# How to prove there exist distinct $a_{i}$ such $f'(a_{1})f'(a_{2})f'(a_{3})\cdots f'(a_{n})=1$

Let $f$ be a continuous map from $[0,1]$ to $R$ that is differentiable on $(0,1)$,with $f(0)=0,f(1)=1$, show that for each postive integer $n$ there exist distinct numbers $a_{1},a_{2},\cdots,a_{n}\in (0,1)$,such that

$$f'(a_{1})f'(a_{2})f'(a_{3})\cdots f'(a_{n})=1$$Thanks in advance.

• First try to prove it for $n=1$ and prove it for $n=2$ $\cdots$ and if you did not find any pattern then post your work in the question and people will certainly help you – Elaqqad Jul 25 '15 at 11:25
• see math.stackexchange.com/q/300424/72031 The argument is given for $n = 2$ and can be continued for any $n$ using $g(x)$ as n-th iterate of $f(x)$. – Paramanand Singh Jul 25 '15 at 11:39
• Are you sure that $f$ maps $[0, 1]$ to $\mathbb{R}$ or is it a function from $[0, 1]$ to $[0, 1]$? The latter case is solved in linked question in my previous comment. Otherwise I guess there could be a counterexample, but I am not sure. – Paramanand Singh Jul 25 '15 at 12:07
• but I think this more hard than $n=2$ – user237685 Jul 25 '15 at 15:01

An inductive proof.

We prove a formally more general (in fact obviously equivalent) statement: If $c>0$, $f:[0,c]\to\mathbb{R}$ is continous and differentiable in $(0,c)$ and $f(0)=0$, then there are distinct numbers $0<a_1<\ldots<a_n<c$ such that $f'(a_1)\cdots f'(a_n)=\left(\dfrac{f(c)}{c}\right)^n$.

The case $n=1$ follows from Lagrange's MVT.

Suppose that $n\ge2$, and the statement is true for $n-1$.

Apply Cauchy's MVT to the functions $(f(x))^n$ and $x^n$. By the MVT, there is some $\xi\in(0,c)$ such that $$\dfrac{n\cdot (f(\xi))^{n-1} \cdot f'(\xi)}{n\cdot (\xi)^{n-1}} = \dfrac{f^n(c)}{c^n}$$ $$\left(\dfrac{f(\xi)}{\xi}\right)^{n-1} \cdot f'(\xi) = \left(\dfrac{f(c)}{c}\right)^n.$$ Choose $a_n=\xi$; then we have $$\left(\dfrac{f(a_n)}{a_n}\right)^{n-1} \cdot f'(a_n) = \left(\dfrac{f(c)}{c}\right)^n. \tag1$$ Now apply the induction hypothesis to the interval $[0,a_n]$: there are some $0<a_1<\ldots<a_{n-1}<a_n$ such that $$f'(a_1)\cdots f'(a_{n-1}) = \left(\dfrac{f(a_n)}{a_n}\right)^{n-1}. \tag2$$ Plugging (2) in (1), the statement follows.

• too good an answer. +1 – Paramanand Singh Aug 9 '15 at 4:25

Some ideas: By the MVT, there is $a_1\in (0,1)$ such that $f'(a_1) = 1.$ Suppose $f(a_1) > a_1.$ Then

$$m_1 = \frac{f(a_1)-f(0)}{a_1-0}>1,\,\, m_2 = \frac{f(a_1)-f(1)}{a_1-1}<1.$$

It's good to draw pictures right about now. By the MVT again, there is $x\in(0,a_1)$ such that $f'(x) = m_1.$ Similarly, there is $y\in(a_1,1)$ such that $f'(y) = m_2.$ Now we can apply Darboux: On $(0,a_1), f'$ assumes every value in $(1,m_1).$ And on $(a_1,1),$ $f'$ assumes every value in $(m_2,1).$

Observe that for small $h> 0,1+h \in (1,m_1), 1/(1+h) \in (m_2,1).$ Fix such an $h.$ Then there are $a_2\in (0,a_1),a_3 \in (a_1,1)$ such that

$$f'(a_2)=1+h,f'(a_3)=1/(1+h) \implies f'(a_2)f'(a_3)=1.$$

Well that's a lot of work for $n=2.$ But now $n=3$ is done too: $a_2,a_1,a_3$ will work. For $n=4,$ we do the same as the above: There are $a_4 \in (a_2,a_1), a_5 \in (a_1,a_3)$ such that $f'(a_4)>1,f'(a_5)<1,$ with $f'(a_4)f'(a_5)=1.$ Thus $a_2,a_4,a_5,a_3$ work for $n=4.$

For $n=5,$ we throw in $a_1$ again. This process can be continued, tossing in $a_1$ for odd $n,$ tossing out $a_1$ for even $n$ and doing the above. So this needs tidying up with an induction proof, and there is also the original assumption $f(a_1)>a_1$ to take care of.

• Great answer. When $f(a_{1}) < a_{1}$ then roles of $m_{1}, m_{2}$ are reversed and argument proceeds in same manner. The case $a_{1} = f(a_{1})$ needs to be handled separately. If there are an infinity of values of $x \in (0, 1)$ for which $f(x) = x$ then our job is already done by applying MVT on $f(x) - x$ to get infinity of points with $f'(x) = 1$. If there are finitely many $x \in (0, 1)$ with $f(x) = x$ then let $a$ be the least of such points. Apply the argument of your answer on $[0, a]$ and you are done. You have my +1. – Paramanand Singh Jul 26 '15 at 4:29
• @ParamanandSingh Thanks for your comment. – zhw. Jul 26 '15 at 16:07