[proofplan] Fix $i \in \{1, \dots, n\}$ and let $u, v \in W^{1,p}(\Omega) \cap L^\infty(\Omega)$. The strategy is to approximate $u$ and $v$ by smooth functions that match them in $W^{1,p}$ while preserving the $L^\infty$ bounds, apply the classical product rule pointwise, and pass to the limit in the distributional definition of the weak derivative. The smooth approximations come from mollification of (locally extended) functions, which inherits the $L^\infty$ bound by Jensen's inequality. The key technical ingredient is that the bilinear product map $(f, g) \mapsto fg$ from $L^p \times L^\infty \to L^p$ is continuous when one factor stays bounded in $L^\infty$ and the other converges in $L^p$. With both pieces of the right-hand side $u\, \partial_{x_i} v + v\, \partial_{x_i} u$ converging in $L^p$ and $u_k v_k \to uv$ in $L^p$, the distributional identity passes to the limit and identifies $u\, \partial_{x_i} v + v\, \partial_{x_i} u$ as the weak derivative of $uv$. [/proofplan]

[step:Reduce the claim to the local distributional identity for one coordinate] Fix $i \in \{1, \dots, n\}$. To prove that $uv \in W^{1,p}(\Omega)$ with $\partial_{x_i}(uv) = u\, \partial_{x_i} v + v\, \partial_{x_i} u$, two things must be checked: 1. $uv \in L^p(\Omega)$ and $w_i := u\, \partial_{x_i} v + v\, \partial_{x_i} u \in L^p(\Omega)$. 2. For every test function $\varphi \in C_c^\infty(\Omega)$, \begin{align*} \int_\Omega (uv)\, \partial_{x_i}\varphi \, d\mathcal{L}^n(x) = -\int_\Omega w_i\, \varphi \, d\mathcal{L}^n(x). \end{align*} For (1): since $u \in L^\infty(\Omega)$ and $v \in L^p(\Omega)$, the product $uv$ satisfies $|uv| \le \|u\|_{L^\infty}|v|$, so $uv \in L^p(\Omega)$ with $\|uv\|_{L^p} \le \|u\|_{L^\infty}\|v\|_{L^p}$. Similarly $u\, \partial_{x_i} v \in L^p$ with $\|u\, \partial_{x_i} v\|_{L^p} \le \|u\|_{L^\infty}\|\partial_{x_i} v\|_{L^p}$, and $v\, \partial_{x_i} u \in L^p$ with $\|v\, \partial_{x_i} u\|_{L^p} \le \|v\|_{L^\infty}\|\partial_{x_i} u\|_{L^p}$. Hence $w_i \in L^p(\Omega)$. The remaining work is establishing the distributional identity (2). [/step]

[step:Localise to a relatively compact subdomain] Fix $\varphi \in C_c^\infty(\Omega)$ and let $K := \operatorname{supp}\varphi \subset\subset \Omega$. Choose an open set $\Omega'$ with $K \subset\subset \Omega' \subset\subset \Omega$ and $\operatorname{dist}(\Omega', \partial\Omega) > 0$; for instance, $\Omega' := \{x \in \Omega : \operatorname{dist}(x, K) < r_0\}$ for any $r_0 \in (0, \operatorname{dist}(K, \partial\Omega))$. Both integrals in the identity (2) depend only on the values of $u$, $v$, and their weak derivatives on $\Omega'$, so it suffices to prove the identity with $\Omega$ replaced by $\Omega'$. We retain the symbol $\Omega$ for notational economy and assume henceforth that the relevant integrals are over $\Omega'$. [/step]

