for all $t > t_0$. It is straightforward to derive from these estimates that

$$ \lim_{t \to \infty} \frac{1}{t} \log \det(\Delta_{\alpha,t}) = 4e(\alpha) - \operatorname{tr}(Q\vartheta^{-1}Q^*). $$

(9) Using (5.40), one rewrites the Riccati equation (5.50) as

$$ \begin{aligned} D_{\alpha}^{*}Y_{\alpha} &= -Y_{\alpha}(D_{\alpha} + BY_{\alpha}) = -Y_{\alpha}(A_{\alpha} + B(Y_{\alpha} - X_{\alpha})) \\ &= -Y_{\alpha}(A_{\alpha} + B\theta X_{1-\alpha}\theta) = -Y_{\alpha}\theta(-A_{1-\alpha} + BX_{1-\alpha})\theta \\ &= Y_{\alpha}\theta D_{1-\alpha}\theta = Y_{\alpha}\tilde{D}_{\alpha}. \end{aligned} $$

Thus, the result immediately follows from the fact that

$$ \frac{d}{dt} e^{tD_{\alpha}^{*}} Y_{\alpha} e^{-t\tilde{D}_{\alpha}} = e^{tD_{\alpha}^{*}} (D_{\alpha}^{*}Y_{\alpha} - Y_{\alpha}\tilde{D}_{\alpha}) e^{-t\tilde{D}_{\alpha}} = 0. $$

(10) For any $u \in \Xi$ we infer from Parts (2) and (4) that the function $\alpha \mapsto (u, W_{\alpha}u)$ is convex, real analytic on the interval $\mathcal{I}_c$, and continuous on its closure. Since it vanishes for $|\alpha - \frac{1}{2}| = \frac{1}{2}$ one has either $(u, W_{\alpha}u) = 0$ for all $\alpha \in \mathcal{I}_c$ or $(u, W_{\alpha}u) < 0$ for $|\alpha - \frac{1}{2}| < \frac{1}{2}$ and $(u, W_{\alpha}u) > 0$ for $\frac{1}{2} < |\alpha - \frac{1}{2}| \le \kappa_c$. This proves the first assertion.

Since $Y_{\alpha}+W_{\alpha} = \alpha X_{1}+\theta X_{1-\alpha}\theta$, we deduce from Part (2) that $Y_{\alpha}+W_{\alpha} > 0$ for $|\alpha-\frac{1}{2}| \le \frac{1}{2}$. Consider now $\frac{1}{2} < |\alpha - \frac{1}{2}| \le \kappa_c$. If $u \in \Xi$ is such that $(u, W_{\alpha}u) > 0$, then Part (7) yields $(u, (Y_{\alpha}+W_{\alpha})u) > 0$. Thus, it remains to consider the case of $u \in \Xi$ such that $(u, W_{\alpha}u) = 0$ for all $\alpha \in \mathcal{I}_c$. Using (5.44) we get that

$$ (u, W_{\alpha}''u) = -(u, X_{\alpha}''u) = 2 \int_{0}^{\infty} |Q^*(X_{\alpha}' - \beta)e^{tD_{\alpha}}u|^2 dt = 0 $$

for $\alpha \in \mathcal{I}_c$. Since $QQ^*(X_{\alpha}' - \beta) = -D_{\alpha}'$, this further implies $D_{\alpha}'e^{tD_{\alpha}}u = 0$ for all $(\alpha, t) \in \mathcal{I}_c \times \mathbb{R}$. Duhamel's formula

$$ \frac{d}{d\alpha} e^{tD_{\alpha}} u = \int_{0}^{t} e^{(t-s)D_{\alpha}} D_{\alpha}' e^{sD_{\alpha}} u \, ds = 0 $$

allows us to conclude that $e^{tD_{\alpha}}u = e^{tD_{0}}u = e^{tA}u$, a relation which extends by continuity to all $(\alpha, t) \in \mathcal{I}_c \times \mathbb{R}$. Thus,

$$ \lim_{t \to \infty} e^{tD_{\alpha}}u = \lim_{t \to \infty} e^{tA}u = 0, $$

which, using (7) again, further implies that $u \notin \operatorname{Ker}(Y_{\alpha})$ and hence $(u, (Y_{\alpha}+W_{\alpha})u) = (u, Y_{\alpha}u) > 0$. (11) For $\lambda \in \mathbb{R}$, one has

$$ \mathcal{R}_{\alpha}(\lambda I) = Q\vartheta^{-1}(\lambda\vartheta - (\alpha - 1))(\lambda\vartheta - \alpha)\vartheta^{-1}Q^{*}, $$

so that $\mathcal{R}_{\alpha}(\lambda I) \le 0$ iff $\alpha - 1 \le \lambda\vartheta \le \alpha$. It follows that $\mathcal{P} = \{(\alpha, \lambda) \in \mathbb{R}^2 | \mathcal{R}_{\alpha}(\lambda I) \le 0\}$ is the closed parallelogram limited by the 4 lines (see Figure 11)

$$ \lambda = \frac{\alpha}{\vartheta_{\max}}, \quad \lambda = \frac{\alpha}{\vartheta_{\min}}, \quad \lambda = \frac{\alpha - 1}{\vartheta_{\max}}, \quad \lambda = \frac{\alpha - 1}{\vartheta_{\min}}. $$

The projection of $\mathcal{P}$ on the $\alpha$-axis is the closed interval $[\frac{1}{2} - \kappa_0, \frac{1}{2} + \kappa_0]$. Thus, Theorem A.5 implies that the Riccati equation has a self-adjoint solution for all $\alpha \in [\frac{1}{2} - \kappa_0, \frac{1}{2} + \kappa_0]$. By Theorem A.6 (2)
<figure><img src="image_14.png" /><figcaption>Figure 11: The parallelogram $\mathcal{P}$.</figcaption></figure>

it also has a maximal solution $X_\alpha$ which, by Theorem A.7 (3), satisfies the lower bound (5.39). From this lower bound we further deduce that for $\alpha \in [0, \frac{1}{2} + \kappa_0[$, the gap satisfies

$$Y_{\alpha} = X_{\alpha} + \theta X_{1-\alpha} \theta \geq \frac{\alpha}{\vartheta_{\max}} + \frac{1-\alpha}{\vartheta_{\min}} = \frac{\Delta}{\vartheta_{\max} \vartheta_{\min}} (\frac{1}{2} + \kappa_0 - \alpha) > 0.$$

Since $\text{Ker } Y_{\frac{1}{2}+\kappa_0} \neq \{0\}$ by Parts (6) and (7), we conclude that $\kappa_c \geq \kappa_0$.

(12) The concavity of $R_\alpha = X_\alpha + (1-\alpha)X_1$ and the fact that $R_0 = R_1 = X_1 > 0$ imply that for $|\alpha - \frac{1}{2}| \leq \frac{1}{2}$ one has $R_\alpha \geq X_1 > 0$. For $\frac{1}{2} < \alpha - \frac{1}{2} \leq \kappa_0$, Part (11) gives $X_\alpha \geq \alpha \vartheta_{\max}^{-1}$. Since $M > \vartheta_{\min}$, Part (2) yields $X_1 = \theta M^{-1} \theta < \vartheta_{\min}^{-1}$ and hence

$$R_{\alpha} > \frac{\alpha}{\vartheta_{\max}} + \frac{1-\alpha}{\vartheta_{\min}} = \frac{\kappa_{0} - (\alpha - \frac{1}{2})}{\Delta(\kappa_{0}^{2} - \frac{1}{4})} \geq 0.$$

The case $-\kappa_0 \leq \alpha - \frac{1}{2} < -\frac{1}{2}$ is similar. $\square$

## 5.7 Proof of Proposition 3.18

### 5.7.1 A Girsanov transformation

By Proposition 5.5, for $\alpha \in \bar{\mathcal{J}}_c$ we have $A = D_\alpha + QQ^*(X_\alpha - \alpha\beta)$, and we can rewrite the equation of motion (3.2) as

$$dx(t) = D_{\alpha}x(t)dt + Qdw_{\alpha}(t), \qquad (5.52)$$

where

$$w_{\alpha}(t) = w(t) - \int_{0}^{t} Q^{*}(\alpha\beta - X_{\alpha})x(s)ds.$$

Let $Z_\alpha(t)$ be the stochastic exponential of the local martingale

$$\eta_{\alpha}(t) = \int_{0}^{t} Q^{*}(\alpha\beta - X_{\alpha})x(s) \cdot dw(s).$$
Combining the Riccati equation with the relations $\beta QQ^* = QQ^*\beta = Q\vartheta^{-1}Q$ and $\beta QQ^*\beta = Q\vartheta^{-2}Q^*$, we derive

$$ \frac{1}{2}|Q^*(\alpha\beta - X_\alpha)x|^2 = -\alpha\sigma_\beta(x) - (\alpha\beta - X_\alpha)x \cdot Ax, $$

and we can write the quadratic variation of $\eta_\alpha$ as

$$ \frac{1}{2}[\eta_\alpha](t) = -\alpha \int_0^t \sigma_\beta(x(s))ds - \int_0^t (\alpha\beta - X_\alpha)x(s) \cdot Ax(s)ds. $$

Hence

$$ \eta_\alpha(t) - \frac{1}{2}[\eta_\alpha](t) = \int_0^t (\alpha\beta - X_\alpha)x(s) \cdot dx(s) + \alpha \int_0^t \sigma_\beta(x(s))ds. $$

The Itô calculus and Proposition 3.5 (3) give

$$ \eta_\alpha(t) - \frac{1}{2}[\eta_\alpha](t) = -(\lambda_\alpha t + \alpha \mathfrak{S}^t + \chi_\alpha(x(t)) - \chi_\alpha(x(0))), $$

with $\lambda_\alpha = \frac{1}{2}\text{tr}(QQ^*(\alpha\beta - X_\alpha))$ and

$$ \chi_\alpha(x) = \frac{1}{2}x \cdot X_\alpha x. $$

Finally, we note that Proposition 5.5 (6) yields

$$ \lambda_{\alpha} = -\frac{1}{2}\text{tr}(Q^*(X_{\alpha} - \alpha\beta)Q) = e(\alpha). $$

**Lemma 5.7** *The process*

$$ Z_{\alpha}(t) = \mathcal{E}(\eta_{\alpha})(t) = e^{-[e(\alpha)t + \alpha \mathfrak{S}^t + \chi_{\alpha}(x(t)) - \chi_{\alpha}(x(0))]} \quad (5.53) $$

is a $\mathbb{P}_x$-martingale for all $x \in \Xi$.

**Proof.** We wish to apply the Girsanov theorem; see Section 3.5 in [KS]. However, it is not clear that the Novikov condition is satisfied on a given finite interval. To overcome this difficulty, we follow the argument used in the proof of Corollary 5.14 in [KS, Chapter 3].

