\chapter{Spectral fluctuations across scales}\label{s:scales}

In this section we prove the results of \zcref{s:intro-results} on the second-order statistics of the limiting process $\Xi_X$, namely \zcref{c:intro-pair} on the pair measure $\nu_X$ and the count covariances, the rows of \zcref{p:endpoints} other than the law of $\mu_X$, the large-scale covariance of \zcref{t:largecov}, and the crossover of \zcref{t:crossover}. The law of $\mu_X$ in \zcref{p:endpoints} is treated in \zcref{s:coarse}.

We follow the plan of \zcref{s:ov-scalar}, which explains how these statistics are read off from the function $I_X$ of \zcref{d:kernels}. The main estimate of this section is \zcref{l:scales-approx}, which shows that $X I_X(Xa)$ is close to a stationary pairing $\mathcal I(a)$ when the wire is long compared with the relaxation time of the diffusion, and we compute $\mathcal I(a)$ in closed form in \zcref{l:scales-pairing}.

Throughout this section, we use the notation of \zcref{s:met-fk}. For functions $f$ and $g$ on $D$ such that $fg$ is integrable, we write $\langle f, g \rangle_D = \int_D f g \dd B \dd F$, and we write $\|f\|_{L^p(D)}$ for the norms with respect to $\dd B \dd F$. For functions on $\R$, the Fourier transform $\hat f(k) = \int_\R e^{-2\pi \iu k s} f(s) \dd s$ is that of \zcref{s:intro-notation}.

\section{The pair measure}\label{s:scales-pair}

We begin with the existence of $\nu_X$. Following \cite[Section~8.1]{daley2003introduction}, where it is written $\breve M_{[2]}$, we call the measure $\nu_X$ of \eqref{e:intro-pair} the \emph{reduced second factorial moment measure} of $\Xi_X$. We also keep the shorter name used in \zcref{s:intro-results}, the pair measure. Recall from \zcref{p:proc-subsequences} that $\Xi_X$ is stationary and simple, with $\E \Xi_X(I) = |I|$ for all bounded intervals $I$, that it satisfies the exponential bound \eqref{e:proc-exp} with the constant $C_{\exp}$, which does not depend on $X$, and that it satisfies \eqref{e:proc-pairs}. The existence of $\nu_X$ is a standard fact about stationary point processes with locally finite second moments \cite[Section~8.1]{daley2003introduction}, and the bound \eqref{e:scales-translation-bound}, which is uniform in $X$, is the part that we need beyond it.

\begin{lemma}\label{l:scales-pair}
Let $X > 0$. There exists an even positive Radon measure $\nu_X$ on $\R$ such that \eqref{e:intro-pair} holds for all nonnegative measurable functions $f$ on $\R^2$. Further, there exists $C_\nu$, depending only on $C_{\exp}$, such that for all $X > 0$,
\begin{equation}\label{e:scales-translation-bound}
\sup_{s \in \R} \nu_X([s, s + 1]) \le C_\nu .
\end{equation}
\end{lemma}

\begin{proof}
For a Borel set $A \subset \R^2$, we let
\[
\mu^{[2]}(A) = \E \sum_{\substack{x, y \in \Xi_X \\ x \neq y}} \1_A(x, y)
\]
denote the second factorial moment measure of $\Xi_X$. By \eqref{e:proc-pairs}, we have $\mu^{[2]}(I \times I) \le (C_{\exp} |I|)^2$ for all bounded open intervals $I$, and $\Xi_X$ has locally finite second moments. For Borel sets $A \subset \R^2$, we also let
\[
M_2(A) = \E \sum_{x, y \in \Xi_X} \1_A(x, y)
\]
denote the second moment measure of $\Xi_X$. Since $\Xi_X$ is stationary, \cite[Proposition~8.1.I(b)]{daley2003introduction} implies that there exists a positive Radon measure $\breve M_2$ on $\R$ such that
\[
\int_{\R^2} f \dd M_2 = \int_\R \int_\R f(v, v + s) \, \breve M_2(\dd s) \dd v
\]
for all bounded measurable functions $f$ with bounded support. We call $\breve M_2$ the reduced second moment measure of $\Xi_X$, as in that reference. It is symmetric by \cite[Proposition~8.1.II(i)]{daley2003introduction}.

Since $\Xi_X$ is simple and has intensity one, $M_2$ is the sum of $\mu^{[2]}$ and the image of the Lebesgue measure under $v \mapsto (v, v)$. Therefore \eqref{e:intro-pair} holds with $\nu_X = \breve M_2 - \delta_0$ for bounded measurable $f$ with bounded support, and monotone convergence extends it to all nonnegative measurable $f$. Taking $f(x, y) = \1_{[0, 1)}(x) \1_S(y - x)$ for bounded Borel sets $S$, we see that $\nu_X$ is positive and that \eqref{e:intro-pair} determines it. It is even, since $\breve M_2$ and $\delta_0$ are.

It remains to prove \eqref{e:scales-translation-bound}. Let $s \in \R$ and $\eta \in (0, 1)$, and let $I_1 = (-\eta, 1 + \eta)$ and $I_2 = (s - \eta, s + 2 + \eta)$. Every pair $(v, v + r)$ with $v \in [0, 1)$ and $r \in [s, s + 1]$ belongs to $I_1 \times I_2$. Now \eqref{e:intro-pair} and the Cauchy--Schwarz inequality give
\[
\nu_X([s, s + 1]) \le \E \big[ \Xi_X(I_1) \Xi_X(I_2) \big] \le \Big( \E \Xi_X(I_1)^2 \, \E \Xi_X(I_2)^2 \Big)^{1/2} .
\]
For a bounded open interval $I$, \eqref{e:proc-pairs} and the unit intensity of $\Xi_X$ give
\[
\E \Xi_X(I)^2 = \E \big( \Xi_X(I) \big)_2 + \E \Xi_X(I) \le (C_{\exp} |I|)^2 + |I| .
\]
Letting $\eta \downarrow 0$ yields \eqref{e:scales-translation-bound} with $C_\nu = ((C_{\exp}^2 + 1)(4 C_{\exp}^2 + 2))^{1/2}$.
\end{proof}

We now identify the Poisson transform of $\nu_X$, using the rank-one derivatives of \zcref{t:process}(c).

\begin{lemma}\label{l:scales-pair-transform}
Let $X > 0$. The measure $\nu_X$ satisfies \eqref{e:intro-pair-transform}, and it is the unique positive Radon measure $\nu$ with $\sup_s \nu([s, s + 1]) < \infty$ that satisfies \eqref{e:intro-pair-transform} for all $\eps > 0$ and $v \in \R$.
\end{lemma}

\begin{proof}
Fix $\eps > 0$, and set
\[
c_\eps(x) = \frac{2\eps}{x^2 + \eps^2} = 2\pi P_\eps(x), \qquad x \in \R .
\]
For finite $N$, the product in \eqref{e:intro-ratio} with $k = 1$, $u = 0$, and $b = \eps + t$ is
\[
T_N^{(1)}(0, \eps + t) = \E \prod_{\alpha = 1}^N \Big( 1 + c_\eps(x_\alpha) t + \frac{t^2}{x_\alpha^2 + \eps^2} \Big),
\]
a polynomial in $t$ with bounded coefficients. Differentiating twice at $t = 0$, we find
\begin{equation}\label{e:scales-second-jet}
\partial_t^2 T_N^{(1)}(0, \eps + t) \big|_{t = 0} = \frac{1}{\eps} \E \Xi_N(c_\eps) + \E \big[ \Xi_N(c_\eps)^2 - \Xi_N(c_\eps^2) \big] .
\end{equation}
The functions $c_\eps$ and $c_\eps^2$ are continuous and bounded by multiples of $P_\eps$. Applying \zcref{l:proc-moments}(b) to $\Xi_N \to \Xi_X$, we see that the expectations on the right side of \eqref{e:scales-second-jet} converge to the same expressions for $\Xi_X$, and $\E \Xi_X(c_\eps) = 2\pi$. Comparing with \eqref{e:intro-jets}, we obtain
\[
4\pi^2 \E \sum_{\substack{x, y \in \Xi_X \\ x \neq y}} P_\eps(x) P_\eps(y) = \E \big[ \Xi_X(c_\eps)^2 - \Xi_X(c_\eps^2) \big] = 4 \pi^2 \big( 1 + I_X(2\pi\eps) \big) - \frac{2\pi}{\eps} .
\]
We now apply \eqref{e:intro-pair} to the left side, and use the identity
\[
\int_\R P_\eps(v) P_\eps(v + s) \dd v = P_{2\eps}(s), \qquad s \in \R ,
\]
which follows from $P_\eps * P_\eps = P_{2\eps}$ and the evenness of $P_\eps$. With $a = 2\eps$, we obtain
\begin{equation}\label{e:scales-balanced}
\int_\R P_a(s) \, \nu_X(\dd s) = 1 + I_X(\pi a) - P_a(0), \qquad a > 0 .
\end{equation}

We now remove the restriction to the center $v = 0$. For $\Re w > 0$, we set
\[
\mathcal Y_X(w) = \int_\R \frac{w}{\pi (w^2 + s^2)} \, \nu_X(\dd s) .
\]
For $w$ in a compact subset of the half-plane, $w^2 + s^2$ does not vanish, and $|w/(w^2 + s^2)| \le C (1 + s^2)^{-1}$. Using \eqref{e:scales-translation-bound}, dominated convergence, and Morera's theorem, we see that $\mathcal Y_X$ is holomorphic. The function $I_X$ is holomorphic in the half-plane as well, and \eqref{e:scales-balanced} shows that $\mathcal Y_X(a) = 1 + I_X(\pi a) - (\pi a)^{-1}$ for $a > 0$. The identity theorem now implies that
\[
\mathcal Y_X(w) = 1 + I_X(\pi w) - \frac{1}{\pi w}, \qquad \Re w > 0 .
\]

For $w = \eps - \iu v$, we have
\[
\Re \frac{w}{\pi (w^2 + s^2)} = \frac{1}{2\pi} \Re \Big( \frac{1}{w + \iu s} + \frac{1}{w - \iu s} \Big) = \frac12 \big( P_\eps(v - s) + P_\eps(v + s) \big) .
\]
Integrating against the even measure $\nu_X$, we obtain $\Re \mathcal Y_X(\eps - \iu v) = (P_\eps * \nu_X)(v)$. Taking real parts in the formula for $\mathcal Y_X$ and using $\Re (\pi (\eps - \iu v))^{-1} = P_\eps(v)$, we arrive at \eqref{e:intro-pair-transform}.

Finally, let $\nu$ be a translation-bounded positive Radon measure that satisfies \eqref{e:intro-pair-transform}. Then $P_\eps * \nu = P_\eps * \nu_X$ for all $\eps > 0$, and \zcref{l:tools-poisson}(b) implies that $\nu = \nu_X$.
\end{proof}

\begin{proof}[Proof of \zcref{c:intro-pair}]
The existence of $\nu_X$, its translation bound, and \eqref{e:intro-pair} are \zcref{l:scales-pair}, and \eqref{e:intro-pair-transform} and the uniqueness are \zcref{l:scales-pair-transform}.

Let $A_1$ and $A_2$ be bounded intervals. Applying \eqref{e:intro-pair} with $f(x, y) = \1_{A_1}(x) \1_{A_2}(y)$, and using that $\Xi_X$ has intensity one, we find
\[
\E \big[ \Xi_X(A_1) \Xi_X(A_2) \big] = |A_1 \cap A_2| + \int_\R |A_1 \cap (A_2 - s)| \, \nu_X(\dd s),
\]
Subtracting $\E \Xi_X(A_1) \, \E \Xi_X(A_2) = |A_1| |A_2|$ from this identity gives the formula for the covariance of $\Xi_X(A_1)$ and $\Xi_X(A_2)$.

