L^{A_1,\hat{a}_{2,A_1,\underline{V}}}\underline{V}&\ \ \{x\;:\;0 < \hat{a}_{2,A_1,\underline{V}}<\tilde{a}_{2}\} =\Big\{x\;:\;\frac{\underline{V}'}{\underline{V}''}<B_2\Big\},\\[9pt] L^{A_1,\tilde{a}_{2}}\underline{V}=L^{0,\tilde{a}_{2}}\underline{V}&\ \ \{x\;:\;\tilde{a}_{2}\le\hat{a}_{2,A_1,\underline{V}}\}=\Big\{x\;:\;B_2\le\frac{\underline{V}'}{\underline{V}''}\Big\},\end{cases}\end{equation}

and

(4.6) \begin{equation}\begin{split}&\underset{a_2\in(\tilde{a}_{2},A_2]}{\sup}\underset{a_1\in[0,A_1]}{\mathrm{inf}}L^{a_1,a_2}\underline{V}\\[5pt] & \quad = \begin{cases}L^{0,\tilde{a}_{2}}\underline{V}=L^{A_1,\tilde{a}_{2}}\underline{V}&\ \ \{x\;:\;\hat{a}_{2,0,\underline{V}}\le\tilde{a}_{2}\}=\Big\{x\;:\;\frac{\underline{V}'}{\underline{V}''}\le B_3\Big\},\\[5pt] L^{0,\hat{a}_{2,0,\underline{V}}}\underline{V}&\ \ \{x\;:\;\tilde{a}_{2}<\hat{a}_{2,0,\underline{V}}<A_2\}=\Big\{x\;:\;B_3<\frac{\underline{V}'}{\underline{V}''}<B_4\Big\},\\[5pt] L^{0,A_2}\underline{V}&\ \ \{x\;:\; A_2\le\hat{a}_{2,0,\underline{V}}\}=\Big\{x\;:\;B_4\le\frac{\underline{V}'}{\underline{V}''}\Big\}\end{cases}\end{split}\end{equation}

where

(4.7) \begin{eqnarray}B_2=\frac{A_1(1/2-\rho^2)}{\rho\theta_2-\theta_1},\quad B_3=\frac{A_1/2}{\rho\theta_2-\theta_1},\quad B_4=\frac{A_2}{\theta_2}.\end{eqnarray}

We note that

(4.8) \begin{align}&L^{A_1,\tilde{a}_{2}}\underline{V}=L^{0,\tilde{a}_{2}}\underline{V}\nonumber \\[5pt] &\quad =\underline{V}''\left\{\left(\frac{\theta_1^2}{2\rho^2}-\frac{\theta_1\theta_2}{\rho}\right)\left(\frac{\underline{V}'}{\underline{V}''}\right)^2+\left[rx+\mu+\left(\frac{\theta_1}{2\rho^2}-\frac{\theta_1}{2\rho}\right)A_1\right]\frac{\underline{V}'}{\underline{V}''}+\frac{1}{2}\left(\frac{A_1^2}{4\rho^2}+\sigma^2\right)\right\}.\end{align}

We now consider the optimization in $\mathcal{R}_{1,-}$ in the case $\rho\le\theta_1/\theta_2$ . Note that

(4.9) \begin{equation}\underset{a_2\in[0,\tilde{a}_{2}]}{\sup}\underset{a_1\in[0,A_1]}{\mathrm{inf}}L^{a_1,a_2}\underline{V}=\begin{cases}L^{A_1,\hat{a}_{2,A_1,\underline{V}}}\underline{V}&\ \ \big\{x\;:\;0<\hat{a}_{2,A_1,\underline{V}}<\tilde{a}_{2}\big\}=\Big\{x\;:\;B_2<\frac{\underline{V}'}{\underline{V}''}\Big\},\\[9pt] L^{A_1,\tilde{a}_{2}}\underline{V}&\ \ \big\{x\;:\;\tilde{a}_{2}\le\hat{a}_{2,A_1,\underline{V}}\big\}=\Big\{x\;:\;\frac{\underline{V}'}{\underline{V}''}\le B_2\Big\},\end{cases}\end{equation}

for $\rho<\theta_1/\theta_2$ , and

(4.10) \begin{equation}\underset{a_2\in[0,\tilde{a}_{2}]}{\sup}\underset{a_1\in[0,A_1]}{\mathrm{inf}}L^{a_1,a_2}\underline{V}=\begin{cases}L^{A_1,\hat{a}_{2,A_1,\underline{V}}}\underline{V}&\ \ \big\{x\;:\;0<\hat{a}_{2,A_1,\underline{V}}<\tilde{a}_{2}\big\}\ \ \mbox{if}\ \rho < \sqrt{1/2},\\[5pt] L^{A_1,\tilde{a}_{2}}{\underline{V}}&\ \ \big\{x\;:\;\tilde{a}_{2}\le\hat{a}_{2,A_1,\underline{V}}\big\}\ \ \mbox{if}\ \sqrt{1/2}\le\rho,\end{cases}\end{equation}

for $\rho=\theta_1/\theta_2$ in the region $\mathcal{R}_{1,-}$ .

Further, from $\rho\le\theta_1/\theta_2$ , the inequality $0<\hat{a}_{2,0,\underline{V}}\le\tilde{a}_{2}$ always holds, and we have

(4.11) \begin{equation}\underset{a_2\in(\tilde{a}_{2},A_2]}{\sup}\underset{a_1\in[0,A_1]}{\mathrm{inf}}L^{a_1,a_2}\underline{V}=\underset{a_2\in(\tilde{a}_{2},A_2]}{\sup}L^{0,a_2}\underline{V}=L^{0,\tilde{a}_{2}}\underline{V}=L^{A_1,\tilde{a}_{2}}\underline{V}\end{equation}

in the region $\mathcal{R}_{1,-}$ .

Comparing (4.9) and (4.10) with (4.11), and noticing that $L^{A_1,\hat{a}_{2,A_1,\underline{V}}}\underline{V}\ge L^{A_1,\tilde{a}_{2}}\underline{V}$ on $\{x\;:\;B_2<\frac{\underline{V}'}{\underline{V}''}\}$ , we obtain the following lemma.

Lemma 4.2. For $x\in\mathcal{R}_{1,-}$ and $\rho<{\theta_1}/{\theta_2}$ , we have

(4.12) \begin{equation}\underset{a_2\in[0,A_2]}{\sup}\underset{a_1\in[0,A_1]}{\mathrm{inf}}L^{a_1,a_2}\underline{V}=\begin{cases}L^{A_1,\hat{a}_{2,A_1,\underline{V}}}\underline{V}&\ \ \big\{x\;:\;0<\hat{a}_{2,A_1,\underline{V}}<\tilde{a}_{2}\big\}=\Big\{x\;:\;B_2<\frac{\underline{V}'}{\underline{V}''}\Big\},\\[9pt] L^{A_1,\tilde{a}_{2}}\underline{V}&\ \ \big\{x\;:\;\tilde{a}_{2}\le\hat{a}_{2,A_1,\underline{V}}\big\}=\Big\{x\;:\;\frac{\underline{V}'}{\underline{V}''}\le B_2\Big\};\end{cases}\end{equation}

for $x\in\mathcal{R}_{1,-}$ and $\rho=\frac{\theta_1}{\theta_2}$ , we have

(4.13) \begin{equation}\underset{a_2\in[0,A_2]}{\sup}\underset{a_1\in[0,A_1]}{\mathrm{inf}}L^{a_1,a_2}\underline{V}=\begin{cases}L^{A_1,\hat{a}_{2,A_1,\underline{V}}}\underline{V}&\ \ \ \rho<\sqrt{1/2},\\[5pt] L^{A_1,\tilde{a}_{2}}\underline{V}&\ \ \ \sqrt{1/2}\le \rho.\end{cases}\end{equation}

For $x\in\mathcal{R}_{2,-}$ , the minimizer in $a_1$ of $L^{a_1,a_2}\underline{V}$ is always $\underline{a}^*(x,a_2)=A_1$ , and the vertex point of $L^{A_1,a_2}\underline{V}$ (in $a_2$ ) is $\hat{a}_{2,A_1,\underline{V}}=\theta_2\frac{\underline{V}'}{\underline{V}''}+\rho A_1 >0$ . Then we obtain the following result.

Lemma 4.3. For $x\in\mathcal{R}_{2,-}$ , we have

(4.14) \begin{equation}\underset{a_2\in[0,A_2]}{\sup}\underset{a_1\in[0,A_1]}{\mathrm{inf}}L^{a_1,a_2}\underline{V}=\begin{cases}L^{A_1,\hat{a}_{2,A_1,\underline{V}}}\underline{V}&\ \ \big\{x\;:\;0<\hat{a}_{2,A_1,\underline{V}}<A_2\big\}=\Big\{x\;:\;\frac{\underline{V}'}{\underline{V}''}<B_5\Big\},\\[9pt] L^{A_1,A_2}\underline{V}&\ \ \big\{x\;:\;A_2\le\hat{a}_{2,A_1,\underline{V}}\big\}=\Big\{x\;:\;B_5\le\frac{\underline{V}'}{\underline{V}''}\Big\},\end{cases}\end{equation}

where

(4.15) \begin{eqnarray}B_5=(A_2-\rho A_1)/{\theta_2}.\end{eqnarray}

Short calculations yield the following lemma.

From Lemma 4.4, we have $B_3<B_4<B_1$ if $\frac{\theta_1}{\theta_2}<\rho$ and $\frac{A_1}{A_2}<2(\rho-\frac{\theta_1}{\theta_2})$ ; and $B_1\le B_4\le B_3$ if $\frac{\theta_1}{\theta_2}<\rho$ and $\frac{A_1}{A_2}\ge 2(\rho-\frac{\theta_1}{\theta_2})$ .

In case 1 with $B_1\leq 0$ , we have $\mathcal{R}_{1,-}=\emptyset$ . We consider two subcases: subcase 11 with $B_5\leq 0$ (i.e. $\frac{1}{\rho}\le\frac{A_1}{A_2}$ ) and subcase 12 with $B_5>0$ .

In subcase 11, we have $B_1\leq 0$ and $B_5\leq 0$ . From Lemma 4.3 (with the first line degenerated) and the assumption that $\underline{V}$ solves (3.1), we have

(4.16) \begin{equation}\underset{a_2\in[0,A_2]}{\sup}\underset{a_1\in[0,A_1]}{\mathrm{inf}}L^{a_1,a_2}\underline{V}=L^{A_1,A_2}\underline{V}\end{equation}

in the concave region, which gives

\begin{align*}L^{A_1,A_2}\underline{V}=(rx+\mu+\theta_1 A_1-\theta_2A_2)\underline{V}'+\frac{1}{2}\big(A_1^2+A_2^2-2\rho A_1A_2+\sigma^2\big)\underline{V}''=0\end{align*}

by the BI equation (3.1), yielding $\frac{\underline{V}''}{\underline{V}'}=g(x)$ , where

(4.17) \begin{equation}g(x)=-\frac{2(rx+\mu+\theta_1 A_1-\theta_2A_2)}{A_1^2+A_2^2-2\rho A_1A_2+\sigma^2}.\end{equation}

Note that inequality $g(x)>0$ is equivalent to $x<x_0$ , where

(4.18) \begin{eqnarray}x_0=-(\theta_1 A_1-\theta_2A_2+\mu)/r,\end{eqnarray}

which is the unique inflection point. Thus, in subcase 11, we have

(4.19) \begin{equation}\frac{\underline{V}''}{\underline{V}'}=g1_{(-\infty,x_0)},\end{equation}

which is a continuous, positive, and decreasing function, with the optimizers given in (4.16), in $\mathcal{R}_{1,-}\cup\mathcal{R}_{2,-}=(\!-\!\infty,x_0)\cap[l, u]$ , which is the concave region, where $1_{\mathcal{S}}$ is the indicator function for a set $\mathcal{S}$ of the real line such that $1_{\mathcal{S}}(x)=1$ if $x\in\mathcal{S}$ and $1_{\mathcal{S}}(x)=0$ if $x\notin\mathcal{S}$ .

If $B_1\leq 0$ and $B_5>0$ (subcase 12), from Lemma 4.3 we have

(4.20) \begin{equation}\underset{a_2\in[0,A_2]}{\sup}\underset{a_1\in[0,A_1]}{\mathrm{inf}}L^{a_1,a_2}\underline{V}=\begin{cases}L^{A_1,\hat{a}_{2,A_1,\underline{V}}}\underline{V}&\ \ \Big\{x\;:\;\frac{\underline{V}'}{\underline{V}''}<B_5\Big\},\\[5pt] L^{A_1,A_2}\underline{V}&\ \ \Big\{x\;:\;B_5\le\frac{\underline{V}'}{\underline{V}''}\Big\}.\end{cases}\end{equation}

Note that

(4.21) \begin{align}L^{A_1,\hat{a}_{2,A_1,\underline{V}}}\underline{V}=-\underline{V}''\left\{\frac{\theta_2^2}{2}\left(\frac{\underline{V}'}{\underline{V}''}\right)^2-[rx+\mu+(\theta_1-\rho\theta_2)A_1]\left(\frac{\underline{V}'}{\underline{V}''}\right)-\frac{1}{2}[(1-\rho^2)A_1^2+\sigma^2]\right\}.\nonumber\\[1pt]\end{align}

The equation $L^{A_1,\hat{a}_{2,A_1,\underline{V}}}\underline{V}=0$ yields $\frac{\underline{V}'}{\underline{V}''}=f_1$ , where

(4.22) \begin{align}f_1(x)=\left[rx+\mu+(\theta_1-\rho\theta_2)A_1+\sqrt{[rx+\mu+(\theta_1-\rho\theta_2)A_1]^2+\theta_2^2[(1-\rho^2)A_1^2+\sigma^2]}\right]/\theta_2^2,\nonumber\\[1pt]\end{align}

which is a positive and increasing function. Thus, the condition $\frac{\underline{V}'}{\underline{V}''}=f_1(x)<B_5$ yields $x<x_{5,-}$ , where

(4.23) \begin{equation}x_{5,-}=-\frac{\theta_2(A_1^2+A_2^2-2\rho A_1A_2+\sigma^2)}{2r(A_2-\rho A_1)}+\frac{-\theta_1A_1+\theta_2A_2-\mu}{r}.\end{equation}

Also, $L^{A_1,A_2}\underline{V}=0$ gives $\frac{\underline{V}'}{\underline{V}''}=1/g$ , and $0<g(x)\le 1/B_5$ yields $x_{5,-}\le x<x_0$ . Then we conclude that

(4.24) \begin{equation}\frac{\underline{V}''}{\underline{V}'}=\frac{1_{(-\infty,x_{5,-})}}{f_1}+g1_{[x_{5,-},x_0)},\end{equation}

which is a continuous, positive, and decreasing function in the concave region $(\!-\!\infty,x_0)\cap[l, u]$ in subcase 12.

In case 2, we compare (4.5) and (4.6) to determine $\sup_{a_2\in[0,A_2]}\inf_{a_1\in[0,A_1]}L^{a_1,a_2}\underline{V}$ . From Lemma 4.4 we have $B_2<B_3<B_4<B_1$ and $B_5<B_1$ . From Lemma 4.3 and $B_5<B_1$ , we have

(4.25) \begin{equation}\underset{a_2\in[0,A_2]}{\sup}\underset{a_1\in[0,A_1]}{\mathrm{inf}}L^{a_1,a_2}\underline{V}=L^{A_1,A_2}\underline{V},\end{equation}

in the region $\mathcal{R}_{2,-}$ . Next we consider the region $\mathcal{R}_{1,-}$ . We note that

(4.26) \begin{equation}L^{0,\hat{a}_{2,0,\underline{V}}}\underline{V}=\underline{V}''\left[\left(-\frac{\theta_2^2}{2}\right)\left(\frac{\underline{V}'}{\underline{V}''}\right)^2+(rx+\mu)\frac{\underline{V}'}{\underline{V}''}+\frac{\sigma^2}{2}\right]\end{equation}

and

(4.27) \begin{eqnarray}L^{0,A_2}\underline{V}=(rx+\mu-\theta_2A_2)\underline{V}'+\frac{1}{2}\big(A_1^2+\sigma^2\big)\underline{V}''.\end{eqnarray}

From (4.8), (4.26), and (4.27), we have

(4.28) \begin{equation}L^{A_1,\tilde{a}_{2}}\underline{V}-L^{0,\hat{a}_{2,0,\underline{V}}}\underline{V}=\frac{\underline{V}''}{2}\left[\left(\frac{\theta_1}{\rho}-\theta_2\right)\frac{\underline{V}'}{\underline{V}''}+\frac{A_1}{2\rho}\right]^2\leq 0\end{equation}

and

(4.29) \begin{align}L^{A_1,\tilde{a}_{2}}\underline{V}-L^{0,A_2}\underline{V} & = \frac{\underline{V}''}{2}\left[\left(\theta_2-\frac{\theta_1}{\rho}\right)\frac{\underline{V}'}{\underline{V}''}-\frac{A_1}{2\rho}\right]^2-\frac{\underline{V}''}{2}\left(\theta_2\frac{\underline{V}'}{\underline{V}''}-A_2\right)^2\nonumber\\[5pt] & =\frac{\underline{V}''}{2}\left[\frac{\theta_1}{\rho}\left(B_1-\frac{\underline{V}'}{\underline{V}''}\right)\right]\left[\left[\left(\theta_2-\frac{\theta_1}{\rho}\right)\frac{\underline{V}'}{\underline{V}''}-\frac{A_1}{2\rho}\right]+\left(\theta_2\frac{\underline{V}'}{\underline{V}''}-A_2\right) \right]\nonumber\\[5pt] & =\frac{\underline{V}''}{2}\left[\frac{\theta_1}{\rho}\left(B_1-\frac{\underline{V}'}{\underline{V}''}\right)\right]\left[\left[\left(\theta_2-\frac{\theta_1}{\rho}\right)\left(\frac{\underline{V}'}{\underline{V}''}-B_3\right)\right]+\theta_2\left(\frac{\underline{V}'}{\underline{V}''}-B_4\right) \right],\nonumber\\[1pt]\end{align}

which is nonpositive when $B_3<B_4<\frac{\underline{V}'}{\underline{V}''}<B_1$ and $\frac{\theta_1}{\theta_2}<\rho$ .

