Let $f,g:\mathbb R\to\mathbb R$ be two continuous but nowhere differentiable functions. By the Denjoy–Young–Saks theorem for almost every point $x\_0\in\mathbb R$ one has 
$$
\limsup\limits\_{x\to x\_0}\frac{|f(x)-f(x\_0)|}{|x-x\_0|}=+\infty.
$$


Is $f\circ g$ still nowhere differentiable? It seems possible that the high frequency parts of $g$ lie in the low frequency parts of $f$, which might lead to a cancellation of the oscillation. 


I am trying to prove non-differentiability through the following: Fix $x\_0$ and consider 
$$
I\_{f,x\_0}^k=\{x\in\mathbb R:|f(x)-f(x\_0)|\ge k|x-x\_0|\},
$$
the set of points outside a double-sided cone with slope $k$. Is it true that
$$
\lim\limits\_{\delta\to0}|I\_{f,x\_0}^k\cap[x\_0-\delta,x\_0+\delta]|/(2\delta)=1?
$$
If yes, then we know that “most” of the points near $x\_0$ have values relatively far away from $f(x\_0)$, and so there will not be much frequency cancellation.



Is the composition of two nowhere differentiable functions still nowhere differentiable?
The composition may have points of differentiability. 


Let $f\_0(x)=x$ for $x\geq 0$ and $f\_0(x)=2x$ for $x<0$. Let $g\_0(x)=2x$ for $x\geq 0$ and $g\_0(x)=x$ for $x<0$. Then none of them is differentiable at $0$, but $f\_0\circ g\_0=2x$ is differentiable everywhere. 


Let $h$ be a function that is nowhere differenitable, except at $0$ where it is very rapidly decaying. E. g. I can take $h(x)=e^{-\frac{1}{x^2}}B\_{|x|}$, where $B\_t$ is a sample of Brownian motion. 


Now just take $f=f\_0+h$ and $g=g\_0+h$.


