R is the reals. f **:** R → R is continuous and for any α > 0, lim_{n→∞} f(nα) = 0. Prove lim_{x→∞} f(x) = 0.

**Solution**

*This is fairly hard.*

Take ε > 0. Let A_{n} = {x : |f(nx)| ≤ ε}. Let B_{n} be the intersection of all A_{m} with m ≥ n. The function f is continuous, so each A_{n} is closed. Hence also each B_{n}. Now suppose we can show that an open interval (a, b) is contained in some B_{N}. 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 B_{n} for n sufficiently large. In other words, the union of the B_{n} 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 B_{n} contain an open interval. Take a point not in B_{1}. Since B_{1} is closed we may take a closed interval C_{1}about the point, not meeting B_{1}. Having chosen C_{n}, there must be a point in it not in B_{n+1} (or B_{n+1} would contain an open interval - the interior of C_{n}). Hence we may take C_{n+1}, a closed subinterval of C_{n} which does not meet B_{n+1}. Now C_{n} is a nested sequence of closed intervals, so there must be a point X in all the C_{n} But B_{n} cover the line, so X must be in some B_{n}. Contradiction.].

© John Scholes

jscholes@kalva.demon.co.uk

5 Feb 2002