[proofplan] The even reflection $E$ is defined explicitly: $Eu(x', x_n) := u(x', x_n)$ for $x_n \ge 0$ and $Eu(x', x_n) := u(x', -x_n)$ for $x_n < 0$. Linearity is by inspection. We verify that $Eu \in W^{1,p}(\mathbb{R}^n)$ by computing the candidate weak derivatives explicitly: tangential ones $\partial_{x_i}$ for $i < n$ are themselves even reflections of $\partial_{x_i} u$, while the normal one $\partial_{x_n}$ is the odd reflection (sign-flipped) of $\partial_{x_n} u$. The substantive claim is that these candidates are the actual weak derivatives — i.e., the distributional derivative on all of $\mathbb{R}^n$ has no singular boundary contribution on $\{x_n = 0\}$. This is checked by a direct calculation: for any $\varphi \in C_c^\infty(\mathbb{R}^n)$, split $\int_{\mathbb{R}^n} Eu\, \partial_{x_i}\varphi\, d\mathcal{L}^n$ into integrals over the two half-spaces, integrate by parts in each, and observe that the boundary contributions cancel for tangential indices (matched normal directions, matched values via even reflection) and for the normal index (sign flip of the derivative makes the boundary integrals cancel). The $L^p$ bounds follow by direct computation: even reflection multiplies the $L^p$ norm by $2^{1/p}$ on each derivative, and so the operator norm is bounded by $2^{1/p}$ on the gradient. [/proofplan] [step:Define the even reflection operator] For $u: \mathbb{R}^n_+ \to \mathbb{R}$, where $\mathbb{R}^n_+ := \{x = (x', x_n) \in \mathbb{R}^{n-1} \times \mathbb{R} : x_n > 0\}$, define \begin{align*} E: W^{1,p}(\mathbb{R}^n_+) &\to L^p(\mathbb{R}^n) \\ u &\mapsto Eu, \qquad Eu(x', x_n) := \begin{cases} u(x', x_n), & x_n > 0, \\ u(x', -x_n), & x_n < 0. \end{cases} \end{align*} On the negligible hyperplane $\{x_n = 0\}$, $Eu$ is undefined; this is irrelevant for $L^p$ representatives. The operator $E$ is linear: if $\alpha, \beta \in \mathbb{R}$ and $u, v \in W^{1,p}(\mathbb{R}^n_+)$, then for $x_n > 0$, $E(\alpha u + \beta v)(x', x_n) = (\alpha u + \beta v)(x', x_n) = \alpha Eu(x', x_n) + \beta Ev(x', x_n)$, and similarly for $x_n < 0$. The restriction property $Eu|_{\mathbb{R}^n_+} = u$ is immediate from the definition. We must verify three quantitative claims: (C1) $Eu \in W^{1,p}(\mathbb{R}^n)$. (C2) The weak derivatives are \begin{align*} \partial_{x_i}(Eu)(x', x_n) &= \begin{cases} \partial_{x_i} u(x', x_n), & x_n > 0, \\ \partial_{x_i} u(x', -x_n), & x_n < 0, \end{cases} \quad i = 1, \dots, n-1, \\ \partial_{x_n}(Eu)(x', x_n) &= \begin{cases} \partial_{x_n} u(x', x_n), & x_n > 0, \\ -\partial_{x_n} u(x', -x_n), & x_n < 0. \end{cases} \end{align*} (C3) The norm bound $\|Eu\|_{W^{1,p}(\mathbb{R}^n)} \le C\|u\|_{W^{1,p}(\mathbb{R}^n_+)}$ with $C = C(n)$. For $p = \infty$, the same definition applies, with the relevant reformulations: $L^\infty$ norms instead of $L^p$, weak derivatives still defined via the distributional bracket, and the cancellation arguments below carry over directly. We carry out the proof for $1 \le p < \infty$; the case $p = \infty$ is then a passage to the limit (or a direct re-execution with $L^\infty$ in place of $L^p$). [/step] [step:$L^p$ norm bound for $Eu$] For $1 \le p < \infty$, by the change of variables $y_n = -x_n$ on the lower half-space (with $|d\mathcal{L}^n(y)| = |d\mathcal{L}^n(x)|$ since the change is an isometry of $\mathbb{R}^n$), \begin{align*} \int_{\mathbb{R}^n_-} |Eu(x', x_n)|^p\, d\mathcal{L}^n(x) = \int_{\mathbb{R}^n_-}|u(x', -x_n)|^p\, d\mathcal{L}^n(x) = \int_{\mathbb{R}^n_+}|u(x', y_n)|^p\, d\mathcal{L}^n(y) = \|u\|_{L^p(\mathbb{R}^n_+)}^p. \end{align*} Adding the upper-half contribution $\int_{\mathbb{R}^n_+}|u|^p\, d\mathcal{L}^n = \|u\|_{L^p(\mathbb{R}^n_+)}^p$, \begin{align*} \|Eu\|_{L^p(\mathbb{R}^n)}^p = 2\|u\|_{L^p(\mathbb{R}^n_+)}^p, \qquad \|Eu\|_{L^p(\mathbb{R}^n)} = 2^{1/p}\|u\|_{L^p(\mathbb{R}^n_+)}. \end{align*} For $p = \infty$, $\|Eu\|_{L^\infty(\mathbb{R}^n)} = \|u\|_{L^\infty(\mathbb{R}^n_+)}$, since the reflection takes essential suprema to essential suprema. [/step] [step:Identify the candidate weak derivatives] For $i = 1, \dots, n-1$ (tangential indices), define the function \begin{align*} g_i: \mathbb{R}^n &\to \mathbb{R}, & g_i(x', x_n) := \begin{cases} \partial_{x_i} u(x', x_n), & x_n > 0, \\ \partial_{x_i} u(x', -x_n), & x_n < 0. \end{cases} \end{align*} For the normal index $i = n$: \begin{align*} g_n: \mathbb{R}^n &\to \mathbb{R}, & g_n(x', x_n) := \begin{cases} \partial_{x_n} u(x', x_n), & x_n > 0, \\ -\partial_{x_n} u(x', -x_n), & x_n < 0. \end{cases} \end{align*} Each $g_i$ is in $L^p(\mathbb{R}^n)$ with norm equal to $2^{1/p}\|\partial_{x_i} u\|_{L^p(\mathbb{R}^n_+)}$, by the same change-of-variables computation as in the previous step (the sign change on $g_n$ does not affect the absolute value). We claim that $g_i$ is the weak partial derivative $\partial_{x_i}(Eu)$ on $\mathbb{R}^n$. [guided] **The candidate derivatives — formula and motivation.** The reflection $Eu$ is defined piecewise: on $\mathbb{R}^n_+$ it is $u$; on $\mathbb{R}^n_-$ it is $u$ composed with the reflection $(x', x_n) \mapsto (x', -x_n)$. By the chain rule (for smooth functions; the formula tells us what to expect for the weak derivative), the partial derivative on $\mathbb{R}^n_-$ equals \begin{align*} \partial_{x_i}\bigl[u(x', -x_n)\bigr] = \begin{cases} \partial_{x_i} u(x', -x_n), & i = 1, \dots, n-1, \\ -\partial_{x_n} u(x', -x_n), & i = n. \end{cases} \end{align*} The tangential coordinates $x_1, \dots, x_{n-1}$ are unchanged by the reflection, so their partial derivatives transport without sign change — leading to the *even* reflection of $\partial_{x_i} u$ as the candidate $g_i$ for tangential $i$. The normal coordinate $x_n$ is mapped to $-x_n$, contributing a Jacobian factor $-1$ — leading to the *odd* (sign-flipped) reflection of $\partial_{x_n} u$ as the candidate $g_n$. This is the key asymmetry between tangential and normal directions. **$L^p$ bounds.** Each candidate $g_i \in L^p(\mathbb{R}^n)$ with $\|g_i\|_{L^p(\mathbb{R}^n)} = 2^{1/p}\|\partial_{x_i} u\|_{L^p(\mathbb{R}^n_+)}$, by direct computation: the integral over the lower half-space, by the change of variables $y_n = -x_n$, equals the integral over the upper half-space. **Why the candidates are not automatically the weak derivatives.** Even when $u$ is smooth on $\mathbb{R}^n_+$ and the chain rule formulas above give continuous functions on each half-space, $Eu$ is only continuous (across $\{x_n = 0\}$ — the even reflection ensures continuity there if $u$ extends continuously to $\overline{\mathbb{R}^n_+}$, and $L^p$-continuous in the trace sense in general), not $C^1$. The first derivative $\partial_{x_n}(Eu)$ may have a jump discontinuity across $\{x_n = 0\}$: on $\mathbb{R}^n_+$ it equals $\partial_{x_n} u(x', x_n)$, while on $\mathbb{R}^n_-$ it equals $-\partial_{x_n} u(x', -x_n)$. As $x_n \to 0^+$, both expressions tend to $\partial_{x_n} u(x', 0^+)$ and $-\partial_{x_n} u(x', 0^+)$ respectively — opposite signs. So $\partial_{x_n}(Eu)$ has a jump discontinuity unless $\partial_{x_n} u(x', 0^+) = 0$. **Will this jump produce a singular distributional contribution?** When a piecewise smooth function has a jump across a hypersurface, its distributional derivative typically picks up a Dirac-mass-like contribution proportional to the jump. For the standard heaviside function $H(x_n)$ on $\mathbb{R}$, $H'(x_n) = \delta_0$ in the sense of distributions. So we should worry that $\partial_{x_n}(Eu)$ might pick up a singular distribution on $\{x_n = 0\}$ when $\partial_{x_n} u(x', 0^+) \ne 0$. The cancellation we will verify in the next step is exactly that no such singular contribution occurs. The reason — heuristically — is that the value of $Eu$ on $\{x_n = 0\}$ is the *same* from both sides (even reflection $\Rightarrow$ no jump in $Eu$), so the "boundary integral" picked up when integrating by parts on $\mathbb{R}^n_+$ is matched (with opposite sign, due to opposite outward normals) by the boundary integral on $\mathbb{R}^n_-$, and they cancel. The jump in the *derivative* is integrated against the test function and produces a finite contribution, not a singular one. [/guided] [/step] [step:Verify