Let $U$ be a bounded domain of $\mathbb{R}^d$, and write $m$ for the Lebesgue measure on $U$. For $k=1,2$, we denote by $H^k(U)$ the set of all locally $m$-integrable functions $u\colon U \to \mathbb{R}$ such that for any multi-index $\alpha$ with $|\alpha|\le k$, the weak derivative $D^\alpha u$ exists and belongs to $L^2(U,m)$.

Define the Neumann Laplacian $(L,\text{Dom}(L))$ on $U$ by \begin{eqnarray*} \text{Dom}(L) & = & \{u \in H^1(U) : H^1(U) \ni v \mapsto \int_{U}\nabla u\cdot \nabla v\,dm \\ & & \text{ is continuous on $L^2(U,m)$}\} \\ -\int_{U}v Lu\,dm &= &\int_{U} \nabla u\cdot \nabla v\,dm,\quad u \in \text{Dom}(L),\,v \in H^1(U). \end{eqnarray*}

I know that if $U$ is a bounded $C^2$ domain, $\text{Dom}(L) \subset H^2(U)$ (see Section~10.6.2 in [1] ). Even if $U$ is a bounded Lipschitz domain, does this inclusion hold ? I don't think this is correct in general.

For example, if $U$ is a $C^{1,1}$ domain or convex domain, does this holds?

I would like to know various conditions for $U$ such that the inclusion $\text{Dom}(L) \subset H^2(U)$ holds?