Fix $\tau > 0$. By Lemma 5.3, $\{x(t) - e^{tA}x\}_{t \in [0, \tau]}$ is a centered Gaussian process under the law $\mathbb{P}_x$. Since

$$ \int_s^{s'} |Q^*(\alpha\beta - X_\alpha)x(t)|^2 dt \le C|s - s'| \left( |x|^2 + \sup_{t \in [0, \tau]} |x(t) - e^{tA}x|^2 \right) $$

for some constant $C$, Fernique's theorem implies that there exists $\delta > 0$ such that

$$ \mathbb{E}_x \left[ \exp \left( \frac{1}{2} \int_s^{s'} |Q^*(\alpha\beta - X_\alpha)x(t)|^2 dt \right) \right] < \infty, $$

provided $0 \le s \le s' \le \tau$ and $s' - s < \delta$. Novikov criterion implies that under the same conditions,

$$ \mathbb{E}_x \left[ \frac{\mathcal{E}(\eta_\alpha)(s')}{\mathcal{E}(\eta_\alpha)(s)} | \mathcal{W}_s \right] = \mathbb{E}_x \left[ \exp \left( \int_s^{s'} Q^*(\alpha\beta - X_\alpha)x(t) \cdot dw(t) - \frac{1}{2} \int_s^{s'} |Q^*(\alpha\beta - X_\alpha)x(t)|^2 dt \right) \bigg| \mathcal{W}_s \right] = 1. $$
For $0 \le s \le s' \le s'' \le \tau$, $s' - s < \delta$ and $s'' - s' < \delta$ we deduce

$$
\begin{align*}
\mathbb{E}_x \left[ \frac{\mathcal{E}(\eta_\alpha)(s'')}{\mathcal{E}(\eta_\alpha)(s)} | \mathcal{W}_s \right] &= \mathbb{E}_x \left[ \frac{\mathcal{E}(\eta_\alpha)(s'')}{\mathcal{E}(\eta_\alpha)(s')} \frac{\mathcal{E}(\eta_\alpha)(s')}{\mathcal{E}(\eta_\alpha)(s)} | \mathcal{W}_s \right] \\
&= \mathbb{E}_x \left[ \mathbb{E}_x \left[ \frac{\mathcal{E}(\eta_\alpha)(s'')}{\mathcal{E}(\eta_\alpha)(s')} | \mathcal{W}_{s'} \right] \frac{\mathcal{E}(\eta_\alpha)(s')}{\mathcal{E}(\eta_\alpha)(s)} | \mathcal{W}_s \right] \\
&= \mathbb{E}_x \left[ \frac{\mathcal{E}(\eta_\alpha)(s')}{\mathcal{E}(\eta_\alpha)(s)} | \mathcal{W}_s \right] = 1,
\end{align*}
$$

and an induction argument gives

$$
\mathbb{E}_x [\mathcal{E}(\eta_\alpha)(\tau)] = \mathbb{E}_x \left[ \frac{\mathcal{E}(\eta_\alpha)(\tau)}{\mathcal{E}(\eta_\alpha)(0)} | \mathcal{W}_0 \right] = 1.
$$

Since $\tau > 0$ is arbitrary, the proof is complete.

<figure><img src="image_15.png" /></figure>

The previous lemma allows us to apply Girsanov theorem and to conclude that {$w_\alpha(t)$}$_{t \in [0, \tau]}$ is a
standard Wiener process under the law $Q_{\alpha, \nu}^\tau[\cdot] = \mathbb{E}_\nu[Z_\alpha(\tau) \cdot]$. This change of measure will be our
main tool in the next section.

**5.7.2 Completion of the proof**

From Eq. (3.35) and the results of the previous section we deduce that for $\alpha \in \bar{\mathcal{J}}_c$,

$$
\begin{align*}
e_t(\alpha) &= \frac{1}{t} \log \mathbb{E}_{\mu} \left[ \left( \frac{d\mu}{dx}(\theta x(t)) \right)^{\alpha} \left( \frac{d\mu}{dx}(x(0)) \right)^{-\alpha} e^{-\alpha \mathfrak{S}^t} \right] \\
&= \frac{1}{t} \log \mathbb{E}_{\mu} \left[ Z_{\alpha}(t) \left( \frac{d\mu}{dx}(\theta x(t)) \right)^{\alpha} \left( \frac{d\mu}{dx}(x(0)) \right)^{-\alpha} e^{-\chi_{\alpha}(x(0)) + \chi_{\alpha}(x(t)) + e(\alpha)t} \right] \\
&= e(\alpha) + \frac{1}{t} \log \mathbb{Q}_{\alpha, \mu}^{t} \left[ \left( \frac{d\mu}{dx}(\theta x(t)) \right)^{\alpha} \left( \frac{d\mu}{dx}(x(0)) \right)^{-\alpha} e^{-\chi_{\alpha}(x(0)) + \chi_{\alpha}(x(t))} \right],
\end{align*}
$$

where $\chi_\alpha(x) = \frac{1}{2}x \cdot X_\alpha x$. Denoting by $Q_\alpha^t$ the Markov semigroup associated with Eq. (5.52), we can
write

$$
e_t(\alpha) = e(\alpha) + \frac{1}{t} \log \langle \eta_\alpha | Q_\alpha^t \xi_\alpha \rangle,
$$

where

$$
\eta_{\alpha}(x) = \left(\frac{d\mu}{dx}(x)\right)^{1-\alpha} e^{-\chi_{\alpha}(x)}, \quad \xi_{\alpha}(x) = \left(\frac{d\mu}{dx}(\theta x)\right)^{\alpha} e^{\chi_{\alpha}(x)}. \qquad (5.54)
$$

Thus, to prove Eq. (3.45) we must show that the “prefactor” $\langle \eta_\alpha | Q_\alpha^t \xi_\alpha \rangle$ satisfies

$$
\lim_{t \to \infty} \frac{1}{t} \log \langle \eta_\alpha | Q_\alpha^t \xi_\alpha \rangle = 0.
$$

To this end, let us note that the Markov semigroup for (3.7) can be written as

$$
(P^t f)(x) = \int_X f(e^{tA} x + M_t^{\frac{1}{2}} y) n(dy), \qquad (5.55)
$$
where n denotes the centered Gaussian measure on $\mathcal{X}$ with covariance $I$. For $\alpha \in \bar{\mathcal{J}}_c$, this yields the representation

$$ (Q_{\alpha}^{t} f)(x) = \det(2\pi M_{\alpha,t})^{-\frac{1}{2}} \int e^{-\frac{1}{2}|M_{\alpha,t}^{-\frac{1}{2}}(y-e^{tD_{\alpha}}x)|^2} f(y)dy. \quad (5.56) $$

Using Eq. (5.54), a simple calculation leads to

$$ \langle \eta_{\alpha} | Q_{\alpha}^{t} \xi_{\alpha} \rangle = \det(2\pi M_{\alpha,t})^{-\frac{1}{2}} \det(2\pi M)^{-\frac{1}{2}} \int e^{-\frac{1}{2}z \cdot N_{\alpha,t}z} dz = \det(M_{\alpha,t}^{-1})^{\frac{1}{2}} \det(M^{-1})^{\frac{1}{2}} \det(N_{\alpha,t})^{-\frac{1}{2}}, $$

provided

$$ N_{\alpha,t} = \begin{bmatrix} \theta(Y_{1-\alpha} + W_{1-\alpha} + \Delta_{1-\alpha,t})\theta & -e^{tD_{\alpha}^*} M_{\alpha,t}^{-1} \\ -M_{\alpha,t}^{-1} e^{tD_{\alpha}} & Y_{\alpha} + W_{\alpha} + \Delta_{\alpha,t} \end{bmatrix} $$

is positive definite. By Schur's complement formula, we have

$$ \det(N_{\alpha,t}) = \det(Y_{\alpha} + W_{\alpha} + \Delta_{\alpha,t}) \det(Y_{1-\alpha} + W_{1-\alpha} + \Delta_{1-\alpha,t} - T_{\alpha,t}), $$

where

$$ T_{\alpha,t} = \theta(e^{tD_{\alpha}^*} M_{\alpha,t}^{-\frac{1}{2}}) M_{\alpha,t}^{-\frac{1}{2}} (Y_{\alpha} + W_{\alpha} + \Delta_{\alpha,t})^{-1} M_{\alpha,t}^{-\frac{1}{2}} (M_{\alpha,t}^{-\frac{1}{2}} e^{tD_{\alpha}}) \theta. $$

It follows that

$$ \langle \eta_{\alpha} | Q_{\alpha}^{t} \xi_{\alpha} \rangle = (\det(M) \det(M_{\alpha,t}) \det(Y_{\alpha} + W_{\alpha} + \Delta_{\alpha,t}) \det(Y_{1-\alpha} + W_{1-\alpha} + \Delta_{1-\alpha,t} - T_{\alpha,t}))^{-\frac{1}{2}}. $$

For any $\alpha \in \bar{\mathcal{J}}_c$, Proposition 5.5 implies that $Y_{\alpha} + W_{\alpha} > 0$ while, as $t \to \infty$, $M_{\alpha,t}^{-\frac{1}{2}} \searrow Y_{\alpha}^{\frac{1}{2}}$, $\Delta_{\alpha,t} \searrow 0$ and $\|M_{\alpha,t}^{-\frac{1}{2}} e^{tD_{\alpha}}\| \searrow 0$ monotonically (and exponentially fast for $\alpha \in \bar{\mathcal{J}}_c$). It follows that

$$ f(\alpha) = \lim_{t \to \infty} \langle \eta_{\alpha} | Q_{\alpha}^{t} \xi_{\alpha} \rangle = \left( \frac{\det(Y_{\alpha})}{\det(M) \det(Y_{\alpha} + W_{\alpha}) \det(Y_{1-\alpha} + W_{1-\alpha})} \right)^{\frac{1}{2}}. \quad (5.57) $$

For $\alpha \in \bar{\mathcal{J}}_c$, $Y_{\alpha} > 0$, and we conclude that

$$ \lim_{t \to \infty} \frac{1}{t} \log \langle \eta_{\alpha} | Q_{\alpha}^{t} \xi_{\alpha} \rangle = 0. \quad (5.58) $$

Consider now the limiting cases $\alpha = \frac{1}{2} \pm \kappa_c$. We shall denote by $C$ and $r$ generic positive constants which may vary from one expression to the other. Since $Y_{\alpha}$ is singular, one has $\log \det(M_{\alpha,t}^{-1}) \to -\infty$. However, the obvious estimate $\|e^{tD_{\alpha}}\| \le C(1+t)^r$ implies $M_{\alpha,t} \le C(1+t)^r$ and hence $M_{\alpha,t}^{-1} \ge C(1+t)^{-r}$ from which we conclude that

$$ \lim_{t \to \infty} \frac{1}{t} \log \det(M_{\alpha,t}^{-1}) = 0. \quad (5.59) $$

It follows that (5.58) also holds in the limiting cases $\alpha = \frac{1}{2} \pm \kappa_c$.

By Hölder's inequality $\mathbb{R} \ni \alpha \mapsto e_t(\alpha)$ is a convex function. The above analysis shows that it is a proper convex function differentiable on $\bar{\mathcal{J}}_c$ for any $t > 0$, and such that $\lim_{t \to \infty} e_t(\alpha) = e(\alpha)$ for $\alpha \in \bar{\mathcal{J}}_c$. Since $\lim_{\alpha \uparrow \frac{1}{2} + \kappa_c} e'(\alpha) = +\infty$ by Theorem 3.13 (3), the fact that

$$ \lim_{t \to \infty} e_t(\alpha) = +\infty $$

for $\alpha \in \mathbb{R} \setminus \bar{\mathcal{J}}_c$ is a consequence of the following lemma and the symmetry (3.39).
**Lemma 5.8** Let $(f_t)_{t>0}$ be a family of proper convex functions $f_t: \mathbb{R} \to [-\infty, \infty]$ with the following properties:

(1) For each $t > 0$, $f_t$ is differentiable on $]a, b[$.

(2) The limit $f(\alpha) = \lim_{t \to \infty} f_t(\alpha)$ exists for $\alpha \in ]a, b[$ and is differentiable on $]a, b[$.

(3) $\lim_{\alpha \uparrow b} f'(\alpha) = +\infty$.

Then, for all $\alpha > b$, one has $\lim_{t \to \infty} f_t(\alpha) = +\infty$.

