Two-dimensional logarithmic edge cutoff estimates

If $E$ has no interior boundary node, then ${\theta }_E=0$. If $\mathrm{\mid }E\mathrm{\mid }\mathrm{/}h\le 4$, scaling $E$ to unit length gives a uniform bound on both terms of $\mathrm{\parallel }{\theta }_E{\mathrm{\parallel }}_{00,E}^2$. Otherwise the first and last mesh intervals have length comparable to $h$, ${\theta }_E$ is linear from $0$ to $1$ and from $1$ to $0$ there, and it equals $1$ between them. The weighted integral on either endpoint interval is bounded by ${\int }_0^h(s\mathrm{/}h{)}^{2}ds\mathrm{/}s\le C$; the plateau contributes at most $C+\log (\mathrm{\mid }E\mathrm{\mid }\mathrm{/}h)\le C+\log (H\mathrm{/}h)$. In the double integral, pairs in the plateau contribute zero. Pairs within one endpoint interval have integrand at most $C{h}^{-2}$ and hence contribute at most a constant. For one point $s$ in an endpoint interval and one outside it, integrating $(1-s\mathrm{/}h{)}^2\mathrm{/}(t-s{)}^2$ in the outside variable and then in $s$ gives a constant; the two endpoint intervals are treated in the same way. Therefore $\mathrm{\parallel }{\theta }_E{\mathrm{\parallel }}_{00,E}^2\le C(1+\log (H\mathrm{/}h))$.

Put $v_h=u_h-{\bar{u}}_{{\mathrm{\Omega }}_i}$, $H=H_i$, $h=h_i$ and $G=\mathrm{\parallel }\mathrm{\nabla }{v}_h{\mathrm{\parallel }}_{L^2({\mathrm{\Omega }}_i)}$. Choose a triangle $K$ containing a point where $\mathrm{\mid }{v}_h\mathrm{\mid }$ is maximal, and let $x_0$ be its centroid. Affine scaling gives $\mathrm{\parallel }{v}_h{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\mid }{v}_h(x_0)\mathrm{\mid }+CG$. The uniform interior-cone condition in [F1] supplies at $x_0$ a cone of radius $R=cH$ and angular interval $J$ of fixed positive length, contained in ${\mathrm{\Omega }}_i$. Quasiuniformity and affine scaling give the global inverse bound $\mathrm{\parallel }\mathrm{\nabla }{v}_h{\mathrm{\parallel }}_{\mathrm{\infty }}\le C{h}^{-1}G$. If $H\mathrm{/}h$ is bounded, the inverse bound $\mathrm{\parallel }{v}_h{\mathrm{\parallel }}_{\mathrm{\infty }}\le C{h}^{-1}\mathrm{\parallel }{v}_h{\mathrm{\parallel }}_{L^2}$ and mean-zero scaled Poincaré prove the claim directly. Otherwise choose $\epsilon =c^{\mathrm{\prime }}h<R\mathrm{/}2$. Along each cone ray, the integral of $\mathrm{\mid }\mathrm{\nabla }{v}_h\mathrm{\mid }$ from $0$ to $\epsilon$ is at most $CG$, while weighted Cauchy–Schwarz gives ${\mid {\int }_{\epsilon }^r{\mathrm{\partial }}_s{v}_h(x_0+s{e}_{\theta })ds\mid }^2\le \log (R\mathrm{/}\epsilon ){\int }_{\epsilon }^R\mathrm{\mid }\mathrm{\nabla }{v}_h(x_0+s{e}_{\theta }){\mathrm{\mid }}^{2}sds\mathrm{.}$ Express $v_h(x_0)$ by its value at $x_0+r{e}_{\theta }$ minus these two radial integrals, square, and average with measure $rdrd\theta$ over $R\mathrm{/}2<r<R$, $\theta \in J$. The annulus has area comparable to $H^2$; the ray-energy term is independent of $r$ and integrates in $\theta$ to at most $G^2$. Hence $\mathrm{\mid }{v}_h(x_0){\mathrm{\mid }}^2\le C{H}^{-2}\mathrm{\parallel }{v}_h{\mathrm{\parallel }}_{L^2}^2+C(1+\log (H\mathrm{/}h))G^2\mathrm{.}$ Mean-zero scaled Poincaré bounds the first term by $C{G}^2$, and the centroid comparison proves the asserted $L^{\mathrm{\infty }}$ estimate. This is the radial argument underlying Widlund's estimate (5), with the centroid-to-maximum comparison included explicitly.

Put $z=u_h-{\bar{u}}_E$. In the first Gagliardo term use $\theta (s)z(s)-\theta (t)z(t)=\theta (s)(z(s)-z(t))+z(t)(\theta (s)-\theta (t))\mathrm{.}$ The weighted term is bounded by $\mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\infty }}^2{\int }_E{\theta }_E^2\mathrm{/}\mathrm{dist}(s,\mathrm{\partial }E)ds$. Together these give $\mathrm{\parallel }{\theta }_{E}z{\mathrm{\parallel }}_{00,E}^2\le C(\mathrm{\mid }z{\mathrm{\mid }}_{H^{1\mathrm{/}2}(E)}^2+(1+\log (H\mathrm{/}h))\mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\infty }}^2)\mathrm{.}$ On the two endpoint mesh intervals, $I_h({\theta }_{E}z)-{\theta }_{E}z$ is a quadratic interpolation error vanishing at both endpoints; scaling the interval to unit length bounds its $00,E$ norm by $C\mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\infty }}^2$. The same estimate therefore holds for $I_h({\theta }_{E}z)$. The floating local Schur energy of the trace of $u_h$ is at most ${\rho }_i\mathrm{\parallel }\mathrm{\nabla }{u}_h{\mathrm{\parallel }}_{L^2({\mathrm{\Omega }}_i)}^2$ because it is the minimum over finite-element lifts; the lower trace-Schur bound in [F2] therefore gives $\mathrm{\mid }z{\mathrm{\mid }}_{H^{1\mathrm{/}2}(E)}^2\le C\mathrm{\parallel }\mathrm{\nabla }{u}_h{\mathrm{\parallel }}_{L^2({\mathrm{\Omega }}_i)}^2$. Also $\mathrm{\mid }{\bar{u}}_E-{\bar{u}}_{{\mathrm{\Omega }}_i}\mathrm{\mid }\le \mathrm{\parallel }{u}_h-{\bar{u}}_{{\mathrm{\Omega }}_i}{\mathrm{\parallel }}_{L^{\mathrm{\infty }}({\mathrm{\Omega }}_i)}$, so $\mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\infty }}\le 2\mathrm{\parallel }{u}_h-{\bar{u}}_{{\mathrm{\Omega }}_i}{\mathrm{\parallel }}_{L^{\mathrm{\infty }}({\mathrm{\Omega }}_i)}$. Combining this with steps 1.1 and 1.2 proves the claimed squared-logarithmic multiplier estimate.