# Functional equation $f : \mathbb{R}^* \to \mathbb{R}$ :

I tried to solve this problem :

Determinate all function $$f: \mathbb{R}^* \to \mathbb{R}$$ such that $$\forall x \in \mathbb{R}^*$$ : $$\frac{1}x f(-x)+f\bigg({1 \over x}\bigg)=x$$ Basically I tried the classical way to substitute so I got this :

Let $$P(x)$$ be the assertion :

$$\bullet P(1)$$ : we obtain $$f(-1)+f(1)=1$$

I don't know if there's another technique to simplify it or reduce it, Some help please !

• Hint : try this substitution $x=-y$ you will have $f(y)-yf(-1/y)=y^2$ – Med-Elf May 5 '20 at 1:56

Take $$x = 1/y$$ to obtain

$$yf\left(- \frac1y \right) + f(y) = \frac 1y.$$

Next, take $$x = -y$$ to get

$$-\frac 1y f(y) + f\left( - \frac1y\right) = -y.$$

This is a system of linear equations in $$f(y)$$ and $$f(-1/y)$$. In particular, multiply both sides of the second equation by $$-y$$ to get

$$f(y) - yf\left(-\frac1y\right) = y^2,$$

from which adding with the first equation yields

$$2f(y) = y^2 + \frac1y,$$

or, substituting back in $$x$$,

$$f(x) = \frac12 \left(x^2 + \frac1x \right),$$

from which it should follow that this is the only function $$f$$ that satisfies the functional equation.

Fix $$a \neq 0$$. Then plugging in respectively $$x = a$$ and $$x = -\frac{1}{a}$$, we get, \begin{align*} \frac{1}{a}f(-a) + f\left(\frac{1}{a}\right) &= a \\ -af\left(\frac{1}{a}\right) + f(-a) &= -\frac{1}{a}. \end{align*} This is a system of linear equations in terms of unknowns $$x = f(-a)$$ and $$y = f\left(\frac{1}{a}\right)$$, represented by the following augmented matrix $$\left[\begin{array}{cc|c}\frac{1}{a} & 1 & a \\ 1 & -a & -\frac{1}{a}\end{array}\right].$$ Row reducing, we get $$\left[\begin{array}{cc|c}1 & 0 & \frac{1}{2}\left(a^2 - \frac{1}{a}\right) \\ 0 & 1 & \frac{1}{2}\left(a + \frac{1}{a^2}\right)\end{array}\right].$$ In particular, this implies $$f(-a) = \frac{1}{2}\left(a^2 - \frac{1}{a}\right) \implies f(x) = \frac{1}{2}\left(x^2 + \frac{1}{x}\right).$$