[step:Construct smooth approximations preserving the $L^\infty$ bounds]Let $M := \max(\|u\|_{L^\infty(\Omega)}, \|v\|_{L^\infty(\Omega)})$. Fix a standard mollifier $\eta \in C_c^\infty(\mathbb{R}^n)$ with $\eta \ge 0$, $\operatorname{supp}\eta \subset B(0,1)$, and $\int_{\mathbb{R}^n} \eta \, d\mathcal{L}^n = 1$. Define \begin{align*} \eta_\varepsilon: \mathbb{R}^n &\to \mathbb{R} \\ x &\mapsto \varepsilon^{-n}\eta(x/\varepsilon). \end{align*} For $\varepsilon \in (0, \operatorname{dist}(\Omega', \partial\Omega))$ and $x \in \Omega'$ define the mollifications \begin{align*} u_\varepsilon: \Omega' &\to \mathbb{R}, & u_\varepsilon(x) &:= \int_{\mathbb{R}^n} \eta_\varepsilon(x - y)\, u(y)\, d\mathcal{L}^n(y), \\ v_\varepsilon: \Omega' &\to \mathbb{R}, & v_\varepsilon(x) &:= \int_{\mathbb{R}^n} \eta_\varepsilon(x - y)\, v(y)\, d\mathcal{L}^n(y), \end{align*} where $u$, $v$ are interpreted as elements of $L^p(\Omega) \cap L^\infty(\Omega)$, and the integrals are well-defined for $x \in \Omega'$ because $\operatorname{supp}\eta_\varepsilon(x - \cdot) \subset B(x,\varepsilon) \subset \Omega$. We record three properties. (i) **Smoothness.** $u_\varepsilon, v_\varepsilon \in C^\infty(\Omega')$. (ii) **$L^\infty$ bound.** For $x \in \Omega'$, \begin{align*} |u_\varepsilon(x)| \le \int_{\mathbb{R}^n} \eta_\varepsilon(x - y)\, |u(y)|\, d\mathcal{L}^n(y) \le \|u\|_{L^\infty(\Omega)} \int_{\mathbb{R}^n} \eta_\varepsilon(x - y)\, d\mathcal{L}^n(y) = \|u\|_{L^\infty(\Omega)} \le M, \end{align*} since $\eta_\varepsilon \ge 0$, $u \in L^\infty(\Omega)$ on the support of $\eta_\varepsilon(x - \cdot) \subset B(x,\varepsilon) \subset \Omega$, and $\eta_\varepsilon$ has unit integral. The same bound holds for $v_\varepsilon$. (iii) **Convergence in $W^{1,p}_{\mathrm{loc}}$.** By [Differentiation Through Convolution](/theorems/3096), $u_\varepsilon \to u$ in $L^p(\Omega'')$ and $\partial_{x_j} u_\varepsilon \to \partial_{x_j} u$ in $L^p(\Omega'')$ for every $j = 1, \dots, n$ and every $\Omega'' \subset\subset \Omega'$, as $\varepsilon \to 0$. The same holds for $v$.[/step]

[guided]The aim of this step is to produce smooth approximations $u_\varepsilon, v_\varepsilon$ that simultaneously (a) converge to $u, v$ in $W^{1,p}_{\mathrm{loc}}$ as $\varepsilon \to 0$ and (b) inherit the $L^\infty$ bound on $\Omega$. The standard mollifier construction delivers (a) by [Differentiation Through Convolution](/theorems/3096) — this is the routine fact. What is less routine is the simultaneous $L^\infty$ control: many smooth approximations of $W^{1,p}$ functions do not preserve essential boundedness, and we will need both controls below. **Why the $L^\infty$ control matters.** The product rule we want to push to the limit is \begin{align*} \partial_{x_i}(u_\varepsilon v_\varepsilon) = u_\varepsilon\, \partial_{x_i} v_\varepsilon + v_\varepsilon\, \partial_{x_i} u_\varepsilon. \end{align*} On the right, we have one factor converging in $L^p$ (the derivative) and one factor that we hope converges in some controlled sense. For the product to converge in $L^p$, the non-derivative factor must stay bounded in $L^\infty$ — convergence merely in $L^p$ is insufficient because the bilinear product $L^p \times L^p \to L^{p/2}$ does not in general converge in $L^p$. The same issue appears in the $L^p$-convergence of the products $u_\varepsilon v_\varepsilon \to uv$ that controls the left side. **Mollifier setup.