We now prove the convergence in \eqref{e:intro-count-cov}. Almost surely, $\Xi_X$ does not charge the endpoints of $A_1$ and $A_2$, since it has intensity one. As in the proof of \zcref{p:proc-subsequences}, the map $\Pi \mapsto (\Pi(A_1), \Pi(A_2))$ on $\mathcal N(\R)$ is continuous at every point measure that does not charge these endpoints, and therefore $(\Xi_N(A_1), \Xi_N(A_2))$ converges in law to $(\Xi_X(A_1), \Xi_X(A_2))$. The bound \eqref{e:proc-counts} with $m = 4$ shows that the variables $\Xi_N(A_1) \Xi_N(A_2)$ are bounded in $L^2$, hence uniformly integrable, and the same holds for $\Xi_N(A_i)$. The means and the covariances converge, which proves \eqref{e:intro-count-cov}.
\end{proof}

\section{The Bartlett spectrum}\label{s:scales-fourier}

We next record the Poisson transform of the pair measure of the sine process, which identifies the metallic endpoint of $\nu_X$.

\begin{lemma}\label{l:scales-gue}
The measure $\nu_{\GUE}(\dd s) = \big( 1 - (\sin \pi s/(\pi s))^2 \big) \dd s$ satisfies, for all $\eps > 0$ and $v \in \R$,
\begin{equation}\label{e:scales-gue}
(P_\eps * \nu_{\GUE})(v) = 1 + \Re I_{\GUE}(\pi \eps - \iu \pi v) - P_\eps(v) .
\end{equation}
\end{lemma}

\begin{proof}
We set
\[
\sigma(s) = \Big( \frac{\sin \pi s}{\pi s} \Big)^2, \qquad s \in \R .
\]
The function $\sin(\pi s)/(\pi s)$ is the inverse Fourier transform of the indicator of $[-\frac12, \frac12]$, and $\sigma$ is the inverse Fourier transform of the convolution of this indicator with itself. The Fourier transforms of $\sigma$ and $P_\eps$ are
\[
\hat \sigma(k) = (1 - |k|)_+, \qquad \hat P_\eps(k) = e^{-2\pi \eps |k|} .
\]
Both functions are integrable, and with $z = \pi \eps - \iu \pi v$, the Fourier inversion formula yields
\[
(P_\eps * \sigma)(v) = \int_\R e^{-2\pi \eps |k| + 2\pi \iu k v} (1 - |k|)_+ \dd k = 2 \Re \int_0^1 e^{-2zk} (1 - k) \dd k .
\]
Integrating by parts, we find
\[
\int_0^1 e^{-2zk}(1 - k) \dd k = \frac{1}{2z} - \frac{1 - e^{-2z}}{4z^2},
\]
and therefore $(P_\eps * \sigma)(v) = \Re z^{-1} - \Re I_{\GUE}(z)$. Inserting $\Re z^{-1} = P_\eps(v)$ and $P_\eps * 1 = 1$, we obtain \eqref{e:scales-gue}.
\end{proof}

We now recall the spectral theory of second-order stationary point processes \cite[Section~8.2]{daley2003introduction}, in the normalization of our Fourier transform. We let $\nu^{\mathrm c}_X$ denote the signed measure $\delta_0 + \nu_X - \dd s$, and we call it the \emph{reduced covariance measure} of $\Xi_X$. By the proof of \zcref{l:scales-pair}, $\delta_0 + \nu_X$ is the reduced second moment measure of $\Xi_X$, and since $\Xi_X$ has intensity one, $\nu^{\mathrm c}_X$ is the reduced covariance measure $\breve C_2$ of \cite[(8.1.6)]{daley2003introduction}. Recall from \cite[Proposition~8.2.I and Definition~8.2.II]{daley2003introduction} that the Bartlett spectrum of $\Xi_X$ is defined as the positive measure on $\R$ with Fourier transform $\nu^{\mathrm c}_X$, in the sense of the Parseval relation. That reference uses the Fourier transform $\int_\R e^{\iu \omega u} f(u) \dd u = \hat f(-\omega/(2\pi))$, and we let $m_X$ denote the image of the Bartlett spectrum of $\Xi_X$ under $\omega \mapsto \omega/(2\pi)$. We also call $m_X$ the \emph{Bartlett spectrum} of $\Xi_X$. In physics, $m_X$ is called the structure factor of $\Xi_X$ (see the discussion after \zcref{t:crossover}).

The following lemma states the properties of $m_X$ that we use. Its first assertion and part (b) are standard, and the Laplace identity \eqref{e:scales-laplace} of part (a), which links $m_X$ to the function $I_X$, is the part specific to $\Xi_X$.

\begin{lemma}\label{l:scales-fourier}
Let $X > 0$. The Bartlett spectrum $m_X$ is an even, positive, and translation-bounded Radon measure on $\R$, and
\begin{equation}\label{e:scales-parseval}
\int_\R f \dd \nu^{\mathrm c}_X = \int_\R \hat f \dd m_X
\end{equation}
for all Schwartz functions $f$ on $\R$. Moreover, the following statements hold.
\begin{enumerate}[label=(\alph*)]
\item For $\Re z > 0$, the function $e^{-2z|k|}$ is integrable with respect to $m_X$, and
\begin{equation}\label{e:scales-laplace}
I_X(z) = \int_\R e^{-2z|k|} m_X(\dd k) .
\end{equation}
\item Let $f$ and $g$ be functions on $\R$, each of which is the sum of a bounded measurable function with bounded support and a Schwartz function. Then
\begin{equation}\label{e:scales-cov-fourier}
\Cov\big( \Xi_X(f), \Xi_X(g) \big) = \int_\R \hat f(k) \overline{\hat g(k)} \, m_X(\dd k) .
\end{equation}
\end{enumerate}
\end{lemma}

\begin{proof}
Recall from \zcref{p:proc-subsequences} and \eqref{e:proc-pairs} that $\Xi_X$ is stationary and has locally finite second moments, so that it is second-order stationary in the sense of \cite[Section~8.1]{daley2003introduction}. By \cite[Proposition~8.2.I(a)]{daley2003introduction}, its Bartlett spectrum is a symmetric and translation-bounded measure that satisfies the Parseval relation \cite[(8.2.1)]{daley2003introduction} for all Schwartz functions. It is positive; see the remarks after \cite[Definition~8.2.II]{daley2003introduction}. Its image $m_X$ under $\omega \mapsto \omega/(2\pi)$ is therefore even, positive, and translation-bounded. After the change of variables $\omega = 2\pi k$, the Parseval relation becomes
\[
\int_\R f \dd \nu^{\mathrm c}_X = \int_\R \hat f(-k) \, m_X(\dd k),
\]
which is \eqref{e:scales-parseval}, since $m_X$ is even. In the same way, \cite[Proposition~8.2.I(c)]{daley2003introduction} implies that
\begin{equation}\label{e:scales-variance-bartlett}
\Var \Xi_X(f) = \int_\R |\hat f|^2 \dd m_X
\end{equation}
for all bounded measurable functions $f$ with bounded support and for all Schwartz functions $f$.

We now prove (a). Let $a > 0$. For $\delta \in (0, 1]$, we set
\[
\psi_\delta(s) = \delta^{-1} e^{-\pi s^2/\delta^2}, \qquad s \in \R,
\]
so that $\hat \psi_\delta(k) = e^{-\pi \delta^2 k^2}$. For $r \in \R$, we apply \eqref{e:scales-parseval} to the Schwartz function $\psi_\delta(\cdot - r)$, with Fourier transform $k \mapsto e^{-2\pi \iu k r} \hat \psi_\delta(k)$. We multiply by $P_a(r)$ and integrate over $r$. Since $\psi_\delta$ is even and $\hat P_a(k) = e^{-2\pi a |k|}$, Fubini's theorem gives
\begin{equation}\label{e:scales-laplace-smoothed}
\int_\R (P_a * \psi_\delta) \dd \nu^{\mathrm c}_X = \int_\R e^{-2\pi a |k|} e^{-\pi \delta^2 k^2} \, m_X(\dd k) .
\end{equation}

To justify the use of Fubini's theorem, we note that
\[
(P_a * \psi_\delta)(s) = \int_\R P_a(s - \delta t) e^{-\pi t^2} \dd t \le 4 P_a(s) + \frac{e^{-\pi s^2/4}}{\pi a}, \qquad s \in \R,
\]
where we split the integral at $|\delta t| = |s|/2$ and used $P_a(s/2) \le 4 P_a(s)$ and $P_a \le 1/(\pi a)$. The total variation of $\nu^{\mathrm c}_X$ is at most $\delta_0 + \nu_X + \dd s$, which is translation-bounded by \eqref{e:scales-translation-bound}, and $m_X$ is translation-bounded. Therefore the integrals in \eqref{e:scales-laplace-smoothed} converge absolutely.

As $\delta \downarrow 0$, we have $P_a * \psi_\delta \to P_a$ pointwise, and the bound above does not depend on $\delta$. Dominated convergence on the left side of \eqref{e:scales-laplace-smoothed} and monotone convergence on the right side give
\[
\int_\R e^{-2\pi a |k|} m_X(\dd k) = \int_\R P_a \dd \nu^{\mathrm c}_X = P_a(0) + (P_a * \nu_X)(0) - 1 = I_X(\pi a),
\]
where we used that $P_a$ is even with integral one, \eqref{e:intro-pair-transform} with $\eps = a$ and $v = 0$, and that $I_X$ is real on the positive axis. In particular, $e^{-2\pi a |k|}$ is integrable with respect to $m_X$ for all $a > 0$. By dominated convergence and Morera's theorem, the right side of \eqref{e:scales-laplace} is holomorphic in $\Re z > 0$, and so is $I_X$. The identity theorem now implies \eqref{e:scales-laplace}.

Finally, we prove (b). The functions in (b) form a vector space, and the left and right sides of \eqref{e:scales-cov-fourier} are sesquilinear forms on it. By the polarization identity, it suffices to prove \eqref{e:scales-variance-bartlett} for all $f$ in this space. Let $f = f_1 + f_2$, where $f_1$ is a bounded measurable function with bounded support and $f_2$ is a Schwartz function. Let $\chi$ be a smooth function with compact support such that $0 \le \chi \le 1$ and $\chi = 1$ on $[-1, 1]$, and set
\[
g_n(s) = \chi(s/n) f_2(s), \qquad s \in \R, \ n \ge 1 .
\]
Then $f_1 + g_n$ has bounded support, and $f_2 - g_n$ is a Schwartz function, so that \eqref{e:scales-variance-bartlett} holds for $f_1 + g_n$ and for $f_2 - g_n$. Since $f_2 - g_n \to 0$ in the Schwartz space, there exist $\eps_n \to 0$ such that $|\widehat{f_2 - g_n}(k)| \le \eps_n (1 + k^2)^{-1}$ for all $k$, and the translation bound of $m_X$ implies that
\[
\Var \Xi_X(f_2 - g_n) = \int_\R |\widehat{f_2 - g_n}|^2 \dd m_X \le \eps_n^2 \int_\R (1 + k^2)^{-2} \, m_X(\dd k) \longrightarrow 0 .
\]
It follows that $\Var \Xi_X(f_1 + g_n) \to \Var \Xi_X(f)$ and that $\hat f_1 + \hat g_n \to \hat f$ in $L^2(m_X)$, which proves \eqref{e:scales-variance-bartlett} for $f$.
\end{proof}

By \eqref{e:scales-cov-fourier} with $f = g = \1_{[0, L]}$, we have
\begin{equation}\label{e:scales-variance-fourier}
\Sigma_X^2(L) = \int_\R \Big( \frac{\sin \pi L k}{\pi k} \Big)^2 m_X(\dd k), \qquad L \ge 0 .
\end{equation}
At the two endpoints of \zcref{p:endpoints}, the corresponding measures are explicit. By the proof of \zcref{l:scales-gue}, the reduced covariance measure $\delta_0 + \nu_{\GUE} - \dd s$ of the sine process has the Bartlett spectrum $\min(|k|, 1) \dd k$, and
\[
\int_\R e^{-2z|k|} \min(|k|, 1) \dd k = I_{\GUE}(z), \qquad \Re z > 0 .
\]
For a Poisson process of intensity one, the reduced covariance measure reduces to the atom $\delta_0$, and the Bartlett spectrum to the Lebesgue measure \cite[Example~8.2(a)]{daley2003introduction}, with Laplace transform $1/z$.

