# preimage of convergence sequence

I came across this problem but could not get any conclusion.

Let $\{f_n\}$ be a sequence of continuous functions from $[0,\infty)$ to itself. We assume each $f_n$ is bijective and the sequence $f_n$ converges to $f$ uniformly over compact sets.  Assume that $f$ is also bijective. Put $x_n=f_n^{-1}(1)$, the preimage of 1 under $f_n$, and $x=f^{-1}(1)$. Prove or disprove the statement $\lim_{n\to\infty}x_n=x$.

Dose your conclusion change if we further assume $f_n$'s are strictly increasing functions?

Edit 1: We also assume $f$ is bijective.

• Do you assume that $f$ is bijective? If not, $f^{-1}(1)$ may be empty or may consist of more than one points.
– 23rd
Jun 10, 2013 at 4:13
• Landscape, thanks, I have edited the question assuming $f$ is bijective.
– Kle
Jun 10, 2013 at 4:25
• You are welcome.
– 23rd
Jun 10, 2013 at 4:49

This was more intented to be a comment but it became to long, hope it answers your question anyway.

This one doesn't work with any further assumptions. A trivial counter example would be $$f_n(x)=\frac{x}{n}$$ If we restrict it to $[0,\infty)$ it will surely be bijective and continuous. Furthermore $x_n=n$ which doesn't converge (or when it does the limit is $\infty$ regarding whether you use the one point compactification). Our $f_n$ converge pointwise to the zero function, and on compact sets the convergence is uniformly (using dinis theorem). But $x$ will be the empty set...

The bonus assumption at the end of your question should be rephrased. Because the functions $f_n(x)$ are always strictly increasing functions, because continuous + injective on a connected domain implies strict monotone, as the function should map to $[0,\infty)$ we can't have monotone decreasing. You surely mean that $f_n$ is monotone in $n$ meaning that $f_n(x)\leq f_{n+1}(x)$ for all $x$. Then we will have surely a convergent subsequence due to Bolzano Weierstraß, and as it is monotone the sequnce will be converging, as we can work in a compact set the equation will hold.

Edit: Didn't see your edit. With the edit it works. As $f$ surely isn't the zero function there is a $x$ such that $f([0,x])= [0,2]$. If we restrict our function we see that $f$ is a continuous bijection, and as we work on a compact set it is even a homoemorphism. Hence by sequence continuity we are done.

• Dominic Michaelis, thank you for the answer. I missed a condition in my original post which assert that the limiting function $f$ should be bijective.
– Kle
Jun 10, 2013 at 4:30
• I think you really should change the preimage to just the inverse fucntion, a preimage is a set and convergence of sets is a bit different than convergence of "normal" sequences Jun 10, 2013 at 4:32
• it's a good point. I was trying to avoid the word "inverse" since one could take inverse in many senses. However, since the post is already online and "preimage" is still OK in this case, I decide to keep it.
– Kle
Jun 10, 2013 at 4:42

After a few days, I now can prove the statement. Let me first restate the problem taking into account all the comments.

Let $f$ and $f_n$, ($n=1,2,\dots$) be bijective functions on $[0,\infty)$ such that they are continuous and increasing. Suppose that $f_n$ converges to $f$ pointwise. Put $x_n=f_n^{-1}(1)$ and $x=f^{-1}(1)$. Then $\lim_{n\to\infty}x_n=x$.

Proof: By contradiction, suppose $x_n$ does not converges to $x$. There exist $\epsilon>0$ and a sequence $\{n_1,n_2,\dots\}$ such that $|x_{n_k}-x|>\epsilon$ for all $n_k$. Then, we can find a subsequence, still denoted by $n_k$ such that either $x_{n_k}>x+\epsilon$ for all $k$ or $x_{n_k}<x-\epsilon$ for all $k$. We consider the former case, the later case can be carried out similarly. Applying $f_{n_k}$ to the first inequality we get $$1=f_{n_k}(x_{n_k})>f_{n_k}(x+\epsilon)$$ Upon sending $n_k$ to infinity, this implies $1\ge f(x+\epsilon)$. This a contradiction because $f(x+\epsilon)>f(x)=1$.