Lemma 21 ([Number of exceptional unit intervals]). The set E meets at most \!(O(L^b+L¹-a L))=Y^o(1) intervals [j,j+1] , uniformly in the bounded coefficients b_s .

Choose a point of E in each occupied interval, and split the interval indices into three residue classes to obtain separated points. Put j_0= Y/(2P) . The polynomial P_s(t)^ j_0 has indices at most Y . Its coefficient-square sum is at most j_0!(∑_ p_s P |b_s(p_s)|^2p_s^ -2 )^ j_0 ≤j_0!(C/P)^ j_0 . Indeed an integer has at most j_0! ordered representations as a product of j_0 primes; Cauchy--Schwarz on each fiber proves the first inequality even with repeated primes. For a polynomial Q(t)=∑_ n≤Y c_nn^ it and unit-separated points in [-O(Y),O(Y)] , the unit-interval Sobolev inequality and bounded overlap imply ∑_j|Q(t_j)|^2∫_ -O(Y) ^ O(Y) (|Q(t)|^2+|Q'(t)|^2)\,dt Y((2Y))^ O(1) ∑_n|c_n|^2. In the last step use Lemma 4 ([Mean squares]); differentiation multiplies a coefficient by i n , so the constants here are independent of j_0 . The range 2L/h is o(Y) , as required. If N_ E is the number of occupied intervals, it follows that N_ E YL^ O(1) j_0! (CL^ 2A_s /P)^ j_0 . Now j_0=O(L^ 1-a ) and Y-j_0 P=O(L^b+j_0) by the definition with 2P . Taking logarithms and using (j_0!)≤j_0 j_0 proves the asserted estimate.