\section{Laplace transforms of positive measures}\label{s:scales-filter}

The following lemma passes from the convergence of Laplace transforms of even positive measures to their convergence against test functions that are not compactly supported. Its first part is a form of the continuity theorem for Laplace transforms \cite[Section~XIII.1]{feller1971introduction}. We give the short proof, since we need the bound \eqref{e:scales-growth} for the second part.

\begin{lemma}\label{l:scales-filter}
Let $(\mu_n)_{n \ge 1}$ be even positive Radon measures on $\R$, and set $L_n(a) = \int e^{-2a|u|} \mu_n(\dd u)$ for $a > 0$. Suppose that there exist $C > 0$ and $\beta \in (0, 2)$ such that $L_n(a) \le C a^{-\beta}$ for all $n \ge 1$ and $a > 0$, and that $L_n(a)$ converges to a limit $L(a)$ for all $a > 0$.
\begin{enumerate}[label=(\alph*)]
\item There exists a unique even positive Radon measure $\mu$ on $\R$ such that $\int e^{-2a|u|} \mu(\dd u) = L(a)$ for all $a > 0$, and $\mu_n$ converges vaguely to $\mu$.
\item Let $\Phi \colon \R \to \C$ be continuous, and suppose that there exists $A > 0$ such that $|\Phi(u)| \le A \min(1, u^{-2})$ for all $u \neq 0$. Then $\int \Phi \dd \mu_n \to \int \Phi \dd \mu$.
\item Let $\mathcal T$ be a compact metric space, and let $(u, t) \mapsto \Phi_t(u)$ be a continuous function on $\R \times \mathcal T$ such that each $\Phi_t$ satisfies the assumption of (b) with the same constant $A$. Then the convergence in (b) is uniform in $t \in \mathcal T$.
\end{enumerate}
\end{lemma}

\begin{proof}
We begin with (a). Bounding $e^{-2|u|/R}$ from below by $e^{-2}$ on $[-R, R]$, we find
\begin{equation}\label{e:scales-growth}
\mu_n([-R, R]) \le e^2 L_n(1/R) \le C e^2 R^\beta, \qquad R > 0 .
\end{equation}
The measures $\mu_n$ are bounded on compact sets uniformly in $n$, and every subsequence has a further subsequence that converges vaguely. Let $\mu$ be such a limit. It is even and positive, and it satisfies \eqref{e:scales-growth}, since $\mu([-R, R]) \le \liminf_n \mu_n((-R - \eta, R + \eta))$ for $\eta > 0$. For $a > 0$ and $R > 0$, the tails satisfy
\[
\int_{|u| > R} e^{-2a|u|} \mu_n(\dd u) \le e^{-aR} L_n(a/2) \le C e^{-aR} (a/2)^{-\beta},
\]
uniformly in $n$. Approximating $e^{-2a|u|}$ by continuous functions with compact support, we obtain
\[
\int_\R e^{-2a|u|} \mu(\dd u) = L(a), \qquad a > 0 .
\]

It remains to show that an even positive Radon measure $\mu$ is determined by its Laplace transform $L$, when $L$ is finite on $(0, \infty)$; then all subsequential limits agree, and the whole sequence converges. Let $\rho$ denote the image of $e^{-2|u|} \mu(\dd u)$ under $u \mapsto |u|$, which is a finite measure on $[0, \infty)$ with
\[
\int_{[0, \infty)} e^{-2\lambda x} \rho(\dd x) = L(1 + \lambda), \qquad \lambda \ge 0 .
\]
The linear span of the functions $x \mapsto e^{-2\lambda x}$, $\lambda \ge 0$, is an algebra that contains the constants and separates the points of the one-point compactification $[0, \infty]$, to which these functions extend continuously. By the Stone--Weierstrass theorem, it is dense in $C([0, \infty])$, and $\rho$ is determined by $L$. The even measure $\mu$ is in turn determined by $\rho$.

For (b), let $R \ge 1$. Covering $\{|u| > R\}$ by the shells $2^j R < |u| \le 2^{j + 1} R$, $j \ge 0$, and using \eqref{e:scales-growth}, we have
\begin{equation}\label{e:scales-filter-tail}
\int_{|u| > R} u^{-2} \mu_n(\dd u) \le \sum_{j \ge 0} (2^j R)^{-2} \mu_n([-2^{j + 1} R, 2^{j + 1} R]) \le C_\beta R^{\beta - 2},
\end{equation}
where
\[
C_\beta = C e^2 2^\beta \sum_{j \ge 0} 2^{(\beta - 2) j} < \infty ,
\]
since $\beta < 2$. The same bound holds for $\mu$. Let $\eta_R$ be continuous, with $0 \le \eta_R \le 1$, $\eta_R = 1$ on $[-R, R]$, and $\eta_R = 0$ outside $[-2R, 2R]$. Part (a) implies that
\[
\int_\R \Phi \eta_R \dd \mu_n \longrightarrow \int_\R \Phi \eta_R \dd \mu .
\]
Using \eqref{e:scales-filter-tail}, we bound the integrals of $\Phi (1 - \eta_R)$ against $\mu_n$ and $\mu$ by $A C_\beta R^{\beta - 2}$, and letting $R \to \infty$ proves (b).

For (c), we note that the tail bound in the proof of (b) is uniform in $t$. For fixed $R$, the functions $\Phi_t \eta_R$ form a compact set in the space of continuous functions supported in $[-2R, 2R]$ with the uniform norm, since $(u, t) \mapsto \Phi_t(u) \eta_R(u)$ is uniformly continuous on $[-2R, 2R] \times \mathcal T$. We cover this set by finitely many balls of radius $\delta$ and use the uniform bound \eqref{e:scales-growth} on $[-2R, 2R]$. We conclude that the convergence of $\int \Phi_t \eta_R \dd \mu_n$ to $\int \Phi_t \eta_R \dd \mu$ is uniform in $t$, and (c) follows.
\end{proof}

\section{The stationary approximation}\label{s:scales-stationary}

In this section and the next, $a > 0$, and we write $\bar a = \max(1, a)$. We consider the semigroup $e^{t(\Delta_D - a d)}$ of \zcref{d:kernels} with $X = 1$ and $z = a$, and the singlet states $\Psi^{1, a}_t$. By the scaling of time in the Feynman--Kac formula, we have
\begin{equation}\label{e:scales-scaling}
e^{t(X \Delta_D - zd)} = e^{Xt(\Delta_D - (z/X) d)}, \qquad \Psi^{X, z}_t = \Psi^{1, z/X}_{Xt}, \qquad X, t > 0, \ \Re z > 0 .
\end{equation}
Inserting \eqref{e:scales-scaling} with $z = Xa$ into \eqref{e:intro-K} and \eqref{e:intro-I}, we obtain
\begin{equation}\label{e:scales-scaled-scalar}
X I_X(Xa) = X \int_0^1 \int_x^1 \big\langle \Psi^{1, a}_{Xx}, e^{X(y - x)(\Delta_D - a d)} \Psi^{1, a}_{X(1 - y)} \big\rangle_D \dd y \dd x .
\end{equation}
Heuristically, the diffusion relaxes on the time scale $1/\sqrt{\bar a}$, and when $X \sqrt{\bar a}$ is large, the two outer singlet states in \eqref{e:scales-scaled-scalar} are close to their limit as $t \to \infty$, except near the faces of the simplex. We make this precise in \zcref{l:scales-approx}, after four preliminary lemmas.

We next introduce the limit of the singlet states $\Psi^{1, a}_t$ as $t \to \infty$, which is identified in \zcref{l:scales-limit} below. For $a > 0$, we define functions on $D$ by
\begin{equation}\label{e:scales-pq}
p_a = 2 \sqrt{a(B + 1)}, \qquad q_a = 2 \sqrt{a(F + 1)},
\end{equation}
and
\begin{equation}\label{e:scales-state}
\Psi^{1, a}_\infty = p_a K_1(p_a) I_0(q_a) + q_a I_1(q_a) K_0(p_a) ,
\end{equation}
where $I_0$, $I_1$, $K_0$, and $K_1$ denote the modified Bessel functions of orders zero and one \cite[Section 10.25]{dlmf}. Since $B \ge 1$ and $F \le 1$, we have $p_a \ge \sqrt{8a} \ge q_a$, with equality at the corner. The function $\Psi^{1, a}_\infty$ was found by Skvortsov and Ostrovsky \cite[(14)]{skvortsov2007local} as the exact zero mode of the transfer-matrix Hamiltonian of a disordered wire in the unitary class; in their notation, $B = \lambda_1$, $F = \lambda$, and $4a = -2\iu \omega/\Delta_\xi$. We verify the properties that we need directly in \zcref{l:scales-state} and \zcref{l:scales-limit}.

\begin{lemma}\label{l:scales-state}
Let $a > 0$. The following statements hold.
\begin{enumerate}[label=(\alph*)]
\item Let $\mathfrak u_a = 2 K_0(p_a) I_0(q_a)$. The functions $\Psi^{1, a}_\infty$ and $\mathfrak u_a$ are smooth, positive, and bounded on $D$, and
\begin{equation}\label{e:scales-bessel-identities}
(ad - \Delta_D) \mathfrak u_a = \Psi^{1, a}_\infty, \qquad \partial_a \Psi^{1, a}_\infty = -d \mathfrak u_a .
\end{equation}
\item Let $\mathfrak w_a = \int_0^a \mathfrak u_{a'} \dd a'$. Then $\mathfrak w_a$ is smooth, nonnegative, and bounded on $D$, with $\nabla_D \mathfrak w_a$ bounded, and
\begin{equation}\label{e:scales-w}
\Psi^{1, a}_\infty = 1 - d \mathfrak w_a, \qquad (ad - \Delta_D) \mathfrak w_a = a .
\end{equation}
In particular, $\Psi^{1, a}_\infty \le 1$, with equality at the corner.
\end{enumerate}
\end{lemma}

\begin{proof}
We begin with (a). By the series \cite[(10.25.2)]{dlmf}, $I_0(q)$ and $q I_1(q)$ are entire functions of $q^2 = 4a(F + 1)$, and in particular smooth functions of $F$. The functions $K_0(p)$ and $p K_1(p)$ are analytic for $p > 0$, and $p_a \ge \sqrt{8a} > 0$; they are therefore smooth functions of $B$. All four functions are nonnegative; only $q I_1(q)$ vanishes, and only at $q = 0$. By \cite[(10.29.3), (10.29.4)]{dlmf}, we have $K_0' = -K_1$, $(pK_1(p))' = -p K_0(p)$, $I_0' = I_1$, and $(q I_1(q))' = q I_0(q)$. It follows that $K_0(p)$ and $pK_1(p)$ are decreasing in $p$, and $I_0(q)$ and $q I_1(q)$ are increasing in $q$. The bounds $p_a \ge \sqrt{8a} \ge q_a$ show that all four functions are bounded by their values at $\sqrt{8a}$, and $\Psi^{1, a}_\infty$ and $\mathfrak u_a$ are bounded.

We now prove the first identity in \eqref{e:scales-bessel-identities}. Let $\zeta_a = \sqrt{8a}$. In the variables $(p, q) = (p_a, q_a)$, we have $B + 1 = p^2/(4a)$, $B - 1 = (p^2 - \zeta_a^2)/(4a)$, $F + 1 = q^2/(4a)$, $1 - F = (\zeta_a^2 - q^2)/(4a)$, and $d = (p^2 - q^2)/(4a)$. Since $\partial_B = (2a/p) \partial_p$ and $\partial_F = (2a/q) \partial_q$, we find
\[
\Delta_D = \frac{1}{4p} \partial_p \big( p (p^2 - \zeta_a^2) \partial_p \big) + \frac{1}{4q} \partial_q \big( q (\zeta_a^2 - q^2) \partial_q \big) .
\]
By the modified Bessel equation \cite[(10.25.1)]{dlmf} with order zero, $K_0'' = K_0 - K_0'/p$ and $I_0'' = I_0 - I_0'/q$. Inserting these identities, we find
\begin{align*}
\frac{1}{4p} \partial_p \big( p (p^2 - \zeta_a^2) K_0'(p) \big) &= \frac{(p^2 - \zeta_a^2) K_0(p) + 2p K_0'(p)}{4}, \\
\frac{1}{4q} \partial_q \big( q (\zeta_a^2 - q^2) I_0'(q) \big) &= \frac{(\zeta_a^2 - q^2) I_0(q) - 2 q I_0'(q)}{4} .
\end{align*}
Multiplying the first identity by $I_0(q)$ and the second by $K_0(p)$, adding, and using $K_0' = -K_1$ and $I_0' = I_1$, we find
\begin{align*}
\Delta_D \big[ K_0(p) I_0(q) \big] &= \frac{p^2 - q^2}{4} K_0(p) I_0(q) - \frac12 \big( p K_1(p) I_0(q) + q I_1(q) K_0(p) \big) \\
&= a d \, K_0(p) I_0(q) - \frac12 \Psi^{1, a}_\infty,
\end{align*}
which is the first identity. For the second, we use $\partial_a p_a = p_a/(2a)$ and $\partial_a q_a = q_a/(2a)$ and the derivatives above, which give
\begin{align*}
\partial_a \Psi^{1, a}_\infty &= \frac{1}{2a} \Big( -p^2 K_0(p) I_0(q) + p q K_1(p) I_1(q) + q^2 I_0(q) K_0(p) - p q I_1(q) K_1(p) \Big) \\
&= -\frac{p^2 - q^2}{2a} K_0(p) I_0(q),
\end{align*}
and $(p^2 - q^2)/(2a) = 2d$.

For (b), we note that $K_0(p_{a'}) \le C(1 + |\log a'|)$ for $0 < a' \le a$ by \cite[(10.30.3)]{dlmf}. The integral defining $\mathfrak w_a$ therefore converges, and $\mathfrak w_a$ is bounded. Its derivatives are given by differentiating under the integral sign.

