# Functional equation - Cyclic Substitutions

Please help solve the below functional equation for a function $$f: \mathbb R \rightarrow \mathbb R$$: \begin{align} &f(-x) = -f(x) , \text{ and } f(x+1) = f(x) + 1, \text{ and } f\left(\frac 1x\right) = \frac{f(x)}{x^2} \\ &\text{ for all } x \in \mathbb R \text{ and } x \ne 0 . \end{align}

I know this will be solved by cyclic substitutions, but I'm unable to figure out the exact working. Can someone explain step wise?

• The functional equation forces $f(q)=q$ for all $q\in\Bbb{Q}$. Is $f$ supposed to be continuous, perhaps? – Servaes Oct 29 '18 at 10:52
• If $f$ is continuous, then it is true. But I don't think it's true otherwise. – TonyK Jul 3 at 10:21
• It seems I was wrong $-$ continuity is not required. See @Michael Rozenberg's answer. – TonyK Jul 3 at 10:27

First, we observe that $$f(0)=0, f(1)=1, f(-1)=-1$$ has to be true. Also, it suffices to determine $$f(x)$$ for $$x>0$$, because the rest follows from the condition $$f(-x)=-f(x)$$.
Let $$x>0$$. Applying some of the conditions, we have \begin{aligned} f(x)+1 &= f(x+1) \\ &= f(1(x+1)^{-1})(x+1)^2 \\ &= f(1-x(x+1)^{-1})(x+1)^2\\ &= (1+f(-x(x+1)^{-1}))(x+1)^2\\ &= (1-f(-x(x+1)^{-1}))(x+1)^2\\ &= (1-f(-x(x+1)^{-1}))(x+1)^2\\ &= (1-f((x+1)x^{-1})x^2(x+1)^{-2})(x+1)^2\\ &= (1-f(1+x^{-1})x^2(x+1)^{-2})(x+1)^2\\ &= (1-(1+f(x^{-1}))x^2(x+1)^{-2})(x+1)^2\\ &= (1-(1+f(x)x^{-2})x^2(x+1)^{-2})(x+1)^2\\ &= (x+1)^2-x^2-f(x)\\ \end{aligned}
This yields $$f(x)=x$$.
For $$x\neq0$$ and $$x\neq1$$ we obtain: $$f(x)=x^2f\left(\frac{1}{x}\right)=x^2f\left(\frac{1}{x}-1+1\right)=x^2\left(f\left(\frac{1}{x}-1\right)+1\right)=$$ $$=x^2+x^2f\left(\frac{1-x}{x}\right)=x^2+x^2\cdot\frac{f\left(\frac{x}{1-x}\right)}{\frac{x^2}{(x-1)^2}}=x^2+(x-1)^2f\left(\frac{x}{1-x}\right)=$$ $$=x^2+(x-1)^2f\left(\frac{1}{1-x}-1\right)=x^2+(x-1)^2\left(f\left(\frac{1}{1-x}\right)-1\right)=$$ $$=x^2-(x-1)^2+(x-1)^2\cdot\frac{f(1-x)}{(1-x)^2}=2x-1-f(x-1).$$ Thus, $$f(x)+f(x-1)=2x-1$$ or $$f(x)+f(x+1)=2x+1,$$ which gives $$f(x)+f(x)+1=2x+1$$ or $$f(x)=x.$$ Now, show that $$f(0)=0$$ and $$f(1)=1.$$