The following is an Exercise 5.44 in Lee's Introduction to Smooth manifolds
Suppose $M$ is a smooth manifold with boundary, $f$ is a boundary defining function, and $p\in\partial M$. Show that a vector $v\in T_pM$ is inward-pointing in and only if $vf>0$, outward-pointing if and only if $vf<0$, and tangent to $\partial M$ if and only if $vf =0$.
Let $(x^i)$ be a smooth coordinate on some neighborhood of $p$ and $t$ be a smooth coordinate on some neighborhood of $f(p)$. Write $v = v^i{\partial\over\partial x^i}\bigg|_p$ then $vf = v^i{\partial f\over\partial x^i}\bigg|_p$ and $df_p(v) = df_p\left(v^i{\partial\over\partial x^i}\bigg|_p\right) = v^i{\partial f\over\partial x^i}(p){\partial\over\partial t}\bigg|_{f(p)}$. I'm trying to use the fact that $v\in T_pM$ is inward-pointing if and only if $v^n>0$, outward-pointing if and only if $v^n<0$ and tangent to $\partial M$ if and only if $v^n = 0$ but I'm stuck. Please help.
Best Answer
The following is a sketch of the proof and notations are abused frequently.
Let $p\in\partial M$ and choose a smooth boundary chart $(U,\varphi)$ with local coordinate $(x^i)$. Since $f$ is a smooth function, $f:U\to [0,\infty)$ is a smooth function. For $v\in T_pM$, write $v = v^i{\partial\over\partial x^i}\bigg|_p$. The action of $v$ on $f$ is \begin{align*} vf & = v^i{\partial\over\partial x^i}\bigg|_p f = v^i{\partial f\over\partial x^i}(p) = v^n{\partial f\over\partial x^n}(p). \end{align*} Here, the last equality follows from the fact that the function $f$ along the axis $x^i$ for $i =0,1,...,n-1$ is contained in $\partial M$ so partial derivative is identically zero by definition. Since $df_p\neq 0$, this implies ${\partial f\over\partial x^n}(p)> 0$ ($\because$ codomain is $[0,\infty)$). Hence the sign of $vf$ is completely determined by the sign of $v^n$ which is the desired result.