To justify this, we bound the derivatives of $\mathfrak u_{a'}$ for $0 < a' \le a$. Let $p = p_{a'}$ and $q = q_{a'}$, so that
\[
\partial_B p = \frac{2a'}{p} = \frac{p}{2(B + 1)}, \qquad \partial_F q = \frac{2a'}{q} .
\]
By \cite[(10.29.2), (10.29.4)]{dlmf}, we have
\[
K_0' = -K_1, \qquad \Big( \frac{K_1(p)}{p} \Big)' = -\frac{p K_0(p) + 2 K_1(p)}{p^2}, \qquad I_0' = I_1, \qquad \Big( \frac{I_1(q)}{q} \Big)' = \frac{I_2(q)}{q} .
\]
These identities give
\[
|\partial_B K_0(p)| = \frac{p K_1(p)}{2(B + 1)} \le \frac{1}{2(B + 1)}, \qquad \partial_B^2 K_0(p) = \frac{p^2 K_0(p) + 2 p K_1(p)}{4 (B + 1)^2} ,
\]
where we used $p K_1(p) \le 1$, a consequence of \cite[(10.30.2)]{dlmf} and the monotonicity of $p K_1(p)$, and
\[
\partial_F I_0(q) = 2a' \, \frac{I_1(q)}{q}, \qquad \partial_F^2 I_0(q) = 4 a'^2 \, \frac{I_2(q)}{q^2} .
\]
The functions $p^2 K_0(p)$ and $p K_1(p)$ are bounded on $(0, \infty)$, and $I_1(q)/q$ and $I_2(q)/q^2$ are bounded on $[0, \sqrt{8a}]$, since they are entire functions of $q^2$ by \cite[(10.25.2)]{dlmf}. The mixed derivative of $K_0(p) I_0(q)$ is the product $\partial_B K_0(p) \, \partial_F I_0(q)$. Together with $K_0(p) \le C(1 + |\log a'|)$ and $I_0(q) \le I_0(\sqrt{8a})$, these bounds show that all derivatives of $\mathfrak u_{a'}$ of order at most two are bounded on $D$ by an integrable function of $a' \in (0, a]$. This justifies the differentiation under the integral sign. Since $(B^2 - 1)/(B + 1)^2 \le 1$, they also show that $\nabla_D \mathfrak w_a$ is bounded.

As $a \downarrow 0$, we have $p_a K_1(p_a) \to 1$, $I_0(q_a) \to 1$, and $q_a I_1(q_a) K_0(p_a) \to 0$, by \cite[(10.30.1)--(10.30.3)]{dlmf} and $q_a \le p_a$. Integrating the second identity in \eqref{e:scales-bessel-identities} over $a$ yields the first identity in \eqref{e:scales-w}. For the second, we note that (a) and the first identity give $(a' d - \Delta_D) \mathfrak u_{a'} = 1 - d \mathfrak w_{a'}$ for $a' > 0$. We write $ad = a'd + (a - a')d$ and integrate over $a' \in (0, a)$, which gives
\begin{align*}
(ad - \Delta_D) \mathfrak w_a &= \int_0^a \big( (a' d - \Delta_D) \mathfrak u_{a'} + (a - a') d \mathfrak u_{a'} \big) \dd a' \\
&= a - d \int_0^a \mathfrak w_{a'} \dd a' + d \int_0^a (a - a') \mathfrak u_{a'} \dd a' = a,
\end{align*}
where the last step uses the identity
\[
\int_0^a \mathfrak w_{a'} \dd a' = \int_0^a \int_0^{a'} \mathfrak u_{a''} \dd a'' \dd a' = \int_0^a (a - a'') \mathfrak u_{a''} \dd a'' ,
\]
which follows from Fubini's theorem.
\end{proof}

\begin{lemma}\label{l:scales-l2}
Let $a > 0$. The following statements hold.
\begin{enumerate}[label=(\alph*)]
\item For nonnegative measurable functions $f$ and $g$ on $D$ and $t \ge 0$, we have $\langle f, e^{t(\Delta_D - ad)} g \rangle_D = \langle e^{t(\Delta_D - ad)} f, g \rangle_D$.
\item The operators $e^{t(\Delta_D - ad)}$, $t \ge 0$, restrict to a semigroup of self-adjoint contractions on $L^2(D)$, and for all $f \in L^2(D)$ and $t \ge 0$,
\begin{equation}\label{e:scales-gap}
\big\| e^{t(\Delta_D - ad)} f \big\|_{L^2(D)} \le e^{-t \sqrt{\bar a}/4} \|f\|_{L^2(D)} .
\end{equation}
\item For $t > 0$,
\begin{equation}\label{e:scales-flat-constant}
\big\| e^{t(\Delta_D - ad)} 1 \big\|_{L^2(D)}^2 \le \frac{1 - e^{-4at}}{4 a^2 t^2} .
\end{equation}
In particular, $\| e^{t(\Delta_D - ad)} 1 \|_{L^2(D)}^2 \le 2/a$ for $t \ge 1/(2\sqrt{\bar a})$.
\item There exists an absolute constant $C$ such that for all $g \in L^2(D)$,
\begin{equation}\label{e:scales-multiplier}
\big\| d \, e^{(\Delta_D - ad)/\sqrt{\bar a}} g \big\|_{L^2(D)} \le \frac{C \sqrt{\bar a}}{a} \|g\|_{L^2(D)} .
\end{equation}
\end{enumerate}
\end{lemma}

\begin{proof}
We write $S_t = e^{t(\Delta_D - ad)}$ in this proof. We begin with (a). By \zcref{l:sigma-scalar}(a) with $X = 1$ and $z = a$, the identity holds for bounded integrable functions $f$ and $g$. For nonnegative measurable $f$ and $g$, we apply it to $\min(f, n) \1_{\{B \le n\}}$ and $\min(g, n) \1_{\{B \le n\}}$, and (a) follows by monotone convergence as $n \to \infty$.

For (b), the Cauchy--Schwarz inequality for the positive kernel of $S_t$, the bound $S_t 1 \le 1$, and (a) give
\[
\|S_t f\|^2_{L^2(D)} \le \int_D S_t(|f|^2) \, S_t 1 \le \int_D |f|^2 \, S_t 1 \le \|f\|^2_{L^2(D)} .
\]
In particular, $S_t$ is a contraction on $L^2(D)$, and (a) shows that it is self-adjoint.

To prove \eqref{e:scales-gap}, we use the following form of Schur's test. Let $S$ be a self-adjoint operator with a nonnegative kernel, and suppose that $S \vartheta \le \theta \vartheta$ for a positive function $\vartheta$ and a constant $\theta > 0$. Then $\|S\|_{L^2 \to L^2} \le \theta$. Indeed, by the Cauchy--Schwarz inequality, we have $(S|f|)^2 \le S(|f|^2/\vartheta) \, S\vartheta$, and integrating with the symmetry of $S$, we find
\[
\|S f\|^2 \le \theta \int \vartheta \, S\big( |f|^2/\vartheta \big) = \theta \int |f|^2 \frac{S\vartheta}{\vartheta} \le \theta^2 \|f\|^2 .
\]

We apply Schur's test with two functions. First, let $\vartheta = (B + 1)^{-1/2}$. A computation gives
\[
\Delta_D \vartheta = -\frac{B + 3}{4(B + 1)} \vartheta \le -\frac14 \vartheta .
\]
By \zcref{l:met-ito}(a) with $X = 1$, the process $e^{t/4 - a \int_0^t \tilde d_s \dd s} \vartheta(\tilde B_t, \tilde F_t)$ is a positive local supermartingale, hence a supermartingale, which implies that $S_t \vartheta \le e^{-t/4} \vartheta$. Second, we use the function $\mathfrak u_a$ of \zcref{l:scales-state}(a), which satisfies $(ad - \Delta_D) \mathfrak u_a = \Psi^{1, a}_\infty$. Using \eqref{e:scales-state}, we find
\[
\frac{\Psi^{1, a}_\infty}{\mathfrak u_a} = \frac{p_a K_1(p_a)}{2 K_0(p_a)} + \frac{q_a I_1(q_a)}{2 I_0(q_a)} \ge \frac{p_a}{2} \ge \sqrt{2a},
\]
where we used $K_1 \ge K_0$, which follows from \cite[(10.32.9)]{dlmf}, and $p_a \ge \sqrt{8a}$. The same supermartingale argument shows that $S_t \mathfrak u_a \le e^{-\sqrt{2a} t} \mathfrak u_a$. Schur's test now implies that $\|S_t\| \le \exp(-t \max(\frac14, \sqrt{2a}))$, and $\max(\frac14, \sqrt{2a}) \ge \sqrt{\bar a}/4$, which proves \eqref{e:scales-gap}.

For (c), we note that (a) and the semigroup property give
\[
\|S_t 1\|^2_{L^2(D)} = \langle S_t 1, S_t 1 \rangle_D = \langle 1, S_{2t} 1 \rangle_D .
\]
By Jensen's inequality for the time average of $\tilde d_s$ on $[0, 2t]$,
\[
\exp\Big( -a \int_0^{2t} \tilde d_s \dd s \Big) \le \frac{1}{2t} \int_0^{2t} e^{-2ta \tilde d_s} \dd s .
\]
The radial diffusion without killing preserves $\dd B \dd F$, since its mass is conserved and its transition kernel is symmetric, by (a) and monotone convergence as $a \downarrow 0$. Taking expectations and integrating over the starting point, we find
\[
\int_D S_{2t} 1 \le \int_D e^{-2tad} \dd B \dd F = \int_0^\infty e^{-2tad} \min(d, 2) \dd d = \frac{1 - e^{-4at}}{4 a^2 t^2},
\]
where we used \eqref{e:met-radial}. The last assertion of (c) follows from the bound
\[
\frac{1 - e^{-4at}}{4 a^2 t^2} \le \min\Big( \frac{1}{at}, \frac{1}{4 a^2 t^2} \Big),
\]
applied with the first term if $a \le 1$ and with the second if $a \ge 1$.

For (d), let $r = 1/\sqrt{\bar a}$ and $\gamma = a/\sqrt{\bar a}$. Positivity of the kernel, the Cauchy--Schwarz inequality, and (a) give, also for complex-valued $g$,
\[
\| d \, S_r g \|^2 \le \int_D d^2 \, S_r(|g|^2) \, S_r 1 \le \langle S_r(d^2), |g|^2 \rangle_D \le \sup_D S_r(d^2) \, \|g\|^2 .
\]
To bound $S_r(d^2)$, we apply \zcref{l:met-envelope} to the radial diffusion with $X = 1$, this $\gamma$, $T = r$, $\lambda = \sqrt{\bar a}$, $n = 1$, $q = 2$, $t_1 = r$, and $r_1 = 0$, so that $\lambda \gamma d = ad$ and $\lambda \tau = 1$. We may take $\Theta = 3$, since $(1 + 2\gamma) r \le 3$ for all $a > 0$. With $R = \gamma d$, the lemma yields
\[
S_r\big( (1 + R)^2 \big) \le e^{36} (1 + R)^2 \mathcal E_r(R) \le C,
\]
where $C$ is absolute, since $(1 + R)^2 \mathcal E_r(R)$ is bounded on $[0, \infty)$. With $d^2 \le \gamma^{-2} (1 + R)^2$, it follows that $S_r(d^2) \le C \gamma^{-2}$. Inserting this into the previous bound and using $\gamma^{-2} = \bar a / a^2$, we obtain \eqref{e:scales-multiplier}.
\end{proof}

\begin{lemma}\label{l:scales-limit}
Let $a > 0$. For all $t \ge 0$,
\begin{equation}\label{e:scales-state-identity}
\Psi^{1, a}_t = \Psi^{1, a}_\infty + d \, e^{t(\Delta_D - ad)} \mathfrak w_a .
\end{equation}
Consequently, $0 < \Psi^{1, a}_\infty \le \Psi^{1, a}_t \le 1$, the function $\Psi^{1, a}_t$ is nonincreasing in $t$ and converges to $\Psi^{1, a}_\infty$ at every point of $D$ as $t \to \infty$, and
\begin{equation}\label{e:scales-state-tail}
\Psi^{1, a}_t - \Psi^{1, a}_\infty = a d \int_t^\infty e^{s(\Delta_D - ad)} 1 \dd s .
\end{equation}
\end{lemma}

\begin{proof}
We apply \zcref{l:met-duhamel} with $X = 1$, $z = a$, and the time-independent function $u = \mathfrak w_a$, which satisfies its assumptions by \zcref{l:scales-state}(b). Using $(\Delta_D - ad) \mathfrak w_a = -a$ in \eqref{e:met-duhamel}, we find
\[
e^{t(\Delta_D - ad)} \mathfrak w_a = \mathfrak w_a - a \int_0^t e^{s(\Delta_D - ad)} 1 \dd s .
\]
Multiplying by $d$ and using \eqref{e:intro-singlet} and \eqref{e:scales-w}, we obtain
\[
d \, e^{t(\Delta_D - ad)} \mathfrak w_a = (1 - \Psi^{1, a}_\infty) - (1 - \Psi^{1, a}_t),
\]
which is \eqref{e:scales-state-identity}. Since $\mathfrak w_a \ge 0$, this identity implies that $\Psi^{1, a}_t \ge \Psi^{1, a}_\infty > 0$, and \eqref{e:intro-singlet} shows that $\Psi^{1, a}_t \le 1$ and that $\Psi^{1, a}_t$ is nonincreasing in $t$.

