First, I need to give some definitions and background information:
Define $h(x)=|x|$ for $x\in [-1,1]$. Extend this function to $\mathbb R$ by defining $h(x+2) = h(x)$. Here is a graph of $h$:
Now if we define $g(x) = \sum_{n=0}^\infty {1 \over 2^n}h(2^nx)$ then $g$ is the Takagi function. One can prove that $g$ is continuous and nowhere differentiable. I did both. Continuity follows easily from the Weierstrass M-test and for non-differentiability one can first show it for dyadic points and then for non-dyadic points.
Now I am interested in two different modified versions of the Takagi function:
$$ g_1(x) = \sum_{n=0}^\infty {1\over 2^n}h(3^n x)$$ and $$ g_2(x) = \sum_{n=0}^\infty {1\over 3^n}h(2^n x)$$
These two are easily seen to be continuous.
Could someone please show me how to determine the differentiability of both $g_1$ and $g_2$? I really tried but failed.