On the complex plane, it is true that if $\{ f_n\}$ is a set of holomorphic injective functions on the complex plane defined on a connected open set $\Omega$, which are convergent to $f$ uniformly on compact subsets of $\Omega$,then either $f$ is a constant function or is injective on $\Omega$ (of course, it is also holomorphic). The proof of the above fact is a few lines with the knowledge of Hurwitz's theorem.

I'd like to ask if this is true for injective functions on $\mathbb{R}$. So for example, does there exist a connected open set in $\mathbb{R}$, and $C^\infty$ functions $\{f_n\}$ on it which are injective, such that they converge to a function $f$ on the standard sup norm metric which is not injective or constant? This would characterize another difference between the complex and real numbers in a comprehensive manner.


The corresponding statement is not true for $C^\infty$ function on $\Bbb R$:

Let $f: \Bbb R \to \Bbb R$ be any (weakly) increasing $C^\infty$ which is not injective, for example $$ f(x) = \begin{cases} 0 & \text{ for } x \le 0 \\ e^{-1/x} & \text{ for } x > 0 \end{cases} $$ Then the functions $$ f_n(x) = f(x) + \frac 1n \arctan x $$ are strictly increasing (and therefore injective), and $f_n \to f$ uniformly on $\Bbb R$.

  • $\begingroup$ That's right! Great example. $\endgroup$ – астон вілла олоф мэллбэрг Apr 3 '16 at 7:52
  • $\begingroup$ @астонвіллаолофмэллбэрг So does this answer your question? Please let me know if you need more information. $\endgroup$ – Martin R Apr 5 '16 at 8:15
  • $\begingroup$ That's all I need, thanks. It is always important to emphasise the difference of holomorphicity and real differentiability using examples. I can use this example to understand the difference more clearly. Once again, nice example. $\endgroup$ – астон вілла олоф мэллбэрг Apr 5 '16 at 23:22

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