Show that f is linear

Question

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.

Answer 1

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.

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

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

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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.