**Proof.** By convexity, for any $\gamma \in ]a, b[$ and any $\alpha \in \mathbb{R}$ one has

$$f_t(\alpha) \ge f_t(\gamma) + (\alpha - \gamma)f'_t(\gamma),$$

and Properties (1) and (2) further imply

$$\lim_{t \to \infty} f'_t(\gamma) = f'(\gamma).$$

It follows that

$$\liminf_{t \to \infty} f_t(\alpha) \ge f(\gamma) + (\alpha - \gamma)f'(\gamma). \qquad (5.60)$$

As a limit of a family of convex functions, $f$ is convex on $]a, b[$ and, hence, $\inf_{\gamma \in ]a, b[} f(\gamma) > -\infty$. Thus, Property (3) and Inequality (5.60) yield

$$\liminf_{t \to \infty} f_t(\alpha) \ge \liminf_{\gamma \uparrow b} f(\gamma) + (\alpha - \gamma)f'(\gamma) = +\infty \quad \text{for all } \alpha > b.$$

□

## 5.8 Proof of Proposition 3.22

(1) The required properties of the function $g_t(\alpha)$ are consequences of more general results concerning integrals of exponentials of quadratic forms with respect to a Gaussian measure on an infinite-dimensional space. However, we shall derive here more detailed information about $g_t(\alpha)$ which will be used later (see the proof of Theorem 3.28).

We shall invoke Lemmata 5.2 and 5.3, and use the notations introduced in their proofs. By Proposition 3.5, we can write

$$g_t(\alpha) = \frac{1}{t} \log \int e^{-\frac{\alpha}{2}(x|\mathcal{L}_t x)} \gamma_t(dx),$$

where $\gamma_t$ is the Gaussian measure on $\mathfrak{H}_t$ with mean $T_t a$ and covariance $\mathcal{K}_t = \mathcal{D}_t \mathcal{D}_t^*$. The convexity of $g_t$ is a consequence of Hölder's inequality. The operator $\mathcal{L}_t$, given by

$$(\mathcal{L}_t x)(s) = -\Sigma_{\beta} x(s) + \delta(s-t)(F + X_1 - \beta)x(t) - \delta(s)(G + \theta X_1 \theta - \beta)x(0), \quad (5.61)$$

maps $\mathfrak{H}_+^*$ to $\mathfrak{H}_-$ in such a way that $(x|\mathcal{L}_t y) = (\mathcal{L}_t x|y)$ for all $x, y \in \text{Ran} \mathcal{D}_t$. It follows that the operator $\mathcal{S}_t = \mathcal{D}_t^* \mathcal{L}_t \mathcal{D}_t$ acting in the space $\Xi \oplus \partial \mathfrak{H}$ is self-adjoint, and a simple calculation shows that $\mathcal{S}_t - \mathcal{D}_t^* [\beta, \Omega] \mathcal{D}_t$ is finite rank, so that $\mathcal{S}_t$ is trace class. Using explicit formulas for Gaussian measures, we derive

$$g_t(\alpha) = -\frac{1}{2t} \log \det(I + \alpha \mathcal{S}_t) - \frac{\alpha}{2t} (T_t a | \mathcal{L}_t T_t a) + \frac{\alpha^2}{2t} (\mathcal{D}_t^* \mathcal{L}_t T_t a | (I + \alpha \mathcal{S}_t)^{-1} \mathcal{D}_t^* \mathcal{L}_t T_t a) \quad (5.62)$$
if $I + \alpha S_t > 0$, and $g_t(\alpha) = +\infty$ otherwise. Set $s_-(t) = \min \text{sp}(S_t) \le 0$, $s_+(t) = \max \text{sp}(S_t) \ge 0$, and

$$ \alpha_-(t) = \begin{cases} -s_+(t)^{-1} & \text{if } s_+(t) > 0; \\ -\infty & \text{if } s_+(t) = 0; \end{cases} \qquad \alpha_+(t) = \begin{cases} -s_-(t)^{-1} & \text{if } s_-(t) < 0; \\ +\infty & \text{if } s_-(t) = 0; \end{cases} $$

so that $I + \alpha S_t > 0$ iff $\alpha \in \mathcal{I}_t =]\alpha_-(t), \alpha_+(t)]$. Analyticity of $g_t$ on $\mathcal{I}_t$ follows from the Fredholm theory (e.g., see [Si1]), and a simple calculation yields

$$ g'_t(\alpha) = -\frac{1}{2t}\text{tr}((I + \alpha S_t)^{-1}S_t) - \frac{1}{2t}(T_t a | \mathcal{L}_t T_t a) + \frac{\alpha}{2t}(\mathcal{D}_t^* \mathcal{L}_t T_t a | ((I + \alpha S_t)^{-1} + (I + \alpha S_t)^{-2}) \mathcal{D}_t^* \mathcal{L}_t T_t a). \quad (5.63) $$

Suppose $\alpha_+(t) < \infty$ and denote by $P_-$ the spectral projection of $S_t$ associated to its minimal eigenvalue $s_-(t) < 0$. By the previous formula, for any $\alpha \in [0, \alpha_+(t)]$ one has

$$ g'_t(\alpha) \ge \frac{1}{2t} \frac{\text{tr}(P_-)}{\alpha_+(t) - \alpha} - \frac{1}{2t} \text{tr}((I + \alpha S_t)^{-1} S_t (I - P_-)) - \frac{1}{2t} (T_t a | \mathcal{L}_t T_t a), $$

which implies that $g'_t(\alpha) \to +\infty$ as $\alpha \to \alpha_+(t)$. The analysis of the lower bound $\alpha_-(t)$ is similar.

(2) Is a simple consequence of the continuity and concavity of the maps

$$ \bar{\mathcal{J}}_c \ni \alpha \mapsto F_\alpha = \theta X_{1-\alpha} \theta + \alpha(X_1 + F), \qquad \bar{\mathcal{J}}_c \ni \alpha \mapsto G_\alpha = \hat{N} + P_\nu(X_\alpha - \alpha(G + \theta X_1 \theta))|_{\text{Ran}(N)}, \\ \text{and the fact that } F_0 = \theta X_1 \theta > 0 \text{ and } G_0 = \hat{N} + P_\nu X_1|_{\text{Ran} N} > 0. $$

(3) If $X_1 + F > 0$ and $\hat{N} + P_\nu(X_1 - G - \theta X_1 \theta)|_{\text{Ran} N} > 0$, then we also have $F_1 > 0$ and $G_1 > 0$ and the result is again a consequence of the concavity of $F_\alpha$ and $G_\alpha$.

(4) Proceeding as in the proof of Proposition 3.18, we start from the expression

$$ g_t(\alpha) = e(\alpha) + \frac{1}{t} \log\langle\eta_\alpha|Q_\alpha^t \xi_\alpha\rangle, $$

where

$$ \eta_{\alpha}(dx) = e^{-\frac{1}{2}x \cdot (X_{\alpha} - \alpha(G + \theta X_{1}\theta))} x_{\nu}(dx), \qquad \xi_{\alpha}(x) = e^{-\frac{1}{2}x \cdot (-X_{\alpha} + \alpha(F + X_{1}))} x. $$

Setting

$$ C_{\alpha,t} = \begin{bmatrix} G_{\alpha} + P_{\nu}\theta\Delta_{1-\alpha,t}\theta|_{\text{Ran} N} & -P_{\nu}e^{tD_{\alpha}^{*}}M_{\alpha,t}^{-1} \\ -M_{\alpha,t}^{-1}e^{tD_{\alpha}}|_{\text{Ran} N} & F_{\alpha} + \Delta_{\alpha,t} \end{bmatrix}, $$

$$ c_{\alpha,t} = \frac{1}{2}a \cdot (X_{\alpha} - \alpha(G + \theta X_{1}\theta) + \theta\Delta_{1-\alpha,t}\theta)a, \quad b_{\alpha,t} = \begin{bmatrix} P_{\nu}(X_{\alpha} - \alpha(G + \theta X_{1}\theta) + \theta\Delta_{1-\alpha,t}\theta)a \\ -M_{\alpha,t}^{-1}e^{tD_{\alpha}}a \end{bmatrix}, $$

evaluation of a Gaussian integral leads to

$$ \begin{aligned} \langle \eta_{\alpha} | Q_{\alpha}^{t} \xi_{\alpha} \rangle &= \det(2\pi \hat{N}^{-1})^{-\frac{1}{2}} \det(2\pi M_{\alpha,t})^{-\frac{1}{2}} \int_{\text{Ran}(N) \oplus \Xi} e^{-\frac{1}{2} z \cdot C_{\alpha,t} z - z \cdot b_{\alpha,t} - c_{\alpha,t}} dz \quad (5.64) \\ &= \det(\hat{N}^{\frac{1}{2}}) \det(M_{\alpha,t})^{-\frac{1}{2}} \det(C_{\alpha,t})^{-\frac{1}{2}} e^{\frac{1}{2} b_{\alpha,t} \cdot C_{\alpha,t}^{-1} b_{\alpha,t} - c_{\alpha,t}}, \end{aligned} $$
provided $C_{\alpha,t} > 0$. By Schur's complement formula, the last condition is equivalent to

$$F_{\alpha} + \Delta_{\alpha,t} > 0, \quad G_{\alpha} + P_{\nu}\theta\Delta_{1-\alpha,t}\theta|_{\text{Ran } N} - T_{\alpha,t} > 0,$$

where

$$T_{\alpha,t} = P_{\nu}(e^{tD_{\alpha}^{*}} M_{\alpha,t}^{-\frac{1}{2}}) M_{\alpha,t}^{-\frac{1}{2}} (F_{\alpha} + \Delta_{\alpha,t})^{-1} M_{\alpha,t}^{-\frac{1}{2}} (M_{\alpha,t}^{-\frac{1}{2}} e^{tD_{\alpha}})|_{\text{Ran } N}.$$

Moreover, one has

$$\det(C_{\alpha,t}) = \det(F_{\alpha} + \Delta_{\alpha,t}) \det(G_{\alpha} + P_{\nu}\theta\Delta_{1-\alpha,t}\theta|_{\text{Ran } N} - T_{\alpha,t}).$$

For $\alpha \in \mathcal{J}_{\infty}$, it follows from Proposition 5.5 that

$$\lim_{t \to \infty} T_{\alpha,t} = 0,$$

and $F_{\alpha} + \Delta_{\alpha,t}$ and $G_{\alpha} + P_{\nu}\theta\Delta_{1-\alpha,t}\theta|_{\text{Ran } N} - T_{\alpha,t}$ are both positive definite for large $t$. As in the proof of Proposition 3.18 we can conclude that

$$\lim_{t \to \infty} \frac{1}{t} \log \langle \eta_{\alpha} | Q_{\alpha}^{t} \xi_{\alpha} \rangle = 0.$$

(5) Suppose that $\alpha_+ < \frac{1}{2} + \kappa_c$. If $\alpha \in ]\alpha_+, \frac{1}{2} + \kappa_c]$, then the matrix $C_{\alpha,t}$ acquires a negative eigenvalue as $t$ increases. Consequently, the integral in (5.64) diverges and $g_t(\alpha) = +\infty$ for large $t$, proving (3.62). The case $\alpha_- > \frac{1}{2} - \kappa_c$ and $\alpha \in [\frac{1}{2} - \kappa_c, \alpha_-[$ is similar. Suppose now that $\alpha_+ = \frac{1}{2} + \kappa_c$. Since $e'(\alpha) \to \infty$ as $\alpha \uparrow \frac{1}{2} + \kappa_c$ by Theorem 3.13 (3), Lemma 5.8 applies to $g_t$ and yields (3.62) again. The same argument works in the case $\alpha_- = \frac{1}{2} - \kappa_c$.

Combined with Parts (1) and (4), the above analysis shows that for any $\alpha < \alpha_+$ one has $\alpha_+(t) \ge \alpha$ for large enough $t$ while for any $\alpha > \alpha_+$, $\alpha_+(t) \le \alpha$ for large enough $t$. We deduce

$$\alpha_+ \le \liminf_{t \to \infty} \alpha_+(t) \le \limsup_{t \to \infty} \alpha_+(t) \le \alpha_+,$$

and (3.61) follows. $\square$

## 5.9 Proof of Theorem 3.28

We use the notation of Proposition 3.22 and its proof. We start with a few technical facts that will be used in the proof.

