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Difference between revisions of "2002 USAMO Problems/Problem 4"

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Revision as of 20:49, 6 April 2013

Problem

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

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

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

Solutions

Solution 1

We first prove that $f$ is odd.

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

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

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

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

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

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

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

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

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

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

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

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

Solution 2

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

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

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 $f(2t) = 2f(t)$, 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 $f$ must be of the form $f(x) = kx$.

Solution 3

Let $y=0$, so that the functional equation becomes $f(x^2)=xf(x)$. For positive $x$, then, $f(x)=x^{\frac{1}{2}}f(x^{\frac{1}{2}})=x^{\frac{1}{2}}x^{\frac{1}{4}}f(x^{\frac{1}{4}})=x^{\frac{1}{2}}x^{\frac{1}{4}}x^{\frac{1}{8}}f(x^{\frac{1}{8}})=\cdots =x^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots}f(x^{\frac{1}{\infty}})$, which reduces to $xf(1)$ for nonzero $x$. For $x=0$, we have $f(0)=0\cdot f(0)=0$. Thus, we have limited $f$ to linear functions of the form $f(x)=kx$ where $k$ is a constant. We can verify that if $f(x)=kx$, then any value of $k$ will work: $k(x^2-y^2)=x\cdot kx-y\cdot ky$, which is always true.

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

See also

2002 USAMO (Problems • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions
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