\chapter{General tools}\label{s:tools}

This appendix collects four general facts that are used in several sections. The first is a stability estimate for an operator that differs from a contraction by a perturbation that is small relative to the dissipation of the contraction; it is used in \zcref{s:tube,s:stab}. The second is a Dirichlet-form bound for integral operators whose kernels take values in operators on a Hilbert space, and is used in \zcref{s:stab}. The third concerns Poisson transforms of positive measures on the line, and is used in Part III to identify limiting pair and occupation measures. The fourth is a submean inequality for holomorphic families of Hilbert-space-valued functions, which is used in \zcref{s:end,s:assembly}. All four are elementary, and we give complete proofs. The letters used in this appendix are local to it, except for the defects $D_R$ and $D_L$ of \eqref{e:tools-defects}, which \zcref{s:tube,s:stab} also use, and for $P_a$ and $\| \cdot \|_{\mathrm{loc}}$ in \zcref{s:tools-poisson}.

\section{Perturbations of a contraction}\label{s:tools-defect}

Let $\mathcal H$ be a Hilbert space, and let $C$ be a contraction on $\mathcal H$ (a bounded operator with $\|C\| \le 1$). We define the \emph{right defect} and the \emph{left defect} of $C$ to be
\begin{equation}\label{e:tools-defects}
 D_R = \Id - C^* C, \qquad D_L = \Id - C C^*.
\end{equation}
These are positive operators of norm at most one, and their square roots are the defect operators of $C$ and $C^*$ in the sense of Sz.-Nagy and Foia\c{s} \cite{sznagy2010harmonic}. For all $f \in \mathcal H$, we have
\[
 \|Cf\|^2 = \|f\|^2 - \|D_R^{1/2} f\|^2.
\]
Thus $D_R$ measures how much $C$ dissipates the norm of $f$, and $D_L$ does the same for $C^*$.

In our applications, $C$ is an explicit model for one step of a transfer chain, and $T$ is the actual step. The difference $T - C$ is not small in norm. However, it is small relative to the defects of $C$, up to an error of the order of the step size $\delta_0$. The following lemma shows that $T$ then keeps half of the dissipation of $C$, up to a loss of order $\delta_0$ per step, so that the powers $T^j$ with $j \delta_0 \le t$ stay bounded.

\begin{lemma}\label{l:tools-defect}
Let $C$ be a contraction on a Hilbert space $\mathcal H$, with defects $D_R$ and $D_L$ as in \eqref{e:tools-defects}.
\begin{enumerate}[label=(\alph*)]
\item Let $0 < \delta_0 \le 1$, $A \ge 1$, and $0 < \theta \le 1/8$ be such that $A \delta_0 \le 1$, and let $T$ be a bounded operator on $\mathcal H$ such that
\begin{equation}\label{e:tools-form-bound}
 |\langle g, (T - C) f \rangle|
 \le \theta \, \|(D_L + A \delta_0)^{1/2} g\| \, \|(D_R + A \delta_0)^{1/2} f\|
 \qquad \text{for all } f, g \in \mathcal H.
\end{equation}
Then
\begin{equation}\label{e:tools-defect-conclusion}
 T^* T \le (1 + 4 \theta A \delta_0) \Id - \tfrac12 D_R,
 \qquad
 T T^* \le (1 + 4 \theta A \delta_0) \Id - \tfrac12 D_L.
\end{equation}
\item Let $0 < \delta_0 \le 1$ and $A_1 \ge 0$, and let $T$ be a bounded operator on $\mathcal H$ such that
\begin{equation}\label{e:tools-defect-B}
 T^* T \le (1 + A_1 \delta_0) \Id - \tfrac12 D_R,
 \qquad
 T T^* \le (1 + A_1 \delta_0) \Id - \tfrac12 D_L.
\end{equation}
Let $n \ge 1$ and $t > 0$ satisfy $n \delta_0 \le t$. Then $\|T^j\| \le e^{A_1 t/2}$ for $0 \le j \le n$, and for all $f, g \in \mathcal H$,
\begin{equation}\label{e:tools-defect-sums}
 \sum_{j = 0}^{n - 1} \|D_R^{1/2} T^j f\|^2
 + \sum_{j = 0}^{n - 1} \|D_L^{1/2} (T^*)^j g\|^2
 \le 2 \bigl(1 + A_1 t e^{A_1 t}\bigr) \bigl(\|f\|^2 + \|g\|^2\bigr).
\end{equation}
\end{enumerate}
In particular, under the hypotheses of part (a), the conclusions of part (b) hold with $A_1 = 4 \theta A$ for all $n$ and $t$ with $n \delta_0 \le t$.
\end{lemma}

\begin{proof}
We begin with part (a). For $f \in \mathcal H$, set
\[
 q(f) = \|(D_R + A \delta_0)^{1/2} f\|^2 = \|D_R^{1/2} f\|^2 + A \delta_0 \|f\|^2.
\]
Expanding $\|Cf + (T - C) f\|^2$ and using $\|Cf\|^2 = \|f\|^2 - \|D_R^{1/2} f\|^2$, we have
\begin{equation}\label{e:tools-defect-expansion}
 \|Tf\|^2 = \|f\|^2 - \|D_R^{1/2} f\|^2 + 2 \Re \langle Cf, (T - C) f \rangle + \|(T - C) f\|^2.
\end{equation}
We will bound the last two terms by multiples of $q(f)$.

