Square function needs [w]^{1/2} for large p: the bounded alternating Haar sum

Lower bound for ‖S‖_{L^p(w)} ≳ [w]_{A_p}^{1/2}, on the line, dyadic. Take v = t^{δ-1}, which has [v]_{A_p} ≲ 1/δ for every p, and g = Σ_{k<N} (-1)^k h_k with h_k = 2·1_{[0,2^{-k-1})} - 1_{[0,2^{-k})} and N = ⌈1/δ⌉. The partial sums are 0 or 1, so |g| ≤ 2 and ∫|g|^p v ≲ 1/δ. The averages over [0,2^{-m}) are Σ_{k<m}(-1)^k, so each martingale difference on [0,2^{-N}) is exactly ±1. Hence Sg = √N there, and v([0,2^{-N})) = 2^{-Nδ}/δ ≥ 1/(4δ), which gives ∫(Sg)^p v ≳ δ^{-p/2}/δ. The same g with signs ε_k = (-1)^k makes the martingale transform equal to N, which shows [w]^1 is needed for T. So one example serves both operators: T adds the differences coherently, S adds their squares. Combined with Buckley's weight (exponent 1/(p-1)) this shows Lerner's max(1/2, 1/(p-1)) is sharp for every p.