Complexity of Gaussian Random Fields with Isotropic Increments

Abstract

We study the energy landscape of a model of a single particle on a random potential, that is, we investigate the topology of level sets of smooth random fields on ${\mathbb {R}}^{N}$ of the form $X_N(x) +\frac{\mu }{2} \Vert x\Vert ^2,$ where $X_{N}$ is a Gaussian process with isotropic increments. We derive asymptotic formulas for the mean number of critical points with critical values in an open set as the dimension N goes to infinity. In a companion paper, we provide the same analysis for the number of critical points with a given index.

Appendices

Covariance Function and Its Derivatives

Let $D_N(r)=D(r/N)$. For $x,y\in {\mathbb {R}}^N$, let $\varphi (x,y) =\frac{1}{2}(D_N(\Vert x\Vert ^2)+D_N(\Vert y\Vert ^2)-D_N(\Vert x-y\Vert ^2))$. Under $X_N(0)=0$, isotropic increments imply that ${\mathbb {E}}X_N(x)=0$; see [Yag87, p.439]. We have

$$\begin{aligned} \textrm{Cov}[H_N(x), H_N(y)]=\textrm{Cov}[X_N(x),X_N(y)]={\mathbb {E}}[X_N(x)X_N(y)]=\varphi (x,y). \end{aligned}$$

Lemma A.1

Assume Assumptions I and II. Then for $x\in {\mathbb {R}}^N$,

$$\begin{aligned} {\mathbb {E}}(\nabla H_N(x))&=\mu x, \ \ {\mathbb {E}}(\nabla ^2 H_N(x)) = \mu I_N,\\ \textrm{Cov}[H_N(x), \partial _i H_N(x)]&= D'\left( \frac{\Vert x\Vert ^2}{N}\right) x_i,\\ \textrm{Cov}[\partial _i H_N(x),\partial _j H_N(x)]&= D'(0)\delta _{ij},\\ \textrm{Cov}[H_N(x),\partial _{ij} H_N(x)]&= 2D''\left( \frac{\Vert x\Vert ^2}{N}\right) \frac{x_ix_j}{N} +\left[ D'\left( \frac{\Vert x\Vert ^2}{N}\right) -D'(0)\right] \delta _{ij}\\ \textrm{Cov}[\partial _k H_N(x), \partial _{ij} H_N(x)]&= 0,\\ \textrm{Cov}[\partial _{lk} H_N(x), \partial _{ij} H_N(x)]&= -2D''(0)[\delta _{jl}\delta _{ik}+\delta _{il}\delta _{kj} +\delta _{kl}\delta _{ij}]/N, \end{aligned}$$

where $\delta _{ij}$ are the Kronecker delta function.

Proof

By [AT07, Theorem 1.4.2], $X_N(x)$ is smooth. We can differentiate inside expectation as in [AT07, (5.5.4)] to find the expectations and

$$\begin{aligned} {\mathbb {E}}[X_N(x)\partial _i X_N(y)]/N&= \partial _{y_i} {\mathbb {E}}(X_N(x) X_N(y))/N\\&=D_N'(\Vert y\Vert ^2)y_i +D_N'(\Vert x-y\Vert ^2)(x_i-y_i),\\ {\mathbb {E}}[\partial _i X_N(x)\partial _j X_N(y)]/N&= \partial _{x_i} [ D_N'(\Vert x-y\Vert ^2)(x_j-y_j)]\\&=2D_N''(\Vert x-y\Vert ^2)(x_i-y_i)(x_j-y_j)+ D_N'(\Vert x-y\Vert ^2)\delta _{ij},\\ {\mathbb {E}}[ X_N(x)\partial _{ij} X_N(y)]/N&= \partial _{y_i} [ D_N'(\Vert y\Vert ^2) y_j +D_N'(\Vert x-y\Vert ^2) (x_j-y_j)]\\&=2D_N''(\Vert y\Vert ^2)y_i y_j+ D_N'(\Vert y\Vert ^2)\delta _{ij} \\&\hspace{5ex} -2D_N''(\Vert x-y\Vert ^2)(x_i-y_i)(x_j-y_j) -D_N'(\Vert x-y\Vert ^2)\delta _{ij},\\ {\mathbb {E}}[\partial _k X_N(x)\partial _{ij} X_N(y)]/N&=-4D_N'''(\Vert x-y\Vert ^2)(x_k-y_k)(x_i-y_i)(x_j-y_j) \\&\hspace{-19ex} -2D_N''(\Vert x-y\Vert ^2)(x_j-y_j)\delta _{ki} -2D_N''(\Vert x-y\Vert ^2)(x_i-y_i)\delta _{kj} -2D_N''(\Vert x-y\Vert ^2)(x_k-y_k)\delta _{ij},\\ {\mathbb {E}}[\partial _{lk} X_N(x)\partial _{ij} X_N(y)]/N&=-8D_N^{(4)}(\Vert x-y\Vert ^2)(x_l-y_l)(x_k-y_k)(x_i-y_i)(x_j-y_j) \\&\hspace{-19ex} -4D_N'''(\Vert x-y\Vert ^2)[(x_i-y_i)(x_j-y_j)\delta _{kl} +(x_k-y_k)(x_j-y_j)\delta _{il}+(x_k-y_k)(x_i-y_i)\delta _{jl}\\&+(x_l-y_l)(x_j-y_j)\delta _{ki} +(x_l-y_l)(x_i-y_i)\delta _{kj} +(x_l-y_l)(x_k-y_k)\delta _{ij}]\\&-2D_N''(\Vert x-y\Vert ^2)[\delta _{jl}\delta _{ik}+\delta _{il}\delta _{kj} +\delta _{kl}\delta _{ij}]. \end{aligned}$$

