contrary that there exists a constant $\eta >0$ such that $\tilde v^*(T,y)\geq \tilde U(y) +2 \eta$. From the definition of $\tilde v^*$, there exists a sequence $((t_n,y_n))_{n\in \N}$ such that $(t_n,y_n)\longrightarrow (T,y)$ and $\tilde v(t_n,y_n)\longrightarrow\tilde v^*(T,y)$ when $n$ tends to infinity, which implies $$\begin{aligned}
E[\tilde
U(y_nY_{T}^{\rho_n,D_n})+\int_{t_n}^{T}y_nY_s^{\rho_n,D_n}(\alpha-\beta+(\beta-\int_C\rho_{n_{s}}(z)\,z\,\pi(dz))_+)ds]\geq
\tilde U(y) +\eta,\end{aligned}$$ for all $Y^{\rho_n,D_n}\in \Yc^0(t_n)$.Choosing $\rho_{n_{s}}=1$ and $D_{n_{s}}=1$ for all $s\in[t_n,T]$, we obtain $$\begin{aligned}
\tilde U(y_n)+\int_{t_n}^{T}y_n(\alpha-\beta+(\beta-\int_C\,z\,\pi(dz))_+)ds\geq \tilde U(y) +\eta. \end{aligned}$$ Sending $n$ to infinity, we have $\tilde U(y) \geq \tilde U(y) +\eta$ which is wrong and so inequality is proved.\
We prove now inequality . From the definition of $\tilde v_*$, there exists a sequence $((t_n,y_n))_{n\in \N}$ such that $(t_n,y_n)\longrightarrow (T,y)$ and $\tilde v(t_n,y_n)\longrightarrow\tilde v_*(T,y)$ when $n$ tends to infinity.\
From the definition of the dynamic version of the value function, we have for all $x\geq (\beta-\alpha)(T-t_n)$ $$\begin{aligned}
\label{dyntn}
v(t_n,x)\geq U(x+(\alpha-\beta)(T-t_n)).\end{aligned}$$ Let $\epsilon>0$, there exists $n_0\in \N$ such that for all $n\geq n_0$, we have $$\begin{aligned}
\label{dyntnn}
 U(x+(\alpha-\beta)(T-t_n))\geq U(x)-\epsilon.\end{aligned}$$ Using the conjugate duality relation of Theorem 5.1 of Mnif and Pham [@mnipham01] and relations and , we obtain $$\begin{aligned}
\label{dyntnnn} 
\tilde v(t_n,y_n)&=&\Max_{x\geq (\beta-\alpha)(T-t_n) }[v(t_n,x)-xy_n]\nonumber\\ &\geq& \Max_{x\geq (\beta-\alpha)(T-t_n) }[U(x)-xy_n]-\epsilon.\end{aligned}$$ Since $(t_n,y_n)\longrightarrow (T,y)$ when $n$ tends to infinity, there exists $n_1\in \N$ such that 