TITLE: Cauchy equation $ f ( x + y ) = f ( x ) + f ( y ) $ with an additional condition $ f \left( \frac 1 x \right) = \frac 1 { x ^ 2 } f ( x ) $
QUESTION [1 upvotes]: Let $ f : \mathbb R \to \mathbb R $ satisfy the following:

$ f ( 1 ) = 1 $;
$ f ( x + y ) = f ( x ) + f ( y ) $, $ \forall x , y \in \mathbb R $;
$ f \left( \frac 1 x \right) = \frac 1 { x ^ 2 } f ( x ) , \forall x \in \mathbb R ^ * $.

Find $ f \left( \sqrt { 2019 } \right) $.

This question was a multiple choice one, so I simply observed that one such function is $ f ( x ) = x , \forall x \in \mathbb R $, so $ f \left( \sqrt { 2019 } \right) = \sqrt { 2019 } $.
This is definitely not a rigorous solution, it is based on pure observation. I wanted to know if there is some way to solve this thoroughly.

REPLY [3 votes]: Let $(1)$ denote the additive condition, and $(2)$ denote the condition that $f\left(\frac{1}{x}\right)=\frac{f(x)}{x^2} $. Now note that for all $x\notin \{-1,0\}$, $$\begin{align*}\frac{f\left(x^2+x\right)}{(x^2+x)^2}\stackrel{(2)}=f\left(\frac{1}{x^2+x}\right)&\stackrel{(1)}=f\left(\frac{1}{x}\right)-f\left(\frac{1}{x+1}\right) \\ &\stackrel{(2)}=\frac{f(x)}{x^2}-\frac{f(x+1)}{(x+1)^2} \\ &= \frac{(x+1)^2f(x)-x^2f(x+1)}{(x^2+x)^2} \\ &\stackrel{(1)}=\frac{(2x+1)f(x)-x^2}{(x^2+x)^2} \end{align*}$$ 
Equating the two, we find that $$\frac{f\left(x^2+x\right)}{(x^2+x)^2}=\frac{(2x+1)f(x)-x^2}{(x^2+x)^2}  \implies f(x^2)+x^2=2xf(x)\tag{3}$$Now set $x+y$ in place of $x$ in the above to find $$f(x^2)+2f(xy)+f(y^2)+x^2+2xy+y^2=2(x+y)(f(x)+f(y))$$ and after simplifying using $(3)$, we find that $f(xy)-xy=y(f(x)-x)+x(f(y)-y)$, or $g(xy)=g(x)+g(y)$ where $g(x)=\frac{f(x)-x}{x},x\neq 0$. By $(2)$, $g(x)=g(\frac{1}{x})$, so $$g(xy)=g(x)+g(y)=g(x)+g(\frac{1}{y})=g\left(\frac{x}{y}\right)\tag{4}$$So for any positive reals $a$ and $b$, setting $x=\sqrt{ab}$ and $y=\sqrt{\frac{a}{b}}$ in $(4)$ gives $f(a)=f(b)$, hence $g$ is constant, and therefore $0$ over $\mathbb{R}^+$. Hence over $\mathbb{R}^+$, $f(x)\equiv x$. But by additivity, $f(-x)=-f(x)$, and $f(0)=0$.
Therefore, the only such function $f$ is indeed $f(x)=x$ for all real $x$, which clearly satisfies all conditions presented.
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