** Fix a standard mollifier $\eta \in C_c^\infty(\mathbb{R}^n)$ with $\eta \ge 0$, $\operatorname{supp}\eta \subset B(0,1)$, and $\int_{\mathbb{R}^n} \eta \, d\mathcal{L}^n = 1$. Define the rescaled family \begin{align*} \eta_\varepsilon: \mathbb{R}^n &\to \mathbb{R} \\ x &\mapsto \varepsilon^{-n}\eta(x/\varepsilon), \end{align*} which satisfies $\eta_\varepsilon \in C_c^\infty(\mathbb{R}^n)$, $\operatorname{supp}\eta_\varepsilon \subset B(0,\varepsilon)$, and $\int_{\mathbb{R}^n} \eta_\varepsilon\, d\mathcal{L}^n = 1$. For $\varepsilon \in (0, \operatorname{dist}(\Omega', \partial\Omega))$ and $x \in \Omega'$, the convolution \begin{align*} u_\varepsilon(x) := \int_{\mathbb{R}^n} \eta_\varepsilon(x - y)\, u(y)\, d\mathcal{L}^n(y) \end{align*} is well-defined: the integrand is supported in $\{y : |x - y| < \varepsilon\} = B(x,\varepsilon)$, and $B(x,\varepsilon) \subset \Omega$ by the choice $\varepsilon < \operatorname{dist}(\Omega', \partial\Omega)$, where $u$ is defined and integrable. The same applies to $v_\varepsilon$. **Smoothness.** That $u_\varepsilon, v_\varepsilon \in C^\infty(\Omega')$ is the standard consequence of the smoothness of $\eta_\varepsilon$ and differentiation under the integral sign. **$L^\infty$ bound — derivation in detail.** For $x \in \Omega'$, \begin{align*} |u_\varepsilon(x)| = \left|\int_{\mathbb{R}^n} \eta_\varepsilon(x - y)\, u(y)\, d\mathcal{L}^n(y)\right| \le \int_{\mathbb{R}^n} \eta_\varepsilon(x - y)\, |u(y)|\, d\mathcal{L}^n(y), \end{align*} where we used the triangle inequality for integrals and the non-negativity $\eta_\varepsilon \ge 0$. The integrand is supported in $B(x,\varepsilon) \subset \Omega$, where $u \in L^\infty(\Omega)$ gives $|u(y)| \le \|u\|_{L^\infty(\Omega)}$ for almost every $y$. Hence \begin{align*} |u_\varepsilon(x)| \le \|u\|_{L^\infty(\Omega)} \int_{\mathbb{R}^n} \eta_\varepsilon(x - y)\, d\mathcal{L}^n(y) = \|u\|_{L^\infty(\Omega)} \le M, \end{align*} since $\eta_\varepsilon$ has unit integral. The same calculation gives $|v_\varepsilon(x)| \le \|v\|_{L^\infty(\Omega)} \le M$. This is the form of Jensen's inequality (or of Young's inequality with exponents $(1, \infty)$) that justifies inheritance of the $L^\infty$ bound by the mollification. **Convergence.** By [Differentiation Through Convolution](/theorems/3096) — applied to $u, v \in W^{1,p}(\Omega)$ on each $\Omega'' \subset\subset \Omega'$ — both $u_\varepsilon \to u$ and $\partial_{x_j} u_\varepsilon \to \partial_{x_j} u$ in $L^p(\Omega'')$ for each $j = 1, \dots, n$, with the analogous statement for $v$. We will only need the convergence on the fixed $\Omega'$, which is itself relatively compact in $\Omega$, so we can take $\Omega'' = \Omega'$.[/guided]

