Liouville metric of star-scale invariant fields: tails and Weyl scaling

Abstract

We study the Liouville metric associated to an approximation of a log-correlated Gaussian field with short range correlation. We show that below a parameter $\gamma _c >0$, the left–right length of rectangles for the Riemannian metric $e^{\gamma \phi _{0,n}} ds^2$ with various aspect ratio is concentrated with quasi-lognormal tails, that the renormalized metric is tight when $\gamma < \min ( \gamma _c, 0.4)$ and that subsequential limits are consistent with the Weyl scaling.

Appendix

1.1 Tail estimates for the supremum of $\phi _{0,n}$

We derive in the following lemma some tail estimates for the field $\phi _{0,n}$. The tail estimates are obtained by controlling a discretization of $\phi _{0,n}$ (by union bound and Gaussian tail estimates) and its gradient.

Lemma 11

The supremum of the field $\phi _{0,n}$ satisfies the following tails estimates

$$\begin{aligned} {\mathbb {P}}\left( \underset{[0,1]^2}{\sup } \left|\phi _{0,n}\right| \ge \alpha ( n + C\sqrt{n}) \right) \le C 4^n e^{- \frac{\alpha ^2}{\log 4} n} \end{aligned}$$

(39)

as well as

$$\begin{aligned} {\mathbb {P}}\left( \underset{[0,1]^2}{\sup } \left|\phi _{0,n}\right| \ge n \log 4 + C \sqrt{n} + C s \right) \le C e^{- s}. \end{aligned}$$

(40)

Proof

First we bound a discretization of the field $\phi _{0,n}$. Since the variance of $\phi _{0,n}(x)$ is equal to $(n+1) \log 2$, by union bound and classical Gaussian tail estimates we have ${\mathbb {P}} ( \max _{[0,1]^2 \cap 2^{-n} {\mathbb {Z}}^2} \left|\phi _{0,n}(x)\right| \ge x ) \le 4^n e^{- \frac{ x^2 }{(n+1)\log 4}}$ hence by introducing $x_n \,{:=}\, \sqrt{n+1} \sqrt{n}$ we get

$$\begin{aligned} {\mathbb {P}} \left( \underset{x \in [0,1]^2 \cap 2^{-n} {\mathbb {Z}}^2}{\max } \left|\phi _{0,n}(x)\right| \ge \alpha x_n \right) \le 4^n e^{- \frac{ \alpha ^2}{\log 4} n}. \end{aligned}$$

(41)

Now we want to bound $\sup _{[0,1]^2} \left|\phi _{0,n}(x)\right|$ for which we want an equivalent of the bound (41). By Fernique’s theorem, we have a tail estimate for the gradient of $\phi _{0}$ i.e. there exists some $C > 0$ so that for every $x > 0$, ${\mathbb {P}} ( \sup _{[0,1]^2} \left|\nabla \phi _0\right| \ge x ) \le C e^{-x^2/2C}$. Then, by scaling, for any dyadic cube $P \in {\mathscr {P}}_k$, ${\mathbb {P}} ( \sup _{P} \left|\nabla \phi _k\right| \ge 2^k x ) \le C e^{-x^2/2C}$ thus, by union bound ${\mathbb {P}} ( \sup _{[0,1]^2} \left|\nabla \phi _k\right| \ge 2^k x ) \le C 4^k e^{-x^2/2C}$. We can now work out the gradient field $\nabla \phi _{0,n}$: ${\mathbb {P}} ( \sup _{[0,1]^2} \left|\nabla \phi _{0,n}\right| \ge 2^{n+1} x ) \le {\mathbb {P}} (\sum _{k=0}^n \sup _{[0,1]^2}\left|\nabla \phi _{k}\right| \ge \sum _{k=0}^n 2^{k} x ) \le C 4^{n} e^{-x^2 / 2 C}$ hence ${\mathbb {P}} ( 2^{-n} \sup _{[0,1]^2}\left|\nabla \phi _{0,n}\right| \ge x ) \le C 4^n e^{-x^2/2C}$. This inequality can be rewritten by introducing $y_n \,{:=}\, C \sqrt{n}$ as:

$$\begin{aligned} {\mathbb {P}} \left( 2^{-n} \sup _{[0,1]^2} \left|\nabla \phi _{0,n}\right| \ge \alpha y_n \right) \le C 4^n e^{-\frac{\alpha ^2}{\log 4}n}. \end{aligned}$$

