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
\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.

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!]



1 Answer


The answer to your question is yes.

A slight generalization of your idea is the following. Let f:[0,+∞)→Rf : [0, +\infty) \to \mathbb{R} be any function with values in [0,+∞)[0, +\infty), and consider the curve
α:R→R2t↦{(f(t),f(t))t≥0(−f(−t),f(−t))t<0 .\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 α\alpha is of class Ck\mathcal{C}^k if and only if ff is of class Ck\mathcal{C}^k and f(0)=f′(0)=⋯=f(k)(0)=0f(0) = f'(0) = \dots = f^{(k)}(0) = 0. (Remark: For functions f:[0,+∞)→Rf : [0,+\infty) \to \mathbb{R}, we define differentiability at 00 as "right-differentiability"). In particular, if you choose a C∞\mathcal{C}^\infty function f:[0,+∞)→Rf \colon [ 0,+\infty) \to \mathbb{R} with f(k)(0)=0f^{(k)}(0) = 0 for all nonnegative integers kk, then this construction of α\alpha provides an answer to your question. One classical example of such a function ff is f:[0,+∞)→Rx↦{e−1/xx>00x=0 .\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.

1 It is a direct application of the following theorem (in French we call this “thأ©orأ¨me de prolongement de la dأ©rivأ©e”, I am not sure whether it has a name in English):

Theorem. Let f:I→Rf : I \to \mathbb{R} be a continuous function such that ff is differentiable on I∖{x0}I \setminus \{x_0\} and f′f’ admits a limit as x→x0x \to x_0. Then ff is differentiable at x0x_0 and f′(x0)=limx→x0f′(x)f'(x_0) = \lim_{x\to x_0} f'(x).

This theorem is itself a straighforward consequence of the mean value theorem.