[step:Apply the classical product rule pointwise on $\Omega'$] Each $u_\varepsilon, v_\varepsilon \in C^\infty(\Omega')$, so the classical product rule from real analysis gives, for every $x \in \Omega'$, \begin{align*} \partial_{x_i}(u_\varepsilon v_\varepsilon)(x) = u_\varepsilon(x)\, \partial_{x_i} v_\varepsilon(x) + v_\varepsilon(x)\, \partial_{x_i} u_\varepsilon(x). \end{align*} Multiplying by $\varphi \in C_c^\infty(\Omega') \subset C_c^\infty(\Omega)$, integrating by parts on the left (legitimate because $u_\varepsilon v_\varepsilon \in C^\infty(\Omega')$ and $\varphi$ has compact support in $\Omega'$ for $\varepsilon$ small enough that $K \subset\subset \Omega'$), we obtain \begin{align*} -\int_{\Omega'} (u_\varepsilon v_\varepsilon)\, \partial_{x_i}\varphi \, d\mathcal{L}^n = \int_{\Omega'} \bigl(u_\varepsilon\, \partial_{x_i} v_\varepsilon + v_\varepsilon\, \partial_{x_i} u_\varepsilon\bigr) \varphi \, d\mathcal{L}^n. \end{align*} Equivalently, \begin{align*} \int_{\Omega'} (u_\varepsilon v_\varepsilon)\, \partial_{x_i}\varphi \, d\mathcal{L}^n = -\int_{\Omega'} \bigl(u_\varepsilon\, \partial_{x_i} v_\varepsilon + v_\varepsilon\, \partial_{x_i} u_\varepsilon\bigr) \varphi \, d\mathcal{L}^n. \tag{$\ast$} \end{align*} [/step]

[step:Pass to the limit in $L^p(\Omega')$ as $\varepsilon \to 0$]We claim \begin{align*} u_\varepsilon v_\varepsilon &\to uv \quad \text{in } L^p(\Omega'), \\ u_\varepsilon\, \partial_{x_i} v_\varepsilon &\to u\, \partial_{x_i} v \quad \text{in } L^p(\Omega'), \\ v_\varepsilon\, \partial_{x_i} u_\varepsilon &\to v\, \partial_{x_i} u \quad \text{in } L^p(\Omega'). \end{align*} For the first claim, decompose \begin{align*} u_\varepsilon v_\varepsilon - uv = (u_\varepsilon - u)v_\varepsilon + u(v_\varepsilon - v). \end{align*} The $L^p(\Omega')$ norm of the first term is bounded by \begin{align*} \|(u_\varepsilon - u)v_\varepsilon\|_{L^p(\Omega')} \le \|v_\varepsilon\|_{L^\infty(\Omega')}\, \|u_\varepsilon - u\|_{L^p(\Omega')} \le M\, \|u_\varepsilon - u\|_{L^p(\Omega')} \to 0 \end{align*} by property (ii) of the previous step and the $L^p$ convergence $u_\varepsilon \to u$ on $\Omega'$. The second term satisfies \begin{align*} \|u(v_\varepsilon - v)\|_{L^p(\Omega')} \le \|u\|_{L^\infty(\Omega')}\, \|v_\varepsilon - v\|_{L^p(\Omega')} \le M\, \|v_\varepsilon - v\|_{L^p(\Omega')} \to 0. \end{align*} The triangle inequality gives $\|u_\varepsilon v_\varepsilon - uv\|_{L^p(\Omega')} \to 0$. For the second claim, \begin{align*} u_\varepsilon\, \partial_{x_i} v_\varepsilon - u\, \partial_{x_i} v = (u_\varepsilon - u)\, \partial_{x_i} v_\varepsilon + u\, (\partial_{x_i} v_\varepsilon - \partial_{x_i} v). \end{align*} Now $\partial_{x_i} v_\varepsilon \to \partial_{x_i} v$ in $L^p(\Omega')$ (property (iii)), so $\|\partial_{x_i} v_\varepsilon\|_{L^p(\Omega')}$ is bounded uniformly in $\varepsilon$ — say by some constant $A$. Hölder's inequality with exponents $(\infty, p)$ gives \begin{align*} \|(u_\varepsilon - u)\, \partial_{x_i} v_\varepsilon\|_{L^p(\Omega')} \le \|u_\varepsilon - u\|_{L^\infty(\Omega')}\, \|\partial_{x_i} v_\varepsilon\|_{L^p(\Omega')}. \end{align*} This bound is unhelpful: $\|u_\varepsilon - u\|_{L^\infty}$ does not generally vanish. We use a different splitting. Hölder's inequality with exponents $(\infty, p)$ gives \begin{align*} \|(u_\varepsilon - u)\, \partial_{x_i} v_\varepsilon\|_{L^p(\Omega')} = \left( \int_{\Omega'} |u_\varepsilon - u|^p\, |\partial_{x_i} v_\varepsilon|^p\, d\mathcal{L}^n \right)^{1/p} \le \|u_\varepsilon - u\|_{L^\infty(\Omega')}\, \|\partial_{x_i} v_\varepsilon\|_{L^p(\Omega')}, \end{align*} which is bounded but does not go to zero. Reorganise: since $|u_\varepsilon - u| \le |u_\varepsilon| + |u| \le 2M$ a.e. on $\Omega'$ and $u_\varepsilon \to u$ in $L^p(\Omega')$, by passing to a subsequence $\varepsilon_k \to 0$ we may assume $u_{\varepsilon_k} \to u$ a.e. on $\Omega'$. Then $|u_{\varepsilon_k} - u|^p \to 0$ a.e. and $|u_{\varepsilon_k} - u|^p \le (2M)^p$ a.e. on $\Omega'$. Together with $|\partial_{x_i} v_{\varepsilon_k}|^p \to |\partial_{x_i} v|^p$ in $L^1(\Omega')$ (which follows from $\partial_{x_i} v_{\varepsilon_k} \to \partial_{x_i} v$ in $L^p$), we apply the Generalised Dominated Convergence Theorem (a.k.a. Vitali) to the integrand $|u_{\varepsilon_k} - u|^p\, |\partial_{x_i} v_{\varepsilon_k}|^p$: it converges to zero a.e., and the dominating sequence $(2M)^p\, |\partial_{x_i} v_{\varepsilon_k}|^p$ converges in $L^1$ to $(2M)^p|\partial_{x_i} v|^p$. Hence \begin{align*} \int_{\Omega'} |u_{\varepsilon_k} - u|^p\, |\partial_{x_i} v_{\varepsilon_k}|^p\, d\mathcal{L}^n \to 0. \end{align*} This shows $(u_{\varepsilon_k} - u)\, \partial_{x_i} v_{\varepsilon_k} \to 0$ in $L^p(\Omega')$ along the subsequence. The second piece $u\, (\partial_{x_i} v_\varepsilon - \partial_{x_i} v)$ is handled by Hölder with exponents $(\infty, p)$: \begin{align*} \|u\, (\partial_{x_i} v_\varepsilon - \partial_{x_i} v)\|_{L^p(\Omega')} \le \|u\|_{L^\infty(\Omega')}\, \|\partial_{x_i} v_\varepsilon - \partial_{x_i} v\|_{L^p(\Omega')} \to 0 \end{align*} along the full family $\varepsilon \to 0$. Combining the two pieces along the subsequence, $u_{\varepsilon_k}\, \partial_{x_i} v_{\varepsilon_k} \to u\, \partial_{x_i} v$ in $L^p(\Omega')$. Since the limit is uniquely determined and any subsequence of the original family $\varepsilon \to 0$ has a further subsequence with this convergence, the full family $u_\varepsilon\, \partial_{x_i} v_\varepsilon \to u\, \partial_{x_i} v$ in $L^p(\Omega')$. The third claim, $v_\varepsilon\, \partial_{x_i} u_\varepsilon \to v\, \partial_{x_i} u$ in $L^p(\Omega')$, is identical with the roles of $u$ and $v$ exchanged.[/step]

[guided]We need three $L^p$ convergences. The first, $u_\varepsilon v_\varepsilon \to uv$, is immediate from the bilinear continuity of $L^p \times L^\infty \to L^p$. The second and third — products of a bounded function with a converging derivative — require care. **Bilinear continuity for $u_\varepsilon v_\varepsilon \to uv$.