### 14th Putnam 1954

**Problem A5**

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/2^{n})| = x δ/2 + |f(x/2^{n})|. But since f(x/2^{n}) tends to zero, for any given x we can take n sufficiently large that |f(x/2^{n})| < x δ/2 and hence |f(x)| < x δ as required.

14th Putnam 1954

© John Scholes

jscholes@kalva.demon.co.uk

26 Nov 1999