consider a sequence of continuous and bijective functions $f_n:\mathbb{R}\rightarrow\mathbb{R}$, such that their inverses $f^{-1}_n:\mathbb{R}\rightarrow\mathbb{R}$ are continuous as well. Furthermore let us assume the function $f(x):=\lim\limits_{n\rightarrow \infty}f_n(x)$ is continuous, bijective and his inverse $f^{-1}$ is continuous.

I want to prove or disprove that in this case the following relation holds:

$$f^{-1}(x):=\lim\limits_{n\rightarrow \infty}f_n^{-1}(x)$$

By now I couldn't prove the statement, but there are a couple of examples which seems to confirm this claim. Do you have any idea?

Best regards


1 Answer 1


Let $\epsilon>0$.

Because $f^{-1}$ is continuous we can choose $\delta$ such that $$|f^{-1}(x+\delta)-f^{-1}(x)|<\epsilon\ \ \ \text{ and }\ \ \ |f^{-1}(x-\delta)-f^{-1}(x)|<\epsilon$$

Because $f_n\to f$ there is $N$ such that for $n>N$,

$$|f_n(f^{-1}(x+\delta))-(x+\delta)|<\delta\ \ \ \ \text{ and }\ \ \ \ |f_n(f^{-1}(x-\delta))-(x-\delta)|<\delta$$


$$f_n(f^{-1}(x+\delta))>x\ \ \ \ \text{ and }\ \ \ \ f_n(f^{-1}(x-\delta))<x$$

Using that the $f_n^{-1}$ must be monotonic, $f_n^{-1}(x)$ is between $f^{-1}(x-\delta)$ and $f^{-1}(x+\delta)$. But these two are in the interval $\left[f^{-1}(x)-\epsilon,f^{-1}(x)+\epsilon\right]$.

Therefore $$|f_n^{-1}(x)-f^{-1}(x)|<\epsilon\ \ \ \ \text{ for }n>N$$

  • $\begingroup$ Thank you very much! This looks very nice! $\endgroup$
    – Braten
    Jan 16, 2015 at 6:24
  • $\begingroup$ Any reference for this? $\endgroup$
    – faceclean
    Oct 15, 2019 at 8:17

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