To bound the cross term, we first note that
\[
 C^* C (\Id - C^* C) \le \Id - C^* C,
\]
since $0 \le C^* C \le \Id$. It follows that
\[
 \|D_L^{1/2} C f\|^2 = \langle Cf, (\Id - C C^*) C f \rangle = \langle f, C^* C (\Id - C^* C) f \rangle \le \|D_R^{1/2} f\|^2.
\]
Together with $\|Cf\| \le \|f\|$, this implies
\[
 \|(D_L + A \delta_0)^{1/2} Cf\|^2 = \|D_L^{1/2} C f\|^2 + A \delta_0 \|Cf\|^2 \le q(f).
\]
Taking $g = Cf$ in \eqref{e:tools-form-bound}, we obtain
\begin{equation}\label{e:tools-defect-cross}
 |\langle Cf, (T - C) f \rangle| \le \theta \, q(f).
\end{equation}

For the last term, we take the supremum in \eqref{e:tools-form-bound} over all $g$ with $\|g\| = 1$. Using $\|D_L\| \le 1$ and $A \delta_0 \le 1$, we find
\begin{equation}\label{e:tools-defect-square}
 \|(T - C) f\|^2 \le 2 \theta^2 q(f).
\end{equation}

Inserting \eqref{e:tools-defect-cross} and \eqref{e:tools-defect-square} into \eqref{e:tools-defect-expansion}, we find
\[
 \|Tf\|^2 \le \|f\|^2 - \|D_R^{1/2} f\|^2 + (2 \theta + 2 \theta^2) q(f)
 \le (1 + 4 \theta A \delta_0) \|f\|^2 - (1 - 4 \theta) \|D_R^{1/2} f\|^2,
\]
where we used $2 \theta^2 \le 2 \theta$ in the second inequality. By the assumption that $\theta \le 1/8$, we have $1 - 4 \theta \ge 1/2$, and the first inequality in \eqref{e:tools-defect-conclusion} follows.

For the second inequality, observe that
\[
 |\langle f, (T^* - C^*) g \rangle| = |\langle g, (T - C) f \rangle| \qquad \text{for all } f, g \in \mathcal H,
\]
and that the right and left defects of the contraction $C^*$ are $D_L$ and $D_R$. This shows that the pair $(T^*, C^*)$ satisfies \eqref{e:tools-form-bound}, with the roles of $f$ and $g$ exchanged. The first inequality in \eqref{e:tools-defect-conclusion} for $(T^*, C^*)$ is the second inequality for $(T, C)$.

We now prove part (b). Let $f \in \mathcal H$, and set $x_j = \|T^j f\|^2$. Applying the first inequality in \eqref{e:tools-defect-B} to $T^j f$, we obtain
\begin{equation}\label{e:tools-defect-step}
 x_{j + 1} \le (1 + A_1 \delta_0) x_j - \tfrac12 \|D_R^{1/2} T^j f\|^2.
\end{equation}
In particular, for $0 \le j \le n$, we have
\[
 x_j \le (1 + A_1 \delta_0)^j x_0 \le e^{A_1 \delta_0 j} \|f\|^2 \le e^{A_1 t} \|f\|^2,
\]
which is the bound $\|T^j\| \le e^{A_1 t/2}$.

Next, we bound the defect sums. Summing \eqref{e:tools-defect-step} over $0 \le j \le n - 1$, we find
\[
 \frac12 \sum_{j = 0}^{n - 1} \|D_R^{1/2} T^j f\|^2
 \le x_0 - x_n + A_1 \delta_0 \sum_{j = 0}^{n - 1} x_j
 \le \bigl(1 + A_1 \delta_0 n e^{A_1 t}\bigr) \|f\|^2
 \le \bigl(1 + A_1 t e^{A_1 t}\bigr) \|f\|^2.
\]
The same argument for $T^*$, with the second inequality in \eqref{e:tools-defect-B}, bounds the second sum in \eqref{e:tools-defect-sums} by $2(1 + A_1 t e^{A_1 t}) \|g\|^2$. The final claim of the lemma follows, since \eqref{e:tools-defect-conclusion} is \eqref{e:tools-defect-B} with $A_1 = 4 \theta A$.
\end{proof}

Most applications verify \eqref{e:tools-form-bound} through a bound with four terms: a term with a defect on each side, two terms that carry one defect and a factor $\sqrt{\delta_0}$, and a term of order $\delta_0$. The following corollary turns such a bound into the hypothesis \eqref{e:tools-defect-B} of \zcref{l:tools-defect}(b), with explicit constants.

