\tilde v_*(t,y)+\hat xy= v_*(t,\hat x)\end{aligned}$$ From inequality and equation , we deduce that $$\begin{aligned}
\tilde v_*(t,y)= \Max_{x\geq (\beta-\alpha)(T-t)}[v_*(t,x)-xy].\end{aligned}$$ and so $\tilde v_*$ is convex in $y$.\
2) Fix $(t,y)\in [0,T)\times (0,\infty)$. Since at time $t$, only $D\in \Dc_t$ could make a jump, we have $y=Y_t\geq Y_{t^+}$. From the dynamic programming principle we have $$\begin{aligned}
\tilde v(t,y)\leq \tilde v(t,y(1-\delta))\,\mbox{ for all }0<\delta<1,\end{aligned}$$ and so the dual value function $\tilde v$ in non-increasing with respect to $y$. This yields that $\tilde v_*$ and $\tilde v^*$ are nonincreasing.\
3) Fix $(t,y)\in [0,T)\times (0,\infty)$. From the definition of $\tilde v^*(t,y)$, there exists a sequence $\big((t_n,y_n)\big)_n$ such that $v(t_n,y_n)\longrightarrow \tilde v_*(t,y)$ when $n$ goes to infinity. From the growth condition of $\tilde v$, we have $$\begin{aligned}
|\tilde v(t_n,y_n)| \leq K(y_n+y_n^{-\gamma}),\end{aligned}$$ where $K$ is a positive constant. Sending $n$ to infinity, we obtain the desired result. We use similar arguments to prove that $\Sup_{y>0} \frac{|\tilde v^*(t,y)|}{y+y^{-\gamma}}<\infty$.\
4) We prove only the first inequality. The second one is obtained by using similar arguments. Fix $(t,y_1,y_2)\in [0,T)\times (0,\infty)\times (0,\infty)$. From the definition of $\tilde v^*(t,y_2)$, there exists a sequence $\big((t_n,y_{2,n})\big)_n$ such that $v(t_n,y_{2,n})\longrightarrow \tilde v_*(t,y_2)$ when $n$ goes to infinity. From inequality , we have $$\begin{aligned}
\tilde v(t_n,y_{1})-\tilde
  