I'm curious to know if the following is true. Suppose $\{\Omega_{\alpha}\}_{\alpha}$ is a family of smooth, bounded domains in $\mathbb{R}^n$ (or more generally $C^1$ domains) with $\Omega_{i}\cap\Omega_{j}\neq\phi$ for any $i,j$. For $s>0$, we have the well known trace inequality $$ \|u\|_{H^{s}(\partial\Omega_{\alpha})}\leq C_{\alpha}\|u\|_{H^{s+\frac{1}{2}}(\Omega_{\alpha})} $$ where $C_{\alpha}$ is a positive constant depending on $\Omega_{\alpha}$. Suppose further that the constants $C_{\alpha}$ are uniformly bounded in $\alpha$ by some constant $C_{max}$. Do we have a trace inequality of the form $$ \|u\|_{H^{s}(\partial(\Omega_{i}\cap\Omega_j))}\leq CC_{max}\|u\|_{H^{s+\frac{1}{2}}(\Omega_{i}\cap\Omega_j)} $$ for some range of $s>0$ and some universal constant $C$ independent of the family $\Omega_{\alpha}$? If this is not true, then are there conditions which would guarantee this? For instance, if the domains don't intersect "too transversally", is this true?
This latter question is motivated by the fact that (for domains which don't intersect "too transversally") the estimate is true for the non-sharp trace inequality $\|u\|_{L^2(\partial(\Omega_{i}\cap\Omega_j))}\leq C\|u\|_{H^1(\Omega_{i}\cap\Omega_j)}$. Roughly speaking, this is shown by taking a smooth vector field $X$ defined on a neighborhood of $\Omega_i\cup\Omega_j$ which is uniformly transversal to the boundary of $\Omega_i$ and $\Omega_j$ and using the divergence theorem on $|u_{|\partial(\Omega_{i}\cap\Omega_j)}|^2X\cdot n$ where $n$ is the outward unit normal of $\partial(\Omega_{i}\cap\Omega_j)$.