Localization crossover for the continuous Anderson Hamiltonian in 1-d

Abstract

We investigate the behavior of the spectrum of the continuous Anderson Hamiltonian $\mathcal{H}_{L}$, with white noise potential, on a segment whose size $L$ is sent to infinity. We zoom around energy levels $E$ either of order 1 (Bulk regime) or of order $1\ll E \ll L$ (Crossover regime). We show that the point process of (appropriately rescaled) eigenvalues and centers of mass converge to a Poisson point process. We also prove exponential localization of the eigenfunctions at an explicit rate. In addition, we show that the eigenfunctions converge to well-identified limits: in the Crossover regime, these limits are universal. Combined with the results of our companion paper (Dumaz and Labbé in Ann. Probab. 51(3):805–839, 2023), this identifies completely the transition between the localized and delocalized phases of the spectrum of $\mathcal{H}_{L}$. The two main technical challenges are the proof of a two-points or Minami estimate, as well as an estimate on the convergence to equilibrium of a hypoelliptic diffusion, the proof of which relies on Malliavin calculus and the theory of hypocoercivity.

Appendix A: Technical results

1.1 A.1 Simple estimate on the Laplace transform

The following elementary lemma provides a quantitative estimate on the difference between the Laplace transform of a r.v. and its Taylor expansion at 0 (here we only push the Taylor expansion to the second order).

Lemma A.1

Let $X$ be a real-valued r.v. Assume that there exist $C_{0}>0$ and $q_{0}>0$ such that $\mathbb{E}[e^{q|X|}] < C_{0}$ for all $q \le q_{0}$. Then, for any $q_{1} < q_{0}$ there exist a constant $C_{1} > 0$, that only depends on $C_{0}$, $q_{0}$ and $q_{1}$, which is such that for all $q\in [-q_{1},q_{1}]$

$$ \big| \mathbb{E}[e^{qX}] - 1 - q\mathbb{E}[X]\big| \le C_{1} q^{2}\;. $$

Proof

The map $q\mapsto \mathbb{E}[e^{qX}]$ is real-analytic on $(-q_{0},q_{0})$ and therefore for any such $q$ we have

$$ \mathbb{E}[e^{qX}] - 1 - q\mathbb{E}[X] = \sum _{k\ge 2} q^{k} \frac{\mathbb{E}[X^{k}]}{k!}\;. $$

By assumption $\sup _{k\ge 2} q_{0}^{k}\frac{|\mathbb{E}[X^{k}]|}{k!} \le C_{0}$ so that for any $|q| < q_{0}$ we have

$$ \big|\mathbb{E}[e^{qX}] - 1 - q\mathbb{E}[X]\big| \le C_{0} \frac{(q/q_{0})^{2}}{1-q/q_{0}}\;, $$

and the bound of the statement follows. □

1.2 A.2 Elementary deviation estimate for general SDEs

Lemma A.2

Let $X$ be the solution of $dX(t) = \alpha (t) dt + \beta (t) dB(t)$ with $X(0) = 0$. Assume that $\alpha$, $\beta$ are adapted processes and that there exist $C_{1},C_{2}>0$ such that almost surely for all $t\ge 0$

$$ |\alpha (t)| \le C_{1}\;,\quad |\beta (t)|^{2} \le C_{2}\;. $$

Then for all $x> 2C_{1} t$ we have

$$ \mathbb{P}(\sup _{s\in [0,t]} |X(s)| > x) \le 2\exp (- \frac{x^{2}}{8 C_{2} t})\;. $$

Proof

Let $M(t) := \int _{0}^{t} \beta (s) dB(s)$. Given the bound on the drift, it suffices to show that for all $y>C_{1} t$

$$ \mathbb{P}(\sup _{s\in [0,t]} |M(s)| > y) \le 2\exp (- \frac{y^{2}}{2 C_{2} t})\;. $$

This is precisely the exponential martingale inequality of [34, p.153-154]. □

1.3 A.3 Moments estimates

Lemma A.3

For every $\alpha \in [0,1)$

$$ \sup _{\lambda \in {{\mathchoice{\textit{\textbf{R}}}{\textit{\textbf{R}}}{\textit{\scriptsize \textbf{R}}}{\textit{\tiny \textbf{R}}}}}} \,\mathbb{E}\Big[\exp \Big( \frac{\alpha}{m_{\lambda }^{(\mathbf{E})}} \zeta _{\lambda }^{( \mathbf{E})}\Big)\Big] \le \frac{1}{1-\alpha}\;. $$

Proof

Since $\zeta _{\lambda }^{(E)}/m_{\lambda }^{(E)}$ is equal in law to $\zeta _{\lambda }/m_{\lambda }$, it suffices to work with the original coordinates. It is shown in [2, Appendix A] that for all $\alpha < 0$ we have

$$ \mathbb{E}[e^{\frac{\alpha}{m_{\lambda }} \zeta _{\lambda }}] = 1 + \alpha + \sum _{n\ge 2} \Big(\frac{\alpha}{m_{\lambda }}\Big)^{n} v(n, \lambda )\;, $$

for some $0 \le v(n,\lambda ) \le m_{\lambda }^{n}$. The function on the r.h.s. is analytic in $\alpha$ on $\{z\in {{\mathchoice{\text{\textbf{C}}}{\text{\textbf{C}}}{\text{\scriptsize \textbf{C}}}{ \text{\tiny \textbf{C}}}}}: |z| < 1\}$ so that a standard analytic continuation argument allows to deduce that the identity actually holds for all $\alpha \in (-\infty ,1)$. The bound of the statement then follows easily. □

Next, we obtain a bound on the exponential moment of the number of eigenvalues of $\mathcal{H}_{L}$ below some energy $E$. Therein, we work in the original coordinates exclusively.

Lemma A.4

Uniform bound of Laplace transform of the number of eigenvalues

There exists $q >0$, such that for every $E\in {{\mathchoice{\textit{\textbf{R}}}{\textit{\textbf{R}}}{\textit{\scriptsize \textbf{R}}}{ \textit{\tiny \textbf{R}}}}}$

$$\begin{aligned} \sup _{{L \geq 1}} \,\mathbb{E}\Big[\exp \Big({ q\Big(\frac{m_{E}}{L}\wedge \frac{1}{2} \Big)} \#\{ \lambda _{i}: \lambda _{i} \le E\}\Big)\Big] < \infty \;. \end{aligned}$$

In particular, for any given $E\in {{\mathchoice{\textit{\textbf{R}}}{\textit{\textbf{R}}}{\textit{\scriptsize \textbf{R}}}{ \textit{\tiny \textbf{R}}}}}$ the collection of r.v. $(\#\{\lambda _{i}: \lambda _{i} \le E\}/L)_{L \geq 1}$ is uniformly integrable.

