Can a "not strong tangent" trace admit a $C^\infty$ curve parameterization? Consider the following set: $A=\{(x,|x|)\in \Bbb{R}^2;x\in \Bbb{R}\}$.

My simple question is: Is there any differentiable curve $\alpha:I\to \Bbb{R}^2$, $I\subset \Bbb{R}$ any interval, such that $trace(\alpha)=\alpha(I)$ (the image of $I$ by $\alpha$) is equal to $A$? Here we have to be careful about what "differentiable" means.
For each $k\in \Bbb{N}$, setting
$$\begin{array}{llll}
\alpha_{k}:&  \Bbb{R}  &  \to  & \Bbb{R}^2\\
             &t          &\mapsto&\left\{\begin{array}{ll} (t^{k+1},t^{k+1})&,t\geq 0\\ 
(-|t|^{k+1},|t|^{k+1})&,t\leq 0\end{array}\right.
\end{array}$$
one can easily verify that $\alpha_k$ is a curve of class $C^k$ (has derivative of order $k$ and it is continuous), but not of class $C^{k+1}$, such that $\alpha_k(\Bbb{R})=A$. Furthermore, none of these curves are regular (since $\alpha'_k(0)=(0,0)$).
So, if "diferentiable" means $C^1$, the answer would be "Yes, we have such curve and not only $C^1$, but $C^k$, with $k$ as big as you wish."
But, and if "differentiable" means $C^\infty$? I could not find such curve, neither prove it does not exist...
The question also can be done for more general sets than $A$: if I have a trace wich has not a strong tangent at some point, may it come from a $C^{\infty}$ curve? We know, in particular, that if such a curve exists, it cannot be regular at such point (since regular$\implies$strong tangent line).
[OFF: Do anybody suggest an edit to centralize this image here? "begin{center} end{center}" did not help!]
 A: The answer to your question is yes.
A slight generalization of your idea is the following. Let $f : [0, +\infty) \to \mathbb{R}$ be any function with values in $[0, +\infty)$, and consider the curve
$$\begin{array}{llll}
\alpha \colon &  \Bbb{R}  &  \to  & \Bbb{R}^2\\
             &t          &\mapsto&\left\{\begin{array}{ll} (f(t),f(t))& t\geq 0\\ 
(-f(-t),f(-t))&t < 0~.\end{array}\right.
\end{array}$$
Then it is quite straightforward 1 to prove that $\alpha$ is of class $\mathcal{C}^k$ if and only if $f$ is of class $\mathcal{C}^k$ and $f(0) = f'(0) = \dots = f^{(k)}(0) = 0$. (Remark: For functions $f : [0,+\infty) \to \mathbb{R}$, we define differentiability at $0$ as "right-differentiability"). 
In particular, if you choose a $\mathcal{C}^\infty$ function $f \colon [
0,+\infty) \to \mathbb{R}$ with $f^{(k)}(0) = 0$ for all nonnegative integers $k$, then this construction of $\alpha$ provides an answer to your question. One classical example of such a function $f$ is $$\begin{array}{llll}
f \colon &  [0, +\infty)  &  \to  & \Bbb{R}\\
             &x          &\mapsto&\left\{\begin{array}{ll} e^{-1/x}& x> 0\\ 
0& x = 0~.\end{array}\right.
\end{array}$$

1 It is a direct application of the following theorem (in French we call this "théorème de prolongement de la dérivée", I am not sure whether it has a name in English):
Theorem. Let $f : I \to \mathbb{R}$ be a continuous function such that $f$ is differentiable on $I \setminus \{x_0\}$ and $f'$ admits a limit as $x \to x_0$. Then $f$ is differentiable at $x_0$ and $f'(x_0) = \lim_{x\to x_0} f'(x)$.
This theorem is itself a straighforward consequence of the mean value theorem.
