Find all real-valued functions f on the reals such that f(x2 - y2) = x f(x) - y f(y) for all x, y.
Solution
Answer: f(x) = kx for some real k.
Putting y = 0, f(x2) = x f(x). Hence f(x2 - y2) = f(x2) - f(y2). So for any non-negative x, y, we have f(x - y) = f(x) - f(y). Hence also f(x) = f(x + y - y) = f(x + y) - f(y), so f(x + y) = f(x) + f(y) for non-negative x, y. Also f(0) = f(02) = 0 f(0) = 0, and for non-negative y, f(-y) = f(0 - y) = f(0) - f(y) = -f(y). Hence also f(-y) = -f(y) for negative y. So we have f(x + y) = f(x) + f(y) for non-negative x and any y. But now if x is negative, f(x + y) = -f(-x - y) = - (f(-x) - f(y) ) = f(x) + f(y). So f(x + y) = f(x) + f(y) for all x and y.
Now for any x we have f(x) + f(x - 1) = f(2x - 1) = f( x2 - (x - 1)2) = x f(x) - (x - 1) f(x - 1) = x f(x - 1) + x f(1) - (x - 1) f(x - 1) = x f(1) + f(x - 1), so f(x) = x f(1). So if f(1) = k, then f(x) = kx. It is trivial to check that this does indeed satsify the equation given for any k.
© John Scholes
jscholes@kalva.demon.co.uk
23 July 2002
Last corrected/updated 3 Dec 03