# Is broken Sobolev space a Sobolev space?

The definition of a broken Sobolev space is as follows. Given infinite-dimensional (but mesh-dependent) spaces on an open bounded domain $$\Omega \in R^3$$ with Lipschitz boundary. The mesh, denoted by $$\Omega_h$$, is a disjoint partitioning of $$\Omega$$ into open elements $$K$$ such that the union of their closures is the closure of $$\Omega$$. The collection of element boundaries $$\partial K$$ for all $$K \in \Omega_h$$, is denoted by $$\partial \Omega_h$$. We assume that each element boundary $$\partial K$$ is Lipschitz. The shape of the elements is otherwise arbitrary for now. The broken Sobolev space is defined as $$\hat{H}^{1}\left(\Omega_{h}\right)=\left\{u \in L^{2}(\Omega):\left.u\right|_{K} \in H^{1}(K), K \in \Omega_{h}\right\},$$ where $$H^1$$ is the standard Sobolev space. According to my understanding, if $$\Omega_h$$ can be partitioned into a finite number of subdomains, $$K$$'s, and because the union of the intersections of $$K$$'s is a set of measure zero, then $$\forall u \in \hat{H}^1, \int_{\Omega_h} \sum_{|\alpha|\leq 1} (D^\alpha u)^2 dx = \sum_{K=1}^n \int_K \sum_{|\alpha|\leq 1} (D^\alpha u)^2 dx$$. This seems mean that $$\hat{H}^{1}$$ is also a Sobolev space.

If the partion of $$\Omega_h$$ consists of an infinite number of $$K$$'s, then the equality of the integration above would not hold. In this case, $$\hat{H}^{1}$$ is not a Sobolev space.

Am I right about the above two statements? Thanks a lot.

• The definition is not clear: is the subdivision of $\Omega_h$ be given before, or is the definition: if $u$ in the broken space then there is a corresponding partition?
– daw
Commented Feb 5, 2021 at 7:14
• @daw Thank you very much. I made my question clearer now. Please let me know if there are other things confusing.
– Jeff
Commented Feb 5, 2021 at 16:32

For the first question, $$\hat{H}^1$$ is not a Sobolev space, since the functions in $$\hat{H}^1$$ do not have a weak derivative belonging to any Lebesgue space, e.g. $$H^1$$ is the space of functions in $$L^2$$ with weak derivatives in $$L^2$$. A broken Sobolev space simply means that the functions belongs to Sobolev spaces on each individual element.
The second question is very interesting, and the paper: https://arxiv.org/abs/2006.07215 considers the notion of limiting broken Sobolev spaces (see the Definition 4.1). Here, the infinite partition $$\Omega_\infty$$ is obtained by iteratively refining the mesh of the domain, so that one does obtain an infinite collection of elements. The corresponding limit space is not a classical Sobolev space (unless the refinement of the domain eventually refines everywhere), but it does have similar characteristics.
In general, the answer is no. Even for Lipchitzs polygonal domains with a finite partition $$\Omega_h$$ the broken space $$H^1(\Omega_h)$$ contains $$H^1(\Omega)$$. But the reverse is false.
To have a gradient in $$L_2$$. We need some conditions over traces and jumps. \begin{align*} \nabla u (\sigma) &=-\int_\Omega u~\mathrm{div} \sigma \\ &=- \sum_{K \in \Omega_h} \int_{K} u~\mathrm{div} \sigma \\ &= \sum_{K \in \Omega_h} \left( \int_{K} \nabla (u |_K) \sigma ~ - \int_{\partial K} u (\sigma \cdot n)dS \right) \end{align*} Hence for any $$u \in H^1(\Omega_h)$$, we have $$u \in H^1(\Omega)$$ if only if for any $$\sigma \in L_2(\Omega)$$ with $$\mathrm{div}~\sigma \in L_2(\Omega)$$ $$\sum_{K \in \Omega_h}\int_{\partial K} u (\sigma \cdot n)dS = 0.$$