Proof

Set $X := \#\{\lambda _{i}: \lambda _{i} \le E\}$. We need exponential tails on the probability that $X\ge k$. Let $\theta _{E}$ start from $\theta _{E}(0)=0$ and denote by $\zeta ^{(k)}_{E}, k\ge 1$ the i.i.d. r.v. which are such that $\sum _{i=1}^{k} \zeta ^{({i})}_{E}$ is the hitting time of $k\pi$ by $\theta _{E}$. By the Sturm-Liouville and the strong Markov properties, we have for all $q >0$,

$$\begin{aligned} \mathbb{P}( X \geq k) &= \mathbb{P}\Big(\sum _{i=1}^{k} \zeta _{E}^{(i)} \leq L\Big) =\mathbb{P}\Big(q \sum _{i=1}^{k} (1 - \zeta _{E}^{(i)}/m_{E}) \geq q k - q L/m_{E}\Big) \\ &\leq \mathbb{E}[\exp (q (1 - \zeta _{E}/m_{E}))]^{k} \exp (-q k + q L/m_{E}) \,. \end{aligned}$$

By Lemma A.1 and A.3, there exists a constant $C_{1}$ (independent of $E$) such that for all $q >0$ small enough,

$$\begin{aligned} \mathbb{P}( X \geq k)&\leq (1+ C_{1} q^{2})^{k} \exp (-q k + q L/m_{E}) \,. \end{aligned}$$

Taking $q_{0} >0$ small enough, we deduce that there exists $C>0$ such that for all for all $q \in [0,q_{0}]$ and all $u\ge 0$

$$\begin{aligned} \mathbb{P}( X \geq u) \leq 1 \wedge C\exp (- q \frac{2}{3} u + q \frac{L}{m_{E}})\;. \end{aligned}$$

Note that

$$ \mathbb{E}\Big[\exp \Big(q \Big(\frac{m_{E}}{L} \wedge \frac{1}{2} \Big) X \Big)\Big] = 1 + \int _{u\ge 0} q \Big(\frac{m_{E}}{L} \wedge \frac{1}{2} \Big) e^{q \Big(\frac{m_{E}}{L} \wedge \frac{1}{2} \Big) u} \mathbb{P}(X\ge u) du\;. $$

Using the estimate on the tail of $X$, one can bound separately the contributions to the integral coming from $u\le \frac{3}{2} \frac{L}{m_{E}}$ and $u > \frac{3}{2} \frac{L}{m_{E}}$, and conclude. □

1.4 A.4 Integral formulas

In this paragraph, we gather useful integral formulas about the invariant measure of the phase function.

1.4.1 A.4.1 Invariant measure

The phase function $\{\theta ^{(\mathbf{E})}_{\lambda }\}_{\pi}$ can be mapped to a simple additive SDE via the function $\text{cotan}$, namely $X^{(\mathbf{E})}_{\lambda }:= \text{cotan}\,\{\theta ^{(\mathbf{E})}_{ \lambda }\}_{\pi}$ satisfies the SDE:

$$\begin{aligned} dX^{(\mathbf{E})}_{\lambda }= - (V_{\lambda }^{(\mathbf{E})})'(X^{( \mathbf{E})}_{\lambda })dt + dB^{(\mathbf{E})}(t),\quad \text{with } V_{ \lambda }^{(\mathbf{E})}(x) := \lambda \sqrt{\mathbf{E}} x + \mathbf{E}^{3/2} \frac{x^{3}}{3}\,, \end{aligned}$$

and where $X^{(\mathbf{E})}_{\lambda }$ immediately restarts from $+\infty$ when it blows up to $-\infty$. The diffusion $X_{\lambda }^{(\mathbf{E})}$ admits a unique invariant measure whose density writes $f_{\lambda }^{(\mathbf{E})}(x)/m_{\lambda }^{(\mathbf{E})}$ where:

$$\begin{aligned} &f_{\lambda }^{(\mathbf{E})}(x) := 2 e^{-2V_{\lambda }^{(\mathbf{E})}(x)} \int _{-\infty}^{x} e^{2V_{\lambda }^{(\mathbf{E})}(y)} dy\;,\quad x \in {{\mathchoice{\text{\textbf{R}}}{\text{\textbf{R}}}{\text{\scriptsize \textbf{R}}}{ \text{\tiny \textbf{R}}}}}\;, \end{aligned}$$

(see [2, 7] for more details). Note that $\int _{-\infty}^{\infty} f_{\lambda }^{(\mathbf{E})}(x) dx$ coincides with $m_{\lambda }^{(\mathbf{E})}$, which is defined as the expectation of $\zeta _{\lambda }^{(\mathbf{E})}$.

Therefore, the unique invariant measure $\mu _{\lambda }^{(\mathbf{E})}$ of the Markov process $\{\theta _{\lambda }^{(\mathbf{E})}\}_{\pi}$ is the image of $f_{\lambda }^{(\mathbf{E})}(x) dx/m_{\lambda }^{(\mathbf{E})}$ through the map $x\mapsto \text{arccotan\,}x$. Its density writes

$$ \mu _{\lambda }^{(\mathbf{E})}(\theta ) = \frac{f_{\lambda }^{(\mathbf{E})}(\text{cotan}\,\theta )}{\sin ^{2} \theta \;m_{\lambda }^{(\mathbf{E})}} =g^{(\mathbf{E})}_{\lambda }(\text{cotan}\,\theta ) \;,\quad \theta \in [0,\pi )\;, $$

where $g^{(\mathbf{E})}_{\lambda }(x) = (1+x^{2}) f_{\lambda }^{(\mathbf{E})}(x) /m_{\lambda }^{(\mathbf{E})}$.

Proof of Lemma 3.1

We concentrate on the distorted coordinates: indeed, since the bounds need to be taken uniformly over the unbounded parameter $E>1$, this is the most involved setting. For simplicity, we make the further assumption that $E$ is large: that is, for fixed $h>0$ we will assume that $E>E_{0}(h)$ for some quantity $E_{0}(h)\ge 1$ to be defined below. The complementary case follows by adapting the arguments (and is actually simpler).

Let us abbreviate $V_{\lambda }^{(E)}$ by $V_{\lambda }$ to ease the notations. Pick $E_{0}(h)\ge 1$ such that for some constant $K>1$ and for all $E\ge E_{0}(h)$ and all $\lambda \in \Delta$

$$ \frac{h}{En(E)} \le \frac{E}{2}\;,\quad K^{-1} < n(E)\sqrt{E} \le K\;, \quad K^{-1} < E^{3/2}m_{\lambda }^{(E)} < K\;. $$

(90)

Since $V_{\lambda }'(x) = \lambda \sqrt {E} + E^{3/2} x^{2}$, the first bound implies that for all $E\ge E_{0}(h)$, all $\lambda \in \Delta$ and all $x\in{{\mathchoice{\text{\textbf{R}}}{\text{\textbf{R}}}{\text{\scriptsize \textbf{R}}}{ \text{\tiny \textbf{R}}}}}$, $V'_{\lambda }(x) > 0$.

We claim that there exists $C>0$ such that for all $E\ge E_{0}(h)$, all $\lambda \in \Delta$ and all $\theta \in [0,\pi ]$

$$ \vert \partial _{\theta }\mu _{\lambda }^{(E)}(\theta ) \vert \le \frac{C}{E^{3/2}}\;. $$

Assume that the claim holds. Up to increasing $E_{0}(h)$, we can assume that for all $E\ge E_{0}(h)$, $\frac{1}{\pi } - \frac{C\pi }{E^{3/2}} >0$. Since $\mu _{\lambda }^{(E)}$ integrates to 1 on $[0,\pi ]$, this is enough to deduce that there exist $0 < c' < C'$ such that for all $E\ge E_{0}(h)$, all $\lambda \in \Delta$ and all $\theta \in [0,\pi ]$

$$ c' \le \mu _{\lambda }^{(E)}(\theta ) < C'\;,\qquad |\partial _{ \theta }\mu _{\lambda }^{(E)}(\theta )| < C'\;, $$

and that $\sup _{\lambda \in \Delta } \sup _{\theta \in [0,\pi )} |\mu _{ \lambda }^{(E)}(\theta ) - \frac{1}{\pi }| \to 0$ as $E \to \infty$. We are left with proving the claim.

