If $g$ is contraction and $f+f'=g(f)$ then f(t) has a limit as $t\to\infty$

Question

I would like to prove the following statement:

Let $g:\mathbb{R}^n\to\mathbb{R}^n$ be contraction mapping. If function $f:\mathbb{R}\to\mathbb{R}^n$ satisfies: $$f(t)+f'(t)=g(f(t))$$ then the limit $\lim_{t\to\infty}f(t)$ exists.

Below I write a sketch of my solution to this problem. My question is: is there some simpler way to prove the theorem and is my proof correct?

Let $L<1$ be Lipschitz constant of $g$. It's not hard to show that if $\alpha\in\mathbb{R}^n$ is a fixed point of $g$ then $|f(t)-\alpha|\leq\frac{1}{1-L}|f'(t)|$ so it's enough to show that $|f'(t)|\to 0$. With some elementary calculation it can be proven that if $a,b\in\mathbb{R}^n$ satisfy $|a+b|\leq L|a|$ then $$\langle a,b\rangle\leq \frac{c}{2}|a|^2$$ where $c=\frac{L-1}{2-L}<0$. Let $h\neq 0$. Substituting $a=f(t+h)-f(t)$, $b=f'(t+h)-f'(t)$ in the formula above we get: $$\langle f(t+h)-f(t),f'(t+h)-f'(t)\rangle\leq \frac{c}{2}|f(t+h)-f(t)|^2$$ Since $\frac{d}{dt}|f(t+h)-f(t)|^2=2\langle f(t+h)-f(t),f'(t+h)-f'(t)\rangle$ we obtain: $$\frac{d}{dt}|f(t+h)-f(t)|^2\leq c|f(t+h)-f(t)|^2$$ $$\frac{d}{dt}\ln\left(|f(t+h)-f(t)|^2\right)\leq c$$ Integrating from t to $0$ we get: $$\ln\left(|f(t+h)-f(t)|^2\right)-\ln\left(|f(h)-f(0)|^2\right)\leq ct$$ $$|f(t+h)-f(t)|^2\leq e^{ct}|f(h)-f(0)|^2$$ $$\left|\frac{f(t+h)-f(t)}{h}\right|^2\leq e^{ct}\left|\frac{f(h)-f(0)}{h}\right|^2$$ Since $h$ was arbitrary we can take $h\to 0$ obtaining: $$\left|f'(t)\right|^2\leq e^{ct}\left|f'(0)\right|^2$$ Having $c<0$ we see that $\lim_{t\to \infty}e^{ct}=0$ which implies that $\lim_{t\to \infty}|f'(t)|^2=0$ and this is what we wanted to prove.

Answer 1

Let $\alpha$ be the fixed point of $g$. Multiplying both sides of the equation by $e^t$ gives $$ (e^t(f(t)-\alpha))'=e^t(g(f)-g(\alpha)). \tag1$$ Now integrating (1) from $0$ to $t$ gives $$ f(t)-\alpha=(f(0)-\alpha)e^{-t}+e^{-t}\int_0^t(g(f)(s)-g(\alpha))e^sds. $$ So $$\begin{eqnarray} |f(t)-\alpha|&\le& |f(0)-\alpha|e^{-t}+Le^{-t}\int_0^t|f(s)-\alpha|e^sdt\\ &\le& |f(0)-\alpha|e^{-t}+L\int_0^t|f(s)-\alpha|dt. \end{eqnarray}$$ By Gronwall's equality https://en.wikipedia.org/wiki/Gr%C3%B6nwall%27s_inequality, one has $$ |f(t)-\alpha|\le|f(0)-\alpha|e^{-t}\exp(\int_0^tLds)=|f(0)-\alpha|e^{-(1-L)t} $$ from which gives the result when $t\to\infty$.