If $B_2>0$ (i.e. $\rho<\sqrt{1/2}$ ), from (4.28), (4.29), (4.25) and Lemma 4.1 by comparing (4.5) and (4.6), we conclude that

(4.30) \begin{equation}\underset{a_2\in[0,A_2]}{\sup}\underset{a_1\in[0,A_1]}{\mathrm{inf}}L^{a_1,a_2}\underline{V}=\begin{cases}L^{A_1,\hat{a}_{2,A_1,\underline{V}}}\underline{V}&\ \ \Big\{x\;:\;\frac{\underline{V}'}{\underline{V}''}<B_2\Big\},\\[5pt] L^{0,\tilde{a}_{2}}\underline{V}=L^{A_1,\tilde{a}_{2}}\underline{V}&\ \ \Big\{x\;:\;B_2\le\frac{\underline{V}'}{\underline{V}''}\le B_3\Big\},\\[5pt] L^{0,\hat{a}_{2,0,\underline{V}}}\underline{V}&\ \ \Big\{x\;:\;B_3<\frac{\underline{V}'}{\underline{V}''}<B_4\Big\},\\[5pt] L^{0,A_2}\underline{V}&\ \ \Big\{x\;:\; B_4\le\frac{\underline{V}'}{\underline{V}''}<B_1\Big\},\\[5pt] L^{A_1,A_2}\underline{V}&\ \ \Big\{x\;:\; B_1\le\frac{\underline{V}'}{\underline{V}''}\Big\},\end{cases}\end{equation}

in the concave region $\mathcal{R}_{1,-}\cup\mathcal{R}_{2,-}=(\!-\!\infty,x_0)\cup[l, u]$ in case 2. In the following, we set (4.30) equal to zero, due to BI equation (3.1), to find $\frac{\underline{V}'}{\underline{V}''}$ in the concave region. Note that the equation $L^{A_1,\hat{a}_{2,A_1,\underline{V}}}\underline{V}=0$ yields $\frac{\underline{V}'}{\underline{V}''}=f_1$ , and the inequality $f_1(x)\le B_2$ is equivalent to $x<x_{2,-}$ , where

(4.31) \begin{equation}x_{2,-}=\frac{B_2^2\theta_2^2-[(1-\rho^2)A_1^2+\sigma^2]}{2rB_2}+\frac{(\rho\theta_2-\theta_1)A_1-\mu}{r}.\end{equation}

Note that from (4.8), the equation $L^{0,\tilde{a}_{2}}\underline{V}=0$ yields $\frac{\underline{V}'}{\underline{V}''}=f_2$ , where

(4.32) \begin{equation}f_2(x)=\frac{\!\left[rx+\mu+\!\left(\frac{\theta_1}{\rho}-\theta_2\right)\frac{A_1}{2\rho}\right]+\sqrt{\left[rx+\mu+\left(\frac{\theta_1}{\rho}-\theta_2\right)\frac{A_1}{2\rho}\right]^2+\frac{\theta_1}{\rho^2}(2\rho\theta_2-\theta_1)\!\left(\frac{A_1^2}{4\rho^2}+\sigma^2\right)}}{\frac{\theta_1}{\rho^2}(2\rho\theta_2-\theta_1)},\end{equation}

which is a positive and increasing function when $\theta_1/\theta_2<2\rho$ . Thus, $B_2\le\frac{\underline{V}'}{\underline{V}''}=f_2\le B_3$ yields $x_{2,-}\le x\le x_{3,-}$ , where $x_{2,-}$ is given in (4.31) and

(4.33) \begin{equation}x_{3,-}=\frac{1}{2rB_3}\left[\frac{\theta_2^2A_1^2}{4\left(\rho\theta_2-\theta_1\right)^2}-\sigma^2\right]-\frac{\mu}{r}.\end{equation}

From (4.26), the equation $L^{0,\hat{a}_{2,0,\underline{V}}}\underline{V}=0$ yields $\frac{\underline{V}'}{\underline{V}''}=f_3$ , where

(4.34) \begin{equation}f_3(x)=\frac{(rx+\mu)+\sqrt{(rx+\mu)^2+\theta_2^2\sigma^2}}{\theta_2^2},\end{equation}

and $B_3<f_3<B_4$ yields $x_{3,-}<x<x_{4,-}$ , where

(4.35) \begin{equation}x_{4,-}=\frac{\theta_2\big(A_2^2-\sigma^2\big)}{2rA_2}-\frac{\mu}{r}.\end{equation}

From (4.27), the equation $L^{0,A_2}\underline{V}=0$ yields $\frac{\underline{V}''}{\underline{V}'}=g_2$ , where

(4.36) \begin{equation}g_2(x)=-\frac{2(rx+\mu-\theta_2A_2)}{A_2^2+\sigma^2},\end{equation}

and $1/B_1<g_2\le 1/B_4$ yields $x_{4,-}\le x<x_{1,-}$ , where $x_{4,-}$ is given in (4.35) and

(4.37) \begin{equation}x_{1,-}=-\frac{A_2^2+\sigma^2}{2rB_1}+\frac{\theta_2A_2-\mu}{r}.\end{equation}

We have $\frac{\underline{V}''}{\underline{V}'}=g$ if $0<g\le 1/B_1$ , which is equivalent to $x_{1,-}\le x<x_0$ . Thus, we obtain

(4.38) \begin{equation}\frac{\underline{V}''}{\underline{V}'}=\frac{1_{(-\infty,x_{2,-})}}{f_1}+\frac{1_{[x_{2,-},x_{3,-}]}}{f_2}+\frac{1_{(x_{3,-},x_{4,-})}}{f_3}+g_21_{[x_{4,-},x_{1,-})}+g1_{[x_{1,-},x_0)},\end{equation}

which is a continuous, positive, and decreasing function in the concave region $\mathcal{R}_{1,-}\cup\mathcal{R}_{2,-}=(\!-\!\infty,x_0)\cap[l, u]$ if $B_2>0$ in case 2.

If $B_2\le 0$ (i.e. $\rho\ge\sqrt{1/2}$ ), the set in the top line of (4.30) degenerates, and we have

(4.39) \begin{equation}\underset{a_2\in[0,A_2]}{\sup}\underset{a_1\in[0,A_1]}{\mathrm{inf}}L^{a_1,a_2}\underline{V}=\begin{cases}L^{0,\tilde{a}_{2}}\underline{V}=L^{A_1,\tilde{a}_{2}}\underline{V}&\ \ \Big\{x\;:\;\frac{\underline{V}'}{\underline{V}''}\le B_3\Big\},\\[5pt] L^{0,\hat{a}_{2,0,\underline{V}}}\underline{V}&\ \ \Big\{x\;:\;B_3<\frac{\underline{V}'}{\underline{V}''}<B_4\Big\},\\[5pt] L^{0,A_2}\underline{V}&\ \ \Big\{x\;:\; B_4\le\frac{\underline{V}'}{\underline{V}''}<B_1\Big\},\\[5pt] L^{A_1,A_2}&\ \ \Big\{x\;:\; B_1\le\frac{\underline{V}'}{\underline{V}''}\Big\},\end{cases}\end{equation}

and

(4.40) \begin{equation}\frac{\underline{V}''}{\underline{V}'}=\frac{1_{(-\infty,x_{3,-}]}}{f_2}+\frac{1_{(x_{3,-},x_{4,-})}}{f_3}+g_21_{[x_{4,-},x_{1,-})}+g1_{[x_{1,-},x_0)},\end{equation}

which is a continuous, positive, and decreasing function in the concave region in case 2.

In case 3, we have $0<B_1\leq B_4\le B_3$ . We consider three subcases: subcase 31 with $B_1\le B_2$ , subcase 32 with $B_2\le 0<B_1$ , and subcase 33 with $0<B_2<B_1$ .

In subcase 31, the assumption $B_1\le B_2$ implies $B_1\le B_5$ by Lemma 4.5. From Lemma 4.1, noticing degenerated sets in (4.5) and (4.6), and in (4.28), we have $\sup{a_2\in[0,A_2]}\inf{a_1\in[0,A_1]}L^{a_1,a_2}\underline{V}=L^{A_1,\hat{a}_{2,A_1,\underline{V}}}\underline{V}$ in $\mathcal{R}_{1,-}$ . From Lemma 4.3, in the region $\mathcal{R}_{2,-}$ we have $\sup{a_2\in[0,A_2]}\inf{a_1\in[0,A_1]}L^{a_1,a_2}\underline{V}=L^{A_1,\hat{a}_{2,A_1,\underline{V}}}\underline{V}$ on $B_1\le\frac{\underline{V}''}{\underline{V}'}<B_5$ and $\sup{a_2\in[0,A_2]}\inf{a_1\in[0,A_1]}L^{a_1,a_2}\underline{V}=L^{A_1,A_2}\underline{V}$ on $B_5\le\frac{\underline{V}''}{\underline{V}'}$ . Thus, (4.20) holds and $\frac{\underline{V}''}{\underline{V}'}$ is given in (4.24) in the concave region.

In subcase 32, we have $B_1\leq B_4\le B_3$ . From $B_2\le 0<B_1$ , we have $B_5<B_1$ . From Lemma 4.1, we have $\sup{a_2\in[0,A_2]}\inf{a_1\in[0,A_1]}L^{a_1,a_2}\underline{V}=L^{A_1,\tilde{a}_{2}}\underline{V}$ in $\mathcal{R}_{1,-}$ . From Lemma (4.3), we have $\sup{a_2\in[0,A_2]}\inf{a_1\in[0,A_1]}L^{a_1,a_2}\underline{V}=L^{A_1,A_2}\underline{V}$ in $\mathcal{R}_{2,-}$ . Thus, we have

(4.41) \begin{equation}\underset{a_2\in[0,A_2]}{\sup}\underset{a_1\in[0,A_1]}{\mathrm{inf}}L^{a_1,a_2}\underline{V}=\begin{cases}L^{0,\tilde{a}_{2}}\underline{V}=L^{A_1,\tilde{a}_{2}}\underline{V}&\ \ \Big\{x\;:\;\le\frac{\underline{V}'}{\underline{V}''}\le B_1\Big\},\\[5pt] L^{A_1,A_2}\underline{V}&\ \ \Big\{x\;:\; B_1\le\frac{\underline{V}'}{\underline{V}''}\Big\},\end{cases}\end{equation}

and

(4.42) \begin{equation}\frac{\underline{V}''}{\underline{V}'}=\frac{1_{(-\infty,x_{1,-})}}{f_2}+g1_{[x_{1,-},x_0)},\end{equation}

which is a continuous, positive, and decreasing function in the concave region.

In subcase 33, we have $B_1\leq B_4\le B_3$ . From $0<B_2<B_1$ , we have $B_5<B_1$ . From Lemma 4.1, we have

\begin{equation*}\underset{a_2\in[0,A_2]}{\sup}\underset{a_1\in[0,A_1]}{\mathrm{inf}}L^{a_1,a_2}\underline{V}=\begin{cases}L^{A_1,\hat{a}_{2,A_1,\underline{V}}}\underline{V}&\ \ \{x\;:\;0<\hat{a}_{2,A_1,\underline{V}}<\tilde{a}_{1,\underline{V}}\}=\Big\{x\;:\;\frac{\underline{V}'}{\underline{V}''}<B_2\Big\},\\[5pt] L^{0,\tilde{a}_{2}}\underline{V}=L^{A_1,\tilde{a}_{2}}\underline{V}&\ \ \{x\;:\;\tilde{a}_{1,\underline{V}}\le\hat{a}_{2,A_1,\underline{V}}\}=\Big\{x\;:\;B_2\le\frac{\underline{V}'}{\underline{V}''}\Big\},\end{cases}\end{equation*}

in $\mathcal{R}_{2,-}$ . From Lemma (4.3), we have $\sup{a_2\in[0,A_2]}\inf{a_1\in[0,A_1]}L^{a_1,a_2}\underline{V}=L^{A_1,A_2}\underline{V}$ in $\mathcal{R}_{1,-}$ . Thus, we obtain

(4.43) \begin{equation}\underset{a_2\in[0,A_2]}{\sup}\underset{a_1\in[0,A_1]}{\mathrm{inf}}L^{a_1,a_2}\underline{V}=\begin{cases}L^{A_1,\hat{a}_{2,A_1,\underline{V}}}\underline{V}&\ \ \Big\{x\;:\;\frac{\underline{V}'}{\underline{V}''}<B_2\Big\},\\[5pt] L^{0,\tilde{a}_{2}}\underline{V}=L^{A_1,\tilde{a}_{2}}\underline{V}&\ \ \Big\{x\;:\;B_2\le\frac{\underline{V}'}{\underline{V}''}\le B_1\Big\},\\[5pt] L^{A_1,A_2}\underline{V}&\ \ \Big\{x\;:\; B_1\le\frac{\underline{V}'}{\underline{V}''}\Big\},\end{cases}\end{equation}

and

(4.44) \begin{equation}\frac{\underline{V}''}{\underline{V}'}=\frac{1_{(-\infty,x_{2,-})}}{f_1}+\frac{1_{(x_{2,-},x_{1,-})}}{f_2}+g1_{[x_{1,-},x_0)},\end{equation}

which is a continuous, positive, and decreasing function in the concave region $\mathcal{R}_{1,-}\cup\mathcal{R}_{2,-}=(\!-\!\infty,x_0)\cap[l, u]$ .

To solve case 4, we consider four subcases. We assume $\rho<\theta_1/\theta_2$ for the first three subcases: subcase 41 with $B_1\le B_2$ , subcase 42 with $B_2\le 0$ , and subcase 43 with $0<B_2<B_1$ . The fourth subcase is subcase 44 with $\rho=\theta_1/\theta_2$ .

In subcase 41, from $B_1\le B_2$ , we have $B_5\le B_1$ . From (4.12) and Lemma 4.3, equation (4.41) holds and the sub-value function satisfies (4.42) in the concave region.

In subcase 42, from $B_2\leq 0<B_1$ , we have $B_1<B_5$ . From (4.12) and Lemma 4.3, equation (4.20) holds and the sub-value function satisfies (4.24) in the concave region.

In subcase 43, from $0<B_2<B_1$ , we have $B_1<B_5$ . From (4.12) and Lemma 4.3, we have

(4.45) \begin{equation}\underset{a_2\in[0,A_2]}{\sup}\underset{a_1\in[0,A_1]}{\mathrm{inf}}L^{a_1,a_2}\underline{V}=\begin{cases}L^{0,\tilde{a}_{2}}\underline{V}=L^{A_1,\tilde{a}_{2}}\underline{V}&\ \ \Big\{x\;:\;\frac{\underline{V}'}{\underline{V}''}<B_2\Big\},\\[5pt] L^{A_1,\hat{a}_{2,A_1,\underline{V}}}\underline{V}&\ \ \Big\{x\;:\;B_2\le\frac{\underline{V}'}{\underline{V}''}\le B_5\Big\},\\[5pt] L^{A_1,A_2}&\ \ \Big\{x\;:\; B_5\le\frac{\underline{V}'}{\underline{V}''}\Big\},\end{cases}\end{equation}

and

(4.46) \begin{equation}\frac{\underline{V}''}{\underline{V}'}=\frac{1_{(-\infty,x_{2,-}]}}{f_2}+\frac{1_{(x_{2,-},x_{5,-})}}{f_1}+g1_{[x_{5,-},x_0)},\end{equation}

which is a continuous, positive, and decreasing function in the concave region.

In subcase 44, we have $\rho=\theta_1/\theta_2$ . If $\rho<\sqrt{1/2}$ , we have $B_1<B_5$ . From (4.13) and Lemma 4.3, equation (4.20) holds and the sub-value function is given by (4.24) in the concave region. If $\sqrt{1/2}\le\rho$ , we have $B_5\le B_1$ . From (4.13) and Lemma 4.3, equation (4.41) holds and $\frac{\underline{V}''}{\underline{V}'}$ is given by (4.42) in the concave region.

We have thus found $\frac{\underline{V}''}{\underline{V}'}$ and the optimizers explicitly under all possible parameter cases for the sub-value function in the concave region. The results are summarized in Table 1 of Appendix D.

4.2. The convex region with $\underline{\boldsymbol{V}}''>0$

In this region, we have $\frac{\underline{V}''}{\underline{V}'}<0$ . The quadratic function $L^{a_1,a_2}\underline{V}$ (in $a_1$ ) has a positive coefficient on the second-degree term, and the vertex point $\hat{a}_{1,a_2,\underline{V}}(x)=-\theta_1\frac{\underline{V}'}{\underline{V}''}+\rho a_2$ is positive. If the vertex point satisfies $\hat{a}_{1,a_2,\underline{V}}(x)=-\theta_1\frac{\underline{V}'}{\underline{V}''}+\rho a_2<A_1$ , which is equivalent to $a_2\in(\!-\!\infty,\acute{a}_{2,\underline{V}})\cap[0,A_2]$ , where

(4.47) \begin{equation}\acute{a}_{2,\underline{V}}=\frac{\theta_1}{\rho}\frac{\underline{V}'}{\underline{V}''}+\frac{A_1}{\rho},\end{equation}

then the minimizer of $L^{a_1,a_2}\underline{V}$ in $a_1$ is $\underline{a}^*(x,a_2)=\hat{a}_{1,a_2,\underline{V}}(x)$ . If $A_1\le \hat{a}_{1,a_2,\underline{V}}(x)$ , that is, $a_2\in[\acute{a}_{2,\underline{V}}, \infty)\cap[0,A_2]$ , the minimizer in $a_1$ is $\underline{a}^*(x,a_2)=A_1$ . Below we write $\acute{a}_2=\acute{a}_{2,\underline{V}}$ for convenience. The minimizer $\underline{a}^*(x,a_2)$ of $L^{a_1,a_2}\underline{V}$ in $a_1$ given $a_2\in[0,A_2]$ takes the form

(4.48) \begin{equation}\underline{a}^*(x,a_2)=\begin{cases}A_1 & \acute{a}_2\le 0\ \ \ \ \ \ \ \ \ \ \ a_2\in[0, A_2],\\[5pt] A_1 & 0<\acute{a}_2<A_2\ \ \ a_2\in[\acute{a}_2, A_2],\\[5pt] \hat{a}_{1,a_2,\underline{V}}& 0<\acute{a}_2<A_2\ \ \ a_2\in[0, \acute{a}_2],\\[5pt] \hat{a}_{1,a_2,\underline{V}}& A_2\le\acute{a}_2\ \ \ \ \ \ \ \ \ a_2\in[0, A_2], \end{cases}\end{equation}

which assumes nonzero values. Note that $\hat{a}_{1,\acute{a}_2,\underline{V}}=A_1$ .