Observe that $(f_{\lambda }^{(E)})' = 2 - 2 f_{\lambda }^{(E)} V'_{\lambda }$. Consequently $\partial _{\theta }\mu _{\lambda }^{(E)}(\theta ) = q_{\lambda }^{(E)}( \text{cotan}\,\theta )$ where

$$\begin{aligned} q_{\lambda }^{(E)}(x) = \frac{(1+x^{2})^{2}}{m_{\lambda }^{(E)}} \Big(-2 + 2 f_{\lambda }^{(E)}(x) V_{\lambda }'(x) - \frac{2x}{1+x^{2}} f_{\lambda }^{(E)}(x)\Big)\;. \end{aligned}$$

To prove the claim, it suffices to show that for all $E\ge E_{0}(h)$, all $\lambda \in \Delta$ and all $x\in{{\mathchoice{\text{\textbf{R}}}{\text{\textbf{R}}}{\text{\scriptsize \textbf{R}}}{ \text{\tiny \textbf{R}}}}}$

$$ \vert q_{\lambda }^{(E)}(x) \vert \le \frac{C}{E^{3/2}}\;. $$

Applying successive integrations by parts we obtain

$$ \begin{aligned}f_{\lambda }^{(E)}(x) ={}&\frac{1}{V_{\lambda }'(x)} + \frac{V_{\lambda }''(x)}{2(V_{\lambda }'(x))^{3}} \\ &{} + e^{-2 V_{\lambda }(x)} \int _{-\infty }^{x} e^{2 V_{\lambda }(y)}\Big( \frac{3 (V_{\lambda }''(y))^{2} - V_{\lambda }'''(y) V_{\lambda }'(y)}{2 (V_{\lambda }'(y))^{4}} \Big)dy\;. \end{aligned}$$

A computation yields

$$ \Big(-2 + 2 f_{\lambda }^{(E)}(x) V_{\lambda }'(x) - \frac{2x}{1+x^{2}} f_{\lambda }^{(E)}(x)\Big) = I_{1}(x) + I_{2}(x) \;, $$

with

$$ I_{1}(x) = \frac{V_{\lambda }''(x) - \frac{2x}{1+x^{2}}V_{\lambda }'(x) - \frac{x}{1+x^{2}} \frac{V_{\lambda }''(x)}{V_{\lambda }'(x)}}{V'_{\lambda }(x)^{2}} \;, $$

and

$$ I_{2}(x) = \Big(2V'_{\lambda }(x) - \frac{2x}{1+x^{2}} \Big) e^{-2V_{ \lambda }(x)} \int _{-\infty }^{x} e^{2 V_{\lambda }(y)}\Big( \frac{3 (V_{\lambda }''(y))^{2} - V_{\lambda }'''(y) V_{\lambda }'(y)}{2 (V_{\lambda }'(y))^{4}} \Big)dy\;. $$

Recall from (90) that $m_{\lambda }^{(E)}$ is of order $E^{-3/2}$. To prove the claim, it suffices to check that $\vert I_{1}(x)\vert$ and $\vert I_{2}(x)\vert$ are bounded from above by a term of order $\frac{1}{E^{3} (1+x^{2})^{2}}$ uniformly over all $E\ge E_{0}(h)$, all $\lambda \in \Delta$ and all $x\in{{\mathchoice{\text{\textbf{R}}}{\text{\textbf{R}}}{\text{\scriptsize \textbf{R}}}{ \text{\tiny \textbf{R}}}}}$. Regarding the first term, using (90), note that $\vert E^{3/2} - \lambda \sqrt{E} \vert \le K$ so that there exists a constant $C_{1}>0$ such that

$$\begin{aligned} E^{3/2} (1+x^{2})^{2} \vert I_{1}(x)\vert &= \frac{E^{3/2}(1+x^{2})}{\big(\lambda \sqrt {E} + E^{3/2} x^{2}\big)^{2}} \Big\vert 2x(E^{3/2} - \lambda \sqrt{E}) - \frac{2E^{3/2} x^{2}}{\lambda \sqrt{E} + E^{3/2} x^{2}}\Big\vert \\ &\le C_{1} \frac{E^{3/2}(1+x^{2})}{E^{3} \big(1 +x^{2}\big)^{2}} \big( \vert x \vert + 1\big) \end{aligned}$$

so that the desired bound follows. Regarding $I_{2}(x)$, assume that there exists $C>0$ such that for all $E\ge E_{0}(h)$, all $\lambda \in \Delta$ and all $x\in{{\mathchoice{\text{\textbf{R}}}{\text{\textbf{R}}}{\text{\scriptsize \textbf{R}}}{ \text{\tiny \textbf{R}}}}}$

$$ \Big\vert e^{-2V_{\lambda }(x)} \int _{-\infty }^{x} e^{2 V_{ \lambda }(y)}\Big( \frac{3 (V_{\lambda }''(y))^{2} - V_{\lambda }'''(y) V_{\lambda }'(y)}{2 (V_{\lambda }'(y))^{4}} \Big)dy \Big\vert \le C \frac{1}{E^{9/2}(1+x^{2})^{3}}\;. $$

(91)

Since there exists $C'>0$ such that uniformly over the same set of parameters we have $\vert 2V'_{\lambda }(x) - \frac{2x}{1+x^{2}} \vert \le C' E^{3/2} (1+x^{2})$, we deduce the desired bound. Let us now prove (91). First of all there exists $c>0$ such that uniformly over all $E\ge E_{0}(h)$, all $\lambda \in \Delta$ and all $x\in{{\mathchoice{\text{\textbf{R}}}{\text{\textbf{R}}}{\text{\scriptsize \textbf{R}}}{ \text{\tiny \textbf{R}}}}}$

$$ \Big\vert \frac{3 (V_{\lambda }''(x))^{2} - V_{\lambda }'''(x) V_{\lambda }'(x)}{2 (V_{\lambda }'(x))^{4}} \Big\vert = \Big\vert \frac{5 E^{3} x^{2} - E^{2} \lambda }{(\lambda \sqrt{E} + E^{3/2} x^{2})^{4}} \Big\vert \le \frac{c}{E^{3}(1+x^{2})^{3}}\;, $$

and such that the following rough bound holds

$$ \Big\vert \frac{3 (V_{\lambda }''(x))^{2} - V_{\lambda }'''(x) V_{\lambda }'(x)}{4 (V_{\lambda }'(x))^{5}} \Big\vert \le c\;. $$

