Is broken Sobolev space a Sobolev space?

Question

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.

Answer 1

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.

Answer 2

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) $ $$ \sum_{K \in \Omega_h}\int_{\partial K} u (\sigma \cdot n)dS = 0.$$

See, for example, Theorem 2.3 and Remark 2.5 in https://doi.org/10.1016/j.camwa.2016.05.004