I was wondering whether or not a sequence with the following properties exists and/or can even be explicitly constructed:
I am looking for a sequence $(f_n)_{n \in \mathbb{N}}$ of bijective, continuously differentiable mappings $$f_n: [0, 1] \rightarrow [0,1]$$ that are uniformly converging to a (necessarily continuous) limit mapping $f: [0,1] \rightarrow \mathbb{R}$ that is not bijective on the interval $[0,1]$ (i.e. either not one-to-one or not onto [or both]).
Alternatively, the requirement of $f_n \in C^1([0,1])$ for each $n \in \mathbb{N}$ may be weakened in the following ways:
The $f_n$ may be Lipschitz continuous with (not necessarily uniformly bounded) Lipschitz constants $L_n$;
The $f_n$ may be bijective piecewise $C^1$ mappings of the interval $[0,1]$ onto itself;
I already figured out that, for example, the sequence $(x^n)_{n \in \mathbb{N}}$ doesn't work since it is not uniformly converging to the constant zero function (only locally uniformly). I'd also be very happy if someone could provide some general information on the described situation, for example a general result on sequences of uniformly convergent surjective/injective/bijective [$C^1$] mappings (something like "The uniform limit of surjective/injective/bijective functions is again surjective/injective/bijective")
Thanks for any help!
Since $[0,1]$ is compact a uniform limit of continuous bijections must be surjective.
But not injective: It's possible to concoct a uniformly convergent sequence $(f_n)$ of strictly increasing smooth bijections such that $f_n(1/3)=1/2$ and $f_n(2/3)=1/2+1/n$; the limit is then constant on $[1/3,2/3]$.