Proof of the two all-sequence span deviations

---

Use the model in L164. All supporting deductions below have full proofs on the linked research pages; their examined portions are retained with the whole-statement record.

For fixed $r,A$, suppose one of the following holds at every sufficiently large integer time, keeping the same alternative throughout:

$$nM_n^{(r)}-r\le A,\qquad\text{or}\qquad r-nm_n^{(r)}\le A.$$

The centered spans sum to zero. Either hypothesis bounds one of their positive or negative masses by $A$, hence their total absolute mass by $2A$. The proof of Lemma 6.1 carries this to $kr$-span displacements at real time, retaining the floor error. Lemma 6.2 then uses label injections to compare integrated count errors; it does not import a pointwise two-sided span bound. The real-lift and covering argument handles coincident labels, empty span arcs and full turns, so the terminal estimate of Lemma 6.3 applies.

The finite scale iteration in Proposition 6.4 therefore gives absolute $C_2,C_3$ such that, for $A\ge1$ and $r\ge C_2A$, all sufficiently large integer $n$ satisfy

$$\int_{\mathbb T}|N_n((x,x+D/n])-D|\,dx\le C_3A \quad(0\le D\le S),\qquad S=\frac{\sqrt{Ar}}{\log^2(r/A)}.$$

There is one onset threshold for every $D$ in that interval. Both alternatives use the same constants.

The complete planar discrepancy proof establishes the lower bound $\sqrt{\log M}/41000$ for every list of $M$ points in either anchored convention used in localization. The tensor Haar coefficient argument includes boundary points; equality of the two discrepancy norms is neither asserted nor needed. Applied through Lemma 7.2, it yields absolute $c_4>0,S_0$ such that a uniform short-arc bound $B\ge1$ on the whole interval $[0,S]$, $S\ge S_0$, must satisfy

$$B\ge c_4\sqrt{\log S}.$$

The local planar sets retain arrival time as a second coordinate, so repeated circle positions cause no loss. No common localization point for different prefix lengths is required.

Choose an absolute $0<c<c_4/(2C_3\sqrt3)$ and put $A=c\sqrt{\log r}$. A strict failure of either claimed limsup inequality supplies its corresponding eventual alternative. For all sufficiently large $r$, the conditions above hold, $C_3A\ge1$, $S\ge S_0$, and $\log S\ge(\log r)/3$. Thus the upper and lower count bounds would imply

$$C_3c\sqrt{\log r}\ge c_4\sqrt{\log S} \ge\frac{c_4}{\sqrt3}\sqrt{\log r},$$

contrary to the choice of $c$. This proves the two sequence inequalities separately, with a common absolute threshold and without a simultaneous-time assertion.

For every sequence the upper inequality gives $\limsup_n nM_n^{(r)}\ge r+c\sqrt{\log r}$; take the infimum only now. Since $0\le nm_n^{(r)}\le r$,

$$\limsup_n(r-nm_n^{(r)})=r-\liminf_n nm_n^{(r)}.$$

Every inner liminf is consequently at most $r-c\sqrt{\log r}$; taking the supremum gives the other deficit bound. Each diverges with its own lower bound $c\sqrt{\log r}$. No exchange of a limit with an extremum, or equality of the distinct-point and all-sequence extrema, is used.