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The definition of broken Sobolev space is: Let $\Omega$ be the domain and $\mathcal{E}_h$ be the finite number of partitions of $\Omega$. For any real number $s \geq 0$, we denote the broken Sobolev space \begin{equation} H^s\left(\mathcal{E}_h\right)=\left\{\mathsf{u} \in L^2(\Omega): ~\text{for all}~ E \in \mathcal{E}_h,~\left.\mathsf{u}\right|_E \in H^s(E)\right\}, \end{equation} equipped with the broken Sobolev norm: \begin{equation} \|\mathsf{u}\|\strut_{H^s\left(\mathcal{E}_h\right)}=\left(\sum_{E \in \mathcal{E}_h}\|\mathsf{u}\|\strut_{H^s(E)}^2\right)^{1/2}. \end{equation} Here, $\|\mathsf{u}\|\strut_{H^s(E)}^2$ is a Sobolev norm on an element E of the domain $\Omega$.

Please give some ideas with an explanation, how I can prove or disprove.

Sachin
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  • By "the finite number of partitions" do you mean "a partition of $\Omega$ in a finite number of open sets", or maybe something else entirely? Also I'm assuming the $I$ is supposed to be a $E$. – Bruno B May 23 '23 at 07:51
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    If the "broken parts" are regular enough (say Lipschitz) then I see no difference between $H^s(\mathcal{E}_h)$ and just the Cartesian product of the $H^s(E)$ for $E \in \mathcal{E}_h$. So it is complete. – cs89 May 23 '23 at 09:11
  • @BrunoB The domain $\Omega$ is bounded and we are discretizing domain into the finite elements and $\mathcal{E}_h$ be the collection of all that elements. Yes, $I$ I wrote by mistake, now I have corrected it, thanks for that. – Sachin May 23 '23 at 09:39
  • @cs89 There is a jump in between the elements. – Sachin May 23 '23 at 09:41
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    What do you mean by "a jump"? Say $\Omega = (0,2)$ and $\mathcal{E}_h = { E_1, E_2 }$ where $E_1 = (0,1)$ and $E_2 = (1,2)$. Take $u_1 \in H^s(E_1)$ and $u_2 \in H^s(E_2)$. Their concatenation belongs to $H^s(\mathcal{E}_h)$ as you define it. The fact that you may have $u_1(1) \neq u_2(1)$ does not matter much as you are only asking for $u \in L^2(\Omega)$. – cs89 May 23 '23 at 10:29
  • @cs89 understood..thanks for the clarification. – Sachin May 23 '23 at 11:06

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