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May 25 at 17:59 comment added user24334 That makes sense, thanks. If we constrain $x_n$ to be in $[0,1]$, can we then conclude that the solution $x_n = f_n(x_n)$ is unique?
May 25 at 16:59 comment added Fedor Petrov If $f_n(x)$ goes to uniformly, and $x_n=f_n(x_n)$ is a value of $f_n$, then $x_n$ also goes to 0. Or is less than $\delta$
May 25 at 16:57 comment added user24334 Thanks a lot for your answer! But how do we conclude with $x_n \rightarrow 0$ from the fact that $f_n \rightarrow 0$ uniformly? Wouldn't we need to know more about $x_n$? If, for example, $x_n = e^{-n}$, we would have $(1-\beta x_n)^{n-1} \rightarrow 1$ and not 0.
May 25 at 2:13 history answered fedja CC BY-SA 4.0