> [!theorem] > > Let $p \in [1, \infty)$, $B = B(0, r) \subset \real^n$, $B^+ = B \cap \mathbb{H}^n$, and $B^- = B \cap (-\mathbb{H}^n)$. Let $f \in C^1_c(B^+)$ be a [[Compactly Supported|compactly supported]] [$C^1$](Space%20of%20Continuously%20Differentiable%20Functions) function, and $x' = (x_1, \cdots, x_{n - 1}) = \bracs{x_n = 0}$, then there exists $C \ge 0$, independent of $f$, such that > $ > \int_{\partial U \cap B} \abs{f}^p dx' \le C\norm{f}_{W^{1, p}(B^+)}^p > $ > *Proof*. Since $f \in UC^1(B^+)$, $\abs{f}^p$ is [[Absolutely Continuous Function|absolutely continuous]] with derivative being equal to the weak derivative [[Almost Everywhere|a.e.]]. As $\supp{f} \subset B^+$, we can use FTC to express > $ > 0 - \abs{f}^p(x', 0) = \int_{B^+ \cap \bracs{y' = x'}}D^{e_n}\abs{f}^p(y)dy > $ > this difference as an integral of the [[Distributional Derivative|weak derivative]]. Thus the integrals can be combined to obtain > $ > \int_{\partial U \cap B}\abs{f}^pdx' = -\int_{B^+}D^{x_n}\abs{f}^p(y)dy > $ > where the weak derivative is $p\sgn(f)\abs{f}^{p - 1} \cdot D^{x_n}f$[^1]. From Young's inequality, > $ > \abs{f}^{p - 1} \cdot \abs{D^{x_n}f} \le \frac{1}{p}\abs{D^{x_n}f}^p + \frac{1}{q}\abs{f}^{(p - 1)q} = \frac{1}{p}\abs{D^{x_n}f}^p + \frac{1}{q}\abs{f}^{p} > $ > because $p + q = pq$ and $p = pq - q$. From here, > $ > \int_{\partial U \cap B}\abs{f}^p \le p\int_{B^+}D^{x_n}\abs{f}^p(y)dy \le p\int \abs{f}^p + \abs{D^{x_n}f}^p \le p\norm{f}_{W^{1, p}(B^+)}^p > $ > [!theorem] > > Let $p \in [1, \infty)$, $B = B(0, 2r) \subset \real^n$, $B^+ = B \cap \mathbb{H}^n$, $B^- = B \cap (-\mathbb{H}^n)$, and $\wh B = B(0, r)$. Let $f \in UC^1(B^+)$ be a [[Space of Uniformly Continuously Differentiable Functions|UC]] function, and $x' = (x_1, \cdots, x_{n - 1}) = \bracs{x_n = 0}$, then there exists $C \ge 0$, independent of $f$, such that > $ > \int_{\partial U \cap \wh B} \abs{f}^p dx' \le C\norm{f}_{W^{1, p}(B^+)}^p > $ > *Proof*. Let $\zeta \in C_c^\infty(B)$ be a [[Smooth Bump Function|smooth bump function]] such that $\zeta|_{\wh B} = 1$, then > $ > \int_{\partial U \cap \wh B}\abs{f}^p dx' \le \int_{\partial U \cap \wh B} \abs{\zeta f}^pdx' \le C\norm{\zeta f}_{W^{1, p}(B^+)}^p \le CC_\zeta\norm{f}_{W^{1, p}(B^+)}^p > $ > [!theorem] > > Let $p \in [1, \infty)$, $U \subset \real^n$ be a bounded [[Open Set|open]] set such that $\partial U$ is [[Differentiable Boundary|of class]] $C^1$, and $x \in \partial U$. Then there exists $C \ge 0$ and $V \in \cn^o(U)$ such that > $ > \int_{\partial U \cap V}\abs{f}^pdS \le C\norm{f}_{W^{1, p}(U)}^p > $ > *Proof*. Let $(V, \varphi)$ be a boundary [[Atlas|chart]] at $x$ such that > 1. $\varphi$ and $\varphi^{-1}$ are $UC$[^2]. > 2. $\wh V = \varphi(V) = B(0, 2r)$ is a half-ball centred at $0$. > > Let $\wh B = B(0, r)$ and $B = \varphi^{-1}(\wh B)$, then by the previous theorem, > $ > \int_{\partial \wh V \cap \wh B}\abs{f \circ \varphi^{-1}}^pdx' \le C_B\norm{f \circ \varphi^{-1}}_{W^{1, p}(\wh V)}^p \le C_BC_{\varphi}\norm{f}_{W^{1, p}(U)}^p > $ > From here, since $\varphi^{-1}|_{\bracs{x = 0}}: \bracs{x = 0} \to \partial V$ is a $C^1$-isomorphism, there exists $C_{\partial}$ such that > $ > \int_{\partial V \cap B}\abs{f}^p dS \le C_\partial \int_{\partial \wh V \cap \wh B}\abs{f \circ \varphi^{-1}}^p dx' > $ > Therefore $\int_{\partial V \cap B}\abs{f}^pdS \le C_\partial C_B C_\varphi \norm{f}_{W^{1, p}(U)}^p$. [^1]: [[Weak Derivatives of Functions Raised to the p-th Power]] [^2]: Achieved by restricting to a smaller coordinate ball if necessary.