Let
- $d\in\mathbb N$
- $\Omega\subseteq\mathbb R^d$ be open
- $\mathcal D(\Omega):=C_c^\infty(\Omega)$
- $f\in L^2(\Omega)$ and $$\langle f\rangle:=\left.\langle\;\cdot\;,f\rangle_{L^2(\Omega)}\right|_{\mathcal D(\Omega)}\in\mathcal D'(\Omega)$$
- $i\in\left\{1,\ldots,d\right\}$
Since $\mathcal D(\Omega)$ is dense in the Sobolev space $H_0^1(\Omega)$, it's easy to see that the distributional derivative $$\frac{\partial\langle f\rangle}{\partial x_i}(\phi)=-\langle\frac{\partial\phi}{\partial x_i},f\rangle_{L^2(\Omega)}\;\;\;\text{for all }\phi\in\mathcal D(\Omega)\tag 1$$ has a unique extension $\overline{\frac{\partial\langle f\rangle}{\partial x_i}}\in H_0^1(\Omega)'$ to $H$ with $$\overline{\frac{\partial\langle f\rangle}{\partial x_i}}(u)=-\langle\frac{\partial u}{\partial x_i},f\rangle_{L^2(\Omega)}\;\;\;\text{for all }u\in H_0^1(\Omega)\;,\tag 2$$ where $\frac{\partial u}{\partial x_i}$ denotes the weak derivative. The mapping $$\overline{\frac\partial{\partial x_i}}:L^2(\Omega)\to H_0^1(\Omega)'\tag 3$$ is linear and continuous.
Now, I've read that we can define $$\Delta:=\sum_{j=1}^d\overline{\frac\partial{\partial x_i}}\:\overline{\frac\partial{\partial x_i}}\tag 4$$ as a continuous operator $H_0^1(\Omega)\to H_0^1(\Omega)'$. Why is the composition of the operators in $(4)$ well-defined? That's not clear to me, cause each operator is $H_0^1(\Omega)'$-valued and takes values from $L^2(\Omega)$.
I've noticed that $$\overline{\frac{\partial\langle f\rangle}{\partial x_i}}=\left.\langle\;\cdot\;,\frac{\partial f}{\partial x_i}\rangle_{L^2(\Omega)}\right|_{L^2(\Omega)}\;,$$ if $f\in H_0^1(\Omega)$. So, maybe $\overline{\frac{\partial\langle f\rangle}{\partial x_i}}$ is identified with the weak derivative $\frac{\partial f}{\partial x_i}$ (which is an element of $L^2(\Omega)$ in that case.
One can define $\Delta$ in $H^1$ because $\Delta$ is in divergence form
$$\Delta u = \text{div}\nabla u$$
Thus we define $\Delta : H^1_0(\Omega) \to H^1_0(\Omega)^*$ by
$$\Delta f (\phi) := -\int_\Omega \nabla f \cdot \nabla \phi.$$
The right hand side is well defined for $f , \phi \in H^1_0(\Omega)$. If $f$ is in $H^2_0$, then integration by part gives
$$\Delta f(\phi) = \int_\Omega (\Delta f) \phi$$