
Re: analysis question on periodic functions
Posted:
Aug 18, 2013 9:59 AM


On Sat, 17 Aug 2013 14:10:59 0500, quasi <quasi@null.set> wrote:
>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.
Ah right, I missed the fact that the middle term must vanish.
>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

