[proofplan] We prove the two directions separately. The forward direction represents each difference quotient as an integral of the gradient via the Fundamental Theorem of Calculus, then applies Jensen's inequality and Fubini's theorem to obtain the $L^p$ estimate; an approximation argument via Meyers-Serrin extends the result from smooth functions to $W^{1,p}(U)$. The reverse direction extracts a weak derivative from bounded difference quotients: weak sequential compactness of bounded sets in reflexive $L^p$ spaces ($1 < p < \infty$) yields weak limits on an exhaustion of $U$, a diagonal argument produces a consistent global function, the Monotone Convergence Theorem establishes its $L^p$ membership, and discrete integration by parts identifies it as the weak derivative. [/proofplan] [step:Represent difference quotients as integrals of the gradient for smooth functions] By the Meyers-Serrin Theorem, there exists a sequence $u_k \in C^\infty(U) \cap W^{1,p}(U)$ with $u_k \to u$ in $W^{1,p}(U)$. Fix $V \subset\subset U$, a direction $i \in \{1, \ldots, n\}$, and $h$ with $0 < |h| < \operatorname{dist}(V, \partial U)$. For $v \in C^\infty(U)$ and $x \in V$, the segment $\{x + the_i : t \in [0,1]\} \subset U$ since $|the_i| \le |h| < \operatorname{dist}(V, \partial U)$. The [Fundamental Theorem of Calculus](/theorems/632) gives: \begin{align*} v(x + he_i) - v(x) = \int_0^1 \frac{d}{dt}\, v(x + the_i) \, d\mathcal{L}^1(t) = h \int_0^1 \partial_i v(x + the_i) \, d\mathcal{L}^1(t). \end{align*} Dividing by $h$: \begin{align*} D_i^h v(x) = \int_0^1 \partial_i v(x + the_i) \, d\mathcal{L}^1(t). \end{align*} [guided] Why can we apply the Fundamental Theorem of Calculus? For $x \in V$, the map $t \mapsto v(x + the_i)$ for $t \in [0,1]$ is well-defined because $x + the_i \in U$ for all $t \in [0,1]$: indeed, $|the_i| = |t| \cdot |h| \le |h| < \operatorname{dist}(V, \partial U)$, so the segment stays within $U$. Since $v \in C^\infty(U)$, the map $t \mapsto v(x + the_i)$ is smooth and the Fundamental Theorem of Calculus applies: \begin{align*} v(x + he_i) - v(x) = \int_0^1 h\, \partial_i v(x + the_i) \, d\mathcal{L}^1(t). \end{align*} Dividing by $h$ yields the integral representation $D_i^h v(x) = \int_0^1 \partial_i v(x + the_i) \, d\mathcal{L}^1(t)$. We approximate in $W^{1,p}$ rather than working directly with $u$ because the Fundamental Theorem of Calculus requires pointwise differentiability, which $u \in W^{1,p}(U)$ need not have. The Meyers-Serrin Theorem guarantees smooth approximants $u_k \to u$ in $W^{1,p}(U)$, and we will pass the resulting estimate to the limit. [/guided] [/step] [step:Bound $\|D^h u_k\|_{L^p(V)}$ via Jensen's inequality and Fubini's theorem] Apply the integral representation to $u_k$. Since $t \mapsto |t|^p$ is convex for $p \ge 1$, Jensen's inequality applied to the probability measure $d\mathcal{L}^1(t)$ on $[0,1]$ gives: \begin{align*} |D_i^h u_k(x)|^p \le \int_0^1 |\partial_i