# Can a “not strong tangent” trace admit a C∞C^\infty curve parameterization?

Consider the following set: A={(x,|x|)∈R2;x∈R}A=\{(x,|x|)\in \Bbb{R}^2;x\in \Bbb{R}\}.

My simple question is: Is there any differentiable curve α:I→R2\alpha:I\to \Bbb{R}^2, I⊂RI\subset \Bbb{R} any interval, such that trace(α)=α(I)trace(\alpha)=\alpha(I) (the image of II by α\alpha) is equal to AA? Here we have to be careful about what “differentiable” means.

For each k∈Nk\in \Bbb{N}, setting
αk:R→R2t↦{(tk+1,tk+1),t≥0(−|t|k+1,|t|k+1),t≤0\begin{array}{llll}
\alpha_{k}:& \Bbb{R} & \to & \Bbb{R}^2\\
&t &\mapsto&\left\{\begin{array}{ll} (t^{k+1},t^{k+1})&,t\geq 0\\
(-|t|^{k+1},|t|^{k+1})&,t\leq 0\end{array}\right.
\end{array}

one can easily verify that αk\alpha_k is a curve of class CkC^k (has derivative of order kk and it is continuous), but not of class Ck+1C^{k+1}, such that αk(R)=A\alpha_k(\Bbb{R})=A. Furthermore, none of these curves are regular (since α′k(0)=(0,0)\alpha’_k(0)=(0,0)).

So, if “diferentiable” means C1C^1, the answer would be “Yes, we have such curve and not only C1C^1, but CkC^k, with kk as big as you wish.”

But, and if “differentiable” means C∞C^\infty? I could not find such curve, neither prove it does not exist…

The question also can be done for more general sets than AA: if I have a trace wich has not a strong tangent at some point, may it come from a C∞C^{\infty} curve? We know, in particular, that if such a curve exists, it cannot be regular at such point (since regular⟹\impliesstrong tangent line).

[OFF: Do anybody suggest an edit to centralize this image here? “begin{center} end{center}” did not help!]

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