Is the graph of $y= |x|$ the graph of a smooth function?

Question

Is there a smooth function $f:\mathbb{R} \to \mathbb{R}^2$ such that the image is the graph of $f(x) =|x|$ ?

Answer 1

Yes: take $f(x) = (|x|x,x^2) = (\mathop{\rm sign}(x)x^2,x^2)$. The key fact is that the function $f_1(x)=|x|x$ is differentiable everywhere.

More generally, let $g\colon [0,\infty)\to[0,\infty)$ be any surjective function with $g(0)=0$; then the image of $f(x) = (\mathop{\rm sign}(x)g(x^2),g(x^2))$ is the graph of $y=|x|$. In particular, if $g$ is chosen to be very smooth everywhere and have derivatives of all order equal to $0$ at $x=0$, then $f$ will be very smooth as well.

Answer 2

If $f(x) = (xe^{-1/x^2},|x|e^{-1/x^2}),$ then $f$ is $C^\infty$ and a homeomorphism of $\mathbb R$ onto the graph of $y=|x|.$