$\kappa \to \infty$ and denote by $U_{\kappa}$, $U \in ({\mathbb{R}}^{N_h^2})^N$ the corresponding solutions of . Set $$E_{\kappa} = \sup_{ n=0,\dots,N-1}  \left\| f_h[M_{\kappa}^{n+1}]  -f_h[M^{n+1}] \right\|_\infty$$ and observe that, by assumption (F1) on $f_h$, $E_{\kappa}\to 0$ for $k\to \infty$. Set $V^n_{\kappa}=U^n_{\kappa}- \frac{\Gamma(2-\alpha)}{\alpha(1-\alpha)}((N-n){{\Delta t}})^\alpha E_{\kappa}$, $n=0,\dots,N$, then $$\begin{aligned}
\bar D_{\Delta t}^{\alpha} V^n_{i,j}  - (\Delta_h V^{n})_{i,j} + g(x_{i,j},[D_h V^{n}]_{i,j})\le  f_h[M_k^{n+1}]_{i,j}-E_{\kappa}= \\
f_h[M^{n+1}]_{i,j}+f_h[M_k^{n+1}]_{i,j}-f_h[M^{n+1}]_{i,j}-  E_{\kappa}
\le f_h[M^{n+1}]_{i,j}.\end{aligned}$$ Moreover $V^N\le U^N$ and therefore, by Prop. \[prop:ex\_HJ\], $V^n_{\kappa}\le U^n$ for any $n=0,\dots,N$. Using a similar argument for $W^n_{\kappa}=U^n_{\kappa}+\frac{\Gamma(2-\alpha)}{\alpha(1-\alpha)}((N-n){{\Delta t}})^\alpha E_{\kappa}$, $n=0,\dots,N$, we get that for any $\kappa$ $$\sup_{n=0,\dots,N}\|U^n_{\kappa} -U^n\|_\infty\le \frac{\Gamma(2-\alpha)}{\alpha(1-\alpha)}T^\alpha    E_{\kappa}$$ and therefore the continuity of the map $\Psi_1$.\
Given $U=\Psi_1(M)$, we first show that, for ${{\Delta t}}$ sufficiently small, $\tilde M=\Psi_2(U)$ is well defined. First observe that, by Prop. \[prop:lip\_U\], $\|D_h U\|_\infty$ is bounded uniformly in $M$. Therefore, since $g$ is $C^1$, there exists a constant $C$, independent of $M$, such that for $k=1,2,3,4,$ $$\label{eq:Fix1}
\left|\frac{\partial g}{\partial q_k} (x_{i,j},[D_hU^n]_{i,j})\right|\le C, \quad 0\le i,j\le N_h,\,0\le n\le N.$$ Hence, arguing as in Prop. \[prop:ex\_FP\], we obtain that, for ${{\Delta t}}$ sufficiently small, there exists a solution $\tilde M$ of . Moreover $\tilde M\in {\mathcal{K}}_h^{N}$. The continuity of the map $\Psi_2$ follows by the regularity of $g$ and the linearity 