Is it true that : 


$\forall f,g \in C([0,1],\mathbb R), \exists h \in C([0,1],[0,1])$ $f,g,h$ strictly increasing and $h([0,1])=[0,1]$ with $(f \circ h, g\circ h) \in C^{\infty}([0,1],\mathbb R)^2$?



Regularize continuous functions with bounded variation
The answer is NO:


Let $\,f:[0;1]\to[0;1]\,$ be the identity function. Let $\,g:[0;1]\to[0;1]\,$ be such that the set of $\,D\subseteq[0;1]\,$ of points $\,x\in[0;1]\,$ for which derivative of $\,g\,$ doesn't exist is dense.


Then, if $\,f\circ h\,$ is smooth then $\,h\,$ is smooth.


But then $\,g\circ h\,$ cannot be smooth. Otherwise, the derivative of $\,h\,$ at every $\,y\in h^{-1}(D)\,$ would be $0,\,$ and $\,h^{-1}(D)\,$ would be dense ($\,h\,$ is really assumed to be a homeomorphism), hence $\,h\,$ would be constant. A contradiction.   ***Great!***



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Assume that $\,g\circ h\,$ is smooth, and that the derivative of smooth $h$ at point $\,h^{-1}(x)$ is not $0\,$ for a certain point $x$. Then, $g$ is differentiable at $x$.


**Indeed**,  the derivative of $\,g\,$ at $\,x\,$ is the derivative at $x$ of $\,g\,=\,(g\circ h)\circ h^{-1}.\,$


Happy New Year!