Consequently there exists $C''>0$ such that

$$\begin{aligned} &\Big\vert e^{-2V_{\lambda }(x)} \int _{x-1}^{x} e^{2 V_{\lambda }(y)} \Big( \frac{3 (V_{\lambda }''(y))^{2} - V_{\lambda }'''(y) V_{\lambda }'(y)}{2 (V_{\lambda }'(y))^{4}} \Big)dy \Big\vert \\ &\le e^{-2V_{\lambda }(x)} \int _{x-1}^{x} \frac{2V'_{\lambda }(y)}{2V'_{\lambda }(y)} e^{2 V_{\lambda }(y)} dy \sup _{y\in [x-1,x]} \frac{c}{E^{3}(1+y^{2})^{3}} \\ &\le e^{-2V_{\lambda }(x)} \Big(e^{2V_{\lambda }(x)} - e^{2V_{ \lambda }(x-1)}\Big) \sup _{y\in [x-1,x]} \frac{1}{2V'_{\lambda }(y)} \sup _{y\in [x-1,x]} \frac{c}{E^{3}(1+y^{2})^{3}} \\ &\le C'' \frac{1}{E^{9/2}(1+x^{2})^{4}}\;. \end{aligned}$$

In addition

$$\begin{aligned} &\Big\vert e^{-2V_{\lambda }(x)} \int _{-\infty }^{x-1} e^{2 V_{ \lambda }(y)}\Big( \frac{3 (V_{\lambda }''(y))^{2} - V_{\lambda }'''(y) V_{\lambda }'(y)}{2 (V_{\lambda }'(y))^{4}} \Big)dy \Big\vert \\ &= \Big\vert e^{-2V_{\lambda }(x)} \int _{-\infty }^{x-1} 2V'_{ \lambda }(y) e^{2 V_{\lambda }(y)}\Big( \frac{3 (V_{\lambda }''(y))^{2} - V_{\lambda }'''(y) V_{\lambda }'(y)}{4 (V_{\lambda }'(y))^{5}} \Big)dy \Big\vert \\ &\le c e^{-2V_{\lambda }(x)} \int _{-\infty }^{x-1} 2V'_{\lambda }(y) e^{2 V_{\lambda }(y)}dy \\ &\le c e^{-2V_{\lambda }(x) + 2V_{\lambda }(x-1)} = c e^{-2\lambda \sqrt{E} - 2E^{3/2}((x-\frac{1}{2})^{2}+\frac{1}{12})}\;, \end{aligned}$$

and this suffices to conclude. □

Let us state an additional regularity result on the invariant measure and the density of states, which is used in Sect. 9.

Lemma A.5

The map $\lambda \mapsto (\mu _{\lambda }(\theta ),\theta \in [0,\pi ])$ is continuous from ${{\mathchoice{\textit{\textbf{R}}}{\textit{\textbf{R}}}{\textit{\scriptsize \textbf{R}}}{ \textit{\tiny \textbf{R}}}}}$ into the set of continuous functions on $[0,\pi ]$. As a consequence, $\lambda \mapsto n(\lambda )$ is continuous.

Proof

The second part of the lemma is a consequence of the first part and of the integral formula for the density of states stated in Corollary 6.2. To prove the first part of the lemma, it suffices to check that $\lambda \mapsto \sup _{\theta }\vert \partial _{\lambda }\mu _{ \lambda }(\theta )\vert$ is locally bounded. Recall that $\mu _{\lambda }(\theta ) = g_{\lambda }(\text{cotan}\,\theta )$ where $g_{\lambda }(x) = (1+x^{2}) f_{\lambda }(x) / m_{\lambda }$ for all $x\in{{\mathchoice{\text{\textbf{R}}}{\text{\textbf{R}}}{\text{\scriptsize \textbf{R}}}{ \text{\tiny \textbf{R}}}}}$. The desired property can then be deduced from direct computations on the integral formula for $f_{\lambda }(x)$ and $m_{\lambda }$. □

1.4.2 A.4.2 Expression of the Lyapunov exponent

Proof of Proposition 7.2

We already saw that $\zeta _{\lambda }^{(\mathbf{E})}$ is integrable. Assume that $\sup _{t\in [0,\zeta _{\lambda }^{(\mathbf{E})}]} \vert \rho _{ \lambda }^{(\mathbf{E})}(t)\vert$ is integrable. Decomposing the trajectory of $\rho _{\lambda }^{(\mathbf{E})}$ into i.i.d. excursions in between the successive hitting times of $\pi{{\mathchoice{\text{\textbf{Z}}}{\text{\textbf{Z}}}{\text{\scriptsize \textbf{Z}}}{ \text{\tiny \textbf{Z}}}}}$ by $\theta _{\lambda }^{(\mathbf{E})}$, one deduces from the law of large numbers that almost surely

$$ \frac{\rho _{\lambda }^{(\mathbf{E})}(t)}{t} \to \frac{\mathbb{E}[\rho _{\lambda }^{(\mathbf{E})}(\zeta _{\lambda }^{(\mathbf{E})})]}{\mathbb{E}[\zeta _{\lambda }^{(\mathbf{E})}]} \;,\quad t\to \infty \;. $$

This deterministic limit is denoted $\nu _{\lambda }^{(\mathbf{E})}$.

Before we check the integrability assumption and compute $\nu _{\lambda }^{(\mathbf{E})}$, let us deal with the adjoint diffusion. Arguments based on the strong Markov property show that the ratio of expectations that we obtained in the previous limit remains unchanged if we start the diffusion from its invariant measure, and the same holds with the adjoint diffusions. Applying the adjunction relations of Sect. 7.1, we deduce that

$$ \mathbb{E}[\rho _{\lambda }^{(\mathbf{E})}(\zeta _{\lambda }^{( \mathbf{E})})] = -\bar {\mathbb{E}}[\bar{\rho}_{\lambda }^{( \mathbf{E})}(\bar{\zeta}_{\lambda }^{(\mathbf{E})})]\;,\quad \mathbb{E}[\zeta _{\lambda }^{(\mathbf{E})}] = \bar {\mathbb{E}}[ \bar{\zeta}_{\lambda }^{(\mathbf{E})}]\;, $$

so that the Lyapunov exponent of the adjoint diffusion is the opposite of $\nu _{\lambda }^{(\mathbf{E})}$.

Let us check the integrability assumption. The drift and diffusions coefficients of the (additive) SDE satisfied by $\rho _{\lambda }^{(\mathbf{E})}$ are bounded. Standard stochastic calculus arguments suffice to conclude and allow to show that

$$\begin{aligned} \mathbb{E}[\rho _{\lambda }^{(\mathbf{E})}(\zeta _{\lambda }^{( \mathbf{E})})] = \mathbb{E}\Big[ \int _{0}^{\zeta _{\lambda }^{( \mathbf{E})}} \Big( -\sqrt {\mathbf{E}}(\lambda - \mathbf{E})\sin 2 \theta _{\lambda }^{(\mathbf{E})} -\frac{1}{2} \sin ^{2} 2\theta _{ \lambda }^{(\mathbf{E})} + \sin ^{2} \theta _{\lambda }^{(\mathbf{E})} \Big) dt \Big]\;. \end{aligned}$$

(92)

We now compute $\nu _{\lambda }^{(\mathbf{E})}$. Recall that $m_{\lambda }^{(\mathbf{E})} = \mathbb{E}[\zeta _{\lambda }^{( \mathbf{E})}]$ admits the explicit integral expression (13). To complete the proof, it suffices to show that

$$ \mathbb{E}[\rho _{\lambda }^{(\mathbf{E})}(\zeta _{\lambda }^{( \mathbf{E})})] = \sqrt{2 \pi} \int _{0}^{+\infty} \sqrt{u} \exp (-2 \lambda u - \frac{u^{3}}{6} ) du\;. $$

