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A two-sided estimate for the Gaussian noise stability deficit

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Abstract

The Gaussian noise-stability of a set \(A \subset {\mathbb R}^n\) is defined by

$$ \begin{aligned} {\mathcal {S}}_\rho (A) = {\mathbb P}\left( X \in A ~ \& ~ Y \in A \right) \end{aligned}$$

where \(X,Y\) are standard jointly Gaussian vectors satisfying \({\mathbb E}[X_i Y_j] = \delta _{ij} \rho \). Borell’s inequality states that for all \(0 < \rho < 1\), among all sets \(A \subset {\mathbb R}^n\) with a given Gaussian measure, the quantity \({\mathcal {S}}_\rho (A)\) is maximized when \(A\) is a half-space. We give a novel short proof of this fact, based on stochastic calculus. Moreover, we prove an almost tight, two-sided, dimension-free robustness estimate for this inequality: by introducing a new metric to measure the distance between the set \(A\) and its corresponding half-space \(H\) (namely the distance between the two centroids), we show that the deficit \({\mathcal {S}}_\rho (H) - {\mathcal {S}}_\rho (A)\) can be controlled from both below and above by essentially the same function of the distance, up to logarithmic factors. As a consequence, we also establish the conjectured exponent in the robustness estimate proven by Mossel-Neeman, which uses the total-variation distance as a metric. In the limit \(\rho \rightarrow 1\), we obtain an improved dimension-free robustness bound for the Gaussian isoperimetric inequality. Our estimates are also valid for a generalized version of stability where more than two correlated vectors are considered.

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Acknowledgments

I am grateful to Elchanan Mossel for inspiring me to work on this problem and for several fruitful discussions in which, in particular, he suggested that the method should work for \(q\)-stability and for the isoperimetric problem. I am deeply thankful to Bo’az Klartag for a very useful discussion in which he gave me the idea of using Talagrand’s theorem in the proof of Lemma 24. Finally, I thank Gil Kalai for introducing me to this topic and Yuval Peres, Joe Neeman, Joseph Lehec and James Lee for useful comments on a preliminary version of this note.

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Correspondence to Ronen Eldan.

Appendix

Appendix

In the appendix we fill in a few technical lemmas whose proofs were omitted from the note.

Proof of Lemma 8

Define

$$\begin{aligned} g_{x,t} (y) := \gamma _{y,\sqrt{1-t}} (x) = \frac{1}{(2 \pi (1-t))^{n/2} } \exp \left( - \frac{|x-y|^2}{2 (1-t)} \right) . \end{aligned}$$

A simple calculation gives

$$\begin{aligned} \nabla g_{x,t}(y) = \frac{(x-y)}{(1-t)} g_{x,t}(y), \end{aligned}$$

and therefore

$$\begin{aligned} \Delta g_{x,t} (y) = \left( \frac{ |x-y|^2}{(1-t)^2 } + \frac{n}{(1-t)} \right) g_{x,t} (y). \end{aligned}$$

Moreover,

$$\begin{aligned} \frac{\partial }{\partial t} g_{x,t} (y) = - \left( \frac{ |x-y|^2}{2 (1-t)^2 } + \frac{n}{2 (1-t)} \right) g_{x,t} (y). \end{aligned}$$

We can therefore calculate, using Itô’s formula,

$$\begin{aligned} d F_t(x)&= d g_{x,t} (W_t) = \frac{\partial }{\partial t} g_{x,t} (W_t) + \nabla g_{x,t} W_t \cdot d W_t + \frac{1}{2} \Delta g_{x,t} (W_t)\\&= \nabla g_{x,t} (W_t) \cdot d W_t = (1-t)^{-1} \langle x - W_t, d W_t \rangle F_t(x) \end{aligned}$$

which proves that \(F_t(x)\) is a local martingale and establishes equation (14).