**Lemma 5.9** Assume that Condition (C) holds and that $\epsilon p > 0$. Then, for some constants $c > 0$ and $T > 0$, the following hold true.

(1) $\|S_t\| \le c$ and $\|S_t\|_1 \le ct$ for $t \ge T$.

(2) The function $g_t(\alpha)$ has an analytic continuation from $\mathcal{J}_t$ to the cut plane $\mathbb{C} \setminus (]-\infty, \alpha_-(t)] \cup [\alpha_+(t), \infty])$. Moreover, for any compact subset $K \subset \mathbb{C} \setminus (]-\infty, \alpha_-] \cup [\alpha_+, \infty])$ there is $T_K > 0$ such that

$$\sup_{\substack{\alpha \in K \\ t \ge T_K}} |g_t(\alpha)| < \infty.$$
(3) For $t \ge T$ the interval $\mathfrak{I}_t$ is finite and is mapped bijectively to $\mathbb{R}$ by the function $g'_t$. In the following, we set

$$ \alpha_{s,t} = (g'_t)^{-1}(s) $$

for $t \ge T$ and $s \in \mathbb{R}$.

(4) Let

$$ s_{\pm} = \lim_{\mathfrak{I}_{\infty} \ni \alpha \to \alpha_{\pm}} e'(\alpha), $$

and suppose that $s \in [-\infty, s_-]$ (resp. $s \in [s_+, +\infty[$). Then we have

$$ \lim_{t \to \infty} \alpha_{s,t} = \alpha_{-} \text{ (resp. } \alpha_{+}), \quad \liminf_{t \to \infty} g_t(\alpha_{s,t}) \ge e(\alpha_{-}) \text{ (resp. } e(\alpha_{+})). \quad (5.65) $$

(5) For $t \ge T$ and $s \in [-\infty, s_-] \cup [s_+, +\infty[$, let

$$ \mathfrak{M}_{s,t} = \frac{1}{t} S_t(I + \alpha_{s,t} S_t)^{-1}, \quad b_{s,t} = \frac{1}{t} (I + \alpha_{s,t} S_t)^{-\frac{3}{4}} \mathcal{D}_t^* \mathcal{L}_t T_t a. $$

The operator $\mathfrak{M}_{s,t}$ is trace class on $\Xi \oplus \partial\mathfrak{H}$, with trace norm

$$ \|\mathfrak{M}_{s,t}\|_1 \le c + |s|, $$

and $b_{s,t} \in \Xi \oplus \partial\mathfrak{H}$ is such that

$$ \lim_{t \to \infty} \|b_{s,t}\| = 0. $$

**Proof.** (1) Writing (5.61) as

$$ (\mathcal{L}_t x)(s) = L^{(1)}x(s) + \delta(s-t)L^{(2)}x(t) + \delta(s)L^{(3)}x(0) $$

with $L^{(j)} \in L(\Xi)$, we decompose $S_t = \mathcal{D}_t^* \mathcal{L}_t \mathcal{D}_t = S_t^{(1)} + S_t^{(2)} + S_t^{(3)}$. Lemma 5.3 (4) yields

$$ \|S_t^{(1)}\| \le \|L^{(1)}\| \|\mathcal{D}_t\|^2 = \|L^{(1)}\| \|\mathcal{K}_t\| \le c_1, $$

$$ \|S_t^{(1)}\|_1 = \text{tr}(\mathcal{D}_t^* |L^{(1)}| \mathcal{D}_t) \le \|L^{(1)}\| \text{tr}(\mathcal{D}_t \mathcal{D}_t^*) = \|L^{(1)}\| \|\mathcal{K}_t\|_1 \le c_1 t, $$

for $t \ge 0$. A simple calculation further gives $S_t^{(2)} = \tilde{D}_t^* L^{(2)} \tilde{D}_t$, $S_t^{(3)} = \tilde{D}_0^* L^{(3)} \tilde{D}_0$, where

$$ \tilde{D}_s = \left[ \begin{array}{cc} e^{sA} N^{\frac{1}{2}} & R_s Q \end{array} \right]. $$

It follows from Lemma 5.2 (3) that

$$ \|S_t^{(2)}\| \le \|S_t^{(2)}\|_1 = \text{tr}(\tilde{D}_t^* |L^{(2)}| \tilde{D}_t) \le \|L^{(2)}\| \text{tr}(\tilde{D}_t \tilde{D}_t^*) = \|L^{(2)}\| \text{tr}(e^{tA} N e^{tA^*} + R_t Q Q^* R_t^*) \le c_2, $$

and

$$ \|S_t^{(3)}\| \le \|S_t^{(3)}\|_1 = \text{tr}(\tilde{D}_0^* |L^{(3)}| \tilde{D}_0) \le \|L^{(3)}\| \text{tr}(\tilde{D}_0 \tilde{D}_0^*) = \|L^{(3)}\| \text{tr}(N) \le c_3, $$

for $t \ge 0$. We conclude that $\|S_t\| \le c_1 + c_2 + c_3$ and $\|S_t\|_1 \le (c_1 + c_2 + c_3)t$ for $t \ge 1$.

(2) Since $g_t(0) = 0$ for all $t > 0$, it suffices to show that the function $g'_t$ has the claimed properties. By definition,

$$ \mathbb{C} \setminus (]-\infty, \alpha_{-}(t)] \cup [\alpha_{+}(t), \infty]) \subset \{\alpha \in \mathbb{C} \mid -\alpha^{-1} \notin \text{sp}(S_t)\}, \quad (5.66) $$
and the analyticity of $g'_t$ on this set follows directly from Eq. (5.63). Let $K \subset \mathbb{C} \setminus (]-\infty, \alpha_-] \cup [\alpha_+, \infty])$ be compact. By Proposition 3.22 (5) (6) there exists $T_K \ge T$ such that

$$ \mathrm{dist}(K, ] - \infty, \alpha_-(t)] \cup [\alpha_+(t), \infty]) \ge \delta > 0 \quad (5.67) $$

for all $t \ge T_K$. By Part (1), $\|\alpha S_t\| \le \frac{1}{2}$ so that $\|(I + \alpha S_t)^{-1}\| \le 2$ for all $t \ge T$ and all $\alpha \in \mathbb{C}$ satisfying $|\alpha| \le (2c)^{-1}$. By the spectral theorem, it follows from (5.66) and (5.67) that

$$ \|(I + \alpha S_t)^{-1}\| \le \frac{2c}{\delta} $$

for all $t \ge T_K$ and all $\alpha \in K$ such that $|\alpha| \ge (2c)^{-1}$. Hence $\|(I + \alpha S_t)^{-1}\|$ is bounded on $K$ uniformly in $t \ge T_K$. The boundedness of $g'_t$ now easily follows from Eq. (5.63) and Part (1).

(3) By Part (5) of Proposition 3.22, if $T > 0$ is large enough then the interval $\mathcal{I}_t$ is finite for all $t \ge T$. By Part (1) of the same Proposition, the function $g'_t$ is strictly increasing on $\mathcal{I}_t$ and maps this interval onto $\mathbb{R}$.

(4) We consider $s \ge s_+$, the case $s \le s_-$ is similar. Since $\alpha_{s,t} \in \mathcal{I}_t$, Part (5) of Proposition 3.22 gives

$$ \underline{\alpha} = \liminf_{t \to \infty} \alpha_{s,t} \le \limsup_{t \to \infty} \alpha_{s,t} \le \lim_{t \to \infty} \alpha_+(t) = \alpha_+. $$

Suppose that $\underline{\alpha} < \alpha_+$. Invoking convexity, we deduce from the definition of $\alpha_{s,t}$ and Part (4) of Proposition 3.22

$$ s = \liminf_{t \to \infty} g'_t(\alpha_{s,t}) \le \liminf_{t \to \infty} g'_t(\underline{\alpha}) = e'(\underline{\alpha}). $$

The strict convexity of $e(\alpha)$ leads to $s \le e'(\underline{\alpha}) < s_+$ which contradicts our hypothesis and yields the first relation in (5.65).

To prove the second one, notice that for any $\gamma \in [0, \alpha_+[ one has $\gamma < \alpha_{s,t} \le \alpha_+(t)$ provided $t$ is large enough. By convexity

$$ g_t(\alpha_{s,t}) \ge g_t(\gamma) + (\alpha_{s,t} - \gamma)g'_t(\gamma) \ge g_t(\gamma) + (\alpha_{s,t} - \gamma)g'_t(0), $$

and letting $t \to \infty$ yields

$$ \liminf_{t \to \infty} g_t(\alpha_{s,t}) \ge e(\gamma) + (\alpha_+ - \gamma)e'(0). $$

Taking $\gamma \to \alpha_+$ gives the desired inequality.

(5) We consider $s \ge s_+$, the case $s \le s_-$ is again similar. By Part (3), if $T > 0$ is large enough then $\alpha_{s,t} \in ]0, \alpha_+(t)[ \subset \mathcal{I}_t$ for all $t \ge T$. Since $I + \alpha S_t > 0$ for $\alpha \in \mathcal{I}_t$, Part (1) allows us to conclude

$$ \|\mathfrak{M}_{s,t}^+\|_1 = \frac{1}{t}\mathrm{tr}(S_t^+(I + \alpha_{s,t}S_t^+)^{-1}) \le \frac{1}{t}\|S_t^+\|_1 \le \frac{1}{t}\|S_t\|_1 \le c. $$

By Eq. (5.63) and the definition of $\alpha_{s,t}$ we have

$$ s = -\frac{1}{2}\mathrm{tr}(\mathfrak{M}_{s,t}) - \frac{1}{2t}a \cdot T_t^* \mathcal{L}_t T_t a + \frac{t\alpha_{s,t}}{2}(b_{s,t}|((I + \alpha_{s,t}\mathcal{S}_t)^{\frac{1}{2}} + (I + \alpha_{s,t}\mathcal{S}_t)^{-\frac{1}{2}})b_{s,t}), $$

from which we deduce

$$ \|\mathfrak{M}_{s,t}^-\|_1 = s + \|\mathfrak{M}_{s,t}^+\|_1 + \frac{1}{t}a \cdot T_t^* \mathcal{L}_t T_t a - t\alpha_{s,t}(b_{s,t}|((I + \alpha_{s,t}\mathcal{S}_t)^{\frac{1}{2}} + (I + \alpha_{s,t}\mathcal{S}_t)^{-\frac{1}{2}})b_{s,t}). \quad (5.68) $$
One easily checks that

$$ \lim_{t \to \infty} \frac{1}{t} \| T_t^* \mathcal{L}_t T_t \| = 0, $$

so that $\|\mathfrak{M}_{s,t}^-\|_1 \le s + 2c$ and hence $\|\mathfrak{M}_{s,t}\|_1 = \|\mathfrak{M}_{s,t}^-\|_1 + \|\mathfrak{M}_{s,t}^+\|_1 \le |s| + 3c$ for $t$ large enough. Finally, from (5.68) we derive

$$ \|b_{s,t}\|^2 \le \frac{1}{2}(b_{s,t}|((I+\alpha_{s,t}\mathcal{S}_t)^{\frac{1}{2}}+(I+\alpha_{s,t}\mathcal{S}_t)^{-\frac{1}{2}})b_{s,t}) \le \frac{1}{2t\alpha_{s,t}} \left( s + \frac{1}{2}\mathrm{tr}(\mathfrak{M}_{s,t}) + \frac{1}{2t}a \cdot T_t^* \mathcal{L}_t T_t a \right), $$

from which we conclude that $\|b_{s,t}\| \to 0$ as $t \to \infty$. $\square$

(1) By Proposition 3.22 (4) one has

$$ \lim_{t \to \infty} \frac{1}{t} \log \mathbb{E}_{\nu}[e^{\alpha \eta_t}] = e(-\alpha), $$

for $-\alpha \in \mathcal{J}_{\infty}$. By the Gärtner-Ellis theorem, the local LDP holds on the interval $]\eta_{-}, \eta_{+}[$ with the rate function

$$ I(s) = \sup_{\alpha \in \mathbb{R}} (\alpha s - e(-\alpha)). $$

Note that $I(s) = \sup_{\alpha \in \mathcal{J}_{\infty}} (\alpha s - e(-\alpha))$ for $s \in ]\eta_{-}, \eta_{+}[$. To prove that the global LDP holds we must show that for all open sets $O \subset \mathbb{R}$

$$ \liminf_{t \to \infty} \frac{1}{t} \log \mathbb{P}_{\nu} \left[ \frac{\eta_t}{t} \in O \right] \ge - \inf_{s \in O} J(s). $$

By a simple and well known argument (see, e.g., [dH, Section V.2]), it suffices to show that for any $s \in \mathbb{R}$

$$ \lim_{\epsilon \downarrow 0} \liminf_{t \to \infty} \frac{1}{t} \log \mathbb{P}_{\nu} [|\hat{\eta}_t| < \epsilon] \ge -J(s), $$

where $\hat{\eta}_t = \frac{\eta_t}{t} - s$. The latter holds for any $s \in ]\eta_{-}, \eta_{+}[$ by the Gärtner-Ellis theorem. Next, we observe that whenever $\alpha_{\pm} = \frac{1}{2} \pm \kappa_c$, then by Proposition 3.22 (4) we have $\eta_{\pm} = \pm\infty$. Thus, it suffices to consider the cases where $\alpha_{-} > \frac{1}{2} - \kappa_c$ or/and $\alpha_{+} < \frac{1}{2} + \kappa_c$. We shall only discuss the second case, the analysis of the first one is similar.

