61st Putnam 2000

Problem B4

f(x) is a continuous real function satisfying f(2x2 - 1) = 2 x f(x). Show that f(x) is zero on the interval [-1, 1].



Set F(t) = f(cos t)/sin t. So F(t) is defined and continuous except possibly at multiples of π. Then F(t + 2π) = F(t). Also F(2mt) = F(t). Hence F(1) = F(2m+1) = F(2m+1 + 2nπ) = F(1 + nπ/2m) for all integers n, m. But the set {1 + nπ/2m} is dense in the reals, so F must be constant on each open interval (nπ, (n+1)π).

But we see directly from the relation f(2x2 - 1) = 2 x f(x), that f(-x) = f(x). Hence F(t + π) = f(-cos t)/(- sin t) = F(t). So the constant value on each open interval is the same. In other words F is constant except at integral multiples of π.

But F(-t) = - F(t). Hence F must be zero, except at integral multiples of π. So f(x) must be zero on the interval on the open interval (-1, 1). But f is continuous, so f(1) and f(-1) are also zero. (Or, directly, from f(2x2 - 1) = 2 x f(x) putting x = 0 and then x = 1).



61st Putnam 2000

© John Scholes
1 Jan 2001