Let $f_1,\dots,f_n,\dots :\mathbb{R}_+\to\mathbb{R}$ be non increasing continuous functions such that $f_k(0)\geq 0$ and $\lim_{x\to+\infty}f_k(x)<\infty$ for all $k$. Let $f$ that have the same property.
Assume that for all $x\in\mathbb{R}_+$, $f_n(x)\to f(x)$ as $n\to\infty$.
I proved that $f_1,\dots,f_n,\dots$ and $f$ have unique fixed points, that we denote $x_1,\dots,x_n,\dots$ and $x_\infty$.
Is it possible to prove that $x_n\to x_\infty$ as $n\to\infty$?
My idea is to define $g_k(x) = f_k(x)-x $ and $g(x)=f(x)-x$ (the functions I used to prove existence of a unique fixed points) which are continuous and strictly decreasing. Then I would like to used some kind of continuous inverse theorem but I do not really see how to do it.
Maybe it is only true if we assume uniform convergence of $f_n$?
All you need is that $f_n$ and $f$ are continuous, nonincreasing functions on $[0,\infty)$ with $f_n \to f$ pointwise and $f_n(0) \ge 0$. Such functions always have unique fixed points on $[0,\infty)$. For convenience, we can define them on all of $\mathbb R$ by taking $f_n(x) = f_n(0)$ and $f(x) = f(0)$ for $x < 0$.
If $x_\infty$ is the unique fixed point of $f$, take any $\epsilon > 0$ and consider $x_\infty-\epsilon$ and $x_\infty + \epsilon$. We have $f(x_\infty - \epsilon) \ge f(x_\infty) = x_\infty$, so for $n$ sufficiently large we have $f_n(x_\infty - \epsilon) > x_\infty - \epsilon $. Similarly, for $n$ sufficiently large we have $f_n(x_\infty + \epsilon) < x_\infty+ \epsilon$. By the Intermediate Value Theorem applied to $f_n(x)-x$, $f_n$ has a fixed point in the interval $(x_\infty - \epsilon, x_\infty + \epsilon)$. By uniqueness, that fixed point is $x_n$. Thus we have proven that $\lim_{n\to \infty} x_n = x_\infty$.