Fix $s \le \eta_{-}$ and set $\alpha_t = -\alpha_{-s,t}$ so that $g'_t(-\alpha_t) = -s$ and, by Lemma 5.9 (3),

$$ \lim_{t \to \infty} \alpha_t = -\alpha_+, \quad \liminf_{t \to \infty} g_t(-\alpha_t) \ge e(\alpha_+). \qquad (5.69) $$

Defining the tilted probability $\hat{\mathbb{P}}_{\nu}^t$ on $C([0, t], \Xi)$ by

$$ \frac{d\hat{\mathbb{P}}_{\nu}^t}{d\mathbb{P}_{\nu}^t} = e^{\alpha_t \eta_t - tg_t(-\alpha_t)}, $$

we immediately get the estimate

$$ \mathbb{P}_{\nu}^{t} [|\hat{\eta}_{t}| < \epsilon] \ge e^{-t(s\alpha_{t}+\epsilon|\alpha_{t}|-g_{t}(-\alpha_{t}))} \hat{\mathbb{P}}_{\nu}^{t} [|\hat{\eta}_{t}| < \epsilon], $$

and hence,

$$ \frac{1}{t} \log \mathbb{P}_{\nu}^{t} [|\hat{\eta}_{t}| < \epsilon] \ge g_{t}(-\alpha_{t}) - s\alpha_{t} - \epsilon|\alpha_{t}| + \frac{1}{t} \log \hat{\mathbb{P}}_{\nu}^{t} [|\hat{\eta}_{t}| < \epsilon]. \qquad (5.70) $$
We claim that for any sufficiently small $\epsilon > 0$,

$$p_{\epsilon} = \liminf_{t \to \infty} \hat{\mathbb{P}}_{\nu}^{t} [|\hat{\eta}_{t}| < \epsilon] > 0. \qquad (5.71)$$

Using (5.69) we derive from (5.70) that

$$\liminf_{t \to \infty} \frac{1}{t} \log \mathbb{P}_{\nu}^{t} [|\hat{\eta}_{t}| < \epsilon] \geq e(\alpha_{+}) + s\alpha_{+} - \epsilon|\alpha_{+}|,$$

provided $\epsilon > 0$ is small enough. Letting $\epsilon \downarrow 0$, we finally get

$$\lim_{\epsilon \downarrow 0} \liminf_{t \to \infty} \frac{1}{t} \log \mathbb{P}_{\nu}^{t} [|\hat{\eta}_{t}| < \epsilon] \geq e(\alpha_{+}) + s\alpha_{+},$$

which, in view of (3.66), is the desired relation.

Thus, it remains to prove our claim (5.71). To this end, note that for $\lambda \in \mathbb{R}$,

$$\hat{\mathbb{E}}_{\nu}^{t} \left[ e^{-i\lambda \hat{\eta}_{t}} \right] = \mathbb{E}_{\nu}^{t} \left[ e^{(\alpha_{t} - i\lambda/t)\eta_{t} - tg_{t}(-\alpha_{t}) - i\lambda g'_{t}(-\alpha_{t})} \right] = e^{t(g_{t}(-\alpha_{t} + i\lambda/t) - g_{t}(-\alpha_{t}) - ig'_{t}(-\alpha_{t})\lambda/t)},$$

and a simple calculation using Eq. (5.62), (5.63) yields

$$\hat{\mathbb{E}}_{\nu}^{t} \left[ e^{-i\lambda \hat{\eta}_{t}} \right] = \left( \det \left( I + i\lambda \mathfrak{M}_{-s,t} \right)^{-1} e^{i\lambda \operatorname{tr}(\mathfrak{M}_{-s,t})} - \lambda^2 (b_{-s,t} |I + i\lambda \mathfrak{M}_{-s,t}|^{-1} b_{-s,t}) \right)^{\frac{1}{2}}. \quad (5.72)$$

Let $S(\mathbb{R}, \Xi)$ be the Schwartz space of rapidly decaying $\Xi$-valued smooth functions on $\mathbb{R}$ and $S'(\mathbb{R}, \Xi)$ its dual w.r.t. the inner product of $\mathfrak{H}$. Denote by $\hat{\gamma}$ the centered Gaussian measure on $\mathcal{K}_{-} = \Xi \oplus S'(\mathbb{R}, \Xi)$ with covariance $I$ and let

$$\tilde{\eta}_{t}(k) = -\frac{1}{2}(k|\mathfrak{M}_{-s,t}k - 2b_{-s,t}|).$$

By Lemma 5.9 (5), $|\hat{\eta}_{t}(k)| < \infty$ for $\hat{\gamma}$-a.e. $k \in \mathcal{K}_{-}$ and

$$\bar{\eta}_{t} = \int \tilde{\eta}_{t}(k) \hat{\gamma}(dk) = -\frac{1}{2} \operatorname{tr}(\mathfrak{M}_{-s,t}).$$

It follows that for $\lambda \in \mathbb{R}$,

$$\int e^{-i\lambda(\hat{\eta}_t(k) - \bar{\eta}_t)} \hat{\gamma}(dk) = \left( \det \left( I + i\lambda \mathfrak{M}_{-s,t} \right)^{-1} e^{i\lambda \operatorname{tr}(\mathfrak{M}_{-s,t})} - \lambda^2 (b_{-s,t} |I + i\lambda \mathfrak{M}_{-s,t}|^{-1} b_{-s,t}) \right)^{\frac{1}{2}},$$

and comparison with (5.72) allows us to conclude that the law of $\hat{\eta}_t$ under $\hat{\mathbb{P}}_\nu^t$ coincides with the one of $\tilde{\eta}_t - \bar{\eta}_t$ under $\hat{\gamma}$, so that

$$p_{\epsilon} = \liminf_{t \to \infty} \hat{\gamma} [|\tilde{\eta}_{t} - \bar{\eta}_{t}| < \epsilon].$$

For $m > 0$ let $P_m$ denote the spectral projection of $\mathfrak{M}_{s,t}$ for the interval $[-m, m]$ and define

$$\zeta_{t}^{<}(k) = -\frac{1}{2}(P_{m}k|\mathfrak{M}_{-s,t}k - 2b_{-s,t}|) + \frac{1}{2}\operatorname{tr}(P_{m}\mathfrak{M}_{-s,t}),$$

$$\zeta_{t}^{>}(k) = -\frac{1}{2}((I - P_{m})k|\mathfrak{M}_{-s,t}k - 2b_{-s,t}|) + \frac{1}{2}\operatorname{tr}((I - P_{m})\mathfrak{M}_{-s,t}),$$
so that $\tilde{\eta}_t - \bar{\eta}_t = \zeta_t^< + \zeta_t^>$ and $\hat{\gamma}[\zeta_t^<] = \hat{\gamma}[\zeta_t^>] = 0$. Since $\zeta_t^<$ and $\zeta_t^>$ are independent under $\hat{\gamma}$, we have

$$p_{\epsilon} \geq \liminf_{t \to \infty} \hat{\gamma} [|\zeta_t^<| < \epsilon/2] \hat{\gamma} [|\zeta_t^>| < \epsilon/2]. \qquad (5.73)$$

The Chebyshev inequality gives

$$\hat{\gamma} [|\zeta_t^<| < \epsilon/2] = 1 - \hat{\gamma} [|\zeta_t^<| \ge \epsilon/2] \ge 1 - \frac{4}{\epsilon^2} \hat{\gamma} [|\zeta_t^<|^2].$$

Choosing $m = \frac{1}{3}(c + |s|)\epsilon^2$, the estimate

$$\hat{\gamma} [|\zeta_t^<|^2] = \frac{1}{2}\mathrm{tr}(P_m \mathfrak{M}_{s,t}^2) + \|P_m b_{s,t}\|^2 \le \frac{1}{2}\|P_m \mathfrak{M}_{s,t}\| \|\mathfrak{M}_{s,t}\|_1 + \|b_{s,t}\|^2 \le \frac{m}{2}(c + |s|) + \|b_{s,t}\|^2,$$

together with Lemma 5.9 (5) shows that

$$\liminf_{t \to \infty} \hat{\gamma} [|\zeta_t^<| < \epsilon/2] \ge 1 - \frac{2m}{(c + |s|)\epsilon^2} = \frac{1}{3}.$$

To deal with the second factor on the right-hand side of (5.73) we first note that $(I - P_m)|\mathfrak{M}_{s,t}| \ge m(I - P_m)$, so that, using again Lemma 5.9 (5),

$$N_m = \mathrm{tr}(I - P_m) \le \frac{1}{m}\mathrm{tr}((I - P_m)|\mathfrak{M}_{s,t}|) \le \frac{\|\mathfrak{M}_{s,t}\|_1}{m} \le \frac{c + |s|}{m} = \frac{3}{\epsilon^2}.$$

Setting

$$\epsilon_j = \frac{|\mu_j|}{c + |s|}, \quad (j = 1, \dots, N_m),$$

where the $\mu_j$ denote the repeated eigenvalues of $(I - P_m)\mathfrak{M}_{s,t}$ we have $\sum_j \epsilon_j \le \epsilon$ and hence, passing to an orthonormal basis of eigenvectors of $(I - P_m)\mathfrak{M}_{s,t}$, we obtain

$$\hat{\gamma} [|\zeta_t^>| < \epsilon/2] = n_{N_m} \left[ \left| \sum_{j=1}^{N_m} \mu_j k_j^2 - 2b_j k_j - \mu_j \right| < \epsilon \right] \ge \prod_{j=1}^{N_m} n_1 \left[ |k^2 - 2\frac{b_j}{\mu_j}k - 1| < \frac{\epsilon_j}{|\mu_j|} \right],$$

where $n_{N_m}$ denotes the centered Gaussian measure of unit covariance on $\mathbb{R}^N$ and the $b_j \in \mathbb{R}$ are such that $|b_j| \le \|b_{s,t}\|$. An elementary analysis shows that if $|b| \le 1$ and $0 < \delta \le 1$, then

$$n_1 [|k^2 - 2bk - 1| < \delta] \ge \frac{\delta}{e\sqrt{6\pi}}.$$

Thus, provided $\epsilon < c + |s|$, we can conclude that

$$\liminf_{t \to \infty} \hat{\gamma} [|\zeta_t^>| < \epsilon/2] \ge \left( \frac{\epsilon}{e(c + |s|)\sqrt{6\pi}} \right)^{3\epsilon^2} > 0,$$

which shows that $p_{\epsilon} > 0$ and concludes the proof of Part (2).