the candidate is the weak derivative — tangential case] Fix $i \in \{1, \dots, n-1\}$ and $\varphi \in C_c^\infty(\mathbb{R}^n)$. We verify \begin{align*} \int_{\mathbb{R}^n} Eu\, \partial_{x_i}\varphi\, d\mathcal{L}^n = -\int_{\mathbb{R}^n} g_i\, \varphi\, d\mathcal{L}^n. \end{align*} Split the LHS into the two half-spaces: \begin{align*} \int_{\mathbb{R}^n} Eu\, \partial_{x_i}\varphi\, d\mathcal{L}^n = \int_{\mathbb{R}^n_+} u(x', x_n)\, \partial_{x_i}\varphi(x', x_n)\, d\mathcal{L}^n + \int_{\mathbb{R}^n_-} u(x', -x_n)\, \partial_{x_i}\varphi(x', x_n)\, d\mathcal{L}^n. \end{align*} In the second integral, substitute $y_n := -x_n$ (with $d\mathcal{L}^n(y) = d\mathcal{L}^n(x)$ since the substitution is an isometry; note that $\partial_{x_i}\varphi(x', x_n) = \partial_{x_i}\tilde\varphi(x', y_n)$ where $\tilde\varphi(x', y_n) := \varphi(x', -y_n)$, since the substitution does not involve the $i$-th coordinate for $i < n$): \begin{align*} \int_{\mathbb{R}^n_-} u(x', -x_n)\, \partial_{x_i}\varphi(x', x_n)\, d\mathcal{L}^n(x) = \int_{\mathbb{R}^n_+} u(x', y_n)\, \partial_{x_i}\tilde\varphi(x', y_n)\, d\mathcal{L}^n(y), \end{align*} where $\tilde\varphi(x', y_n) := \varphi(x', -y_n)$. Note that $\tilde\varphi \in C_c^\infty(\mathbb{R}^n)$ — but it need not have compact support **inside** $\mathbb{R}^n_+$. We address this momentarily. Combining the two integrals, \begin{align*} \int_{\mathbb{R}^n} Eu\, \partial_{x_i}\varphi\, d\mathcal{L}^n = \int_{\mathbb{R}^n_+} u(x', y_n)\bigl[\partial_{x_i}\varphi(x', y_n) + \partial_{x_i}\tilde\varphi(x', y_n)\bigr]d\mathcal{L}^n(y). \end{align*} Define $\Psi: \mathbb{R}^n_+ \to \mathbb{R}$ by $\Psi(x', y_n) := \varphi(x', y_n) + \tilde\varphi(x', y_n) = \varphi(x', y_n) + \varphi(x', -y_n)$. Then $\Psi$ is $C^\infty$ on $\mathbb{R}^n_+$, with $\partial_{x_i}\Psi = \partial_{x_i}\varphi(x', y_n) + \partial_{x_i}\tilde\varphi(x', y_n)$ for $i = 1, \dots, n-1$. We have \begin{align*} \int_{\mathbb{R}^n} Eu\, \partial_{x_i}\varphi\, d\mathcal{L}^n = \int_{\mathbb{R}^n_+} u\, \partial_{x_i}\Psi\, d\mathcal{L}^n. \end{align*} To apply the weak-derivative identity for $u$ on $\mathbb{R}^n_+$, we need $\Psi$ to be a test function on $\mathbb{R}^n_+$ — i.e., to have compact support contained in $\mathbb{R}^n_+$. It does not (its support extends to $\overline{\mathbb{R}^n_+}$, including the boundary $\{y_n = 0\}$). However, we can use a slightly stronger version of the weak-derivative identity that holds for $u \in W^{1,p}(\mathbb{R}^n_+)$ and any $\Psi$ with compact support in $\overline{\mathbb{R}^n_+}$ that is smooth and tangentially-supported away from the boundary: this is essentially the statement that the weak derivative satisfies the integration-by-parts identity, and we must justify the absence of a boundary contribution. We argue as follows. By approximation: pick a sequence $\Psi_\varepsilon \in C_c^\infty(\mathbb{R}^n_+)$ with $\Psi_\varepsilon \to \Psi$ in $W^{1,p'}_{\mathrm{loc}}$ on a fixed neighbourhood of $\operatorname{supp}\Psi \cap \overline{\mathbb{R}^n_+}$, where $p'$ is the Hölder conjugate of $p$. Specifically, multiply $\Psi$ by a cutoff $\chi_\varepsilon(y_n)$ with $\chi_\varepsilon = 1$ for $y_n \ge \varepsilon$, $\chi_\varepsilon = 0$ for $y_n \le \varepsilon/2$, and $|\chi_\varepsilon'| \le 4/\varepsilon$. Set $\Psi_\varepsilon := \chi_\varepsilon\Psi$. Then $\Psi_\varepsilon \in C_c^\infty(\mathbb{R}^n_+)$ for the appropriate small $\varepsilon$. By the weak-derivative identity for $u$ on $\mathbb{R}^n_+$, \begin{align*} \int_{\mathbb{R}^n_+} u\, \partial_{x_i}\Psi_\varepsilon\, d\mathcal{L}^n = -\int_{\mathbb{R}^n_+} \partial_{x_i} u\, \Psi_\varepsilon\, d\mathcal{L}^n. \end{align*} For $i \le n-1$ (tangential), $\partial_{x_i}\Psi_\varepsilon = \chi_\varepsilon \partial_{x_i}\Psi$ — the cutoff $\chi_\varepsilon$ depends only on $y_n$, so it does not contribute to $\partial_{x_i}$ for $i < n$. As $\varepsilon \to 0$: \begin{align*} \int_{\mathbb{R}^n_+}u\, \chi_\varepsilon\, \partial_{x_i}\Psi\, d\mathcal{L}^n \to \int_{\mathbb{R}^n_+}u\, \partial_{x_i}\Psi\, d\mathcal{L}^n, \end{align*} by dominated convergence ($\chi_\varepsilon \to \mathbb{1}_{y_n > 0}$ a.e., $|\chi_\varepsilon u\partial_{x_i}\Psi| \le |u||\partial_{x_i}\Psi| \in L^1$ since $u \in L^p$ and $\partial_{x_i}\Psi \in L^{p'}$ with compact support intersected with $\mathbb{R}^n_+$). Similarly, $\int_{\mathbb{R}^n_+}\partial_{x_i} u\, \chi_\varepsilon\, \Psi\, d\mathcal{L}^n \to \int_{\mathbb{R}^n_+}\partial_{x_i} u\, \Psi\, d\mathcal{L}^n$. Hence \begin{align*} \int_{\mathbb{R}^n_+}u\, \partial_{x_i}\Psi\, d\mathcal{L}^n = -\int_{\mathbb{R}^n_+}\partial_{x_i} u\, \Psi\, d\mathcal{L}^n. \end{align*} Substituting back, \begin{align*} \int_{\mathbb{R}^n} Eu\, \partial_{x_i}\varphi\, d\mathcal{L}^n = -\int_{\mathbb{R}^n_+}\partial_{x_i} u(x', y_n)\bigl[\varphi(x', y_n) + \varphi(x', -y_n)\bigr]d\mathcal{L}^n(y). \end{align*} Reverse the substitution on the second piece (substitute $x_n := -y_n$ on the term involving $\varphi(x', -y_n)$): \begin{align*} \int_{\mathbb{R}^n_+}\partial_{x_i} u(x', y_n)\, \varphi(x', -y_n)\, d\mathcal{L}^n(y) = \int_{\mathbb{R}^n_-}\partial_{x_i} u(x', -x_n)\, \varphi(x', x_n)\, d\mathcal{L}^n(x) = \int_{\mathbb{R}^n_-}g_i(x', x_n)\varphi(x', x_n)\, d\mathcal{L}^n(x). \end{align*} And the first piece is, of course, $\int_{\mathbb{R}^n_+}g_i\, \varphi\, d\mathcal{L}^n$. Combining, \begin{align*} \int_{\mathbb{R}^n} Eu\, \partial_{x_i}\varphi\, d\mathcal{L}^n = -\int_{\mathbb{R}^n_+}g_i\, \varphi\, d\mathcal{L}^n - \int_{\mathbb{R}^n_-}g_i\, \varphi\, d\mathcal{L}^n = -\int_{\mathbb{R}^n}g_i\, \varphi\, d\mathcal{L}^n. \end{align*} This is the weak-derivative identity for $\partial_{x_i}(Eu) = g_i$ for tangential $i$. [/step] [step:Verify the candidate is the weak derivative — normal case] Now $i = n$. We verify \begin{align*} \int_{\mathbb{R}^n} Eu\, \partial_{x_n}\varphi\, d\mathcal{L}^n = -\int_{\mathbb{R}^n} g_n\, \varphi\, d\mathcal{L}^n. \end{align*} Split the LHS: \begin{align*} \int_{\mathbb{R}^n} Eu\, \partial_{x_n}\varphi\, d\mathcal{L}^n = \int_{\mathbb{R}^n_+} u(x', x_n)\, \partial_{x_n}\varphi(x', x_n)\, d\mathcal{L}^n + \int_{\mathbb{R}^n_-} u(x', -x_n)\, \partial_{x_n}\varphi(x', x_n)\, d\mathcal{L}^n. \end{align*} On the second integral, substitute $y_n := -x_n$: \begin{align*} \int_{\mathbb{R}^n_-} u(x', -x_n)\, \partial_{x_n}\varphi(x', x_n)\, d\mathcal{L}^n(x) = \int_{\mathbb{R}^n_+}u(x', y_n)\, \partial_{x_n}\varphi(x', -y_n)\, d\mathcal{L}^n(y). \end{align*} Now $\partial_{x_n}\varphi(x', -y_n) = -\partial_{y_n}[\varphi(x', -y_n)] = -\partial_{y_n}\tilde\varphi(x', y_n)$ — note the **sign flip** from the chain rule, since $\varphi$ is differentiated with respect to its second argument $x_n$, while the substitution $y_n = -x_n$ has Jacobian $-1$ in the $y_n$ derivative. So \begin{align*} \int_{\mathbb{R}^n_+}u(x', y_n)\, \partial_{x_n}\varphi(x', -y_n)\, d\mathcal{L}^n(y) = -\int_{\mathbb{R}^n_+}u\, \partial_{y_n}\tilde\varphi\, d\mathcal{L}^n. \end{align*} Combining, \begin{align*} \int_{\mathbb{R}^n} Eu\, \partial_{x_n}\varphi\, d\mathcal{L}^n = \int_{\mathbb{R}^n_+}u\, \partial_{x_n}\varphi\, d\mathcal{L}^n - \int_{\mathbb{R}^n_+}u\, \partial_{y_n}\tilde\varphi\, d\mathcal{L}^n = \int_{\mathbb{R}^n_+}u\, \partial_{y_n}\bigl[\varphi - \tilde\varphi\bigr](x', y_n)\, d\mathcal{L}^n(y). \end{align*} Define $\Theta(x', y_n) := \varphi(x', y_n) - \tilde\varphi(x', y_n) = \varphi(x', y_n) - \varphi(x', -y_n)$ on $\mathbb{R}^n_+$. Then $\Theta \in C^\infty(\overline{\mathbb{R}^n_+})$ and, importantly, **$\Theta(x', 0) = \varphi(x', 0) - \varphi(x', 0) = 0$**. The vanishing of $\Theta$ on the boundary $\{y_n = 0\}$ is the cancellation that makes the whole proof work — it is the consequence of the *odd* combination $\varphi - \tilde\varphi$ that arises from the normal index, where the substitution introduces a sign flip. Apply the weak-derivative identity for $u$ on $\mathbb{R}^n_+$ with the function $\Theta$. Since $\Theta$ vanishes on the boundary $\{y_n = 0\}$, the cutoff approximation $\Theta_\varepsilon := \chi_\varepsilon\Theta$ satisfies $\Theta_\varepsilon \in C_c^\infty(\mathbb{R}^n_+)$ for $\varepsilon$ small, and also $\partial_{y_n}\Theta_\varepsilon = \chi_\varepsilon \partial_{y_n}\Theta + \chi_\varepsilon'\Theta$. The dangerous term is $\chi_\varepsilon'\Theta$: $\chi_\varepsilon'$ is supported in $\{y_n \in [\varepsilon/2, \varepsilon]\}$ with $|\chi_\varepsilon'| \le 4/\varepsilon$, but $|\Theta(x', y_n)| \le |y_n| \cdot 2\|\partial_{y_n}\varphi\|_{L^\infty}$ near $y_n = 0$ (by the mean value theorem applied to $\varphi(x', y_n) - \varphi(x', -y_n) = 2y_n\, \partial_{y_n}\varphi(x', \xi)$ for some $\xi \in (-y_n, y_n)$, hence for $|y_n| \le \varepsilon$, $|\Theta| \le 2\varepsilon\|\partial_{y_n}\varphi\|_{L^\infty}$). So $|\chi_\varepsilon'\Theta| \le (4/\varepsilon) \cdot 2\varepsilon\|\partial_{y_n}\varphi\|_{L^\infty} = 8\|\partial_{y_n}\varphi\|_{L^\infty}$ uniformly in $\varepsilon$. Better, $\chi_\varepsilon'\Theta$ is supported in $\{y_n \in [\varepsilon/2, \varepsilon]\}$ which has $\mathcal{L}^n$-measure of order $\varepsilon$ on a fixed compact set. Therefore $\|\chi_\varepsilon'\Theta\|_{L^{p'}(\mathbb{R}^n_+)} \to 0$ as $\varepsilon \to 0$, and the IBP identity \begin{align*} \int_{\mathbb{R}^n_+}u\, \partial_{y_n}\Theta_\varepsilon\, d\mathcal{L}^n = -\int_{\mathbb{R}^n_+}\partial_{y_n} u\, \Theta_\varepsilon\, d\mathcal{L}^n \end{align*} passes to the limit $\varepsilon \to 0$: \begin{align*} \int_{\mathbb{R}^n_+}u\, \partial_{y_n}\Theta\, d\mathcal{L}^n = -\int_{\mathbb{R}^n_+}\partial_{y_n} u\, \Theta\, d\mathcal{L}^n. \end{align*} The boundary contribution from $\chi_\varepsilon'\Theta$ — which would be the singular Dirac-like contribution if $\Theta$ did not vanish on $\{y_n = 0\}$ — is killed because $\Theta(x', 0) = 0$. This is the cancellation. Substituting back, \begin{align*} \int_{\mathbb{R}^n} Eu\, \partial_{x_n}\varphi\, d\mathcal{L}^n = \int_{\mathbb{R}^n_+}u\, \partial_{y_n}\Theta\, d\mathcal{L}^n = -\int_{\mathbb{R}^n_+}\partial_{y_n} u\, \Theta\, d\mathcal{L}^n = -\int_{\mathbb{R}^n_+}\partial_{y_n} u(x', y_n)\bigl[\varphi(x', y_n) - \varphi(x', -y_n)\bigr]d\mathcal{L}^n(y). \end{align*} Reverse the substitution on the second piece: \begin{align*} \int_{\mathbb{R}^n_+}\partial_{y_n} u(x', y_n)\, \varphi(x', -y_n)\, d\mathcal{L}^n = \int_{\mathbb{R}^n_-}\partial_{y_n} u(x', -x_n)\, \varphi(x', x_n)\, d\mathcal{L}^n(x). \end{align*} Hence \begin{align*} \int_{\mathbb{R}^n} Eu\, \partial_{x_n}\varphi\, d\mathcal{L}^n &= -\int_{\mathbb{R}^n_+}\partial_{y_n} u(x', y_n)\, \varphi(x', y_n)\, d\mathcal{L}^n + \int_{\mathbb{R}^n_-}\partial_{y_n} u(x', -x_n)\, \varphi(x', x_n)\, d\mathcal{L}^n \\ &= -\int_{\mathbb{R}^n_+}g_n\, \varphi\, d\mathcal{L}^n - \int_{\mathbb{R}^n_-}g_n\, \varphi\, d\mathcal{L}^n \\ &= -\int_{\mathbb{R}^n}g_n\, \varphi\, d\mathcal{L}^n, \end{align*} using the definition $g_n(x', x_n) = -\partial_{y_n} u(x', -x_n)$ for $x_n < 0$, which makes $\int_{\mathbb{R}^n_-}\partial_{y_n} u(x', -x_n)\varphi\, d\mathcal{L}^n = -\int_{\mathbb{R}^n_-}g_n\varphi\, d\mathcal{L}^n$. So $\partial_{x_n}(Eu) = g_n$ in the weak sense. [guided] **The goal restated.** Fix $\varphi \in C_c^\infty(\mathbb{R}^n)$. We aim to verify the weak-derivative identity \begin{align*} \int_{\mathbb{R}^n} Eu(x', x_n)\, \partial_{x_n}\varphi(x', x_n)\, d\mathcal{L}^n(x) = -\int_{\mathbb{R}^n} g_n(x', x_n)\, \varphi(x', x_n)\, d\mathcal{L}^n(x), \end{align*} which would identify $g_n$ as the distributional partial derivative $\partial_{x_n}(Eu)$ on the whole space $\mathbb{R}^n$. Recall $g_n(x', x_n) = \partial_{x_n} u(x', x_n)$ for $x_n > 0$ and $g_n(x', x_n) = -\partial_{x_n} u(x', -x_n)$ for $x_n < 0$ — the *odd* reflection of $\partial_{x_n} u$. The asymmetry between the tangential candidates $g_i$ ($i < n$, even reflection) and the normal candidate $g_n$ (odd reflection) is the subtle point. **Splitting the integral.