# Function bounded by exponential functions has a bounded derivative?

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?

• 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. – Michael Jan 3 at 8:18