> [!definition] > > Let $U \subset \real^n$ be [[Open Set|open]] and bounded. The [[Boundary|boundary]] $\partial U$ is of class $C^k$ if for each $x \in \partial U$, there exists $r > 0$ and a $C^k$ map $\gamma: \real^{n - 1} \to \real$ such that (reorienting if necessary), > $ > U \cap B(x, r) = \bracs{y \in B(x, r): y_n > \gamma(y_1, \cdots, y_{n - 1})} > $ > > This is equivalent to saying that $\ol U$ is a $C^1$-[[Manifold|manifold]]. > [!theorem] > > Let $U \subset \real^n$ with $\partial U \in C^1$. Let $x_0 \in \partial U$, $r > 0$ and $\gamma: \real^{n - 1} \to \real$ such that > $ > U \cap B(x_0, r) = \bracs{y \in B(x, r): y_n > \gamma(y_1, \cdots, y_{n - 1})} > $ > then there exists $\lambda, t' > 0$ such that for any $x \in B(x_0, r/2)$ and $t \in (0, t')$, > $ > x_t = x + \lambda t e_n \quad B(x_t, t) \subset U \cap B(x_0, r) = V > $ > *Proof*. There are two constraints, $B(x_t, t) \subset B(x_0, r)$ and $y_n > \gamma(y_1, \cdots, y_{n - 1})$ for all $y \in B(x_t, t)$. For any $y \in B(x_t, t)$, > $ > \begin{align*} > \norm{y - x_0} &\le \norm{y - x_t} + \norm{x_t - x} + \norm{x - x_0} \\ > &\le t + \lambda t + r/2 > \end{align*} > $ > For the distance constraint to work, we must have $(\lambda + 1)t < r/2$. For the second constraint, first let $y \in B(0, t)$, then > $ > (x_t + y)_n = x_n + t\lambda + y_n > x_n + t(\lambda - 1) > $ > On the other hand, since $\gamma \in C^1$, $\gamma|_{\pi_{1, \cdots, n - 1}(\ol{V})}$ is Lipschitz continuous. Let $L \ge 0$ be its Lipschitz constant, then since $x_t$ has the same first $n - 1$ components as $x$, > $ > \begin{align*} > \abs{\gamma(x_t + y)_{1, \cdots, n - 1} - \gamma(x_{1, \cdots, n - 1})} &= \abs{\gamma(x + y)_{1, \cdots, n - 1} - \gamma(x_{1, \cdots, n - 1})} \\ > &\le L \cdot \norm{y} < tL \\ > \gamma((x_t + y)_{1, \cdots, n - 1}) &< tL + x_n > \end{align*} > $ > This yields the constraints > $ > \lambda - 1 > L \quad t(\lambda + 1) < r/2 > $ > Let $\lambda > L + 1$ and $t' < r/\braks{2(\lambda + 1)}$ yields the desired result.