What can we say if the gradient at the boundary has constant norm?

Question

Let $(M^n,g)$ be a Riemannian manifold and consider $\Omega$ a smooth and bounded domain in $M$. Let $u : \overline{\Omega} \to \mathbb{R}$ be a smooth function that satisfies both $u = 0$ and $\Vert \nabla u \Vert = 1$ along $\partial \Omega$. Is it correct to say that, under these assumptions, $\nabla u$ is the unit normal to $\partial \Omega$ in $M$? In other words, is $\nabla u(x)$ orthogonal to $T_x (\partial \Omega)$ for every $x \in \partial \Omega$?

Answer 1

Fix $x \in \partial \Omega$. Let $f (t) = u(\gamma(t))$, where $\gamma : \mathbb{R} \to \partial \Omega$ is a curve in $\partial \Omega$ such that $\gamma(0) = x$. Since $f \equiv 0$, $$ 0 = f'(0) = \nabla u(\gamma(0)) \cdot \gamma ' (0) = \nabla u(x) \cdot \gamma ' (0) $$ We conclude that $\nabla u (x)$ is orthogonal to $\gamma ' (0)$.

Every tangent vector is given by $\gamma ' (0)$ for some curve $\gamma : \mathbb{R} \to \partial \Omega$ (this is essentially by definition, though there are many ways to define tangent vectors), so $\nabla u(x)$ must be orthogonal to $T_x (\partial \Omega)$. Since $||\nabla u|| \equiv 1$, $\nabla u$ must be a unit normal to $\partial \Omega$. It may not, however, be the unit normal to $\partial \Omega$. For example, a line in $\mathbb{R}^3$ has infinitely many unit normals at every point.