Let R be the reals. f **:** R → R is twice-differentiable and we can find g **:** R → R such that g(x) >= 0 and f(x) + f ''(x) = -xg(x)f '(x) for all x. Prove that f(x) is bounded.

**Solution**

This needs a trick. Multiply through by f '(x). Then the lhs is (half) the derivative of f ^{2} + f '^{2}. The rhs is non-positive for x > 0 and non-negative for x < 0. Hence f ^{2} + f '^{2} has a maximum at x = 0. A fortiori, it is bounded. Hence f is bounded.

© John Scholes

jscholes@kalva.demon.co.uk

12 Dec 1998