to $g'(0)>0$.

Then we can prove the following fundamental theorem.

\[smallmaturity\] Suppose that the condition (\[integralefinie\]) is satisfied and that the function $x\mapsto
x\beta(m,x)$ is Lipschitz continuous for all $m\in[0,1]$. Set $$\tau_{0} =\sup_{m>0} \Big( \int_{m}^{g^{-1}(m)} \frac{ds}{V(s)} \Big).$$ Suppose that $N_{1}(t,m)$ and $N_{2}(t,m)$ are two solutions of Problem (\[mildformulation\])-(\[icmf\]) for the initial conditions $\varphi_{1}$ and $\varphi_{2}$ respectively. If $$\underline{\tau} > \tau_{0},$$ and if there exists $b \in (0,1]$ such that $$\varphi_{1}(t,m)=\varphi_{2}(t,m),$$ for $m \in [0,b]$ and $t \in [0,\overline{\tau}]$, then there exists $\overline{t} > \overline{\tau}$ such that $$N_{1}(t,m)=N_{2}(t,m),$$ for $m \in [0,1]$ and $t \geq \overline{t}$.

Since $\underline{\tau} > \tau_{0}$, then, from (\[delta\]), $\Delta(a,m)<m$ for all $a\in[\underline{\tau},\overline{\tau}]$.

We first show, using the Gronwall’s lemma and a method of steps, that $N_{1}(t,m)=N_{2}(t,m)$ for $m\in[0,b]$ and $t\geq 0$.

Let $ m \in [0,b]$ and $t\geq\overline{\tau}$ (if $t\in[0, \overline{\tau}]$, the result is obvious). Then $$\pi_{-(t-\overline{\tau})}(m) < m \leq b.$$ So $$\varphi_{1}(\overline{\tau},\pi_{-(t-\overline{\tau})}(m))=\varphi_{2}(\overline{\tau},\pi_{-(t-\overline{\tau})}(m)).$$ Let $t \in [\overline{\tau},\overline{\tau}+\underline{\tau}]$. Then, $$\Delta(a,\pi_{-(t-s)}(m)) < \pi_{-(t-s)}(m) < m \leq b,$$ and consequently $$G(N_{1})(t,m)=G(N_{2})(t,m).$$ As the mapping $x\mapsto x\beta(m,x)$ is Lipschitz continuous and $K$ is bounded, we can write $$\begin{aligned}
|N_{1}(t,m)-N_{2}(t,m)| & = & |J(N_{1})(t,m)-J(N_{2})(t,m)| \nonumber \\
            & \leq & c \int_{\overline{\tau}}^{t} |N_{1}(s,\pi_{-(t-s)}(m))-N_{2}(s,\pi_{-(t-s)}(m))| ds, \nonumber\end{aligned}$$ where $c$ is a nonnegative constant. Then, 