\begin{corollary}\label{c:tools-placements}
Let $C$ be a contraction on a Hilbert space $\mathcal H$, with defects $D_R$ and $D_L$ as in \eqref{e:tools-defects}. Let $0 \le a \le 1/16$, $A_0 \ge 0$, and $0 < \delta_0 \le 1$, and let $T$ be a bounded operator on $\mathcal H$ such that for all $f, g \in \mathcal H$,
\begin{multline}\label{e:tools-placements}
 |\langle g, (T - C) f \rangle|
 \le a \|D_L^{1/2} g\| \, \|D_R^{1/2} f\| \\
 + A_0 \sqrt{\delta_0} \bigl(\|g\| \, \|D_R^{1/2} f\| + \|D_L^{1/2} g\| \, \|f\|\bigr)
 + A_0 \delta_0 \|g\| \, \|f\|.
\end{multline}
Set
\begin{equation}\label{e:tools-placements-B}
 A = \max\bigl\{1, 2304 A_0^2\bigr\},
 \qquad
 A_1 = A \bigl((1 + a + 3 A_0)^2 + 1\bigr).
\end{equation}
Then $T$ satisfies \eqref{e:tools-defect-B} with this $A_1$. If moreover $A \delta_0 \le 1$, then $T$ satisfies \eqref{e:tools-defect-B} with $A/2$ in place of $A_1$. In particular, the conclusions of \zcref{l:tools-defect}(b) hold for all $n$ and $t$ with $n \delta_0 \le t$, with $A/2$ in place of $A_1$ when $A \delta_0 \le 1$.
\end{corollary}

The bound \eqref{e:tools-placements} holds, for example, if
\[
 T - C = D_L^{1/2} Y_0 D_R^{1/2} + \sqrt{\delta_0} \bigl(Y_1 D_R^{1/2} + D_L^{1/2} Y_2\bigr) + \delta_0 Y_3
\]
for bounded operators $Y_0, \dots, Y_3$ with $\|Y_0\| \le a$ and $\|Y_1\|, \|Y_2\|, \|Y_3\| \le A_0$.

\begin{proof}
Suppose first that $A \delta_0 \le 1$. Let $f, g \in \mathcal H$, and set
\[
 q_R = \|(D_R + A \delta_0)^{1/2} f\|, \qquad q_L = \|(D_L + A \delta_0)^{1/2} g\|.
\]
Then
\[
 \|D_R^{1/2} f\| \le q_R, \qquad \sqrt{\delta_0} \, \|f\| \le A^{-1/2} q_R,
 \qquad
 \|D_L^{1/2} g\| \le q_L, \qquad \sqrt{\delta_0} \, \|g\| \le A^{-1/2} q_L.
\]

Inserting these bounds into \eqref{e:tools-placements}, we obtain
\[
 |\langle g, (T - C) f \rangle|
 \le \bigl(a + 2 A_0 A^{-1/2} + A_0 A^{-1}\bigr) q_L q_R
 \le \bigl(a + 3 A_0 A^{-1/2}\bigr) q_L q_R
 \le \tfrac18 \, q_L q_R,
\]
where we used $A \ge 1$ in the second inequality, and $a \le 1/16$ and $A^{1/2} \ge 48 A_0$, which follows from $2304 = 48^2$, in the third. Thus \eqref{e:tools-form-bound} holds with $\theta = 1/8$, and \zcref{l:tools-defect}(a) implies \eqref{e:tools-defect-B} with $4 \theta A = A/2$ in place of $A_1$. Since $A/2 \le A_1$, the bound \eqref{e:tools-defect-B} also holds with $A_1$ itself.

Suppose next that $A \delta_0 > 1$. We take the supremum in \eqref{e:tools-placements} over unit vectors $f$ and $g$. Since the defects have norm at most one and $\delta_0 \le 1$, we find
\[
 \|T - C\| \le a + 3 A_0, \qquad \|T\| \le 1 + a + 3 A_0.
\]
The second bound implies that
\[
 T^* T \le (1 + a + 3 A_0)^2 \Id \le \bigl(1 + A \delta_0 ((1 + a + 3 A_0)^2 + 1)\bigr) \Id - \tfrac12 D_R,
\]
where we used $A \delta_0 > 1$ and $D_R \le \Id$ in the second inequality. The same bound holds for $T T^*$, with $D_L$ in place of $D_R$, and \eqref{e:tools-defect-B} follows. The final claim is now a consequence of \zcref{l:tools-defect}(b).
\end{proof}

\section{An operator-valued Dirichlet-form bound}\label{s:tools-dirichlet}

We will need the following bound for a symmetric integral operator whose kernel takes values in bounded operators on a Hilbert space. In \zcref{s:stab}, these values are coefficient operators that need not commute with one another. The lemma bounds the difference between the integral operator and the multiplication by its marginal by the Dirichlet form of the underlying scalar kernel, uniformly in the operator values.

