Divergence operator: is it a linear continuous map on Sobolev Spaces?

Question

Let $\Omega $ be a bounded open set in $\mathbb{R^n}$

Let $H_{}^{1}(\Omega)=\{v\in L^2(\Omega) ; \frac{\partial v}{\partial x_i}\in L^2(\Omega)\}$

and $H_{0}^{1}(\Omega)=\{v\in H_{}^{1}(\Omega) ; v=0 \quad\text{on } \partial\Omega\}$ . Basically Sobolev Spaces with their norms.

I'm trying to prove that the subspace $V$ of the product space $(H_{0}^{1}(\Omega))^n$ is closed, with $$V=\{\vec v \in (H_{0}^{1}(\Omega))^n ; \sum_{i=1}^{n}\frac{\partial v_i}{\partial x_i}=0 \}$$

My idea : I defined the map $\psi :(H_{0}^{1}(\Omega))^n \rightarrow L^2(\Omega)$, such that $\psi(\vec v)=\sum_{i=1}^{n}\frac{\partial v_i}{\partial x_i}$ and I'm trying to prove that this map is a linear continuous map. The fact that $\psi$ is linear is obvious, but I'm stuck with continuity.

Answer 1

Show: $\psi:(H_0^1(\Omega))^n \to L^2(\Omega),\ v \mapsto \sum_{i=1}^n \partial_i v_i$ is continuous.

$$\|\psi(v)\|_{L^2}^2=\int_\Omega \Big|\sum_{i=1}^n \partial_i v_i(x) \Big|^2dx \leq C \sum_{i=1}^n \int_\Omega |\partial_i v_i(x)|^2 dx \leq C \sum_{i,j=1}^n \int_\Omega |\partial_i v_j(x)|^2dx=C\|v\|_{H_0^1}^2 $$

since

$$\|v\|_{H_0^1}^2\equiv\sum_{j=1}^n \int_\Omega |\partial_j v(x)|^2_{\mathscr{l}^2} dx =\sum_{j=1}^n \int_\Omega (|\partial_j v_1(x)|^2+\dots|\partial_j v_n(x)|^2) dx.$$

You can also include the $L^2$ of the function in the $H_0^1$ norm if you want to. These are equivalent, see here.