3
$\begingroup$

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 $(f \circ h, g\circ h) \in C^{\infty}([0,1],\mathbb R)^2$?

Improve this question
$\endgroup$
1
  • $\begingroup$ It was kind of you to add the homeomorphism assumption about $h$. $\endgroup$
    – Wlod AA
    Commented Jan 5, 2020 at 13:26

1 Answer 1

Reset to default
5
$\begingroup$

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!

 

  For the sake of mine and everybody's peace of mind let me add the ultimate detail:

 

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!

Improve this answer
$\endgroup$
5
  • 1
    $\begingroup$ why the derivative of $h$ at every $y\in h^{-1}(D)$ would be $0$ ? $\endgroup$
    – Dattier
    Commented Jan 5, 2020 at 17:37
  • $\begingroup$ For instance, if $h$ were constant then $\,g\circ h\,$ would be smooth. $\endgroup$
    – Wlod AA
    Commented Jan 5, 2020 at 20:45
  • $\begingroup$ it does not seem to me that you are answering my question. $\endgroup$
    – Dattier
    Commented Jan 5, 2020 at 21:51
  • $\begingroup$ @Dattier, I've misread your question (have imagined a question which could have been "natural" to ask). You already know at the stage of the proof that $h$ is smooth. Thus at the critical points that derivate (would exist and) being different from zero would not compose with $g$ to a smooth function. $\endgroup$
    – Wlod AA
    Commented Jan 5, 2020 at 22:03
  • $\begingroup$ I may add a small but truly detailed note at the end of the Answer if you still would like me to do so. $\endgroup$
    – Wlod AA
    Commented Jan 5, 2020 at 22:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .