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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.

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    $\begingroup$ A comment on your idea for a proof. The graph of $f$ is not purely unrectifiable. More precisely, a substantial part of the graph is contained in the graph of a $1$-Lipschitz function defined on the quadrant bisector. I would expect that therefore (using Lebesgue differentiation) there exists a set $E\subset\mathbb{R}$ of positive measure and $k>0$ such that for all $x_0\in E$ the set $\mathbb{R}\setminus I^k_{f,x_0}$ has positive lower density at $x_0$. So to me it seems that your idea might not work. $\endgroup$ – Manfred Sauter May 27 '19 at 16:02
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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 $g=g_0+h$.

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The composition can be differentiable.

Example: Let $f(x):=1$ for rational $x$ and $f(x):=0$ for irrational $x$. Then $f$ is nowhere differentiable. Let $g:=f$. Then $f\circ g(x)=1$ for all $x$, thus $f\circ g$ is differentiable everywhere.

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    $\begingroup$ Thank you, but sorry I forgot to mention the continuity assumption. $\endgroup$ – yaoliding May 26 '19 at 8:56

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