\overline{\tau}$, and $$\label{icdde}
x(t)=\psi(t):=\varphi(t,0), \ \ \textrm{for } t\in[0,\overline{\tau}],$$ where $$\left\{ \begin{array}{l}
I_{0}=\delta(0)+V'(0),\\
\beta_{0}(x)=\beta(0,x), \quad \textrm{for all } x\in\mathbb{R},\\
Z(a)=\zeta(a,0)=e^{-\nu a}k(0,a), \quad \textrm{with } \nu=\gamma(0)+V'(0).
\end{array} \right.$$ It is clear that $I_{0}\geq 0$.

Since $\varphi\in C([0,\overline{\tau}]\times[0,1])$, then $\psi\in C[0,\overline{\tau}]$, where $C[0,\overline{\tau}]$ is the space of all continuous functions on $[0,\overline{\tau}]$. Under classical Lipschitz conditions on the function $\beta_{0}$, one can prove the existence and uniqueness of solutions of Equation (\[dde\]) (see Hale and Lunel [@halelunel]).

As in [@mackey1978], it is natural to add the assumption that the function $\beta_{0}$ is decreasing for $x\geq 0$ and satisfies $$\lim_{x\to +\infty} \beta_{0}(x)=0.$$ A first property of the solutions of Problem (\[dde\])-(\[icdde\]) is given in the following proposition.

Let $\psi\in C[0,\overline{\tau}]$. If $\psi$ is non-negative on $[0,\overline{\tau}]$, then the solution of Equation (\[dde\]) for the initial condition $\psi$, denoted by $x^{\psi}$, is non-negative on $[\overline{\tau}, +\infty)$.

Suppose that the assumption of the proposition is false. First, suppose that there exist $t_{0}\in[\overline{\tau},\overline{\tau}+\underline{\tau})$ and $\varepsilon>0$ such that $ t_{0}+\varepsilon\leq\overline{\tau}+\underline{\tau}$, $x^{\psi}(t_{0})=0$ and $x^{\psi}(t)<0$ for $t\in(t_{0},t_{0}+\varepsilon)$. Then, for all $t\in[t_{0},t_{0}+\varepsilon)$, $$-(I_{0}+\beta_{0}(x^{\psi}(t)))x^{\psi}(t) \geq 0,$$ and $$\int_{\underline{\tau}}^{\overline{\tau}} Z(a)\beta_{0}(x^{\psi}(t-a))x^{\psi}(t-a)da=\int_{\underline{\tau}}^{\overline{\tau}} Z(a)\beta_{0}(\psi(t-a))\psi(t-a)da \geq 0,$$ because, for all $t\in[\overline{\tau},\overline{\tau}+\underline{\tau}]$ and all $a\in[\underline{\tau},\overline{\tau}]$, $t-a\in[0,\overline{\tau}]$. Hence, for $t\in[t_{0},t_{0}+\varepsilon)$, $(x^{\psi})'(t)\geq 0$, and we obtain a contradiction. So 