function bounded by an exponential has a bounded derivative?

Question

here's the question. I want to be sure of that. Let $v:[0,\infty) \rightarrow \mathbb{R}_+$ a positive function satisfying $$\forall t \ge 0,\qquad v(t)\le kv(0) e^{-c t}$$ for some positive constants $c$ and $k$. Can I conclude that $$\dot{v}(t) \le -c v(t)$$ ?

Answer 1

I don't think you can. Look at something like $v(t) = e^{-t}\sin(g(t) t)$. Then $v'(t) = e^{-t}(\sin(g(t)t) + g(t)\sin(g(t)t)$. If you choose $g(t)$ so that it grows fast enough, your original function will be bounded from above by exponential decay, but you can make the derivative grow as fast as you like.

Edit: This reply doesn't look at the positivity condition, but this should be fixable by replacing $\sin(g(t) t)$ with something like $2 + sin(g(t) t)$

Answer 2

No. I will show a non-negative function which does the job (it's easy enough to turn it into a positive one). Take $v(x)=e^{-x}\cos^2(e^x)$. Clearly we have $v(x)\leq v(0)e^{-x}$. The derivative, which is $-e^x\sin(2e^x)$, is not bounded.

Answer 3

The answer is no.

Take $v(t) = e^{-t} (1+\sin(e^x) )$

$v'(t) = \cos(e^t)-e^{-t} (1+\sin(e^t))$

and v'(t) doesn't converge to 0