5
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ I really need help in solving this $\endgroup$ – user10024395 Apr 20 '14 at 4:01
2
$\begingroup$

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 $$f(x)=\lim\limits_{n\to\infty} f(x_{n+1})=\lim\limits_{n\to\infty} g(x_n)=g(x).$$

$\endgroup$
  • $\begingroup$ why is f(x) = n, g(x) = n+1 not a counterexample? $\endgroup$ – user10024395 Apr 26 '14 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.) $\endgroup$ – Martin Sleziak Apr 26 '14 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.) $\endgroup$ – Martin Sleziak Apr 30 '14 at 13:04
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ I know how to prove that but how does it help? Using Bolzano theorem, this fact is obvious $\endgroup$ – user10024395 Apr 20 '14 at 6:37

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.