### 28th Putnam 1967

Problem B3

R is the reals. f, g are continuous functions R → R with period 1. Show that limn→∞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/nr/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/nr/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/nr/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.