Let
- $d\in\mathbb N$
- $\Omega\subseteq\mathbb R^d$ be open
- $\langle\;\cdot\;,\;\cdot\;\rangle$ denote the $L^2(\Omega)$- or $L^2(\Omega,\mathbb R^d)$-inner product (depending on the context)
- $\mathcal D:=C_c^\infty(\Omega)$ and $$H:=\overline{\mathcal D}^{\langle\;\cdot\;,\;\cdot\;\rangle_H}$$ with $$\langle\phi,\psi\rangle_H:=\langle\phi,\psi\rangle+\langle\nabla\phi,\nabla\psi\rangle\;\;\;\text{for }\phi,\psi\in\mathcal D$$
Let $$\frac{\partial p}{\partial x_i}(\phi):=-p\left(\frac{\partial\phi}{\partial x_i}\right)\;\;\;\text{for }\phi\in\mathcal D$$ and $\nabla p:=\left(\frac{\partial p}{\partial x_1},\ldots,\frac{\partial p}{\partial x_d}\right)^T$ for $p\in\mathcal D'$ and $$(\nabla\cdot p)(\phi):=\sum_{i=1}^d\frac{\partial p_i}{\partial x_i}(\phi)\;\;\;\text{for }\phi\in\mathcal D$$ for $p\in(\mathcal D')^d$. Then, $$\Delta p:=\nabla\cdot\nabla p$$ is the distributional Laplacian of $p\in\mathcal D$ and it's easy to see that $$(\Delta p)(\phi)=p(\Delta\phi)\;\;\;\text{for all }\phi\in\mathcal D\;.\tag 1$$
Now, each $f\in L_{\text{loc}}^1(\Omega)$ can be identified with some unique $\langle f\rangle\in\mathcal D'$ via $$\langle f\rangle:=\left.\langle\;\cdot\;,f\rangle\right|_{\mathcal D}\;.$$ Let $\phi\in\mathcal D$. We can show that $\Delta\langle\phi\rangle$ has a unique extension $F\in H'$ with $$F(v)=-\langle\nabla v,\nabla\phi\rangle\;\;\;\text{for all }v\in H\;.\tag 2$$ We continue to denote $F$ by $\Delta\langle\phi\rangle$. In the same way $\Delta\langle\;\cdot\;\rangle:\mathcal D\to H'$ has a unique extension $L\in\mathfrak L(H,H')$ with $$(Lu)v=-\langle\nabla v,\nabla u\rangle\;\;\;\text{for all }u,v\in H\;.\tag 3$$
If $u\in L_{\text{loc}}^1(\Omega)$ is weakly differentiable, then $v\in L_{\text{loc}}^1(\Omega)$ is called weak Laplacian of $u$, if $$\langle\phi,v\rangle=-\langle\nabla u,\nabla\phi\rangle\;\;\;\text{for all }\phi\in\mathcal D\tag 4\;.$$ In that case, we write $\Delta u:=v$ and $$(\Delta\langle u\rangle)\psi=\langle\Delta u\rangle(\psi)=\langle\psi,\Delta u\rangle=-\langle\nabla\psi,\nabla u\rangle=(Lu)\psi\;\;\;\text{for all }\psi\in\mathcal D\;.\tag 5$$
If each $u\in H$ would admit a weak Laplacian $\Delta u$, it would make sense to continue to denote $L$ by $\Delta\langle\;\cdot\;\rangle$ and the relationship to the weak Laplacian would be clear. So, does each $u$ admit a weak Laplacian in the sense of $(4)$?