### 25th Putnam 1964

Problem B3

R is the reals. f : R → R is continuous and for any α > 0, limn→∞ f(nα) = 0. Prove limx→∞ f(x) = 0.

Solution

This is fairly hard.

Take ε > 0. Let An = {x : |f(nx)| ≤ ε}. Let Bn be the intersection of all Am with m ≥ n. The function f is continuous, so each An is closed. Hence also each Bn. Now suppose we can show that an open interval (a, b) is contained in some BN. That means that for any x in (a, b) and any n > N we have |f(nx)| < ε. Hence |f(x)| < ε for any x in (na, nb). But the intervals (na, nb) expand in size as they move to the right, so for n sufficiently large they overlap and we have that their union includes all sufficiently large x. In other words, for any sufficiently large x we have |f(x)| < ε. But ε was arbitrary, so we have established that lim f(x) = 0 as x tends to infinity.

It remains to show that we can find such an open interval (a, b). Now we are given that for any given x, lim f(nx) = 0 (as n tends to infinity), so x belongs to some Bn for n sufficiently large. In other words, the union of the Bn is the entire line.

We now need the Baire category theorem which states that if a union of closed sets covers the line, then one of the sets contains an open interval. This is bookwork. [But straightforward. Assume none of the Bn contain an open interval. Take a point not in B1. Since B1 is closed we may take a closed interval C1about the point, not meeting B1. Having chosen Cn, there must be a point in it not in Bn+1 (or Bn+1 would contain an open interval - the interior of Cn). Hence we may take Cn+1, a closed subinterval of Cn which does not meet Bn+1. Now Cn is a nested sequence of closed intervals, so there must be a point X in all the Cn But Bn cover the line, so X must be in some Bn. Contradiction.].