u_k(x + the_i)|^p \, d\mathcal{L}^1(t). \end{align*} Integrating over $x \in V$ and applying Fubini's theorem (the integrand is non-negative, so Tonelli's theorem applies): \begin{align*} \int_V |D_i^h u_k(x)|^p \, d\mathcal{L}^n(x) \le \int_0^1 \int_V |\partial_i u_k(x + the_i)|^p \, d\mathcal{L}^n(x) \, d\mathcal{L}^1(t). \end{align*} For each $t \in [0,1]$, the substitution $y = x + the_i$ is a translation preserving Lebesgue measure: $d\mathcal{L}^n(x) = d\mathcal{L}^n(y)$. The image $V + the_i \subseteq U$ by the constraint on $|h|$. Enlarging the domain of integration from $V + the_i$ to $U$: \begin{align*} \int_V |\partial_i u_k(x + the_i)|^p \, d\mathcal{L}^n(x) \le \|\partial_i u_k\|_{L^p(U)}^p. \end{align*} Since this bound is independent of $t$, integrating over $[0,1]$ gives $\|D_i^h u_k\|_{L^p(V)}^p \le \|\partial_i u_k\|_{L^p(U)}^p$. Summing over $i = 1, \ldots, n$ and taking the $p$-th root: \begin{align*} \|D^h u_k\|_{L^p(V)} \le \|\nabla u_k\|_{L^p(U)}. \end{align*} [/step] [step:Pass to the limit $k \to \infty$ to establish the forward direction] The convergence $u_k \to u$ in $W^{1,p}(U)$ implies $u_k \to u$ in $L^p(U)$ and $\nabla u_k \to \nabla u$ in $L^p(U; \mathbb{R}^n)$. The difference quotient operator $D_i^h$ is a bounded linear map on $L^p$: for fixed $h$, \begin{align*} \|D_i^h u_k - D_i^h u\|_{L^p(V)} \le \frac{2}{|h|}\|u_k - u\|_{L^p(U)} \to 0. \end{align*} Taking $k \to \infty$ in the estimate from the previous step: \begin{align*} \|D^h u\|_{L^p(V)} = \lim_{k \to \infty} \|D^h u_k\|_{L^p(V)} \le \lim_{k \to \infty} \|\nabla u_k\|_{L^p(U)} = \|\nabla u\|_{L^p(U)}. \end{align*} [/step] [step:Extract weak limits on an exhaustion via weak compactness and diagonal extraction] Fix a direction $i \in \{1, \ldots, n\}$. Since $1 < p < \infty$, the space $L^p(V)$ is reflexive for any open $V \subset\subset U$. Choose a countable exhaustion $V_1 \subset\subset V_2 \subset\subset \cdots$ with $\bigcup_{m=1}^\infty V_m = U$. Fix $m \in \mathbb{N}$ and a sequence $h_j \to 0$ with $0 < |h_j| < \operatorname{dist}(V_m, \partial U)$ for all $j$. By hypothesis, $\|D_i^{h_j} u\|_{L^p(V_m)} \le C$ for all $j$. Since bounded sets in the reflexive space $L^p(V_m)$ are weakly sequentially compact (Banach-Alaoglu), there exist a subsequence and a function $w_m \in L^p(V_m)$ with $D_i^{h_{j_\ell}} u \rightharpoonup w_m$ weakly in $L^p(V_m)$. Apply the Cantor diagonal argument across $\{V_m\}_{m=1}^\infty$: extract a subsequence converging weakly on $V_1$, then a further subsequence on $V_2$, and so on. The diagonal subsequence (relabelled $h_k \to 0$) satisfies $D_i^{h_k} u \rightharpoonup w_m$ weakly in $L^p(V_m)$ for every $m$. The local weak limits are consistent: for $\phi \in L^q(V_m)$ (extended by zero to $V_{m+1}$), weak convergence on both $V_m$ and $V_{m+1}$ gives $\int_{V_m} w_{m+1}\, \phi \, d\mathcal{L}^n = \int_{V_m} w_m\, \phi \, d\mathcal{L}^n$ for all such $\phi$, hence $w_{m+1} = w_m$ $\mathcal{L}^n$-a.e. on $V_m$. Define $v_i: U \to \mathbb{R}$ by $v_i(x) := w_m(x)$ for $x \in V_m$; this is well-defined a.e. [guided] The reverse direction is the deeper result. Given bounded difference quotients, we need to produce a weak derivative. The strategy is: 1. Use weak compactness to extract weak limits on compact subdomains. 