Throughout this section, $(\mathcal Y, \mu)$ is a $\sigma$-finite measure space and $\mathcal K$ is a separable Hilbert space. Let $L^2(\mathcal Y; \mathcal K)$ denote the space of square-integrable measurable functions from $\mathcal Y$ to $\mathcal K$. We call a measurable function $p \colon \mathcal Y \times \mathcal Y \to [0, \infty)$ a \emph{symmetric Markov kernel} if $p(x, y) = p(y, x)$ for all $x, y \in \mathcal Y$ and $\int p(x, y) \dd \mu(y) = 1$ for all $x \in \mathcal Y$. For such a kernel, we define
\begin{equation}\label{e:tools-Pi}
 \Pi f(x) = \int p(x, y) f(y) \dd \mu(y).
\end{equation}
Using the symmetry and the normalization of $p$, the Schur test shows that $\Pi$ is a self-adjoint operator on $L^2(\mathcal Y; \mathcal K)$ with $\|\Pi\| \le 1$. We call a family $(Y(x, y))_{x, y \in \mathcal Y}$ of bounded operators on $\mathcal K$ \emph{strongly measurable} if $(x, y) \mapsto Y(x, y) e$ is measurable for all $e \in \mathcal K$.

\begin{lemma}\label{l:tools-dirichlet}
Let $p$ be a symmetric Markov kernel on $\mathcal Y$, and let $(R(x, y))_{x, y \in \mathcal Y}$ be a strongly measurable family of bounded operators on $\mathcal K$. Let $c_R \ge 0$ be such that
\begin{equation}\label{e:tools-row-hyp}
 \|R(x, y)\| \le c_R \qquad \text{for all } x, y \in \mathcal Y.
\end{equation}
Define the operator $T$ on $L^2(\mathcal Y; \mathcal K)$ and the bounded operators $R_{\mathrm{marg}}(x)$ on $\mathcal K$ by
\begin{equation}\label{e:tools-row-T}
 T f(x) = \int p(x, y) R(x, y) f(y) \dd \mu(y),
 \qquad
 R_{\mathrm{marg}}(x) = \int p(x, y) R(x, y) \dd \mu(y),
\end{equation}
where the second integral is taken strongly. We call $R_{\mathrm{marg}}$ the \emph{marginal} of the kernel, and we let $R_{\mathrm{marg}}$ also denote the multiplication by $R_{\mathrm{marg}}(x)$ on $L^2(\mathcal Y; \mathcal K)$.
\begin{enumerate}[label=(\alph*)]
\item Suppose that $R(x, y) = R(y, x)$ for all $x, y \in \mathcal Y$. Then for all $f, g \in L^2(\mathcal Y; \mathcal K)$,
\begin{equation}\label{e:tools-row-identity}
 \langle g, (T - R_{\mathrm{marg}}) f \rangle
 = - \frac12 \int_{\mathcal Y^2} p(x, y) \bigl\langle g(x) - g(y), R(x, y) \bigl(f(x) - f(y)\bigr) \bigr\rangle \dd \mu(x) \dd \mu(y)
\end{equation}
and
\begin{equation}\label{e:tools-row-bound}
 |\langle g, (T - R_{\mathrm{marg}}) f \rangle|
 \le c_R \langle g, (\Id - \Pi) g \rangle^{1/2} \langle f, (\Id - \Pi) f \rangle^{1/2}.
\end{equation}
If moreover $\Pi \ge 0$, then for all $f \in L^2(\mathcal Y; \mathcal K)$,
\begin{equation}\label{e:tools-row-bound-square}
 \langle f, (\Id - \Pi) f \rangle \le \|(\Id - \Pi^2)^{1/2} f\|^2.
\end{equation}
\item \emph{Transported kernels.} Let $(\tau_{xy})_{x, y \in \mathcal Y}$ be a strongly measurable family of unitary operators on $\mathcal K$ such that $\tau_{yx} = \tau_{xy}^*$ for all $x, y$, and suppose that $R(y, x) = \tau_{yx} R(x, y) \tau_{xy}$ for all $x, y$. Define $T_\tau$ and $\Pi_\tau$ by replacing $f(y)$ with $\tau_{xy} f(y)$ in the definitions of $T$ and $\Pi$. Then part (a) holds for $T_\tau$ and $\Pi_\tau$ in place of $T$ and $\Pi$, with $f(x) - f(y)$ and $g(x) - g(y)$ replaced by $f(x) - \tau_{xy} f(y)$ and $g(x) - \tau_{xy} g(y)$ in \eqref{e:tools-row-identity}.
\end{enumerate}
\end{lemma}

Suppose that the operators $R(x, y)$ act on the fibers $\mathcal K_x$ of a Hilbert bundle over $\mathcal Y$, with unitary transports from $\mathcal K_y$ to $\mathcal K_x$, and that the fibers are identified with $\mathcal K$ by a measurable choice of unitary frames. Then the transports become operators $\tau_{xy}$ as in part (b). This is the form in which part (b) is applied.

