> [!theorem] > > Split $\real^n = \real^{n - 1} \times \real$. Let $r > 0$, $B = B(0, r)$, $B^+ = B \cap \mathbb{H}^n$ and $B^- = B \cap (-\mathbb{H}^n)$. Let $f \in UC^1(B^+)$ be a function with [[Space of Uniformly Continuously Differentiable Functions|uniformly continuous derivative]], then there exists an extension $F \in C^1(B)$ such that $F|_{B^+} = f$, known as a **higher order reflection** of $f$, and the map $f \mapsto F$ is linear. > > Moreover, there exists $C \ge 0$, independent of $f$, such that > $ > \norm{F}_{W^{1, p}(B)} \le C\norm{f}_{W^{1, p}(B^+)} > $ > the [[Sobolev Space|Sobolev norm]] of the extension is preserved. > > > *Proof*. Let $g, h: B^- \to \real$ be defined by > $ > g(x) = f(x_{\ol{n - 1}}, -x_n) \quad h(x) = f(x_{\ol{n - 1}}, -x_n/2) > $ > and define the extension as > $ > F(x) = \begin{cases} > F^+(x) = f(x) &x \in B^+ \\ > F^-(x) = -3g(x) + 4h(x) &x \in B^- > \end{cases} > $ > On the boundary $\bracs{x_n = 0} = B^+ \cap B^-$, the two definitions agree. As differentiability above and below the boundary is given, $F \in C^1(B)$ if the [[Partial Derivative|partial derivatives]] agree on the boundary as well. Since for all other coordinates, the partial derivatives can be taken as a limit from both sides, $\partial^{x_j}F^+|_{\bracs{x_n = 0}} = \partial^{x_j}F^-|_{\bracs{x_n = 0}}$ for all $j < n$. It's sufficient to verify that $\partial^{x_n}F^+|_{\bracs{x_n = 0}} =\partial^{x_n}F^-|_{\bracs{x_n = 0}}$. To this end, applying the chain rule yields that > $ > \begin{align*} > \partial^{x_n}F^-(x) &= 3\partial^{x_n}g(x) - 4\partial^{x_n}h(x) \\ > &= 3\partial^{x_n}f(x) - 2\partial^{x_n}f(x) = \partial^{x_n}F^+(x) > \end{align*} > $ > Therefore $DF^+|_{\bracs{x_n = 0}} = DF^-|_{\bracs{x_n = 0}}$. As differentiability above and below the boundary is given, $F \in C^1(B)$. > > Now, we can express > $ > \norm{F}_{W^{1, p}(B)} \le \norm{f}_{W^{1, p}(B^+)} + \norm{F^-}_{W^{1, p}(B^-)} > $ > Firstly, > $ > \norm{F^-}_{W^{1, p}} \le 3\norm{f}_{p} + 4 \cdot 2\norm{f}_p \le 11 \norm{f}_p > $ > For any $j < n$, over scaling, > $ > \norm{\partial^{x_j} F^-}_p \le 3\norm{\partial^{x_j} f}_{p} + 4 \cdot 2\norm{\partial^{x_j}f}_p = 11\norm{f}_{W^{1, p}} > $ > On the other hand, > $ > \norm{\partial^{x_n}F^-}_p \le 3\norm{\partial^{x_n}f}_p + 4\norm{\partial^{x_n}f}_p \le 7 \norm{f}_{W^{1, p}} > $ > So $\norm{F^-}_{W^{1, p}(B^-)} \le 11\norm{f}_{W^{1, p}(B^+)}$, and by summing over the coordinates, there exists $C \ge 0$ such that $\norm{F}_{W^{1, p}(B)} \le C\norm{f}_{W^{1, p}(B^+)}$. > [!theorem] > > Let $U \subset \real^n$ be a bounded [[Open Set|open]] set such that $\partial U$ is [[Differentiable Boundary|of class]] $C^1$, and $p \in U$. Let $f \in UC^1(U)$, then there exists a neighbourhood $V \in \cn^o(p)$ such that $f$ admits a local extension $F \in C^1(V)$, and the map $f \mapsto F$ is linear. > > Moreover, there exists $C \ge 0$, independent of $f$, such that > $ > \norm{F}_{W^{1, p}(V)} \le C\norm{f}_{W^{1, p}(U)} > $ > > *Proof*. Since $\partial U \in C^1$, there exists $V_0 \in \cn^o(p)$, $r > 0$, and a $C^1$-isomorphism $\varphi: V_0 \to B(0, 2r)$ such that $V_0 \cap U = \varphi^{-1}(B(0, 2r) \cap (\mathbb{H}^n)^o)$[^1]. > > Let $B = B(0, r)$, $V = \varphi^{-1}(B)$ and $B^+, B^-$ as in the previous theorem, then $\varphi|_V \in UC^1(V)$ and $\varphi^{-1}|_B \in UC^1(B)$. We now identify $\varphi$ with this restriction. > > From here, $f \circ \varphi^{-1} \in UC^1(B^+)$, and admits an extension $F_0 \in UC^1(B)$, thus $F = F_0 \circ \varphi \in C^1(V)$, and we have found a desired extension. > > Since composing with isomorphisms is bounded[^2], > $ > \begin{align*} > \norm{F}_{W^{1, p}(V)} &\le C_\varphi\norm{F_0}_{W^{1, p}(B)} \le C_\varphi C_r \norm{f \circ \varphi^{-1}}_{W^{1, p}(B^+)} \\ > &\le C_\varphi C_r C_{\varphi^{-1}} \norm{f}_{W^{1, p}(U)} > \end{align*} > $ [^1]: This arises from a local extension of a boundary chart at $p$. [^2]: [[Sobolev Norm Over Charts]]