2. Paste together via a diagonal argument. 3. Show the resulting function is in $L^p(U)$. 4. Identify it as the weak derivative via discrete integration by parts. Why do we need $1 < p < \infty$? Because we invoke weak sequential compactness of bounded sets in $L^p(V_m)$. This requires $L^p(V_m)$ to be reflexive, which holds if and only if $1 < p < \infty$. For $p = 1$, $L^1$ is not reflexive: bounded sequences need not have weakly convergent subsequences. This is why the theorem fails for $p = 1$. Fix a direction $i$ and a countable exhaustion $V_1 \subset\subset V_2 \subset\subset \cdots$ with $\bigcup_m V_m = U$. Take any sequence $h_j \to 0$. On $V_1$, the sequence $D_i^{h_j}u$ is bounded in $L^p(V_1)$ by $C$, so Banach-Alaoglu gives a subsequence $h_{j}^{(1)}$ with $D_i^{h_j^{(1)}}u \rightharpoonup w_1$ in $L^p(V_1)$. On $V_2$, extract a further subsequence $h_j^{(2)} \subset h_j^{(1)}$ with $D_i^{h_j^{(2)}}u \rightharpoonup w_2$ in $L^p(V_2)$. Continue for all $m$. The diagonal sequence $h_k := h_k^{(k)}$ converges weakly on every $V_m$. Are the limits consistent? Yes: if $\phi \in L^q(V_m)$, extend $\phi$ by zero to $V_{m+1}$. Then \begin{align*} \int_{V_m} w_m\, \phi \, d\mathcal{L}^n = \lim_k \int_{V_m} D_i^{h_k}u\, \phi \, d\mathcal{L}^n = \lim_k \int_{V_{m+1}} D_i^{h_k}u\, \phi \, d\mathcal{L}^n = \int_{V_m} w_{m+1}\, \phi \, d\mathcal{L}^n. \end{align*} Since this holds for all $\phi \in L^q(V_m)$, we get $w_m = w_{m+1}$ a.e. on $V_m$, so we can define $v_i$ globally on $U$. [/guided] [/step] [step:Establish $v_i \in L^p(U)$ with $\|v_i\|_{L^p(U)} \le C$ via the Monotone Convergence Theorem] By lower semicontinuity of the norm under weak convergence, $\|w_m\|_{L^p(V_m)} \le \liminf_{k \to \infty} \|D_i^{h_k} u\|_{L^p(V_m)} \le C$ for every $m$. Define the non-negative measurable functions $g_m := |v_i|^p \cdot \mathbb{1}_{V_m}$. Since $V_m \subseteq V_{m+1}$ and $\bigcup_m V_m = U$, the sequence $(g_m)$ is non-decreasing and converges pointwise a.e. to $|v_i|^p$. By the Monotone Convergence Theorem: \begin{align*} \int_U |v_i|^p \, d\mathcal{L}^n = \lim_{m \to \infty} \int_{V_m} |v_i|^p \, d\mathcal{L}^n = \lim_{m \to \infty} \|v_i\|_{L^p(V_m)}^p \le C^p. \end{align*} Therefore $v_i \in L^p(U)$ with $\|v_i\|_{L^p(U)} \le C$. [/step] [step:Identify $v_i$ as the weak derivative $\partial_i u$ via discrete integration by parts] Let $\phi \in C_c^\infty(U)$. Choose $m$ large enough that $\operatorname{supp}(\phi) \subset V_m$. For $k$ large enough that $|h_k| < \operatorname{dist}(V_m, \partial U)$, perform discrete integration