We now identify the limit, and we first show that $e^{t(\Delta_D - ad)} 1 \to 0$ at every point $p \in D$. The radial diffusion escapes to infinity in the coordinate $B$. Indeed, we have $\Delta_D \log B = 1 + B^{-2}$ and $|\nabla_D \log B|^2 = 1 - B^{-2}$. Applying \zcref{l:met-ito}(a) with $X = 1$ and $f = \log B$, which is smooth on a neighborhood of $D$, we find
\[
\log \tilde B_t = \log \tilde B_0 + \int_0^t \big( 1 + \tilde B_s^{-2} \big) \dd s + \tilde M_t \ge t + \tilde M_t ,
\]
where $\tilde M$ is a continuous local martingale with $\langle \tilde M \rangle_t \le 2t$. For $\theta > 0$, the process $\exp(-\theta \tilde M_t - \theta^2 \langle \tilde M \rangle_t/2)$ is a supermartingale, as in the proof of \zcref{l:met-ito}(d), and Doob's maximal inequality with $\theta = 1/8$ implies that
\[
\Prob_p \Big( \inf_{s \le t} \tilde M_s \le -\frac t4 \Big) \le e^{-t/64} .
\]
On the complementary event, we have for all $s \in [t/2, t]$
\[
\log \tilde B_s \ge \frac s2 \ge \frac t4, \qquad \tilde d_s \ge \tilde B_s - 1 \ge e^{t/4} - 1 .
\]
It follows that
\[
\big( e^{t(\Delta_D - ad)} 1 \big)(p) \le \exp\Big( -\frac{at}{2} \big( e^{t/4} - 1 \big) \Big) + e^{-t/64} ,
\]
which tends to zero as $t \to \infty$.

Next, by \eqref{e:scales-state-identity}, we have
\[
0 \le \Psi^{1, a}_t - \Psi^{1, a}_\infty = d \, e^{t(\Delta_D - ad)} \mathfrak w_a \le d \, \|\mathfrak w_a\|_\infty \, e^{t(\Delta_D - ad)} 1 ,
\]
and the right side tends to zero at every point of $D$. Finally, for $t < t'$, \eqref{e:intro-singlet} implies that
\[
\Psi^{1, a}_t - \Psi^{1, a}_{t'} = ad \int_t^{t'} e^{s(\Delta_D - ad)} 1 \dd s,
\]
and letting $t' \to \infty$ with monotone convergence gives \eqref{e:scales-state-tail}.
\end{proof}

\begin{lemma}\label{l:scales-bounds}
There exists an absolute constant $C$ such that for all $a > 0$ the following statements hold.
\begin{enumerate}[label=(\alph*)]
\item For $t \ge 1/\sqrt{\bar a}$, we have $\|\Psi^{1, a}_t\|_{L^1(D)} \le C/a$. Moreover, $\|\Psi^{1, a}_\infty\|^2_{L^2(D)} \le \|\Psi^{1, a}_\infty\|_{L^1(D)} \le C/a$.
\item For $t \ge 2/\sqrt{\bar a}$,
\begin{equation}\label{e:scales-transient}
\big\| \Psi^{1, a}_t - \Psi^{1, a}_\infty \big\|_{L^2(D)} \le \frac{C}{\sqrt a} e^{-t \sqrt{\bar a}/4} .
\end{equation}
\item For $s, v, t \ge 0$ with $s + v + t \ge 3/\sqrt{\bar a}$,
\begin{equation}\label{e:scales-kernel-bound}
0 \le \big\langle \Psi^{1, a}_s, e^{v(\Delta_D - ad)} \Psi^{1, a}_t \big\rangle_D \le \frac{C}{a} e^{-v \sqrt{\bar a}/4} .
\end{equation}
\item For $X \ge 3$, we have $X I_X(Xa) \le C/a$.
\end{enumerate}
\end{lemma}

\begin{proof}
Below, $C$ denotes an absolute constant whose value may change from line to line, and we write $S_t = e^{t(\Delta_D - ad)}$. We begin with (a). As $\Psi^{1, a}_t$ is nonincreasing in $t$ (\zcref{l:scales-limit}), we may assume that $t = 1/\sqrt{\bar a}$. By \zcref{l:res-singlet}(c) with $X = 1$ and $z = a$, we have
\[
\Psi^{1, a}_t(p) = \E^{\mathrm s}_p \exp\Big( -a \int_0^t d_s \dd s \Big), \qquad p \in D^\circ,
\]
where the singlet diffusion has generator $\Delta^{\mathrm s}_D$. We apply \zcref{l:met-envelope} to the singlet diffusion with $X = 1$, $\gamma = a/\sqrt{\bar a}$, $T = t$, $\lambda = \sqrt{\bar a}$, $n = 1$, $q = 0$, $t_1 = t$, and $r_1 = 0$, so that $\lambda \gamma d = ad$ and $\lambda \tau = 1$. We may take $\Theta = 3$, since $(1 + 2\gamma) t \le 3$. With $R = \gamma d$, we obtain
\[
\Psi^{1, a}_t \le \mathcal E_t(R) = e^{-((1 + R)^{1/2} - 1)} + e^{-(\frac12 \log(1 + R) - 3)_+^2/24} \qquad \text{on } D^\circ .
\]
The boundary of $D$ is a null set, and this bound suffices for the $L^1$ norm.

Using \eqref{e:met-radial}, we find
\[
\|\Psi^{1, a}_t\|_{L^1(D)} \le \int_0^\infty \mathcal E_t(\gamma d) \min(d, 2) \dd d .
\]
If $a \le 1$, then $\gamma = a$, and we bound $\min(d, 2) \le 2$. If $a \ge 1$, then $\gamma = \sqrt a$, and we bound $\min(d, 2) \le d$. In both cases, the change of variables $R = \gamma d$ gives
\[
\|\Psi^{1, a}_t\|_{L^1(D)} \le \frac2a \int_0^\infty (1 + R) \, \mathcal E_t(R) \dd R \le \frac Ca ,
\]
since $\mathcal E_t(R)$ decays faster than every power of $1 + R$. The second assertion of (a) follows from $0 < \Psi^{1, a}_\infty \le \Psi^{1, a}_t \le 1$.

We next prove (b). Let $r = 1/\sqrt{\bar a}$ and $r' = 1/(2\sqrt{\bar a})$, and let $t \ge 2/\sqrt{\bar a}$. The identity \eqref{e:scales-state-tail}, the semigroup property, and Tonelli's theorem give
\[
\Psi^{1, a}_t - \Psi^{1, a}_\infty = a d \, S_r u, \qquad u = \int_{t - r}^\infty S_s 1 \dd s .
\]
For $s \ge r'$, \eqref{e:scales-gap} and \zcref{l:scales-l2}(c) give
\[
\|S_s 1\|_{L^2(D)} \le e^{-(s - r')\sqrt{\bar a}/4} \|S_{r'} 1\|_{L^2(D)} \le e^{-(s - r')\sqrt{\bar a}/4} \Big( \frac2a \Big)^{1/2} .
\]
Integrating over $s \ge t - r$ and using $t - r \ge r'$, we find
\[
\|u\|_{L^2(D)} \le \Big( \frac2a \Big)^{1/2} \frac{4}{\sqrt{\bar a}} e^{-(t - r - r')\sqrt{\bar a}/4} \le \frac{C}{\sqrt{a \bar a}} e^{-t \sqrt{\bar a}/4} .
\]
Finally, by \eqref{e:scales-multiplier}, we have $\|a d \, S_r u\| \le C \sqrt{\bar a} \|u\|$, which proves \eqref{e:scales-transient}.

For (c), the lower bound follows from $\Psi^{1, a}_t \ge 0$ and the positivity of $S_v$. Suppose first that $v \ge 1/\sqrt{\bar a}$. The bound $0 \le \Psi^{1, a}_t \le 1$ for all $t$, \zcref{l:scales-l2}(a), \eqref{e:scales-gap}, and \zcref{l:scales-l2}(c) give
\[
\big\langle \Psi^{1, a}_s, S_v \Psi^{1, a}_t \big\rangle_D \le \langle 1, S_v 1 \rangle_D = \|S_{v/2} 1\|^2 \le e^{-(v - 2r')\sqrt{\bar a}/4} \|S_{r'} 1\|^2 \le \frac{2 e^{1/4}}{a} e^{-v \sqrt{\bar a}/4} .
\]
Suppose next that $v < 1/\sqrt{\bar a}$. Then $s + t > 2/\sqrt{\bar a}$, and $s \ge 1/\sqrt{\bar a}$ or $t \ge 1/\sqrt{\bar a}$. In the first case, we use $S_v \Psi^{1, a}_t \le S_v 1 \le 1$ and (a), which give
\[
\big\langle \Psi^{1, a}_s, S_v \Psi^{1, a}_t \big\rangle_D \le \|\Psi^{1, a}_s\|_{L^1(D)} \le \frac Ca \le \frac{C e^{1/4}}{a} e^{-v\sqrt{\bar a}/4} .
\]
In the second case, we use \zcref{l:scales-l2}(a) to move $S_v$ to the other side, and conclude in the same way. This proves \eqref{e:scales-kernel-bound}.

For (d), we use \eqref{e:scales-scaled-scalar} and (c) with $s = Xx$, $v = X(y - x)$, and $t = X(1 - y)$, whose sum $X$ satisfies $X \ge 3 \ge 3/\sqrt{\bar a}$. Integrating, we obtain
\[
X I_X(Xa) \le X \int_0^1 \int_x^1 \frac{C}{a} e^{-X(y - x)\sqrt{\bar a}/4} \dd y \dd x \le \frac{4C}{a \sqrt{\bar a}},
\]
which proves (d).
\end{proof}

We define the \emph{stationary pairing} by
\begin{equation}\label{e:scales-pairing}
\mathcal I(a) = \int_0^\infty \big\langle \Psi^{1, a}_\infty, e^{v(\Delta_D - ad)} \Psi^{1, a}_\infty \big\rangle_D \dd v, \qquad a > 0,
\end{equation}
which is finite by \eqref{e:scales-gap} and \zcref{l:scales-bounds}(a).

The stationary pairing and its integrand have a meaning in physics, which we explain in \zcref{r:scales-physics}.

The next lemma is the main estimate of this section. The difficulty is that the bounds must be uniform in $a \in (0, \infty)$: the relaxation time $1/\sqrt{\bar a}$ and the size $1/a$ of the stationary state both vary with $a$, and the three regimes of \zcref{p:scales-scalar} below use small $a$, large $a$, and fixed $a$.

\begin{lemma}\label{l:scales-approx}
There exists an absolute constant $C$ such that for all $a > 0$ and $X > 0$ with $X \sqrt{\bar a} \ge 3$,
\begin{equation}\label{e:scales-approx}
\big| X I_X(Xa) - \mathcal I(a) \big| \le \frac{C}{a \bar a X} .
\end{equation}
\end{lemma}

\begin{proof}
Below, $C$ denotes an absolute constant whose value may change from line to line, and we write $S_v = e^{v(\Delta_D - ad)}$ and $\Psi_t = \Psi^{1, a}_t$ for $t \in [0, \infty]$. In \eqref{e:scales-scaled-scalar}, we change variables to $b = x$ and $v = X(y - x)$, which gives
\begin{equation}\label{e:scales-simplex}
X I_X(Xa) = \int_0^X \int_0^{1 - v/X} \big\langle \Psi_{Xb}, S_v \Psi_{X(1 - b) - v} \big\rangle_D \dd b \dd v .
\end{equation}
The three times $Xb$, $v$, and $X(1 - b) - v$ add up to $X \ge 3/\sqrt{\bar a}$, and \eqref{e:scales-kernel-bound} bounds the integrand by $(C/a) e^{-v\sqrt{\bar a}/4}$.

We first remove the region where one of the times $Xb$ and $X(1 - b) - v$ of the two end states is less than $2/\sqrt{\bar a}$. For fixed $v$, the set of $b$ in this region has measure at most $4/(X \sqrt{\bar a})$. Its contribution to \eqref{e:scales-simplex} is therefore at most
\[
\int_0^X \frac{4}{X \sqrt{\bar a}} \frac{C}{a} e^{-v \sqrt{\bar a}/4} \dd v \le \frac{C}{a \bar a X} .
\]

On the remaining region, both of these times are at least $2/\sqrt{\bar a}$, and we replace the two end states by $\Psi_\infty$. Writing $s = Xb$ and $t = X(1 - b) - v$, we have
\[
\big| \langle \Psi_s, S_v \Psi_t \rangle_D - \langle \Psi_\infty, S_v \Psi_\infty \rangle_D \big| \le \|\Psi_s - \Psi_\infty\| \, \|S_v \Psi_t\| + \|\Psi_\infty\| \, \|S_v (\Psi_t - \Psi_\infty)\| .
\]
We bound the right side with \eqref{e:scales-gap}, \eqref{e:scales-transient}, and the bounds $\|\Psi_t\|^2 \le \|\Psi_t\|_{L^1(D)} \le C/a$ of \zcref{l:scales-bounds}(a). The right side is then at most
\[
\frac{C}{a} e^{-v \sqrt{\bar a}/4} \big( e^{-s \sqrt{\bar a}/4} + e^{-t \sqrt{\bar a}/4} \big) .
\]
The integral of $e^{-s\sqrt{\bar a}/4} + e^{-t\sqrt{\bar a}/4}$ over $b$ is bounded by $8/(X \sqrt{\bar a})$, and the integral of $e^{-v\sqrt{\bar a}/4}$ over $v$ by $4/\sqrt{\bar a}$. The error of the replacement is therefore bounded by $C/(a \bar a X)$.

