Extending a $C^2$-function from a $C^{1,1}$-curve to some neighbourhood Suppose I have a simple, compact $C^{1,1}$-curve $L$ in $\mathbb{R}^3$ and a $C^2$-function $f$ on it ($C^2$ meaning with two continuous arclength derivatives). Can it be extended to a $C^2$-function on some neighborhood of the curve?
I read about the Whitney Extension Theorem, but I'm not sure if it can be applied here. I think it suffices to construct an extension in a neighborhood of each point of the curve. Then I can assume that the tangent direction at that point coincides with one of the coordinate axes. But the tangent is possibly changing, so I would have to prescribe partial derivatives in such a way that they add up to the arclength derivative at each point of the curve, depending on the coordinates of the tangent vector at that point, and the same for the second derivative. But do these conditions together with the compatibility conditions have a solution? I don't know how to proceed.
Edit: The answer I received settles the question for plane curves, but I am interested in curves in $\mathbb{R}^3$ (now added above). The remark in the answer doesn't seem to work out, but I would be glad if it did.
 A: No, this cannot be done in general. Instead of a counterexample I will prove something "positive".  
Claim. Let $f$ be the arclength function of $C^{1,1} $ curve $L$; that is, $f(\gamma(t))\equiv t$ for $t\in [0,b]$ where $\gamma:[0,b]\to\mathbb R^2$ is the arclength parametrization of $L$. Suppose that $f$ has a $C^2$ extension to a neighborhood of $L$. Then $\gamma$ is $C^2$ on a dense open subset of $[0,b]$.  
Proof. Let $u$ be such an extension. Differentiate $u\circ \gamma \equiv t$ with respect to $t$ to obtain $$\langle (\nabla u)\circ \gamma ,\gamma' \rangle \equiv 1\tag1$$ which can be written as $$ |(\nabla u)\circ \gamma|\, \cos \theta  \equiv 1\tag2$$ where $\theta=\theta(t)$ is the angle between the two vectors in (1). Since $(\nabla u)\circ \gamma $ is $C^1$-smooth and nonvanishing, so is $1/|(\nabla u)\circ \gamma|$. Therefore, $\cos \theta $ is   $C^1$-smooth, which implies that  $\theta $ is $C^1$-smooth on the  set $U=\{t: \theta(t)\ne 0\}$. 
The vector $\gamma'(t)$ is obtained from $\dfrac{\nabla u(\gamma(t))}{|\nabla u(\gamma(t))|}$ by rotating it by $\theta(t)$ either clockwise or counterclockwise. Since $\gamma'$ is continuous, the choice of clockwise/counterclockwise is the same within each connected component of $U$. Thus, $\gamma'$ is $C^1$-smooth on $U$. 
On the set $N=\{t: \theta(t)= 0\}$  we   have $\gamma' =(\nabla u)\circ \gamma $, and therefore $\gamma'$ is $C^1$ in the interior of $N$. 
The union of $U$ and of the interior of $N$ is an open sense subset of $[0,b]$. $\Box$

Remarks: 


*

*it is not hard to construct $\gamma\in C^{1,1}$ which is not $C^2$ on any nontrivial interval, starting from a nowhere continuous element of $L^\infty$ and integrating it twice. 

*the claim is not true for higher-dimensional curves, but the answer stays the same. Indeed, we can differentiate (1) a.e. and  find that $\gamma''$ (which is a vector-valued  $L^\infty$ function) is a.e. orthogonal to the continuous function $((\nabla u)\circ \gamma)'$. It is not hard to find an $L^\infty$ function that does not admit any nonvanishing continuous orthogonal function. (E.g., let every coordinate attain the values $1$ and $-1$ on a positive measure subset of every nontrivial interval.) 

