Is the sum of two Sobolev spaces defined on two different sets the Sobolev space defined on the union of these two sets?

Question

Is it true that $H^1(\Omega_1 \cup \Omega_2 )=H^1(\Omega_1)+H^1(\Omega_2)$?

Below, we already have a counterexample. Let me ask further. If I impose $\Omega_1 \cap \Omega_2=\emptyset$, is there still a counter example?

Answer 1

Consider $\Omega_1 = (0,2)$ and $\Omega_2 = (1,3)$.

Then the constant functions $x\mapsto 1$ (one the respective intervals) are in $H^1(\Omega_i)$, but their sum (if extended by 0, otherwise, what do you mean by the sum?) is not continuous (has no continuous representative), hence is not in $H^1(\Omega_1 \cup \Omega_2)$.