# Convergence of a sequence of functions and their inverses

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

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 $$n>N$$,

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

Therefore

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

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

• Thank you very much! This looks very nice! Jan 16, 2015 at 6:24
• Any reference for this? Oct 15, 2019 at 8:17