2002 USAMO Problems/Problem 4

Problem

Let $\mathbb{R}$ be the set of real numbers. Determine all functions $f : \mathbb{R} \rightarrow \mathbb{R}$ such that

$\displaystyle f(x^2 - y^2) = xf(x) - yf(y)$

for all pairs of real numbers $\displaystyle x$ and $\displaystyle y$ .

Solutiona

Solution 1

We first prove that $\displaystyle f$ is odd.

Note that $\displaystyle f(0) = f(x^2 - x^2) = xf(x) - xf(x) = 0$ , and for nonzero $\displaystyle y$ , $\displaystyle xf(x) + yf(-y) = f(x^2 - y^2) = xf(x) - yf(y)$ , or $\displaystyle yf(-y) = -yf(y)$ , which implies $\displaystyle f(-y) = -f(y)$ . Therefore $\displaystyle f$ is odd. Henceforth, we shall assume that all variables are non-negative.

If we let $\displaystyle y = 0$ , then we obtain $\displaystyle f(x^2) = xf(x)$ . Therefore the problem's condition becomes

$\displaystyle f(x^2 - y^2) + f(y^2) = f(x^2)$ .

But for any $\displaystyle a,b$ , we may set $x = \sqrt{a}$ , $y = \sqrt{b}$ to obtain

$\displaystyle f(a-b) + f(b) = f(a)$ .

(It is well known that the only continuous solutions to this functional equation are of the form $\displaystyle f(x) = kx$ , but there do exist other solutions to this which are not solutions to the equation of this problem.)

We may let $\displaystyle a = 2t$ , $\displaystyle b = t$ to obtain $\displaystyle 2f(t) = f(2t)$ .

Letting $\displaystyle x = t+1$ and $\displaystyle y = t$ in the original condition yields

$\displaystyle \begin{matrix}f(2t+1) &=& (t+1)f(t+1) - tf(t) \qquad \\ &=& (t+1)[f(t) + f(1) ] - tf(t) \\ &=& f(t) + (t+1)f(1) \qquad \qquad \end{matrix}$

But we know $\displaystyle f(2t + 1) = f(2t) + f(1) = 2f(t) + f(1)$ , so we have $\displaystyle 2f(t) + f(1) = f(t) + tf(1) + f(1)$ , or

$\displaystyle f(t) = tf(1)$ .

Hence all solutions to our equation are of the form $\displaystyle f(x) = kx$ . It is easy to see that real value of $\displaystyle k$ will suffice.

Solution 2

As in the first solution, we obtain the result that $\displaystyle f$ satisfies the condition

$\displaystyle f(a) + f(b) = f(a+b)$ .

We note that

$f(x) = f\left[ \left(\frac{x+1}{2}\right)^2 - \left( \frac{x-1}{2} \right)^2 \right] = \frac{x+1}{2} f \left( \frac{x+1}{2} \right) - \frac{x-1}{2} f \left( \frac{x-1}{2} \right)$ .

Since $\displaystyle f(2t) = 2f(t)$ , this is equal to

$\frac{(x+1)[f(x) +f(1)]}{4} - \frac{(x-1)[f(x) - f(1)]}{4} = \frac{xf(1) + f(x)}{2}$

It follows that $\displaystyle f$ must be of the form $\displaystyle f(x) = kx$ .

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

Resources

Page

Toolbox

Search