The stationary integrand $\langle \Psi_\infty, S_v \Psi_\infty \rangle_D$ is also bounded by $(C/a) e^{-v \sqrt{\bar a}/4}$, by \eqref{e:scales-gap} and \zcref{l:scales-bounds}(a). Adding it back on the removed region costs at most $C/(a \bar a X)$, as in the first step. We have shown that
\[
\Big| X I_X(Xa) - \int_0^X \Big( 1 - \frac vX \Big) \langle \Psi_\infty, S_v \Psi_\infty \rangle_D \dd v \Big| \le \frac{C}{a \bar a X} .
\]
Finally, we compare this integral with $\mathcal I(a)$. Their difference is bounded by
\[
\frac{C}{a} \int_0^X \frac vX e^{-v \sqrt{\bar a}/4} \dd v + \frac{C}{a} \int_X^\infty e^{-v \sqrt{\bar a}/4} \dd v \le \frac{16 C}{a \bar a X} + \frac{4C}{a \sqrt{\bar a}} e^{-X \sqrt{\bar a}/4} \le \frac{C}{a \bar a X},
\]
where we used $e^{-y} \le 1/y$ for $y > 0$ in the last step. This completes the proof.
\end{proof}

\section{The stationary pairing}\label{s:scales-pairing}

We now compute $\mathcal I(a)$. The computation is elementary once the stationary state is explicit, and we will need a Laplace transform of the product $I_0 K_0$.

\begin{lemma}\label{l:scales-bessel}
For $a > 0$,
\begin{equation}\label{e:scales-bessel}
I_0(\sqrt{8a}) K_0(\sqrt{8a}) = \int_0^\infty e^{-at} \frac{e^{-4/t} I_0(4/t)}{2t} \dd t .
\end{equation}
\end{lemma}

\begin{proof}
For $r, \rho \ge 0$, let
\[
\mathrm{Av}(r, \rho) = \frac{1}{2\pi} \int_0^{2\pi} K_0\big( \sqrt a \, |r e^{\iu \theta} - \rho| \big) \dd \theta ,
\]
which is the average of $K_0(\sqrt a |x - y|)$ over $x$ on the circle of radius $r$ centered at zero, for $|y| = \rho$. We first show that
\begin{equation}\label{e:scales-circle}
\mathrm{Av}(r, \rho) = I_0\big( \sqrt a \min(r, \rho) \big) K_0\big( \sqrt a \max(r, \rho) \big), \qquad r, \rho > 0 .
\end{equation}
Fix $\rho > 0$. For $y \in \R^2$, the function $x \mapsto K_0(\sqrt a |x - y|)$ is smooth on $\R^2 \setminus \{y\}$ and satisfies $\Delta u = a u$ there, by the modified Bessel equation \cite[(10.25.1)]{dlmf} in polar coordinates centered at $y$. Averaging over the rotations of $y$ about zero, we see that $\mathrm{Av}(\cdot, \rho)$ is smooth on $[0, \rho)$ and satisfies the radial equation
\[
\partial_r^2 \mathrm{Av}(r, \rho) + \frac1r \partial_r \mathrm{Av}(r, \rho) = a \, \mathrm{Av}(r, \rho), \qquad 0 < r < \rho .
\]
The solutions of this equation are the combinations of $I_0(\sqrt a r)$ and $K_0(\sqrt a r)$, and $K_0(\sqrt a r)$ is unbounded as $r \downarrow 0$ by \cite[(10.30.3)]{dlmf}. The function $\mathrm{Av}(\cdot, \rho)$ is continuous at $r = 0$, with $\mathrm{Av}(0, \rho) = K_0(\sqrt a \rho)$. With $I_0(0) = 1$, this gives \eqref{e:scales-circle} for $r < \rho$.

The identity $|r e^{\iu \theta} - \rho| = |\rho e^{\iu\theta} - r|$ shows that $\mathrm{Av}$ is symmetric, which implies \eqref{e:scales-circle} for $r > \rho$. It remains to treat $r = \rho$. For $r \ge \rho/2$, we have
\[
|r e^{\iu \theta} - \rho|^2 = (r - \rho)^2 + 2 r \rho (1 - \cos\theta) \ge \rho^2 (1 - \cos\theta),
\]
and $K_0(\sqrt a \rho (1 - \cos\theta)^{1/2})$ is integrable in $\theta$, since $K_0$ is decreasing and has a logarithmic singularity at zero. Dominated convergence shows that $\mathrm{Av}(\cdot, \rho)$ is continuous at $\rho$, and \eqref{e:scales-circle} follows for $r = \rho$.

We now take $r = \rho = \sqrt 8$. By \cite[(10.32.1)]{dlmf}, we have
\[
I_0(4/t) = \frac{1}{2\pi} \int_0^{2\pi} e^{(4/t) \cos\theta} \dd \theta, \qquad t > 0,
\]
and since $|\sqrt 8 e^{\iu\theta} - \sqrt 8|^2 = 16 (1 - \cos\theta)$, we obtain
\[
\frac{e^{-4/t} I_0(4/t)}{2t} = \frac{1}{2\pi} \int_0^{2\pi} \frac{1}{2t} \exp\Big( - \frac{|\sqrt 8 e^{\iu\theta} - \sqrt 8|^2}{4t} \Big) \dd \theta .
\]
For $R > 0$, the substitution $s = at$ and \cite[(10.32.10)]{dlmf} with order zero give
\[
\int_0^\infty e^{-at} \frac{1}{2t} e^{-R^2/(4t)} \dd t = \frac12 \int_0^\infty \exp\Big( -s - \frac{a R^2}{4s} \Big) \frac{\dd s}{s} = K_0(\sqrt a R) .
\]
Integrating over $t$, using Tonelli's theorem and \eqref{e:scales-circle}, we obtain \eqref{e:scales-bessel}.
\end{proof}

Recall the function $m_\infty(u) = e^{-2/|u|} I_0(2/|u|)$ of \zcref{t:crossover} (with $m_\infty(0) = 0$). By \cite[(10.32.1)]{dlmf},
\begin{equation}\label{e:scales-angular}
m_\infty(u) = \frac1\pi \int_0^\pi \exp\Big( -\frac{2(1 - \cos\theta)}{|u|} \Big) \dd \theta, \qquad u \neq 0 .
\end{equation}

\begin{lemma}\label{l:scales-pairing}
The following statements hold.
\begin{enumerate}[label=(\alph*)]
\item The function $m_\infty$ is even and continuous, $0 < m_\infty(u) < 1$ for $u \neq 0$, and $m_\infty(u)$ is strictly increasing in $|u|$. Moreover, $m_\infty(u) \to 1$ as $|u| \to \infty$,
\begin{equation}\label{e:scales-m-endpoints}
m_\infty(u) \sim \frac{|u|^{1/2}}{2\sqrt\pi} \quad (u \to 0), \qquad m_\infty(u) \le \min\Big( 1, \frac{\sqrt{\pi |u|}}{4} \Big) \quad (u \in \R) .
\end{equation}
\item For $a > 0$, with $\zeta_a = \sqrt{8a}$,
\begin{equation}\label{e:scales-pairing-explicit}
\begin{aligned}
\mathcal I(a) &= \frac{8}{\zeta_a} \big( I_0(\zeta_a) K_1(\zeta_a) - I_1(\zeta_a) K_0(\zeta_a) \big) \\
&= -2 \frac{\dd}{\dd a} \Big( I_0(\sqrt{8a}) K_0(\sqrt{8a}) \Big) = \int_\R e^{-2a|u|} m_\infty(u) \dd u .
\end{aligned}
\end{equation}
\item As $a \downarrow 0$, we have $a \mathcal I(a) \to 1$, and as $a \to \infty$, we have $a^{3/2} \mathcal I(a) \to 1/(4\sqrt 2)$.
\end{enumerate}
\end{lemma}

\begin{proof}
We begin with (a). The integrand in \eqref{e:scales-angular} lies in $(0, 1)$ for $\theta \in (0, \pi]$ and is strictly increasing in $|u|$, which implies the first assertions, and the limit as $|u| \to \infty$ follows by dominated convergence. For the behavior at zero, we substitute $\theta = |u|^{1/2} v$, which gives
\[
m_\infty(u) = \frac{|u|^{1/2}}{\pi} \int_0^{\pi |u|^{-1/2}} \exp\Big( - \frac{2(1 - \cos(|u|^{1/2} v))}{|u|} \Big) \dd v .
\]
The inequality $1 - \cos\theta \ge 2\theta^2/\pi^2$ for $\theta \in [0, \pi]$ bounds the integrand by $e^{-4v^2/\pi^2}$, and the integrand tends to $e^{-v^2}$ as $u \to 0$. Dominated convergence and the integral $\int_0^\infty e^{-v^2} \dd v = \sqrt\pi/2$ give the asymptotic in \eqref{e:scales-m-endpoints}. The same bound implies that
\[
m_\infty(u) \le \frac{|u|^{1/2}}{\pi} \int_0^\infty e^{-4v^2/\pi^2} \dd v = \frac{\sqrt{\pi |u|}}{4},
\]
which completes the proof of (a).

We now prove (b). By \zcref{l:scales-state}(a), the function $\mathfrak u_a$ is smooth and bounded, $(\Delta_D - ad) \mathfrak u_a = -\Psi^{1, a}_\infty$, and $\nabla_D \mathfrak u_a$ is bounded, by the argument in the proof of \zcref{l:scales-state}(b). Applying \zcref{l:met-duhamel} with $X = 1$, $z = a$, and the time-independent function $u = \mathfrak u_a$, we obtain
\[
\mathfrak u_a = e^{t(\Delta_D - ad)} \mathfrak u_a + \int_0^t e^{v(\Delta_D - ad)} \Psi^{1, a}_\infty \dd v, \qquad t \ge 0 .
\]
The function $\mathfrak u_a$ belongs to $L^2(D)$, since $K_0$ decays exponentially by \cite[(10.40.2)]{dlmf} and $p_a \ge 2 \sqrt{aB}$. We pair this identity with $\Psi^{1, a}_\infty$ and let $t \to \infty$. The contribution of the first term on the right tends to zero by \eqref{e:scales-gap}, and that of the second tends to $\mathcal I(a)$ by monotone convergence. We conclude that
\[
\mathcal I(a) = \langle \Psi^{1, a}_\infty, \mathfrak u_a \rangle_D .
\]

We change variables to $(p, q) = (p_a, q_a)$, which range over $p \ge \zeta_a$ and $0 \le q \le \zeta_a$, with $\dd B \dd F = pq \dd p \dd q/(4a^2)$. This gives
\[
\mathcal I(a) = \frac{1}{2a^2} \bigg( \int_{\zeta_a}^\infty p^2 K_0 K_1 \dd p \int_0^{\zeta_a} q I_0^2 \dd q + \int_{\zeta_a}^\infty p K_0^2 \dd p \int_0^{\zeta_a} q^2 I_0 I_1 \dd q \bigg),
\]
where the arguments of $K_0$, $K_1$ are $p$ and those of $I_0$, $I_1$ are $q$. Using $K_0' = -K_1$ and $I_0' = I_1$, we integrate by parts in the mixed integrals, which gives
\[
\int_{\zeta_a}^\infty p^2 K_0 K_1 \dd p = \frac{\zeta_a^2}{2} K_0(\zeta_a)^2 + \int_{\zeta_a}^\infty p K_0^2 \dd p, \qquad \int_0^{\zeta_a} q^2 I_0 I_1 \dd q = \frac{\zeta_a^2}{2} I_0(\zeta_a)^2 - \int_0^{\zeta_a} q I_0^2 \dd q .
\]
We also use
\[
\int_{\zeta_a}^\infty p K_0(p)^2 \dd p = \frac{\zeta_a^2}{2} \big( K_1(\zeta_a)^2 - K_0(\zeta_a)^2 \big), \qquad \int_0^{\zeta_a} q I_0(q)^2 \dd q = \frac{\zeta_a^2}{2} \big( I_0(\zeta_a)^2 - I_1(\zeta_a)^2 \big),
\]
which follow by differentiation in $\zeta_a$ from the derivative formulas \cite[(10.29.2), (10.29.3)]{dlmf}. Inserting these four identities and using $\zeta_a^4 = 64 a^2$, we obtain
\[
\mathcal I(a) = \frac{\zeta_a^4}{8 a^2} \Big( K_0^2 \big( I_0^2 - I_1^2 \big) + I_0^2 \big( K_1^2 - K_0^2 \big) \Big)(\zeta_a) = 8 \Big( (I_0 K_1)^2 - (I_1 K_0)^2 \Big)(\zeta_a) .
\]
By the Wronskian \cite[(10.28.2)]{dlmf}, $I_0 K_1 + I_1 K_0 = 1/\zeta_a$, and the first identity in \eqref{e:scales-pairing-explicit} follows. The second follows from $\dd \zeta_a/\dd a = 4/\zeta_a$ and $(I_0 K_0)' = I_1 K_0 - I_0 K_1$.