We now define three regions which form a partition of the convex region:

(4.49) \begin{equation}\begin{split}&\mathcal{R}_{1,+}=\{x\in(l,u)\;:\;\underline{V}''>0, \acute{a}_{2}\le 0\}=\Bigg\{x\in(l,u)\;:\;\underline{V}''>0, \frac{\underline{V}'}{\underline{V}''}\le B_6\Bigg\},\\[5pt] &\mathcal{R}_{2,+}=\{x\in(l,u)\;:\;\underline{V}''>0, 0<\acute{a}_{2}<A_2\}=\Bigg\{x\in(l,u)\;:\;\underline{V}''>0, B_6<\frac{\underline{V}'}{\underline{V}''}<B_7\Bigg\},\\[5pt] &\mathcal{R}_{3,+}=\{x\in(l,u)\;:\;\underline{V}''>0, A_2\le\acute{a}_{2}\}=\Bigg\{x\in(l,u)\;:\;\underline{V}''>0, B_7\le\frac{\underline{V}'}{\underline{V}''}\Bigg\},\end{split}\end{equation}

where

(4.50) \begin{equation}B_6=-\frac{A_1}{\theta_1},\quad B_7=\frac{A_2\rho-A_1}{\theta_1}.\end{equation}

We note that $B_6<B_7$ and the region $\mathcal{R}_{3,+}$ degenerates when $B_7\geq 0$ . Next we show results to determine the maximizer in $a_2$ of $L^{\underline{a}^*(x,a_2),a_2}\underline{V}$ in these three regions.

For $x\in\mathcal{R}_{1,+}$ , the minimizer in $a_1$ is equal to $\underline{a}^*(x,a_2)=A_1$ . The vertex point of $L^{A_1,a_2}$ in $a_2$ is $\hat{a}_{2,A_1,\underline{V}}=\theta_2\frac{\underline{V}'}{\underline{V}''}+\rho A_1$ , and the maximizer in $a_2$ is 0 or $A_2$ . We obtain the following lemma.

Lemma 4.6. For $x\in\mathcal{R}_{1,+}$ , we have

(4.51) \begin{equation}\underset{a_2\in[0,A_2]}{\sup}\underset{a_1\in[0,A_1]}{\mathrm{inf}}L^{a_1,a_2}\underline{V}=\begin{cases}L^{A_1,A_2}\underline{V}&\ \ \Big\{x\;:\;\hat{a}_{2,A_1,\underline{V}}\le\frac{A_2}{2}\Big\}=\Big\{x\;:\;\frac{\underline{V}'}{\underline{V}''}\le B_8\Big\},\\[5pt] L^{A_1,0}\underline{V}&\ \ \Big\{x\;:\;\frac{A_2}{2}<\hat{a}_{2,A_1\underline{V}}\Big\}=\Big\{x\;:\;B_8<\frac{\underline{V}'}{\underline{V}''}\Big\},\end{cases}\end{equation}

where

(4.52) \begin{equation}B_8=(A_2/2-\rho A_1)/\theta_2.\end{equation}

In $\mathcal{R}_{2,+}$ , from (4.48), if $a_2\in[0,\acute{a}_{2})$ , the minimizer in $a_1$ is $\underline{a}^*(x,a_2)=\hat{a}_{1,a_2,\underline{V}}$ . The function $L^{\hat{a}_{1,a_2,\underline{V}},a_2}\underline{V}$ simplifies to

(4.53) \begin{equation}L^{\hat{a}_{1,a_2,\underline{V}},a_2}\underline{V}=[rx+\mu+(\rho\theta_1-\theta_2)a_2)]\underline{V}'-\frac{\theta_1^2}{2}\frac{\underline{V}'^2}{\underline{V}''}+\frac{1}{2}\big[(1-\rho^2)a_2^2+\sigma^2\big]\underline{V}'',\end{equation}

which is a quadratic function in $a_2$ with a positive coefficient on the second-degree term, and its vertex point in $a_2$ is given by

(4.54) \begin{equation}\breve{a}_{2,\underline{V}}=-\frac{\rho\theta_1-\theta_2}{1-\rho^2}\frac{\underline{V}'}{\underline{V}''},\end{equation}

which is positive when $\frac{1}{\rho}<\frac{\theta_1}{\theta_2}$ in the convex region. The maximizer in $a_2\in[0,\acute{a}_{2}]$ of $L^{\hat{a}_{1,a_2,\underline{V}},a_2}\underline{V}$ is 0 or $\acute{a}_{2}$ . If $a_2\in [\acute{a}_{2},A_2]$ , the minimizer in $a_1$ is $\underline{a}^*(x,a_2)=A_1$ . The vertex point of $L^{A_1,a_2}\underline{V}$ in $a_2$ is $\hat{a}_{2,A_1,\underline{V}}$ , and the maximizer in $a_2\in[\acute{a}_{2}, A_2]$ of $L^{A_1,a_2}\underline{V}$ is $\acute{a}_{2}$ or $A_2$ . We obtain the following lemma.

Lemma 4.7. For $x\in\mathcal{R}_{2,+}$ , if $\frac{1}{\rho}<\frac{\theta_1}{\theta_2}$ , we have

(4.55) \begin{align}\underset{a_2\in[0,\acute{a}_2)}{\sup}\underset{a_1\in[0,A_1]}{\mathrm{inf}}L^{a_1,a_2}\underline{V}=\begin{cases}L^{\hat{a}_{1,0,\underline{V}},0}\underline{V}&\ \ \{x\;:\;\acute{a}_2/2\le\breve{a}_{2,\underline{V}}\}=\Big\{x\;:\;\frac{\underline{V}'}{\underline{V}''}\le B_9\Big\},\\[9pt] L^{\hat{a}_{1,\acute{a}_2,\underline{V}},\acute{a}_2}\underline{V}=L^{A_1,\acute{a}_2}\underline{V}&\ \ \{x\;:\;\breve{a}_{2,\underline{V}}<\acute{a}_2/2\}=\Big\{x\;:\;B_9<\frac{\underline{V}'}{\underline{V}''}\Big\},\end{cases}\nonumber\\[2pt]\end{align}

where

(4.56) \begin{equation}B_9=-\frac{(1-\rho^2)A_1}{\theta_1(1+\rho^2)-2\rho\theta_2};\end{equation}

for $x\in\mathcal{R}_{2,+}$ , if $\frac{\theta_1}{\theta_2}\le\frac{1}{\rho}$ , we have

(4.57) \begin{equation}\underset{a_2\in[0,\acute{a}_2]}{\sup}\underset{a_1\in[0,A_1]}{\mathrm{inf}}L^{a_1,a_2}\underline{V}=L^{\hat{a}_{1,\acute{a},\underline{V}},\acute{a}_2}\underline{V}=L^{A_1,\acute{a}_2}\underline{V}.\end{equation}

For $x\in\mathcal{R}_{2,+}$ , if $2\rho<\frac{\theta_1}{\theta_2}$ , we have

(4.58) \begin{align}\underset{a_2\in[\acute{a}_2,A_2]}{\sup}\underset{a_1\in[0,A_1]}{\mathrm{inf}}L^{a_1,a_2}\underline{V}=\begin{cases}L^{A_1,\acute{a}_2}\underline{V}&\ \ \{x\;:\;(\acute{a}_2+A_2)/2\le\hat{a}_{2,A_1,\underline{V}}\}=\Big\{x\;:\;\frac{\underline{V}'}{\underline{V}''}\le B_{10}\Big\},\\[5pt] L^{A_1,A_2}\underline{V}&\ \ \{x\;:\;\hat{a}_{2,A_1,\underline{V}}<(\acute{a}_2+A_2)/2\}=\Big\{x\;:\;B_{10}<\frac{\underline{V}'}{\underline{V}''}\Big\},\end{cases}\nonumber\\[2pt]\end{align}

where

(4.59) \begin{equation}B_{10}=-\frac{\rho A_2+(1-2\rho^2)A_1}{\theta_1-2\rho\theta_2};\end{equation}

for $x\in\mathcal{R}_{2,+}$ , if $\frac{\theta_1}{\theta_2}<{2\rho}$ , we have

(4.60) \begin{align}\underset{a_2\in[\acute{a}_2,A_2]}{\sup}\underset{a_1\in[0,A_1]}{\mathrm{inf}}L^{a_1,a_2}\underline{V}=\begin{cases}L^{A_1,\acute{a}_2}\underline{V}&\ \ \{x\;:\;(\acute{a}_2+A_2)/2\le\hat{a}_{2,A_1,\underline{V}}\}=\Big\{x\;:\;B_{10}\le\frac{\underline{V}'}{\underline{V}''}\Big\},\\[8pt] L^{A_1,A_2}\underline{V}&\ \ \{x\;:\;\hat{a}_{2,A_1,\underline{V}}<(\acute{a}_2+A_2)/2\}=\Big\{x\;:\;\frac{\underline{V}'}{\underline{V}''}<B_{10}\Big\};\end{cases}\nonumber\\[2pt]\end{align}

for $x\in\mathcal{R}_{2,+}$ , if $\frac{\theta_1}{\theta_2}={2\rho}$ , we have

(4.61) \begin{equation}\underset{a_2\in[\acute{a}_2,A_2]}{\sup}\underset{a_1\in[0,A_1]}{\mathrm{inf}}L^{a_1,a_2}\underline{V}=\begin{cases}L^{A_1,\acute{a}_2}\underline{V}&\ \ \ A_2+\big(\frac{1}{\rho}-2\rho\big)A_1\le 0,\\[5pt] L^{A_1,A_2}\underline{V}&\ \ \ A_2+\big(\frac{1}{\rho}-2\rho\big)A_1>0.\end{cases}\end{equation}

To show (4.57), we note that $B_9\le B_6$ if $\frac{2\rho}{1+\rho^2}<\frac{\theta_1}{\theta_2}\le\frac{1}{\rho}$ and the set $\{x\;:\;\acute{a}_2/2\le\breve{a}_{2,\underline{V}}\}$ degenerates in $\mathcal{R}_{2,+}$ .

In $\mathcal{R}_{3,+}$ , from (4.48), the minimizer of $L^{a_1,a_2}\underline{V}$ in $a_1$ is $\underline{a}^*(x,a_2)=\hat{a}_{1,a_2,\underline{V}}$ for all $a_2\in[0,A_2]$ . From (4.53), we note that $L^{\hat{a}_{1,a_2,\underline{V}},a_2}\underline{V}$ is a quadratic function of $a_2$ and its vertex point is $\breve{a}_{2,\underline{V}}$ in (4.54). The maximizer in $a_2\in[0,A_2]$ of $L^{\hat{a}_{1,a_2,\underline{V}},a_2}\underline{V}$ is 0 or $A_2$ . We obtain the following lemma.

Lemma 4.8. For $x\in\mathcal{R}_{3,+}$ , if $\frac{\theta_1}{\theta_2}>\frac{1}{\rho}$ , we have

(4.62) \begin{equation}\underset{a_2\in[0,A_2]}{\sup}\underset{a_1\in[0,A_1]}{\mathrm{inf}}L^{a_1,a_2}\underline{V}=\begin{cases}L^{\hat{a}_{1,0,\underline{V}},0}\underline{V}&\ \ \Big\{x\;:\;\breve{a}_{1,\underline{V}}\ge\frac{A_2}{2}\Big\}=\Big\{x\;:\;\frac{\underline{V}'}{\underline{V}''}\le B_{11}\Big\},\\[5pt] L^{\hat{a}_{1,A_2,\underline{V}},A_2}\underline{V}&\ \ \Big\{x\;:\;\breve{a}_{1,\underline{V}}<\frac{A_2}{2}\Big\}=\Big\{x\;:\;B_{11}<\frac{\underline{V}'}{\underline{V}''}\Big\},\end{cases}\end{equation}

where

(4.63) \begin{equation}B_{11}=-\frac{(1-\rho^2)A_2}{2(\rho\theta_1-\theta_2)};\end{equation}

for $x\in\mathcal{R}_{3,+}$ , if $\frac{\theta_1}{\theta_2}\le\frac{1}{\rho}$ , we have

(4.64) \begin{equation}\underset{a_2\in[0,A_2]}{\sup}\underset{a_1\in[0,A_1]}{\mathrm{inf}}L^{a_1,a_2}\underline{V}=L^{\hat{a}_{1,A_2,\underline{V}},A_2}\underline{V}.\end{equation}

In the sequel, we find $\frac{\underline{V}''}{\underline{V}'}$ in the convex region via the following parameter cases:

1. 1. $2\rho<\frac{\theta_1}{\theta_2}\le\frac{1}{\rho}$ ;

2. 2. $\frac{\theta_1}{\theta_2}\le\frac{1}{\rho}, 2\rho$ and $A_2+(\frac{1}{\rho}-2\rho)A_1>0$ ;

3. 3. $\frac{\theta_1}{\theta_2}\le\frac{1}{\rho}, 2\rho$ and $A_2+(\frac{1}{\rho}-2\rho)A_1\le 0$ ;

4. 4. $\frac{1}{\rho}<\frac{\theta_1}{\theta_2}\le 2\rho$ and $A_2+(\frac{1}{\rho}-2\rho)A_1>0$ ;

5. 5. $\frac{1}{\rho}<\frac{\theta_1}{\theta_2}\le 2\rho$ and $A_2+(\frac{1}{\rho}-2\rho)A_1\le 0$ ;

6. 6. $\frac{1}{\rho}, 2\rho<\frac{\theta_1}{\theta_2}$ and $A_2+(\frac{1}{\rho}-2\rho)A_1>0$ ;

7. 7. $\frac{1}{\rho}, 2\rho<\frac{\theta_1}{\theta_2}$ and $A_2+(\frac{1}{\rho}-2\rho)A_1\le 0$ .

In case 1, the parameter conditions imply $B_6<B_8$ and $B_{10}<B_6$ . From Lemma 4.6 and $B_6<B_8$ , we have $\sup{a_2\in[0,A_2]}\inf{a_1\in[0,A_1]}L^{a_1,a_2}\underline{V}=L^{A_1,A_2}\underline{V}$ in $\mathcal{R}_{1,+}$ . From (4.57), (4.58) and $B_{10}<B_6$ , we have $\sup{a_2\in[0,A_2]}\inf{a_1\in[0,A_1]}L^{a_1,a_2}\underline{V}=L^{A_1,A_2}\underline{V}$ in $\mathcal{R}_{2,+}$ . In addition, from (4.64), if $B_7<0$ , we have $\sup{a_2\in[0,A_2]}\inf{a_1\in[0,A_1]}L^{a_1,a_2}\underline{V}=L^{\hat{a}_{1,A_2,\underline{V}},A_2}\underline{V}$ in $\mathcal{R}_{3,+}=\{x\;:\;B_{7}<\frac{\underline{V}'}{\underline{V}''}\}$ . We obtain

(4.65) \begin{equation}\underset{a_2\in[0,A_2]}{\sup}\underset{a_1\in[0,A_1]}{\mathrm{inf}}L^{a_1,a_2}\underline{V}=\begin{cases}L^{A_1,A_2}\underline{V}&\ \ \mathcal{R}_{1,+}\cup\mathcal{R}_{2,+}=\Big\{x\;:\;\frac{\underline{V}'}{\underline{V}''}\le B_{7}\Big\},\\[8pt] L^{\hat{a}_{1,A_2,\underline{V}},A_2}\underline{V}&\ \ \mathcal{R}_{3,+}=\Big\{x\;:\;B_{7}<\frac{\underline{V}'}{\underline{V}''}\Big\},\end{cases}\end{equation}

in the convex region. From (4.53), this yields

(4.66) \begin{align}L^{\hat{a}_{1,A_2,\underline{V}},A_2}\underline{V}=-\underline{V}''\left\{\frac{\theta_1^2}{2}\left(\frac{\underline{V}'}{\underline{V}''}\right)^2-[rx+\mu+(\rho\theta_1-\theta_2)A_2]\frac{\underline{V}'}{\underline{V}''}-\frac{1}{2}\big[\big(1-\rho^2\big)A_2^2+\sigma^2\big]\right\}.\nonumber\\[2pt]\end{align}

Thus, the equation $L^{\hat{a}_{1,A_2,\underline{V}},A_2}\underline{V}=0$ yields $\frac{\underline{V}'}{\underline{V}''}=f_5$ , where

(4.67) \begin{equation}f_5(x)=\frac{rx+\mu+(\rho\theta_1-\theta_2)A_2-\sqrt{[rx+\mu+(\rho\theta_1-\theta_2)A_2]^2+\theta_1^2\big[(1-\rho^2)A_2^2+\sigma^2\big]}}{\theta_1^2},\end{equation}

which is a negative and increasing function. The inequality $f_5(x)>B_7$ gives $x>x_{3,+}$ , where

(4.68) \begin{equation}x_{3,+}=\frac{B_7^2\theta_1^2-\sigma^2-(1-\rho^2)A_2^2}{2rB_7}-\frac{(\rho\theta_1-\theta_2)A_2+\mu}{r}.\end{equation}

The inequality $1/B_7\le g(x)<0$ gives $x_0<x<x_{3,+}$ . Thus, in case 1, if $B_7<0$ (i.e. $\frac{A_1}{A_2}<\rho$ ), we have

(4.69) \begin{equation}\frac{\underline{V}''}{\underline{V}'}=g1_{(x_0,x_{3,+}]}+\frac{1_{(x_{3,+},\infty)}}{f_5},\end{equation}

which is a continuous, negative, and decreasing function in the convex region $\mathcal{R}_{1,+}\cup\mathcal{R}_{2,+}\cup\mathcal{R}_{3,+}=(x_0,\infty)\cap[u,l]$ . If $B_7\ge 0$ (i.e. $\rho\le\frac{A_1}{A_2}$ ), we have $\mathcal{R}_{3,+}=\emptyset$ ; so we have

(4.70) \begin{equation}\underset{a_2\in[0,A_2]}{\sup}\underset{a_1\in[0,A_1]}{\mathrm{inf}}L^{a_1,a_2}\underline{V}=L^{A_1,A_2}\underline{V}\end{equation}

and

(4.71) \begin{equation}\frac{\underline{V}''}{\underline{V}'}=g1_{(x_0,\infty)},\end{equation}

which is a continuous, negative, and decreasing function in the convex region.