By the standard characterization of the invariant probability measure of positive recurrent Markov processes, we have

$$\begin{aligned} \mathbb{E}[\rho _{\lambda }^{(\mathbf{E})}(\zeta _{\lambda }^{( \mathbf{E})})] &= \int _{0}^{\pi }\Big( -\sqrt {\mathbf{E}}(\lambda - \mathbf{E})\sin 2\theta -\frac{1}{2} \sin ^{2} 2\theta + \sin ^{2} \theta \Big)\mu _{\lambda }^{(\mathbf{E})}(\theta ) m_{\lambda }^{( \mathbf{E})}d\theta \;. \end{aligned}$$

Applying the change of variable $x = \text{cotan}\,\theta$, we get

$$ \mathbb{E}[\rho _{\lambda }^{(\mathbf{E})}(\zeta _{\lambda }^{( \mathbf{E})})] = \int \Big( \frac{2x \sqrt {\mathbf{E}}(\mathbf{E}-\lambda )}{1+x^{2}} + \frac{1-x^{2}}{(1+x^{2})^{2}}\Big) f_{\lambda}^{(\mathbf{E})}(x)dx\;. $$

(93)

We compute the r.h.s. of (93). By the Dominated Convergence Theorem, we have

$$ \mathbb{E}[\rho _{\lambda }^{(\mathbf{E})}(\zeta _{\lambda }^{( \mathbf{E})})] = \lim _{A\to \infty} \int _{-A}^{A} \Big(\sqrt {\mathbf{E}}(\mathbf{E}-\lambda )\frac{2x}{1+x^{2}} + \frac{1-x^{2}}{(1+x^{2})^{2}}\Big) f_{\lambda}^{(\mathbf{E})}(x)dx\;. $$

Fix $A>0$. We have (from now on, we abbreviate $f_{\lambda }^{(\mathbf{E})}$, $V_{\lambda }^{(\mathbf{E})}$ in $f_{\lambda }$, $V_{\lambda }$)

$$ \sqrt {\mathbf{E}}(\mathbf{E}-\lambda )\frac{2x}{1+x^{2}} = 2x \mathbf{E}^{3/2} - \frac{2x}{1+x^{2}} V'_{\lambda }(x)\;. $$

Using the identity $f_{\lambda }'(x) = -2V'_{\lambda }(x) f_{\lambda }(x) + 2$, an integration by parts yields

$$\begin{aligned} \int _{-A}^{A} - \frac{2x}{1+x^{2}} V'_{\lambda }(x) f_{\lambda }(x) dx &= \int _{-A}^{A} \frac{x}{1+x^{2}} (f'_{\lambda }(x)-2) dx \\ &= \Big[\frac{x}{1+x^{2}} f_{\lambda }(x)\Big]_{-A}^{A} - \int _{-A}^{A} \frac{1-x^{2}}{(1+x^{2})^{2}} f_{\lambda }(x) dx \\ &\quad - \int _{-A}^{A} \frac{2x}{1+x^{2}}\;. \end{aligned}$$

The last term vanishes by symmetry. We deduce from the previous computation that

$$ \begin{aligned}\int _{-A}^{A} \Big(\sqrt {\mathbf{E}}(\mathbf{E}-\lambda ) \frac{2x}{1+x^{2}} + \frac{1-x^{2}}{(1+x^{2})^{2}}\Big) f_{\lambda}(x)dx ={}& \int _{-A}^{A} 2x \mathbf{E}^{3/2} f_{\lambda }(x) dx \\ &{} + \Big[ \frac{x}{1+x^{2}} f_{\lambda }(x)\Big]_{-A}^{A}\;. \end{aligned}$$

The second term vanishes as $A\to \infty$. Indeed Lemma 3.1 ensures that $f_{\lambda }(x) \le C m_{\lambda }^{(\mathbf{E})} / (1+x^{2})$. Regarding the first term, we have

$$\begin{aligned} \int _{-A}^{A} 2x f_{\lambda }(x) dx &= \int _{-A}^{A} 4x e^{-2V_{ \lambda }(x)} \int _{-\infty}^{x} e^{2V_{\lambda }(y)} dy dx\;. \end{aligned}$$

We then apply the change of variables $(u,v) := (x-y,x+y)$ and get

$$\begin{aligned} \int _{-A}^{A} 2x f_{\lambda }(x) dx &= \int _{u \in (0,\infty )} \int _{v\in {{\mathchoice{\text{\textbf{R}}}{\text{\textbf{R}}}{\text{\scriptsize \textbf{R}}}{\text{\tiny \textbf{R}}}}}} \mathbf{1}_{\{u+v \in [-2A,2A]\}} u e^{-2 \lambda \sqrt {\mathbf{E}}u - \frac{\mathbf{E}^{3/2}}{6}u^{3}} e^{- \frac{\mathbf{E}^{3/2}}{2} u v^{2}} dv du \\ &+ \int _{u \in (0,\infty )} \int _{v\in {{\mathchoice{\text{\textbf{R}}}{ \text{\textbf{R}}}{\text{\scriptsize \textbf{R}}}{\text{\tiny \textbf{R}}}}}} \mathbf{1}_{\{u+v \in [-2A,2A]\}} v e^{-2 \lambda \sqrt {\mathbf{E}}u - \frac{\mathbf{E}^{3/2}}{6}u^{3}} e^{- \frac{\mathbf{E}^{3/2}}{2} u v^{2}} dv du\;. \end{aligned}$$

It is straightforward to check that the second term on the r.h.s. goes to 0 as $A\to \infty$. By the Dominated Convergence Theorem, the first term converges as $A\to \infty$ to

$$ \int _{0}^{+\infty} u e^{-2 \lambda \sqrt {\mathbf{E}}u - \frac{\mathbf{E}^{3/2}}{6} u^{3}} (\int _{-\infty}^{+\infty}e^{- \frac{\mathbf{E}^{3/2}}{2} u v^{2}} dv) du = \frac{\sqrt{2 \pi}}{ \mathbf{E}^{3/2}} \int _{0}^{+\infty} \sqrt{u} e^{-2 \lambda u - \frac{u^{3}}{6}} du \;, $$

as required. □

1.5 A.5 Control of the time spent near $\pi \mathbf{Z}$

In the previous sections, we needed a moment bound on the r.v.

$$ \Big(\int _{0}^{1} F(\theta _{\lambda }^{(\mathbf{E})}(t))dt\Big)^{-1} \quad \text{and}\quad \Big(\int _{0}^{1} F(\bar{\theta}_{\lambda }^{( \mathbf{E})}(t))dt\Big)^{-1}\;, $$

where $F$ is either $F(x) =\sin ^{2}(x)$ or $F(x) = \sin ^{4}(x)$. To prove such a bound, one needs to show that $\theta _{\lambda }^{(\mathbf{E})}$ does not spend too much time near $\pi {{\mathchoice{\text{\textbf{Z}}}{\text{\textbf{Z}}}{\text{\scriptsize \textbf{Z}}}{ \text{\tiny \textbf{Z}}}}}$ (since the function $F$ vanishes there).

