\section{Quantitative purity and sharp structural stability}\label{sec:stability}
\begin{lemma}[A two-multiplier switching margin]\label{lem:margin}
Suppose $K=\sin\theta\cos\theta+\eta\cos^2\theta-\xi\sin^2\theta$
has $\Delta=1+4\eta\xi>0$. Let $Z$ be its two zeros on
$\R/\pi\mathbb Z$, and let $d_Z$ be distance to this set. Then
\begin{equation}\label{eq:margin}
|K(\theta)|\ge\frac{2\sqrt\Delta}{\pi^2}d_Z(\theta).
\end{equation}
For the bounds-only calibration $\xi=0$, this is exactly the uniform v1.0
constant $2/\pi^2$, independent of $\eta$.
\end{lemma}
\begin{proof}
Factor the indefinite quadratic as
$K=A\sin(\theta-\theta_1)\sin(\theta-\theta_2)$.
Let $\sigma\in(0,\pi/2]$ be the shorter distance between its zero lines.
At either zero, differentiation gives $|K'|=|A|\sin\sigma=\sqrt\Delta$;
the last equality also follows by writing
$K=(\eta-\xi)/2+(\eta+\xi)\cos2\theta/2+\sin2\theta/2$.
Let $d_1,d_2$ be the distances to the two individual zero lines.
The circle triangle inequality gives $d_1+d_2\ge\sigma$, so
$d_1d_2\ge\sigma\min(d_1,d_2)/2$. Using
$|\sin x|\ge2\dist(x,\pi\mathbb Z)/\pi$ on both factors,
\[
|K|\ge\frac{2|A|\sigma}{\pi^2}\min(d_1,d_2)
\ge\frac{2\sqrt\Delta}{\pi^2}d_Z,
\]
because $\sigma\ge\sin\sigma$.
\end{proof}

\begin{theorem}[Phase and clock stability from the exact defect]\label{thm:phasestability}
For a growing fixed-resource profile with winding $n$, put
$d=B-1$, $E=nG_m(S/n)-L(u)$, and $\Delta_n=1+4\eta_n\xi_n$.
With the canonical selector $b=b_{\eta_n,\xi_n}$,
\begin{align}
Q:=\int_{\theta_0}^{\theta_0+n\pi}|\widehat u-b|\dd\theta
&\le2\pi B\sqrt{ndE}\,\Delta_n^{-1/4},\label{eq:phaseL1}\\
\inf_{t_0}\int_0^S|u(s)-u^*_{m,n}(s-t_0)|\dd s
&\le(1+2nd)\,2\pi B\sqrt{ndE}\,\Delta_n^{-1/4}.
\label{eq:timeL1}
\end{align}
Here $u^*_{m,n}$ denotes all $n$ repeated cells. The same assertions hold
for a bounds-only calibrated cell with $\xi=0$, its branch defect, and
$\Delta_n=1$.
\end{theorem}
\begin{proof}
Let $e=|\widehat u-b|\le d$. Over $n$ turns the set with $d_Z\le a$ has
measure at most $4na$. Layer-cake integration gives
\[
\int d_Z e\dd\theta
=\int_0^\infty\int_{d_Z>a}e\dd\theta\dd a
\ge\int_0^{Q/(4nd)}(Q-4nda)\dd a=\frac{Q^2}{8nd}.
\]
The denominators of the defect are at most $B^2$. Combining
\eqref{eq:resourceDefect} with Lemma~\ref{lem:margin} gives
$E\ge\sqrt\Delta_n Q^2/(4\pi^2B^2nd)$, proving the first estimate.

