in my research I saw this argument used in an article and I don't know if it's true (I believe yes).
Let $\Omega \subset \mathbb{R}^3$ be an open, bounded and simply connected domain and let $\mathbf{v}: \Omega\rightarrow \mathbb{R}^3$ be a smooth vector field. Then exist a smooth extension $\mathbf{V}: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ of $\mathbf{v}$ such that it vanishes at infinity and $\mathbf{V}(\mathbf{x}) = \mathbf{v}(\mathbf{x}), \; \forall \mathbf{x} \in \Omega$ .
Where "vanish at infinity" mean: $\| f(\mathbf{x})\| \to 0$ as $\| \mathbf{x} \| \to \infty$.
Can someone help me?
It is wrong. Let $\Omega = (0,1)^3$ and $\mathbf{v}(x_1,x_2,x_3) = (x_1,x_2, \frac{1}{x_3})$. Then $\lVert \mathbf{v}(\frac{1}{2},\frac{1}{2},x_3) \rVert \to \infty$ as $x_3 \to 0$. Hence there cannot exist a continuous extension to $\mathbb{R}^3$.