Suppose $f(x)$ and $g(x)$ are continuous functions on $[a,b]$ with $f$ monotone increasing. Assume there exists a sequence $x_n \in [a, b]$ such that for all $n \in N$ , $g(x_n) = f(x_{n+1})$. Show that there exists $x_0 \in [a,b]$ such that $g(x_0) = f(x_0)$.

Can someone provide an example of functions that fulfills this condition?

  • $\begingroup$ I really need help in solving this $\endgroup$ Apr 20, 2014 at 4:01

2 Answers 2


If there is an $n$ such that $g(x_n)=f(x_n)$, we are done. So let us assume that $g(x_n)\ne f(x_n)$ for each $n$.

Now if $g(x_n)>f(x_n)$ and $g(x_{n+1})<f(x_{n+1})$, then the continuity of $f$ and $g$ implies that and $x$ such that $f(x)=g(x)$ exists somewhere between $x_n$ and $x_{n+1}$.

The case that $g(x_n)<f(x_n)$ and $g(x_{n+1})>f(x_{n+1})$ is basically the same.

So the only two remaining cases are:
A. $(\forall n) g(x_n)>f(x_n)$
B. $(\forall n) g(x_n)<f(x_n)$

Let us discuss the case A. (The case B is similar.)

For each $n$ we have $f(x_n)<g(x_n)=f(x_{n+1})$. Since $f$ is increasing, this implies $x_n<x_{n+1}$, i.e. the sequence $(x_n)$ is monotone.

Every monotone bounded sequence must have a limit, so there exists an $x$ such that $$\lim\limits_{n\to\infty} x_n=x.$$ Now we get, using the continuity of $f$ and $g$, that $$f(x)=\lim\limits_{n\to\infty} f(x_{n+1})=\lim\limits_{n\to\infty} g(x_n)=g(x).$$

  • $\begingroup$ why is f(x) = n, g(x) = n+1 not a counterexample? $\endgroup$ Apr 26, 2014 at 0:17
  • $\begingroup$ @user136266 Do you mean constant functions? They do not fulfill the assumptions. (You cannot find $x_n$ with the required properties.) $\endgroup$ Apr 26, 2014 at 4:22
  • $\begingroup$ I will repost also here link to the chat where some examples and clarifications related to this were given. (I have already mentioned this in comments to the duplicate question.) $\endgroup$ Apr 30, 2014 at 13:04

Hint Prove that there is a monotonic subsequence $\{a_{n_k}\}$ and let be $n_0$ the limit of that subsequence.

  • $\begingroup$ I know how to prove that but how does it help? Using Bolzano theorem, this fact is obvious $\endgroup$ Apr 20, 2014 at 6:37

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