** Decompose \begin{align*} u_\varepsilon v_\varepsilon - uv = (u_\varepsilon - u)v_\varepsilon + u(v_\varepsilon - v). \end{align*} The $L^p(\Omega')$ norm of the first summand is, by Hölder with exponents $(p, \infty)$, \begin{align*} \|(u_\varepsilon - u)v_\varepsilon\|_{L^p(\Omega')} \le \|u_\varepsilon - u\|_{L^p(\Omega')}\, \|v_\varepsilon\|_{L^\infty(\Omega')} \le M\, \|u_\varepsilon - u\|_{L^p(\Omega')} \to 0, \end{align*} using the uniform $L^\infty$ bound from the previous step and the $L^p$ convergence from [Differentiation Through Convolution](/theorems/3096). The second summand likewise satisfies $\|u(v_\varepsilon - v)\|_{L^p} \le \|u\|_{L^\infty}\|v_\varepsilon - v\|_{L^p} \to 0$. Hence $u_\varepsilon v_\varepsilon \to uv$ in $L^p(\Omega')$ by the triangle inequality. **Why the same trick fails for $u_\varepsilon\, \partial_{x_i} v_\varepsilon$.** Decompose \begin{align*} u_\varepsilon\, \partial_{x_i} v_\varepsilon - u\, \partial_{x_i} v = (u_\varepsilon - u)\, \partial_{x_i} v_\varepsilon + u\, (\partial_{x_i} v_\varepsilon - \partial_{x_i} v). \end{align*} The second summand is harmless: $\|u(\partial_{x_i} v_\varepsilon - \partial_{x_i} v)\|_{L^p} \le \|u\|_{L^\infty}\|\partial_{x_i} v_\varepsilon - \partial_{x_i} v\|_{L^p} \to 0$. The first summand is the trouble: pairing $(u_\varepsilon - u)$ with $\partial_{x_i} v_\varepsilon$, neither factor is in $L^\infty$ (we have $u_\varepsilon - u \to 0$ in $L^p$ but not in $L^\infty$, and $\partial_{x_i} v_\varepsilon$ is in $L^p$ not $L^\infty$). Hölder with exponents $(p, \infty)$ would require $\|\partial_{x_i} v_\varepsilon\|_{L^\infty}$, which we do not have. **The fix: a.e. convergence plus uniform domination.** Since $u_\varepsilon \to u$ in $L^p(\Omega')$, there is a subsequence $u_{\varepsilon_k} \to u$ a.e. on $\Omega'$ (a standard consequence of $L^p$ convergence). Both sides are uniformly bounded by $M$ on $\Omega'$: \begin{align*} |u_{\varepsilon_k} - u| \le |u_{\varepsilon_k}| + |u| \le M + M = 2M \quad \text{a.e. on } \Omega'. \end{align*} Now apply the Generalised Dominated Convergence Theorem (Vitali's form): if $f_k \to f$ a.e. and there exist non-negative $g_k$ with $|f_k| \le g_k$ a.e. and $g_k \to g$ in $L^1$ with $g \ge |f|$, then $\int f_k \to \int f$. Here, set $f_k := |u_{\varepsilon_k} - u|^p\, |\partial_{x_i} v_{\varepsilon_k}|^p$, $f := 0$, $g_k := (2M)^p\, |\partial_{x_i} v_{\varepsilon_k}|^p$, and $g := (2M)^p|\partial_{x_i} v|^p$. We have $f_k \to 0$ a.e. (because $u_{\varepsilon_k} \to u$ a.e. on $\Omega'$ and the second factor is finite a.e.) and $g_k \to g$ in $L^1(\Omega')$ — this last is the statement that $\|\partial_{x_i} v_\varepsilon\|_{L^p}^p \to \|\partial_{x_i} v\|_{L^p}^p$ together with the $L^p$ convergence of $\partial_{x_i} v_\varepsilon$ to $\partial_{x_i} v$, which means $|\partial_{x_i} v_{\varepsilon_k}|^p \to |\partial_{x_i} v|^p$ in $L^1(\Omega')$. The conclusion is \begin{align*} \int_{\Omega'} |u_{\varepsilon_k} - u|^p\, |\partial_{x_i} v_{\varepsilon_k}|^p\, d\mathcal{L}^n \to 0, \end{align*} i.e., $(u_{\varepsilon_k} - u)\, \partial_{x_i} v_{\varepsilon_k} \to 0$ in $L^p(\Omega')$. **Subsequence to the full family.** Convergence $u_\varepsilon\, \partial_{x_i} v_\varepsilon \to u\, \partial_{x_i} v$ in $L^p(\Omega')$ follows by the standard "every subsequence has a further subsequence" trick: the limit candidate $u\, \partial_{x_i} v$ is fixed independent of the subsequence, and the argument above produces an a.e.-convergent further subsequence from any starting subsequence. Hence the full family converges. The third claim is the same with $u$ and $v$ exchanged.[/guided]

