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.

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.