(2) According to Bryc's lemma (see [Br] or [JOPP, Section 4.8.4]) the Central Limit Theorem for the family $(\eta_t)_{t>0}$ holds, provided that the generating function $g_t$ has an analytic continuation to the disc $D_{\epsilon} = \{\alpha \in \mathbb{C} \mid |\alpha| < \epsilon\}$ for some $\epsilon > 0$ and satisfies the estimate

$$\sup_{\substack{\alpha \in D_{\epsilon} \\ t > t_0}} |g_t(\alpha)| < \infty,$$

for some $t_0 > 0$. These properties clearly follow from Lemma 5.9 (2).
## 5.10 Proof of Lemma 4.1

(1) Let

$$C = \bigvee_{j \ge 0} \mathrm{Ran} (\omega^* \omega)^{j \iota}$$

be the controllable subspace of $(\omega^*\omega, \iota)$. From (3.1) and (3.4) we derive

$$\Omega^{2j} Q = (-1)^j \begin{bmatrix} (\omega^*\omega)^{j\iota} & 0 \\ 0 & 0 \end{bmatrix} \vartheta^{\frac{1}{2}},$$

and hence

$$\bigvee_{j \ge 0} \mathrm{Ran} (\Omega^{2j} Q) = C \oplus \{0\}.$$

The last relation and $\Omega(C \oplus \{0\}) = \{0\} \oplus \omega^*C$ yield that the controllable subspace of $(\Omega, Q)$ is $C \oplus \omega^*C$. Since $A = \Omega - \frac{1}{2}Q\vartheta^{-1}Q^*$, $(A, Q)$ has the same controllable subspace. Finally, since $\mathrm{Ker} \omega = \{0\}$, we conclude that $C \oplus \omega^*C = \Xi$ iff $C = \mathbb{R}^I$.

(2) The same argument yields $C(\Omega, Q\pi_i) = C_i \oplus \omega^*C_i$. Thus if $0 \ne u \in C_i \cap C_j$, we have $0 \ne u \oplus 0 \in C(\Omega, Q\pi_i) \cap C(\Omega, Q\pi_j)$ and the result follows from Proposition 3.7 (2).

## 5.11 Proof of Theorem 4.2

(1) By assumption (J), the Jacobi matrix

$$\omega^2 = \begin{bmatrix} b_1 & a_1 & 0 & 0 & \cdots & 0 & 0 \\ a_1 & b_2 & a_2 & 0 & & 0 & 0 \\ 0 & a_2 & b_3 & a_3 & & 0 & 0 \\ \vdots & \ddots & \ddots & \ddots & & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & a_{L-2} & 0 \\ 0 & 0 & 0 & 0 & \cdots & b_{L-1} & a_{L-1} \\ 0 & 0 & 0 & 0 & \cdots & a_{L-1} & b_L \end{bmatrix},$$

is positive and $a_i \ne 0$ for all $i \in \mathcal{I}$. Denote by $\{\delta_i\}_{i \in \mathcal{I}}$ the canonical basis of $\mathbb{R}^I$. Starting with the obvious fact that $\mathrm{Ran}(\iota) = \mathrm{span}(\{\delta_i | i \in \partial \mathcal{I}\})$, a simple induction yields

$$\bigvee_{0 \le j \le k} \mathrm{Ran} (\omega^{2j} \iota) = \mathrm{span}(\{\delta_i | \mathrm{dist}(i, \partial \mathcal{I}) \le k\}).$$

Hence the pair $(\omega^2, \iota)$ is controllable.

(2) The argument in the proof of Part (1) yields $C_1 = C_L = \mathbb{R}^I$ and the first statement follows directly from Proposition 3.7 (2). To prove the second one, we may assume that $\vartheta_{\min} = \vartheta_1$ and $\vartheta_{\max} = \vartheta_L$. From Theorem 3.2 (3) we already know that $\vartheta_1 \le M \le \vartheta_L$ and that

$$M - \vartheta_1 = \vartheta_1 \int_0^\infty e^{tA} Q (\vartheta_1^{-1} - \vartheta^{-1}) Q^* e^{tA^*} dt.$$
Since $\vartheta_1^{-1} - \vartheta^{-1} \ge 0$ it follows that

$$ \mathrm{Ker} (M - \vartheta_1) \subset \bigcap_{n \ge 0} \mathrm{Ker} (\vartheta_1^{-1} - \vartheta^{-1}) Q^* A^{*n} = \left( \bigvee_{n \ge 0} A^n Q \delta_1 \right)^\perp = C_1^\perp = \{0\}, $$

which implies $M - \vartheta_1 > 0$. A similar argument shows that $\vartheta_2 - M > 0$.

(3) Set $\kappa = \alpha - \frac{1}{2}$ and $\kappa_0 = \frac{\bar{\vartheta}}{\Delta} > \frac{1}{2}$. Writing

$$ i\nu - K_{\alpha} = \begin{bmatrix} \Omega + i\nu & 0 \\ 0 & \Omega + i\nu \end{bmatrix} + \begin{bmatrix} Q\vartheta^{-\frac{1}{2}} & 0 \\ 0 & Q\vartheta^{-\frac{1}{2}} \end{bmatrix} \begin{bmatrix} \kappa & -\vartheta \\ (\kappa^2 - \frac{1}{4})\vartheta^{-1} & -\kappa \end{bmatrix} \begin{bmatrix} \vartheta^{-\frac{1}{2}} Q^* & 0 \\ 0 & \vartheta^{-\frac{1}{2}} Q^* \end{bmatrix}, $$

one derives $\det(i\nu - K_\alpha) = \det(\Omega + i\nu)^2 \det(I + \Sigma(i\nu))$, where

$$ \Sigma(z) = \begin{bmatrix} \kappa R(z) & -R(z)\vartheta \\ (\kappa^2 - \frac{1}{4})R(z)\vartheta^{-1} & -\kappa R(z) \end{bmatrix}, \quad R(z) = \vartheta^{-\frac{1}{2}} Q^* (\Omega + z)^{-1} Q \vartheta^{-\frac{1}{2}}. \quad (5.74) $$

A simple calculation further gives

$$ \det(\Omega + i\nu) = \det(\omega^2 - \nu^2), \quad R(i\nu) = i\nu i^* (\omega^2 - \nu^2)^{-1} i. $$

Denote by $D(\nu^2)$ the adjugate of $\omega^2 - \nu^2$. Expressing $(\omega^2 - \nu^2)^{-1}$ with Cramer's formula and observing that $D_{1L}(\nu^2) = D_{L1}(\nu^2) = \hat{a}$, we get

$$ R(i\nu) = \frac{2\bar{\gamma}i\nu}{d(\nu^2)} \begin{bmatrix} b(\nu^2)e^{\frac{1}{2}\delta} & \hat{a} \\ \hat{a} & c(\nu^2)e^{-\frac{1}{2}\delta} \end{bmatrix}, \quad (5.75) $$

where

$$ b(\nu^2) = D_{11}(\nu^2), \quad c(\nu^2) = D_{LL}(\nu^2), \quad d(\nu^2) = \det(\omega^2 - \nu^2), $$

are polynomials in $\nu^2$ with real coefficients. Inserting (5.75) into (5.74), an explicit calculation of $\det(I + \Sigma(i\nu))$ yields

$$ \det(i\nu - K_{\alpha}) = \left( d(\nu^2) + (\bar{\gamma}\nu)^2 \frac{b(\nu^2)c(\nu^2) - \hat{a}^2}{d(\nu^2)} \right)^2 + (\bar{\gamma}\nu)^2 \left( e^{\delta/2}b(\nu^2) - e^{-\delta/2}c(\nu^2) \right)^2 - 4\hat{a}^2 \frac{\kappa^2 - \kappa_0^2}{\kappa_0^2 - \frac{1}{4}} (\bar{\gamma}\nu)^2. $$

By the Desnanot-Jacobi identity,

$$ \frac{b(\nu^2)c(\nu^2) - \hat{a}^2}{d(\nu^2)} = \det(\tilde{\omega}^2 - \nu^2) = \tilde{d}(\nu^2), $$

where $\tilde{\omega}^2$ is the matrix obtained from $\omega^2$ by deleting its first and last rows and columns. Thus, we finally obtain

$$ \det(i\nu - K_{\alpha}) = \left( d(\nu^2) + (\bar{\gamma}\nu)^2 \tilde{d}(\nu^2) \right)^2 + (\bar{\gamma}\nu)^2 \left( e^{\delta/2}b(\nu^2) - e^{-\delta/2}c(\nu^2) \right)^2 - 4\hat{a}^2 \frac{\kappa^2 - \kappa_0^2}{\kappa_0^2 - \frac{1}{4}} (\bar{\gamma}\nu)^2, $$
where $b, c, d$ and $\tilde{d}$ are polynomials with real coefficients. Since $d(0) = \det(\omega^2) > 0$, $K_\alpha$ is regular for all $\alpha \in \mathbb{R}$ and we can rewrite the eigenvalue equation as

$$g(\nu^2) = \frac{\kappa^2 - \kappa_0^2}{\kappa_0^2 - \frac{1}{4}}, \qquad (5.76)$$

where the rational function

$$g(x) = \frac{1}{4\hat{a}^2} \left[ \frac{(d(x) + \bar{\gamma}^2 x \tilde{d}(x))^2}{\bar{\gamma}^2 x} + \left( e^{\delta/2} b(x) - e^{-\delta/2} c(x) \right)^2 \right]$$

has real coefficients, a simple pole at 0, a pole of order $2L$ at infinity and is non-negative on $]0, \infty[$. It follows that

$$\kappa_c = \sqrt{\kappa_0^2 + g_0(\kappa_0^2 - \frac{1}{4})},$$

where

$$g_0 = \min_{x \in ]0, \infty[} g(x).$$

Since $\kappa_0 > \frac{1}{2}$, we conclude that $\kappa_c \ge \kappa_0$, with equality iff $g_0 = 0$.

Under Assumption (S) the polynomials $b$ and $c$ coincide and $\delta = 0$. Thus, $g_0 = 0$ iff the polynomial

$$f(x) = d(x) + \bar{\gamma}^2 x \tilde{d}(x) \qquad (5.77)$$

has a positive zero. If $L$ is odd, then this property follows immediately from the fact that

$$f(0) = \det(\omega^2) > 0, \quad f(x) = (-x)^L + \mathcal{O}(x^{L-1}) < 0 \quad (x \to \infty).$$

A more elaborate argument is needed in the case of even $L$. We shall invoke the deep connection between spectral analysis of Jacobi matrices and orthogonal polynomials. We refer the reader to [Si2] for a detailed introduction to this vast subject.