In case 2, we have $B_6<B_8$ and $B_{10}< 0$ (if $\frac{\theta_1}{\theta_2}<2\rho$ ). From Lemma 4.6, (4.57), (4.60) or (4.61), and (4.64), we obtain the same expression for $\frac{\underline{V}''}{\underline{V}'}$ as in case 1.

In case 3, we have $0<A_2/A_1\leq 2\rho-1/\rho$ and $\frac{\theta_1}{\theta_2}\le 1/\rho<2\rho$ . We can show that $B_6<B_8$ , $B_6<B_{10}$ , and $B_7<B_{10}\le 0$ . From Lemma 4.6 and $B_6<B_8$ , we have $\sup{a_2\in[0,A_2]}\inf{a_1\in[0,A_1]}L^{a_1,a_2}\underline{V}=L^{A_1,A_2}\underline{V}$ in $\mathcal{R}_{1,+}$ . From (4.57) and (4.60), we obtain $\sup{a_2\in[0,A_2]}\inf{a_1\in[0,A_1]}L^{a_1,a_2}\underline{V}=L^{A_1,A_2}\underline{V}$ in $\mathcal{R}_{2,+}$ . From (4.64), we have $\sup{a_2\in[0,A_2]}\inf{a_1\in[0,A_1]}L^{a_1,a_2}\underline{V}=L^{\hat{a}_{1,A_2,\underline{V}},A_2}\underline{V}$ in $\mathcal{R}_{3,+}$ . Thus, we conclude that $\frac{\underline{V}''}{\underline{V}'}$ takes the form (4.69) with the optimizers given in (4.65) in the convex region in case 3.

For the remaining parameter cases, we have $\frac{1}{\rho}<\frac{\theta_1}{\theta_2}$ , under which the following quantities are all positive:

(4.72) \begin{align}\xi_1=\frac{\theta_1}{2(\rho\theta_1-\theta_2)},\quad\xi_2=\frac{(1+\rho^2)\theta_1-2\rho\theta_2}{2(\rho\theta_1-\theta_2)},\quad\xi_3=\frac{\theta_1-\rho\theta_2}{\rho\theta_1-\theta_2},\quad\xi_4=\frac{(1+\rho^2)\theta_1-2\rho\theta_2}{2\rho^2(\rho\theta_1-\theta_2)}.\nonumber\\[2pt]\end{align}

We give the following lemma which can be shown by simple algebra.

In case 4, from Lemma 4.9, we have $\xi_2<\xi_4<\xi_3<\xi_1$ . From $A_2+(\frac{1}{\rho}-2\rho)A_1>0$ , we have $\frac{A_1}{A_2}<\frac{1}{2\rho-1/\rho}<\xi_4$ , where the second inequality is equivalent to $\frac{\theta_1}{\theta_2}<2\rho$ . We consider two subcases: subcase 41 with $\xi_2<\frac{A_1}{A_2}({<}\xi_4<\xi_3<\xi_1)$ and subcase 42 with $\frac{A_1}{A_2}\le\xi_2({<}\xi_4<\xi_3<\xi_1)$ .

In subcase 41, applying Lemma 4.9, we have $B_6<B_8$ , $B_7\le B_{11}$ , and $B_6<B_7<B_9<0<B_{10}$ . From Lemma 4.6 and $B_6<B_8$ , we have

(4.73) \begin{equation}\underset{a_2\in[0,A_2]}{\sup}\underset{a_1\in[0,A_1]}{\mathrm{inf}}L^{a_1,a_2}\underline{V}=L^{A_1,A_2}\underline{V},\end{equation}

in $\mathcal{R}_{1,+}$ . From (4.5), (4.11), and $B_6<B_7<B_9<0<B_{10}$ , we have $\sup{a_2\in[0,A_2]}\inf{a_1\in[0,A_1]}L^{a_1,a_2}\underline{V}=\max\{L^{A_1,A_2}\underline{V}, L^{\hat{a}_{1,0,\underline{V}},0}\underline{V}\}$ in $\mathcal{R}_{2,+}$ . From

(4.74) \begin{equation}L^{\hat{a}_{1,0,\underline{V}},0}\underline{V}=\underline{V}''\left[-\frac{\theta_1^2}{2}\left(\frac{\underline{V}'}{\underline{V}''}\right)^2+(rx+\mu)\frac{\underline{V}'}{\underline{V}''}-\frac{1}{2}\sigma^2\right]\end{equation}

we obtain

(4.75) \begin{equation}L^{\hat{a}_{1,0,\underline{V}},0}\underline{V}-L^{A_1,A_2}\underline{V}=\underline{V}''G\left(\frac{\underline{V}'}{\underline{V}''}\right)\!,\end{equation}

where G is a quadratic function given by

(4.76) \begin{equation}G(A)\;:\!=\;-\frac{\theta_1^2}{2}A^2+(\theta_2A_2-\theta_1A_1)A-\frac{1}{2}\big(A_1^2+A_2^2-2\rho A_1A_2\big).\end{equation}

We have $L^{\hat{a}_{1,0,\underline{V}},0}\underline{V}-L^{A_1,A_2}\underline{V}>0$ if and only if $A_-<\frac{\underline{V}'}{\underline{V}''}<A_+$ , where

(4.77) \begin{equation}A_+,A_-=\frac{\theta_2A_2-\theta_1A_1\pm\sqrt{(\theta_2A_2-\theta_1A_1)^2-\theta_1^2\big(A_1^2+A_2^2-2\rho A_1A_2\big)}}{\theta_1^2}\end{equation}

are roots of the quadratic equation $G(A)=0$ , whose discriminant is positive if and only if $\frac{A_1}{A_2}>\xi_5$ (provided $\frac{1}{\rho}<\frac{\theta_1}{\theta_2}$ ), where

(4.78) \begin{equation}\xi_5=\frac{\theta_1^2-\theta_2^2}{2\theta_1(\rho\theta_1-\theta_2)}.\end{equation}

By simple computations, we give the following lemmas.

To show (ii), we note that $\frac{A_1}{A_2}>\xi_5>\frac{\theta_2}{\theta_1}$ , which yields $\theta_2A_2-\theta_1A_1<0$ .

The inequality $\frac{A_-+A_+}{2}<B_9$ is equivalent to $\frac{\theta_2\xi_2}{\rho\theta_1}<\frac{A_1}{A_2}$ . It holds by showing $\frac{\theta_2\xi_2}{\rho\theta_1}<\xi_5$ , which is equivalent to $(\frac{\theta_1}{\theta_2}-\rho)(\frac{\theta_1}{\theta_2}-\frac{1}{\rho})>0$ . The other two results in Lemma 4.11 immediately follow:

\begin{equation*}\begin{split}G(B_9)&=-\frac{2\rho^2(\rho\theta_1-\theta_2)^2A_2^2}{[\theta_1(1+\rho^2)-2\rho\theta_2]^2}\left(\frac{A_1}{A_2}-\xi_2\right)\left(\frac{A_1}{A_2}-\xi_4\right)\!,\\[5pt] G(B_{10})&=-\frac{2\rho^2(\rho\theta_1-\theta_2)^2A_2^2}{(\theta_1-2\rho\theta_2)^2}\left(\frac{A_1}{A_2}-\xi_1\right)\left(\frac{A_1}{A_2}-\xi_4\right)\!.\end{split}\end{equation*}

From Lemma 4.10, we see that $A_-, A_+$ exist and we have $B_6 < A_- < B_7 < A_+$ in subcase 41. Thus, we have

(4.79) \begin{align}\underset{a_2\in[0,A_2]}{\sup}\underset{a_1\in[0,A_1]}{\mathrm{inf}}L^{a_1,a_2}\underline{V}=\max\big\{L^{A_1,A_2}\underline{V}, L^{\hat{a}_{1, 0, \underline{V}}, 0}\underline{V}\big\}=\begin{cases}L^{A_1,A_2}\underline{V}&\ \ \Big\{x\;:\;B_6<\frac{\underline{V}'}{\underline{V}''}\le A_-\Big\},\\[8pt] L^{\hat{a}_{1,0,\underline{V}},0}\underline{V}&\ \ \Big\{x\;:\;A_-<\frac{\underline{V}'}{\underline{V}''}<B_7\Big\},\end{cases}\nonumber\\[2pt]\end{align}

in $\mathcal{R}_{2,+}$ . From (4.62) and $B_7<B_{11}$ , we have

(4.80) \begin{equation}\underset{a_2\in[0,A_2]}{\sup}\underset{a_1\in[0,A_1]}{\mathrm{inf}}L^{a_1,a_2}\underline{V}=\begin{cases}L^{\hat{a}_{1,0,\underline{V}},0}\underline{V}&\ \ \Big\{x\;:\; B_7\le\frac{\underline{V}'}{\underline{V}''}\le B_{11}\Big\},\\[8pt] L^{\hat{a}_{1,A_2,\underline{V}},A_2}\underline{V}&\ \ \Big\{x\;:\;B_{11}<\frac{\underline{V}'}{\underline{V}''}\Big\},\end{cases}\end{equation}

in $\mathcal{R}_{3,+}$ . Thus, from (4.73), (4.79), and (4.80), we obtain

(4.81) \begin{equation}\underset{a_2\in[0,A_2]}{\sup}\underset{a_1\in[0,A_1]}{\mathrm{inf}}L^{a_1,a_2}\underline{V}=\begin{cases}L^{A_1,A_2}\underline{V}&\ \ \Big\{x\;:\;\frac{\underline{V}'}{\underline{V}''}\le A_-\Big\},\\[5pt] L^{\hat{a}_{1,0,\underline{V}},0}\underline{V}&\ \ \Big\{x\;:\; A_-\le\frac{\underline{V}'}{\underline{V}''}\le B_{11}\Big\},\\[5pt] L^{\hat{a}_{1,A_2,\underline{V}},A_2}\underline{V}&\ \ \Big\{x\;:\;B_{11}<\frac{\underline{V}'}{\underline{V}''}\Big\},\end{cases}\end{equation}

in the convex region. The inequality $1/A_-<g(x)<0$ gives $x_0<x<x_{7,+}$ , where

(4.82) \begin{equation}x_{7,+}=-\frac{A_1^2+A_2^2-2\rho A_1A_2+\sigma^2}{2rA_-}-\frac{\theta_1A_1-\theta_2A_2+\mu}{r}.\end{equation}

From (4.74), the equation $L^{\hat{a}_{1,0,u{V}},0}=0$ yields $\frac{\underline{V}'}{\underline{V}''}=f_4$ , where

(4.83) \begin{equation}f_4(x)=\frac{rx+\mu-\sqrt{(rx+\mu)^2+\sigma^2\theta_1^2}}{\theta_1^2}.\end{equation}

The inequality $A_-\le f_4(x)<B_{11}$ gives $x_{7,+}\le x<x_{1,+}$ , where

(4.84) \begin{equation}x_{1,+}=-\frac{\sigma^2+(1-\rho^2)A_2^2-\theta_1^2B_{11}^2}{2B_{11}}-(\rho\theta_1-\theta_2)A_2.\end{equation}

And the inequality $f_5\ge B_{11}$ gives $x\ge x_{1,+}$ , where $f_5$ is given by (4.67). Thus, we obtain

(4.85) \begin{equation}\frac{\underline{V}''}{\underline{V}'}=g1_{(x_0,x_{7,+})}+\frac{1_{[x_{7,+},x_{1,+})}}{f_4}+\frac{1_{[x_{1,+},\infty)}}{f_5},\end{equation}

which is a continuous, negative, and decreasing function in the convex region in subcase 41.

In subcase 42 with $\frac{A_1}{A_2}\le\xi_2(<\xi_4<\xi_3<\xi_1)$ , applying Lemmas 4.9 and 4.10, we have $B_6<B_8$ , $B_{11}<B_7$ , and $B_6<B_9<B_7<B_{10}$ . We also have $B_6<A_-<A_+<B_9<B_7$ (if $\xi_5<\frac{A_1}{A_2}$ ) by Lemma 4.11. From Lemma 4.6, (4.55), (4.60), (4.62), and (4.75), we have

(4.86) \begin{equation}\underset{a_2\in[0,A_2]}{\sup}\underset{a_1\in[0,A_1]}{\mathrm{inf}}L^{a_1,a_2}\underline{V}=\begin{cases}L^{A_1,A_2}\underline{V}&\ \ \Big\{x\;:\; \frac{\underline{V}'}{\underline{V}''}\le A_-\Big\},\\[5pt] L^{\hat{a}_{1,0,\underline{V}},0}\underline{V}&\ \ \Big\{x\;:\;A_-<\frac{\underline{V}'}{\underline{V}''}<A_+\Big\},\\[5pt] L^{A_1,A_2}\underline{V}&\ \ \Big\{x\;:\; A_+\le\frac{\underline{V}'}{\underline{V}''}< B_7\Big\},\\[5pt] L^{\hat{a}_{1,A_2,\underline{V}},A_2}\underline{V}&\ \ \Big\{x\;:\; B_7\le\frac{\underline{V}'}{\underline{V}''}\Big\},\end{cases}\end{equation}

and

(4.87) \begin{equation}\frac{\underline{V}''}{\underline{V}'}=g1_{(x_0,x_{7,+})}+\frac{1_{[x_{7,+},x_{8,+})}}{f_4}+g1_{[x_{8,+},x_{3,+})}+\frac{1_{[x_{3,+},\infty)}}{f_5},\end{equation}

which is a continuous, negative, and decreasing function in the convex region if $\frac{A_1}{A_2}>\xi_5$ and $B_7<0$ in subcase 42, where

(4.88) \begin{equation}x_{8,+}=-\frac{A_1^2+A_2^2-2\rho A_1A_2+\sigma^2}{2rA_+}-\frac{\theta_1A_1-\theta_2A_2+\mu}{r}.\end{equation}

If $\frac{A_1}{A_2}>\xi_5$ and $B_7\ge 0$ , the region $\mathcal{R}_{3,+}$ degenerates and we obtain

(4.89) \begin{equation}\underset{a_2\in[0,A_2]}{\sup}\underset{a_1\in[0,A_1]}{\mathrm{inf}}L^{a_1,a_2}\underline{V}=\begin{cases}L^{A_1,A_2}\underline{V}&\ \ \Big\{x\;:\; \frac{\underline{V}'}{\underline{V}''}\le A_-\Big\},\\[5pt] L^{\hat{a}_{1,0,\underline{V}},0}\underline{V}&\ \ \Big\{x\;:\;A_-<\frac{\underline{V}'}{\underline{V}''}<A_+\Big\},\\[5pt] L^{A_1,A_2}\underline{V}&\ \ \Big\{x\;:\; A_+\le\frac{\underline{V}'}{\underline{V}''}\Big\},\end{cases}\end{equation}

and

(4.90) \begin{equation}\frac{\underline{V}''}{\underline{V}'}=g1_{(x_0,x_{7,+})}+\frac{1_{[x_{7,+},x_{8,+})}}{f_4}+g1_{[x_{8,+},\infty)},\end{equation}

which is a continuous, negative, and decreasing function in the convex region if $\frac{A_1}{A_2}>\xi_5$ and $B_7\ge 0$ in subcase 42. We note that if $\frac{A_1}{A_2}\le \xi_5$ , we have $L^{A_1,A_2}\underline{V}\ge L^{\hat{a}_{1,0,\underline{V}},0}$ . Thus, we have the same expression for $\frac{\underline{V}''}{\underline{V}'}$ as in (4.69) when $B_7<0$ and as in (4.71) when $B_7\ge 0$ .

In case 5, we have $\xi_5<\xi_2<\xi_4<\xi_3<\xi_1$ . From $A_2+(\frac{1}{\rho}-2\rho)A_1\le 0$ we obtain $B_{10}\le 0$ , and from $\frac{A_1}{A_2}>\frac{\rho}{2\rho^2-1}>\rho$ we have $B_7<0$ . We consider six subcases (with different ordering relations between $\frac{A_1}{A_2}$ and $\xi_j$ , $j=1,\ldots,5$ ) as follows: subcase 51 with $\frac{A_1}{A_2}\le\xi_5$ ; subcase 52 with $\xi_5<\frac{A_1}{A_2}\le\xi_2$ ; subcase 53 with $\xi_2<\frac{A_1}{A_2}\le\xi_4$ ; subcase 54 with $\xi_4<\frac{A_1}{A_2}\le\xi_3$ ; subcase 55 with $\xi_3<\frac{A_1}{A_2}<\xi_1$ ; and subcase 56 with $\xi_1\le\frac{A_1}{A_2}$ .

In subcase 51, we have $B_6<B_8$ , $B_6<B_9<B_7<B_{10}$ , and $B_{11}<B_7$ . The inequality $L^{\hat{a}_{1,0,\underline{V}},0}\underline{V}\le L^{A_1,A_2}\underline{V}$ holds due to $\frac{A_1}{A_2}\le\xi_5$ . We conclude (4.65) and the expression for $\frac{\underline{V}''}{\underline{V}'}$ is the same as in (4.69).

In subcase 52, we have $B_6<B_8$ , $B_6<B_9\le B_7<B_{10}$ , and $B_{11}\le B_7$ . We also have $B_6<A_-<A_+<B_9<B_7$ . Thus, the expression for $\frac{\underline{V}''}{\underline{V}'}$ is the same as in (4.87) with the optimizers given in (4.86).

In subcase 53, we have $B_6<B_8$ , $B_6<B_7<B_9\le B_{10}$ , $B_{7}<B_{11}$ , and $B_6<A_-<B_7<A_+$ . The expression for $\frac{\underline{V}''}{\underline{V}'}$ is the same as in (4.85) with the optimizers given in (4.81).

In subcase 54, from Lemmas 4.9 and 4.10, we have $B_6<B_8$ , $B_6<B_7\le B_{10}<B_9<0$ , $B_7<B_{11}$ , and $B_6<A_-<B_7<A_+$ . We conclude that the expression for $\frac{\underline{V}''}{\underline{V}'}$ is the same as in (4.85) with the optimizers given in (4.81).