\begin{proof}
We first check that all integrals converge absolutely. By \eqref{e:tools-row-hyp}, we have
\begin{equation}\label{e:tools-row-schur}
 \int p(x, y) \|R(x, y)\| \dd \mu(y) \le c_R \qquad \text{for all } x \in \mathcal Y,
\end{equation}
and the same bound holds for the integral in $x$, since $p$ is symmetric. By the Schur test, we have $\|T\| \le c_R$, and \eqref{e:tools-row-schur} also implies that $\|R_{\mathrm{marg}}(x)\| \le c_R$ for all $x$. Moreover,
\[
 \int_{\mathcal Y^2} p(x, y) \|R(x, y)\| \, \|g(x)\| \bigl(\|f(x)\| + \|f(y)\|\bigr) \dd \mu(x) \dd \mu(y) \le 2 c_R \|g\| \, \|f\|,
\]
which justifies the uses of Fubini's theorem below. Since the operators $\tau_{xy}$ in part (b) are unitary, the same bounds hold for the kernel $p(x, y) R(x, y) \tau_{xy}$ of $T_\tau$.

We now prove part (a). By the definitions of $T$ and $R_{\mathrm{marg}}$, we have
\[
 \langle g, (T - R_{\mathrm{marg}}) f \rangle
 = \int_{\mathcal Y^2} p(x, y) \bigl\langle g(x), R(x, y) \bigl(f(y) - f(x)\bigr) \bigr\rangle \dd \mu(x) \dd \mu(y).
\]
We exchange the names of $x$ and $y$. Using $p(x, y) = p(y, x)$ and $R(x, y) = R(y, x)$, we see that the same integral equals
\[
 \int_{\mathcal Y^2} p(x, y) \bigl\langle g(y), R(x, y) \bigl(f(x) - f(y)\bigr) \bigr\rangle \dd \mu(x) \dd \mu(y).
\]
Averaging the two expressions, we obtain \eqref{e:tools-row-identity}.

Next, we prove \eqref{e:tools-row-bound}. For $f \in L^2(\mathcal Y; \mathcal K)$, define
\begin{equation}\label{e:tools-dirichlet-form}
 \mathcal E_p(f) = \frac12 \int_{\mathcal Y^2} p(x, y) \|f(x) - f(y)\|^2 \dd \mu(x) \dd \mu(y).
\end{equation}

Expanding the square and using the symmetry and the normalization of $p$ and the self-adjointness of $\Pi$, we find
\begin{equation}\label{e:tools-dirichlet-identity}
 \mathcal E_p(f) = \|f\|^2 - \Re \langle f, \Pi f \rangle = \langle f, (\Id - \Pi) f \rangle.
\end{equation}

By the Cauchy--Schwarz inequality, \eqref{e:tools-row-identity} implies
\[
\begin{aligned}
 |\langle g, (T - R_{\mathrm{marg}}) f \rangle|
 &\le \frac{c_R}{2} \int_{\mathcal Y^2} p(x, y) \|g(x) - g(y)\| \, \|f(x) - f(y)\| \dd \mu(x) \dd \mu(y) \\
 &\le c_R \, \mathcal E_p(g)^{1/2} \mathcal E_p(f)^{1/2},
\end{aligned}
\]
which is \eqref{e:tools-row-bound}. If $\Pi \ge 0$, then $0 \le \Pi \le \Id$, since $\|\Pi\| \le 1$. The spectral theorem then yields $\Pi^2 \le \Pi$, which is \eqref{e:tools-row-bound-square}.

We now prove part (b). We have
\[
 \langle g, (T_\tau - R_{\mathrm{marg}}) f \rangle
 = \int_{\mathcal Y^2} p(x, y) \bigl\langle g(x), R(x, y) \bigl(\tau_{xy} f(y) - f(x)\bigr) \bigr\rangle \dd \mu(x) \dd \mu(y).
\]

We again exchange the names of $x$ and $y$. Recall that $\tau_{yx} = \tau_{xy}^*$, so that $\tau_{yx} = \tau_{xy}^{-1}$. Using also $R(y, x) = \tau_{yx} R(x, y) \tau_{xy}$, we find
\[
 \bigl\langle g(y), R(y, x) \bigl(\tau_{yx} f(x) - f(y)\bigr) \bigr\rangle
 = \bigl\langle \tau_{xy} g(y), R(x, y) \bigl(f(x) - \tau_{xy} f(y)\bigr) \bigr\rangle.
\]
Averaging the two expressions gives \eqref{e:tools-row-identity} with the transported differences.

