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Let $f : \mathbb R \to \mathbb R$ be a solution of the additive Cauchy functional equation satisfying the condition $$f(x) = x^2 f(1/x)\quad \forall x \in \mathbb R\setminus \{0\}.$$ Then show that $f(x) = cx,$ where $c$ is an arbitrary constant.

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  • $\begingroup$ If we show that f is continuouse we finish. Is this true ? And if this true how we can prove this ? $\endgroup$ – user217960 Feb 28 '15 at 13:47
  • $\begingroup$ Please let me know if I've transcribed your question correctly, particularly the part where you wrote $\forall x \in \mathbb R$ r {0}. $\endgroup$ – Namaste Feb 28 '15 at 13:48
  • $\begingroup$ Yes Correct ............................................ $\endgroup$ – user217960 Feb 28 '15 at 13:50
  • $\begingroup$ What do you think of my proof ? $\endgroup$ – Gabriel Romon Mar 1 '15 at 9:27
  • $\begingroup$ @LeGrandDODOM I think that its a convincing proof . $\endgroup$ – user217960 Mar 1 '15 at 17:54
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Let $F(x)=f(x)-xf(1)$

For some $x\neq 0$, $F(\frac{1}{x})=f(\frac{1}{x})-\frac{1}{x}f(1)$.

Hence for $x\neq 0$ $$\begin{align} x^2F(\frac{1}{x})&=x^2f(\frac{1}{x})-xf(1)\\&=f(x)-xf(1)\\&=F(x)\end{align}$$ and of course $F(1)=0$ and $F$ is additive.


Let us prove that $\forall x\in \mathbb R, F(x)=-F(-x)$

Indeed, $0=F(1)=F(x+1-x)=F(x)+F(1)+F(-x)=F(x)+F(-x)$


Also, for some $x\neq -1$,

$$\begin{align} F(x)=F(x+1) &=(x+1)^2F\left(\frac{1}{x+1}\right)\\ &=(x+1)^2F\left(1-\frac{x}{x+1}\right)\\ &=-(x+1)^2F\left(\frac{x}{x+1}\right)\\ &=-(x+1)^2\left(\frac{x}{x+1}\right)^2F\left(\frac{x+1}{x}\right)\\ &=-x^2F\left(1+\frac{1}{x}\right)\\ &=-x^2F\left(\frac{1}{x}\right)\\ &=-F(x)\\\end{align}$$

This also holds for $x=-1$ since $F(-1)=-F(1)$.

Hence $2F=0$.

Hence $F=0$ and we're done.

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