Let $\rho$ be the spectral measure of $\omega^2$ for the vector $\delta_1$. The argument in the proof of Part (1) shows that $\delta_1$ is cyclic for $\omega^2$. Thus, $\omega^2$ is unitarily equivalent to multiplication by $x$ on $L^2(\mathbb{R}, \rho(dx))$ and in this Hilbert space $\delta_1$ is represented by the constant polynomial $p_0 = 1$. Starting with $\delta_2 = a_1^{-1}(\omega^2 - b_1)\delta_1 = p_1(\omega^2)\delta_1$, a simple induction shows that there are real polynomials $\{p_k\}_{k \in \{0, \dots, L-1\}}$ satisfying the recursion

$$a_k p_{k-1}(x) + (b_{k+1} - x)p_k(x) + a_{k+1} p_{k+1}(x) = 0, \quad (k \in \{0, \dots, L-2\}, p_{-1} = 0, p_0 = 1), \qquad (5.78)$$

and such that $\delta_k = p_{k-1}(\omega^2)\delta_1$. Thus, these polynomials form an orthonormal basis of $L^2(\mathbb{R}, \rho(dx))$ such that

$$\langle \delta_k | (\omega^2 - x)^{-1} \delta_j \rangle = \int \frac{p_{k-1}(\lambda) p_{j-1}(\lambda)}{\lambda - x} \rho(d\lambda). \qquad (5.79)$$

For $1 \le j \le k \le L$, define

$$d_{[j,k]}(x) = \det(x - J_{[j,k]}), \quad J_{[j,k]} = \begin{bmatrix} b_j & a_j & 0 & 0 & \cdots & 0 & 0 \\ a_j & b_{j+1} & a_{j+1} & 0 & & 0 & 0 \\ 0 & a_{j+1} & b_{j+2} & a_{j+2} & & 0 & 0 \\ \vdots & & \vdots & \vdots & & \vdots & & \vdots \\ 0 & 0 & 0 & \cdots & \cdots & a_{k-2} & 0 \\ 0 & 0 & 0 & 0 & \cdots & b_{k-1} & a_{k-1} \\ 0 & 0 & 0 & 0 & \cdots & a_{k-1} & b_k \end{bmatrix}.$$
Laplace expansion of the determinant $P_{k+1}(x) = d_{[1,k+1]}(x)$ on its last row yields the recursion

$$P_{k+1}(x) = (x - b_{k+1})P_k(x) - a_k^2 P_{k-1}(x).$$

Comparing this relation with (5.78) one easily deduces

$$a_1 \cdots a_k p_k(x) = P_k(x), \quad (k \in \{1, \dots, L-1\}), \quad d(x) = P_L(x). \tag{5.80}$$

Polynomials of the second kind $\{q_k\}_{k \in \{0, \dots, L-1\}}$ associated to the measure $\rho$ are defined by

$$q_k(x) = \int \frac{p_k(\lambda) - p_k(x)}{\lambda - x} \rho(d\lambda). \tag{5.81}$$

Note in particular that $q_0(x) = 0$ and $q_1(x) = a_1^{-1}$. Applying the recursion relation (5.78) to both sides of this definition, we obtain

$$a_k q_{k-1}(x) + (b_{k+1} - x)q_k(x) + a_{k+1}q_{k+1}(x) = \int p_k(\lambda)\rho(d\lambda) = 0, \quad (k \in \{1, \dots, L-2\}).$$

Set $\tilde{q}_k(x) = a_1 q_{k+1}(x)$ and observe that these polynomials satisfy the recursion

$$a_{k+1} \tilde{q}_{k-1}(x) + (b_{k+2} - x) \tilde{q}_k(x) + a_{k+2} \tilde{q}_{k+1}(x) = 0, \quad (k \in \{0, \dots, L-3\}, \tilde{q}_{-1} = 0, \tilde{q}_0 = 1).$$

Comparing this Cauchy problem with (5.78) and repeating the argument leading to (5.80) we deduce that $a_2 \cdots a_{k+1} \tilde{q}_k(x) = d_{[2,k+1]}(x)$, so that

$$a_1 \cdots a_k q_k(x) = d_{[2,k]}(x), \quad (k \in \{2, \dots, L-1\}).$$

In particular, we can rewrite Definition (5.77) as

$$f(x) = P_L(x) + \bar{\gamma}^2 x q_{L-1}(x). \tag{5.82}$$

Taking now Assumption (S) into account we derive from (5.79) that for any $z \in \mathbb{C} \setminus \text{sp}(\omega^2)$,

$$\int \frac{|p_{L-1}(\lambda)|^2}{\lambda - z} \rho(d\lambda) = \langle \delta_L | (\omega^2 - z)^{-1} \delta_L \rangle \\ = \langle S \delta_1 | (\omega^2 - z)^{-1} S \delta_1 \rangle = \langle \delta_1 | (\omega^2 - z)^{-1} \delta_1 \rangle = \int \frac{\rho(d\lambda)}{\lambda - z},$$

from which we conclude that $|p_{L-1}(\lambda)| = 1$ for all $\lambda \in \text{sp}(\omega^2)$. Denote by $\lambda_L \ge \lambda_{L-1} \ge \cdots \ge \lambda_1$ the eigenvalues of $\omega^2 = J_{[1,L]}$ and by $\mu_{L-1} \ge \mu_{L-2} \ge \cdots \ge \mu_1$ that of $J_{[1,L-1]}$. It is a well known property of Jacobi matrices (or equivalently of orthogonal polynomials) that

$$\lambda_L < \mu_{L-1} < \lambda_{L-1} < \cdots < \mu_1 < \lambda_1$$

(see Figure 12). These interlacing inequalities and the previously established property allow us to conclude that

$$p_{L-1}(\lambda_j) = (-1)^j, \quad p'_{L-1}(\lambda_1) < 0.$$
<figure><img src="image_16.png" /><figcaption>Figure 12: The zeros of the polynomials $P_L$ and $p_{L-1}$ interlace.</figcaption></figure>

From Eq. (5.82) and Definition (5.81), we deduce

$$ 
\begin{aligned}
f(\lambda_1) &= \bar{\gamma}^2 \lambda_1 \left( p'_{L-1}(\lambda_1) \rho(\{\lambda_1\}) + \sum_{j=2}^{L-1} \frac{p_{L-1}(\lambda_1) - p_{L-1}(\lambda_j)}{\lambda_1 - \lambda_j} \rho(\{\lambda_j\}) \right) \\
&= \bar{\gamma}^2 \lambda_1 \left( p'_{L-1}(\lambda_1) \rho(\{\lambda_1\}) - 2 \sum_{j=1}^{\frac{1}{2}L-1} \frac{\rho(\{\lambda_{2j}\})}{\lambda_1 - \lambda_{2j}} \right) < 0,
\end{aligned}
$$

which, together with $f(0) > 0$, shows that $f$ has a positive root.

By Proposition 5.5 (12), the validity of Condition (R) follows from Part (2) and the fact that $\kappa_c = \kappa_0$.

## Appendix: Basic theory of the algebraic Riccati equation

In this appendix, for the reader convenience we briefly expose the basic results on algebraic Riccati equation used in this work. We refer the reader to [LR, AFIJ, Sc] for detailed expositions and proofs.

Let $\mathfrak{h}$ be a $d$-dimensional complex Hilbert space. We denote by $(\cdot, \cdot)$ the inner product of $\mathfrak{h}$. We equip the vector space $\mathcal{H} = \mathfrak{h} \oplus \mathfrak{h}$ with the Hilbertian structure induced by $\mathfrak{h}$ and the symplectic form

$$ \omega(x \oplus y, x' \oplus y') = (x \oplus y, J(x' \oplus y')) = (x, y') - (y, x'). $$

The symplectic complement of $\mathcal{V} \subset \mathcal{H}$ is the subspace $\mathcal{V}^{\omega} = \{v | \omega(u, v) = 0 \text{ for all } u \in \mathcal{V}\}$. A subspace $\mathcal{V} \subset \mathcal{H}$ is isotropic if $\mathcal{V} \subset \mathcal{V}^{\omega}$ and Lagrangian if $\mathcal{V} = \mathcal{V}^{\omega}$. $\mathcal{V}$ is Lagrangian iff it is isotropic and $d$-dimensional. For $Y, Z \in L(\mathfrak{h})$, we denote by $Y \oplus Z$ the element of $L(\mathfrak{h}, \mathcal{H})$ defined by $(Y \oplus Z)x = Yx \oplus Zx$. In the block-matrix notation,

$$ Y \oplus Z = \begin{bmatrix} Y \\ Z \end{bmatrix}, \quad (Y \oplus Z)^* = \begin{bmatrix} Y^* & Z^* \end{bmatrix}. $$
The graph of $X \in L(\mathfrak{h})$ is the $d$-dimensional subspace of $\mathcal{H}$ defined by

$$ \mathcal{G}(X) = \mathrm{Ran} G_X, \quad G_X = I \oplus X. $$

A subspace $\mathcal{V} \subset \mathcal{H}$ is a graph iff $\mathcal{V} \cap \{\{0\} \oplus \mathfrak{h}\} = \{0 \oplus 0\}$.

The algebraic Riccati equation associated to the triple $(A, B, C)$ of elements of $L(\mathfrak{h})$ is the following quadratic equation for the unknown self-adjoint $X \in L(\mathfrak{h})$:

$$ \mathcal{R}(X) = XBX - XA - A^*X - C = 0. \qquad (A.1) $$

In the following, we shall assume that $C$ is self-adjoint, that $B \ge 0$ and that the pair $(A, B)$ is controllable. We denote by $\mathfrak{R}(A, B, C)$ the set of self-adjoint elements of $L(\mathfrak{h})$ satisfying Eq. (A.1), which we can also write as

$$ \mathcal{R}(X) = G_X^* L G_X = 0, \quad L = \begin{bmatrix} C & A^* \\ A & -B^* \end{bmatrix}. $$

## A.1 Existence of self-adjoint solutions

The Hamiltonian associated to the Riccati equation (A.1) is the unique element of $L(\mathcal{H})$ such that $(u, Lv) = \omega(u, Kv)$ for all $u, v \in \mathcal{H}$. One easily checks that

$$ K = \begin{bmatrix} -A & B \\ C & A^* \end{bmatrix}. $$

Note that since $L = L^*$, $K$ is $\omega$-skew adjoint:

$$ \omega(u, Kv) + \omega(Ku, v) = \omega(u, Kv) - \omega(v, Ku) = (u, Lv) - (v, Lu) = 0. \qquad (A.2) $$

The first result we recall is a characterization of the set $\mathfrak{R}(A, B, C)$.

**Theorem A.1 (Theorem 7.2.4 in [LR])** The map $X \mapsto \mathcal{G}(X)$ is a bijection from $\mathfrak{R}(A, B, C)$ onto the set of $K$-invariant Lagrangian subspaces of $\mathcal{H}$.

The following are elementary symplectic geometric properties of projections:

**Lemma A.2 (1)** The range of a projection $P \in L(\mathcal{H})$ is isotropic iff $P^*JP = 0$ and Lagrangian iff $I - P = J^*P^*J$.

(2) Denote by $P_\kappa$ the spectral projection of $K$ for $\kappa \in \mathrm{sp}(K)$. Then $JP_\kappa J^* = P_{-\bar{\kappa}}^*$ and in particular $\mathrm{Ran} P_\kappa$ is isotropic iff $\kappa \notin i\mathbb{R}$.

(3) Let $\Sigma \subset \mathrm{sp}(K)$ be such that $\Sigma \cap (-\bar{\Sigma}) = \emptyset$. Then the spectral subspace of $K$ for $\Sigma$ is isotropic.

Note that $JK + K^*J = 0$, which implies that the spectrum of $K$, including multiplicities, is symmetric w.r.t. the imaginary axis. If $\mathrm{sp}(K) \cap i\mathbb{R} = \emptyset$, then the spectral subspace of $K$ for $\Sigma = \mathrm{sp}(K) \cap C_+$ is $d$-dimensional and hence, by Lemma A.2 (3), Lagrangian. Thus, Theorem A.1 yields (see Theorems 7.2.4 and 7.5.1 in [LR])

**Corollary A.3** If $\mathrm{sp}(K) \cap i\mathbb{R} = \emptyset$, then $\mathfrak{R}(A, B, C) \neq \emptyset$.
**Remark A.4** In cases where $sp(K) \cap i\mathbb{R} \neq \emptyset$, and under our controllability assumption, a necessary and sufficient condition for the existence of self-adjoint solution is that all Jordan blocks of $K$ corresponding to eigenvalues in $i\mathbb{R}$ are even-dimensional. For the Riccati equations arising in our analysis of harmonic networks, the singular case $sp(K_\alpha) \cap i\mathbb{R} \neq \emptyset$ only occurs at the boundary points $\alpha = \frac{1}{2} \pm \kappa_c$. There, the existence of solutions follows by continuity (Part (4) of Theorem 5.5).

