$|x|$ is not differentiable at zero. However, if we compose it with another function that "eases up" very gently to zero, for example, $x=t^3$, we can obtain a function $|t^3|$ which is differentiable everywhere. I imagine this as "stretching out" $|x|$ to make it smooth. It seems like any continuous function could be turned in to a differentiable function by composing it with another function, that momentarily comes down to a derivative of 0 everywhere the continuous function is not differentiable.
As I have posed it, this question has a trivial solution: $f\circ 0$ is always differentiable (even if $f$ is not continuous!). So, an additional requirement should be imposed, to fit the intuition of "stretching $f$." The function being composed should be monotonically increasing and should cover the entire domain of $f$.
Given a continuous $f(x)$, can such a $g(t)$ always be found so that $f(g(t))$ is differentiable everywhere?
Best Answer
There is a broad class of functions for which this reparametrization works at a point, say the origin. A function $f$ is in $C^\alpha$, $0<\alpha$, when $$|f(x)-f(y)|\le k|x-y|^\alpha$$
Let $g_n(x):=x^n$ for $n$ odd, and $g_n(x):=\begin{cases}x^n&x\ge0\\-x^n&x<0\end{cases}$ for $n$ even.
Given $f\in C^\alpha$, let $n>1/\alpha$ , then $$\left|\frac{f(g_n(x))-f(g_n(0))}{x}\right|\le k\frac{|g_n(x)-g_n(0)|^\alpha}{|x|}= k|x|^{n\alpha-1}\to0,\quad x\to0$$ so $f(g_n(x))=f(0)+o(x)$.
This means functions like $|x|^{\beta}$, $\beta\ge0$, and the Weierstrass functions, can be smoothed out (at one point). But functions like $1/\ln|x|$, though continuous, cannot be smoothed with this reparametrization (though it can with $g(x)=sgn(x)e^{-1/|x|}$).