Substituting $x=y$,

$$\begin{aligned} {\mathbb {E}}[X_N(x)\partial _i X_N(x)]/N&=D'_N(\Vert x\Vert ^2)x_i,\\ {\mathbb {E}}[\partial _i X_N(x)\partial _j X_N(x)]/N&= D'_N(0)\delta _{ij},\\ {\mathbb {E}}[X_N(x)\partial _{ij} X_N(x)]/N&= 2D_N''(\Vert x\Vert ^2)x_ix_j +D_N'(\Vert x\Vert ^2)\delta _{ij}-D'_N(0)\delta _{ij}\\ {\mathbb {E}}[\partial _k X_N(x)\partial _{ij} X_N(x)]/N&= 0,\\ {\mathbb {E}}[\partial _{lk} X_N(x)\partial _{ij} X_N(x)]/N&= -2D_N''(0)[\delta _{jl}\delta _{ik}+\delta _{il}\delta _{kj} +\delta _{kl}\delta _{ij}]. \end{aligned}$$

Then we note that $D'_N(r)=D'(r/N)/N$ and $D_N''(r)=D''(r/N)/N^2$. $\quad \square$

Auxiliary Lemmas

For the integral ${\mathbb {E}}\int _{\mathbb {R}}\exp \big (-\frac{1}{2}(N+1)x^2 -\frac{\sqrt{N(N+1)}\mu x}{\sqrt{-D''(0)}}\big ) L_{N+1}(\textrm{d}x)$, we have the following elementary fact which is used in Sect. 2.

Lemma B.1

Let $\nu _N$ be probability measures on ${\mathbb {R}}$ and $\mu \ne 0$. Suppose

$$\begin{aligned} \lim _{N\rightarrow \infty } \frac{1}{N}\log \int _{\mathbb {R}}e^{-\frac{1}{2}(N+1)x^2 -\frac{(N+1)\mu x}{\sqrt{-D''(0)}} }\nu _{N+1}(\textrm{d}x) >-\infty . \end{aligned}$$

Then we have

$$\begin{aligned} \lim _{N\rightarrow \infty } \frac{1}{N} \Big (\log \int _{\mathbb {R}}e^{-\frac{1}{2}(N+1)x^2 -\frac{\sqrt{N(N+1)}\mu x}{\sqrt{-D''(0)}} } \nu _{N+1}(\textrm{d}x) - \log \int _{\mathbb {R}}e^{-\frac{1}{2}(N+1)x^2 -\frac{(N+1)\mu x}{\sqrt{-D''(0)}} }\nu _{N+1}(\textrm{d}x)\Big )=0. \end{aligned}$$

Proof

Let

$$\begin{aligned} a_N&= \int _{\mathbb {R}}e^{-\frac{1}{2}(N+1)x^2 -\frac{(N+1)\mu x}{\sqrt{-D''(0)}} }\nu _{N+1}(\textrm{d}x),\\ b_N&= \int _{\mathbb {R}}e^{-\frac{1}{2}(N+1)x^2 -\frac{\sqrt{N(N+1)}\mu x}{\sqrt{-D''(0)}} } \nu _{N+1}(\textrm{d}x) ,\\ c_N&=\int _{\mathbb {R}}e^{-\frac{1}{2}Nx^2 -\frac{N\mu x}{\sqrt{-D''(0)}} }\nu _{N+1}(\textrm{d}x). \end{aligned}$$

