Discrete Schur energy and the fractional trace norm

On the upper half-plane, Fourier transformation in the tangential variable gives, for a smooth lift $V$ with trace $g$, $\mathrm{\mid }\xi \mathrm{\mid }\mathrm{\mid }\widehat{g}(\xi ){\mathrm{\mid }}^2\le {\int }_0^{\mathrm{\infty }}(\mathrm{\mid }{\mathrm{\partial }}_t\widehat{V}(\xi ,t){\mathrm{\mid }}^2+\mathrm{\mid }\xi {\mathrm{\mid }}^2\mathrm{\mid }\widehat{V}(\xi ,t){\mathrm{\mid }}^2)dt\mathrm{.}$ Indeed, integrate $-{\mathrm{\partial }}_t\mathrm{\mid }\widehat{V}{\mathrm{\mid }}^2$ and apply $2ab\le a^2+b^2$. Integration in $\xi$ proves the trace seminorm bound. Conversely, $\widehat{V}(\xi ,t)=e^{-t\mathrm{\mid }\xi \mathrm{\mid }}\widehat{g}(\xi )$ has gradient energy $\int \mathrm{\mid }\xi \mathrm{\mid }\mathrm{\mid }\widehat{g}(\xi ){\mathrm{\mid }}^{2}d\xi$. In a bounded chart, multiply the half-plane extension by a fixed cutoff in the inward coordinate; the added gradient term is controlled by the localized $L^2$ boundary norm, hence by the full H1/2 norm after constants are removed. A finite collection of Lipschitz graph charts covers the polygon boundary, including its corners. Flattening each chart is bi-Lipschitz, so it changes the $H^1$ gradient and boundary Gagliardo seminorm by bounded factors. Applying a partition of unity and the half-plane trace inequality to arbitrary chart-localized lifts gives $\mathrm{\mid }g{\mathrm{\mid }}_{1\mathrm{/}2,\mathrm{\partial }{\mathrm{\Omega }}_i}\le C\mathrm{\parallel }V{\mathrm{\parallel }}_{H^1({\mathrm{\Omega }}_i)}$. Subtract the domain mean of $V$ and use scaled Poincar'e to replace the full $H^1$ norm by $\mathrm{\parallel }\mathrm{\nabla }V{\mathrm{\parallel }}_{L^2}$. In the other direction, apply a partition of unity to boundary data, use the half-plane extensions, and sum to obtain a right inverse of the trace. First subtract a boundary constant; the fractional Poincaré bound on the whole boundary controls the localized full norms: for the boundary mean $c$, $\mathrm{\parallel }g-c{\mathrm{\parallel }}_{L^2(\mathrm{\partial }{\mathrm{\Omega }}_i)}^2\le H_i^2\mathrm{\mid }g{\mathrm{\mid }}_{1\mathrm{/}2,\mathrm{\partial }{\mathrm{\Omega }}_i}^2\mathrm{/}(2\mathrm{\mid }\mathrm{\partial }{\mathrm{\Omega }}_i\mathrm{\mid })$, by integrating $\mathrm{\mid }s-t\mathrm{\mid }\le H_i$. Add that constant back as a constant lift, which has zero gradient. The chart count, scaled radii, Lipschitz constants and mean-zero Poincaré bounds are precisely the fixed shape parameters in [F1]; a pointwise interior-cone condition alone would not justify this uniformity.

Construct an interpolant $I_h:H^1({\mathrm{\Omega }}_i)\to V_h$ as follows. For each interior mesh vertex choose one incident triangle and set its coefficient to the average of $V$ on that triangle. For each boundary vertex choose one incident boundary edge and set its coefficient to the integral of the trace of $V$ against the $L^2$-dual function to the endpoint nodal basis on that edge. These dual functionals reproduce every affine edge trace, so the trace of $I_{h}V$ is exactly $g$ whenever the trace of $V$ is $g$. All coefficients reproduce constants. For a triangle $K$, the chosen elements and edges for its vertices lie in a connected patch ${\omega }_K$ of diameter at most $Ch$ containing a uniformly bounded number of triangles; these patches have bounded overlap. Let $c_K$ be the mean of $V$ on ${\omega }_K$. Scaled Poincaré gives $\mathrm{\parallel }V-c_K{\mathrm{\parallel }}_{L^2({\omega }_K)}\le Ch\mathrm{\parallel }\mathrm{\nabla }V{\mathrm{\parallel }}_{L^2({\omega }_K)}$, and the scaled trace inequality gives $\mathrm{\parallel }V-c_K{\mathrm{\parallel }}_{L^2(e)}\le C{h}^{1\mathrm{/}2}\mathrm{\parallel }\mathrm{\nabla }V{\mathrm{\parallel }}_{L^2({\omega }_K)}$ on a selected boundary edge $e$. Thus each interior average coefficient of $V-c_K$ is bounded by $C\mathrm{\parallel }\mathrm{\nabla }V{\mathrm{\parallel }}_{L^2({\omega }_K)}$. A boundary dual function has $L^2(e)$ norm at most $C{h}^{-1\mathrm{/}2}$, so its coefficient has the same bound. Since $I_h{c}_K=c_K$ and the affine basis functions have uniformly bounded $L^2(K)$ gradient norms in two dimensions, $\mathrm{\parallel }\mathrm{\nabla }{I}_{h}V{\mathrm{\parallel }}_{L^2(K)}\le C\mathrm{\parallel }\mathrm{\nabla }V{\mathrm{\parallel }}_{L^2({\omega }_K)}$. Summing and using bounded overlap proves $\mathrm{\parallel }\mathrm{\nabla }{I}_{h}V{\mathrm{\parallel }}_{L^2({\mathrm{\Omega }}_i)}\le C\mathrm{\parallel }\mathrm{\nabla }V{\mathrm{\parallel }}_{L^2({\mathrm{\Omega }}_i)}$.

For any finite-element lift $v_h$ of $g$, step 1.1 gives $\mathrm{\mid }g{\mathrm{\mid }}_{1\mathrm{/}2,\mathrm{\partial }{\mathrm{\Omega }}_i}^2\le C\mathrm{\parallel }\mathrm{\nabla }{v}_h{\mathrm{\parallel }}_{L^2({\mathrm{\Omega }}_i)}^2$. For the floating matrix, put $H^{\mathrm{f}\mathrm{l}}g=(-K_{II}^{-1}{K}_{I\mathrm{\Gamma }}g,g{)}^T$. Symmetry of the local Poisson form and its zero interior row give $a_i(H^{\mathrm{f}\mathrm{l}}g+(z,0),H^{\mathrm{f}\mathrm{l}}g+(z,0))=g^T{S}_i^{\mathrm{f}\mathrm{l}}g+z^T{K}_{II}z$. The positive interior block from [F2] therefore makes $g^T{S}_i^{\mathrm{f}\mathrm{l}}g$ the unique fixed-trace minimum, by the same identity as [F3]. Taking that minimum gives the lower Schur bound. For the upper bound, step 1.1 supplies a continuous lift $V$ with gradient energy at most $C\mathrm{\mid }g{\mathrm{\mid }}_{1\mathrm{/}2,\mathrm{\partial }{\mathrm{\Omega }}_i}^2$, and step 1.2 gives a boundary-preserving finite-element lift $I_{h}V$ with the same bound up to a uniform factor. The just-proved floating minimality bounds its energy by this lift. Multiplication by ${\rho }_i$ proves both inequalities. If $g$ is admissible, the fixed-trace feasible lifts already vanish on the outer Dirichlet portion; [F2]–[F3] identify the same minimum with $g^T{S}_{i}g$.