(42)

Using the discrete bound (41) and the gradient one (42), since

$$\begin{aligned} \sup _{[0,1]^2}\left|\phi _{0,n}\right| \le \max _{[0,1]^2 \cap 2^{-n} {\mathbb {Z}}^2} \left|\phi _{0,n}\right| + 2^{-n} \sup _{[0,1]^2} \left|\nabla \phi _{0,n}\right|, \end{aligned}$$

we get the result (39) by union bound. Indeed, with $z_n \,{:=}\, x_n + y_n$. ${\mathbb {P}} ( \underset{[0,1]^2}{\sup } \left|\phi _{0,n}\right| \ge \alpha z_n ) \le {\mathbb {P}}( X_n \ge \alpha x_n ) + {\mathbb {P}} ( Y_n \ge \alpha Y_n ) \le C 4^n e^{- \frac{\alpha ^2}{\log 4} n}$. Taking $\alpha = \log 4 \sqrt{1+\frac{s}{n \log 4}} \le \log 4+\frac{s}{n}$ gives the second part (40).

The following lemma is a corollary of the previous one: using the tail estimates we control exponential moments.

Lemma 12

We have the following upper bounds for the exponential moments of the field $\phi _{0,n}$: for $\gamma < 2$ and $n \ge 0$, ${\mathbb {E}} \left( e^{ \gamma \sup _{[0,1]^2} \left|\phi _{0,n}\right|} \right) \le C 4^{ \gamma n (1 + o(1)) }$, where o(1) is of the form $O(n^{-1/2})$.

Proof

Fix $0< \gamma < 2$. We use the bound (39) as follows. By introducing $s_n \,{:=}\, n + C \sqrt{n}$ we have, by using the elementary bound ${\mathbb {E}}(e^{\gamma X}) \le e^{\gamma x} + \int _{x}^{\infty } \gamma e^{\gamma t} {\mathbb {P}}(X \ge t) dt$ and for $\alpha$ to be specified:

$$\begin{aligned} {\mathbb {E}} \left( e^{\gamma \underset{[0,1]^2}{\sup } \left|\phi _{0,n}\right|} \right) \le e^{\gamma \alpha s_n} + \gamma \int _{\alpha s_n}^{\infty } e^{\gamma t} {\mathbb {P}} \left( \underset{[0,1]^2}{\sup } \left|\phi _{0,n}\right| \ge t \right) dt. \end{aligned}$$

Setting $t = s_nu$, $\int _{\alpha s_n}^{\infty } e^{\gamma t} {\mathbb {P}} ( \sup _{[0,1]^2} \left|\phi _{0,n}\right| \ge t ) dt = s_n \int _{\alpha }^{\infty } e^{\gamma s_n u} {\mathbb {P}} ( \sup _{[0,1]^2} \left|\phi _{0,n}\right| \ge s_n u) du$ and by using the bound (39)

$$\begin{aligned} \int _{\alpha }^{\infty } e^{\gamma s_n u} {\mathbb {P}} \left( \sup _{[0,1]^2} \left|\phi _{0,n}\right| \ge s_n u \right) du \le C 4^n \int _{\alpha }^{\infty } e^{\gamma s_n u} e^{-\frac{u^2}{\log 4}n} du. \end{aligned}$$

By introducing $r_n \,{:=}\, n^{-1} s_n$, by a change of variables we obtain:

$$\begin{aligned} \int _{\alpha }^{\infty } e^{\gamma s_n u} e^{-\frac{u^2}{\log 4}n} du \le 4^{ \frac{\gamma ^2 r_n^2}{4} n} \int _{\alpha - \gamma r_n \frac{\log 4 }{2}}^{\infty } e^{- \frac{n}{\log 4} u^2} du. \end{aligned}$$

Taking $\alpha \,{:=}\, r_n \log 4$, the integral in the right-hand side becomes

$$\begin{aligned} \int _{\alpha - \gamma r_n \frac{\log 4 }{2}}^{\infty } e^{- \frac{n}{\log 4} u^2} du = \int _{\left( 1 - \gamma /2 \right) r_n \log 4 }^{\infty } e^{- \frac{n}{\log 4} u^2} du \le \frac{4^{-n \left( 1- \frac{\gamma }{2} \right) ^2 r_n^2 }}{\left( 2-\gamma \right) n r_n}, \end{aligned}$$

by using the inequality $\int _a^{\infty } e^{-b x^2} dx \le (2ab)^{-1} e^{-ba^2}$ valid for $a> 0$ and $b >0$. Gathering the pieces we get ${\mathbb {E}} ( e^{\gamma \sup _{[0,1]^2} \left|\phi _{0,n}\right|} ) \le (1 + C \frac{\gamma }{2-\gamma } )4^{\gamma r_n^2 n}$ hence the result.

