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Let

  • $d\in\mathbb N$
  • $\lambda$ denote the Lebesgue measure on $\mathbb R^d$
  • $\Lambda\subseteq\mathbb R^d$ be open
  • $\mathcal D(\Lambda):=C_c^\infty(\Lambda)$ denote the test function space on $\Lambda$
  • $H_0^1(\Lambda)$ denote the Sobolev space $W_0^{1,\:2}(\Lambda)$ and $H^{-1}(\Lambda)$ denote the topological dual space of $H_0^1(\Lambda)$
  • $\eta_1,\ldots,\eta_d\in\mathcal D'(\Lambda)$ with $$\sum_{i=1}^d\langle\phi_i,\eta_i\rangle_{\mathcal D(\Lambda),\:\mathcal D'(\Lambda)}=0\;\;\;\text{for all }\phi\in\mathcal V:=\left\{\phi\in\mathcal D(\Lambda)^d:\nabla\cdot\phi=0\right\}\tag 1$$

We can show (see Temam, Proposition 1.1) that $$\eta_i=\frac{\partial p}{\partial x_i}\;\;\;\text{for all }i\in\left\{1,\ldots,d\right\}\tag 2$$ for some $p\in\mathcal D'(\Lambda)$. If "$\eta_1,\ldots,\eta_d\in H^{-1}(\Lambda)$", i.e. $\exists\overline\eta_1,\ldots,\overline\eta_d\in H^{-1}(\Lambda)$ with $$\left.\overline\eta_i\right|_{\mathcal D(\Lambda)}=\eta_i\;\;\;\text{for all }i\in\left\{1,\ldots,d\right\}\,,\tag 3$$ then "$p\in L_{\text{loc}}^2(\Lambda)$", i.e. $\exists f\in L^2_{\text{loc}}(\Lambda)$ with $$p(\psi)=\int_\Lambda\psi f\;{\rm d}\lambda\;\;\;\text{for all }\psi\in\mathcal D(\Lambda)\;,\tag 4$$ and hence $$\eta_i(\psi)=-\int_\Lambda\frac{\partial\psi}{\partial x_i}f\;{\rm d}\lambda\;\;\;\text{for all }i\in\left\{1,\ldots,d\right\}\text{ and }\psi\in\mathcal D(\Lambda)\;.\tag 5$$

Can we conclude that $$\overline\eta_i(u)=-\int_\Lambda\frac{\partial u}{\partial x_i}f\;{\rm d}\lambda\;\;\;\text{for all }i\in\left\{1,\ldots,d\right\}\text{ and }u\in H_0^1(\Lambda)\tag 6$$ or do we need $f\in L^2(\Lambda)$ in order for $(6)$ to hold?

I guess the latter is the case, because the integral on the right-hand side of $(6)$ is not well-defined, unless $f\in L^2(\Lambda)$.

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