In both cases, $F:{{\mathchoice{\text{\textbf{R}}}{\text{\textbf{R}}}{\text{\scriptsize \textbf{R}}}{ \text{\tiny \textbf{R}}}}}\to {{\mathchoice{\text{\textbf{R}}}{\text{\textbf{R}}}{\text{\scriptsize \textbf{R}}}{\text{\tiny \textbf{R}}}}}_{+}$ is a $\pi$-periodic smooth function that satisfies, for some $k\ge 1$ and some $c>0$

$$ F(x) \ge c\, d(x,\pi {{\mathchoice{\text{\textbf{Z}}}{\text{\textbf{Z}}}{\text{\scriptsize \textbf{Z}}}{\text{\tiny \textbf{Z}}}}})^{k}\;,\quad x\in {{\mathchoice{\text{\textbf{R}}}{ \text{\textbf{R}}}{\text{\scriptsize \textbf{R}}}{\text{\tiny \textbf{R}}}}}\;,$$

(94)

where $d$ is the Euclidean distance. Our proof will only rely on this property, and therefore applies to a large class of functions $F$.

Lemma A.6

Control of the time spent near $\pi {{\mathchoice{\text{\textbf{Z}}}{\text{\textbf{Z}}}{\text{\scriptsize \textbf{Z}}}{\text{\tiny \textbf{Z}}}}}$ by the phase function

Original coordinates. For any compact interval $\Delta \subset {{\mathchoice{\textit{\textbf{R}}}{\textit{\textbf{R}}}{\textit{\scriptsize \textbf{R}}}{\textit{\tiny \textbf{R}}}}}$, there exists a constant $q>0$ such that

$$ \sup _{\lambda \in \Delta} \sup _{\theta \in [0,\pi )} \mathbb{E}_{(0, \theta )}\Big[\exp \Big(q \big(\int _{0}^{1} F(\theta _{\lambda }(t)) dt \big)^{-\frac{1}{k+1}} \Big)\Big] < \infty \;. $$

Distorted coordinates. Fix $h>0$ and set $\Delta := [E-h/(n(E) E) , E + h/(n(E)E)]$. There exists a constant $q>0$ such that

$$ \sup _{E>1} \sup _{\lambda \in \Delta} \sup _{\theta \in [0,\pi )} \mathbb{E}_{(0,\theta )}\Big[\exp \Big(q \big(\int _{0}^{1} F(\theta _{ \lambda }^{(E)}(t)) dt\big)^{-\frac{1}{k+1}} \Big)\Big] < \infty \;. $$

The same holds with $\theta _{\lambda }$, $\theta _{\lambda }^{(E)}$ replaced by $\bar{\theta}_{\lambda }$, $\bar{\theta}_{\lambda }^{(E)}$, the adjoint diffusions defined in (55).

Proof

From now on, “all the parameters” will refer to $\lambda$ and $\theta$ when working with the original coordinates, and $E$, $\lambda$, $\theta$ when working with the distorted coordinates.

We only need to prove that there exists a constant $c' >0$ such that for all $\varepsilon >0$ small enough, and for all the parameters

$$\begin{aligned} \mathbb{P}_{(0,\theta )}\Big(\int _{0}^{1} F(\theta ^{(\mathbf{E})}_{ \lambda }(t))dt < \epsilon \Big) \leq \exp \big(- c' \;\varepsilon ^{- \frac{1}{k+1}}\big)\;. \end{aligned}$$

(95)

The idea of the proof is to consider the solution $\gamma _{\lambda }$ of the ODE corresponding to the deterministic part of our diffusion, to estimate the integral of $F(\gamma _{\lambda })$ on some interval $[0,T]$, with $T<1$, and to control the probability that $\gamma _{\lambda }$ and $\theta _{\lambda }^{(\mathbf{E})}$ differ on $[0,T]$.

Let $\gamma _{\lambda }$ be the solution of

$$ d\gamma _{\lambda }(t) = \Big(\mathbf{E}^{3/2} + \sqrt{\mathbf{E}} ( \lambda -\mathbf{E}) \sin ^{2} \gamma _{\lambda }+ \sin ^{3} \gamma _{ \lambda }\cos \gamma _{\lambda }\Big) dt\;, $$

starting from $\gamma _{\lambda }(0)=\theta _{\lambda }^{(\mathbf{E})}(0) = \theta$. We would like to bound from below the time spent by $\gamma _{\lambda }$ near $\pi {{\mathchoice{\text{\textbf{Z}}}{\text{\textbf{Z}}}{\text{\scriptsize \textbf{Z}}}{ \text{\tiny \textbf{Z}}}}}$ on the time interval $[0,T]$.

Let us introduce $a := \sup _{t\in [0,T]} d(\gamma _{\lambda }(t),\pi {{\mathchoice{ \text{\textbf{Z}}}{\text{\textbf{Z}}}{\text{\scriptsize \textbf{Z}}}{\text{\tiny \textbf{Z}}}}})$. We claim that:

• The distance $a$ is not too small i.e. there exists a constant $c_{0}\in (0,1)$ independent of all the parameters and of $T$ such that

  $$ a \ge c_{0} \min (T\mathbf{E}^{3/2},1) \ge c_{0} T\;.$$

  (96)

• The function $\gamma _{\lambda }$ spends some time above $a/2$: There exists a constant $c_{1}>0$ independent of all the parameters and of $T$ such that

  $$ \int _{0}^{T} \mathbf{1}_{\{d(\gamma _{\lambda }(t),\pi {{\mathchoice{ \text{\textbf{Z}}}{\text{\textbf{Z}}}{\text{\scriptsize \textbf{Z}}}{\text{\tiny \textbf{Z}}}}}) > a/2\}} dt \ge c_{1} T\;.$$

  (97)

Given these two claims, we now control the difference between $\theta _{\lambda }^{(\mathbf{E})}$ and $\gamma _{\lambda }$. By Grönwall’s Lemma there exists a constant $C>0$ such that for all $T\in [0,1]$,

$$ \sup _{t\in [0,T]} |\theta _{\lambda }^{(\mathbf{E})}(t) - \gamma _{ \lambda }(t)| \le C \sup _{t\in [0,T]} |M(t)|\;, $$

where $M(t) := - \int _{0}^{t} \sin ^{2} \theta _{\lambda }^{(\mathbf{E})} dB(s)$. Let $\tau := \inf \{t\ge 0: |\theta _{\lambda }^{(\mathbf{E})}(t)-\gamma _{ \lambda }(t)| \ge a/4\}$. We deduce that

$$\begin{aligned} \mathbb{P}(\tau \le T) \leq \mathbb{P}\Big( \sup _{t\in [0,T\wedge \tau ]} |M(t)| \ge \frac{a}{4C}\Big)\;. \end{aligned}$$