Next, let \(\phi : {\mathbb R}^n \rightarrow {\mathbb R}\) satisfy \(|\phi (x)| < C_1 + C_2 |x|^p\) for some constants \(C_1,C_2,p>0\). Remarking that the integral \(\int _{{\mathbb R}^n} \phi (x) \exp (-\alpha |x-x_0| ^2)\) is absolutely convergent for all \(\alpha > 0\) and \(x_0 \in {\mathbb R}^n\), we deduce that for all \(0<t<1\) and all \(y \in {\mathbb R}^n\), we have

$$\begin{aligned} \nabla \int _{{\mathbb R}^n} \phi (x) g_{x,t} (y) dx = \int _{{\mathbb R}^n} \phi (x) \nabla g_{x,t}(y) dx \end{aligned}$$

and

$$\begin{aligned} \left( \frac{\partial }{\partial t} - \frac{1}{2} \Delta \right) \int _{{\mathbb R}^n} \phi (x) g_{x,t}(y) dx = 0. \end{aligned}$$

Formula (15) follows. Finally, the fact that the process \(t \rightarrow \int _{{\mathbb R}^n} \phi (x) F_t(x) dx\) is a martingale follows immediately from the fact that

$$\begin{aligned} \int _{{\mathbb R}^n} \phi (x) F_t(x) dx = {\mathbb E}[\phi (W_1) | {\mathcal {F}}_t]. \end{aligned}$$

\(\square \)

Proof of Lemma 21

We begin with formula (63). By equation (88) we have

$$\begin{aligned} \frac{q(s)}{s} = \frac{e^{-\Psi (s)^2 / 2}}{\int _{- \Psi (s)}^\infty e^{-x^2/2}dx }. \end{aligned}$$

Denote \(y = - \Psi (s)\). Since, by (85), \(q'(s)\) is a decreasing function, we may assume that \(s < \frac{1}{2}\) and thus \(y > 0\). The inequality \(\left( y + \frac{1}{y+1} \right) ^2 \le y^2 + 3\) suggests that

$$\begin{aligned} \int _{y}^\infty e^{-x^2 / 2} \ge \int _{y}^{y + 1/(y+1)} e^{-(y + 1/(y+1))^2 / 2} dx \ge e^{-3} \frac{1}{y+1} e^{-y^2/2} \end{aligned}$$

so

$$\begin{aligned} \frac{q(s)}{s} = \frac{e^{-y^2 / 2}}{\int _{y}^\infty e^{-x^2/2}dx } \le e^{3} (y + 1) = - e^{3} (\Psi (s) + 1) \end{aligned}$$

for all \(s < 1/2\). But a well known fact about the Gaussian distribution is that for \(s < 1/2\)

$$\begin{aligned} - \Psi (s) \le C \sqrt{|\log s|} \end{aligned}$$

for some a universal constant \(C>0\). Formula (63) follows.

The upper bound of formula (62) now follows immediately from the symmetry of the function \(q(s)\) around \(s=1/2\), and we are left with proving the lower bound. Consider the function

$$\begin{aligned} h(s) = 4 s(1-s) q(1/2) = \frac{4}{\sqrt{2 \pi }} s(1-s). \end{aligned}$$

We know that \(h(s) = q(s)\) for \(s \in \{0,1/2,1\}\). Moreover, \(h(s)\) is tangent to \(q(s)\) at \(s = 1/2\), and lastly, according to formula (86), we see that \(q'(s)\) is a convex function in \(s \in [0,1/2]\). Consequently, the convex function \(g(s) = q'(s) - h'(s)\) intersects the x-axis exactly once in the interval \((0,1/2)\), say at the point \(s_0\) (since it is equal to zero at \(s=1/2\) and since its integral on that interval is equal zero). Now, we have

$$\begin{aligned} q''(1/2) = - \sqrt{2 \pi } > - \frac{8}{\sqrt{2 \pi }} = h''(1/2), \end{aligned}$$

which implies that \(g'(1/2) > 0\). We conclude that \(g(s)(s-s_0) < 0\) for \(0 < s < 1/2\). By the fact that \(q(0) = h(0)\) and \(q(1/2) = h(1/2)\) we know that

$$\begin{aligned} \int _0^{1/2} g(s) ds = 0 \end{aligned}$$

and therefore

$$\begin{aligned} q(s) - h(s) = - \int _{s}^{1/2} g(x) dx \ge 0, ~~ \forall 0 < s < 1/2 \end{aligned}$$

so \(q(s) \ge h(s)\) in \(0 < s < 1/2\). Since both functions are symmetric around \(s=1/2\), we have established that

$$\begin{aligned} q(s) \ge h(s) = \frac{4}{\sqrt{2 \pi }} s(1-s) \end{aligned}$$

and the upper bound is proven. \(\square \)