For the third identity, we differentiate the identity of \zcref{l:scales-bessel} in $a$. This is justified by dominated convergence, since the integrand of \eqref{e:scales-bessel} can be written as $e^{-at} m_\infty(t/2)/(2t)$ and $m_\infty \le 1$. We obtain
\[
-2 \frac{\dd}{\dd a} \Big( I_0(\sqrt{8a}) K_0(\sqrt{8a}) \Big) = \int_0^\infty e^{-at} m_\infty(t/2) \dd t = \int_\R e^{-2a|u|} m_\infty(u) \dd u ,
\]
where we substituted $t = 2|u|$ and used that $m_\infty$ is even.

For (c), we substitute $u = v/a$ in the last expression in \eqref{e:scales-pairing-explicit}, which gives
\[
a \mathcal I(a) = \int_\R e^{-2|v|} m_\infty(v/a) \dd v, \qquad a^{3/2} \mathcal I(a) = \int_\R e^{-2|v|} \sqrt a \, m_\infty(v/a) \dd v .
\]
As $a \downarrow 0$, we find the limit $\int e^{-2|v|} \dd v = 1$ by dominated convergence with $m_\infty \le 1$. As $a \to \infty$, dominated convergence with the bound $\sqrt a \, m_\infty(v/a) \le \sqrt{\pi |v|}/4$ of \eqref{e:scales-m-endpoints} gives the limit
\[
\frac{1}{2\sqrt\pi} \int_\R e^{-2|v|} |v|^{1/2} \dd v = \frac{1}{2\sqrt\pi} \cdot \frac{2 \Gamma(3/2)}{2^{3/2}} = \frac{1}{4\sqrt 2} .
\]
This completes the proof.
\end{proof}

\begin{remark}\label{r:scales-physics}
The stationary pairing and its inverse Laplace transform $m_\infty$ appear in the physics literature on long wires in the unitary class. We begin with the integrand of \eqref{e:scales-pairing}. In the notation of Ivanov, Ostrovsky, and Skvortsov \cite{ivanov2009correlations}, where $2\tilde H = ad - \Delta_D$ with $a = 2\Omega$, the integrand in \eqref{e:scales-pairing} at $v = t$ is the bilinear expression $\langle \Psi_0 | e^{-2\tilde H t} | \Psi_0 \rangle$ of \cite[(16)]{ivanov2009correlations}, which enters their formula for the local density of states correlation at two points of an infinite wire at distance $t$, in units of the localization length. In \cite[(16)--(22)]{micklitz2016spectral}, Micklitz computed the correlation of levels in a long wire from the zero mode and from the solution of the transfer-matrix equation with the zero mode as its source. The stationary pairing corresponds to the integral of the bilinear expression over the distance, and he evaluated it in closed form.

We now compare the closed forms precisely. Micklitz \cite[(8)--(9)]{micklitz2016spectral} and Marinho and Micklitz \cite[(12)--(13)]{marinho2018spectral} computed the connected correlation function $K$ of the density of states at a separation $\omega$ of energies, for a wire whose length $L$ is large compared with its localization length $\xi$. We use the localization length $\xi$ and the level spacing $\Delta_\xi$ of a localization volume as defined in \cite[after (2)]{micklitz2016spectral}, we let $\Delta$ denote the mean level spacing, and we set
\[
\frac L\xi = X, \qquad \Delta_\xi = \frac{X \Delta}{2\pi}, \qquad s = \frac \omega\Delta .
\]
This is the identification $g_{\rm KM} = 1/(2\pi X)$ of \zcref{r:met-km}, since in these conventions the conductance $g_{\rm KM}$ of \cite{kravtsov1994level}, the ratio of the Thouless energy to $\Delta$, equals $\xi/(2\pi L)$. The localization length and the level spacing in \cite{marinho2018spectral} are those of the orthogonal class, and they correspond to $\xi/2$ and $4\Delta_\xi$ in our notation \cite[notes 31 and 43]{marinho2018spectral}. With $a = -\iu \pi s/X$, the arguments of the Bessel functions in both formulas are $\sqrt{8a}$, and the first identity in \eqref{e:scales-pairing-explicit} shows that the formula of \cite[(12)--(13)]{marinho2018spectral} for the unitary class is
\begin{equation}\label{e:scales-physics}
K = \frac1X \Re \mathcal I\Big( -\frac{\iu \pi s}{X} \Big),
\end{equation}
where $\mathcal I$ is continued to $\{\Re a \ge 0, \ a \neq 0\}$ by its closed form. The formula of \cite[(8)--(9)]{micklitz2016spectral} is twice the right side of \eqref{e:scales-physics}, and \cite[note 34]{marinho2018spectral} records the correction of a factor $4$ in \cite{micklitz2016spectral}, in the relation between the zero mode and the generating function.

In our setting, \eqref{e:scales-physics} arises as follows. Formally, \eqref{e:intro-pair-transform} at $\eps = 0$ identifies the density of $\nu_X - \dd s$ at $s \neq 0$ with $\Re I_X(-\iu \pi s)$, and \zcref{p:scales-scalar}(c) gives $I_X(z) \approx X^{-1} \mathcal I(z/X)$ for large $X$. The right side of \eqref{e:scales-physics} is therefore the pair function of $\Xi_X$ on the scale $X$ that our results suggest, although we do not prove a pointwise statement of this kind. As $s/X \to \infty$, the asymptotic expansions \cite[(10.40.1), (10.40.2)]{dlmf} give $\mathcal I(a) \sim a^{-3/2}/(4\sqrt 2)$ also for $a$ on the imaginary axis, and the right side of \eqref{e:scales-physics} behaves like $-X^{1/2} s^{-3/2}/(8 \pi^{3/2})$. This is the regime of \zcref{t:largecov}, whose constant agrees with the law of Altshuler and Shklovskii (see the discussion after \zcref{t:largecov}).

By \zcref{l:scales-pairing}(b), the function $m_\infty$ is the inverse Laplace transform of $\mathcal I$. In the physics literature, $m_\infty$ is computed in \cite[(12)--(13)]{micklitz2014strong}, as the forward-scattering contrast discussed after \zcref{t:crossover}, and in \cite[(19)]{marinho2018spectral}, as a form factor, where $u$ is the time in units of the Heisenberg time $1/\Delta_\xi$ of a localization volume. At finite $X$, the corresponding object is the correlation function of the density of states of a wire of finite length. Altland and Fuchs \cite{altland1995spectral} computed it in the supersymmetric $\sigma$-model for practically all lengths and energy separations, from numerical solutions of the transfer-matrix equations according to \cite[Section~V]{micklitz2016spectral}, and Micklitz \cite[(25)--(26)]{micklitz2016spectral} gives a closed formula for separations large compared with $\Delta_\xi$ at every ratio $L/\xi$.
\end{remark}

We now collect the consequences of \zcref{l:scales-approx} and \zcref{l:scales-pairing} for the function $I_X$.

\begin{proposition}\label{p:scales-scalar}
The following statements hold.
\begin{enumerate}[label=(\alph*)]
\item As $X \to \infty$, $I_X(z) \to 1/z$ locally uniformly in $\Re z > 0$.
\item For fixed $X > 0$, as $z \to \infty$ along the positive axis,
\begin{equation}\label{e:scales-large-damping}
I_X(z) = \frac{\sqrt X}{4 \sqrt 2 \, z^{3/2}} \big( 1 + o(1) \big) .
\end{equation}
\item For fixed $a > 0$, $X I_X(Xa) \to \mathcal I(a)$ as $X \to \infty$. Moreover, $X I_X(Xa) \le C/a$ for all $X \ge 3$ and $a > 0$, with an absolute constant $C$.
\end{enumerate}
\end{proposition}

\begin{proof}
We begin with (a), first for $z$ in a compact interval $[z_0, z_1] \subset (0, \infty)$. For $X \ge \max(z_1, 3)$, we apply \zcref{l:scales-approx} with $a = z/X \le 1$, for which $\bar a = 1$, and we find
\[
I_X(z) = \frac1X \mathcal I\Big( \frac zX \Big) + O\Big( \frac{1}{zX} \Big) = \frac1z \Big( \frac zX \, \mathcal I\Big( \frac zX \Big) \Big) + O\Big( \frac{1}{zX} \Big) .
\]
The right side tends to $1/z$ uniformly for $z \in [z_0, z_1]$, by \zcref{l:scales-pairing}(c).

We now treat complex $z$. For $\Re z \ge \alpha > 0$, \eqref{e:scales-laplace} implies that
\[
|I_X(z)| \le I_X(\Re z) \le I_X(\alpha),
\]
and the right side is bounded for large $X$ by what we just proved. The functions $I_X$ are therefore locally bounded on the half-plane, uniformly for large $X$. By Montel's theorem, every sequence $X_n \to \infty$ has a subsequence along which $I_{X_n}$ converges locally uniformly to a holomorphic function, which equals $1/z$ on the positive axis. The identity theorem shows that this limit coincides with $1/z$ on the whole half-plane. This proves (a).

For (b), let $a = z/X$. When $z \ge X$, we have $\bar a = a$ and $X \sqrt{\bar a} = \sqrt{Xz}$, which is at least three for large $z$. Applying \zcref{l:scales-approx}, we find
\[
I_X(z) = \frac1X \mathcal I\Big( \frac zX \Big) + O\Big( \frac{1}{a^2 X^2} \Big) = \frac1X \mathcal I\Big( \frac zX \Big) + O(z^{-2}),
\]
and \zcref{l:scales-pairing}(c) gives $X^{-1} \mathcal I(z/X) \sim X^{-1} (4\sqrt 2)^{-1} (X/z)^{3/2}$, which is \eqref{e:scales-large-damping}.

The first assertion of (c) is \zcref{l:scales-approx} with $a$ fixed, and the second is \zcref{l:scales-bounds}(d).
\end{proof}

\section{The endpoints}\label{s:scales-endpoints}

We now prove the rows of \zcref{p:endpoints} for the pair measure, the number variance, and the kernel. The rows for the law of $\mu_X$ and for the eigenvector correlation are \zcref{p:coarse-endpoints}.

\begin{proof}[Proof of the rows of \zcref{p:endpoints} for $\nu_X$, $\Sigma_X^2$, and $K_X$]
We begin with the pair measure. By \eqref{e:scales-translation-bound}, the measures $\nu_X$ have unit-interval masses bounded by $C_\nu$ for all $X > 0$. As $X \downarrow 0$, \zcref{p:metallic} implies that $I_X \to I_{\GUE}$ pointwise in $\Re z > 0$. Using \eqref{e:intro-pair-transform} and \eqref{e:scales-gue}, we find
\[
(P_\eps * \nu_X)(v) \longrightarrow 1 + \Re I_{\GUE}(\pi\eps - \iu\pi v) - P_\eps(v) = (P_\eps * \nu_{\GUE})(v), \qquad \eps > 0, \ v \in \R .
\]
By \zcref{l:tools-poisson}(d), applied along every sequence $X_n \downarrow 0$, the measures $\nu_{X_n}$ converge vaguely to a translation-bounded measure $\nu$ with $P_\eps * \nu = P_\eps * \nu_{\GUE}$ for all $\eps > 0$, and \zcref{l:tools-poisson}(b) gives $\nu = \nu_{\GUE}$.

As $X \to \infty$, \zcref{p:scales-scalar}(a) implies that $I_X(z) \to 1/z$. Using $\Re (\pi \eps - \iu \pi v)^{-1} = P_\eps(v)$, we see that the right side of \eqref{e:intro-pair-transform} tends to one (the Poisson transform of the Lebesgue measure), and the same argument gives $\nu_X \to \dd s$ vaguely.

For the number variance, we note that $s \mapsto (L - |s|)_+$ is continuous with compact support, and $\int (L - |s|)_+ \dd s = L^2$. By \eqref{e:intro-variance} and the vague convergence just proved, $\Sigma_X^2(L)$ tends to
\[
L + L^2 - 2 \int_0^L (L - s) \Big( \frac{\sin \pi s}{\pi s} \Big)^2 \dd s - L^2 = \Sigma^2_{\GUE}(L)
\]
as $X \downarrow 0$, and to $L + L^2 - L^2 = L$ as $X \to \infty$.

For the kernel, \zcref{c:met-kernel} implies that
\[
\frac12 K_X(x, y; z) \longrightarrow \frac{1 - e^{-2z}}{2 z^2} = I_{\GUE}(z) \qquad (X \downarrow 0),
\]
uniformly for $z$ in compact subsets of $\{\Re z > 0\}$.

