quasi
Posts:
11,192
Registered:
7/15/05


Re: analysis question on periodic functions
Posted:
Aug 17, 2013 2:30 PM


dullrich wrote: >David Bernier wrote: >>David Bernier wrote: >>> >>> Suppose we have an odd continuous function f: R > R >>> with period 1 so that f(x+1) = f(x), and f(x) = f(x). >>> >>> Suppose f has mean zero over the unit interval, so >>> int_{0, 1} f(x) dx = 0, but that >>> int_{0, 1}  f(x)  dx > 0 >>> (so it's not constantly zero). >>> >>> Given a real number a> 0, consider the series: >>> >>> f(a) + f(2a) + f(3a) + ... >>> >>> with partial sums >>> >>> S_k (a) = sum_{j = 1 ... k} f(ka). >>> >>> (i) If a is rational, is it necessarily true that the S_k (a) >>> are bounded in absolute value? >> >>How about: >> >>(i) YES > >??? Are you sure about that?
As I see it, the answer "yes" is correct.
Let Z,R denote the usual sets (integers, reals).
Assume the domain of f is R.
The mix of
f is odd f has period 1
clinches it.
Since f is odd, f(0) = 0.
Since f has period 1, f(x) = 0 for all x in Z.
Suppose x,y in R are such that x + y is in Z.
Claim f(x) + f(y) = 0.
If x in Z, then y in Z, so f(x) + f(y) = 0 + 0 = 0.
Suppose x is not in Z.
Let s = x  floor(x) and t = y  floor(y).
Since x is not in Z, neither is y, hence s,t are positive,
Since x + y is in Z, it follows that s + t = 1. Hence
f(x) + f(y) = f(s) + f(t) = f(s) + f(1s) = f(s) + f(s) = 0
as claimed.
As a special case, f(1/2) = 0. Now suppose a is a positive rational.
Write a = b/c as a fraction in lowest terms, with b,c in N.
Now for j in N with j*a < (1/2)*c,
f(j*a) + f((cj)*a) = 0
In the sum
f(a) + f(2*a) + ... + f((c1)*a)
all terms except a possible middle term (which is present iff c is even) pair up and cancel. The middle term, if present, is also equal to 0. Noting that c*a = b which is in Z, it follows that
f(a) + f(2*a) + ... + f(c*a) = 0.
It's then easily seen that the partial sums
S_1, S_2, S_3, ...
repeat every c terms, that is, S_(k+c) = S_k.
It follows that the partial sums are bounded, as claimed.
quasi

