$$\begin{aligned}
A^\rho\left(t,y,\tilde v, \Dy {\tilde v}\right) :=\Sum_{i=1}^d \pi_i\left( \tilde v(t,\rho_i y)- 
\tilde v(t,y)-(\rho_i -1)y \Dy {\tilde v}(t,y)\right).\end{aligned}$$ The following lemma sates some properties of the functions $\tilde v_*$ and $\tilde v^*$.

\[conv\] We assume that there exists a solution the the dual problem . The following properties hold :\
1) The functions $\tilde v^*$ and $\tilde v_*$ are convex in $y$.\
2) The functions $\tilde v^*$ and $\tilde v_*$ are nonincreasing on $(0,\infty)$.\
3) The functions $\tilde v_*$ and $\tilde v^*$ satisfy the following growth condition $$\begin{aligned}
\label{growthc}
\Sup_{y>0} \frac{|\tilde v_*(t,y)|}{y+y^{-\gamma}},\,\,\Sup_{y>0} \frac{|\tilde v^*(t,y)|}{y+y^{-\gamma}}<\infty\end{aligned}$$ 4) The functions $\tilde v_*$ and $\tilde v^*$ satisfy $$\begin{aligned}
\Sup_{y_1>0,y_2>0} 
\frac{|\tilde v_*(t,y_1)-\tilde v_*(t,y_2)|}
{|y_1-y_2|(1+y_{1}^{-(\gamma+1)}+y_{2}^{-(\gamma+1)})}<\infty.\end{aligned}$$ $$\begin{aligned}
\Sup_{y_1>0,y_2>0} 
\frac{|\tilde v^*(t,y_1)-\tilde v^*(t,y_2)|}
{|y_1-y_2|(1+y_{1}^{-(\gamma+1)}+y_{2}^{-(\gamma+1)})}<\infty.
 \end{aligned}$$

[**Proof.**]{} 1) Fix $\lambda \in (0,1)$ and $(y,y^{'},y^{''})\in (0,\infty)^3$ such that $y=\lambda y^{'}+(1-\lambda)y^{''}$. 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. We set $y_n=\lambda y^{'}+(1-\lambda)y^{''}_n$, then we have $y^{''}_n\longrightarrow y^{''}$ when $n$ goes to infinity. Since $\tilde v$ is convex in $y$, then we have $$\begin{aligned}
\label{v*conv}
\tilde v(t_n,y_n)&\leq& \lambda\tilde v(t_n,y^{'})+(1-\lambda)\tilde v(t_n,y^{''}_n)\end{aligned}$$ Sending $n$ to infinity, inequality implies $$\begin{aligned}
\tilde v^*(t,y)\leq  \lambda\tilde v^*(t,y^{'})+(1-\lambda)\tilde v^*(t,y^{''}),\end{aligned}$$ which is the desired result.\
We turn to 