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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!

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    See also [this question](http://math.stackexchange.com/questions/67120/) -- not an exact duplicate but with much the same properties.2011-10-31

1 Answers 1

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(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.

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    @HenningMakholm Nice answer! Thanks!2011-11-01