Dear friends,
Let me introduce you the problem killing me.
Assume that $\displaystyle x,y\in\mathrm{C}\big([0,\infty),(-\infty,\infty)\big)$ and consider the following equation
$\displaystyle x(t)=\int_{0}^{t}\mathrm{e}^{-\textstyle\int_{\eta}^{t}p(\zeta)\mathrm{d}\zeta}x (\eta)\mathrm{d}\eta+y(t)$ for $\displaystyle t\geq0$, where $\displaystyle p\in\mathrm{C}\big([0,\infty),[0,\infty)\big)$.
Prove: $\displaystyle y\geq0$ implies $\displaystyle x\geq0$.
Note: We do not know whether y is differentiable or not.
The first method comes into my mind is proving by contradiction.
Suppose the contrary that $\displaystyle y\geq0$ but $\displaystyle x<0$ for some values in $\displaystyle [0,\infty)$.
Then we may pick $\displaystyle s\in[0,\infty)$ satisfying $\displaystyle x\geq0$ on $\displaystyle [0,s]$ and $\displaystyle x<0$ on $\displaystyle (s,s+\varepsilon)$ for some $\displaystyle \varepsilon>0$.
For simplicity, set $\displaystyle \mu:=\int_{0}^{s}\mathrm{e}^{-\textstyle\int_{\eta}^{s+\varepsilon}p(\zeta)\math rm{d}\zeta}x(\eta)\mathrm{d}\eta$ and $\displaystyle x_{\min}:=\min\nolimits_{\eta\in[s,s+\varepsilon]}\{x(\eta)\}$.
Then, for all $\displaystyle t\in(s,s+\varepsilon)$, we have
$\displaystyle 0>x(t)=\int_{0}^{s}\mathrm{e}^{-\textstyle\int_{\eta}^{t}p(\zeta)\mathrm{d}\zeta}x (\eta)\mathrm{d}\eta$$\displaystyle +\int_{s}^{t}\mathrm{e}^{-\textstyle\int_{\eta}^{t}p(\zeta)\mathrm{d}\zeta}x (\eta)\mathrm{d}\eta+y(t)$
.............$\displaystyle \geq\mu+x_{\min}\int_{s}^{t}\mathrm{e}^{-\textstyle\int_{s}^{t}p(\zeta)\mathrm{d}\zeta}\mat hrm{d}\eta$
.............$\displaystyle \geq\mu+x_{\min}\int_{s}^{t}\mathrm{d}\eta$$\displaystyle =\mu+(t-s)x_{\min}$.
If $\displaystyle \mu>0$, then letting $\displaystyle \delta>0$ with $\displaystyle \delta\leq\min\{-\mu/x_{\min},\varepsilon/2\}$, we get $\displaystyle s+\delta\in(s,s+\varepsilon)$ and thus $\displaystyle 0>x(s+\delta)\geq\mu+\delta x_{\min}\geq0$.
This is a contradiction.
But my method
fails to deliver the case $\displaystyle \mu=0$.
I need help at this point.
Your help is highly appreciated!