Two-dimensional scaled averaging energy bound

Fix an open edge $E_{ij}$ and let $U_i=H_i{w}_i$, $U_j=H_j{w}_j$ be the local discrete harmonic extensions. Write $e_i=w_i^T{S}_i{w}_i={\rho }_i\mathrm{\parallel }\mathrm{\nabla }{U}_i{\mathrm{\parallel }}_{L^2({\mathrm{\Omega }}_i)}^2$ and similarly for $e_j$, by harmonic minimality and the definition of $S_i$. The $i$-copy of $Pw$ has nodal values ${\alpha }_j(w_i-w_j)$ at interior edge nodes and zero at its primal endpoints, so it is $I_h({\theta }_E{\alpha }_j(w_i-w_j))$; the $j$-copy has the complementary coefficient ${\alpha }_i$. The $00,E$ norm controls the full boundary seminorm of an edge trace extended by zero off $E$: the cross-edge integrals are bounded by the endpoint weight in that norm, with constants controlled by the model's scaled Lipschitz boundary-chart bounds, polygon angles and side count. Thus the trace-Schur upper bound [F3] applies to these supported traces.

Set $m_i={\bar{w}}_{i,E}$, $m_j={\bar{w}}_{j,E}$, and $d_E=m_i-m_j$. At a primal endpoint $v$, $w_i(v)=w_j(v)$, so $\mathrm{\mid }{d}_E\mathrm{\mid }\le \mathrm{\mid }{m}_i-w_i(v)\mathrm{\mid }+\mathrm{\mid }{w}_j(v)-m_j\mathrm{\mid }$. Applying the discrete Sobolev estimate [F4] to $U_i$ and $U_j$, and noting that a boundary mean differs from the domain mean by at most the same $L^{\mathrm{\infty }}$ deviation, gives $\mathrm{\mid }{d}_E{\mathrm{\mid }}^2\le C(1+\log (H\mathrm{/}h))(\frac{e_i}{{\rho }_i}+\frac{e_j}{{\rho }_j})\mathrm{.}$ The factors $1\mathrm{/}{\rho }_i$ and $1\mathrm{/}{\rho }_j$ appear because [F4] is unweighted while $S_i$ includes ${\rho }_i$.

Write $z_i=w_i-m_i$ and $z_j=w_j-m_j$ on $E$, and set $L=1+\log (H\mathrm{/}h)$. The multiplier estimate [F4] bounds the squared $00,E$ norm of each $I_h({\theta }_E{z}_k)$ by $C{L}^2{e}_k\mathrm{/}{\rho }_k$. Also [F4] bounds $\mathrm{\parallel }{\theta }_E{\mathrm{\parallel }}_{00,E}^2$ by $CL$. For the $i$-copy use $w_i-w_j=z_i-z_j+d_E$ and the triangle inequality in the squared norm. After multiplication by its Schur coefficient ${\rho }_i{\alpha }_j^2$, the mean-zero terms are bounded by $C{L}^2({\alpha }_j^2{e}_i+\frac{{\rho }_i{\alpha }_j^2}{{\rho }_j}{e}_j),$ and the constant term has the same bound by step 1.2. Since $\frac{{\rho }_i{\alpha }_j^2}{{\rho }_j}=\frac{{\rho }_i{\rho }_j}{({\rho }_i+{\rho }_j{)}^2}\le \frac{1}{4},{\alpha }_j^2\le 1,$ the $i$-side contribution is at most $C{L}^2(e_i+e_j)$. Interchanging $i,j$ gives the same estimate for the $j$-side contribution, using ${\rho }_j{\alpha }_i^2\mathrm{/}{\rho }_i\le 1\mathrm{/}4$.

Sum over all interface sides. Each polygon has uniformly bounded side count, so each local energy occurs a bounded number of times. On each subdomain the complementary trace is the sum of these edge-supported traces; the squared norm of that sum is bounded by the side count times the sum of the squared norms. The resulting bound is $C{L}^2{\sum }_i{e}_i=C{L}^2\mathrm{\parallel }w{\mathrm{\parallel }}_{\tilde{S}}^2$, with no coefficient-contrast or subdomain-count factor.