[step:Take the limit in the distributional identity to identify the weak derivative] In identity $(\ast)$ from the step "Apply the classical product rule pointwise on $\Omega'$", \begin{align*} \int_{\Omega'} (u_\varepsilon v_\varepsilon)\, \partial_{x_i}\varphi \, d\mathcal{L}^n = -\int_{\Omega'} \bigl(u_\varepsilon\, \partial_{x_i} v_\varepsilon + v_\varepsilon\, \partial_{x_i} u_\varepsilon\bigr) \varphi \, d\mathcal{L}^n, \end{align*} we let $\varepsilon \to 0$. Since $\partial_{x_i}\varphi \in L^{p'}(\Omega')$ (where $p'$ is the Hölder conjugate of $p$, with $p' = \infty$ when $p = 1$), and $u_\varepsilon v_\varepsilon \to uv$ in $L^p(\Omega')$ from the previous step, Hölder's inequality applied to the difference gives \begin{align*} \left| \int_{\Omega'} (u_\varepsilon v_\varepsilon - uv)\, \partial_{x_i}\varphi\, d\mathcal{L}^n \right| \le \|u_\varepsilon v_\varepsilon - uv\|_{L^p(\Omega')}\, \|\partial_{x_i}\varphi\|_{L^{p'}(\Omega')} \to 0, \end{align*} so the left-hand side converges to $\int_{\Omega'} (uv)\, \partial_{x_i}\varphi\, d\mathcal{L}^n$. The right-hand side converges similarly: using $\varphi \in L^{p'}(\Omega')$ and the $L^p$ convergences from the previous step, \begin{align*} \int_{\Omega'} \bigl(u_\varepsilon\, \partial_{x_i} v_\varepsilon + v_\varepsilon\, \partial_{x_i} u_\varepsilon\bigr) \varphi\, d\mathcal{L}^n \to \int_{\Omega'} \bigl(u\, \partial_{x_i} v + v\, \partial_{x_i} u\bigr) \varphi\, d\mathcal{L}^n. \end{align*} The limiting identity is \begin{align*} \int_{\Omega'} (uv)\, \partial_{x_i}\varphi \, d\mathcal{L}^n = -\int_{\Omega'} \bigl(u\, \partial_{x_i} v + v\, \partial_{x_i} u\bigr) \varphi \, d\mathcal{L}^n. \end{align*} Because $\varphi \in C_c^\infty(\Omega)$ was arbitrary with $\operatorname{supp}\varphi = K \subset\subset \Omega' \subset\subset \Omega$, and because both integrals over $\Omega'$ equal the same integrals over $\Omega$ (the integrand vanishes outside $\operatorname{supp}\varphi \subset \Omega'$), \begin{align*} \int_{\Omega} (uv)\, \partial_{x_i}\varphi \, d\mathcal{L}^n = -\int_{\Omega} \bigl(u\, \partial_{x_i} v + v\, \partial_{x_i} u\bigr) \varphi \, d\mathcal{L}^n. \end{align*} This is the distributional identity defining $u\, \partial_{x_i} v + v\, \partial_{x_i} u$ as the weak partial derivative $\partial_{x_i}(uv)$. Since both $uv \in L^p(\Omega)$ and $u\, \partial_{x_i} v + v\, \partial_{x_i} u \in L^p(\Omega)$ from the first step, $uv \in W^{1,p}(\Omega)$ and $\partial_{x_i}(uv) = u\, \partial_{x_i} v + v\, \partial_{x_i} u$. As $i \in \{1, \dots, n\}$ was arbitrary, this completes the proof. [/step]