The book Understanding Analysis by Stephen Abbott asserts that
$$
g(x)=\sum_{n=0}^{\infty}\frac{1}{2^n}h(2^nx),
$$
where $h(x)=\left|x\right|$, $h:\left[-1,1\right]\to\mathbb{R}$, and $h(x+2)=h(x)$, is continuous on all of $\mathbb{R}$ but fails to be differentiable at any point.
However, if I am not mistaken, can the $2^n$'s be cancelled out in $g$? I tried plotting this and could not obtain a nowhere-differentiable function.
Best Answer
The plot looks like this. It's hard to tell by a casual look that this is nowhere differentiable.
For a better view of what's going on, here's an animation that zooms into the graph. You can see that the picture looks similar at different length scales. A differentiable curve, on the other hand, would look more and more like a straight line as you zoomed in.
EDIT: I plotted this in Maple. I don't have the code any more, but it might have been something like this.