where $\varepsilon$ is given by (\[hypbeta\]), then, the solution $N^{\varphi}$ of Problem (\[mildformulation\])-(\[icmf\]) satisfies $$|N^{\varphi}(t,m)|\leq \varepsilon,$$ for all $(t,m)\in[\overline{\tau},+\infty)\times[0,1]$.

It is clear that the assumption in this proposition implies that $I>0$. Let $\varepsilon >0$ be given by (\[hypbeta\]) and $\varphi\in C([0,\overline{\tau}]\times[0,1])$ such that $||\varphi||\leq\varepsilon$.\
We define the sequence $(N_{n})_{n\in\mathbb{N}}$ by $$\label{N0def}
N_{0}(t,m)=\left\{ \begin{array}{ll}
\varphi(\overline{\tau},\pi_{-(t-\overline{\tau})}(m))K(t-\overline{\tau},m), &  \textrm{ if } t\geq\overline{\tau}, \\
\varphi(t,m), &\textrm{ if } t\in[0,\overline{\tau}],
\end{array} \right.$$ and, for $n\geq 1$, $$\label{Nndef}
N_{n}(t,m)=\left\{ \begin{array}{ll}
N_{0}(t,m)+G(N_{n-1})(t,m)-J(N_{n-1})(t,m), & \textrm{ if } t\geq\overline{\tau}, \\
\varphi(t,m), &\textrm{ if } t\in[0,\overline{\tau}].
\end{array} \right.$$ Since $\pi_{-(t-\overline{\tau})}(m) < m \leq g(1)$, then $|\varphi(\overline{\tau},\pi_{-(t-\overline{\tau})}(m))| \leq \varepsilon$. Moreover, $$|K(t-\overline{\tau},m)|\leq e^{-I(t-\overline{\tau})},$$ therefore $$\label{hypN0}
|N_{0}(t,m)| \leq \varepsilon e^{-I(t-\overline{\tau})} \leq \varepsilon.$$ Let $T>\overline{\tau}$ be given. We show by induction that for all $n\in\mathbb{N}$ and all $(t,m)\in[\overline{\tau},T]\times[0,1]$, $$\label{hyprecurrence}
|N_{n}(t,m)| \leq \varepsilon.$$ For $n=0$, we just showed it in (\[hypN0\]). Suppose that (\[hyprecurrence\]) is true for $n\geq 0$. Then $$|N_{n+1}(t,m)| \leq |N_{0}(t,m)|+|G(N_{n})(t,m)|+|J(N_{n})(t,m)|,$$ with $$|G(N_{n})(t,m)|\leq 2\tilde{\zeta}\varepsilon L\int_{\overline{\tau}}^{t} e^{-I(t-s)}ds,$$ and $$|J(N_{n})(t,m)|\leq \varepsilon L\int_{\overline{\tau}}^{t}e^{-I(t-s)}ds.$$ So $$\begin{aligned}
|N_{n+1}(t,m)| & \leq & \varepsilon \bigg( e^{-I(t-\overline{\tau})} +L(1+2\tilde{\zeta})\int_{\overline{\tau}}^{t}e^{-I(t-s)}ds \bigg) \\
        & \leq & \varepsilon \bigg( e^{-I(t-\overline{\tau})} +L(1+2\tilde{\zeta})\frac{1-e^{-I(t-\overline{\tau})}}{I} \bigg).\end{aligned}$$ Set $$\alpha=\frac{L(2\tilde{\zeta}+1)}{I}.$$ Noting that $\alpha < 1$, we obtain $$\begin{aligned}
|N_{n+1}(t,m)| & \leq  & \varepsilon \Big( e^{-I(t-\overline{\tau})}(1-\alpha) + \alpha \Big), \\
        & \leq  