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Define $\phi: \ \mathbb{R} \rightarrow \mathbb{R}$ by $$ \phi(x) = \begin{cases} x\ :\ 0\le x\le \frac{1}{2}\\ 1-x :\ \frac{1}{2} \le x \le 1 \end{cases}$$

And then extend periodically to all of $\mathbb{R}$ by $\phi(x) = \phi(x+1)$ so that it just repeats itself. Now define $$S_m(x) = \sum\limits_{i=0}^m \left(\frac{3}{4}\right)^i \phi\left(4^ix\right)$$ Where the limit function can be written as, for $k,n\in\mathbb{N}$ $$S\left(k4^{-n}\right) = \sum\limits_{i=0}^{n-1}\phi\left(4^{i-n}k\right)$$

Show that S (the limit function) is not differentiable at any point in $\mathbb{R}$. You are allowed to define functions somewhat implicitly, such as "let $a_n$ be the largest multiple of $4^{-n}$ which is less than or equal to $\pi$.

I'm not sure where to start. My guess is that we can choose two sequences $a_n$ and $b_n$ where $a_n$ is monotone increasing and $b_n$ is monotone decreasing and they both converge to an arbitrary point $c$ in $\mathbb{R}$ and show that $\lim \limits_{n\rightarrow\infty} \dfrac{f(b_n) - f(a_n)}{b-a} \neq f'(c)$. So I want to find two sequences that converge from above and below, but then show that the expression above doesn't hold so that it can't be differentiable. But I'm not sure how to define those sequences. Or if that would even be sufficient.

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  • $\begingroup$ I edited your post to add brackets for cases, you can look to see how I did it (for future reference). $\endgroup$ – Najib Idrissi Apr 9 '14 at 14:11
  • $\begingroup$ . what is $f$? . $\endgroup$ – user127.0.0.1 Apr 9 '14 at 14:15
  • $\begingroup$ Your limit function seems wrong in two ways: firstly, I think you have left out the factor $(\frac34)^i$; and secondly, the limit is only valid for arguments of the form $k4^{-n}$, which doesn't even exhaust the rationals. $\endgroup$ – TonyK Apr 9 '14 at 14:55

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