A positive far-panel quadrature residual and normalized cluster combination

Fix a target node $x_i$, source node $y_j$, and points $x\in T_i$, $y\in S_j$ in their subtriangles. The paths $x(t)=x_i+t(x-x_i)$ and $y(t)=y_j+t(y-y_j)$ stay in $T$ and $S$ by convexity. Every pair on either path with the other variable fixed has distance at least $d$. Apply [F3] to $t\mapsto G_3(x(t),y)$ and then to $t\mapsto G_3(x_i,y(t))$; using [F2] and $\mathrm{\mid }x-x_i\mathrm{\mid }\le h_T$, $\mathrm{\mid }y-y_j\mathrm{\mid }\le h_S$ gives $\mathrm{\mid }{G}_3(x,y)-G_3(x_i,y_j)\mathrm{\mid }\le \frac{h_T+h_S}{4\pi d^2}\mathrm{.}$

One midpoint refinement divides a triangle into four triangles with half-length sides; the determinant of each affine map is $1\mathrm{/}4$, so each new area is one quarter of the old area. Induction gives ${\sum }_i{w}_i=\mathrm{\mid }T\mathrm{\mid }$ and ${\sum }_j{v}_j=\mathrm{\mid }S\mathrm{\mid }$. Integrate the pointwise bound of step 1.1 over each $T_i\times S_j$ and sum; the area sums give $\mathrm{\mid }{A}_{TS}-B_{TS}\mathrm{\mid }\le \mathrm{\mid }T\mathrm{\mid }\mathrm{\mid }S\mathrm{\mid }(h_T+h_S)\mathrm{/}(4\pi d^2)$.

After $m$ midpoint refinements every subtriangle diameter is at most $2^{-m}$ times its initial diameter. Since $d>0$ and the right side of step 2.1 is proportional to $h_T+h_S$, it tends to zero; for every prescribed positive quadrature tolerance a finite $m$ suffices. With exact algebraic coordinates, finite arithmetic and a certified strict lower bound for $d$ can be enclosed outward to any requested positive radius. If the geometry or arithmetic certificate is absent, no certified numerical result follows from this lemma.

For each target node the hypotheses in [F4] give the displayed uniform single-cluster error. Multiplication by its positive weight and summation bounds the target sum error by $W_T$ times that tail. The source charges must be $q_j=v_j\mathrm{/}(4\pi )$ because the evaluator sums $q_j\mathrm{/}\mathrm{\mid }x-y_j\mathrm{\mid }$ while the physical kernel is $1\mathrm{/}(4\pi \mathrm{\mid }x-y\mathrm{\mid })$. Adding the centroid residual from step 2.1 and the separately certified node-evaluation and outward-arithmetic radii by the triangle inequality proves the stated total entry bound. No cluster conclusion is asserted when $R\le a+b$.