Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $$F(x)=\sum_{n=1}^\infty \frac{(nx)}{n^2},$$ where $(x)=x$ for $x\in (-1/2,1/2]$ and $(x)$ is continued to $\mathbb{R}$ by periodicity, that is $(x+1)=(x)$. Prove that $F$ is discontinuous whenever $x=m/2n$, where $m,n\in \mathbb{Z}$ with $m$ odd and $n\neq 0$. This does not seem obvious to me.

share|improve this question
    
Isn't this just the difference between $\sum -n^{-2}$ vs. $\sum n^{-2}$? –  user4143 Dec 3 '10 at 21:43
    
It suffices to find a sequence x_k such that x_k converges to m/2n but F(x_k) does not converge to F(m/2n). Think about how you would do this. –  Qiaochu Yuan Dec 3 '10 at 21:45
add comment

2 Answers

up vote 1 down vote accepted

Note that $\lim_{x \rightarrow a}(x) = a$ except when $a$ is an odd multiple of ${1 \over 2}$, in which case $\lim_{x \rightarrow a^-}(x) - \lim_{x \rightarrow a^+}(x) = 1$. Another way of saying this is that $\lim_{x \rightarrow a^-}(x) - \lim_{x \rightarrow a^+}(x) = 0$ unless $a$ is an odd multiple of ${1 \over 2}$ in which case it is 1.

So for the term $f_n(x) = {(nx) \over n^2}$, $\lim_{x \rightarrow a^-}f_n(x) - \lim_{x \rightarrow a^+}f_n(x) = 0$ unless $x = {m \over 2n}$ for some odd integer $m$, in which case it is ${1 \over n^2}$. So for your arbitrary ${m \over 2n}$ with $m$ odd, it will incur a jump of ${1 \over n^2}$ due to $f_n(x)$, and other terms can only increase or keep the same the size of the jump. For any $\epsilon > 0$, if enough terms have been added, the absolute value of the sum of the remaining terms will be at most $\epsilon$, so that no matter how these remaining terms add up there will still be a jump of at least ${1 \over n^2} - \epsilon$ at ${m \over 2n}$. Letting $\epsilon \rightarrow 0$ this means the infinite sum has a jump of at least ${1 \over n^2}$ at ${m \over 2n}$. (By jump I now mean the $\liminf$ from the left minus the $\limsup$ from the right.) In particular it's discontinuous there.

share|improve this answer
    
You meant at $\frac{m}{2n}$ in the last two sentences? –  TCL Dec 4 '10 at 1:17
    
yes, corrected that –  Zarrax Dec 4 '10 at 1:44
add comment

A direct computation shows that $$F(x+)=F(x)-\frac{1}{2n^2}\sum_{k=0}^{\infty}\frac{1}{(2k+1)^2}=F(x)-\frac{\pi^2}{16n^2},$$ $$F(x-)=F(x)+\frac{1}{2n^2}\sum_{k=0}^{\infty}\frac{1}{(2k+1)^2}=F(x)+\frac{\pi^2}{16n^2},$$ when $x=m/2n$, with $m$ and $2n$ being relatively prime. (The infinite sum is easily derived from the Euler result, of course.)


The function $F(x)$ was suggested by Riemann in his thesis as an example of a (Riemann)-integrable function which has a point of discontinuity in every interval (see "Gesammelte Mathematische Werke und Wissenschaftlicher Nachlass", Springer -Verlag 1990, pp. 259-296).

share|improve this answer
    
For your version do you need $(x) = 0$ at ${1 \over 2}$ to have the symmetric limits? I'm thinking for the way he wrote it $F(x-) = F(x)$ and $F(x+) = F(x) - {\pi \over 16n^2}$. –  Zarrax Dec 4 '10 at 16:05
    
Correction: $F(x-) = F(x)$ and $F(x+) = F(x) - {\pi \over 8n^2}$ –  Zarrax Dec 4 '10 at 16:23
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.