We claim $\lim _{N\rightarrow \infty } \frac{1}{N} \log \frac{ a_N}{c_{N}}=0$. Indeed, note that

$$\begin{aligned} a_N =\int _{\mathbb {R}}e^{-\frac{1}{2}(N+1)(x+\frac{\mu }{\sqrt{-D''(0)}})^2 +\frac{(N+1)\mu ^2}{-2D''(0)}} \nu _{N+1}(\textrm{d}x)\le e^{\frac{\mu ^2}{-2D''(0)}} c_N. \end{aligned}$$

By Jensen’s inequality,

$$\begin{aligned} \frac{\mu ^2}{2D''(0)}\le \log \frac{ c_{N}}{ a_{N}}\le \log \frac{ a_N^{N/(N+1)}}{ a_{N}}=-\frac{1}{N+1}\log a_N. \end{aligned}$$

Then the claim follows from the assumption that $\lim _{N\rightarrow \infty } \frac{1}{N}\log a_N >-\infty$. From the elementary inequality $a\wedge b \le (a+b)/2 \le a\vee b$, we have $\lim _{N\rightarrow \infty } \frac{1}{N} (\log ( a_{N} + c_{N})-\log a_N )=0$. It remains to prove that

$$\begin{aligned} \lim _{N\rightarrow \infty } \frac{1}{N} (\log ( a_{N} + c_{N}) - \log b_N)=0. \end{aligned}$$

Note that

$$\begin{aligned} b_N&\le \int _{-\infty }^{0} e^{-\frac{1}{2}(N+1)x^2 -\frac{(N+1)\mu x}{\sqrt{-D''(0)}} } \nu _{N+1}(\textrm{d}x) + \int _{0}^{\infty } e^{-\frac{1}{2} Nx^2 -\frac{N\mu x}{\sqrt{-D''(0)}} } \nu _{N+1}(\textrm{d}x) \le a_N + c_N. \end{aligned}$$

Let t be a large constant (independent of N) such that

$$\begin{aligned} \lim _{N\rightarrow \infty } \frac{1}{N}\log \int _{\mathbb {R}}e^{-\frac{1}{2}(N+1)x^2 -\frac{(N+1)\mu x}{\sqrt{-D''(0)}} }\nu _{N+1}(\textrm{d}x) > - \frac{t^2}{8} \end{aligned}$$

and that

$$\begin{aligned} \int _{|x|>t} e^{-\frac{1}{2}(N+1)x^2 -\frac{(N+1)\mu x}{\sqrt{-D''(0)}} } \nu _{N+1}(\textrm{d}x) \le e^{-(N+1)t^2/4}. \end{aligned}$$

It follows that

$$\begin{aligned} \int _{|x|>t} e^{-\frac{1}{2} Nx^2 -\frac{N\mu x}{\sqrt{-D''(0)}} } \nu _{N+1}(\textrm{d}x) \le e^{-Nt^2/4}, \end{aligned}$$

and since $\frac{1}{N} \log \frac{a_N}{c_N}\rightarrow 0$ as $N\rightarrow \infty$,

$$\begin{aligned}&\lim _{N\rightarrow \infty } \frac{1}{N} \log \int _{-t}^t e^{-\frac{1}{2} Nx^2 -\frac{N\mu x}{\sqrt{-D''(0)}} } \nu _{N+1}(\textrm{d}x)\\&= \lim _{N\rightarrow \infty } \frac{1}{N} \log \int _{-\infty }^\infty (1-{\varvec{1}}\{|x|>t\})e^{-\frac{1}{2} Nx^2 -\frac{N\mu x}{\sqrt{-D''(0)}} } \nu _{N+1}(\textrm{d}x)\\&=\lim _{N\rightarrow \infty } \frac{1}{N} \log \int _{-\infty }^\infty e^{-\frac{1}{2} Nx^2 -\frac{N\mu x}{\sqrt{-D''(0)}} } \nu _{N+1}(\textrm{d}x). \end{aligned}$$

