0
$\begingroup$

I want to prove the following:

(1) There is a parameter transformation from the continuous curve $c: \mathbb R \to \mathbb R^2; t \mapsto (t, \lvert t \rvert)$ into a smooth curve.
(2) Every continuous parameter transformation from $c$ into a smooth curve is not regular.

Attempt:
(1) I noticed that every bijective function $f$ that is smooth with $f^{(n)}(0) = 0$ works. Since polynomials become eventually constant with $f^{(k)} \neq 0$ for some $k\in \mathbb N$ we need some other function. From basic analysis we know that $f: \mathbb R \to \mathbb R$ defined by $$f(t) := \begin{cases}e^{-1/t} & t \neq 0 \\ 0 & t=0\end{cases}$$ is smooth with $f^{(n)} = 0$ for each $n\in \mathbb N$. But to make it bijective and satisfy $f(t) = -f(-t)$ we modify it:$$f(t) := \begin{cases}te^{-1/|t|} & t \neq 0 \\ 0 & t=0\end{cases}$$

It's easy to see that this function is indeed bijective and smooth. Finally, applying the transformation $f$ we get our smooth curve: $$d:\mathbb R \to \mathbb R^2; t \mapsto(f(t), \lvert f(t) \rvert)$$ Is this correct so far?
(2) Let $d = c \circ f$ be a continuous parameter transformation into a smooth curve. We need to show $\exists t\in \mathbb R$ such that $d'(t) = 0$. Claim: $d'(0) = 0$. But how do prove this? Using differential quotients I don't quiet get to what I want to show...

$\endgroup$
1
$\begingroup$

We have $d’=\bigl(f’,\frac{f}{|f|}\cdot f’\bigr)=f’\cdot(1,f/|f|)$. Now if $f’(0)$ wouldn’t be zero, $d$ wouldn’t be continuous in zero.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.