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 $f(-1)+f(1)=1$

All what I know about this function is this information !

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

  • $\begingroup$ Hint : try this substitution $x=-y$ you will have $f(y)-yf(-1/y)=y^2$ $\endgroup$ – 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 $$f(-a) = \frac{1}{2}\left(a^2 - \frac{1}{a}\right) \implies f(x) = \frac{1}{2}\left(x^2 + \frac{1}{x}\right).$$


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