Recall that $a = \sup _{t\in [0,T]} d(\gamma _{\lambda }(t),\pi {{\mathchoice{ \text{\textbf{Z}}}{\text{\textbf{Z}}}{\text{\scriptsize \textbf{Z}}}{\text{\tiny \textbf{Z}}}}})$. There exists a constant $C'>0$ such that for all $t \leq T\wedge \tau$, $\partial _{t}\langle M \rangle _{t} \le C' a^{4}$. Consequently Lemma A.2 yields the existence of $C_{0}>0$ such that

$$ \mathbb{P}(\tau \le T) \le 2e^{-\frac{C_{0}}{T}}\;. $$

Combining (94), (96) and (97), we deduce that on the event $\{\tau > T\}$ we have

$$ \int _{0}^{T} F(\theta _{\lambda }^{(\mathbf{E})}(t))dt = \int _{0}^{T} F(\gamma _{\lambda }(t) + \theta _{\lambda }^{(\mathbf{E})}(t)- \gamma _{\lambda }(t)) dt \ge c_{1} c \; T (a/4)^{k} \ge c_{2} T^{k+1} \;, $$

for some constant $c_{2}>0$ independent of all parameters and of $T$. For any $\varepsilon \in (0,c_{2}]$, one can choose $T\in [0,1]$ such that $\varepsilon = c_{2} T^{k+1}$. Then we have

$$ \mathbb{P}\Big(\int _{0}^{T} F(\theta _{\lambda }^{(\mathbf{E})}(t))dt < \varepsilon \Big) \le \mathbb{P}(\tau \le T) < 2\exp \Big(- C_{0} \Big(\frac{c_{2}}{\varepsilon }\Big)^{\frac{1}{k+1}}\Big)\;, $$

as required. It remains to prove the two claims stated above.

For the first point, there exists a constant $c'>0$ such that uniformly over the parameters the derivative of $\gamma _{\lambda }$ is larger than $c'\mathbf{E}^{3/2}$ whenever $\gamma _{\lambda }$ lies in some fixed neighborhood of $\pi {{\mathchoice{\text{\textbf{Z}}}{\text{\textbf{Z}}}{\text{\scriptsize \textbf{Z}}}{ \text{\tiny \textbf{Z}}}}}$, say $[-\delta ,\delta ] + \pi {{\mathchoice{\text{\textbf{Z}}}{\text{\textbf{Z}}}{ \text{\scriptsize \textbf{Z}}}{\text{\tiny \textbf{Z}}}}}$ with $\delta \leq {\pi /2}$. Set $c_{0} := \min (c'/4,\delta )$. We now distinguish two cases. If $d(\theta ,\pi {{\mathchoice{\text{\textbf{Z}}}{\text{\textbf{Z}}}{\text{\scriptsize \textbf{Z}}}{\text{\tiny \textbf{Z}}}}})$ is larger than $\min ( (c'/4)T\mathbf{E}^{3/2} , \delta )$ then

$$ a \ge d(\theta ,\pi {{\mathchoice{\text{\textbf{Z}}}{\text{\textbf{Z}}}{\text{\scriptsize \textbf{Z}}}{\text{\tiny \textbf{Z}}}}}) \ge \min ( (c'/4)T\mathbf{E}^{3/2} , \delta ) \ge c_{0} \min (T\mathbf{E}^{3/2},1)\;. $$

If $d(\theta ,\pi {{\mathchoice{\text{\textbf{Z}}}{\text{\textbf{Z}}}{\text{\scriptsize \textbf{Z}}}{\text{\tiny \textbf{Z}}}}})$ is smaller than $\min ( (c'/4)T\mathbf{E}^{3/2} , \delta )$, then until $\gamma _{\lambda }$ reaches $\delta$ mod $[\pi ]$, its derivative is larger than $c'\mathbf{E}^{3/2}$. Therefore by time $T$, the distance of $\gamma _{\lambda }$ to $\pi {{\mathchoice{\text{\textbf{Z}}}{\text{\textbf{Z}}}{\text{\scriptsize \textbf{Z}}}{ \text{\tiny \textbf{Z}}}}}$ passes above $\min ((c'/2)T\mathbf{E}^{3/2},\delta )$, and this yields $a \ge c_{0} \min (T\mathbf{E}^{3/2},1)$. Finally, since $T\in [0,1]$ and $\mathbf{E}\ge 1$, the bound $c_{0} \min (T\mathbf{E}^{3/2},1) \ge c_{0} T$ is immediate.

For the second point, the absolute value of the derivative of $\gamma _{\lambda }$ is bounded from above by $K \mathbf{E}^{3/2}$ for some $K>0$. Consequently (the distance to $\pi {{\mathchoice{\text{\textbf{Z}}}{\text{\textbf{Z}}}{\text{\scriptsize \textbf{Z}}}{ \text{\tiny \textbf{Z}}}}}$ of) $\gamma _{\lambda }$ takes a time at least $a/(2K \mathbf{E}^{3/2})$ to go from $a/2$ to $a$ (or from $a$ to $a/2$). With the original coordinates, using (96) we thus get $a/(2K) \ge c_{0} T/(2K)$ as required.

Let us now consider the distorted coordinates: for convenience, we work under the further assumption that $E>E_{0}(h)$ where $E_{0}(h)\ge 1$ is a constant that depends only on $h$; the complementary case follows from an easy adaptation. Then the derivative of $\gamma _{\lambda }$ is larger than $E^{3/2}/2$ everywhere. First assume that $TE^{3/2} < 4\pi$. Then by (96) we have $a \ge c_{0} T E^{3/2}/(4\pi )$ and thus we find $a/(2K E^{3/2}) \ge c_{0} T /(8 \pi K)$ as required. We now assume that $TE^{3/2}\ge 4\pi$. The function $\gamma _{\lambda }$ makes a number of rotations over the circle which is at least ${\lfloor TE^{3/2}/(2\pi ) \rfloor } \ge 2$. Necessarily $a=\pi /2$. On each rotation, it spends a time at least $a/(2KE^{3/2})$ in the favorable region. We easily conclude to the second point of our claim. □

1.6 A.6 A bound on the centers of mass

Let $f$, $g$ be two functions with $L^{2}({{\mathchoice{\text{\textbf{R}}}{\text{\textbf{R}}}{\text{\scriptsize \textbf{R}}}{ \text{\tiny \textbf{R}}}}})$-norm equal to one such that $\int |t| f^{2}(t) dt < \infty$ and $\int |t| g^{2}(t) dt < \infty$. Let $V_{f}$, $V_{g}$ be their centers of mass, that is, $V_{f} := \int t f^{2}(t) dt$ and $V_{g}:=\int t g^{2}(t)dt$, and let $w_{f}$, $w_{g}$ be the associated probability measures on ${{\mathchoice{\text{\textbf{R}}}{\text{\textbf{R}}}{\text{\scriptsize \textbf{R}}}{ \text{\tiny \textbf{R}}}}}$ recentered at $V_{f}$, $V_{g}$, that is, $w_{f}(dt) = f^{2}(V_{f} +t)dt$ and $w_{g}(dt) = g^{2}(V_{g}+t)dt$.

Lemma A.7

Assume that there exist some constants $q,c>0$ and some $\delta \in (0,1/10)$ such that

$$ \begin{aligned}&\int |f(t)-g(t)|^{2} dt \le \delta ^{2}\;, \\ &\quad \textit{ and for all }t \in {{\mathchoice{\textit{\textbf{R}}}{\textit{\textbf{R}}}{\textit{\scriptsize \textbf{R}}}{ \textit{\tiny \textbf{R}}}}}\;,\quad |f(t)| \le c\, e^{-q|t-V_{f}|}\;,\quad |g(t)| \le c\, e^{-q|t-V_{g}|}\;. \end{aligned}$$

Then, there exists a function $C := C(q,c)>0$ depending only on $q$ and $c$, and which grows at most polynomially in $c$, such that $|V_{f} - V_{g}| \le C \delta ^{1/2}$ and $d_{\mathcal{M}}(w_{f},w_{g}) \le C\delta ^{1/2}$, where $d_{\mathcal{M}}$ is the Lévy-Prokhorov distance introduced in (87).

