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Math Help - Series of continuous functions, continued

  1. #1
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    Series of continuous functions, continued

    "Letting <x> denote the fractional part of the real number x (i.e., <x> = x - floor(x)), consider the function:

    f(x) = \sum_{n=1}^{\infty} \frac{<nx>}{n^2}, x \in \mathbb{R}

    Find all discontinuities of f, and show that the form a countable dense set."

    OK here is what I tried: I set $a_n(x) = \frac{<nx>}{n^2}$ and said that clearly, a discontinuity results at all $nx = m, m \in \mathbb{Z}$; i.e., $x = m/n $. Now this is the part where I'm a little unsure of whether I did enough: I said that if we let n, m be any integer, n <> 0, then \{ \frac{m}{n} \} \to \mathbb{Q}, which is countably dense in the reals. Can I do that? I looked up the answer and found a very different approach, but it was VERY long and confusing...
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    Re: Series of continuous functions, continued

    First, you did not actually show that there is a discontinuity at all values of $x$ for which there exists an $n\in \mathbb{N}$ such that $nx \in \mathbb{Z}$. Does your professor typically accept statements like "clearly this is true"? You also need to prove that there are no irrational points of discontinuity. Suppose $x$ is irrational. Can you show that the function is continuous at that point?
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    Re: Series of continuous functions, continued

    Quote Originally Posted by SlipEternal View Post
    First, you did not actually show that there is a discontinuity at all values of $x$ for which there exists an $n\in \mathbb{N}$ such that $nx \in \mathbb{Z}$. Does your professor typically accept statements like "clearly this is true"? You also need to prove that there are no irrational points of discontinuity. Suppose $x$ is irrational. Can you show that the function is continuous at that point?
    Crap. Alright, step one is to prove that continuity. Can I say that, clearly, a_n(x) is continuous at x if and only if floor(nx) is continuous at x?
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    Re: Series of continuous functions, continued

    Quote Originally Posted by phys251 View Post
    Crap. Alright, step one is to prove that continuity. Can I say that, clearly, a_n(x) is continuous at x if and only if floor(nx) is continuous at x?
    Yes
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    Re: Series of continuous functions, continued

    Quote Originally Posted by SlipEternal View Post
    Yes
    OK. What I did was to, wlog, consider y on (0, 1] and prove that floor(y) is continuous on (0, 1) and discontinuous at 1. Turns out, this wasn't too hard to do.

    From there, I concluded that for some continuous, nonempty interval $\mathbb{I} \subset \mathbb{R}, \lfloor y \rfloor \in C(\mathbb{I}) \iff y \notin \mathbb{Z} \iff nx \notin \mathbb{Z} \iff x \notin \mathbb{Q}$ (we originally assumed x is real).

    Observe that Q is countably dense on (0, 1), and then I'm pretty much done, right? Or do I have to show THAT as well?
    Last edited by phys251; March 17th 2014 at 07:23 PM.
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    Re: Series of continuous functions, continued

    You need to show that if $a_n(x)$ is discontinuous at some $x \in \mathbb{R}$ then $\sum_{n\ge 1} a_n(x)$ is also discontinuous at $x$. So far, you have no proof that $a_n(x)$ is discontinuous at $x$ implies $f(x)$ is discontinuous at $x$. What happens if $a_m(x)$ and $a_n(x)$ are both discontinuous at $x$, but $a_m(x) + a_n(x)$ is actually continuous? (I am pretty sure you can prove that never happens, but you still need to check for it).

    Edit: An example of two discontinuous functions whose sum is continuous: $\lfloor x \rfloor$ and $-\lfloor x \rfloor$. Both are discontinuous at all integers, yet their sum is the constant zero function which is continuous for all $x$.
    Last edited by SlipEternal; March 17th 2014 at 07:43 PM.
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    Re: Series of continuous functions, continued

    Quote Originally Posted by SlipEternal View Post
    You need to show that if $a_n(x)$ is discontinuous at some $x \in \mathbb{R}$ then $\sum_{n\ge 1} a_n(x)$ is also discontinuous at $x$. So far, you have no proof that $a_n(x)$ is discontinuous at $x$ implies $f(x)$ is discontinuous at $x$. What happens if $a_m(x)$ and $a_n(x)$ are both discontinuous at $x$, but $a_m(x) + a_n(x)$ is actually continuous? (I am pretty sure you can prove that never happens, but you still need to check for it).

    Edit: An example of two discontinuous functions whose sum is continuous: $\lfloor x \rfloor$ and $-\lfloor x \rfloor$. Both are discontinuous at all integers, yet their sum is the constant zero function which is continuous for all $x$.
    Oh man. I'm lost, then. Doesn't your counterexample negate the very thing I need to prove--i.e., if $a_n(x)$ is discontinuous at some $x \in \mathbb{R}$ then $\sum_{n\ge 1} a_n(x)$ is also discontinuous at $x$? The solutions manual goes into this long-winded proof about considering what happens when x is irrational vs. when it is rational, but wow is it hard for me to follow...
    Last edited by phys251; March 17th 2014 at 08:28 PM.
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    Re: Series of continuous functions, continued

    Obviously, $0 \le <nx> < 1$ for all $n\in \mathbb{N}, x \in \mathbb{R}$. So, $0 \le a_n(x) < \dfrac{1}{n^2}$. Define $S_n(x) = \sum_{k = 1}^n a_k(x)$. Then $|f(x) - S_n(x)| = f(x) - S_n(x) = \sum_{k > n} a_k(x)$. Hence, $0 \le f(x) - S_n(x) < \sum_{k>n} \dfrac{1}{k^2}$. Can you show that $\sum_{k>n} \dfrac{1}{k^2} < \dfrac{1}{n}$? (Hint: consider upper and lower sums of the function $\dfrac{1}{x^2}$ where you consider rectangles of width 1).

