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Let $v:\mathbb{R}_{+}\mapsto \mathbb{R}_{+}$ and $v(t)$ is non-increasing, i.e., $\dot{v}(t)\le 0$. If there exist positive constants $k_{1}$, $k_{2}$, $c_{1}$, $c_{2}$, such that \begin{align} k_{1}e^{-c_{1}t}v(0) \le v(t)\le k_{2}e^{-c_{2}t}v(0). \end{align} Can I conclude that \begin{align} \dot{v}(t)\le -c_{3}v(t) \end{align} for some positive constant $c_{3}$?

Under a weaker assumption, a similar question has been asked, but with counterexamples: function bounded by an exponential has a bounded derivative?

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    $\begingroup$ No. Essentially the same counter-example type holds. You can have a curve that starts near the bottom curve, but then takes a flat part (with slope 0 or as close to 0 as we want) to go over to the top curve. $\endgroup$ – Michael Jan 3 at 8:18

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