To prove the transported versions of \eqref{e:tools-row-bound} and \eqref{e:tools-row-bound-square}, let $\mathcal E_{p, \tau}(f)$ be defined by \eqref{e:tools-dirichlet-form}, with $f(x) - \tau_{xy} f(y)$ in place of $f(x) - f(y)$. The operator $\Pi_\tau$ is self-adjoint, since $p$ is symmetric and $\tau_{yx} = \tau_{xy}^*$, and the Schur test gives $\|\Pi_\tau\| \le 1$. The expansion that gave \eqref{e:tools-dirichlet-identity}, together with $\|\tau_{xy} f(y)\| = \|f(y)\|$, yields
\[
 \mathcal E_{p, \tau}(f) = \langle f, (\Id - \Pi_\tau) f \rangle,
\]
and the rest of the proof is unchanged.
\end{proof}

\section{Poisson transforms of translation-bounded measures}\label{s:tools-poisson}

For $a > 0$, let $P_a(u) = a / (\pi (u^2 + a^2))$ denote the Poisson kernel; for $a = \eps$, this is the kernel $P_\eps$ of \zcref{s:intro}. For a positive Radon measure $\nu$ on $\R$, we set
\begin{equation}\label{e:tools-loc}
 \|\nu\|_{\mathrm{loc}} = \sup_{s \in \R} \nu([s, s + 1]),
\end{equation}
and when $\|\nu\|_{\mathrm{loc}} < \infty$, we define the \emph{Poisson transform} $P_a * \nu$ of $\nu$ by
\[
 (P_a * \nu)(u) = \int P_a(u - r) \, \nu(\dd r), \qquad u \in \R.
\]
This integral is finite by part (a) of the following lemma, which is used to identify limiting pair and occupation measures from their Poisson transforms.

\begin{lemma}\label{l:tools-poisson}
Let $\nu$ be a positive Radon measure on $\R$ with $\|\nu\|_{\mathrm{loc}} < \infty$.
\begin{enumerate}[label=(\alph*)]
\item For all $a > 0$, $R \ge 1$, and $u \in \R$ with $|u| \le R/2$,
\[
 \int_{|r| > R} P_a(u - r) \, \nu(\dd r) \le \frac{16 \, a \, \|\nu\|_{\mathrm{loc}}}{\pi R}.
\]
\item For all $f \in C_c(\R)$,
\[
 \lim_{a \downarrow 0} \int f(u) (P_a * \nu)(u) \dd u = \int f \dd \nu.
\]
In particular, $\nu$ is determined by the functions $P_a * \nu$ along every sequence $a \downarrow 0$.
\item For all $u \in \R$,
\[
 \nu(\{u\}) = \lim_{a \downarrow 0} \pi a \, (P_a * \nu)(u).
\]
\item Let $(\nu_n)_{n \ge 1}$ be positive Radon measures on $\R$ such that $\sup_n \|\nu_n\|_{\mathrm{loc}} \le C_0$, and suppose that there exists a function $\vartheta$ on $(0, \infty) \times \R$ such that $(P_a * \nu_n)(u) \to \vartheta(a, u)$ for all $a > 0$ and $u \in \R$. Then $\nu_n$ converges vaguely to a positive Radon measure $\nu$ with $\|\nu\|_{\mathrm{loc}} \le 2 C_0$ and $P_a * \nu = \vartheta(a, \cdot)$ for all $a > 0$.
\end{enumerate}
\end{lemma}

\begin{proof}
We begin with part (a). Let $|r| > R$. Since $R \ge 2|u|$, we have
\[
 |u - r| \ge |r| - |u| \ge |r|/2,
\]
and therefore
\begin{equation}\label{e:tools-poisson-decay}
 P_a(u - r) \le \frac{a}{\pi (u - r)^2} \le \frac{4a}{\pi r^2}.
\end{equation}

Covering $(R, \infty)$ by the intervals $[R + k, R + k + 1]$ with $k \ge 0$, we have
\begin{equation}\label{e:tools-poisson-tail}
 \int_{r > R} \frac{\nu(\dd r)}{r^2}
 \le \|\nu\|_{\mathrm{loc}} \sum_{k = 0}^\infty \frac{1}{(R + k)^2}
 \le \|\nu\|_{\mathrm{loc}} \Bigl(\frac{1}{R^2} + \frac1R\Bigr)
 \le \frac{2 \|\nu\|_{\mathrm{loc}}}{R},
\end{equation}
where we used $R \ge 1$ in the last step. The same bound holds for the integral over $r < -R$. Combining \eqref{e:tools-poisson-decay} and \eqref{e:tools-poisson-tail} proves part (a).