** Since $\mathbb{R}^n = \mathbb{R}^n_+ \sqcup \{x_n = 0\} \sqcup \mathbb{R}^n_-$ and $\{x_n = 0\}$ is $\mathcal{L}^n$-negligible, we split \begin{align*} \int_{\mathbb{R}^n} Eu\, \partial_{x_n}\varphi\, d\mathcal{L}^n = \int_{\mathbb{R}^n_+} u(x', x_n)\, \partial_{x_n}\varphi(x', x_n)\, d\mathcal{L}^n + \int_{\mathbb{R}^n_-} u(x', -x_n)\, \partial_{x_n}\varphi(x', x_n)\, d\mathcal{L}^n, \end{align*} using $Eu(x', x_n) = u(x', x_n)$ on the upper half-space and $Eu(x', x_n) = u(x', -x_n)$ on the lower half-space. **Substitution on the lower half-space.** Apply the change of variables $y_n := -x_n$ on the second integral. Since the map $x \mapsto (x', -x_n)$ is an isometry of $\mathbb{R}^n$, $d\mathcal{L}^n(y) = d\mathcal{L}^n(x)$, and the domain $\mathbb{R}^n_-$ maps to $\mathbb{R}^n_+$. The integrand transforms: $u(x', -x_n)$ becomes $u(x', y_n)$, and $\partial_{x_n}\varphi(x', x_n)$ becomes $\partial_{x_n}\varphi(x', -y_n)$. Thus \begin{align*} \int_{\mathbb{R}^n_-} u(x', -x_n)\, \partial_{x_n}\varphi(x', x_n)\, d\mathcal{L}^n(x) = \int_{\mathbb{R}^n_+} u(x', y_n)\, \partial_{x_n}\varphi(x', -y_n)\, d\mathcal{L}^n(y). \end{align*} **The chain-rule sign flip.** Set $\tilde\varphi(x', y_n) := \varphi(x', -y_n)$. Differentiating with respect to $y_n$ via the chain rule, $\partial_{y_n}\tilde\varphi(x', y_n) = -\partial_{x_n}\varphi(x', -y_n)$, since the inner map $y_n \mapsto -y_n$ contributes a Jacobian factor of $-1$. Equivalently, \begin{align*} \partial_{x_n}\varphi(x', -y_n) = -\partial_{y_n}\tilde\varphi(x', y_n). \end{align*} Substituting this into the previous display, \begin{align*} \int_{\mathbb{R}^n_+} u(x', y_n)\, \partial_{x_n}\varphi(x', -y_n)\, d\mathcal{L}^n(y) = -\int_{\mathbb{R}^n_+} u(x', y_n)\, \partial_{y_n}\tilde\varphi(x', y_n)\, d\mathcal{L}^n(y). \end{align*} This sign flip is the source of the asymmetry: in the tangential case ($i < n$), no Jacobian factor appears, so the analogous combination is the *sum* $\varphi + \tilde\varphi$; here we get the *difference* $\varphi - \tilde\varphi$. **Combined integrand on the upper half-space.** Plugging the two pieces back together (and renaming the integration variable $y_n$ as a dummy variable; we keep $y_n$ to avoid collision with the original $x_n$), \begin{align*} \int_{\mathbb{R}^n} Eu\, \partial_{x_n}\varphi\, d\mathcal{L}^n &= \int_{\mathbb{R}^n_+} u\, \partial_{x_n}\varphi\, d\mathcal{L}^n - \int_{\mathbb{R}^n_+} u\, \partial_{y_n}\tilde\varphi\, d\mathcal{L}^n \\ &= \int_{\mathbb{R}^n_+} u(x', y_n)\, \partial_{y_n}\bigl[\varphi(x', y_n) - \tilde\varphi(x', y_n)\bigr]d\mathcal{L}^n(y). \end{align*} **The odd combination $\Theta$.** Define $\Theta: \overline{\mathbb{R}^n_+} \to \mathbb{R}$ by \begin{align*} \Theta(x', y_n) := \varphi(x', y_n) - \tilde\varphi(x', y_n) = \varphi(x', y_n) - \varphi(x', -y_n). \end{align*} Then $\Theta \in C^\infty(\overline{\mathbb{R}^n_+})$, with all derivatives bounded (since $\varphi$ has compact support and bounded derivatives of all orders). The decisive property is that $\Theta$ vanishes on the boundary hyperplane $\{y_n = 0\}$: \begin{align*} \Theta(x', 0) = \varphi(x', 0) - \varphi(x', 0) = 0. \end{align*} This boundary vanishing is the algebraic shadow of the desired cancellation: when we attempt integration by parts on $\mathbb{R}^n_+$ with the variable $y_n$, the boundary contribution at $\{y_n = 0\}$ is proportional to $\Theta(x', 0)$, which is zero. (Compare: in the tangential case, the analogous function is $\Psi = \varphi + \tilde\varphi$, which satisfies $\Psi(x', 0) = 2\varphi(x', 0)$ and does *not* vanish on the boundary; the tangential proof handled this differently because tangential differentiation of a cutoff $\chi_\varepsilon(y_n)$ gives zero, not a $1/\varepsilon$ blow-up.) **Approximation by an admissible test function.** The function $\Theta$ is not a valid test function on $\mathbb{R}^n_+$ in the strict sense — its support extends up to $\{y_n = 0\}$ rather than being compactly contained in the open half-space. To apply the weak-derivative identity for $u$ on $\mathbb{R}^n_+$, we approximate $\Theta$ by $\Theta_\varepsilon := \chi_\varepsilon(y_n)\Theta(x', y_n)$, where $\chi_\varepsilon: [0, \infty) \to [0, 1]$ is a smooth cutoff satisfying $\chi_\varepsilon(y_n) = 0$ for $y_n \le \varepsilon/2$, $\chi_\varepsilon(y_n) = 1$ for $y_n \ge \varepsilon$, and $|\chi_\varepsilon'(y_n)| \le 4/\varepsilon$. Then $\Theta_\varepsilon$ has compact support inside the open half-space $\mathbb{R}^n_+$ (since $\Theta$ has compact support in $\overline{\mathbb{R}^n_+}$ inherited from $\varphi$, and $\chi_\varepsilon$ vanishes near $\{y_n = 0\}$), and $\Theta_\varepsilon \in C_c^\infty(\mathbb{R}^n_+)$ — an admissible test function. **Weak-derivative identity for $\Theta_\varepsilon$.