Modified Takagi function I am interested in a modified version of the Takagi function. Particularly, I am interested in showing that it is not differentiable at non dyadic triadic points.
Previously I proved (for the Takagi function $g$ where in the notation I denote the $s$ on Wikipedia by $h$) that $g'(x)$ does not exist where 
$$ g(x) = \sum_{n=0}^\infty {1\over 2^n }h(2^n x) = \sum_{n \ge 0}h_n(x)$$
and  $h_n(x) = (1/2)^n h(2^n x)$ and  $h(x)=|x|$ on $[-1,1]$ and extend it to $\mathbb R$ by defining $h(2+x) = h(x)$. ($h_n$ is a sawtooth function)
It is not difficult: If $x$ is not dyadic it is possible to show that $g'(x)$ does not exist by showing
(i) that $|g_m' (x) - g_{m+1}'(x)|=1$ and
(ii) that $$  {g(y_m) - g(x) \over y_m - x} < g'_m (x) < {g(x_m) - g(x) \over x_m - x}$$
Here $g_m$ denotes the finite sum $\sum_{n=1}^m h_n(x)$.
Now consider the modifed Takagi function: $$g(x) = \sum_{n=0}^\infty {1\over 2^n }h(\color{red}{3}^n x)  $$
Again I want to show that $g'(x)$ does not exist if $x$ is not dyadic. It is easy to see that $|g_m' (x) - g_{m+1}'(x)|={3^{m+1} \over 2^{m+1}}$ therefore $g_m'(x)$ is not a Cauchy sequence and does not converge. 
I am having trouble showing 
$$  {g(y_m) - g(x) \over y_m - x} < g'_m (x) < {g(x_m) - g(x) \over x_m - x}$$
for this modified $g$ because $h_n$ is not $0$ if $n$ is large enough. Nonetheless $g_m$ is still piecewise linear and linear on $[x_m,y_m]$ so that 
$$ g'_m (x) = {g_m(y_m) - g_m (x) \over y_m - x} > {g_m(y_m) - g (x) \over y_m - x} $$
What to do about $g_m (y_m)$?
 A: Let $\langle \langle x\rangle \rangle = \min_{n \in \mathbb{Z}}|x-n|$.   Lagarias defines the Takagi function as 
$$ \tau(x) = \sum_{n=0}^\infty \frac{1}{2^n} \langle \langle 2^n x \rangle \rangle$$
However we could replace $2$ with $3$ or any real number $a \geq 2$, and get another self-similar function:
$$ \sum_{n=0}^\infty \frac{1}{a^n} \langle \langle a^n x \rangle \rangle$$
The introduction says that Knopp analyzed functions connection various scales
$$ F(x) =\sum_{n=0}^\infty \frac{1}{a^n} \phi ( b^n x )\rangle$$
and showed in 1918 $F(x)$ has no finite derivative  when $\frac{b}{a} \geq 1$  which is your case, $a=2,b=3$.

 Look at Theorem 8.1 to 8.3 - a result proven by Kohno in 1987 - that 

$$|\tau(x+h) - \tau(x)| \approx h \log_2 \tfrac{1}{h}$$
Let's try to estimate $ F(x+h) - F(x)$  in some kind of way. Let $h = \tfrac{1}{3^n}$. If $m > n$ s a finite sum 
$$ \langle \langle 3^m (x+h) \rangle \rangle - \langle \langle 3^m x \rangle \rangle = 0$$
If $m < n$   we get non-zero contributions to the difference:
$$ \tfrac{1}{2^m} \big|\langle \langle 3^m (x+h) \rangle \rangle - \langle \langle 3^m x \rangle \rangle \big| 
= \frac{1}{2^m}\frac{1}{3^{n-m}}$$
These terms can't possibly cancel each other out.  However, maybe you can use a Hölder exponent and have something like:
$$ |F(x+h)-F(x)| \approx h \log_3 \tfrac{1}{h}$$
which accounts for the oscillatory behavior of your function. 

Theorem (or
see here) If $\sum h'_n(x)$ is uniformly  convergent on $ [a,b]$ and $\sum h_n(x)$ converges at one point, then $\frac{d}{dx} \sum h_n(x) = \sum \frac{dh_n}{dx} (x)$.  
Here $ |h_n(3^n x) | \leq 1$ and your sum is uniformly bounded by $2$ and all the terms are positive:
$$ \left|\sum \frac{1}{2^n}h_n(3^nx)\right| \leq \sum_{n=0}^\infty  \frac{1}{2^n} =2 $$
So your series is uniformly convergent.  All the $h_n(x)$ are differentiable off the ternary points, $\mathbb{Z}[\tfrac{1}{3}]$.  
The derivatives are not uniformly convergent:
$$ \left| \frac{d}{dx} \frac{1}{2^n}h_n(3^nx)\right|
 = \left| \left(\frac{3}{2}\right)^n h'_n(3^n x)\right| = \left(\frac{3}{2}\right)^n \to \infty $$
Actually this works to your advantage... since we have shown this modified function has divergent terms in the derivative.

Wikipedia: Takagi function
arXiv:1112.4205 The Takagi Function and Its Properties

