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

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.