In subcase 55, we have $B_6\le B_8$ , $B_6\le B_{10}<B_7<B_9<0$ , $B_7<B_{11}$ , and $B_6\le A_-<B_7<A_+$ . The expression for $\frac{\underline{V}''}{\underline{V}'}$ is the same as in (4.85) with the optimizers given in (4.81) in the convex region.

In subcase 56, we have $B_8<B_6$ , $B_{10}<B_6<B_7<B_9$ , and $B_7<B_{11}$ . The inequality $A_-<B_6<B_7<A_+$ holds. From Lemma 4.6, (4.55), (4.60), and (4.62), we have

(4.91) \begin{equation}\underset{a_2\in[0,A_2]}{\sup}\underset{a_1\in[0,A_1]}{\mathrm{inf}}L^{a_1,a_2}\underline{V}=\begin{cases}L^{A_1,A_2}\underline{V}&\ \ \Big\{x\;:\; \frac{\underline{V}'}{\underline{V}''}\le B_8\Big\},\\[5pt] L^{A_1,0}\underline{V}&\ \ \Big\{x\;:\;B_8<\frac{\underline{V}'}{\underline{V}''}\le B_6\Big\},\\[5pt] L^{\hat{a}_{1,0,\underline{V}},0}\underline{V}&\ \ \Big\{x\;:\; B_6<\frac{\underline{V}'}{\underline{V}''}< B_{11}\Big\},\\[5pt] L^{\hat{a}_{1,A_2,\underline{V}},A_2}\underline{V}&\ \ \Big\{x\;:\; B_{11}\le\frac{\underline{V}'}{\underline{V}''}\Big\},\end{cases}\end{equation}

and

(4.92) \begin{equation}\frac{\underline{V}''}{\underline{V}'}=g1_{(x_0,x_{5,+})}+g_11_{[x_{5,+},x_{4,+})}+\frac{1_{[x_{4,+},x_{1,+})}}{f_4}+\frac{1_{[x_{1,+},\infty)}}{f_5},\end{equation}

which is a continuous, negative, and decreasing function in the convex region in subcase 56, where

(4.93) \begin{equation}g_1(x)=-\frac{2(rx+\mu+\theta_1A_1)}{A_1^2+\sigma^2},\end{equation}

which is yielded by the equation $L^{A_1,0}\underline{V}=0$ , and

(4.94) \begin{equation}x_{5,+}=-\frac{A_1^2+\sigma^2}{2rB_8}-\frac{\theta_1A_1+\mu}{r},\quad x_{4,+}=\frac{\theta_1\big(A_1^2+\sigma^2\big)}{2rA_1}-\frac{\theta_1A_1+\mu}{r},\end{equation}

which are computed by solving $g_1(x)=1/B_8$ and $g_1(x)=1/B_6$ .

In case 6, we have $B_{10}<0$ and $B_{11}<0$ . From Lemma 4.9, we have $\xi_1,\xi_2<\xi_3<\xi_4$ . We consider the following subcases: subcase 61 with $(\xi_1,\xi_2<\xi_3{<})\xi_4\le\frac{A_1}{A_2}$ ; subcase 62 with $(\xi_1,\xi_2{<})\xi_3\le\frac{A_1}{A_2}<\xi_4$ ; subcase 63 with $\xi_1, \xi_2\le\frac{A_1}{A_2}<\xi_3({<}\xi_4)$ ; subcase 64 with $\xi_1\le\frac{A_1}{A_2}<\xi_2({<}\xi_3<\xi_4)$ ; subcase 65 with $\xi_2\le\frac{A_1}{A_2}<\xi_1({<}\xi_3<\xi_4)$ ; subcase 66 with $\frac{A_1}{A_2}\le\xi_1,\xi_2({<}\xi_3<\xi_4)$ ; and subcase 67 with $\frac{A_1}{A_2}\le\xi_5$ .

In subcase 61, from Lemma 4.9, we have $B_8<B_6$ , $B_6<B_7<B_9\le B_{10}$ , and $B_7<B_{11}$ . From Lemma 4.6, (4.55), (4.58), and (4.62), equation (4.91) holds. We have the same expression for $\frac{\underline{V}''}{\underline{V}'}$ as in (4.92) in the convex region.

In subcase 62, we have $B_8<B_6$ , $B_6<B_7\le B_{10}<B_9$ , and $B_7<B_{11}$ . We have the same expression for $\frac{\underline{V}''}{\underline{V}'}$ as in (4.92) in the convex region.

In subcase 63, from Lemma 4.9, we have $B_8\le B_6$ , $B_6\le B_{10}<B_7\le B_9$ , and $B_7\le B_{11}$ . From Lemma 4.10, we have $A_-<B_6<B_7<A_+$ . We also have the same expression for $\frac{\underline{V}''}{\underline{V}'}$ as in (4.92) with the optimizers given in (4.91) in the convex region.

In subcase 64, from Lemma 4.9, we have $B_8\le B_6$ , $B_6\le B_{10}<B_9<B_7$ , and $B_{11}<B_7$ . From Lemma 4.10, we have $A_-<B_6<A_+<B_9<B_7$ . Thus, if $B_7<0$ , from Lemma 4.6, (4.55), (4.58), and (4.62), we obtain

(4.95) \begin{equation}\underset{a_2\in[0,A_2]}{\sup}\underset{a_1\in[0,A_1]}{\mathrm{inf}}L^{a_1,a_2}\underline{V}=\begin{cases}L^{A_1,A_2}\underline{V}&\ \ \Big\{x\;:\; \frac{\underline{V}'}{\underline{V}''}\le B_8\Big\},\\[5pt] L^{A_1,0}\underline{V}&\ \ \Big\{x\;:\;B_8<\frac{\underline{V}'}{\underline{V}''}\le B_6\Big\},\\[5pt] L^{\hat{a}_{1,0,\underline{V}},0}\underline{V}&\ \ \Big\{x\;:\; B_6<\frac{\underline{V}'}{\underline{V}''}< A_+\Big\},\\[5pt] L^{A_1,A_2}\underline{V}&\ \ \Big\{x\;:\; A_+\le\frac{\underline{V}'}{\underline{V}''}< B_7\Big\},\\[5pt] L^{\hat{a}_{1,A_2,\underline{V}},A_2}\underline{V}&\ \ \Big\{x\;:\; B_7\le\frac{\underline{V}'}{\underline{V}''}\Big\},\end{cases}\end{equation}

and consequently,

(4.96) \begin{equation}\frac{\underline{V}''}{\underline{V}'}=g1_{(x_0,x_{5,+})}+g_1 1_{[x_{5,+},x_{4,+})}+\frac{1_{[x_{4,+},x_{8,+})}}{f_4}+g1_{[x_{8,+},x_{2,+})}+\frac{1_{[x_{2,+},\infty)}}{f_5},\end{equation}

which is a continuous, negative, and decreasing function, where

(4.97) \begin{equation}x_{2,+}=\frac{\theta_1^2B_7^2-\sigma^2}{2rB_7}-\frac{\mu}{r}.\end{equation}

If $B_7\ge 0$ , the region $\mathcal{R}_{3,+}$ degenerates and we have

(4.98) \begin{equation}\underset{a_2\in[0,A_2]}{\sup}\underset{a_1\in[0,A_1]}{\mathrm{inf}}L^{a_1,a_2}\underline{V}=\begin{cases}L^{A_1,A_2}\underline{V}&\ \ \Big\{x\;:\; \frac{\underline{V}'}{\underline{V}''}\le B_8\Big\},\\[5pt] L^{A_1,0}\underline{V}&\ \ \Big\{x\;:\;B_8<\frac{\underline{V}'}{\underline{V}''}\le B_6\Big\},\\[5pt] L^{\hat{a}_{1,0,\underline{V}},0}\underline{V}&\ \ \Big\{x\;:\; B_6<\frac{\underline{V}'}{\underline{V}''}< A_+\Big\},\\[5pt] L^{A_1,A_2}\underline{V}&\ \ \Big\{x\;:\; A_+\le\frac{\underline{V}'}{\underline{V}''}\Big\},\end{cases}\end{equation}

and

(4.99) \begin{equation}\frac{\underline{V}''}{\underline{V}'}=g1_{(x_0,x_{5,+})}+g_1 1_{[x_{5,+},x_{4,+})}+\frac{1_{[x_{4,+},x_{8,+})}}{f_4}+g1_{[x_{8,+},\infty)},\end{equation}

which is a continuous, negative, and decreasing function in the convex region in subcase 64.

In subcase 65, we have $B_6<B_8$ , $B_{10}<B_6<B_7\le B_9$ , and $B_{7}\le B_{11}$ . We also have $B_6<A_-<B_7\le A_+$ . Thus, from Lemma 4.6, (4.55), (4.58), and (4.62), equation (4.81) holds and we obtain the same expression for $\frac{\underline{V}''}{\underline{V}'}$ as in (4.85).

In subcase 66, we have $B_6<B_8$ , $B_{10}<B_6<B_9<B_7$ , and $B_{11}<B_{7}$ . The inequality $B_6<A_-<A_+<B_9<B_7$ holds. From Lemma 4.6, (4.55), (4.58), and (4.62), the function $\frac{\underline{V}''}{\underline{V}'}$ is the same as in (4.87) when $B_7<0$ , and is the same as in (4.90) when $B_7\ge 0$ .

In subcase 67, we have $B_6<B_8$ , $B_{10}<B_6<B_9<B_7$ , and $B_{11}<B_{7}$ . The inequality $L^{\hat{a}_{1,0,\underline{V}},0}\le L^{A_1,A_2}$ holds. The function $\frac{\underline{V}''}{\underline{V}'}$ is the same as in (4.69) when $B_7<0$ , and is the same as in (4.71) when $B_7\ge 0$ .

In case 7, we have $\xi_1,\xi_2<\xi_3<\xi_4$ . From $A_2+(\frac{1}{\rho}-2\rho)\,A_1\le 0$ , we have $2\rho^2-$ $1>0$ and $\frac{A_1}{A_2}\ge\frac{\rho}{2\rho^2-1}>\xi_4$ , using $2\rho<\frac{\theta_1}{\theta_2}$ . Then we have $B_8<B_6$ , $B_6<B_7<B_9<B_{10}$ , and $B_7<B_{11}<0$ . Thus, equation (4.91) holds and we obtain the same expression for $\frac{\underline{V}''}{\underline{V}'}$ as in (4.92).

We have thus found $\frac{\underline{V}''}{\underline{V}'}$ and the optimizers explicitly in the convex region for all possible parameter cases. The results are summarized in Table 2 of Appendix D.

4.3. Explicit sub-value function

In this subsection, we use the boundary conditions (3.4) and combine the two regions to give an explicit form of the sub-value function $\underline{V}$ in the entire region [l, u].

Let $\underline{v}_-$ and $\underline{v}_+$ denote the function $\frac{\underline{V}''}{\underline{V}'}$ in the concave region and convex region, respectively. From the previous subsections, we see that the concave region is $(\!-\!\infty, x_0)\cap[l, u]$ and the convex region is $(x_0, \infty)\cap[l, u]$ . Special properties of the functions $\underline{v}_-$ and $\underline{v}_+$ can be seen from their explicit forms, which are summarized in the following remarks.

For $x\in [l, u]$ , define the function

(4.100) \begin{equation}\underline{v}=\underline{v}_-1_{(-\infty,x_0)}+\underline{v}_+1_{(x_0,\infty)},\end{equation}

and let $\underline{v}(x_0)=g(x_0)=0$ if $x_0\in[l, u]$ . Then the function v is a continuous and decreasing function which satisfies $\frac{\underline{V}''}{\underline{V}'}=\underline{v}$ on [l, u]. The point $x_0$ is the only inflection point of the function $\underline{V}$ , that is, $\underline{V}''(x_0)=0$ , if $x_0\in (u,l)$ . With the boundary conditions (3.4) we obtain the following theorem.

Theorem 4.1. The sub-value function of the game is given by

(4.101) \begin{equation}\underline{V}(x)=\frac{\int_x^u\exp\!\big\{\!\int_l^y\underline{v}(z)\,{\mathrm{d}} z\big\}\,{\mathrm{d}} y}{\int_l^u\exp\!\big\{\!\int_l^y\underline{v}(z)\,{\mathrm{d}} z\big\}\,{\mathrm{d}} y},\end{equation}

where $\underline{v}$ is given in (4.100) which can be determined, given any parameter case, by using one of the seven forms in (4.19), (4.24), (4.38), (4.40), (4.42), (4.44), and (4.46) in the concave region, and one of the eight forms in (4.71), (4.69), (4.85), (4.87), (4.90), (4.92), (4.96), and (4.99) in the convex region. The associated optimizers can be found in explicit forms using one of the seven forms in (4.16), (4.20), (4.30), (4.39), (4.41), (4.43), and (4.45) in the concave region, and one of the eight forms in (4.70), (4.65), (4.81), (4.86), (4.89), (4.91), (4.95), and (4.98) in the convex region.

The function $\underline{V}$ given in (4.101) has a negative first derivative. Its second derivative satisfies $\underline{V}''(x)=\underline{v}(x)\underline{V}'(x)$ , which is continuous on [l, u]. The function $\underline{V}$ solves the BI equation (3.5) with optimizers which are bounded and piecewise continuous (thus Borel measurable). Hence, the function given in (4.101) is the sub-value function by the verification theorem, Theorem 3.1.

Below we show some game results when parameter conditions are specified, under which the sub-value function and the optimizers can be obtained explicitly via the concave and convex regions. The following theorems are special cases of Theorem 4.1.

Theorem 4.2. If $\rho<\frac{A_1}{A_2}<2(\rho-\frac{\theta_1}{\theta_2})$ and $\rho<\sqrt{1/2}$ then the sub-value function is given by (4.101) with

(4.102) \begin{equation}\underline{v}=\frac{1_{(-\infty,x_{2,-}]}}{f_1}+\frac{1_{(x_{2,-},x_{3,-}]}}{f_2}+\frac{1_{(x_{3,-},x_{4,-})}}{f_3}+g_21_{[x_{4,-},x_{1,-})}+g1_{[x_{1,-},x_{3,+})}+\frac{1_{[x_{3,+},\infty)}}{f_5}\end{equation}

on [l, u], and the optimizers of the game are given by

(4.103) \begin{equation}\left(\underline{a}_1^*(x), \underline{a}_2^*(x)\right)=\begin{cases}(A_1, \theta_2\,f_1(x)+\rho A_1)&\ \ (\!-\!\infty, x_{2,-})\cap[l, u],\\[5pt] \Big(0, \frac{\theta_1}{\rho}f_2(x)+\frac{A_1}{2\rho}\Big) \ \ or \ \ \Big(A_1, \frac{\theta_1}{\rho}f_2(x)+\frac{A_1}{2\rho}\Big)&\ \ [x_{2,-}, x_{3,-})\cap[l, u],\\[5pt] (0, \theta_2\,f_3(x))&\ \ [x_{3,-}, x_{4,-})\cap[l, u],\\[5pt] (0, A_2)&\ \ [x_{4,-},x_{1,-})\cap[l, u],\\[5pt] (A_1, A_2)&\ \ [x_{1,-}, x_{3,+})\cap[l, u],\\[5pt] (\!-\!\theta_1\,f_5(x)+\rho A_2, A_2)&\ \ [x_{3,+}, \infty)\cap[l, u].\end{cases}\end{equation}

Theorem 4.3. If $\frac{\theta_1}{\theta_2}>\frac{2}{\rho}$ , $\max\{\rho, \xi_1\}<\frac{A_1}{A_2}<\min\,\{\xi_2, \frac{2(\theta_1-\rho\theta_2)}{2\rho\theta_1-\theta_2}\}$ , and $\rho>\sqrt{1/2}$ then the sub-value function is given by (4.101) with

(4.104) \begin{align}\underline{v}&=\frac{1_{(-\infty,x_{2,-})}}{f_2}+\frac{1_{[x_{2,-},x_{5,-})}}{f_1}+g1_{[x_{5,-},x_{5,+})}+g_11_{[x_{5,+},x_{4,+})}+\frac{1_{[x_{4,+},x_{8,+})}}{f_4}+g1_{[x_{8,+},x_{2,+})}\nonumber \\&\quad +\frac{1_{[x_{2,+},\infty)}}{f_5}\end{align}

on [l, u], and the optimizers of the game are given by

(4.105) \begin{equation}\left(\underline{a}_1^*(x), \underline{a}_2^*(x)\right)=\begin{cases}\Big(0, \frac{\theta_1}{\rho}f_2(x)+\frac{A_1}{2\rho}\Big) \ \ or \ \ \Big(A_1, \frac{\theta_1}{\rho}f_2(x)+\frac{A_1}{2\rho}\Big)&\ \ (\!-\!\infty, x_{2,-})\cap[l, u],\\[5pt] (A_1, \theta_2f_1(x)+\rho A_1)&\ \ [x_{2,-}, x_{5,-})\cap[l, u],\\[5pt] (A_1, A_2)&\ \ [x_{5,-}, x_{5,+})\cap[l, u],\\[5pt] (A_1, 0)&\ \ [x_{5,+},x_{4,+})\cap[l, u],\\[5pt] (\!-\!\theta_1f_4(x), 0)&\ \ [x_{4,+}, x_{8,+})\cap[l, u],\\[5pt] (A_1, A_2)&\ \ [x_{8,+}, x_{2,+})\cap[l, u],\\[5pt] (\!-\!\theta_1f_5(x)+\rho A_2, A_2)&\ \ [x_{2,+}, \infty)\cap[l, u].\end{cases}\end{equation}

5.1. The concave region with $\overline{\boldsymbol{V}}''(\boldsymbol{x})<0$

In this region, we have $\frac{\overline{V}''}{\overline{V}'}>0$ . To find $\overline{V}$ , we note that the maximization of $L^{a_1,a_2}\overline{V}$ in $a_2$ now comes first. The quadratic function $L^{a_1,a_2}\overline{V}$ (in $a_2$ ) has a negative coefficient on the second-degree term. The vertex point (in $a_2$ ) is $\hat{a}_{2,a_1,\overline{V}}(x)=\theta_2\frac{\overline{V}'}{\overline{V}''}+\rho a_1$ , which is positive for all $a_1\in [0,A_1]$ . If the vertex point satisfies $\hat{a}_{2,a_1,\overline{V}}(x)=\theta_2\frac{\overline{V}'}{\overline{V}''}+\rho a_1<A_2$ , or equivalently, $a_1\in(\!-\!\infty,\acute{a}_{1,\overline{V}})\cap[0,A_1]$ , where

