monotonic property of the solution to an optimization problem $v\in[0,v_0]$ is a parameter, and I am trying to solve the following optimization problem that is parametrized by $v$.
\begin{align*}
y^*_v\triangleq\underset{y\geq 0}{\arg\max}~G(y;v)\triangleq\int_0^{w_0}F\bigg(y+p-h-x(w)-w,~v+x(w)-p-y\bigg)dw,
\end{align*} 
where
\begin{align*}
F(z_1,z_2)=\begin{cases}
&z_2,~\mbox{if }z_1\geq 0~\mbox{and }z_2\geq 0\\
&-\infty,~\mbox{otherwise}
\end{cases}
\end{align*}
And $x(w)$ is a non-negative function of $w\in[0,w_0]$ and $p>h>0$ are constants. And $v\in[0,v_0]$ for some constant $v_0>w_0$.
As $F$ is increasing in $z_2$ which contains a linear term of $v$, in addition, $z_1$ is independent of $v$. Hence, the whole integral is increasing in $z_2$. Therefore, the optimal solution $y^*_v$ is increasing in $v$. Note that $z_1$ in this function determines the integral domain. 
My question is: is $y^*_v$ a linear function of $v$? My hunch is that: depends on the value of the parameters of $w_0,v_0,p,h$, either $y^*_v=0,~\forall~v\in[0,v_0]$ or $y^*_v=0,~\forall~v\in[0,\bar{v}]$ for some $\bar{v}<v_0$ which is a function of $p,h,w_0,v_0$ and $v$ is linearly increasing in $v$ for $v\in[\bar{v},v_0]$ and at $v=\bar{v}$, $y^*_{\bar{v}}=0$. I believe that $y^*_v$ has this general form, i.e., there first should be a $\underline{v}$ such that for all $v\geq \underline{v}$, we have $G(y^*_v;v)\geq 0$, then $y^*_v=0$ for $v\in[\underline{v},\bar{v}]$ and $y^*_v$ is linearly increasing in $v$ for $v\in[\bar{v},v_0]$. Note that if $\bar{v}>v_0$, then this just means that for all $v\in[\underline{v},v_0]$, $y^*_v=0$. Similarly, if $\bar{v}<0$, then it means $y^*_v$ is linearly increasing in $v$ for all $v\in[0,v_0]$.
Further, can we show that if there exists $\bar{v}\in[0,v_0]$ such that $y^*_{\bar{v}}>0$, then there must be some $\tilde{v}\in[0,\bar{v}]$ such that $y^*_{\tilde{v}}=0$? (If my hunch is true, then the answer to this question is obvious)
What if the optimization problem is changed slightly to the following form (basically, $F(z_1,z_2)=0$ instead of $-\infty$ if at least one of $z_1$ and $z_1$ is negative)
\begin{align*}
y^*_v\triangleq
\underset{y\geq 0}{\arg\max}~G(y;v)=\int_0^{w_0}\mathbf{1}_{\{y+p-h-x(w)-w\geq 0\}}[v+x(w)-p-y]^+dw
\end{align*}
 A: I'm going to assume $w_0>0$ and $x$ is continuous. The optimization problem is equivalent to:
\begin{array}{lll}
\text{maximize}_y & \int_0^{w_0} v + x(w) - p - y \, dw \\
\text{subject to} & y + p - h - x(w) - w \geq 0 & \forall w\in[0,w_0] \\
                  & v + x(w) - p - y \geq 0 & \forall w\in[0,w_0] \\
                  & y \geq 0
\end{array}
Thus must be the case because if either of the first two inequalities is violated over a set of measure greater than zero, the integral will be $-\infty$. So the nonnegativity of $z_1$ and $z_2$ are implicit constraints.
Pulling the independent terms out of the integral and rearranging the inequalities yields
\begin{array}{lll}
\text{maximize}_{y,v} & - w_0 y + w_0 (v-p) + \int_0^{w_0} x(w) \, dw \\
\text{subject to}     & y \geq \max_{w\in[0,w_0]} x(w) - p  + w + h \\
                      & y \leq \min_{w\in[0,w_0]} x(w) - p + v \\
                      & y \geq 0
\end{array}
Now let us define the constants:
$$\alpha = \max\{0,\max_{w\in[0,w_0]} x(w) - p + w + h\}, \quad 
 \beta = \min_{w\in[0,w_0]} x(w) - p$$
Substituting these in, dropping the constants from the objective, and reversing its sense yields:
\begin{array}{lll}
\text{minimize}   & y \\
\text{subject to} & \alpha \leq y \leq v + \beta
\end{array}
From this, we can read the solution by inspection:


*

*If $v<\alpha-\beta$, then the problem is infeasible (the argmax is empty).

*Otherwise, $y=\alpha$.


It does not appear, then that $y_v$ is a linear function of $v$. Rather, it is either infeasible or equal to a constant function of $x,w_0,p,h$.
So the range of $x$ (or more specifically, $x-p$) plays an important part in determining the feasibility of the problem. Note that
$$\alpha - \beta = \max\{p, h + \max_{w\in[0,w_0]} (x(w) + w) \} - \min_{w\in[0,w_0]} x(w) $$
Since $h+\max_w x(w) > \min_w x(w)$, it must be the case that $\alpha - \beta > 0$. If, in addition, $\alpha-\beta\leq v_0$, then there is indeed a value of $v$ which attains $y=\alpha$. And of course, it is possible for $\alpha=0$ depending upon the precise values of $p$ and $h$; specifically,
$$\max_{w\in[0,w_0]} x(w) + w \leq p - h \quad\Longrightarrow\quad \alpha=0.$$