** Since $\Theta_\varepsilon \in C_c^\infty(\mathbb{R}^n_+)$, the definition of the weak partial $\partial_{y_n} u \in L^p(\mathbb{R}^n_+)$ yields \begin{align*} \int_{\mathbb{R}^n_+} u(x', y_n)\, \partial_{y_n}\Theta_\varepsilon(x', y_n)\, d\mathcal{L}^n(y) = -\int_{\mathbb{R}^n_+} \partial_{y_n} u(x', y_n)\, \Theta_\varepsilon(x', y_n)\, d\mathcal{L}^n(y). \end{align*} By the product rule, $\partial_{y_n}\Theta_\varepsilon = \chi_\varepsilon'(y_n)\Theta + \chi_\varepsilon(y_n)\partial_{y_n}\Theta$, so the left-hand side decomposes as \begin{align*} \int_{\mathbb{R}^n_+} u\, \chi_\varepsilon\, \partial_{y_n}\Theta\, d\mathcal{L}^n + \int_{\mathbb{R}^n_+} u\, \chi_\varepsilon'\, \Theta\, d\mathcal{L}^n = -\int_{\mathbb{R}^n_+} \partial_{y_n} u\, \chi_\varepsilon\, \Theta\, d\mathcal{L}^n. \end{align*} **Passage to the limit.** As $\varepsilon \to 0^+$, $\chi_\varepsilon(y_n) \to \mathbb{1}_{(0, \infty)}(y_n)$ pointwise a.e., and $|\chi_\varepsilon| \le 1$. The integrands $u\, \chi_\varepsilon\, \partial_{y_n}\Theta$ and $\partial_{y_n} u\, \chi_\varepsilon\, \Theta$ are dominated by $|u||\partial_{y_n}\Theta|$ and $|\partial_{y_n} u||\Theta|$ respectively, both of which lie in $L^1(\mathbb{R}^n_+)$ (Hölder: $u \in L^p$, $\partial_{y_n}\Theta \in L^{p'}$ with compact support; similarly for the other term). By dominated convergence, \begin{align*} \int_{\mathbb{R}^n_+} u\, \chi_\varepsilon\, \partial_{y_n}\Theta\, d\mathcal{L}^n \to \int_{\mathbb{R}^n_+} u\, \partial_{y_n}\Theta\, d\mathcal{L}^n, \qquad \int_{\mathbb{R}^n_+} \partial_{y_n} u\, \chi_\varepsilon\, \Theta\, d\mathcal{L}^n \to \int_{\mathbb{R}^n_+} \partial_{y_n} u\, \Theta\, d\mathcal{L}^n. \end{align*} **The error term vanishes.** The remaining term $\int_{\mathbb{R}^n_+} u\, \chi_\varepsilon'\, \Theta\, d\mathcal{L}^n$ is the dangerous one. Here $\chi_\varepsilon'$ is supported in the strip $S_\varepsilon := \{y_n \in [\varepsilon/2, \varepsilon]\}$ with $|\chi_\varepsilon'| \le 4/\varepsilon$. The boundary vanishing $\Theta(x', 0) = 0$ now plays its role: by the mean value theorem applied to $\varphi(x', y_n) - \varphi(x', -y_n)$ as a function of $y_n$, there exists $\xi \in (-y_n, y_n)$ with $\varphi(x', y_n) - \varphi(x', -y_n) = 2 y_n\, \partial_{y_n}\varphi(x', \xi)$, hence \begin{align*} |\Theta(x', y_n)| \le 2 |y_n|\, \|\partial_{y_n}\varphi\|_{L^\infty(\mathbb{R}^n)}. \end{align*} On the strip $S_\varepsilon$, $|y_n| \le \varepsilon$, so $|\Theta(x', y_n)| \le 2\varepsilon\|\partial_{y_n}\varphi\|_{L^\infty}$, and consequently \begin{align*} |\chi_\varepsilon'(y_n)\Theta(x', y_n)| \le \frac{4}{\varepsilon} \cdot 2\varepsilon\|\partial_{y_n}\varphi\|_{L^\infty} = 8\|\partial_{y_n}\varphi\|_{L^\infty}, \end{align*} uniformly in $\varepsilon$ on $S_\varepsilon \cap \operatorname{supp}\varphi$. The Lebesgue measure of $S_\varepsilon \cap \operatorname{supp}\varphi$ is at most $C(\operatorname{supp}\varphi)\cdot \varepsilon$, so $\|\chi_\varepsilon'\Theta\|_{L^{p'}(\mathbb{R}^n_+)}^{p'} \le 8^{p'}\|\partial_{y_n}\varphi\|_{L^\infty}^{p'}\cdot C\varepsilon \to 0$ as $\varepsilon \to 0$. By Hölder's inequality, \begin{align*} \left|\int_{\mathbb{R}^n_+} u\, \chi_\varepsilon'\, \Theta\, d\mathcal{L}^n\right| \le \|u\|_{L^p(\mathbb{R}^n_+)}\, \|\chi_\varepsilon'\Theta\|_{L^{p'}(\mathbb{R}^n_+)} \to 0. \end{align*} **The limiting identity for $\Theta$.