    Then, $|f(x) - S_n(x)| < \sum_{k > n}\dfrac{1}{k^2} < \dfrac{1}{n}$, so given any $\varepsilon>0$ let $N = 1+ \lceil \varepsilon^{-1}\rceil$. Then for any $x\in \mathbb{R}$ and any $n\ge N$, you have $|f(x) - S_n(x)| \le |f(x) - S_N(x)| < \dfrac{1}{n} < \varepsilon$, so $S_n(x)$ converges uniformly to $f(x)$.

    You already showed that for all $x \in \mathbb{R} \setminus \mathbb{Q}$, $a_n(x)$ is continuous at $x$. Hence, $S_n(x)$ must also be continuous at $x$ (since it is a finite sum of continuous functions). Then, show that $f(x)$ is continuous:

    Given $\varepsilon>0$, choose $N\in \mathbb{N}$ such that $|f(x) - S_n(x)| < \dfrac{\varepsilon}{3}$ for all $x \in \mathbb{R}, n\ge N$. Since $S_N(x)$ is continuous, choose $\delta>0$ such that $|y-x|<\delta$ implies $|S_N(y) - S_N(x)| < \dfrac{\varepsilon}{3}$. Then you have:

    $|f(y) - f(x)| \le |f(y) - S_N(y)| + |S_N(y) - S_N(x)| + |S_N(x) - f(x)| < \dfrac{\varepsilon}{3}+\dfrac{ \varepsilon }{3}+\dfrac{ \varepsilon }{3} = \varepsilon$

    Hence, $f$ is continuous at $x$ for all $x \in \mathbb{R}\setminus \mathbb{Q}$.

    Does this help? Now, you just need to show that $f(x)$ is discontinuous for all $x \in \mathbb{Q}$, which you were well on your way to proving.
    Last edited by SlipEternal; March 17th 2014 at 09:24 PM.
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    Re: Series of continuous functions, continued

    Hi,
    This problem has been puzzling me. It's pretty straight forward to see that the function f is continuous at any irrational, but to see that it's discontinuous at any rational is kind of tricky. I gave up and looked for a solution. Here's a link to a correct solution: real analysis - Series with fractional part of $nx$ - Mathematics Stack Exchange (maybe the same solution you referenced in the OP). If you have any questions about this solution, I can try and help.
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    Re: Series of continuous functions, continued

    If you want an epsilon-delta proof that $f$ is discontinuous on the set of all rationals, here it is:

    Suppose $x \in \mathbb{Q}$. Then suppose $x = \dfrac{p}{q}$ where $p,q \in \mathbb{Z}, q>0$, and $\dfrac{p}{q}$ is irreducible.

    If $q=1$, then $a_n(x)=0$ for all $n$. Hence, for any $y\in (x-1,x)$, we know $f(y)-f(x)>0$ and $\displaystyle \lim_{y \to x^-} |f(y) - f(x)| = \sum_{k\ge 1}\dfrac{1}{k^2} > 0$. So, $f$ is discontinuous on the set of integers.

    If $q>1$, then for each $k = 1,\ldots, q^4$, if $a_k$ is continuous at $x$, choose $\delta_k$ such that $|y-x| < \delta_k$ implies $|a_k(y) - a_k(x)|<\dfrac{1}{q^8}$ and $\delta_k = 1$ otherwise. Let $\displaystyle \delta = \min\left\{\dfrac{3}{4q^4},\delta_1,\ldots, \delta_{q^4}\right\}$. For any $y \in \left(x - \delta, x\right)$, we have:

    $\displaystyle \begin{align*}|f(y) - f(x)| & \ge f(y) - f(x) \\ & = \sum_{k\ge 1}(a_k(y) - a_k(x)) \\ & = \left(\sum_{k=1}^{q^4}(a_k(y) - a_k(x)) - \sum_{k = 1}^{q^3}\left( a_{kq}(y) - a_{kq}(x) \right) \right) + \sum_{k=1}^{q^3}\left( a_{kq}(y) - a_{kq}(x) \right) + \sum_{k > q^4} (a_k(y) - a_k(x)) \\ & = \left(\sum_{k=1}^{q^4}(a_k(y) - a_k(x)) - \sum_{k = 1}^{q^3}\left( a_{kq}(y) - a_{kq}(x) \right) \right) + \sum_{k=1}^{q^3}\left( a_{kq}(y) - 0 \right) + \sum_{k > q^4} (a_k(y) - a_k(x)) \\ & \ge \left(-\sum_{k=1}^{q^4}|a_k(y) - a_k(x)| + \sum_{k=1}^{q^3}|a_{kq}(y) - a_{kq}(x)|\right) + \sum_{k=1}^{q^3}\dfrac{3}{4(kq)^2} + \sum_{k > q^4}(0 - a_k(x)) \\ & > -q^4\dfrac{1}{q^8} + \dfrac{3}{4q^2} - \sum_{k>q^4} \dfrac{1}{k^2} \\ & > -\dfrac{1}{q^4} + \dfrac{3}{4q^2} - \dfrac{1}{q^4} \\ & = \dfrac{3q^2-8}{4q^4} > 0\end{align*}$

    Hence, for any $0 < \varepsilon < \dfrac{3q^2-8}{4q^4}$ and any $\delta>0$, there exists $y \in (x-\delta,x+\delta)$ such that $|f(y) - f(x)| > \varepsilon$. This clearly shows that $f$ is discontinuous at any rational number. Note: $3q^2-8 \ge 4$ for $q\ge 2$.
    Last edited by SlipEternal; March 19th 2014 at 10:10 PM.
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