Note that

$$\begin{aligned} b_N&\ge e^{-\frac{t^2}{2}}\int _{-t}^{0} e^{-\frac{1}{2} Nx^2 -\frac{N\mu x}{\sqrt{-D''(0)}} } \nu _{N+1}(\textrm{d}x) + e^{-\frac{t^2}{2} -\frac{\mu t}{\sqrt{-D''(0)}}}\int _{0}^{t} e^{-\frac{1}{2} Nx^2 -\frac{N\mu x}{\sqrt{-D''(0)}} } \nu _{N+1}(\textrm{d}x)\\&\ge e^{-\frac{t^2}{2} -\frac{\mu t}{\sqrt{-D''(0)}}} \int _{-t}^t e^{-\frac{1}{2} Nx^2 -\frac{N\mu x}{\sqrt{-D''(0)}} } \nu _{N+1}(\textrm{d}x). \end{aligned}$$

Since

$$\begin{aligned} \lim _{N\rightarrow \infty } \frac{1}{N}\Big ( \log (a_N+c_N)- \log \int _{-t}^t e^{-\frac{1}{2} Nx^2 -\frac{N\mu x}{\sqrt{-D''(0)}} } \nu _{N+1}(\textrm{d}x)\Big )=0, \end{aligned}$$

we have $\lim _{N\rightarrow \infty } \frac{1}{N} (\log ( a_{N} + c_{N}) - \log b_N)=0$. $\quad \square$

The following discussion is about Assumption IV.

Proof of Lemma 3.2

1. Since $y\mapsto D'(y)$ is a strictly decreasing convex function and $D'''(y)>0$ for any $y>0$, $|D''(y)|< \frac{D'(0)-D'(y)}{y}$. By assumption,

$$\begin{aligned} (\alpha \rho ^2)^2 = \frac{4D''(\rho ^2)^2 \rho ^4}{D(\rho ^2)-\frac{\rho ^2 D'(\rho ^2)^2}{D'(0)}} \le -\frac{8 D''(\rho ^2)^2 D''(0)}{ 3[D'(\rho ^2)-D'(0)]^2/\rho ^4}< -\frac{8}{3} D''(0). \end{aligned}$$

It follows that

$$\begin{aligned} (\alpha \rho ^2+\beta )\beta&< \sqrt{-\frac{2}{3} D''(0)} \sqrt{-\frac{8}{3} D''(0)} -\frac{2}{3} D''(0) =-2D''(0),\\ (\alpha \rho ^2+\beta )\alpha \rho ^2&< -\frac{8}{3} D''(0)+ \sqrt{-\frac{2}{3} D''(0)} \sqrt{-\frac{8}{3} D''(0)}=-4D''(0). \end{aligned}$$

2. We verify (3.10). If (3.11) holds, then $y\mapsto \beta (y)^2$ is a decreasing function and (3.10) follows from Lemma 3.1.

3. By item 1, it suffices to check (3.10). Consider the function

$$\begin{aligned} f(y)=-D''(0)[D'(0)D(y)-D'(y)^2y]-\frac{3}{2}D'(0)[D'(y)-D'(0)]^2. \end{aligned}$$

Condition (3.10) is equivalent to $f(y)\ge 0$. Note that $f(0)=0$ and that

$$\begin{aligned} f'(y)= & {} [D'(0)-D'(y)][D'(0)D''(y)-D''(0)D'(y)]+2D''(y)(D''(0)D'(y)y\\{} & {} -D'(0)[D'(y)-D'(0)]). \end{aligned}$$

By convexity, $\frac{D'(y)-D'(0)}{y} \le D''(y)\le 0$. If (3.12) holds, $D''(0)D'(y)y -D'(0)[D'(y)-D'(0)]\le 0$ and

$$\begin{aligned} \frac{D'(y)}{D'(0)}-\frac{D''(y)}{D''(0)}\ge 0. \end{aligned}$$

Then (3.10) follows from here since $D'(0)\ge D'(y)$ and we have $f'(y)\ge 0$.

4. By Cauchy’s mean value theorem, condition (3.12) is equivalent to (3.13).

5. Direct calculation yields

$$\begin{aligned} \frac{\textrm{d}}{\textrm{d}y} \frac{D'(y)}{- D''(y)} =\frac{-D''(y)^2+D'''(y)D'(y)}{D''(y)^2}. \end{aligned}$$

Then (3.14) implies (3.13).

