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 $[1/3,2/3]$.

  • 2
    $\begingroup$ To the proposer: We can also arrange for $\sup_{n\in \Bbb N} \sup_{x\in [0,1]} f'_n(x)<$ $\infty,$ so that $\{f_n:n\in \Bbb N\}$ is a uniformly-Lipschitz family. $\endgroup$ – DanielWainfleet Apr 6 '18 at 22:43
  • $\begingroup$ Thanks for your answer David C. Ullrich! Can you give a reference to a textbook etc. containing the mentioned statement? And how would the functions $f_n$ of the sequence look like that give a counterexample to injectivity? @DanielWainfleet : Thank you very much as well! Same question for you: Can we actually write down these mappings $f_n$? ;-) And can you name some reference (textbook etc.)? $\endgroup$ – ComplexFlo Apr 7 '18 at 6:53
  • 1
    $\begingroup$ I don't have a reference. What would $f_n$ look like? I said what $f_n$ would, or could, look like. Imagine a graph consisting of line segments connecting the points $(0,0)$, $(1/3,1/2)$, $(2/3,1/2+1/n)$, $(1,1)$. That gives an example except it's not $C^1$. So smooth out the corners a little. (Uniform convergence comes for free: As Wainfleet pointed out, the Lipschitz constant is bounded, so Arzela-Ascoli implies there is a uniformly convergent subsequence.) $\endgroup$ – David C. Ullrich Apr 7 '18 at 11:59
  • $\begingroup$ @DavidC.Ullrich thanks for your reply and please excuse my late answer! I think I understand your reasoning with the surjectivity due to the compactness of $[0,1]$. Also thanks for the hint with the piecewise linear mappings, I'll definitely have a look at it. Is there any standard way for "smoothing out corners" (e.g. spline functions)? I think I also get your point using the Arzela-Ascoli theorem - the equicontinuity follows from the uniformly bounded Lipschitz constant, right? $\endgroup$ – ComplexFlo Apr 9 '18 at 14:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.