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.

This function is continuous, it follows by M-Weierstrass Test. But proving non-differentiability, I think it's too hard. Does someone know how can I prove this? Or at least have a paper with the proof? The function is $$ f(x) = \sum_{k=1}^\infty \frac{\sin((k + 1)!\;x )}{k!}$$ Thanks!

share|improve this question
    
I wonder if it can be proven using induction and the fundamental theorem of calculus. –  Ahmed Masud Oct 31 '11 at 19:44
1  
See also this question -- not an exact duplicate but with much the same properties. –  Henning Makholm Oct 31 '11 at 19:57
add comment

2 Answers 2

(Edited: handwaving replaced by rigor)

For conciseness, define the helper functions $\gamma_k(x)=\sin((k+1)!x)$. Then $f(x)=\sum_k \frac{\gamma_k(x)}{k!}$.

Fix an arbitrary $x\in\mathbb R$. We will construct a sequence $(x_n)_n$ such that $$\lim_{n\to\infty} x_n = x \quad\land\quad \lim_{n\to\infty} \left|\frac{f(x_n)-f(x)}{x_n-x}\right| = \infty$$ Such a sequence will directly imply that $f$ is not differentiable at $x$.

Let $x'_n$ be the largest number less than $x$ such that $|\gamma_n(x'_n)-\gamma_n(x)|=1$. Let $x''_n$ be the smallest number larger than $x$ such that $\gamma_n(x''_n)=\gamma_n(x'_n)$. One of these, to be determined later, will become our $x_n$.

No matter which of these two choices of $x_n$ we have $|x_n-x|<\frac{2\pi}{(n+1)!}$ so $\lim x_n=x$.

To estimate the difference quotient, write $$f(x) = \underbrace{\sum_{k=1}^{n-1}\frac{\gamma_k(x)}{k!}}_{p(x)}+ \underbrace{\frac{\gamma_n(x)}{n!}}_{q(x)}+ \underbrace{\sum_{k=n+1}^{\infty} \frac{\gamma_k(x)}{k!}}_{r(x)}$$ and so, $$\underbrace{f(x_n)-f(x)}_{\Delta f} = \underbrace{p(x_n)-p(x)}_{\Delta p} + \underbrace{q(x_n)-q(x)}_{\Delta q} + \underbrace{r(x_n)-r(x)}_{\Delta r}$$ Of these, by construction of $x_n$ we have $|\Delta q| = \frac{1}{n!}$.

Also, $r(x)$ is globally bounded by the remainder term in the series $\sum 1/n! = e$, which by Taylor's theorem is at most $\frac{e}{(n+1)!}$. So $|\Delta r| \le \frac{2e}{(n+1)!}$.

$\Delta p$ is not dealt with as easily. In some cases it may be numerically larger than $\Delta q$, ruining a simple triange-equality based estimate. But it can be tamed by a case analysis:

  • If $p$ is strictly monotonic on $[x'_n, x''_n]$, then $p(x'_n)-p(x)$ and $p(x''_n)-p(x)$ will have opposite signs. Since $q(x'_n)=q(x''_n)$, we can choose $x_n$ such that $\Delta p$ and $\Delta q$ has the same sign. Therefore $|\Delta p+\Delta q|\ge|\Delta q|=\frac{1}{n!}$.

  • Otherwise, $p$ has an extremum between $x'_n$ and $x''_n$; select $x_n$ such that the extremum is between $x$ and $x_n$. Because $p$ is a finite sum of $C^\infty$ functions, we can bound its second derivative separately for each of its terms: $$\forall t: |p''(t)| \le \sum_{k=1}^{n-1}\left|\frac{\gamma''_k(t)}{k!}\right| \le \sum_{k=1}^{n-1}\frac{(k+1)!^2}{k!} \le \sum_{k=1}^{n-1} (k+1)!(k+1) \le 2n!n $$ Therefore the maximal variation of $p$ in an interval of length $\le\frac{2\pi}{(n+1)!}$ that contains a stationary point must be $\left(\frac{2\pi}{(n+1)!}\right)^2 2n!n = \frac{8\pi^2n}{(n+1)^2}\frac{1}{n!}$. The $\frac{8\pi^2n}{(n+1)^2}$ factor is less than $1/2$ for $n>16\pi^2$, so for large enough $n$ we have $|\Delta p+\Delta q|\ge \frac{1}{2n!}$.

Thus, for large $n$ we always have $$|\Delta f| \ge \frac{1}{2n!} - \frac{2e}{(n+1)!} = \frac{1}{n!}\left(\frac{1}{2}-\frac{2e}{n+1}\right)$$ and therefore $$\left|\frac{f(x_n)-f(x)}{x_n-x}\right| \ge \frac{(n+1)!}{2\pi}|\Delta f| \ge \frac{n+1}{2\pi}\left(\frac{1}{2}-\frac{2e}{n+1}\right) = \frac{n+1}{4\pi}-\frac{e}{\pi} \to \infty$$ as promised.

share|improve this answer
    
this site should have a +10 arrow :) Nice argument! Thanks for sharing this. –  Leandro Nov 1 '11 at 0:24
    
@Henning Makholm Sorry for my stupid question Henning, but I only see that $ \Delta p $ it´s bigger than $ \frac{1} {{n!}}\left( {1 - \frac{{n^2 }} {{n + 1}} - \frac{{2e}} {{n + 1}}} \right) $ but the factor $ \left( {1 - \frac{{n^2 }} {{n + 1}} - \frac{{2e}} {{n + 1}}} \right) $ it´s negative. The other step that i don´t understand it´s why also it´s bigger than $$ \frac{1} {{n!}}\left( {1 - \frac{{2e}} {{n + 1}}} \right) $$ since $$ \left( {1 - \frac{{2e}} {{n + 1}}} \right) > \left( {1 - \frac{{2e}} {{n + 1}} - \frac{{n^2 }} {{n + 1}}} \right) $$ –  Susuk Nov 1 '11 at 2:58
    
@Susuk, that's what the handwaving at the end was for. I will edit the proof to make it more explicit how this part works. –  Henning Makholm Nov 1 '11 at 12:27
    
@Susuk, now de-handwaved. That was harder than I had expected ... –  Henning Makholm Nov 1 '11 at 13:53
    
@HenningMakholm Nice answer! Thanks! –  Susuk Nov 1 '11 at 16:38
add comment

Here :

http://epubl.ltu.se/1402-1617/2003/320/LTU-EX-03320-SE.pdf

Look paragraph : 3.5 Darboux function on page 32. The proof is given in a reference which I can't find on web.

share|improve this answer
    
Yes ! It happened, as I also saw that paper, but not the reference –  Susuk Oct 31 '11 at 21:16
    
Here is the reference: Darboux, Gaston, Addition au mémoire sur les fonctions discontinues. Annales scientifiques de l'École Normale Supérieure, Sér. 2, 8 (1879), p. 195-202. Direct link to the pdf file. –  Pierre-Yves Gaillard Nov 1 '11 at 14:45
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.