Show that the following quantity is super-additive over set. Let $\Omega$ be a bounded regular domain of $\mathbb{R}^3$ split into two disjoint subdomains $\Omega = \Omega_1 \cup \Omega_2$. Consider a source term $f \in L^{6/5}(\mathbb{R}^3$) and the following energy :
$$H(\Omega)= \min_{u \in H^1(\Omega)}\int_{\Omega} |\nabla u|^2 + |u|^2 - f u \ dx.$$
We will denote $F_\Omega(u):= \int_{\Omega} |\nabla u|^2 + |u|^2 - f u \ dx.$
This energy corresponds to the following well-posed PDE :
$$- \Delta u + u =f \text{ on } \Omega \text{ with } \partial_n u =0 \text{ on } \partial \Omega \quad \quad \quad(\star)$$
I would like to prove the following super-additive property over set :
$$H(\Omega) \geq H(\Omega_1) + H(\Omega_2).$$
I tried the following : let $\bar{u}$ the solution of $(\star)$, then split it into $\bar{u}= u_1 +u_2$ with $u_1 := u|_{\Omega_1}$ and $u_2:=u|_{\Omega_2}$. Then I would compute $F_\Omega(u)$ and try to make appear $F_{\Omega_1}(u_1)$ and $F_{\Omega_2}(u_2)$ but it didn't work like I expected.
Do you have any ideas on how to proceed ? Feel free to ask questions, thanks for your help.
 A: I will ignore the boundary conditions since Neumann boundary conditions are ill-defined for $u \in H^1(\Omega)$ (not up the boundary). Recovering boundary conditions will require some additional work and probably additional regularity and boundary regularity assumptions on $u$ and $\partial \Omega \cap  \partial \Omega_{1,2}$ respectively.
Define
$$
U=\{u:\Omega \mapsto \mathbb{R}| \; u|_{\Omega_1} \in H^1(\Omega_1), u|_{\Omega_2} \in H^1(\Omega_2)   \}
$$
Clearly, we have $H^1(\Omega) \subset U$, since restricting a Sobolev function to a smaller domain makes it a Sobolev function on a smaller domain - exactly what is appearing in the definition of $U$.
On the other hand:
$$
H(\Omega)\geq\min_{u \in U} \big(F_{\Omega_1}(u|_{\Omega_1})+F_{\Omega_2}(u|_{\Omega_2})\big)\geq H(\Omega_1)+H(\Omega_2)
$$
Those estimates just boil down to the basic properties of the minimum.
$$
\min_{a \in A} f(a) \geq \min_{a \in A'} f(a)   \; \; \; A \subset A'
$$
and
$$
\min_{a \in A}(f(a)+g(a)) \geq \min_{a \in A} f(a)+\min_{a \in A} g(a)
$$
On a sidenote: Your assumptions ensure the minimum exists.