6. By the representation (3.7) of Thorin–Bernstein functions, we have

$$\begin{aligned} D''(x)= -\int _{(0,\infty )} \frac{1}{(x+t)^2} \sigma (\textrm{d}t), \qquad D'''(x)= \int _{(0,\infty )} \frac{2}{(x+t)^3} \sigma (\textrm{d}t). \end{aligned}$$

By the Cauchy–Schwarz inequality, we have

$$\begin{aligned} 2D''(x)^2\le D'(x)D'''(x). \end{aligned}$$

It follows that $\frac{\textrm{d}}{\textrm{d}y} \frac{D'(y)}{- D''(y)} \ge 1$ and (3.14) holds. $\quad \square$

If $A=0$ in the representation (1.2), using the Cauchy–Schwarz inequality, we can see

$$\begin{aligned} \frac{\textrm{d}}{\textrm{d}y} \frac{D'(y)}{- D''(y)} =\frac{-D''(y)^2+D'''(y)D'(y)}{D''(y)^2}\ge 0, \end{aligned}$$

compared with (3.14). It is easy to check that for any ${\varepsilon }>0, 0<\gamma <1$, our major examples $D(r)=\log (1+r/{\varepsilon })$ and $D(r)=(r+{\varepsilon })^\gamma -{\varepsilon }^{\gamma }$ satisfy (3.13). With more work, one can check that these functions satisfy (3.11).

On the other hand, according to [SSV12, p. 332],

$$\begin{aligned} D(x)=\frac{\sqrt{x}\sinh ^2(\sqrt{x})}{\sinh (2\sqrt{x})} \end{aligned}$$

is a complete Bernstein function which is not Thorin–Bernstein. One can check (at least numerically) that it violates (3.13) but still verifies (3.10). We suspect that (3.8) and (3.9) always hold for any structure function D. The following shows that this is the case at least in a neighborhood of 0.

Lemma B.2

Assume $A=0$ in (1.2). We have

$$\begin{aligned} \lim _{y\rightarrow 0+} \frac{\textrm{d}}{\textrm{d}y} [\alpha (y) y+\beta (y)]\beta (y)&<0, \\ \lim _{y\rightarrow 0+} \frac{\textrm{d}}{\textrm{d}y} [\alpha (y) y+\beta (y)]\alpha (y)y&<0. \end{aligned}$$

Consequently, there exists $\delta >0$ such that $-2D''(0)>[\alpha (y) y+\beta (y)]\beta (y)$ and $-4D''(0)>[\alpha (y) y+\beta (y)]\alpha (y)y$ for $y\in (0,\delta )$.

Proof

We only prove the first inequality as the second is similar. Write

$$\begin{aligned} (\alpha y+\beta )\beta = \frac{[2D''(y)+\frac{D'(y)-D'(0)}{y}] \frac{D'(y)-D'(0)}{y}}{\frac{D(y)}{y^2}-\frac{D'(y)^2}{D'(0)y}} =:\frac{T}{B}. \end{aligned}$$

Since $[(\alpha y+\beta )\beta ]'=\frac{T'B-B'T}{B^2}$ and $\lim _{y\rightarrow 0+} B=-\frac{3}{2} D''(0) \ne 0$, it suffices to show that $\lim _{y\rightarrow 0+}T'B-B'T<0$. By calculation, we have $\lim _{y\rightarrow 0+} T= 3D''(0)^2$ and

$$\begin{aligned} T'&= [2D'''(y) +\frac{D''(y)y-D'(y) +D'(0)}{y^2}] \frac{D'(y)-D'(0)}{y}\\&\quad ~ +[2D''(y)+\frac{D'(y)-D'(0)}{y}] \frac{D''(y)y-D'(y) +D'(0)}{y^2},\\ B'&=\frac{D'(0)D'(y)y-2D'(0)D(y)-2D'(y)D''(y)y^2+y D'(y)^2}{D'(0) y^3}. \end{aligned}$$

After some tedious computation, we find $\lim _{y\rightarrow 0+} T'=4D'''(0) D''(0)$ and $\lim _{y\rightarrow 0+} B'=-\frac{5}{6} D'''(0)-\frac{D''(0)^2}{D'(0)}$. Then

$$\begin{aligned} \lim _{y\rightarrow 0+} T'B-B'T = D''(0)^2 \Big [\frac{3D''(0)^2}{D'(0)}-\frac{7}{2} D'''(0)\Big ]. \end{aligned}$$

By the Cauchy–Schwarz inequality,

$$\begin{aligned} D''(0)^2 = \Big (\int _0^\infty t^4 \nu (\textrm{d}t) \Big )^2\le \int _0^\infty t^2 \nu (\textrm{d}t) \int _0^\infty t^6 \nu (\textrm{d}t)=D'(0)D'''(0). \end{aligned}$$

From here the conclusion follows. $\quad \square$