Proof of Fact 26

The upper bound follows immediately from the fact that, according to formula (86) one has \(q''(s) < q''(1/2) < - 2\) for all \(0 < s < 1\). Let us prove the lower bound. By the symmetry of \(q(s)\) around \(s=1/2\), we may assume without loss of generality that \(h < 1/2\). Define

$$\begin{aligned} f(s) = q(s) - q(h) - q'(h) (s-h) \end{aligned}$$

and

$$\begin{aligned} g(s) = h^{-2} f(0) (s-h)^2. \end{aligned}$$

Note that by definition, the functions \(f(0) = g(0)\), \(f(h) = g(h) = 0\) and \(f'(h) = g'(h) = 0\). Now, according to formula (86), the function \(q'(s)\) is convex in \([0,1/2]\) (here, we use the assumption that \(h < 1/2\)). Therefore, the function \(w(s) = f'(s) - g'(s)\) is also convex in this interval. Now, we know that \(w(h) = 0\) and that \(\int _0^h w(s) ds = 0\), so from the convexity of \(w(s)\) we conclude that there exists \(s_0 \in (0,h)\) such that

$$\begin{aligned} w(h) = 0 \text{ and } w(s) (s-s_0) \le 0, ~~ \forall 0 < s < h \end{aligned}$$
(132)

and therefore

$$\begin{aligned} \int _s^h w(x) dx \le 0, ~~ \forall 0<s<h. \end{aligned}$$

It follows that \(g(s) < f(s)\) for all \(0 < s < h\). Moreover, since \(w(s)\) is convex up to \(s=1/2\), necessarily we have \(w(s) > 0\) for \(h<s<1/2\) and it follows that

$$\begin{aligned} g(s) \le f(s), ~~ \forall 0 \le s \le 1/2. \end{aligned}$$

Next, we show that \(g(s) \le f(s)\) also for \(1/2 < s < 1\), or in other words we will show that

$$\begin{aligned} p(s) \le q(s), ~~ \forall 0 < s < 1 \end{aligned}$$

where

$$\begin{aligned} p(s) = g(s) + q(h) + q'(h) (s-h). \end{aligned}$$

Indeed, the fact that \(w(s)\) is convex up to \(s=1/2\) and by (132), we know that \(w(1/2) > 0\), which means that \(p'(1/2) < q'(1/2) = 0\), and therefore the parabola \(p(s)\) attains a maximum at some point \(b \le 1/2\) which means that \(p(1-s) \le p(s)\) for all \(s < 1/2\). So by the symmetry of \(q(s)\) around \(s=1/2\) we get

$$\begin{aligned} q(1-s) = q(s) \ge p(s) \ge p(1-s) , ~~ \forall 0 < s < 1/2. \end{aligned}$$

We finally have \(f(s) \ge g(s)\) for all \(0 \le s \le 1\). In order to prove the lower bound, it therefore suffices to show that

$$\begin{aligned} - \frac{4}{h^2(1-h)^2} (s-h)^2 \le g(s)&= h^{-2} f(0) (s-h)^2\\&= h^{-2} (s-h)^2 (-q(h) + h q'(h)) \end{aligned}$$

or in other words, using the assumption \(h<1/2\),

$$\begin{aligned} 1 \ge q(h) - h q'(h) \end{aligned}$$

a combination of (85) with the fact that \(q(h) \le q(1/2) < 1\) finishes the proof. \(\square \)

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Eldan, R. A two-sided estimate for the Gaussian noise stability deficit. Invent. math. 201, 561–624 (2015). https://doi.org/10.1007/s00222-014-0556-6

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