PRH | Aux | 4.3.2 • Attaching Blur–Consistent Step Functions to 3x and 3x+1

We give a blur-consistent decomposition of the accelerated Collatz step that makes the descent mechanism explicit. Using the Mellin/Fourier lens distinction and a soft lens switch with a blur floor $\theta \in(0, \delta+\rho)$, we attach two nonnegative functions: a residue function $m_\theta$ on odd classes modulo $2^k$ (multiplicative side) and an integer function $a_\theta$ on odd $n$ (additive nudge), where

$$
a_\theta(n)=\max \left\{\log \left(1+\frac{1}{3 n}\right)-\theta, 0\right\}, \quad \Rightarrow \quad a_\theta(n)=0 \text { for all } n>N_\theta=\left\lceil\frac{1}{3\left(e^\theta-1\right)}\right\rceil
$$

Assuming a finite max-plus residue certificate $a(r)+\rho+\phi\left(F_k(r)\right) \leq \phi(r)-\delta$ with potential $\phi$ (and the tight bar $\rho+\delta<\log (4 / 3)$ ), the scale functional $S(n)=\log n+\phi\left(n \bmod 2^k\right)$ satisfies the split

$$
S(n)-S(T(n)) \geq m_\theta\left(n \bmod 2^k\right)-a_\theta(n)
$$

hence for all $n>N_\theta$ one has the uniform drift $S(T(n)) \leq S(n)-(\delta-\theta)$. Past this threshold the additive contribution vanishes and the decrease is carried purely by the multiplicative component, making the proof strategy transparent; the construction is finite, checkable, and reflects the unavoidable "channel switch toll" when mixing addition with multiplication.