Let $x(t):[0,\infty) \to \mathbb{R}$ be a differentiable function. Suppose $|x'(t)| \le Ce^{-t}$ (which implies that $\sup_t |x(t)| < \infty$).
Can we claim that $\lim_{t\to \infty} x(t) = x^*$ for some $x^*$?
I think this is true, however, I cannot rigorously prove or disprove it.
Any answers/comments/suggestions will be very appreciated.
On
I'm going to use Cantor's Intersection Theorem:
Suppose $(A_n)_{n \in \Bbb{N}}$ is a sequence of closed subsets of $\Bbb{R}$ (or indeed, any complete metric space) such that $A_{n+1} \subseteq A_n$, and $$\operatorname{diam} A_n := \sup_{x, y \in A_n} |x - y| \to 0$$ as $n \to \infty$. Then, $$\left|\bigcap_{n \in \Bbb{N}} A_n\right| = 1.$$
Let $A_n = \overline{x[n, \infty)}$. Then clearly $A_n$ is closed, and $A_{n+1} \subseteq A_n$ (this is true for any real function). We can use the bound on the derivative to show that $\operatorname{diam} A_n \to 0$.
Suppose $t \in (n, \infty)$. Then, \begin{align*} x(t) - x(n) &= \int_n^t x'(s) \, \mathrm{d}s \\ &\le \int_n^t |x'(s)| \, \mathrm{d}s \\ &\le \int_n^t Ce^{-s} \, \mathrm{d}s \\ &= C(e^{-n} - e^{-t}) \\ &\le Ce^{-n} \to 0. \end{align*}
This shows us that elements of $x[n, \infty)$ can be no more than $Ce^{-n}$ apart, which is also true of the closure $A_n$. Hence, by the Cantor Intersection Theorem, there exists a unique $x^* \in \Bbb{R}$ such that
$$x^* \in \bigcap_{n \in \Bbb{N}} A_n.$$
I claim that this is the limit we need. For all $\varepsilon > 0$, there exists some $N \in \Bbb{N}$ such that $n \ge N \implies |Ce^{-n}| < \varepsilon$, hence $$t \ge N \implies x(t), x^* \in A_N \implies |x(t) - x^*| \le |Ce^{-N}| < \varepsilon.$$
Hint: Using Mean Value Theorem, show $x(n)$ is a Cauchy sequence...