We add here a Lemma which is in the same vein as the previous one.

Lemma 13

Suppose that we have the following tail estimate on a sequence of positive random variables $(X_k)_{k \ge 0}$: for $k \ge 0$ and $s > 2$,

$$\begin{aligned} {\mathbb {P}}\left( X_k \ge e^{s} \right) \le 4^k e^{- c \frac{s^2}{\log s}}. \end{aligned}$$

Then, we have the following moment estimate: there exists $C >0$ depending only on c such that for k large,

$$\begin{aligned} {\mathbb {E}}\left( X_k \right) \le e^{C \sqrt{k \log k}}. \end{aligned}$$

Proof

Fix $x_k > 2$ to be specified. Note that

$$\begin{aligned} {\mathbb {E}}(X_k) - e^{x_k}\le & {} \int _{e^{x_k}}^{\infty } {\mathbb {P}}\left( X_k \ge x \right) dx = \int _{x_k}^{\infty } {\mathbb {P}}\left( X_k \ge e^s \right) e^s ds \\\le & {} 4^k \int _{x_k}^{\infty } e^{-c \frac{s^2}{\log s}} e^s ds \le 4^k e^{x_k} \int _{x_k}^{\infty } e^{-c\frac{s^2}{\log x_k}} ds. \end{aligned}$$

By using the bound $\int _{a}^{\infty } e^{-bx^2} dx \le (2ab)^{-1} e^{-b a^2}$, we get ${\mathbb {E}}(X_k) \le e^{x_k} + 4^k e^{x_k} (2 x_k \frac{c}{\log x_k})^{-1} e^{-c \frac{x_k^2}{\log x_k}}$. Taking $x_k$ such that $k \log 4 = c \frac{x_k^2}{\log x_k}$ gives $\log k \sim 2 \log x_k$ and $x_k \sim C \sqrt{k \log k}$.

1.2 Upper bound for F(s)

In this subsection, we derive two lemmas that allow us to bound the term F(s) which appears in the proof of Proposition 10. The first one corresponds to $a_{t_s}$, the second one to $\int _{0}^{\infty } a_t dt$.

Lemma 14

If $a,b,c >0$ and $\alpha \in (0,1/2)$ then the function $f_s(t) \,{:=}\, -at + b t^{1/2 + \alpha } + c s \sqrt{t}$ in increasing on $[0,t_s]$, decreasing on $[t_s,\infty ]$ for some $t_s > 0$ which satisfy $a t_s^{1/2} = \frac{1}{2} cs + O(s^{2\alpha })$. In particular, we have: $\exp (f_s(t_s)) \le e^{ \frac{c^2 s^2}{4 a} + C s^{1+ 2 \alpha } }$.

Proof

First, notice that $f_s'(t) = -a + (\frac{1}{2}+\alpha ) b t^{-1/2+\alpha } + \frac{1}{2} cs t^{-1/2}$. Since $f_s'(t_s) = 0$ we obtain $a = (\frac{1}{2} + \alpha ) b t_s^{-1/2 + \alpha } + \frac{1}{2} c s t_s^{-1/2}$ which we write:

$$\begin{aligned} a t_s^{1/2} = \frac{cs}{2} + \left( \frac{1}{2} + \alpha \right) b t_s^{\alpha }. \end{aligned}$$

(43)

Thus $a t_s^{1/2} \ge cs/2$. In particular, $\lim _{s \rightarrow \infty } t_s = + \infty$. Using (43), we obtain $a t_s^{1/2} \sim _{s \rightarrow \infty } \frac{1}{2}cs$. Using again (43), we have $a t_s^{1/2} = \frac{1}{2} cs + O(s^{2\alpha })$. Using again (43) we conclude by noticing that: $f_s(t_s) = -a t_s + b t_s^{1/2 + \alpha } + c s t_s^{1/2} = a t_s - 2b \alpha t_s^{1/2+\alpha }$.

