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 then $f\circ g$ 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 nondifferentiability 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 that 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.

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.

Let $f,g:\mathbb R\to\mathbb R$ be two continuous but nowhere differentiable function, that means for almost point $x_0\in\mathbb R$, the limsup $\varlimsup\limits_{x\to x_0}\frac{|f(x)-f(x_0)|}{|x-x_0|}=+\infty$. Must it be true that $f\circ g$ is still nowhere differentiable? Though it's seemingly possible that the high frequency of $g$ lays in the low frequency of $f$ and thus make the cancellation of the oscillation.

I am trying prove it through the following: Fix $x_0$, consider $I_{f,x_0}^k=\{x\in\mathbb R:|f(x)-f(x_0)|\ge k|x-x_0|\}$ be the set of point outside cone with slope $k$. Is that 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 ''most'' of the points near $x_0$ are far away from $x_0$ and so there is no much frequency cancellation.

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 then $f\circ g$ 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 nondifferentiability 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 that 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.

Let $f,g:\mathbb R\to\mathbb R$ be two nowhere differentiable function, that means for almost point $x_0\in\mathbb R$, the limsup $\varlimsup\limits_{x\to x_0}\frac{|f(x)-f(x_0)|}{|x-x_0|}=+\infty$. Must it be true that $f\circ g$ is still nowhere differentiable? Though it's seemingly possible that the high frequency of $g$ lays in the low frequency of $f$ and thus make the cancellation of the oscillation.

I am trying prove it through the following: Fix $x_0$, consider $I_{f,x_0}^k=\{x\in\mathbb R:|f(x)-f(x_0)|\ge k|x-x_0|\}$ be the set of point outside cone with slope $k$. Is that 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 ''most'' of the points near $x_0$ are far away from $x_0$ and so there is no much frequency cancellation.

Let $f,g:\mathbb R\to\mathbb R$ be two continuous but nowhere differentiable function, that means for almost point $x_0\in\mathbb R$, the limsup $\varlimsup\limits_{x\to x_0}\frac{|f(x)-f(x_0)|}{|x-x_0|}=+\infty$. Must it be true that $f\circ g$ is still nowhere differentiable? Though it's seemingly possible that the high frequency of $g$ lays in the low frequency of $f$ and thus make the cancellation of the oscillation.

I am trying prove it through the following: Fix $x_0$, consider $I_{f,x_0}^k=\{x\in\mathbb R:|f(x)-f(x_0)|\ge k|x-x_0|\}$ be the set of point outside cone with slope $k$. Is that 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 ''most'' of the points near $x_0$ are far away from $x_0$ and so there is no much frequency cancellation.