Does it exist $F\in \mathcal C^1(\Omega ,\mathbb R^3)$ s.t. $Curl(F)=0$ and $F(x,y,0)=\left(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2},0\right)$

Question

Let $\Omega =\mathbb R^3\backslash \{(0,0,0)\}$. Does there exist $F\in \mathcal C^1(\Omega ,\mathbb R^3)$ s.t. $Curl(F)=0$ and $F(x,y,0)=G(x,y)$ whenever $(x,y)\neq (0,0)$, where $G(x,y)=\left(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2},0\right)$.

I proved by contradiction that it's not possible using Stokes theorem on $\{(x,y,z)\mid x^2+y^2+z^2=1, z\geq 0\}$. But I also found a candidate that would be $$F(x,y,z)=\left(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2},z\right),$$ that is $\mathcal C^1(\Omega ,\mathbb R^3)$ s.t. $Curl(F)=0$ and $F(x,y,0)=G(x,y)$. So, why it doesn't work ?

Answer 1

Your candidate function $F$ is not defined anywhere on the $z$-axis, and in particular it is not continuous on $\Omega := \Bbb R^3 - \{0\}$.