Lemma 15

Let $\alpha , a,b >0$ with $\alpha < 1/2$. For every $s > 0$ the following inequality holds

$$\begin{aligned} \int _0^{\infty } e^{-t + a t^{1/2+\alpha } + b s \sqrt{t}} dt \le C_{\alpha ,a} (2 + bs) e^{\frac{(bs)^2}{4}} e^{C_{\alpha } (bs)^{1+2\alpha }}, \end{aligned}$$

where $C_{\alpha ,a} < \infty$ just depends on a and $C_{\alpha }$ just depends on $\alpha$.

Proof

By writing $-t + bs \sqrt{t} = \frac{(bs)^2}{4} - (\sqrt{t} - \frac{bs}{2})^2$ and the change of variable $u = \sqrt{t}$,

$$\begin{aligned} \int _0^{\infty } e^{-t + a t^{1/2+\alpha } + b s \sqrt{t}} dt = e^{\frac{(bs)^2}{4}} \int _0^{\infty } e^{-\left( u - \frac{bs}{2}\right) ^2 + a u^{1+2\alpha }} 2 u du . \end{aligned}$$

Now, by the change of variables $v = u - bs/2$, we get

$$\begin{aligned} \int _0^{\infty } e^{-\left( u - \frac{bs}{2}\right) ^2 + a u^{1+2\alpha }} 2 u du = \int _{-\frac{bs}{2}}^{\infty } e^{-v^2 + a \left( v+ \frac{bs}{2}\right) ^{1+2\alpha }} (2v + bs) dv. \end{aligned}$$

Finally, by Jensen’s inequality, $(v + \frac{bs}{2})^{1+2\alpha } \le C_{\alpha } (\left|v\right|^{1+2\alpha } + (bs)^{1+2\alpha })$ thus

$$\begin{aligned}&\int _{-\frac{bs}{2}}^{\infty } e^{-v^2 + a \left( v+ \frac{bs}{2}\right) ^{1+2\alpha }} (2v + bs) dv \\&\quad \le e^{C_{\alpha } a (bs)^{1+2\alpha }} \int _{-\frac{bs}{2}}^{\infty } e^{-v^2 + C_{\alpha } a |v|^{1+2\alpha }} (2v + bs) dv \\&\quad \le e^{C_{\alpha } a (bs)^{1+2\alpha }} (2 + bs) \int _{-\infty }^{\infty } e^{-v^2 + C_{\alpha } a |v|^{1+2\alpha }} \left( 1+|v|\right) dv. \end{aligned}$$

Now, we bound F(s). Recall first that $F(s) \le 2 a_{t_s} + \int _{0}^{\infty } a_{t} dt$ where $a_t = \exp (f_s(t))$, $f_s(t) \,{:=}\, - t(1-\lambda ) \log 2 + C t^{1/2+ \alpha } + \beta s \sqrt{t}$, $\lambda \,{:=}\, (1+a_{\varepsilon }) \gamma$, $\alpha \,{:=}\, \frac{\delta }{2}$ and $\beta \,{:=}\, \frac{\gamma }{2} \sqrt{\log 4}$. By Lemma 14, $a_{t_s} \le e^{\frac{\beta ^2 s^2}{4 (1-\lambda ) \log 2} + C s^{1+2\alpha }} = e^{\frac{\gamma ^2 \log 4 s^2}{16 (1-(1+a_{\varepsilon })\gamma ) \log 2} + C s^{1+\delta }} = e^{\frac{\gamma ^2 s^2}{8(1-(1+a_{\varepsilon })\gamma )} + Cs^{1+ \delta }}$. By the change of variable $u = t (1-\lambda )\log 2$ and Lemma 15, we obtain the integral bound $\int _0^{\infty } a_t dt \le C e^{\frac{\gamma ^2 s^2}{8(1-(1+a_{\varepsilon })\gamma )}} e^{C s^{1+ \delta }}$. Altogether we get $F(s) \le C e^{\frac{\gamma ^2 s^2}{8(1-(1+a_{\varepsilon })\gamma )}} e^{C s^{1+\delta }}$.