Linear contraction of AFEM energy error and oscillation

By [A1, F5] or [A1, F6], as appropriate to the chosen branch, the weak solution exists uniquely under the stated choice assumptions. By [F7], the finite-dimensional P1 Galerkin solution on each mesh also exists uniquely. Write $d_k=U_{k+1}-U_k$. The completed algorithm refines every original edge neighbor of $M_E$ through three generations, so the marked-correction premise in [F1] holds, including its adjacent-cell tests. By [F1] and the Dörfler property, ${\theta }_E{\eta }_k^2\le {\eta }_k(M_E{)}^2\le C_{\mathrm{e}\mathrm{f}\mathrm{f}}(\mathrm{\parallel }\mathrm{\nabla }{d}_k{\mathrm{\parallel }}_2^2+{\mathrm{osc}}_k^2),{\eta }_k^2\ge e_k^2\mathrm{/}{C}_{\mathrm{r}\mathrm{e}\mathrm{l}}\mathrm{.}$ Thus $\mathrm{\parallel }\mathrm{\nabla }{d}_k{\mathrm{\parallel }}_2^2\ge \kappa e_k^2-{\mathrm{osc}}_k^2$. By [F2], $e_{k+1}^2\le (1+\nu )e_k^2-\mathrm{\parallel }\mathrm{\nabla }{d}_k{\mathrm{\parallel }}_2^2$, where $\nu =0$ in the symmetric case and $\nu =\mu \mathrm{/}(1-\mu )$ otherwise. Consequently $e_{k+1}^2\le (1-\delta )e_k^2+{\mathrm{osc}}_k^2,\delta =\kappa -\nu >0.$

Combine step 1.1 with [F3]: $e_{k+1}^2+\beta {\mathrm{osc}}_{k+1}^2\le (1-\delta )e_k^2+(1+\beta {\gamma }_O){\mathrm{osc}}_k^2\mathrm{.}$ Choose $\beta >1\mathrm{/}(1-{\gamma }_O)$. Then both coefficients $1-\delta$ and ${\gamma }_O+1\mathrm{/}\beta$ are strictly less than one. Their maximum is a valid $\alpha <1$, proving the first contraction.

Set $\epsilon ={\theta }_E\mathrm{/}4$ in [F4]. Dörfler marking gives ${\eta }_{k+1}^2\le {\rho }_E{\eta }_k^2+C_E\mathrm{\parallel }\mathrm{\nabla }{d}_k{\mathrm{\parallel }}_2^2,{\rho }_E=1+\epsilon -\frac{1}{2}{\theta }_E=1-\frac{1}{4}{\theta }_E<1,$ where $C_E$ is finite. Also [F2] implies $\mathrm{\parallel }\mathrm{\nabla }{d}_k{\mathrm{\parallel }}_2^2\le (1+\nu )e_k^2$. Choose $\sigma >0$ so $\sigma C_E(1+\nu )<\delta \mathrm{/}2$. Adding $\sigma$ times this estimator inequality to the error and oscillation bounds gives coefficients strictly below one on $e_k^2$ and ${\eta }_k^2$. Choose ${\beta }^{\mathrm{\prime }}$ large enough that the coefficient on ${\mathrm{osc}}_k^2$, divided by ${\beta }^{\mathrm{\prime }}$, is below one; this is possible by ${\gamma }_O<1$. The maximum of the three resulting coefficients is ${\alpha }^{\mathrm{\prime }}<1$, which proves the second inequality. The case of zero error or zero oscillation satisfies the same inequalities without division.