Proof of the all-sequence consecutive-span ratio bound

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Use the conventions of L163. The complete supporting deductions are the canonical research pages linked below; their exact examined portions also survive in the whole-statement record. No distinct-point theorem is used to infer the repeated-point assertion.

Fix $\alpha=1/200$, let $r$ be large and put $A=\alpha\log r$. If a sequence violated the proposed limsup bound, then for some $0<C<A$, eventually $M_n/m_n\le1+C/r$. The mean identity gives $n(M_n-m_n)\le C$ and $nM_n\le r+C$. For $n=\lfloor t\rfloor$ set $a_t=r-nm_n$ and $b_t=tM_n-r$. These are nonnegative; all $r$-spans lie in $[(r-a_t)/t,(r+b_t)/t]$ and

$$a_t+b_t\le C+(r+C)/n<A$$

at every sufficiently large real $t$.

The two label injections in Lemma 2.1 and the label, terminal-count and real-lift transfers apply to this hypothesis, with every coincident term counted separately. The finite scale iteration of Proposition 3.1 therefore supplies absolute $C_0,C_1$ such that, when $r\ge C_0A$, all late integer $n$, all $x$ and all $0\le D\le S$ satisfy

$$\left|N_n((x,x+D/n])-D\right|\le B,\qquad S=\frac{\sqrt{Ar}}{\log^2(r/A)},\quad B=3A+\frac{C_1A}{\log(r/A)}.$$

One time threshold covers the entire range of $D$. Proposition 1221.1 of the extraction argument chooses a short arc avoiding the early terms and reads later terms in arrival order. For $B\ge1$ and $S\ge2$ it yields a list of length $L'\ge\lfloor S\rfloor-B$ in $(0,1)$, with repetitions allowed, whose maximum prefix star discrepancy $H_{L'}$ is at most $B$. The complementary arc errors have coefficients $1-u$ and $-u$, so this is $B$, not $2B$.

The complete four-block finite-list proof gives, for every nonempty list,

$$H_{L'}\ge\frac{\lfloor\log_4L'\rfloor}{16} \ge\frac{\log L'}{16\log4}-\frac1{16}.$$

Its range and jump argument includes repeated entries and the right jump at zero; it consumes no optimized discrepancy theorem from an unprovided source. Now $B=3\alpha\log r+O(1)$ and $\log S=\frac12\log r-\frac32\log\log r+O(1)$. Thus $L'\ge S/2>0$ for all sufficiently large $r$, and the two discrepancy bounds imply

$$3\alpha\log r+O(1)\ge \frac{\log r}{32\log4}-O(\log\log r).$$

This is impossible eventually, since $3/200<1/64<1/(32\log4)$. All constants in this comparison are absolute. The sequence-dependent time threshold only changes how late the extraction is made. Hence one absolute $r_0$ gives the claimed inequality for every sequence. Infinite limsup ratios satisfy it directly.

For fixed $r\ge r_0$, every value in the infimum defining $\mu_r$ is at least $1+\log r/(200r)$. Taking that infimum, then subtracting one and multiplying by $r$, proves the quantitative consequence. Its lower bound $\log r/200$ tends to infinity. No limit and extremum are interchanged.