Lemma 10 ([Width tails control hitting differences]). For either sign and all integers n'≥n≥0 , u_n^σ-u_ n' ^σ ≤C(n'-n)F_σ(n), ∑_ l≥0 2^lF_σ(2^l)<∞.

Fix a sign and suppress it. Partition according to the last renewal at or before m . The next width exceeds the distance from that renewal to m , so ∑_ k=0 ^m F(k) u_ m-k =1. Subtracting the identity at m+1 , and using u_0=1 , gives ∑_ k=0 ^m F(k) ( u_ m-k - u_ m+1-k )=F(m+1). By, every adjacent difference in this sum is nonnegative except the one with m-k=M-1 , which equals -ξ . Take m=n+M . The exceptional index is k=n+1 , while the term k=0 equals ξ(u_n-u_ n+1 ) . Dropping the other nonnegative terms yields ξ(u_n-u_ n+1 ) ≤F(n+M+1)+ξF(n+1) ≤(1+ξ)F(n). Summation from n to n'-1 , using that F decreases, proves the first bound. The second follows by grouping the ordinary tail sum ∑_ k≥0 F(k)=E_ ν_σ L<∞ into dyadic intervals.