28th Putnam 1967

R is the reals. f, g are continuous functions R → R with period 1. Show that lim_n→∞ ∫₀¹ f(x) g(nx) dx = (∫₀¹ f(x) dx) (∫₀¹ g(x) dx).

Solution

The idea is to split the integration range into n equal parts. Thus we get ∫₀¹ f(x) g(nx) dx = ∑ ∫_r/n^r/n+1/n f(x) g(nx) dx. For large n, f is roughly constant over the range, so we get S f(r/n) ∫_r/n^r/n+1/n g(nx) dx. Changing the integration variable to t = nx, gives ∑ f(r/n) 1/n ∫₀¹ g(t) dt since g is periodic. But lim ∑ f(r/n) 1/n is just ∫₀¹ f(x) dx, so we get the required (∫₀¹ f(x) dx) (∫₀¹ g(x) dx).

It remains to look at the error involved in approximating f. The function f is continuous and [0, 1] is compact, so it must be uniformly continuous on [0, 1]. Thus we given any ε > 0, we can find N such that for n > N, we have |f(x) - f(r/n)| < ε on [r/n, r/n + 1/n] for each of r = 0, 1, 2, ... , n-1. So the error is at most ∑ ∫_r/n^r/n+1/n |f(x) - f(r/n)| |g(nx)| dx < ∑ ε 1/n ∫₀¹ |g(t)| dt = ε ∫₀¹ |g(t)| dt, which can be made arbitrarily small.

28th Putnam 1967

© John Scholes
jscholes@kalva.demon.co.uk
14 Jan 2002