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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?

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

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 $f(x)=x$.

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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 $f(1)=1.$

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