Sum of sawtooth function not differentiable at dyadic rational points This question is related to this question.
Define $h(x)=|x|$ on $[-1,1]$ and extend it to $\mathbb R$ by defining $h(2+x) = h(x)$. This is a sawtooth function that is $0$ at even and $1$ at odd integers. 
Furthermore define $h_n(x) = (1/2)^n h(2^n x)$ and
$$ g(x) = \sum_{n=0}^\infty {1\over 2^n }h(2^n x) = \sum_{n \ge 0}h_n(x)$$
Now I
(a) showed that $g'(1)$ does not exist and that $g'({1\over 2})$ does not exist. 
(b) Showed that $g'(x)$ does not exist for any rational number of the form $x={p \over 2^k}$ where $p\in \mathbb Z$ and $k \in \mathbb N $.
Please can you check my work:
(a) Define $x_m = 1 + {1\over 2^m}$. Then 
$$\begin{aligned} 
{g(x_m) - g(1) \over x_m - 1} &= 2^m \sum_{n \ge 0} {1\over 2^n} h(2^n + 2^{n-m}) - 2^mg(1) \\
&=  2^mh(1 + 2^{-m}) + \sum_{n \ge 1} 2^{m-n} h(2^n + 2^{n-m}) - 2^mg(1) \\
&=  2^mh(2^{-m}) + \sum_{n \ge 1} 2^{m-n} h(2^n + 2^{n-m}) - 2^mg(1) \\
&= \sum_{n \ge 0} 2^{m-n} h(2^{n-m}) - 2^mg(1) \\
&= \sum_{n=0}^m 2^{m-n} h(2^{n-m}) - 2^mg(1) \\
&= m + 1 - 2^m g(1) \to - \infty
\end{aligned}$$
hence $g'(1)$ does not exist. Next define $x_m = {1\over 2} + {1 \over 2^m}$. Then 
$$\begin{aligned} 
{g(x_m) - g({1\over 2}) \over x_m - {1\over 2}} = {g(x_m) - g({1\over 2}) \over {1\over 2^m} - {1\over 2}}  &= {2^{m+1}\over 2-2^m} \left ( \sum_{n \ge 0} {1\over 2^n} h(2^{n-1} + 2^{n-m}) - 2^mg({1\over 2}) \right )\\
&=  {2^{m+1}\over 2-2^m} \left ( \sum_{n = 0}^m {1\over 2^n} h(2^{n-m}) - 2^mg({1\over 2}) \right )\\
&=  {2 \over 2-2^m} \left ( \sum_{n = 0}^m 2^{m-n} h(2^{n-m}) - 2^m 2^mg({1\over 2}) \right )\\
&=  {2 \over 2-2^m} \left ( m+1 - 2^{2m} g({1\over 2}) \right )\\
&\to  \infty
\end{aligned}$$
(b) Let $x = {p \over 2^k}$ and $x_m = x + {1 \over 2^m}$. Then 
$$\begin{aligned} 
{g(x_m) - g(x) \over x_m -x} = {g(x_m) - g(x) \over {1\over 2^m}} &= 2^m \sum_{n \ge 0} {1\over 2^n} h(x2^n + 2^{n-m}) - 2^mg(x) \\
&= 2^m \sum_{n \ge 0} {1\over 2^n} h(2^{n-k} + 2^{n-m}) - 2^mg(x)  \\
&= 2^m \sum_{n = 0}^{m+k} {1\over 2^n} h(2^{n-k} + 2^{n-m}) - 2^mg(x)  \\
&= \left ( \sum_{n = 0}^{m+k}  2^{m-k} + 1 \right ) - 2^mg(x)  \\
&= (m+ k + 1) 2^{m-k} + (m+ k + 1)  - 2^mg(x)  \\
&\xrightarrow{m \to \infty} \infty
\end{aligned}$$
hence the derivative of $g$ does not exist at $x$. 
 A: In the first part of (a) you replaced $h(1+2^{-m})$ by $h(2^{-m})$ going from second to third line. This is incorrect; the two values are not equal. I think there are similar mistakes in other parts too. In all cases, the divergence of divided differences to infinity should be of order $m$, not exponential in $m$. 
I suggest writing a proof in a modular way, using some sentences along the formulas. A proof that is only a long chain of formulas is hard to read and to debug. I would write something like:
Since $h$ is $2$-periodic, the function $h_n(x) = 2^{-n}h(2^nx)$ is $2^{1-n}$-periodic. Therefore, the tail of the series $\sum_{n=m+1}^\infty h_n(x)$ is $2^{-m}$-periodic. For any $x\in\mathbb R$, the divided difference of $h(x+2^{-m})$ and $h(x)$ simplifies to
$$\frac{g(x+2^{-m})-g(x)}{2^{-m}} =  \sum_{n=0}^m 2^{m-n}\bigg(h (2^nx+2^{n-m})-h (2^nx)\bigg) \tag{1}$$
Suppose $x=p/2^k$ here, with $p$, $k$ as in the problem statement. Then for $n> k$ the number $2^nx$ is an even integer, hence $h(2^nx)=0$. Also, for  $n=k+1,\dots,  m$ we have $$h (2^nx+2^{n-m}) = h(2^{n-m}) = 2^{n-m}$$ Using the above, rewrite the right side of (1) as 
$$ \sum_{n=0}^k 2^{m-n}\bigg(h (2^nx+2^{n-m})-h (2^nx)\bigg)  +\sum_{n=k+1}^m 2^{m-n} 2^{n-m}  \tag{2}$$
The first sum is bounded by $2(k+1)\max|g|$, the second is equal to $(m-k)$ and thus  tends to infinity as $m\to\infty$.