Another powerful criterion for the existence of self-adjoint solutions is the following

**Theorem A.5 (Theorem 9.1.1 in [LR])** If there exists a self-adjoint $X \in L(\mathfrak{h})$ such that $\mathcal{R}(X) \le 0$, then $\mathcal{R}(A, B, C) \neq \emptyset$.

## A.2 Extremal solutions

The set $\mathcal{R}(A, B, C)$ inherits the partial order of $L(\mathfrak{h})$. A minimal/maximal solution of (A.1) is a minimal/maximal element of $\mathcal{R}(A, B, C)$. Clearly, a minimal/maximal solution, if it exists, is unique.

**Theorem A.6** Assume that $\mathcal{R}(A, B, C) \neq \emptyset$.

(1) $\mathcal{R}(A, B, C)$ is compact.

(2) $\mathcal{R}(A, B, C)$ contains a minimal element $X_-$ and a maximal element $X_+$. In the following, we set

$$D_{\mp} = A - B X_{\mp}.$$

(3) $X \in \mathcal{R}(A, B, C)$ is minimal/maximal iff $sp(A - B X) \subset \bar{\mathbb{C}}_{\pm}$.

(4) $\mathcal{R}(A, B, C) = X_- + \mathcal{R}(D_-, B, 0) = X_+ - \mathcal{R}(-D_+, B, 0)$.

Parts (2) and (3) are stated as Theorems 7.5.1 in [LR]. Part (4) follows from simple algebra. Since $X \mapsto \mathcal{R}(X)$ is continuous, $\mathcal{R}(A, B, C)$ is closed. Its boundedness follows from Part (4) and the fact that

$$\|X - X_-\|_1 = \text{tr}(X - X_-) \le \text{tr}(X_+ - X_-),$$

for all $X \in \mathcal{R}(A, B, C)$. The Heine-Borel theorem thus yields Part (1).

## A.3 The gap

In this section, we assume that $\mathcal{R}(A, B, C) \neq \emptyset$ and use the notations introduced in Theorem A.6.

The gap of the Riccati equation (A.1) is the non-negative element of $L(\mathfrak{h})$ defined by

$$Y = X_+ - X_-.$$

We set $\mathcal{K} = \text{Ker } Y$, so that $\mathcal{K}^{\perp} = \text{Ran } Y$. For $X \in L(\mathfrak{h})$, we define

$$D_X = A - B X.$$

**Theorem A.7** (1) For any $X \in \mathcal{R}(A, B, C)$, $\mathcal{K}$ is the spectral subspace of $D_X$ for $sp(D_X) \cap i\mathbb{R}$ and $\mathcal{K}^{\perp}$ is the spectral subspace of $D_X^*$ for $sp(D_X^*) \setminus i\mathbb{R}$. Moreover, $D_X|_{\mathcal{K}}$ is independent of $X \in \mathcal{R}(A, B, C)$.
(2) The map $X \mapsto \text{Ker } X$ is a bijection from $\mathfrak{R}(D_-, B, 0)$ onto the set of all $D_-$-invariant subspaces containing the spectral subspace of $D_-$ to the part of its spectrum in $i\mathbb{R}$. Moreover, $X \le X'$ iff $\text{Ker } X' \subset \text{Ker } X$.

(3) If $\mathcal{R}(X) \le 0$ for some self-adjoint $X \in L(\mathfrak{h})$, then $X_- \le X \le X_+$.

(4) If $\mathcal{R}(X) < 0$ for some self-adjoint $X \in L(\mathfrak{h})$, then $\text{sp}(K) \cap i\mathbb{R} = \emptyset$.

The first and last Assertions of Part (1) is Theorem 7.5.3 in [LR]. The second Assertion is dual to the first one. Part (2) is a special case of Theorem 1 and Part (3) is Theorem 14(b) in [Sc]. Part (4) is the first assertion of Theorem 9.1.3 in [LR].

Note that Theorem A.1 implies that for $X \in \mathfrak{R}(A, B, C)$ one has

$$ -KG_X = G_X D_X, $$

so that $\text{sp}(D_X) = \text{sp}(-K|_{G(X)})$. Whenever $\text{sp}(K) \cap i\mathbb{R} = \emptyset$, it follows that $\text{sp}(D_X) \cap i\mathbb{R} = \emptyset$ and hence $K = \{0\}$ and $Y > 0$. By Part (3) of Theorem A.6, we further have $\text{sp}(D_+) \subset \mathbb{C}_-$ so that $G_{X_+}$ is the spectral subspace of $K$ to the part of its spectrum in $\mathbb{C}_-$.

## A.4 Real Riccati equations and real solutions

In this section, we assume that $\mathcal{E}$ is a $d$-dimensional real Hilbert space and $(A, B, C)$ a triple of elements of $L(\mathcal{E})$ such that $(A, B)$ is controllable, $B \ge 0$, and $C$ self-adjoint.

Denote by $\mathfrak{h} = \mathbb{C}\mathcal{E}$ the complexification of $\mathcal{E}$ equipped with its natural Hilbertian structure and conjugation $\mathcal{C}$. The $\mathbb{C}$-linear extensions of $A, B$ and $C$ to $\mathfrak{h}$ (which we denote by the same symbols) are such that $(A, B)$ is controllable, $B \ge 0$, and $\mathcal{C}$ is self-adjoint on $\mathfrak{h}$. Let $\mathfrak{R}(A, B, C)$ be the set of self-adjoint solutions of (A.1), interpreted as a Riccati equation in $L(\mathfrak{h})$, and define

$$ \mathfrak{R}(A, B, C) = \{X \in \mathfrak{R}(A, B, C) | X = \bar{X}\}. $$

Clearly, $\mathfrak{R}(A, B, C)$ is the set of real self-adjoint solutions of (A.1) viewed as a Riccati equation on $L(\mathcal{E})$.

**Theorem A.8** (1) If $\mathfrak{R}(A, B, C) \neq \emptyset$, then its minimal/maximal element is real and hence coincides with the minimal/maximal element of $\mathfrak{R}(A, B, C)$.

(2) Under the same assumption, the gap $Y = X_+ - X_-$ is real and so is $K = \text{Ker } Y$.

(3) For any $X \in \mathfrak{R}(A, B, C)$, $K$ is the spectral subspace of $D_X$ for $\text{sp}(D_X) \cap i\mathbb{R}$ and $K^\perp$ is the spectral subspace of $D_X^*$ for $\text{sp}(D_X^*) \setminus i\mathbb{R}$. Moreover, $D_X|_K$ is independent of $X \in \mathfrak{R}(A, B, C)$.

To prove Part (1), note that $\bar{X} \in \mathfrak{R}(A, B, C)$ whenever $X \in \mathfrak{R}(A, B, C)$. In particular, one has $\bar{X}_+ \in \mathfrak{R}(A, B, C)$ and hence $X_+ - \bar{X}_+ \ge 0$. It follows that

$$ \|X_+ - \bar{X}_+\|_1 = \text{tr}(X_+ - \bar{X}_+) = \text{tr}(X_+ - X_+^*) = 0. $$

The remaining statements are simple consequences of the reality of $X_\pm$.
Index

<table>
  <tr>
    <td>A, 15</td>
    <td>generalized detailed balance, 19</td>
  </tr>
  <tr>
    <td>B, 17</td>
    <td>Jarzynski identity, 5</td>
  </tr>
  <tr>
    <td>E(&omega;), 24</td>
    <td>Jarzynski's inequality, 5</td>
  </tr>
  <tr>
    <td>F, G, 29</td>
    <td>Rényi entropy, 4</td>
  </tr>
  <tr>
    <td>Q, 15</td>
    <td>stochastic entropy, 21</td>
  </tr>
  <tr>
    <td>S<sup>&tau;</sup>, 23</td>
    <td>structural constraints, 16</td>
  </tr>
  <tr>
    <td>S<sub>&nu;</sub><sup>t</sup>, 30</td>
    <td>symmetry function, 4</td>
  </tr>
  <tr>
    <td>S<sub>GS</sub>, 18</td>
    <td>thermodynamic entropy, 14</td>
  </tr>
  <tr>
    <td>Ent, 3, 18</td>
    <td></td>
  </tr>
  <tr>
    <td>Ent<sub>&alpha;</sub>, 4, 18</td>
    <td></td>
  </tr>
  <tr>
    <td>Ep(&mu;, t), 30</td>
    <td></td>
  </tr>
  <tr>
    <td>Ep(&nu;, t), 20</td>
    <td></td>
  </tr>
  <tr>
    <td>&Gamma;, 15</td>
    <td></td>
  </tr>
  <tr>
    <td>&Omega;, 15</td>
    <td></td>
  </tr>
  <tr>
    <td>P<sub>&nu;</sub><sup>&tau;</sup>, 22</td>
    <td></td>
  </tr>
  <tr>
    <td>&Phi;, &Psi;, 29</td>
    <td></td>
  </tr>
  <tr>
    <td>&Pi;<sub>&tau;</sub>, 45</td>
    <td></td>
  </tr>
  <tr>
    <td>&Theta;, 16</td>
    <td></td>
  </tr>
  <tr>
    <td>&alpha;<sub>+</sub>, &alpha;<sub>-</sub>, 31</td>
    <td></td>
  </tr>
  <tr>
    <td>&beta;, 18</td>
    <td></td>
  </tr>
  <tr>
    <td>A<sup>p</sup>, 47</td>
    <td></td>
  </tr>
  <tr>
    <td>D<sub>&tau;</sub>, 45</td>
    <td></td>
  </tr>
  <tr>
    <td>K<sub>&tau;</sub>, 46</td>
    <td></td>
  </tr>
  <tr>
    <td>R<sub>&alpha;</sub>, 30</td>
    <td></td>
  </tr>
  <tr>
    <td>ep, 21</td>
    <td></td>
  </tr>
  <tr>
    <td>H, H+, H-, 44</td>
    <td></td>
  </tr>
  <tr>
    <td>J+, J-, 31</td>
    <td></td>
  </tr>
  <tr>
    <td>J<sub>&infin;</sub>, 31</td>
    <td></td>
  </tr>
  <tr>
    <td>J<sub>c</sub>, 24</td>
    <td></td>
  </tr>
  <tr>
    <td>S<sup>t</sup>, 14, 30</td>
    <td></td>
  </tr>
  <tr>
    <td>S<sup>t</sup><sub>QM</sub>, 30</td>
    <td></td>
  </tr>
  <tr>
    <td>&iota;, 14</td>
    <td></td>
  </tr>
  <tr>
    <td>&kappa;<sub>c</sub>, 24</td>
    <td></td>
  </tr>
  <tr>
    <td>&mu;<sub>&beta;</sub>, 18</td>
    <td></td>
  </tr>
  <tr>
    <td>&omega;, 13</td>
    <td></td>
  </tr>
  <tr>
    <td>&part;H, 45</td>
    <td></td>
  </tr>
  <tr>
    <td>&sigma;<sub>&beta;</sub>, 18</td>
    <td></td>
  </tr>
  <tr>
    <td>&theta;, 13</td>
    <td></td>
  </tr>
  <tr>
    <td>&epsilon;<sub>+</sub>, &epsilon;<sub>-</sub>, 24</td>
    <td></td>
  </tr>
  <tr>
    <td>&vartheta;, 14</td>
    <td></td>
  </tr>
  <tr>
    <td>P&#773;<sub>&nu;</sub><sup>&tau;</sup>, 22</td>
    <td></td>
  </tr>
</table>

de Bruijn relation, 19

Gallavotti–Cohen symmetry, 4
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