Periodic packings are packings, so the periodic constant is at most \Delta_d; by Lemma 2.1.7 it suffices to prove the second statement. Let X be the centers of S. For L > 0 consider the cube C_L = [0,L)^d (axisAlignedCell), the cubic lattice \Lambda_L = L\mathbb{Z}^d (axisCellLattice), whose translates of C_L tile space, and the inset cube [1/2, L-1/2]^d (insetAxisAlignedCell). For a translate g + C_L, g \in \Lambda_L, keep the centers x \in X \cap (g + C_L) whose ball B(x,1/2) lies inside g + C_L, and repeat them \Lambda_L-periodically (latticeReplicatedCenters): since the balls of the kept centers lie in one tile and the tiles are disjoint, distinct replicated centers are still at distance at least 1, so this is a periodic packing of separation 1 (replicateToPeriodicPacking). By Lemma 2.1.8 its density is |X_{\mathrm{kept}}|\operatorname{vol}(B(0,1/2))/L^d, and the discarded centers lie in the boundary shell [-1/2, L+1/2]^d \setminus [1, L-1]^d (boundaryShell), whose volume is (L+1)^d - (L-2)^d = o(L^d) (volume_boundaryShell, tendsto_volume_boundaryShell_div_cell). Averaging over the translates g + C_L meeting a large ball B(0,R) (lattice_count_mul_volume_cell_le_volume_ball) and using that the covered proportion of B(0,R) exceeds b for arbitrarily large R, one finds L and g for which the kept centers alone give density above b (exists_replicated_packing_density_eq).

By Theorem 2.4.1 and Lemma 2.1.7, \Delta_d is the supremum of the upper densities of the periodic packings of separation 1, so it suffices to bound each of them. A periodic packing without centers has density 0. For a periodic packing P with centers, choose a \mathbb{Z}-basis of its lattice (PeriodicSpherePacking.canonicalPackingLatticeBasis); its fundamental domain is bounded (PeriodicSpherePacking.fundamental_region_admits_norm_bound) and its lattice translates tile space (PeriodicSpherePacking.basis_region_translates_cover_uniquely), so Theorem 2.3.1 applies to P and gives the bound.