(5.1) \begin{equation}\acute{a}_{1,\overline{V}}=-\frac{\theta_2}{\rho}\frac{\overline{V}'}{\overline{V}''}+\frac{A_2}{\rho},\end{equation}

then the maximizer in $a_2$ is $\overline{a}^*(x,a_1)=\hat{a}_{2,a_1,\overline{V}}(x)$ . If the vertex point satisfies $\hat{a}_{2,a_1,\overline{V}}(x)\ge A_2$ , that is, $a_1\in[\acute{a}_{1,\overline{V}},\infty)\cap[0,A_1]$ , then the maximizer in $a_2$ is $\overline{a}^*(x,a_1)=A_2$ . For convenience, we write $\acute{a}_{1}=\acute{a}_{1,\overline{V}}$ . Then the maximizer in $a_2$ is given in the following form:

(5.2) \begin{equation}\overline{a}^*(x,a_1)=\begin{cases}A_2 & \acute{a}_1\le 0\ \ \ \ \ \ \ \ \ \ \ a_1\in[0, A_1],\\[5pt] A_2 & 0<\acute{a}_1<A_1\ \ \ a_1\in[\acute{a}_1, A_1],\\[5pt] \hat{a}_{2,a_1,\underline{V}}& 0<\acute{a}_1<A_1\ \ \ a_1\in[0, \acute{a}_1],\\[5pt] \hat{a}_{2,a_1,\underline{V}}& A_1\le\acute{a}_1\ \ \ \ \ \ \ \ \ a_1\in[0, A_1]. \end{cases}\end{equation}

We now define three regions which form a partition of the concave region:

(5.3) \begin{equation}\begin{split}&\mathcal{R}_1^-=\{x\in(l,u)\;:\;\overline{V}''<0, A_1\le\acute{a}_{1,\overline{V}}\}=\Bigg\{x\in(l,u)\;:\;\overline{V}''<0, \frac{\overline{V}'}{\overline{V}''}\le B_5\Bigg\},\\[5pt] &\mathcal{R}_2^-=\{x\in(l,u)\;:\;\overline{V}''<0, 0<\acute{a}_{1,\overline{V}}<A_1\}=\Bigg\{x\in(l,u)\;:\;\overline{V}''<0, B_5<\frac{\underline{V}'}{\underline{V}''}<B_4\Bigg\},\\[5pt] &\mathcal{R}_3^-=\{x\in(l,u)\;:\;\overline{V}''<0,\acute{a}_{1,\overline{V}}\le 0\}=\Bigg\{x\in(l,u)\;:\;\overline{V}''<0, B_4\le\frac{\overline{V}'}{\overline{V}''}\Bigg\},\end{split}\end{equation}

where $B_4$ and $B_5$ (with $B_5 < B_4$ and $0<B_4$ ) are given in (4.7) and (4.15). We note that the region $\mathcal{R}_1^-$ degenerates when $B_5\leq 0$ .

For $x\in\mathcal{R}_1^-$ , the maximizer of $L^{a_1,a_2}\overline{V}$ in $a_2$ is $\overline{a}^*(x,a_1)=\hat{a}_{2,a_1,\overline{V}}$ for all $a_1\in[0,A_1]$ . Note that

(5.4) \begin{equation}L^{a_1,\hat{a}_{2,a_1,\overline{V}}}\overline{V}=[rx+\mu+(\theta_1-\rho\theta_2)a_1]\overline{V}'-\frac{\theta_2^2}{2}\frac{\overline{V}'^2}{\overline{V}''}+\frac{1}{2}\big[(1-\rho^2)a_1^2+\sigma^2\big]\overline{V}'',\end{equation}

which is a quadratic function of $a_1$ . The vertex point of $L^{a_1,\hat{a}_{2,a_1,\overline{V}}}\overline{V}$ in $a_1$ is

(5.5) \begin{equation}\breve{a}_{1,\overline{V}}=\frac{\rho\theta_2-\theta_1}{1-\rho^2}\frac{\overline{V}'}{\overline{V}''},\end{equation}

which is positive when $\frac{1}{\rho}<\frac{\theta_2}{\theta_1}$ in the concave region. We obtain the following lemma to find the minimizer in $a_1$ (equal to 0 or $A_1$ ) of $L^{a_1,\hat{a}_{2,a_1,\overline{V}}}\overline{V}$ in $\mathcal{R}_1^-$ .

Lemma 5.1. For $x\in\mathcal{R}_1^-$ , if $\frac{1}{\rho}<\frac{\theta_2}{\theta_1}$ , we have

(5.6) \begin{equation}\underset{a_1\in[0,A_1]}{\mathrm{inf}}\underset{a_2\in[0,A_2]}{\sup}L^{a_1,a_2}\overline{V}=\begin{cases}L^{0,\hat{a}_{2,0,\overline{V}}}\overline{V}&\ \ \Big\{x\;:\;\breve{a}_{1,\overline{V}}>\frac{A_1}{2}\Big\}=\Big\{x\;:\;C_1<\frac{\overline{V}'}{\overline{V}''}\Big\},\\[5pt] L^{A_1,\hat{a}_{2,A_1,\overline{V}}}\overline{V}&\ \ \Big\{x\;:\;\breve{a}_{1,\overline{V}}\le\frac{A_1}{2}\Big\}=\Big\{x\;:\;\frac{\overline{V}'}{\overline{V}''}\le C_1\Big\},\end{cases}\end{equation}

where

(5.7) \begin{equation}C_1=\frac{(1-\rho^2)A_1}{2(\rho\theta_2-\theta_1)};\end{equation}

and for $x\in\mathcal{R}_1^-$ , if $\frac{\theta_2}{\theta_1}\le\frac{1}{\rho}$ then

(5.8) \begin{equation}\underset{a_1\in[0,A_1]}{\mathrm{inf}}\underset{a_2\in[0,A_2]}{\sup}L^{a_1,a_2}\overline{V}=L^{A_1,\hat{a}_{2,A_1,\overline{V}}}\overline{V}.\end{equation}

For $x\in\mathcal{R}^-_{2}$ , the maximizer of $L^{a_1,a_2}\overline{V}$ in $a_2$ is $\overline{a}^*(x,a_1)=\hat{a}_{2,a_1,\overline{V}}$ if $a_1\in[0,\acute{a}_{1})$ . We note that $\hat{a}_{2,\acute{a}_1,\overline{V}}=A_2$ . We also note that the quadratic function $L^{a_1,\hat{a}_{2,a_1,\underline{V}}}\underline{V}$ (in $a_1$ ) has the vertex $\breve{a}_{1,\overline{V}}$ in (5.5), which is positive if $\frac{1}{\rho}<\frac{\theta_2}{\theta_1}$ in the concave region. If $a_1\in [\acute{a}_{1},A_1]$ , the maximizer of $L^{a_1,a_2}\overline{V}$ in $a_2$ is $\overline{a}^*(x,a_1)=A_2$ . The vertex point of $L^{a_1,A_2}$ in $a_1$ is $\hat{a}_{1,A_2,\underline{V}}$ . Then we obtain the following lemma to find the minimizer in $a_1\in[0,\acute{a}_{1})$ of $L^{a_1,\hat{a}_{2,a_1,\underline{V}}}\underline{V}$ and in $a_1\in [\acute{a}_{1},A_1]$ of $L^{a_1,A_2}$ in $\mathcal{R}_{2}^-$ .

Lemma 5.2. For $x\in\mathcal{R}_{2}^-$ , if $\frac{1}{\rho}<\frac{\theta_2}{\theta_1}$ , we have

(5.9) \begin{align}\underset{a_1\in[0,\acute{a}_1)}{\mathrm{inf}}\underset{a_2\in[0,A_2]}{\sup}L^{a_1,a_2}\overline{V}=\begin{cases}L^{0,\hat{a}_{2,0,\overline{V}}}\overline{V}&\ \ \{x\;:\;\acute{a}_1/2\le\breve{a}_{1,\underline{V}}\}=\Big\{x\;:\;C_2\le\frac{\overline{V}'}{\overline{V}''}\Big\},\\[6pt] L^{\acute{a}_1,\hat{a}_{2,\acute{a}_1,\overline{V}}}\overline{V}=L^{\acute{a}_1,A_2}\overline{V}&\ \ \{x\;:\;\breve{a}_{1,\underline{V}}<\acute{a}_1/2\}=\Big\{x\;:\;\frac{\overline{V}'}{\overline{V}''}<C_2\Big\},\end{cases}\nonumber\\[1pt]\end{align}

where

(5.10) \begin{equation}C_2=\frac{(1-\rho^2)A_2}{\theta_2(1+\rho^2)-2\rho\theta_1};\end{equation}

for $x\in\mathcal{R}_{2}^-$ , if $\frac{\theta_2}{\theta_1}\le\frac{1}{\rho}$ , we have

(5.11) \begin{equation}\underset{a_1\in[0,\acute{a}_1]}{\mathrm{inf}}\underset{a_2\in[0,A_2]}{\sup}L^{a_1,a_2}\overline{V}=L^{\acute{a}_1,\hat{a}_{2,\acute{a}_1,\overline{V}}}\overline{V}=L^{\acute{a}_1,A_2}\overline{V}.\end{equation}

For $x\in\mathcal{R}_{2}^-$ , if $2\rho<\frac{\theta_2}{\theta_1}$ , we have

(5.12) \begin{equation}\underset{a_1\in[\acute{a}_1,A_1]}{\mathrm{inf}}\underset{a_2\in[0,A_2]}{\sup}L^{a_1,a_2}\overline{V}=\begin{cases}L^{\acute{a}_1,A_2}\overline{V}&\ \ \{x\;:\;(\acute{a}_1+A_1)/2\le\hat{a}_{1,A_2,\overline{V}}\}=\Big\{x\;:\;C_3\le\frac{\overline{V}'}{\overline{V}''}\Big\},\\[5pt] L^{A_1,A_2}\overline{V}&\ \ \{x\;:\;\hat{a}_{1,A_2,\underline{V}}<(\acute{a}_2+A_2)/2\}=\Big\{x\;:\;\frac{\overline{V}'}{\overline{V}''}<C_3\Big\},\end{cases}\end{equation}

where

(5.13) \begin{equation}C_3=\frac{\rho A_1+(1-2\rho^2)A_2}{\theta_2-2\rho\theta_1};\end{equation}

for $x\in\mathcal{R}_{2}^-$ , if $\frac{\theta_2}{\theta_1}<{2\rho}$ , we have

(5.14) \begin{equation}\underset{a_1\in[\acute{a}_1,A_1]}{\mathrm{inf}}\underset{a_2\in[0,A_2]}{\sup}L^{a_1,a_2}\overline{V}=\begin{cases}L^{\acute{a}_1,A_2}\overline{V}&\ \ \{x\;:\;(\acute{a}_1+A_1)/2\le\hat{a}_{1,A_2,\overline{V}}\}=\Big\{x\;:\;\frac{\overline{V}'}{\overline{V}''}\le C_3\Big\},\\[5pt] L^{A_1,A_2}\overline{V}&\ \ \{x\;:\;\hat{a}_{1,A_2,\underline{V}}<(\acute{a}_2+A_2)/2\}=\Big\{x\;:\;C_3<\frac{\overline{V}'}{\overline{V}''}\Big\};\end{cases}\end{equation}

for $x\in\mathcal{R}_{2}^-$ , if $\frac{\theta_2}{\theta_1}={2\rho}$ , we have

(5.15) \begin{equation}\underset{a_1\in[\acute{a}_1,A_1]}{\mathrm{inf}}\underset{a_2\in[0,A_2]}{\sup}L^{a_1,a_2}\overline{V}=\begin{cases}L^{\acute{a}_1,A_2}\overline{V}&\ \ \{x\;:\;(\acute{a}_1+A_1)/2\le\hat{a}_{1,A_2,\overline{V}}\}\ if \ A_1+\left(\frac{1}{\rho}-2\rho\right)A_2\le 0,\\[5pt] L^{A_1,A_2}\overline{V}&\ \ \{x\;:\;\hat{a}_{1,A_2,\underline{V}}<(\acute{a}_2+A_1)/2\} \ if \ A_1+\left(\frac{1}{\rho}-2\rho\right)A_2>0.\end{cases}\end{equation}

For $x\in\mathcal{R}_{3}^-$ , the maximizer of $L^{a_1,a_2}\overline{V}$ in $a_2$ is $\overline{a}^*(x,a_1)=A_2$ for all $a_1\in[0,A_1]$ . We note that $L^{a_1,A_2}\overline{V}$ has the vertex $\hat{a}_{1,A_2,\overline{V}}$ . Then we obtain the following lemma to find the minimizer in $a_1\in[0, A_1]$ (equal to 0 or $A_1$ ) of $L^{a_1,A_2}\overline{V}$ .

Lemma 5.3. For $x\in\mathcal{R}_{3}^-$ , we have

(5.16) \begin{equation}\underset{a_1\in[0,A_1]}{\mathrm{inf}}\underset{a_2\in[0,A_2]}{\sup}L^{a_1,a_2}\overline{V}=\begin{cases}L^{0,A_2}\overline{V}&\ \ \Big\{x\;:\;\hat{a}_{1,A_2,\overline{V}}\ge\frac{A_1}{2}\Big\}=\Big\{x\;:\;\frac{\overline{V}'}{\overline{V}''}\le B_1\Big\},\\[8pt] L^{A_1,A_2}\overline{V}&\ \ \Big\{x\;:\;\hat{a}_{1,A_2,\overline{V}}<\frac{A_1}{2}\Big\}=\Big\{x\;:\;B_1<\frac{\overline{V}'}{\overline{V}''}\Big\},\end{cases}\end{equation}

where $B_1$ is given in (4.4).

Next we find $\frac{\overline{V}''}{\overline{V}'}$ in the concave region via the following parameter cases:

1. 1. $2\rho<\frac{\theta_2}{\theta_1}\le\frac{1}{\rho}$ ;

2. 2. $\frac{\theta_2}{\theta_1}\le\frac{1}{\rho}, 2\rho$ and $A_1+(\frac{1}{\rho}-2\rho)\,A_2>0$ ;

3. 3. $\frac{\theta_2}{\theta_1}\le\frac{1}{\rho}, 2\rho$ and $A_1+(\frac{1}{\rho}-2\rho)\,A_2\le 0$ ;

4. 4. $\frac{1}{\rho}<\frac{\theta_2}{\theta_1}\le 2\rho$ and $A_1+(\frac{1}{\rho}-2\rho)\,A_2>0$ ;

5. 5. $\frac{1}{\rho}<\frac{\theta_2}{\theta_1}\le 2\rho$ and $A_1+(\frac{1}{\rho}-2\rho)\,A_2\le 0$ ;

6. 6. $\frac{1}{\rho}, 2\rho<\frac{\theta_2}{\theta_1}$ and $A_1+(\frac{1}{\rho}-2\rho)\,A_2>0$ ;

7. 7. $\frac{1}{\rho}, 2\rho<\frac{\theta_2}{\theta_1}$ and $A_1+(\frac{1}{\rho}-2\rho)\,A_2\le 0$ .

We note that the condition $A_1+(\frac{1}{\rho}-2\rho)A_2\le0$ implies $\frac{A_1}{A_2}\le 2\rho-\frac{1}{\rho}<\frac{1}{\rho}$ , which gives $B_5 > 0$ .

In case 1, we have $B_4<C_3$ and $B_1<B_4$ . From (5.8), (5.11), (5.12) and (5.16), we have $\inf{a_1\in[0,A_1]}\sup{a_2\in[0,A_2]}L^{a_1,a_2}\overline{V}=L^{A_1,\hat{a}_{2,A_2,\overline{V}}}\overline{V}$ in $\mathcal{R}_{1}^-$ , and $\inf{a_1\in[0,A_1]}$ $\sup{a_2\in[0,A_2]}L^{a_1,a_2}\overline{V}=L^{A_1,A_2}\overline{V}$ in $\mathcal{R}_{2}^-\cup\mathcal{R}_3^-$ , if $B_5 > 0$ . Thus, we have

(5.17) \begin{equation}\underset{a_1\in[0,A_1]}{\mathrm{inf}}\underset{a_2\in[0,A_2]}{\sup}L^{a_1,a_2}\overline{V}=\begin{cases}L^{A_1,\hat{a}_{2,A_2,\overline{V}}}\overline{V}&\ \ \Big\{x\;:\;\frac{\overline{V}'}{\overline{V}''}\le B_5\Big\},\\[5pt] L^{A_1,A_2}\overline{V}&\ \ \Big\{x\;:\;B_5<\frac{\overline{V}'}{\overline{V}''}\Big\},\end{cases}\end{equation}

and

(5.18) \begin{equation}\frac{\overline{V}''}{\overline{V}'}=\frac{(\!-\!\infty,x_{5,-}]}{f_1}+g1_{(x_{5,-},x_0)},\end{equation}

which is a continuous, positive, and decreasing function, where $x_{5,-}$ is given in (4.23), $f_1$ is in (4.22) and g is in (4.17), in the concave region if $B_5 > 0$ . And if $B_5\le 0$ , the region $\mathcal{R}_{1}^-$ degenerates and we obtain

(5.19) \begin{equation}\underset{a_1\in[0,A_1]}{\mathrm{inf}}\underset{a_2\in[0,A_2]}{\sup}L^{a_1,a_2}\overline{V}=L^{A_1,A_2}\overline{V}\end{equation}

and

(5.20) \begin{equation}\frac{\overline{V}''}{\overline{V}'}=g1_{(-\infty,x_0)},\end{equation}

which is a continuous, positive, and decreasing function in the concave region.

In case 2, we have $C_3<0$ (if $\frac{\theta_2}{\theta_1}< 2\rho$ ), $B_1<B_4$ . The expression for $\frac{\overline{V}''}{\overline{V}'}$ is the same as in case 1.

