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Let $f\colon\mathbb R \to \mathbb R$ be a continuous odd function such that

1) $f(1+x)=1+f(x)$

2) $x^2f(1/x)=f(x)$ for $x\ne0$.

Prove that $f(x)=x$.

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closed as off-topic by Cameron Buie, user63181, Thomas Andrews, Nick Peterson, Davide Giraudo Jan 22 '14 at 16:11

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    $\begingroup$ What are your thoughts? What have you tried? Is this homework? $\endgroup$ – Carsten S Jan 22 '14 at 14:13
  • $\begingroup$ Using the 1 st eqn i am getting f(1+x)+f(1-x)=2 but after that I am not able to use the 2nd eqn and yes it's homework $\endgroup$ – user118899 Jan 22 '14 at 14:17
  • $\begingroup$ Is there a continuity assumption that you forgot to mention? These often help. $\endgroup$ – Carsten S Jan 22 '14 at 14:22
  • $\begingroup$ It's continuous $\endgroup$ – user118899 Jan 22 '14 at 14:25
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If we do it in a more "functional equation"-ish approach, we notice that $$\left(\frac{x+1}{x}\right)^2f\left(\frac{x}{x+1}\right) = f\left(\frac{x+1}{x}\right)$$ On the other hand, $$\begin{align}f\left(\frac{x}{x+1}\right) &= f\left(1 - \frac{1}{x + 1}\right)\\ &= 1 + \left(-\frac{1}{x+1}\right)\\ &= 1 - \left(\frac{1}{x+1}\right)\\ &= 1 - \frac{f(x+1)}{(x+1)^2} \end{align}$$ Substituting back: $$\left(\frac{x+1}{x}\right)^2\left(1 - \frac{f(x+1)}{(x+1)^2}\right) = f\left(\frac{x+1}{x}\right)$$ $$\begin{align}\frac{1}{x^2}\left((x+1)^2 - f(x+1)\right) &= f\left(1 + \frac{1}{x}\right)\\ &= 1 + f\left(\frac{1}{x}\right) \\ &= 1 + \frac{f(x)}{x^2}\end{align}$$ Dividing throughout by $\frac{1}{x^2}$ (valid since $x \neq 0$): $$\begin{align}(x+1)^2 - f(x+1) &= x^2 + f(x)\\ x^2 + 2x + 1 - x^2 &= f(x) + f(x + 1)\\ 2x + 1 &= 1 + 2f(x)\end{align}$$ $$2x = 2f(x)$$ Which proves the desired statement: $$f(x) = x$$ This is a rather long winded approach and I'm pretty sure it can be simplified a bit.

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  • $\begingroup$ Very nicely done! $\endgroup$ – Carsten S Jan 22 '14 at 14:44
  • $\begingroup$ @CarstenSchultz Thank you sir! Though I must admit, I kinda used a rather brute force approach to get the correct substitution. $\endgroup$ – Yiyuan Lee Jan 22 '14 at 14:46
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    $\begingroup$ If one would like to extract the key idea, it might be to use $\frac1{1+\frac1x}=1-\frac1{x+1}$. $\endgroup$ – Carsten S Jan 22 '14 at 14:57
  • $\begingroup$ Nice and without assuming continuity +1. $\endgroup$ – Macavity Jan 22 '14 at 15:33
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Hint: (i) As $x$ is odd, what can you say about $f(0) = -f(-0)$?

(ii) Using 1), compute $f$ on the integers (use induction).

(iii) Using 2) compute $f$ for numbers of the form $\frac 1n$, $n \in \mathbb N - \{0\}$.

(iv) Using 1) again, this gives $f$ for rational numbers.

(v) Now use continuity to conclude.

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    $\begingroup$ Am I missing something if I do not see (iv) immediately? (I would know how to solve the problem using continued fractions.) $\endgroup$ – Carsten S Jan 22 '14 at 14:35
  • $\begingroup$ Is part (iv) related to the fact that every rational number has a terminating continued fraction expression? $\endgroup$ – peterwhy Jan 22 '14 at 14:37
  • $\begingroup$ For (i), I believe you mean "$f$ is odd" rather than $x$. $\endgroup$ – Brian Jan 22 '14 at 14:46
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This is some kind of addendum to a Hint in the answers.


The function is odd and therefore it is zero at zero: $f(0)=0$. Therefore from two equalities in the question, we respectively get $f(n)=n$ and $f(\frac 1n)=\frac 1n$ for all integers $n$.

For even numbers, we have $f(\frac n2)=\frac n2$ and $f(\frac 2n)=\frac 2n$. For odd numbers we have the following: $$ f(\frac n2)=f(\frac{n-1}2+\frac{1}2)=\frac{n-1}2+\frac{1}2=\frac n2 $$ Therefore for all $n$, we have $f(\frac n2)=\frac n2$ and therefore $f(\frac 2n)=\frac 2n$. So far we can say that for all $0 \leq r\leq 2$, we have: $$ f(\frac nr)=\frac nr \\ f(\frac rn)=\frac rn $$ Now we can use induction to prove the previous equalities for all positive integers $r$. Suppose that for all $0\leq r\leq m-1$ we have: $$ f(\frac nr)=\frac nr \\ f(\frac rn)=\frac rn $$ We want to prove that : $$ f(\frac nm)=\frac nm \\ f(\frac mn)=\frac mn. $$ There are two numbers $q$ and $0 \leq r<m$ such that, we can write the integer $n$ as $qm+r$. We can then write the following: $$ f(\frac nm)=f(\frac{qm}m+\frac{r}m)=q+f(\frac{r}m)=\frac nm $$. where the last equality comes from the assumption of induction, namely that for $0\leq r\leq m-1$ and all $n$ we have $f(\frac{r}m)=\frac rm$. This finishes the induction proof and says that for all integers $m$ and $n$: $$ f(\frac nm)=\frac nm \\ f(\frac mn)=\frac mn. $$

The function is identity function over all rational $x$. For irrational number $x$, we consider a sequence of rational numbers $r_n$ converging to $x$ and then use the continuity argument to show $f(x)=x$.

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  • $\begingroup$ You don't need to assume continuity, as another answer shows. $\endgroup$ – Macavity Jan 22 '14 at 15:32
  • $\begingroup$ I know, but the question assumes it anyway. $\endgroup$ – Arash Jan 22 '14 at 15:36
  • $\begingroup$ Could you elaborate on the part where you claim $f(n/m)=n/m$? $\endgroup$ – Carsten S Jan 22 '14 at 17:07
  • $\begingroup$ I tried to add something to the answer. $\endgroup$ – Arash Jan 22 '14 at 19:35

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