Let $x(t)\in C^1(\mathbb{R}_+,\mathbb{R}^m)$ be a vector-function such that 1) $\|x(t)\|+\|\dot x(t)\|\to 0$ as $t\to\infty$ and 2) for some real $c_1>0$ and all $t>0$ one has $\|x(t)\|\le c_1\|\dot x(t)\|$
Is it true that $\|x(t)\|\le c_2 e^{-c_3t}$ all $t>0$? Here $c_i$ are positive constants.
The answer is no. E.g., let $m=2$, $x(t)=a(t)(\sin t,\cos t)$, where e.g. $a(t):=\frac1{1+t^2}$. Then $\|x(t)\|^2=a(t)^2\le a(t)^2+\dot a(t)^2=\|\dot x(t)\|^2$, but $\|x(t)\|/e^{-ct}\to\infty$ for any $c>0$. By using coordinates of $x$ identically equal $0$, one can easily see that the conjecture fails to hold for any $m\ge2$.
Yet, the answer is yes for $m=1$. Indeed, there may be two cases: (i) $x(s)=0$ for some real $s$ or (ii) otherwise.
In case (i), without loss of generality (wlog), $s=0$. Then $x(t)=\int_0^t\dot x(u)du$, whence $$|x(t)|\le c_1\int_0^t du_1|x(u_1)|$$ $$\le c_1^2\int_0^t du_1\int_0^{u_1} du_2|x(u_2)|$$ $$\le\dots \le c_1^k\int_0^t du_1\int_0^{u_1} du_2\dots\int_0^{u_{k-1}}du_k |x(u_k)| $$ $$\le c_1^k\int_0^t du_1\int_0^{u_1} du_2\dots\int_0^{u_{k-1}}du_k M_t =M_tc_1^k t^k/k!\to0$$ as $k\to\infty$ for each real $t>0$, where $M_t:=\max_{0\le u\le t}|x(u)|<\infty$. So, $x(t)=0$ for all $t\ge0$, in case (i).
In case (ii), in view of the condition $\|x(t)\|\le c_1\|\dot x(t)\|$, one has $\dot x(t)\ne0$ for any real $t$. So,
wlog, $\dot x(t)<0$ for all real $t$ and hence $\dot x(t)+cx(t)\le\dot x(t)+c{c_1}|\dot x(t)|=0$ if $c:=1/c_1$.
So, for $y(t):=e^{ct}x(t)$ one has $\dot y(t)=e^{ct}(\dot x(t)+cx(t))\le0$, so that $y$ is nonincreasing.
But the conditions $\dot x<0$ and $\|x(t)\|\to 0$ as $t\to\infty$ imply $x>0$ and hence $y>0$. So, $y(t)\le y(0)$ for $t\ge0$, that is, $0\le x(t)\le y(0)e^{-ct}$ for $t\ge0$.
So, in both cases $\|x(t)\|\le c_2 e^{-c_3t}$ for some positive real $c_2$ and $c_3$ and all $t>0$.