In case 3, from $0<\frac{A_1}{A_2}\le 2\rho-\frac{1}{\rho}$ , we see that $\frac{\theta_2}{\theta_1}\leq \frac{1}{\rho}<2\rho$ . It can be shown that $0\le C_3<B_5<B_4$ and $B_1<B_4$ . From (5.8), (5.11), (5.14), and (5.16), equation (5.17) holds and we obtain the same expression for $\frac{\overline{V}''}{\overline{V}'}$ as in (5.18) in the concave region.

For the remaining cases, the parameter condition $\frac{1}{\rho}<\frac{\theta_2}{\theta_1}$ holds. Define

(5.21) \begin{align}\zeta_1=\frac{\theta_2}{2(\rho\theta_2-\theta_1)},\quad\zeta_2=\frac{(1+\rho^2)\theta_2-2\rho\theta_1}{2(\rho\theta_2-\theta_1)},\quad\zeta_3=\frac{\theta_2-\rho\theta_1}{\rho\theta_2-\theta_1},\quad\zeta_4=\frac{(1+\rho^2)\theta_2-2\rho\theta_1}{2\rho^2(\rho\theta_2-\theta_1)},\nonumber\\[2pt]\end{align}

which are all positive if $\frac{1}{\rho}<\frac{\theta_2}{\theta_1}$ . By simple calculations, we have the following lemma.

In case 4, from Lemma 5.4, we have $\zeta_5<\zeta_2<\zeta_4<\zeta_3<\zeta_1$ and $C_3<0$ (if $\frac{\theta_2}{\theta_1}<2\rho$ ), where $\zeta_5$ is given in (5.27). We also have $\frac{A_2}{A_1}<\frac{1}{2\rho-1/\rho}<\zeta_4$ . We consider three subcases: subcase 41 with $\zeta_5<\frac{A_2}{A_1}<\zeta_2$ ; subcase 42 with $\zeta_2\le\frac{A_2}{A_1}<\zeta_4$ ; and subcase 43 with $\frac{A_2}{A_1}\le\zeta_5.$

In subcase 41, we have $B_5 < C_1$ , $B_5 < C_2<B_4$ , $C_3<0$ , and $B_1<B_4$ . From (5.6), we have

(5.22) \begin{equation}\underset{a_1\in[0,A_1]}{\mathrm{inf}}\underset{a_2\in[0,A_2]}{\sup}L^{a_1,a_2}\overline{V}=L^{A_1,\hat{a}_{2,A_1,\overline{V}}}\overline{V}\end{equation}

in $\mathcal{R}_{1}^-$ . From (5.9) and (5.14) (with the first line degenerated due to $C_3<0$ ), we have

(5.23) \begin{equation}\underset{a_1\in[0,A_1]}{\mathrm{inf}}\underset{a_2\in[0,A_2]}{\sup}L^{a_1,a_2}\overline{V}=\begin{cases}\min\,\{L^{0,\hat{a}_{2,0,\overline{V}}}\overline{V}, L^{A_1,A_2}\overline{V}\} &\ \ \Big\{x\;:\;C_2\le\frac{\overline{V}'}{\overline{V}''}<B_4\Big\},\\[8pt] L^{A_1,A_2}\overline{V}&\ \ \Big\{x\;:\;B_5<\frac{\overline{V}'}{\overline{V}''}<C_2\Big\},\end{cases}\end{equation}

in $\mathcal{R}_2^-$ . From (5.16), we have $\inf{a_1\in[0,A_1]}\sup{a_2\in[0,A_2]}L^{a_1,a_2}\overline{V}=L^{A_1,A_2}\overline{V}\ \ \{x\;:\;B_4\le\frac{\overline{V}'}{\overline{V}''}\}$ in $\mathcal{R}_3^-$ . From (4.26), we have

(5.24) \begin{equation}L^{0,\hat{a}_{2,0,\overline{V}}}\overline{V}-L^{A_1,A_2}\overline{V}=-\overline{V}''H\left(\frac{\overline{V}'}{\overline{V}''}\right),\end{equation}

which is negative if and only if $A^-<\frac{\overline{V}'}{\overline{V}''}<A^+$ , where H is a quadratic function given by

(5.25) \begin{equation}H(A)=\frac{\theta_2^2}{2}A^2-(\theta_2A_2-\theta_1A_1)A+\frac{1}{2}\big(A_1^2+A_2^2-2\rho A_1A_2\big),\end{equation}

and

(5.26) \begin{equation}A^+,A^-=\frac{\theta_2A_2-\theta_1A_1\pm\sqrt{(\theta_2A_2-\theta_1A_1)^2-\theta_2^2\big(A_1^2+A_2^2-2\rho A_1A_2\big)}}{\theta_2^2},\end{equation}

which are both positive roots of $H(A)=0$ when $\frac{A_2}{A_1}>\zeta_5$ under which the discriminant of H is positive (provided $\frac{1}{\rho}<\frac{\theta_2}{\theta_1}$ ), where

(5.27) \begin{equation}\zeta_5=\frac{\theta_2^2-\theta_1^2}{2\theta_2(\rho\theta_2-\theta_1)}.\end{equation}

We now give the following lemmas.

To show $C_2<\frac{A^-+A^+}{2}$ , which is equivalent to $\frac{\theta_1\zeta_2}{\rho\theta_2}<\frac{A_2}{A_1}$ , we only need to show $\frac{\theta_1\zeta_2}{\rho\theta_2}<\zeta_5$ which is equivalent to $(\frac{\theta_2}{\theta_1}-\rho)(\frac{\theta_2}{\theta_1}-\frac{1}{\rho})>0$ . For the other two results in Lemma 5.6, we note the following observations:

\begin{equation*}\begin{split}H(C_2)&=\frac{2\rho^2(\rho\theta_2-\theta_1)^2A_1^2}{[\theta_2(1+\rho^2)-2\rho\theta_1]^2}\left(\frac{A_2}{A_1}-\zeta_2\right)\left(\frac{A_2}{A_1}-\zeta_4\right)\!,\\[5pt] H(C_3)&=\frac{2\rho^2(\rho\theta_2-\theta_1)^2A_1^2}{(\theta_2-2\rho\theta_1)^2}\left(\frac{A_2}{A_1}-\zeta_1\right)\left(\frac{A_2}{A_1}-\zeta_4\right)\!.\end{split}\end{equation*}

From Lemma 5.5, we have $B_5 < A^- < A^+ < B_4$ in subcase 41. From Lemma 5.6, we have $B_5 < C_2 < A^- < A^+ < B_4$ . From (5.22) and (5.23), we obtain

(5.28) \begin{equation}\underset{a_1\in[0,A_1]}{\mathrm{inf}}\underset{a_2\in[0,A_2]}{\sup}L^{a_1,a_2}\overline{V}=\begin{cases}L^{A_1,\hat{a}_{2,A_1,\overline{V}}}\overline{V}&\ \ \Big\{x\;:\;\frac{\overline{V}'}{\overline{V}''}\le B_5\Big\},\\[5pt] L^{A_1,A_2}\overline{V}&\ \ \Big\{x\;:\;B_5<\frac{\overline{V}'}{\overline{V}''}\le A^-\Big\},\\[5pt] L^{0,\hat{a}_{2,0,\overline{V}}}\overline{V}&\ \ \Big\{x\;:\;A^-<\frac{\overline{V}'}{\overline{V}''}\le A^+\Big\},\\[5pt] L^{A_1,A_2}\overline{V}&\ \ \Big\{x\;:\;A^+<\frac{\overline{V}'}{\overline{V}''}\Big\},\end{cases}\end{equation}

and hence

(5.29) \begin{equation}\frac{\overline{V}''}{\overline{V}'}=\frac{(\!-\!\infty,x_{5,-}]}{f_1}+g1_{(x_{5,-},x_6^-]}+\frac{1_{(x_6^-,x_7^-]}}{f_3}+g1_{(x_7^-,x_0)},\end{equation}

which is a continuous, positive, and decreasing function in the concave region when $B_5>0$ in subcase 41, where

(5.30) \begin{equation}\begin{split}x_6^-&=-\frac{A_1^2+A_2^2-2\rho A_1A_2+\sigma^2}{2rA^-}-\frac{\theta_1A_1-\theta_2A_2+\mu}{r},\\[5pt] x_7^-&=-\frac{A_1^2+A_2^2-2\rho A_1A_2+\sigma^2}{2rA^+}-\frac{\theta_1A_1-\theta_2A_2+\mu}{r}.\end{split}\end{equation}

If $B_5\le 0$ ( $\mathcal{R}_1^-$ degenerates), we obtain

(5.31) \begin{equation}\underset{a_1\in[0,A_1]}{\mathrm{inf}}\underset{a_2\in[0,A_2]}{\sup}L^{a_1,a_2}\overline{V}=\begin{cases}L^{A_1,A_2}\overline{V}&\ \ \Big\{x\;:\;\frac{\overline{V}'}{\overline{V}''}\le A^-\Big\},\\[5pt] L^{0,\hat{a}_{2,0,\overline{V}}}\overline{V}&\ \ \Big\{x\;:\;A^-<\frac{\overline{V}'}{\overline{V}''}\le A^+\Big\},\\[5pt] L^{A_1,A_2}\overline{V}&\ \ \Big\{x\;:\;A^+<\frac{\overline{V}'}{\overline{V}''}\Big\},\end{cases}\end{equation}

and

(5.32) \begin{equation}\frac{\overline{V}''}{\overline{V}'}=g1_{(-\infty,x_6^-]}+\frac{1_{(x_6^-,x_7^-]}}{f_3}+g1_{(x_7^-,x_0)},\end{equation}

which is a continuous, positive, and decreasing function in the concave region in subcase 41.

In subcase 42 with $\zeta_2\le\frac{A_2}{A_2}<\zeta_4$ , from Lemmas 5.4 and 5.5, we have $C_1\le B_5$ , $C_2\le A^-\le B_5 < A^+ < B_4$ , $C_3<0$ , and $B_1 < B_4$ . Then we obtain

\begin{equation*}\underset{a_1\in[0,A_1]}{\mathrm{inf}}\underset{a_2\in[0,A_2]}{\sup}L^{a_1,a_2}\overline{V}=\begin{cases}L^{A_1,\hat{a}_{2,A_1,\overline{V}}}\overline{V}&\ \ \Big\{x\;:\;\frac{\overline{V}'}{\overline{V}''}\le C_1\Big\},\\[5pt] L^{0,\hat{a}_{2,0,\overline{V}}}\overline{V}&\ \ \Big\{x\;:\;C_1<\frac{\overline{V}'}{\overline{V}''}\le B_5\Big\},\end{cases}\end{equation*}

in $\mathcal{R}_1^-$ , and

\begin{equation*}\underset{a_1\in[0,A_1]}{\mathrm{inf}}\underset{a_2\in[0,A_2]}{\sup}L^{a_1,a_2}\overline{V}=\begin{cases}L^{0,\hat{a}_{2,0,\overline{V}}}\overline{V}&\ \ \Big\{x\;:\;B_5<\frac{\overline{V}'}{\overline{V}''}\le A^+\Big\},\\[5pt] L^{A_1,A_2}\overline{V}&\ \ \Big\{x\;:\;A^+<\frac{\overline{V}'}{\overline{V}''}\Big\},\end{cases}\end{equation*}

in $\mathcal{R}_2^-\cup\mathcal{R}_3^-$ . Thus, we have

(5.33) \begin{equation}\underset{a_1\in[0,A_1]}{\mathrm{inf}}\underset{a_2\in[0,A_2]}{\sup}L^{a_1,a_2}\overline{V}=\begin{cases}L^{A_1,\hat{a}_{2,A_1,\overline{V}}}\overline{V}&\ \ \Big\{x\;:\;\frac{\overline{V}'}{\overline{V}''}\le C_1\Big\},\\[5pt] L^{0,\hat{a}_{2,0,\overline{V}}}\overline{V}&\ \ \Big\{x\;:\;C_1<\frac{\overline{V}'}{\overline{V}''}\le A^+\Big\},\\[5pt] L^{A_1,A_2}\overline{V}&\ \ \Big\{x\;:\;A^+<\frac{\overline{V}'}{\overline{V}''}\Big\},\end{cases}\end{equation}

and we obtain

(5.34) \begin{equation}\frac{\overline{V}''}{\overline{V}'}=\frac{1_{(-\infty,x_{1}^-]}}{f_1}+\frac{1_{(x_{1}^-,x_{7}^-]}}{f_3}+g1_{(x_7^-,x_0)},\end{equation}

which is a continuous, positive, and decreasing function in the concave region in subcase 42, where

(5.35) \begin{equation}x_1^-=\frac{\theta_2^2C_1^2-\sigma^2}{2rC_1}-\frac{\mu}{r}.\end{equation}

In subcase 43, we have $B_5<C_1$ , $B_5 < C_2 < B_4$ , $C_3<0$ , $A^- < B_5 < B_4\le A^+$ , and $B_4\le B_1$ . Thus, we obtain the same expression for $\frac{\overline{V}''}{\overline{V}'}$ as (5.18) when $B_5>0$ and as (5.20) when $B_5\le 0$ in the convex region.

In case 5, we have $B_5>0$ . From Lemma 5.4, we have $\zeta_5<\zeta_2<\zeta_4\le\zeta_3\le\zeta_1$ and $C_3<0$ . We consider six subcases: subcase 51 with $\zeta_1\le\frac{A_2}{A_1}$ ; subcase 52 with $\zeta_3\le\frac{A_2}{A_1}<\zeta_1$ ; subcase 53 with $\zeta_4\le\frac{A_2}{A_1}<\zeta_3$ ; subcase 54 with $\zeta_2\le\frac{A_2}{A_1}<\zeta_4$ ; subcase 55 with $\zeta_5<\frac{A_2}{A_1}<\zeta_2$ ; and subcase 56 with $\frac{A_2}{A_2}\le\zeta_5$ .

In subcase 51, we have $C_1 < B_5$ , $C_2 < B_5 < B_4\le C_3$ , and $B_4\le B_1$ . We obtain

(5.36) \begin{equation}\underset{a_1\in[0,A_1]}{\mathrm{inf}}\underset{a_2\in[0,A_2]}{\sup}L^{a_1,a_2}\overline{V}=\begin{cases}L^{A_1,\hat{a}_{2,A_1,\overline{V}}}\overline{V}&\ \ \Big\{x\;:\;\frac{\overline{V}'}{\overline{V}''}\le C_1\Big\},\\[5pt] L^{0,\hat{a}_{2,0,\overline{V}}}\overline{V}&\ \ \Big\{x\;:\;C_1<\frac{\overline{V}'}{\overline{V}''}<B_4\Big\},\\[5pt] L^{0,A_2}\overline{V}&\ \ \Big\{x\;:\;B_4\le\frac{\overline{V}'}{\overline{V}''}\le B_1\Big\},\\[5pt] L^{A_1,A_2}\overline{V}&\ \ \Big\{x\;:\;B_1<\frac{\overline{V}'}{\overline{V}''}\Big\},\end{cases}\end{equation}

and

(5.37) \begin{equation}\frac{\overline{V}''}{\overline{V}'}=\frac{1_{(-\infty,x_{1}^-]}}{f_1}+\frac{1_{(x_{1}^-,x_{3,-}]}}{f_3}+g_21_{[x_{3,-},x_{4,-}]}+g1_{(x_{4,-},x_0)},\end{equation}

which is a continuous, positive, and decreasing function in the concave region.

In subcase 52, we have $C_1 < B_5$ , $C_2<B_5\le C_3<B_4$ , $A^-<B_5\le C_3<A^+<B_4$ , and $B_1<B_4$ . Thus, (5.33) holds and we obtain the same expression for $\frac{\overline{V}'}{\overline{V}''}$ as in (5.34).

In subcase 53, we have $C_1 < B_5$ , $C_2\le C_3<B_5<B_4$ , $A^-<B_5<A^+<B_4$ , and $B_1<B_4$ . We obtain the same expression for $\frac{\overline{V}'}{\overline{V}''}$ as in (5.34).

In subcase 54, we have $C_1\le B_5$ , $C_3<C_2\le B_5<B_4$ , $A^-\le B_5<A^+<B_4$ , and $B_1<B_4$ . We also obtain the same expression for $\frac{\overline{V}'}{\overline{V}''}$ as in (5.34).

In subcase 55, we have $B_5<C_1$ , $C_3<B_5<C_2<B_4$ , $B_5<C_2<A^-<A^+<B_4$ , and $B_1<B_4$ . Thus, we obtain the same expression for $\frac{\overline{V}'}{\overline{V}''}$ as in (5.18).

In subcase 56, we have $B_5<C_1$ , $C_3<B_5<C_2<B_4$ , and $B_1<B_4$ . We obtain the same expression for $\frac{\overline{V}'}{\overline{V}''}$ as in (5.18).

In case 6, we have $\zeta_1,\zeta_2<\zeta_3<\zeta_4$ . Note that $C_1>0$ and $C_3>0$ . We consider the following subcases: subcase 61 with $(\zeta_1,\zeta_2<\zeta_3{<})\zeta_4\le\frac{A_2}{A_1}$ ; subcase 62 with $(\zeta_1,\zeta_2{<})\zeta_3\le\frac{A_2}{A_1}<\zeta_4$ ; subcase 63 with $\zeta_1, \zeta_2\le\frac{A_2}{A_1}<\zeta_3({<}\zeta_4)$ ; subcase 64 with $\zeta_1\le\frac{A_2}{A_1}<\zeta_2({<}\zeta_3<\zeta_4)$ ; subcase 65 with $\zeta_2\le\frac{A_2}{A_1}<\zeta_1({<}\zeta_3<\zeta_4)$ ; subcase 66 with $\zeta_5<\frac{A_2}{A_1}\le\zeta_1,\zeta_2({<}\zeta_3<\zeta_4)$ ; and subcase 67 with $\frac{A_2}{A_1}\le\zeta_5$ .

In subcase 61, we have $C_1<B_5$ , $C_3\le C_2<B_5<B_4$ , and $B_4<B_1$ . From (5.6), (5.9), (5.12), and (5.16), equation (5.36) holds and we obtain the same expression for $\frac{\overline{V}'}{\overline{V}''}$ as in (5.37).

In subcase 62, we have $C_1<B_5$ , $C_2<C_3\le B_5<B_4$ , and $B_4<B_1$ . We obtain the same expression for $\frac{\overline{V}'}{\overline{V}''}$ as in (5.37).