Next, we prove part (b). Let $\supp f \subset [-R_0, R_0]$, and set $R_1 = 2R_0 + 2$. If $|r| \ge R_1$ and $|u| \le R_0$, then $|r - u| \ge |r|/2$, which implies that
\begin{equation}\label{e:tools-poisson-test}
 |(P_a * f)(r)| \le \|f\|_{L^1} \sup_{|u| \le R_0} P_a(r - u) \le \frac{4 a \|f\|_{L^1}}{\pi r^2}.
\end{equation}

The bound \eqref{e:tools-poisson-tail} shows that the right side of \eqref{e:tools-poisson-test}, with $|f|$ in place of $f$, is integrable against $\nu$ over $|r| > R_1$. Therefore Fubini's theorem applies, and we obtain
\begin{equation}\label{e:tools-poisson-fubini}
 \int f(u) (P_a * \nu)(u) \dd u = \int (P_a * f) \dd \nu.
\end{equation}

We now let $a \downarrow 0$ in \eqref{e:tools-poisson-fubini}. As $a \downarrow 0$, we have $P_a * f \to f$ uniformly on $\R$, since $f$ is bounded and uniformly continuous and $(P_a)_{a > 0}$ is an approximate identity. Since $\nu([-R_1, R_1]) < \infty$, this implies
\[
 \lim_{a \downarrow 0} \int_{[-R_1, R_1]} (P_a * f - f) \dd \nu = 0.
\]

Outside $[-R_1, R_1]$, the function $f$ vanishes. By \eqref{e:tools-poisson-test} and by \eqref{e:tools-poisson-tail} with $R = R_1$, applied on both sides of the origin, we have
\[
 \Bigl| \int_{|r| > R_1} (P_a * f) \dd \nu \Bigr|
 \le \frac{4 a \|f\|_{L^1}}{\pi} \int_{|r| > R_1} \frac{\nu(\dd r)}{r^2}
 \le \frac{16 a \|f\|_{L^1} \|\nu\|_{\mathrm{loc}}}{\pi R_1},
\]
which tends to zero as $a \downarrow 0$. This proves the limit in part (b). The second claim of part (b) follows, since a positive Radon measure is determined by its integrals against $C_c(\R)$.

We now prove part (c). The functions
\[
 \pi a P_a(u - r) = \frac{a^2}{a^2 + (u - r)^2}
\]
are bounded by one and decrease to $\1_{\{r = u\}}$ as $a \downarrow 0$. Dominated convergence on $[u - 1, u + 1]$ shows that
\[
 \lim_{a \downarrow 0} \int_{[u - 1, u + 1]} \pi a P_a(u - r) \, \nu(\dd r) = \nu(\{u\}).
\]
Outside this interval, the covering argument of part (a) gives
\[
 \int_{|r - u| > 1} \pi a P_a(u - r) \, \nu(\dd r) \le a^2 \int_{|r - u| > 1} \frac{\nu(\dd r)}{(u - r)^2} \le 4 a^2 \|\nu\|_{\mathrm{loc}},
\]
which tends to zero.

Finally, we prove part (d). Note that $\nu_n([-m, m]) \le 2m C_0$ for all $m \ge 1$ and $n$. By Helly's selection theorem and a diagonal argument, every subsequence of $(\nu_n)$ has a further subsequence that converges vaguely to a positive Radon measure. Let $\nu$ be such a limit along a subsequence $(\nu_{n'})$. For every open interval $G$, we have
\[
 \nu(G) \le \liminf_{n' \to \infty} \nu_{n'}(G).
\]

Applying this bound with $G = (s - \gamma, s + 1 + \gamma)$, where $s \in \R$ and $0 < \gamma < 1/2$, we find
\[
 \nu([s, s + 1]) \le \nu((s - \gamma, s + 1 + \gamma)) \le \liminf_{n'} \nu_{n'}([s - \gamma, s + 1 + \gamma]) \le 2 C_0,
\]
so that $\|\nu\|_{\mathrm{loc}} \le 2 C_0$.

It remains to identify the Poisson transform of $\nu$. Fix $a > 0$ and $u \in \R$, and let $R \ge \max(1, 2|u|)$. Let $\chi \in C_c(\R)$ satisfy $0 \le \chi \le 1$, $\chi = 1$ on $[-R, R]$ and $\supp \chi \subset [-R - 1, R + 1]$. By vague convergence, we have
\[
 \lim_{n' \to \infty} \int P_a(u - r) \chi(r) \, \nu_{n'}(\dd r) = \int P_a(u - r) \chi(r) \, \nu(\dd r).
\]

By part (a), with $\|\nu_{n'}\|_{\mathrm{loc}} \le C_0$ and $\|\nu\|_{\mathrm{loc}} \le 2 C_0$, we have
\[
 \int P_a(u - r) (1 - \chi(r)) \, \nu_{n'}(\dd r) \le \frac{16 a C_0}{\pi R},
 \qquad
 \int P_a(u - r) (1 - \chi(r)) \, \nu(\dd r) \le \frac{32 a C_0}{\pi R}.
\]
Letting $n' \to \infty$ and then $R \to \infty$, we conclude that $(P_a * \nu)(u) = \vartheta(a, u)$. By part (b), all subsequential vague limits therefore coincide, and the full sequence converges vaguely to $\nu$.
\end{proof}