Translate the actual profile to place its expanding angle at the start
of the canonical high arc at $s=0$. Both clocks now start at that angle
and end at time $S$. Their difference obeys
\[
\sup_\theta|s(\theta)-s_*(\theta)|
\le\int\left|\frac1{v_{\widehat u}}-\frac1{v_b}\right|\dd\theta
\le Q.
\]
The intermediate binary signal $b(\theta(s))$ differs from the actual
profile in temporal $L^1$ by at most $Q$, since $v_{\widehat u}\ge1$.
Its ordered switches agree in number and state ordering with the ideal
signal; each of the at most $2n$ switch times is displaced by at most $Q$.
The symmetric difference of two intervals is bounded by the sum of the
absolute displacements of their endpoints. Summing high-state intervals
bounds the remaining temporal $L^1$ error by $2ndQ$, proving the result.
\end{proof}

\begin{theorem}[Global resource-constrained square-root rigidity]\label{thm:globalstability}
Suppose $F=F(R,S,m)>0$. Let $F_n=nG_m(S/n)$ and let $\mathcal O_*$ be
the union of all translates of every winning architecture. Define
\begin{align}
\delta_b&=\min\bigl(\{F\}\cup\{F-F_n:n\in\mathcal N_m\setminus\mathcal W_m\}\bigr)>0,\nonumber\\
C_0&=\max_{n\in\mathcal W_m}
\frac{2\pi B(1+2nd)\sqrt{nd}}{S\Delta_n^{1/4}},\qquad
C=\max\{C_0,d/\sqrt{\delta_b}\}.\label{eq:resourceC}
\end{align}
Every admissible measurable profile satisfies
\begin{equation}\label{eq:globalstability}
\boxed{\dist_1(u,\mathcal O_*):=
\inf_{v\in\mathcal O_*}\frac1S\int_0^S|u-v|\dd s
\le C\sqrt{F-L(u)}.}
\end{equation}
Replacing the distance by mean absolute permittivity error is valid after
multiplying $C$ by $\varepsilon_+$.
\end{theorem}
\begin{proof}
Set $\delta=F-L$. If $\delta<\delta_b$, then $L>0$ and the actual winding
must win; every other branch would force
$\delta\ge F-F_n\ge\delta_b$. Apply Theorem~\ref{thm:phasestability}.
If $\delta\ge\delta_b$, the normalized $L^1$ distance is at most $d$,
which gives the second term in $C$. Finally
$|\varepsilon_+/u-\varepsilon_+/v|\le\varepsilon_+|u-v|$ and
$\dd t/T=\dd s/S$.
\end{proof}
The estimate is global for each fixed parameter triple, not uniform as
means approach endpoints, as branches approach degeneracy, or as contrast
collapses. The explicit constants display those possible deteriorations.

\begin{proposition}[Sharpness while preserving the resource exactly]\label{prop:resourceSharp}
The exponent $1/2$ in \eqref{eq:globalstability} cannot generally be improved,
even when period, contrast, and mean are all fixed.
\end{proposition}
\begin{proof}
Take $R=2$, $m=2$, $S=3\pi/4$. The only possible growing winding is one,
and its unique optimizer has high time $h_0=\pi/4$, low time $\ell_0=\pi/2$,
and $F=\log2$. For small $a>0$, use the four states and durations
\[
(4,h_0-a),\quad(1,a),\quad(4,a),\quad(1,\ell_0-a).
\]
Their total high time, low time, period, and mean are unchanged. Exact
multiplication of the four segment matrices followed by Taylor expansion gives
\[
D(a)=-\frac54+\frac94a^2-\frac{15}{2}a^4+O(a^6),\qquad
F-L(u_a)=3a^2+O(a^4).
\]
The high-state set is $[0,h_0-a]\cup[h_0,h_0+a]$. For sufficiently small
$a$, its intersection with any circular interval of length $h_0$ has measure
at most $h_0-a$: an interval covering both components loses their intervening
gap of length $a$, whereas an interval covering only one component contains
at most $h_0-a$. The long complementary gap prevents another circular case.
This maximum is attained by $[0,h_0]$. Hence the minimum symmetric difference
is $2a$ and
\[
\dist_1(u_a,\mathcal O_*)=6a/S.
\]
For $\alpha>1/2$, its ratio to $(F-L(u_a))^\alpha$ diverges as $a\downarrow0$.
The symbolic expansion and independent floating-point checks are supplied
in the verification suite.
\end{proof}