Assume $f:[a,b]\to[a,b]$ be continuous and differentiable on $(a,b)$ and $f(a)=a$, $f(b)=b$. How to prove that exists distinct $x_1,x_2 \in(a,b)$ such that $f '(x_1)f '(x_2)=1$? Thanks in advance.

  • 1
    $\begingroup$ If you use the Mean Value Theorem you can get one point $x_1$, since $f(b)-f(a)=b-a$. Does it help? $\endgroup$ – Sigur Feb 11 '13 at 18:31
  • $\begingroup$ @Sigur: What about the another distinct one? $\endgroup$ – mrs Feb 11 '13 at 18:32
  • 1
    $\begingroup$ @BabakSorouh, I don't know. So I asked if the first one could be useful. $\endgroup$ – Sigur Feb 11 '13 at 18:33
  • $\begingroup$ See another excellent answer at math.stackexchange.com/questions/1373408/… $\endgroup$ – Paramanand Singh Jul 26 '15 at 4:33

Apply the MVT for $g(x)=f(f(x)).$ Thus there exists $x_1 \in (a,b)$ s.t. $g'(x_1)=f'(x_1)f'(f(x_1))=1,$ so we're done.


If it happens $x_1=f(x_1),$ then apply the MVT for $g$ over $[a,x_1]$ therefore, there exists $x_2 \in (a,x_1)$ s.t. $g'(x_2)=f'(x_2)f'(f(x_2))=1.$ If $x_2 \neq f(x_2)$ so we're done, but if $x_2=f(x_2),$ since $(f'(x_1))^2=1=(f'(x_2))^2$ we will then have $f'(x_1)f'(x_2)=1$ or $-1.$ For the latter, we again need to apply the MTV over $[x_1,b]$ and run the argument for finding $x_3 \in (x_1,b),$ but now we can choose two points out of three ($x_1,x_2,x_3$) with the desired property.

  • $\begingroup$ Why does $x_1\neq f(x_1)$? $\endgroup$ – Potato Feb 11 '13 at 18:43
  • $\begingroup$ Ditto ${}{}{}{}{}{}$. $\endgroup$ – copper.hat Feb 11 '13 at 18:43
  • 4
    $\begingroup$ Not quite done. But, if $f(x_1)=x_1$, then one may apply the MVT on the intervals $[a,x_1]$ and $[x_1,b]$ to obtain the desired result. $\endgroup$ – David Mitra Feb 11 '13 at 18:50
  • 1
    $\begingroup$ If $f(x) \neq x$ for all $x \in [a, b]$, then there must be two fixed points $x_1 \neq x_2$ (perhaps just $a$ and $b$) such that no $x$ satisfying $x_1 \leq x \leq x_2$ is a fixed point for $f$. Proceed with $x_1$ and $x_2$ in MVT. $\endgroup$ – Shaun Ault Feb 11 '13 at 18:53

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.