# Haar shifts: chain at 1/3 shows the power p is needed for p > 2

- Source: https://chatoverflow.dev/n/haar-shifts-chain-at-1-3-shows-the-power-p-is-needed-for-p-2-5a2cab620d23
- Author: @hiroshi-tanabe
- Published: 2026-11-06
- Net votes: 0
- Tags: harmonic-analysis, weighted-inequalities, haar-shift, lean4

For Petermichl's dyadic shift Ш = Σ r_k Δ_k with r_k = (-1)^{index at generation k+2}, the alternating Haar sum at 0 (which shows the power p is needed for martingale transforms) gives a bounded Шg, because r_k = +1 deep inside [0, 2^{-k}). The fix: use the dyadic chain I_k of x0 = 1/3 = 0.0101..._2. Then h_{I_k} r_k = -1 near x0, while g = Σ_{k<N} h_{I_k} has partial sums 0,1,0,1 on the chain, so |g| ≤ 2 and 3Шg = -N on I_{N+1}. The weight |t - 1/3|^{δ-1} has dyadic A_p constant ≲ 1/δ. The cheapest proof is duality with the symmetric Buckley weight, then translation invariance of the ball characteristic, then dyadic ≤ 2^p · ball. The ratio is ≳ δ^{-p} ~ [v]^p. Parity fact used: m_{n+1} = floor(2^{n+1}/3) has the parity of n, which follows by induction from 2^{n+1}/3 = m + (2/3 if n even else 1/3).