Weak differentiation has a closed graph on its natural domains

By [F2] and [F4], weak-derivative value classes are unique and weak differentiation is linear. Thus $W^{1,p}(\mathrm{\Omega };\mathbb{K})$ and each maximal domain $V_i(\mathrm{\Omega })$ are linear subspaces of $L^p(\mathrm{\Omega };\mathbb{K})$, and the full-gradient and partial-derivative graphs are well-defined linear graphs in the stated normed spaces

Suppose $(u_j,D{u}_j)$ in the full-gradient graph converges to $(u,v)$ in $L^p(\mathrm{\Omega };\mathbb{K})\times (L^p(\mathrm{\Omega };\mathbb{K}){)}^n$ with the maximum product norm. Each coordinate derivative converges in $L^p$, hence locally in $L^p$; applying [F3] with $\alpha =e_i$ and $q=p$ gives $D_{i}u=v_i$ for every $i$. Therefore $u\in W^{1,p}$ and $Du=v$, so the graph contains every convergent limit

Fix $i$ and suppose $(u_j,D_i{u}_j)$ in the maximal graph converges to $(u,v)$ in $L^p(\mathrm{\Omega };\mathbb{K}{)}^2$. The same local convergence and [F3] give $D_{i}u=v$ weakly; because $u,v\in L^p$, $u\in V_i$ and the limit pair is $(u,D_{i}u)$. Thus this graph is sequentially closed

If $\mathrm{\Omega }=\mathrm{\varnothing }$, the spaces in both graph assertions contain only the zero class; for any $\mathrm{\Omega }$, the zero class has zero weak derivatives by the test identity. Hence both graphs contain their zero pairs and, on the empty domain, are closed

Suppose $n\ge 2$, choose distinct coordinates $i,k$, and set $Q=(-1,1{)}^n$

For finite $p$, define $u(x)={\mathbf{1}}_{(0,1)}(x_k)$ and $u_j(x)=\sigma (j{x}_k)$, using the coordinates and cube fixed in step 1.5. Each $u_j$ is smooth and bounded with bounded first partials for fixed $j$, so it belongs to $W^{1,p}(Q)$ and $D_i{u}_j=0$. It converges pointwise to $u$ and is bounded by $1$, so [F6] and [F7] give $u_j\to u$ in $L^p(Q)$; consequently $(u_j,D_i{u}_j)\to (u,0)$ in the ambient graph space

For $p=\mathrm{\infty }$, put $w(x)=\mathrm{\mid }{x}_k{\mathrm{\mid }}^{1\mathrm{/}2}$ and $w_j(x)=(x_k^2+1\mathrm{/}(j+1){)}^{1\mathrm{/}4}$ for $j\in \mathbb{N}$, using step 1.5. Each $w_j$ is smooth with bounded first partials for fixed $j$ and $D_i{w}_j=0$. Since $0\le (x_k^2+1\mathrm{/}(j+1){)}^{1\mathrm{/}4}-\mathrm{\mid }{x}_k{\mathrm{\mid }}^{1\mathrm{/}2}\le (j+1{)}^{-1\mathrm{/}4}$, $w_j\to w$ in $L^{\mathrm{\infty }}(Q)$ and $(w_j,D_i{w}_j)\to (w,0)$. If $w\in W^{1,\mathrm{\infty }}(Q)$, then on the half-cube $x_k>0$, [F4] and [F9] identify its weak $k$th derivative with $1\mathrm{/}(2\sqrt{x_k})$. For every $M>0$, choose $0<\delta <\min \{1,1\mathrm{/}(4M^2)\}$; this derivative exceeds $M$ on the positive-measure box $0<x_k<\delta$, $x_{\widehat{k}}\in (-1\mathrm{/}2,1\mathrm{/}2{)}^{n-1}$ by [F6]. Thus $w\mathrm{\notin }{W}^{1,\mathrm{\infty }}(Q)$ and the displayed limit point lies outside the restricted graph

For finite $p$, the limit $u$ of step 2.1 is not in $W^{1,p}(Q)$. If it had weak derivative $D_{k}u=h\in L^p(Q)$, then [F6] gives $h\in L^1(Q)$. Choose a nonnegative smooth bump $\rho \in C_c^{\mathrm{\infty }}((-1\mathrm{/}2,1\mathrm{/}2{)}^{n-1})$ equal to $1$ on a smaller ball, with $m=\int \rho >0$ by [F8], and choose a smooth bump $\eta \in C_c^{\mathrm{\infty }}((-1\mathrm{/}2,1\mathrm{/}2))$ with $0\le \eta \le 1$ and $\eta (0)=1$. The test ${\phi }_{\epsilon }(x)=\eta (x_k\mathrm{/}\epsilon )\rho (x_{\widehat{k}})$ is supported in $Q$ for $0<\epsilon <1\mathrm{/}2$. Fubini and Newton–Leibniz [F8] give ${\int }_{Q}u{\mathrm{\partial }}_k{\phi }_{\epsilon }dx=m(\eta (1\mathrm{/}\epsilon )-\eta (0))=-m$, so the weak identity forces $\mathrm{\mid }{\int }_{Q}h{\phi }_{\epsilon }dx\mathrm{\mid }=m$. Its support lies in $A_{\epsilon }=\{\mathrm{\mid }{x}_k\mathrm{\mid }<\epsilon \mathrm{/}2\}\cap ((-1\mathrm{/}2,1\mathrm{/}2{)}^{n-1})$, whose measure tends to zero; [F6] gives ${\int }_{A_{\epsilon }}\mathrm{\mid }h\mathrm{\mid }\to 0$, while $\mathrm{\mid }{\phi }_{\epsilon }\mathrm{\mid }\le \mathrm{\parallel }\rho {\mathrm{\parallel }}_{\mathrm{\infty }}$. This contradiction proves $u\mathrm{\notin }{W}^{1,p}(Q)$ and, with step 2.1, that the restricted graph is not closed for finite $p$

The finite-p counterexample is established by steps 2.1 and 3.1, and the $p=\mathrm{\infty }$ counterexample by step 2.2: for each $n\ge 2$ and coordinate $i$, the distinct $k$ fixed in step 1.5 gives a limit point outside $W^{1,p}(Q)$, so the restricted graph is not closed. Steps 1.2 and 1.3 prove closedness of the full-gradient and maximal partial-derivative graphs. When $n=1$, $W^{1,p}$ requires only the sole derivative $D_{1}u\in L^p$, hence $V_1=W^{1,p}$ and the restricted graph is the maximal closed graph from step 1.3. Step 1.4 handles the zero pairs and empty domain; step 2.1 includes $p=1$, and step 2.2 treats $p=\mathrm{\infty }$. There is no biconditional assertion.