Reduction of the P1 residual estimator after marked refinement

Write $d=U_{\ast }-U$. On a fine child $K^{\mathrm{\prime }}\subset K$, $R_{K^{\mathrm{\prime }}}(U_{\ast })=R_K(U)-b\cdot \mathrm{\nabla }d$. Young's inequality gives $h_{K^{\mathrm{\prime }}}^2\mathrm{\parallel }{R}_{K^{\mathrm{\prime }}}(U_{\ast }){\mathrm{\parallel }}_2^2\le (1+\epsilon )h_{K^{\mathrm{\prime }}}^2\mathrm{\parallel }{R}_K(U){\mathrm{\parallel }}_2^2+(1+{\epsilon }^{-1})\mathrm{\mid }b{\mathrm{\mid }}^2{h}_{K^{\mathrm{\prime }}}^2\mathrm{\parallel }\mathrm{\nabla }d{\mathrm{\parallel }}_{L^2(K^{\mathrm{\prime }})}^2\mathrm{.}$ Unmarked cells retain the same old volume weight. On marked parents, [F2] makes the old residual part at most one quarter its old value. Summing, and charging closure-refined cells to their old parent, bounds the volume contribution by $(1+\epsilon )$ times the old volume sum minus $\frac{3}{4}$ of its marked volume sum, plus $C_{\epsilon }\mathrm{\mid }b{\mathrm{\mid }}^2{h}_{\max }^{}\mathrm{\parallel }\mathrm{\nabla }d{\mathrm{\parallel }}_2^2$.

On every split old edge, each subedge length is at most half the parent length and the sum of their lengths is the parent length; hence the sum of squared subedge lengths is at most one half the squared parent length. The old jump is constant along each parent edge, so its inherited contribution is reduced by at least one half whenever its marked cell is refined. On new interior edges within one old parent the old affine gradient has zero jump. The new jump is the old jump plus the jump of $\mathrm{\nabla }d$; Young's inequality and [F3] give a remainder at most $C_{\epsilon }\mathrm{\parallel }\mathrm{\nabla }d{\mathrm{\parallel }}_2^2$. Assign each old edge half to each incident cell as in the definition. Combining this with step 1.1, the marked cell and edge reductions are at least one half of ${\eta }_{\mathcal{T}}(U,M_E{)}^2$; absorb the remaining nonnegative reduction into the left inequality to obtain the displayed bound.