14th Putnam 1954

R is the reals. f : (0, 1) → R satisfies lim_x→0 f(x) = 0, and f(x) - f(x/2) = o(x) as x→0. Show that f(x) = o(x) as x→0.

Solution

We wish to show that given δ > 0, |f(x)| < x δ for sufficiently small x. Certainly we can find ε > 0, such that for x < ε, |f(x/2^r) - f(x/2^r+1)| < x/2^r+2 δ. Summing, |f(x)| < x δ (1/4 + 1/8 + ... ) + |f(x/2ⁿ)| = x δ/2 + |f(x/2ⁿ)|. But since f(x/2ⁿ) tends to zero, for any given x we can take n sufficiently large that |f(x/2ⁿ)| < x δ/2 and hence |f(x)| < x δ as required.

14th Putnam 1954

© John Scholes
jscholes@kalva.demon.co.uk
26 Nov 1999