A disordered fixed-neighbor stencil can fail derivative consistency

The adjacent gaps in one period are 3ε/5, 7ε/5, 3ε/2, and ε/2, so the fill radius is h=3ε/4. The four Voronoi-cell lengths per period are 11ε/20, ε, 29ε/20, and ε; hence every volume is at most 2h. If a cell meets B(x,r), its node lies in B(x,r+h). Nodes are separated by at least ε/2, so at most 1+4(r+h)/ε such nodes can lie in that interval. Each cell volume is at most 29ε/20, and r+h≥3ε/4; therefore the total active volume is at most (29/20)ε+(29/5)(r+h)≤(116/15)(r+h)≤8(r+h). The local-volume bound holds with C_L=8. In particular, the cells at −ε/2 and 3ε/5 are [−5ε/4,−ε/4] and [3ε/10,13ε/10], each of volume ε.

The nearest-node intervals tile R, repeat with period 4ε, and have gaps between their endpoints only at shared cell boundaries. Every compact support meets finitely many intervals. For any compactly supported continuous g, choose a finite consecutive block of cells containing supp(g); [F4] makes g integrable on its compact union, and repeated [F5] splits that integral into the cell integrals. Cells outside the block have zero integral because they do not meet supp(g). Thus the integration-compatibility identity in [F2] holds for every compactly supported continuous g.

On |s|<1, W′(s)=−(105/16)s(1−s²)². For f(y)=y at x=0, set g(y)=y,\partial_xW_\varepsilon(x-y)|_{x=0}. By [F1], g vanishes outside [−ε,ε], and [F2] defines its particle sum as Q_hg=Σ_jV_jg(x_j). The node at 0 contributes zero; only y=−ε/2 and y=3ε/5 contribute, with values −W′(1/2)/2=945/1024 and −(3/5)W′(3/5)=3024/3125. Thus Q_hg=945/1024+3024/3125, independent of ε.

Here f′(0)=1 and f‴=0, so [F3] gives D_εf(0)=1 exactly. Since each summand in the value from step 3.1 exceeds 1/2, that value is greater than 1 for every ε. The derivative error therefore does not tend to zero although h/ε=3/4 remains fixed.