\section{A submean inequality for holomorphic families}\label{s:tools-submean}

We say that a function $f$ from an open set $\mathcal U \subset \C^n$ to a Hilbert space $\mathcal H$ is \emph{holomorphic} if $\zeta \mapsto \lambda(f(\zeta))$ is holomorphic on $\mathcal U$ for every continuous complex-linear functional $\lambda$ on $\mathcal H$. With the inner product linear in its first argument, this is equivalent to requiring $\zeta \mapsto \langle f(\zeta), e \rangle$ to be holomorphic for all $e \in \mathcal H$. Let $\dd^{2n} \zeta$ denote Lebesgue measure on $\C^n = \R^{2n}$, let $|S|$ denote the Lebesgue measure of a set $S \subset \C^n$, and let $B(\zeta, r)$ denote the open Euclidean ball of radius $r$ about $\zeta$. Recall that $|B(\zeta, r)| = \pi^n r^{2n}/n!$.

\begin{lemma}\label{l:tools-submean}
Let $\mathcal U \subset \C^n$ be open, let $\mathcal H$ be a separable Hilbert space, and let $(\mathcal Y, m)$ be a $\sigma$-finite measure space. Let $f \colon \mathcal U \times \mathcal Y \to \mathcal H$ be measurable and such that $\zeta \mapsto f(\zeta, y)$ is holomorphic on $\mathcal U$ for all $y \in \mathcal Y$. Let $\zeta_* \colon \mathcal Y \to \mathcal U$ be measurable, and let $r > 0$ be such that $B(\zeta_*(y), r) \subset \mathcal U$ for all $y \in \mathcal Y$. Then
\begin{equation}\label{e:tools-submean}
 \int_{\mathcal Y} \|f(\zeta_*(y), y)\|^2 \, m(\dd y)
 \le \frac{n!}{\pi^n r^{2n}} \int_{\mathcal Y} \int_{B(\zeta_*(y), r)} \|f(\zeta, y)\|^2 \dd^{2n} \zeta \, m(\dd y).
\end{equation}
\end{lemma}

\begin{proof}
Fix $y \in \mathcal Y$ and an orthonormal basis $(e_k)_{k \ge 1}$ of $\mathcal H$, and define the holomorphic functions
\[
 f_k(\zeta) = \langle f(\zeta, y), e_k \rangle, \qquad \zeta \in \mathcal U.
\]
Their real and imaginary parts are harmonic on $\mathcal U \subset \R^{2n}$. Let $0 < r' < r$, and write $B' = B(\zeta_*(y), r')$ and $B = B(\zeta_*(y), r)$, so that the closure of $B'$ lies in $\mathcal U$. The mean value property of harmonic functions and the Cauchy--Schwarz inequality give
\[
 |f_k(\zeta_*(y))|^2
 = \Bigl| \frac{1}{|B'|} \int_{B'} f_k(\zeta) \dd^{2n} \zeta \Bigr|^2
 \le \frac{1}{|B'|} \int_{B'} |f_k(\zeta)|^2 \dd^{2n} \zeta
 \le \frac{1}{|B'|} \int_{B} |f_k(\zeta)|^2 \dd^{2n} \zeta.
\]
Summing over $k$ and using the monotone convergence theorem, we obtain
\[
 \|f(\zeta_*(y), y)\|^2 \le \frac{1}{|B'|} \int_{B} \|f(\zeta, y)\|^2 \dd^{2n} \zeta.
\]
Letting $r' \uparrow r$ gives \eqref{e:tools-submean} without the integral over $y$.

It remains to integrate in $y$, and for this we check that the two sides are measurable functions of $y$. The left side is the composition of $f$ with the measurable map $y \mapsto (\zeta_*(y), y)$. After the change of variables $\zeta = \zeta_*(y) + w$, the right side is the integral over $w \in B(0, r)$ of the jointly measurable function $(w, y) \mapsto \|f(\zeta_*(y) + w, y)\|^2$, and Tonelli's theorem shows that it is measurable in $y$. Integrating in $y$ completes the proof.
\end{proof}

\begin{remark}\label{r:tools-ellipsoid}
The ball in \zcref{l:tools-submean} may be replaced by an ellipsoid. Let $A$ be an invertible complex-linear map of $\C^n$, and suppose that $\zeta_*(y) + A B(0, r) \subset \mathcal U$ for all $y$. We apply \zcref{l:tools-submean} to the function $(w, y) \mapsto f(A w, y)$ on $A^{-1} \mathcal U$, with the center $A^{-1} \zeta_*(y)$ and the same $r$. The change of variables $\zeta = A w$ multiplies Lebesgue measure by $|\det A|^2$, which is also the ratio of $|A B(0, r)|$ to $|B(0, r)|$. We obtain
\begin{equation}\label{e:tools-ellipsoid}
 \int_{\mathcal Y} \|f(\zeta_*(y), y)\|^2 \, m(\dd y)
 \le \frac{1}{|A B(0, r)|} \int_{\mathcal Y} \int_{\zeta_*(y) + A B(0, r)} \|f(\zeta, y)\|^2 \dd^{2n} \zeta \, m(\dd y).
\end{equation}
The same argument applies when $A = A(y)$ depends measurably on $y$, since the proof of \zcref{l:tools-submean} is pointwise in $y$ before the final integration.
\end{remark}