Let R be the reals. Let f : [a, b] → R have a continuous derivative, and suppose that if f(x) = 0, then f '(x) ≠ 0. Show that we can find g : [a, b] → R with a continuous derivative, such that f(x)g'(x) > f '(x)g(x) for all x ∈ [a, b].
Solution
We note that the derivative of f/g is - (fg' - f 'g)/g2 so if we take g so that f/g is decreasing, then we are almost home. Not quite, because of the difficulty that f may be zero. For example, if we take g(x) = x f(x), then fg' - f 'g is zero whenever f(x) = 0.
f has only finitely many zeros, for otherwise the zeros would have a limit point c and then by continuity we would have f(c) = f '(c) = 0. So we may take a polynomial p(x) such that p(x) f '(x) = -1 at each zero of f. Now consider g(x) = x f(x) + k p(x). We have f(x) g'(x) - f '(x) g(x) = f(x)2 + k(f(x) p'(x) - f '(x) p(x) ). Now f(x) p'(x) - f '(x) p(x) = 1 at each zero of f, so we can find an open set K containing all these zeros such that f(x) p'(x) - f '(x) p(x) > 1/2 on K. Now [a, b] - K is compact and contains no zeros of f(x)2, which is continuous, so we can find ε > 0, so that f(x)2 > ε on [a, b] - K. But [a, b] is compact, so |f(x) p'(x) - f '(x) p(x)| < some M on [a, b]. Take k < ε/M. Then on [a, b] - K, f(x)2 > ε and k|f(x) p'(x) - f '(x) p(x)| < ε, so f(x) g'(x) - f '(x) g(x) > 0. On K, f(x)2 is non-negative and k( f(x) p'(x) - f '(x) p(x) ) is positive, so f(x) g'(x) - f '(x) g(x) > 0.
The motivation for the problem is probably that if f and g are linearly independent solutions of y'' + a(x)y' + b(x)y = 0, then the Wronskian fg' - f 'g is non-zero. But this requires f to have a continuous second derivative, which is not necessarily the case.
© John Scholes
jscholes@kalva.demon.co.uk
15 Feb 2002