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A function discontinuous only on Q

  1. Nov 24, 2007 #1
    Let [tex]f(x)=\sum_{n=1}^{\infty}\frac{(nx)}{n^2}[/tex], where (x) denotes the fractional part of x. [tex]|\frac{(nx)}{n^2}| \leq \frac{1}{n^2}[/tex], so [tex]\sum \frac{(nx)}{n^2}[/tex] converges uniformly to f on all of R. (x) is discontinuous precisely at the integers, so [tex]s_k(x) = (x) + \frac{(2x)}{2^2} + ... + \frac{(kx)}{k^2}[/tex] is discontinuous precisely when [tex]x = \frac{p}{m}[/tex], where m = 1,...,k, and p is any integer. It follows that if [tex]x_0[/tex] is irrational, then [tex]\lim_{x \rightarrow x_0}s_n(x) = s_n(x_0)[/tex] for all n. So by theorem 7_11 in Rudin, [tex]\lim_{x \rightarrow x_0}f(x) = \lim_{n \rightarrow \infty}s_n(x_0) = f(x_0)[/tex], which shows f is continuous at every irrational point.

    Another thing to add is that it is clear f(x+1) = f(x) for all x. It is also clear by uniform convergence that f is Riemann integrable on any bounded interval, since each s_n is as such. That's pretty much most of Problem 10 in Chapter 7, but not all.

    Is the limit function discontinuous on Q (the rationals)? Or equivalently, by the f(x+1)=f(x) property, is f discontinuous at all the rationals in [0,1]? I don't see off hand the direct reason WHY the limit function is discontinuous, despite the fact that every rational number q is a discontinuity point of s_n, for all n >= some N. (In particular, the problem asks to show the discontinuous points are a countable dense subset, which would be implied if that subset is Q...) I.e., just because g_n(x) = 1/n, x rational, g(x) = 0, x irrational, is discontinuous everywhere, doesn't imply it's uniform limit function g(x)=0 is discontinuous anywhere.
    Last edited: Nov 24, 2007
  2. jcsd
  3. Nov 24, 2007 #2
    The Solution

    It's not hard to see that f(x) is not continuous at 0. For instance, f(0) = 0. And if 0 < e < 1/2, then f(-e) > 1/2.

    Now suppose a rational number p/q is given, where p and q have no common factors, and we wish to show f is not continuous at p/q. By absolute convergence, we can write [tex]f(x) = f_1(x) + f_2(x) = \sum_{n \neq kq}^{\infty}\frac{(nx)}{n^2} + \sum_{k = 1}^{\infty}\frac{(qkx)}{q^2 k^2}[/tex]. By the argument given above for continuity at irrational numbers, we have [tex]f_1[/tex] is continuous at p/q (note that the fact that p and q have no common factors is required for [tex]f_1[/tex] to be continuous there). But [tex]f_2(p/q) = 0[/tex], and for 0 < e < 1/2, we have [tex]f_2(p/q - e/q) \geq \frac{(p-e)}{q^2} = \frac{1-e}{q^2} > \frac{1}{2q^2}[/tex]. Since q is fixed, it follows that [tex]f_2[/tex] is discontinuous at p/q from the left. Hence [tex]f = f_1 + f_2[/tex] is discontinuous at p/q from the left.
    Last edited: Nov 25, 2007
  4. Nov 25, 2007 #3


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    How can f not be discontinuous everywhere if it's discontinuous on a dense subset?

    Suppose the set A of points where f is discontinuous is dense in R.
    Then for each a in A there exists an [itex]\epsilon>0[/itex], such that for all [itex]\delta>0[/itex] there's an x such that [itex]0<|x-a|<\delta[/itex] and [itex]|f(x)-f(a)|\geq \epsilon[/itex].

    So take any point b in R and suppose f is continuous in b. Then there exists for any [itex]\epsilon>0[/itex] a [itex]\delta>0[/itex] such that [itex]|x-b|<\delta \Rightarrow |f(x)-f(b)|<\epsilon[/itex].
    But [itex](b-\delta,b+\delta)[/itex] contains a point a in A no matter how small delta, so there's an interval centered on a which is contained in [itex](b-\delta,b+\delta)[/itex] and contains a point x such that [itex]|f(x)-f(a)|\geq \epsilon[/itex].

    I don't imagine Rudin is wrong. I just want to know the error in my reasoning.

    EDIT: Ok. Found the error. Ofcourse |f(x)-f(a)| is too big, but |f(x)-f(b)| shouldn't be.
    Last edited: Nov 25, 2007
  5. Nov 25, 2007 #4


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    The function f(x)= 0 if x is irrational, f(x)= 1/n if x is rational, x= m/n reduced to lowest terms. It's easy to show that that [itex]\lim_{x\rightarrow a} f(x)= 0[/itex] for any a and so is continous at every irrational number, discontinuous at every rational number (and, of course, the rationals are dense in the reals).
  6. Nov 26, 2007 #5
    can you explain this more thoroughly?

    this is the first non-trivial problem from chapter 7.
  7. Nov 26, 2007 #6
    If 0 < e < 1/2, then (-e) = 1 - e > 1 - 1/2 = 1/2, then
    [tex]f(-e)=\sum_{n=1}^{\infty}\frac{(n\cdot -e)}{n^2} = (-e) + ... \geq (-e) > 1/2[/tex].

    But [tex]f(0)=\sum_{n=1}^{\infty}\frac{(n \cdot 0)}{n^2} = 0[/tex],
    so |f(-e)-f(0)| = |f(-e)| >= 1/2. -1/2 < -e < 0 was arbitrary, hence f is discontinuous from the left.
  8. Nov 26, 2007 #7
    Just wanted to add.. Another example to keep in mind is [tex]f(x) = exp(\frac{-1}{x^2})[/tex], x irrational, [tex]f(x) = 0[/tex], x rational. This function is discontinuous everywhere except 0. It's infinitely differentiable at 0.
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