Proof

Let $X$, $Y$ be two r.v. with densities $f^{2}$ and $g^{2}$ respectively. Note that $V_{f} = \mathbb{E}[X]$ and $V_{g} = \mathbb{E}[Y]$. By assumption, we have for any $A>0$

$$ \mathbb{P}(|X-\mathbb{E}[X]| > A) < \frac{c^{2}}{q} e^{-{ 2}qA}\;,\quad \mathbb{P}(|Y-\mathbb{E}[Y]| > A) < \frac{c^{2}}{q} e^{-{2}qA}\;. $$

Furthermore

$$\begin{aligned} &\big| \mathbb{P}(|X-\mathbb{E}[Y]| \le A) - \mathbb{P}(|Y-\mathbb{E}[Y]| \le A) \big| \\ &\quad = \Big|\int _{[\mathbb{E}[Y]-A,\mathbb{E}[Y]+A]} (f^{2}(t) - g^{2}(t))dt\Big| \\ &\quad \le \big(\int |f(t) - g(t)|^{2}dt\big)^{1/2} \big(\int |f(t) + g(t)|^{2}dt \big)^{1/2} \\ &\quad \le 2{\delta}\;. \end{aligned}$$

Choose $A>0$ such that (it can be chosen in such a way that it grows logarithmically in $c$)

$$ 2\frac{c^{2}}{q} e^{-{2}qA} + 2{\delta} < 2 \frac{c^{2}}{q} e^{-{2}qA} + 2\frac{1}{10}< 1\;. $$

Assume that $|\mathbb{E}[X]-\mathbb{E}[Y]|>2A$. Then

$$ \mathbb{P}(|X-\mathbb{E}[Y]| \le A) \le \mathbb{P}(|X-\mathbb{E}[X]| > A) < \frac{c^{2}}{q} e^{-{2}qA}\;, $$

while

$$ \mathbb{P}(|Y-\mathbb{E}[Y]| \le A) \ge 1-\frac{c^{2}}{q} e^{-qA}\;, $$

so that

$$ \big| \mathbb{P}(|X-\mathbb{E}[Y]| \le A)-\mathbb{P}(|Y-\mathbb{E}[Y]| \le A)\big| > 1 - 2\frac{c^{2}}{q} e^{-qA} > 2{\delta}\;, $$

thus raising a contradiction. Consequently, $|\mathbb{E}[X]-\mathbb{E}[Y]|\le 2A$.

This being given, take $A' >0$ and define the interval $I := [\min (\mathbb{E}[X],\mathbb{E}[Y]) -A', \max (\mathbb{E}[X], \mathbb{E}[Y]) +A']$, and let $m$ be its midpoint. We have

$$\begin{aligned} \mathbb{E}[X] - m = \int _{I} (t-m) f^{2}(t) dt + \int _{{{ \mathchoice{\text{\textbf{R}}}{\text{\textbf{R}}}{\text{\scriptsize \textbf{R}}}{\text{\tiny \textbf{R}}}}}\backslash I} (t-m) f^{2}(t) dt\;, \\ \mathbb{E}[Y] - m = \int _{I} (t-m) g^{2}(t) dt + \int _{{{ \mathchoice{\text{\textbf{R}}}{\text{\textbf{R}}}{\text{\scriptsize \textbf{R}}}{\text{\tiny \textbf{R}}}}}\backslash I} (t-m) g^{2}(t) dt\;. \end{aligned}$$

Since the length of $I$ is smaller than $2(A+A')$, we deduce that

$$ \begin{aligned}\Big| \int _{I} (t-m) f^{2}(t) dt - \int _{I} (t-m) g^{2}(t) dt\Big| &\le (A+A') \int _{{{\mathchoice{\text{\textbf{R}}}{\text{\textbf{R}}}{\text{\scriptsize \textbf{R}}}{\text{\tiny \textbf{R}}}}}}|f^{2}(t)-g^{2}(t)| dt \\ &\le 2(A+A') \delta \;. \end{aligned}$$

Moreover,

$$\begin{aligned} \Big|\int _{{{\mathchoice{\text{\textbf{R}}}{\text{\textbf{R}}}{\text{\scriptsize \textbf{R}}}{\text{\tiny \textbf{R}}}}}\backslash I} (t-m) f^{2}(t) dt \Big| & \le \int _{{{\mathchoice{\text{\textbf{R}}}{\text{\textbf{R}}}{\text{\scriptsize \textbf{R}}}{\text{\tiny \textbf{R}}}}}\backslash [\mathbb{E}[X]-A',\mathbb{E}[X]+A'])}(|t- \mathbb{E}[X]| + A) c^{2} e^{-2q|t-\mathbb{E}[X]|} \\ &\le \Big(\frac{A'}{q} + \frac{1}{2q^{2}} + \frac{A}{q}\Big)c^{2}e^{-2qA'} \;. \end{aligned}$$

The same bound holds for $\big|\int _{{{\mathchoice{\text{\textbf{R}}}{\text{\textbf{R}}}{\text{\scriptsize \textbf{R}}}{\text{\tiny \textbf{R}}}}}\backslash I} (t-m) g^{2}(t) dt\big|$. This ensures that

$$ \big| \mathbb{E}[X]-\mathbb{E}[Y]\big| \le 2(A+A') \delta + 2\Big( \frac{A'}{q} + \frac{1}{2q^{2}} + \frac{A}{q}\Big)c^{2}e^{-2qA'}\;. $$

Choosing $A'=\delta ^{-1/2}$, we get the desired bound on $|V_{f}-V_{g}|$. Let $w'_{f}$ be the probability measure associated with $f^{2}$ recentered at $V_{g}$, that is, $w'_{f}(dt) = f^{2}(V_{g}+t)dt$. Then, it is easy to check that

$$ d_{\mathcal{M}}(w_{g},w'_{f}) \le \int |f^{2}-g^{2}|(t) dt \le 2 ( \int |f-g|^{2}dt)^{1/2} \le 2 \delta \;. $$

Furthermore for any Borel set $B\subset {{\mathchoice{\text{\textbf{R}}}{\text{\textbf{R}}}{\text{\scriptsize \textbf{R}}}{\text{\tiny \textbf{R}}}}}$, note that for any $\epsilon > |V_{f}-V_{g}|$ we have $V_{f} + B \subset V_{g} + B^{\epsilon}$ and $V_{g} + B \subset V_{f} + B^{\epsilon}$ so that

$$\begin{aligned} w_{f}(B) = \int _{V_{f}+B} f^{2}(t) dt \le \int _{V_{g}+B^{\epsilon}} f^{2}(t) dt = w_{f}'(B^{\epsilon})\;, \\ w_{f}'(B) = \int _{V_{g}+B} f^{2}(t) dt \le \int _{V_{f}+B^{\epsilon}} f^{2}(t) dt = w_{f}(B^{\epsilon})\;, \end{aligned}$$

so that $d_{\mathcal{M}}(w_{f},w'_{f}) \le |V_{f}-V_{g}|$, thus concluding the proof. □