Suppose we have a bounded smooth domain $\Omega$ in $\mathbb{R}^n$, so there exists an outer unit normal vector field $\eta$ everywhere on the boundary. Can we extend it to the interior satisfying some conditions? To be more specific, for example, can we construct a vector field $ X$ on $\Omega$ satisfying


(1) $X|\_{\partial \Omega}=\eta;$


(2) $X$ is compactly supported in $\delta$-neighborhood of boundary $\Omega\_\delta=\{x\in\Omega: d(x,\partial\Omega)\leq \delta\}$ for very small $\delta$.


(3) For example,the vector field $X$ is bounded in C^1: $\|X\|\_{C^1(\bar{\Omega})}\leq C/\delta$ where C is a universal constant. 


As it is pointed out, I should let the boundary be at least $C^2$ such that (3) makes sense. So for simplicity, let's just consider domain with smooth boundary. My thought was to think $\eta=(\eta\_1,\cdots,\eta\_n)$, so $\eta\_i$ are smooth functions on the boundary, we can use the partition of unity to extend these functions to functions $X\_i$ in some neighborhood of boundary. Then certainly we will get bounds in $C^k$ for $X\_i$. I am just wondering if we can extend it to $\delta$-neighborhood and make those bounds (such as $C^1$ norm) depending on $\delta$, for example, bounds like $C\delta; C/\delta; C\delta^2;C\log\delta...$ where $C$ is a universal constant. How do we make this work exactly? 


Edit: Maybe those bounds such as $C\delta,C\delta^2,..$ don't look reasonable since we expect that the derivative should become greater when $\delta$ is smaller. So it only makes sense to expect bounds like $C/\delta;C/\log\delta..$



Extension of outer unit normal vector to interior
If $\partial\Omega\in C^2$, then the answer is yes. We need $C^2$ as this condition implies that the vector normal to the boundary is $C^1$ (since the normal vector id defined through derivatives).


According to the collar neighborhood theorem, there is $d>0$ and a diffeomorphism of class $C^1$:
$$
\Phi:\partial\Omega\times(-1,1)\to (\partial\Omega)\_d=\{x:\operatorname{dist}(x,\partial\Omega)<d\}
$$
Let 
$$
\pi\_1:\partial\Omega\times(-1,1)\to\partial\Omega,\ \ \pi\_1(x,t)=x, 
$$
and
$$\pi\_2:\partial\Omega\times(-1,1)\to\partial\Omega,\ \ \pi\_2(x,t)=t, 
$$
be projections on components. 
Then $\tilde{X}=\eta\circ\pi\_1\circ\Phi^{-1}$
is a $C^1$ extension of the vector field $\eta$ to a neighborhood of the boundary. 
Let
$$
\phi\in C\_0^\infty(-1,1),
\quad
\text{with}
\quad
\text{$\phi(t)=1$ for $t\in [-1/2,2]$}
\quad
\text{and $\phi\_\epsilon(t)=\phi(t/\epsilon)$.}
$$
Then $|(\phi\_\epsilon)'(t)|\leq C/\epsilon$ and hence
$$
X=\tilde{X}\cdot (\phi\_\epsilon\circ\pi\_2\circ\Phi^{-1})
$$
satisfies the required condition (with $\epsilon$ in place of $\delta$).