by parts. Substituting $y = x + h_k e_i$ (translation preserving $d\mathcal{L}^n$) in the first term: \begin{align*} \int_U (D_i^{h_k} u)(x)\, \phi(x) \, d\mathcal{L}^n(x) &= \int_U u(y)\, \frac{\phi(y - h_k e_i) - \phi(y)}{h_k} \, d\mathcal{L}^n(y) \\ &= -\int_U u(y)\, (D_i^{-h_k}\phi)(y) \, d\mathcal{L}^n(y). \end{align*} **Left side as $k \to \infty$:** Since $\phi \in L^q(V_m)$ and $D_i^{h_k}u \rightharpoonup v_i$ weakly in $L^p(V_m)$: \begin{align*} \lim_{k \to \infty} \int_U (D_i^{h_k}u)\, \phi \, d\mathcal{L}^n = \int_U v_i\, \phi \, d\mathcal{L}^n. \end{align*} **Right side as $k \to \infty$:** Since $\phi \in C_c^\infty(U)$, the Mean Value Theorem gives $|D_i^{-h_k}\phi(y) - \partial_i\phi(y)| \le |h_k| \cdot \|\partial_{ii}\phi\|_{L^\infty} \to 0$ uniformly. Since $u \in L^p(U)$ and $\operatorname{supp}(\phi)$ is compact, the Dominated Convergence Theorem gives: \begin{align*} \lim_{k \to \infty} \int_U u\, (D_i^{-h_k}\phi) \, d\mathcal{L}^n = \int_U u\, \partial_i\phi \, d\mathcal{L}^n. \end{align*} Combining: \begin{align*} \int_U v_i\, \phi \, d\mathcal{L}^n = -\int_U u\, \partial_i\phi \, d\mathcal{L}^n \quad \text{for all } \phi \in C_c^\infty(U). \end{align*} This is the definition of $v_i = \partial_i u$ in the weak sense. [guided] The key identity is a discrete analogue of integration by parts. Starting from the definition of $D_i^{h_k}u$: \begin{align*} \int_U \frac{u(x + h_k e_i) - u(x)}{h_k}\, \phi(x) \, d\mathcal{L}^n(x). \end{align*} In the first term, substitute $y = x + h_k e_i$, so $x = y - h_k e_i$ and $d\mathcal{L}^n(x) = d\mathcal{L}^n(y)$: \begin{align*} \frac{1}{h_k}\int_U u(y)\, \phi(y - h_k e_i) \, d\mathcal{L}^n(y) - \frac{1}{h_k}\int_U u(x)\, \phi(x) \, d\mathcal{L}^n(x). \end{align*} Combining (replacing $x$ by $y$ in the second integral): \begin{align*} \int_U u(y)\, \frac{\phi(y - h_k e_i) - \phi(y)}{h_k} \, d\mathcal{L}^n(y) = -\int_U u(y)\, (D_i^{-h_k}\phi)(y) \, d\mathcal{L}^n(y). \end{align*} This is the discrete integration by parts formula: the difference quotient moves from $u$ to $\phi$ at the cost of a sign change and a reversal of the shift direction. Now we pass to the limit. On the left, we use the weak convergence $D_i^{h_k}u \rightharpoonup v_i$ in $L^p(V_m)$ paired against $\phi \in L^q(V_m)$. On the right, $D_i^{-h_k}\phi \to \partial_i\phi$ uniformly (since $\phi$ is smooth), and $u \in L^p$ provides the integrability needed for the Dominated Convergence Theorem. The result is $\int_U v_i\, \phi \, d\mathcal{L}^n = -\int_U u\, \partial_i\phi \, d\mathcal{L}^n$, which is the defining relation for $v_i = \partial_i u$ weakly. [/guided] [/step] [step:Conclude $u \in W^{1,p}(U)$ with the norm bound] Since $v_i = \partial_i u \in L^p(U)$ for each $i \in \{1, \ldots, n\}$, we have $u \in W^{1,p}(U)$. From the $L^p$ bound established above, $\|\partial_i u\|_{L^p(U)} = \|v_i\|_{L^p(U)} \le C$ for each $i$, so $\|\nabla u\|_{L^p(U)} \le C$. [/step]