In subcase 63, we have $C_1\le B_5$ , $C_2\le B_5<C_3\le B_4$ , $A^-\le B_5<B_4\le A^+$ , and $B_4\le B_1$ . From (5.6), (5.9), (5.12), and (5.16), we also conclude (5.36) and obtain the same expression for $\frac{\overline{V}'}{\overline{V}''}$ as in (5.37).

In subcase 64, we have $B_5<C_1$ , $B_5<C_2<A^-<C_3\le B_4\le A^+$ , and $B_4\le B_1$ . From (5.6), (5.9), (5.12), and (5.16), we obtain

(5.38) \begin{equation}\underset{a_1\in[0,A_1]}{\mathrm{inf}}\underset{a_2\in[0,A_2]}{\sup}L^{a_1,a_2}\overline{V}=\begin{cases}L^{A_1,\hat{a}_{2,A_1,\overline{V}}}\overline{V}&\ \ \Big\{x\;:\;\frac{\overline{V}'}{\overline{V}''}\le B_5\Big\},\\[5pt] L^{A_1,A_2}\overline{V}&\ \ \Big\{x\;:\;B_5<\frac{\overline{V}'}{\overline{V}''}\le A^-\Big\},\\[5pt] L^{0,\hat{a}_{2,0,\overline{V}}}\overline{V}&\ \ \Big\{x\;:\;A^-<\frac{\overline{V}'}{\overline{V}''}<B_4\Big\},\\[5pt] L^{0,A_2}\overline{V}&\ \ \Big\{x\;:\;B_4\le\frac{\overline{V}'}{\overline{V}''}\le B_1\Big\},\\[5pt] L^{A_1,A_2}\overline{V}&\ \ \Big\{x\;:\;B_1<\frac{\overline{V}'}{\overline{V}''}\Big\},\end{cases}\end{equation}

and then

(5.39) \begin{equation}\frac{\overline{V}''}{\overline{V}'}=\frac{1_{(-\infty,x_{5,-}]}}{f_1}+g1_{(x_{5,-},x_6^-]}+\frac{1_{(x_{6}^-,x_{4,-}]}}{f_3}+g_21_{[x_{4,-},x_{1,-}]}+g1_{(x_{1,-},x_0)},\end{equation}

which is a continuous, positive, and decreasing function in the concave region if $B_5>0$ . When $B_5\le 0$ , we obtain

(5.40) \begin{equation}\underset{a_1\in[0,A_1]}{\mathrm{inf}}\underset{a_2\in[0,A_2]}{\sup}L^{a_1,a_2}\overline{V}=\begin{cases}L^{A_1,A_2}\overline{V}&\ \ \Big\{x\;:\;\frac{\overline{V}'}{\overline{V}''}\le A^-\Big\},\\[5pt] L^{0,\hat{a}_{2,0,\overline{V}}}\overline{V}&\ \ \Big\{x\;:\;A^-<\frac{\overline{V}'}{\overline{V}''}<B_4\Big\},\\[5pt] L^{0,A_2}\overline{V}&\ \ \Big\{x\;:\;B_4\le\frac{\overline{V}'}{\overline{V}''}\le B_1\Big\},\\[5pt] L^{A_1,A_2}\overline{V}&\ \ \Big\{x\;:\;B_1<\frac{\overline{V}'}{\overline{V}''}\Big\},\end{cases}\end{equation}

and

(5.41) \begin{equation}\frac{\overline{V}''}{\overline{V}'}=g1_{(-\infty,x_6^-]}+\frac{1_{(x_{6}^-,x_{4,-}]}}{f_3}+g_21_{[x_{4,-},x_{1,-}]}+g1_{(x_{1,-},x_0)},\end{equation}

which is a continuous, positive, and decreasing function in the concave region in subcase 64.

In subcase 65, we have $C_1\le B_5$ , $C_2\le B_5<B_4<C_3$ , $A^-\le B_5<A^+<B_4$ , and $B_1<B_4$ . From (5.6), (5.9), (5.12), and (5.16), we see that (5.33) holds and we obtain the same expression for $\frac{\overline{V}'}{\overline{V}''}$ as in (5.34).

In subcase 66, we have $B_5<C_1$ , $B_5<C_2<B_4<C_3$ , $B_5<C_2<A^-<A^+<B_4$ , and $B_1\le B_4$ . From (5.6), (5.9), (5.12), and (5.16), we obtain the same expression for $\frac{\overline{V}'}{\overline{V}''}$ as in (5.29) if $B_5>0$ and as in (5.32) if $B_5\le 0$ in the concave region.

In subcase 67, we have $B_5<C_1$ , $B_5<C_2<B_4<C_3$ , and $B_1\le B_4$ . From (5.6), (5.9), (5.12), and (5.16), we obtain the same expression for $\frac{\overline{V}'}{\overline{V}''}$ as in (5.18) if $B_5>0$ , and as in (5.20) if $B_5\le 0$ in the concave region.

In case 7, note that $C_3\le 0$ and $B_5>0$ . From $\frac{A_2}{A_1}>\frac{\rho}{2\rho^2-1}>\zeta_4$ due to $2\rho<\frac{\theta_2}{\theta_4}$ , we have $C_1<B_5$ , $C_3\le C_2<B_5<B_4$ , and $B_4<B_1$ . Thus, (5.36) holds and we obtain the same expression for $\frac{\overline{V}'}{\overline{V}''}$ as in (5.37) in the concave region.

We have thus found the expression for $\frac{\overline{V}'}{\overline{V}''}$ in the concave region $(\!-\!\infty,x_0)\cap[l, u]$ for all possible parameter cases. The results are summarized in Table 3 of Appendix D.

5.2. The convex region with $\overline{\boldsymbol{V}}''(\boldsymbol{x})>0$

In this region, we have $\frac{\overline{V}'}{\overline{V}''}<0$ . Since $L^{a_1,a_2}\underline{V}(x)$ is a quadratic function in $a_2$ with a positive coefficient on the second-degree term, the maximizer of $L^{a_1,a_2}\overline{V}(x)$ in $a_2$ is $\overline{a}^*(x,a_1)=A_2$ if the vertex point $\hat{a}_{2,a_1,\overline{V}}$ satisfies $\hat{a}_{2,a_1,\overline{V}}=\theta_2\frac{\overline{V}'}{\overline{V}''}+\rho a_1\le\frac{A_2}{2}$ , which is equivalent to $a_1\in (\!-\!\infty,\tilde{a}_{1,\overline{V}}]\cap[0,A_1]$ , where

(5.42) \begin{eqnarray}\tilde{a}_{1,\underline{V}}(x)=-\frac{\theta_2}{\rho}\frac{\overline{V}'}{\overline{V}''}+\frac{A_2}{2\rho},\end{eqnarray}

which is positive in the convex region. And the maximizer in $a_2$ is $\overline{a}^*(x,a_1)=0$ if $\hat{a}_{2,a_1,\overline{V}}=\theta_2\frac{\overline{V}'}{\overline{V}''}+\rho a_1\ge\frac{A_2}{2}$ , which is equivalent to $a_1\in [\tilde{a}_{1,\overline{V}}(x),\infty)\cap[0,A_1]$ .

For convenience, below we write $\tilde{a}_1=\tilde{a}_{1,\underline{V}}$ . Then the maximizer of $L^{a_1,a_2}\overline{V}(x)$ in $a_2$ (given $a_1\in[0,A_1]$ ) can be expressed as

(5.43) \begin{equation}\overline{a}^*(x,a_1)=\begin{cases}0 & \tilde{a}_1<A_1\ \ a_1\in[\tilde{a}_1, A_1],\\[5pt] A_2 & \tilde{a}_1<A_1\ \ a_1\in[0, \tilde{a}_1),\\[5pt] A_2 & \tilde{a}_1\ge A_1\ \ a_1\in[0, A_1], \end{cases}\end{equation}

which takes only the boundary values 0 and $A_2$ .

We now define two regions which form a partition of the convex region:

(5.44) \begin{equation}\begin{split}&\mathcal{R}_{1}^+=\{x\in (l,u)\;:\; \overline{V}''>0,\ \tilde{a}_1\ge A_1\}=\bigg\{x\in (l,u)\;:\; \overline{V}''>0,\ \frac{\overline{V}'}{\overline{V}''}\le B_8\bigg\},\\[5pt] &\mathcal{R}_{2}^+=\{x\in (l,u)\;:\; \overline{V}''>0,\ \tilde{a}_1<A_1\}=\bigg\{x\in (l,u)\;:\; \overline{V}''>0,\ \frac{\overline{V}'}{\overline{V}''}>B_8\bigg\},\end{split}\end{equation}

where $B_8$ is given in (4.52).

For $x\in\mathcal{R}^+_1$ , the maximizer of $L^{a_1,a_2}\overline{V}$ in $a_2$ is $\overline{a}^*(x,a_1)=A_2$ for all $a_1\in [0,A_1]$ . The vertex point of $L^{a_1,A_2}\overline{V}$ in $a_1$ is $\hat{a}_{1,A_2,\overline{V}}=-\theta_1\frac{\overline{V}'}{\overline{V}''}+\rho A_2$ , which is positive in the convex region. We obtain the following lemma to find the minimizer of $L^{a_1,A_2}\overline{V}$ in $a_1$ .

Lemma 5.7. For $x\in\mathcal{R}^+_1$ , we have

(5.45) \begin{equation}\underset{a_1\in[0,A_1]}{\mathrm{inf}}\underset{a_2\in[0,A_2]}{\sup}L^{a_1,a_2}\overline{V}=\begin{cases}L^{\hat{a}_{1,A_2,\overline{V}}, A_2}\overline{V}&\ \ \big\{x\;:\;0<\hat{a}_{1,A_2,\overline{V}}<A_1\big\}=\Big\{x\;:\;B_7<\frac{\overline{V}'}{\overline{V}''}\Big\},\\[5pt] L^{A_1,A_2}\overline{V}&\ \ \big\{x\;:\;\hat{a}_{1,A_2,\overline{V}}\ge A_1\big\}=\Big\{x\;:\;\frac{\overline{V}'}{\overline{V}''}\le B_7\Big\},\end{cases}\end{equation}

where $B_7$ is given in (4.50).

For $x\in\mathcal{R}^+_2$ , the maximizer of $L^{a_1,a_2}\overline{V}$ in $a_2$ is $\overline{a}^*(x,a_1)=A_2$ if $a_1\in (0,\tilde{a}_1]$ , and $\overline{a}^*(x,a_1)=0$ if $a_1\in (\tilde{a}_1,A_1)$ . The vertex point of $L^{a_1,0}\overline{V}$ in $a_1$ is $\hat{a}_{1,0,\overline{V}}=-\theta_1\frac{\overline{V}'}{\overline{V}''}$ , which is positive in the convex region. We obtain the following lemma to find the minimizer in $a_1\in (0,\tilde{a}_1]$ of $L^{a_1,A_2}\overline{V}$ and $a_1\in (\tilde{a}_1,A_1)$ of $L^{a_1,0}\overline{V}$ in the region $\mathcal{R}^+_{2}$ .

Lemma 5.8. For $x\in\mathcal{R}^+_2$ , if $\frac{\theta_2}{\theta_1}<\rho$ , we have

(5.46) \begin{equation}\underset{a_1\in[0,\tilde{a}_1]}{\mathrm{inf}}\underset{a_2\in[0,A_2]}{\sup}L^{a_1,a_2}\overline{V}=\begin{cases}L^{\hat{a}_{1,A_2,\overline{V}}, A_2}\overline{V}&\ \ \big\{x\;:\;0<\hat{a}_{1,A_2,\overline{V}}<\tilde{a}_1\big\}=\Big\{x\;:\;C_4<\frac{\overline{V}'}{\overline{V}''}\Big\},\\[5pt] L^{\tilde{a}_1,A_2}\overline{V}&\ \ \{x\;:\;\hat{a}_{1,A_2,\overline{V}}\ge\tilde{a}_1\}=\Big\{x\;:\;\frac{\overline{V}'}{\overline{V}''}\le C_4\Big\},\end{cases}\end{equation}

and

(5.47) \begin{equation}\underset{a_1\in(\tilde{a}_1,A_1]}{\mathrm{inf}}\underset{a_2\in[0,A_2]}{\sup}L^{a_1,a_2}\overline{V}=\begin{cases}L^{\tilde{a}_1,0}\overline{V}&\ \ \{x\;:\;\hat{a}_{1,0,\overline{V}}\le\tilde{a}_1\}=\Big\{x\;:\;C_5\le\frac{\overline{V}'}{\overline{V}''}\Big\},\\[5pt] L^{\hat{a}_{1,0,\overline{V}}, 0}\overline{V}&\ \ \{x\;:\;\tilde{a}_1<\hat{a}_{1,0,\overline{V}}<A_1\}=\Big\{x\;:\;B_6<\frac{\overline{V}'}{\overline{V}''}<C_5\Big\},\\[5pt] L^{A_1,0}\overline{V}&\ \ \{x\;:\;\hat{a}_{1,0,\overline{V}}\ge A_1\}=\Big\{x\;:\;\frac{\overline{V}'}{\overline{V}''}\le B_6\Big\},\end{cases}\end{equation}

where $B_6$ is given in (4.50) and

(5.48) \begin{equation}C_4=\frac{(2\rho^2-1)A_2}{2(\rho\theta_1-\theta_2)},\quad C_5=-\frac{A_2}{2(\rho\theta_1-\theta_2)}.\end{equation}

For $x\in\mathcal{R}^+_2$ , if $\rho<\frac{\theta_2}{\theta_1}$ , we have

(5.49) \begin{equation}\underset{a_1\in[0,A_1]}{\mathrm{inf}}\underset{a_2\in[0,A_2]}{\sup}L^{a_1,a_2}\overline{V}=\begin{cases}L^{\hat{a}_{1,A_2,\overline{V}}, A_2}\overline{V}&\ \ \big\{x\;:\;0<\hat{a}_{1,A_2,\overline{V}}<\tilde{a}_1\big\}=\Big\{x\;:\;\frac{\overline{V}'}{\overline{V}''}<C_4\Big\},\\[5pt] L^{\tilde{a}_1,A_2}\overline{V}&\ \ \big\{x\;:\;\hat{a}_{1,A_2,\overline{V}}\ge\tilde{a}_1\big\}=\Big\{x\;:\;\frac{\overline{V}'}{\overline{V}''}\ge C_4\Big\}.\end{cases}\end{equation}

For $x\in\mathcal{R}^+_2$ , if $\rho=\frac{\theta_2}{\theta_1}$ , we have

(5.50) \begin{equation}\underset{a_1\in[0,A_1]}{\mathrm{inf}}\underset{a_2\in[0,A_2]}{\sup}L^{a_1,a_2}\overline{V}=\begin{cases}L^{\hat{a}_{1,A_2,\overline{V}}, A_2}\overline{V}&\ \ \Big\{x\;:\;0<\hat{a}_{1,A_2,\overline{V}}<\tilde{a}_1\Big\}\ if \ \rho<\sqrt{1/2},\\[5pt] L^{\tilde{a}_1,A_2}\overline{V}&\ \ \Big\{x\;:\;\hat{a}_{1,A_2,\overline{V}}\ge\tilde{a}_1\Big\}\ if\ \rho\ge\sqrt{1/2}.\end{cases}\end{equation}

Short calculations yield the following lemma.

From Lemma 5.9, we have $B_8<B_6<C_5<C_4$ if $\frac{\theta_2}{\theta_1}<\rho$ and $\frac{A_2}{A_1}<2(\rho-\frac{\theta_2}{\theta_1})$ ; on the other hand, we have $C_5\le B_6\le B_8$ if $\frac{\theta_2}{\theta_1}<\rho$ and $\frac{A_2}{A_1}\ge 2(\rho-\frac{\theta_2}{\theta_1})$ .

Next we consider the following parameter cases to find the expression for $\frac{\overline{V}''}{\overline{V}'}$ in the convex region:

1. 1. $\frac{A_2}{A_1}\ge 2\rho$ ( $B_8\geq 0$ );

2. 2. $\frac{A_2}{A_1}<2\rho$ , $\frac{\theta_2}{\theta_1}<\rho$ , $\frac{A_2}{A_1}<2(\rho-\frac{\theta_2}{\theta_1})$ ;

3. 3. $\frac{A_2}{A_1}<2\rho$ , $\frac{\theta_2}{\theta_1}<\rho$ , $\frac{A_2}{A_1}\ge 2(\rho-\frac{\theta_2}{\theta_1})$ ;

4. 4. $\frac{A_2}{A_1}<2\rho$ , $\frac{\theta_2}{\theta_1}\ge\rho$ .

In case 1, from $B_8\geq 0$ , the region $\mathcal{R}_2^+$ degenerates. If $B_7<0$ , from (5.45), we have

(5.51) \begin{equation}\underset{a_1\in[0,A_1]}{\mathrm{inf}}\underset{a_2\in[0,A_2]}{\sup}L^{a_1,a_2}\overline{V}=\begin{cases}L^{\hat{a}_{1,A_2,\overline{V}}, A_2}\overline{V}&\ \ \Big\{x\;:\;0<\hat{a}_{1,A_2,\overline{V}}<\tilde{a}_1\Big\}=\Big\{B_7<\frac{\overline{V}'}{\overline{V}''}\Big\},\\[5pt] L^{A_1,A_2}\overline{V}&\ \ \Big\{x\;:\;\hat{a}_{1,A_2,\overline{V}}\ge\tilde{a}_1\Big\}=\Big\{\frac{\overline{V}'}{\overline{V}''}\le B_7\Big\},\end{cases}\end{equation}

and

(5.52) \begin{equation}\frac{\overline{V}''}{\underline{V}'}=g1_{(x_0,x_{3,+}]}+\frac{1_{(x_{3,+},\infty)}}{f_5},\end{equation}

which is a continuous, negative, and decreasing function where $x_{3,+}$ is defined in (4.68) and $f_5$ in (4.67) in the convex region. If $B_7\geq 0$ , we obtain

(5.53) \begin{equation}\underset{a_1\in[0,A_1]}{\mathrm{inf}}\underset{a_2\in[0,A_2]}{\sup}L^{a_1,a_2}\overline{V}=L^{A_1,A_2}\overline{V}\end{equation}