We now consider the limit $X \to \infty$. Let $\mathcal K$ be such a compact set, and let $\alpha > 0$ and $\alpha' < \infty$ be the minimum and the maximum of $\Re z$ over $\mathcal K$. By \zcref{r:intro-singlet} and \eqref{e:met-fk-domination}, we have $|K_X(x, y; z)| \le K_X(x, y; \Re z)$. Using \eqref{e:scales-scaling} with $a = \Re z/X$ and \eqref{e:scales-kernel-bound} with $s = Xx$, $v = X(y - x)$, and $t = X(1 - y)$, we obtain for $X \ge \max(\alpha', 3)$, so that $a \le 1$,
\[
|K_X(x, y; z)| \le \big\langle \Psi^{1, a}_{Xx}, e^{X(y - x)(\Delta_D - ad)} \Psi^{1, a}_{X(1 - y)} \big\rangle_D \le \frac{C X}{\alpha} e^{-X(y - x)/4},
\]
which tends to zero uniformly for $z \in \mathcal K$.
\end{proof}

The decay of the bound on $K_X$ is not uniform as $y - x \downarrow 0$; this is why the last two rows of \zcref{p:endpoints} concern fixed separated sites.

\section{Large-scale covariance}\label{s:scales-largecov}

We now prove \zcref{t:largecov}. For $X > 0$ and $L > 0$, we define $m_{X, L}$ as the image of $L^{3/2} m_X$ under $k \mapsto Lk$.

\begin{proof}[Proof of \zcref{t:largecov}]
Fix $X > 0$, and let $f$ and $g$ be as in the statement. With $\hat f_L(k) = L \hat f(Lk)$, \eqref{e:scales-cov-fourier} gives
\begin{equation}\label{e:scales-largecov-exact}
\Cov\big( \Xi_X(f_L), \Xi_X(g_L) \big) = L^2 \int_\R \hat f(Lk) \overline{\hat g(Lk)} \, m_X(\dd k) = \sqrt L \int_\R \hat f(u) \overline{\hat g(u)} \, m_{X, L}(\dd u) .
\end{equation}
We apply \zcref{l:scales-filter} to the measures $m_{X, L}$ along an arbitrary sequence $L \to \infty$, with $\beta = 3/2$. By \eqref{e:scales-laplace}, their Laplace transforms are
\[
\int_\R e^{-2a|u|} m_{X, L}(\dd u) = L^{3/2} I_X(aL), \qquad a > 0 .
\]
\zcref[S]{p:scales-scalar}(b) shows that these converge to $\sqrt X/(4\sqrt 2 a^{3/2})$.

We next verify the growth bound of \zcref{l:scales-filter}. Summing the integrand of \eqref{e:scales-laplace} over unit intervals and using the translation bound of $m_X$ in \zcref{l:scales-fourier}, we find $I_X(s) \le C_X s^{-1}$ for $0 < s \le 1$. By \eqref{e:scales-large-damping} and the continuity of $I_X$ on $[1, \infty)$, we have $I_X(s) \le C_X s^{-3/2}$ for $s \ge 1$. For $L \ge 1$ and $a > 0$, these two bounds give
\[
L^{3/2} I_X(aL) \le
\begin{cases}
C_X L^{1/2} a^{-1} \le C_X a^{-3/2} & \text{if } aL \le 1, \\
C_X a^{-3/2} & \text{if } aL \ge 1 .
\end{cases}
\]
The even measure $\sqrt X |u|^{1/2} \dd u/(2\sqrt\pi)$ has the Laplace transform $\sqrt X/(4\sqrt 2 a^{3/2})$, since
\[
\int_\R e^{-2a|u|} \frac{\sqrt X |u|^{1/2}}{2\sqrt\pi} \dd u = \frac{\sqrt X}{\sqrt\pi} \cdot \frac{\Gamma(3/2)}{(2a)^{3/2}} = \frac{\sqrt X}{4 \sqrt 2 \, a^{3/2}} .
\]
The uniqueness in \zcref{l:scales-filter}(a) shows that $m_{X, L}$ converges vaguely to this measure.

If $f$ is of bounded variation with compact support, then $f$ is bounded and integrable, and an integration by parts in the Stieltjes sense gives
\begin{equation}\label{e:scales-bv-fourier}
|\hat f(u)| \le \min\Big( \|f\|_{L^1}, \frac{\operatorname{TV}(f)}{2\pi |u|} \Big), \qquad u \neq 0,
\end{equation}
where $\operatorname{TV}(f)$ is the total variation of $f$ on $\R$. Schwartz functions satisfy $|\hat f(u)| \le C (1 + u^2)^{-1}$. Therefore the function $\Phi = \hat f \overline{\hat g}$ is continuous, and there exists $A > 0$ such that $|\Phi(u)| \le A \min(1, u^{-2})$ for all $u \neq 0$. \zcref[S]{l:scales-filter}(b) and \eqref{e:scales-largecov-exact} give
\[
\frac{\Cov\big( \Xi_X(f_L), \Xi_X(g_L) \big)}{\sqrt{XL}} \longrightarrow \frac{1}{2\sqrt\pi} \int_\R |u|^{1/2} \hat f(u) \overline{\hat g(u)} \dd u = \frac{1}{2\sqrt\pi} \langle f, g \rangle_{\dot H^{1/4}},
\]
which is the first limit in \eqref{e:intro-largecov}, and the bound on $\Phi$ shows that this integral converges absolutely.

We next prove the second expression in \eqref{e:intro-largecov}. For $c > 0$, an integration by parts and the integral $\int_0^\infty x^{-1/2} \sin x \dd x = \sqrt{\pi/2}$ give
\[
\int_0^\infty \frac{1 - \cos(c y)}{y^{3/2}} \dd y = \sqrt c \int_0^\infty \frac{1 - \cos x}{x^{3/2}} \dd x = 2 \sqrt c \int_0^\infty \frac{\sin x}{x^{1/2}} \dd x = \sqrt{2\pi c} .
\]
Taking $c = 2\pi |u|$, we find
\begin{equation}\label{e:scales-fractional}
\int_\R \big| e^{2\pi \iu u y} - 1 \big|^2 \frac{\dd y}{|y|^{3/2}} = 2 \int_\R \big( 1 - \cos(2\pi u y) \big) \frac{\dd y}{|y|^{3/2}} = 8 \pi |u|^{1/2}, \qquad u \in \R .
\end{equation}
By the Plancherel theorem and Tonelli's theorem,
\[
\int_\R \int_\R \frac{|f(s) - f(s')|^2}{|s - s'|^{3/2}} \dd s \dd s' = \int_\R \frac{\|f(\cdot + y) - f\|_{L^2}^2}{|y|^{3/2}} \dd y = 8\pi \int_\R |u|^{1/2} |\hat f(u)|^2 \dd u,
\]
which is finite. Both sides are Hermitian forms in $f$, and polarization gives the identity for $f \neq g$, and the Cauchy--Schwarz inequality shows that the double integral in \eqref{e:intro-largecov} converges absolutely. Dividing by $16 \pi^{3/2}$, we arrive at the second expression in \eqref{e:intro-largecov}.

Finally, $\Sigma_X^2(L) = \Var \Xi_X(f_L)$ for $f = \1_{[0, 1]}$, and for this function the double integral equals
\begin{align*}
2 \int_0^1 \Big( \int_{-\infty}^0 (x - y)^{-3/2} \dd y + \int_1^\infty (y - x)^{-3/2} \dd y \Big) \dd x &= 2 \int_0^1 \big( 2 x^{-1/2} + 2 (1 - x)^{-1/2} \big) \dd x \\
&= 16 .
\end{align*}
Inserting this value into the second expression in \eqref{e:intro-largecov}, we find
\[
\frac{\Sigma_X^2(L)}{\sqrt{XL}} \longrightarrow \frac{16}{16 \pi^{3/2}} = \frac{1}{\pi^{3/2}} .
\]
\end{proof}

\section{The crossover}\label{s:scales-crossover}

We now prove \zcref{t:crossover}. For $X > 0$, we define $\bar m_X$ as the image of $X m_X$ under $k \mapsto Xk$. The next proposition describes the covariance of linear statistics on the scale $X$; it shows that $m_\infty$ is the limit of the Bartlett spectra $m_X$ in the scaling of \zcref{t:crossover}.

\begin{proposition}\label{p:scales-wire}
As $X \to \infty$, the measures $\bar m_X$ converge vaguely to $m_\infty(u) \dd u$. Moreover, for compactly supported functions $f$ and $g$ of bounded variation, or Schwartz functions,
\begin{equation}\label{e:scales-wire}
\frac1X \Cov\big( \Xi_X(f_X), \Xi_X(g_X) \big) \longrightarrow \int_\R m_\infty(u) \hat f(u) \overline{\hat g(u)} \dd u .
\end{equation}
\end{proposition}

\begin{proof}
By \eqref{e:scales-laplace}, the Laplace transforms of $\bar m_X$ are
\[
\int_\R e^{-2a|u|} \bar m_X(\dd u) = X I_X(Xa), \qquad a > 0 .
\]
\zcref[S]{p:scales-scalar}(c) shows that they converge to $\mathcal I(a)$, and they are bounded by $C/a$ for $X \ge 3$. \zcref[S]{l:scales-pairing}(b) shows that $\mathcal I$ is the Laplace transform of the even measure $m_\infty(u) \dd u$. Applying \zcref{l:scales-filter}(a) with $\beta = 1$ along every sequence $X \to \infty$, we obtain the vague convergence. As in \eqref{e:scales-largecov-exact}, \eqref{e:scales-cov-fourier} gives
\[
\frac1X \Cov\big( \Xi_X(f_X), \Xi_X(g_X) \big) = X \int_\R \hat f(Xk) \overline{\hat g(Xk)} \, m_X(\dd k) = \int_\R \hat f(u) \overline{\hat g(u)} \, \bar m_X(\dd u),
\]
and \zcref{l:scales-filter}(b), applied with $\Phi = \hat f \overline{\hat g}$ and the bound $|\Phi(u)| \le A \min(1, u^{-2})$ from the proof of \zcref{t:largecov}, gives \eqref{e:scales-wire}.
\end{proof}

\begin{proof}[Proof of \zcref{t:crossover}]
For $\ell \ge 0$, we let
\[
\Phi_\ell(u) = \Big( \frac{\sin \pi \ell u}{\pi u} \Big)^2, \quad u \neq 0, \qquad \Phi_\ell(0) = \ell^2,
\]
so that $\Phi_\ell = |\hat f|^2$ for $f = \1_{[0, \ell]}$. By \eqref{e:scales-variance-fourier} with $L = X\ell$, we have
\[
\frac{\Sigma_X^2(X\ell)}{X} = \int_\R \Phi_\ell \dd \bar m_X .
\]
Let $\ell_0 > 0$. The function $(u, \ell) \mapsto \Phi_\ell(u)$ is continuous on $\R \times [0, \ell_0]$, and $\Phi_\ell(u) \le \max(\ell_0^2, \pi^{-2}) \min(1, u^{-2})$. \zcref[S]{p:scales-wire} and \zcref{l:scales-filter}(c) give $\Sigma_X^2(X\ell)/X \to V(\ell)$ uniformly for $\ell \in [0, \ell_0]$, and the same lemma shows that $V$ is continuous, with $V(0) = 0$.

We now prove the properties of $V$. The substitution $v = \ell u$ gives
\begin{equation}\label{e:scales-V-ratio}
\frac{V(\ell)}{\ell} = \int_\R m_\infty(v/\ell) \Big( \frac{\sin \pi v}{\pi v} \Big)^2 \dd v, \qquad \ell > 0 .
\end{equation}
By \zcref{l:scales-pairing}(a), the function $\ell \mapsto m_\infty(v/\ell)$ is strictly decreasing for $v \neq 0$, with values in $(0, 1)$. It follows from \eqref{e:scales-V-ratio} that $V(\ell)/\ell$ is strictly decreasing in $\ell > 0$, and that
\[
0 < V(\ell) < \ell \int_\R \Big( \frac{\sin \pi v}{\pi v} \Big)^2 \dd v = \ell .
\]
As $\ell \downarrow 0$, we have $m_\infty(v/\ell) \to 1$ for $v \neq 0$, and dominated convergence gives $V(\ell)/\ell \to 1$. As $\ell \to \infty$, the bound $\sqrt\ell \, m_\infty(v/\ell) \le \sqrt{\pi |v|}/4$ of \eqref{e:scales-m-endpoints} and the asymptotic there give, by dominated convergence,
\[
\frac{V(\ell)}{\sqrt \ell} = \int_\R \sqrt \ell \, m_\infty(v/\ell) \Big( \frac{\sin \pi v}{\pi v} \Big)^2 \dd v \longrightarrow \frac{1}{2\sqrt\pi} \int_\R |v|^{1/2} \Big( \frac{\sin \pi v}{\pi v} \Big)^2 \dd v = \frac{1}{\pi^{3/2}} .
\]
In the last step, we used $(\sin \pi v)^2 = (1 - \cos 2\pi v)/2$ and \eqref{e:scales-fractional} with $u = 1$, which give
\[
\int_\R |v|^{1/2} \Big( \frac{\sin \pi v}{\pi v} \Big)^2 \dd v = \frac{1}{2\pi^2} \int_\R \frac{1 - \cos 2\pi v}{|v|^{3/2}} \dd v = \frac{2}{\pi} .
\]
This completes the proof.
\end{proof}