I'm trying to prove that for the system $x' \leq -x$, if the system starts with $x \leq 0$, then it is always the case that $x \leq 0$. I'm sure that there are easy ways to prove this, but I'm particularly trying to understand how to prove this using barrier certificates. Barrier certificates are a way of proving reachability theorems. In particular, if one can build a function $V(x)$ such that $V(x) \leq 0$ within the safe region ($x \leq 0$), $V(x) > 0$ outside the safe region ($x > 0$) and has non-positive time derivative $\frac{dV(x)}{dx}x' \leq 0$, then one can conclude that if the system starts in the safe region, it stays in the safe region.
However, I can't build such a function for $x' \leq -x$ with safe region $x \leq 0$. Is there such a function?
Assuming $x$ is scalar, the following (smooth non-analytic) function may work:
$$ V(x) = \left\{\begin{matrix} \exp( - \frac{1}{x}), x>0,\\ 0, x \leq 0, \end{matrix}\right. $$
It satisfies:
$$ V(x) > 0, x >0, \\ V(x) \leq 0, x \leq 0 $$
While the derivative (exists everywhere) and is:
$$ \frac{\partial}{\partial x } V(x) = \left\{\begin{matrix} \exp( - \frac{1}{x})\frac{1}{x^2}, x>0,\\ 0, x \leq 0, \end{matrix}\right. $$
And so:
$$ \frac{\partial}{\partial x } V(x) \dot{x} = \left\{\begin{matrix} \exp( - \frac{1}{x})\frac{1}{x^2} \dot{x} < 0, x>0,\\ 0, x \leq 0 \end{matrix}\right. $$