** Passing $\varepsilon \to 0$ in the equation for $\Theta_\varepsilon$, the dangerous term drops out, and we obtain \begin{align*} \int_{\mathbb{R}^n_+} u(x', y_n)\, \partial_{y_n}\Theta(x', y_n)\, d\mathcal{L}^n(y) = -\int_{\mathbb{R}^n_+} \partial_{y_n} u(x', y_n)\, \Theta(x', y_n)\, d\mathcal{L}^n(y). \end{align*} This is the integration-by-parts identity for $u$ tested against $\Theta$, with no boundary contribution — the boundary contribution that *would* have appeared had $\Theta$ been a generic smooth function on $\overline{\mathbb{R}^n_+}$ is absent because $\Theta(x', 0) = 0$. **Substituting back into the original integral.** Recalling $\int_{\mathbb{R}^n} Eu\, \partial_{x_n}\varphi\, d\mathcal{L}^n = \int_{\mathbb{R}^n_+} u\, \partial_{y_n}\Theta\, d\mathcal{L}^n$, \begin{align*} \int_{\mathbb{R}^n} Eu\, \partial_{x_n}\varphi\, d\mathcal{L}^n &= -\int_{\mathbb{R}^n_+} \partial_{y_n} u(x', y_n)\, \Theta(x', y_n)\, d\mathcal{L}^n(y) \\ &= -\int_{\mathbb{R}^n_+} \partial_{y_n} u(x', y_n)\bigl[\varphi(x', y_n) - \varphi(x', -y_n)\bigr]d\mathcal{L}^n(y). \end{align*} **Reversing the substitution.** Split the integral on the right and reverse the substitution $y_n = -x_n$ (i.e., set $x_n = -y_n$) on the term involving $\varphi(x', -y_n)$: \begin{align*} \int_{\mathbb{R}^n_+} \partial_{y_n} u(x', y_n)\, \varphi(x', -y_n)\, d\mathcal{L}^n(y) = \int_{\mathbb{R}^n_-} \partial_{y_n} u(x', -x_n)\, \varphi(x', x_n)\, d\mathcal{L}^n(x). \end{align*} By the definition of $g_n$ on the lower half-space, $g_n(x', x_n) = -\partial_{y_n} u(x', -x_n)$ for $x_n < 0$, hence $\partial_{y_n} u(x', -x_n) = -g_n(x', x_n)$ for $x_n < 0$, and the integral equals $-\int_{\mathbb{R}^n_-} g_n(x', x_n)\, \varphi(x', x_n)\, d\mathcal{L}^n(x)$. **Final assembly.** Substituting, \begin{align*} \int_{\mathbb{R}^n} Eu\, \partial_{x_n}\varphi\, d\mathcal{L}^n &= -\int_{\mathbb{R}^n_+} \partial_{y_n} u(x', y_n)\, \varphi(x', y_n)\, d\mathcal{L}^n(y) + \int_{\mathbb{R}^n_+} \partial_{y_n} u(x', y_n)\, \varphi(x', -y_n)\, d\mathcal{L}^n(y) \\ &= -\int_{\mathbb{R}^n_+} g_n(x', y_n)\, \varphi(x', y_n)\, d\mathcal{L}^n(y) - \int_{\mathbb{R}^n_-} g_n(x', x_n)\, \varphi(x', x_n)\, d\mathcal{L}^n(x) \\ &= -\int_{\mathbb{R}^n} g_n(x', x_n)\, \varphi(x', x_n)\, d\mathcal{L}^n(x), \end{align*} using $g_n(x', y_n) = \partial_{y_n} u(x', y_n)$ on $\mathbb{R}^n_+$ in the first equality, and the rewriting from the previous paragraph for the second. **Conclusion.** The identity $\int_{\mathbb{R}^n} Eu\, \partial_{x_n}\varphi\, d\mathcal{L}^n = -\int_{\mathbb{R}^n} g_n\, \varphi\, d\mathcal{L}^n$ holds for every $\varphi \in C_c^\infty(\mathbb{R}^n)$. By the definition of the weak partial derivative on $\mathbb{R}^n$, this identifies $g_n$ as $\partial_{x_n}(Eu)$ in the weak sense. The whole proof hinges on a single algebraic fact — $\Theta(x', 0) = \varphi(x', 0) - \varphi(x', 0) = 0$ — which encodes the cancellation of the would-be boundary contributions from the two half-spaces, with the cancellation made possible by the *odd* sign convention in the definition of the candidate $g_n$ (as opposed to the *even* convention for tangential $g_i$). [/guided] [/step] [step:Combine the bounds and conclude] From the previous steps, $Eu \in W^{1,p}(\mathbb{R}^n)$ with $\partial_{x_i}(Eu) = g_i$ for $i = 1, \dots, n$. The $L^p$ norms are \begin{align*} \|Eu\|_{L^p(\mathbb{R}^n)} = 2^{1/p}\|u\|_{L^p(\mathbb{R}^n_+)}, \qquad \|g_i\|_{L^p(\mathbb{R}^n)} = 2^{1/p}\|\partial_{x_i} u\|_{L^p(\mathbb{R}^n_+)}. \end{align*} Combining (with the $W^{1,p}$ norm taken as the sum, or any equivalent variant), \begin{align*} \|Eu\|_{W^{1,p}(\mathbb{R}^n)} = \|Eu\|_{L^p(\mathbb{R}^n)} + \sum_{i=1}^n \|\partial_{x_i}(Eu)\|_{L^p(\mathbb{R}^n)} = 2^{1/p}\left(\|u\|_{L^p(\mathbb{R}^n_+)} + \sum_{i=1}^n\|\partial_{x_i} u\|_{L^p(\mathbb{R}^n_+)}\right) = 2^{1/p}\|u\|_{W^{1,p}(\mathbb{R}^n_+)}. \end{align*} For $1 \le p < \infty$, the constant in the operator norm is $C = 2^{1/p}$, which depends only on $p$ (independent of $n$). For $p = \infty$, $\|Eu\|_{L^\infty(\mathbb{R}^n)} = \|u\|_{L^\infty(\mathbb{R}^n_+)}$ and likewise for derivatives, so $C = 1$. The restriction property $Eu|_{\mathbb{R}^n_+} = u$ is immediate from the definition. Linearity of $E$ was established at the beginning. This completes the proof. [/step]