# Prove that $f(x)=x$ if the following holds true [closed]

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 $f(x)=x$.

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Cameron Buie, Community, Thomas Andrews, Nick Peterson, Davide Giraudo
If this question can be reworded to fit the rules in the help center, please edit the question.

• What are your thoughts? What have you tried? Is this homework? – Carsten S Jan 22 '14 at 14:13
• 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 – user118899 Jan 22 '14 at 14:17
• Is there a continuity assumption that you forgot to mention? These often help. – Carsten S Jan 22 '14 at 14:22
• It's continuous – user118899 Jan 22 '14 at 14:25

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.

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

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.

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

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 $f(x)=x$.

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