## Gaussian Noise Sensitivity and Fourier Tails (2011)

Citations: | 3 - 0 self |

### BibTeX

@MISC{Kindler11gaussiannoise,

author = {Guy Kindler},

title = {Gaussian Noise Sensitivity and Fourier Tails},

year = {2011}

}

### OpenURL

### Abstract

We observe a subadditivity property for the noise sensitivity of subsets of Gaussian space. For subsets of volume 1 2, this leads to an almost trivial proof of Borell’s Isoperimetric Inequality for ρ = cos ( π), ℓ ∈ N. In turn this can be used to obtain the Gaussian Isoperimetric Inequality for

### Citations

967 | Improved Approximation Algorithms for Maximum Cut and Satisfiability Problems Using Semidefinite Programming
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(Show Context)
Citation Context ...1 2 cos(ɛ),1− +δ)-approximating Max-Cut. π + o(1))�δ)-hardness. Taking ɛ = ɛ0 ≈ 3.876 to minimize the ratio yields factor-.8786 hardness. Both results are optimal, by the Goemans–Williamson algorithm =-=[8]-=-. Using our Theorem 2.5 in place of Borell’s Theorem, we can also obtain the (1−δ,1−( 2 π +o(1))�δ) hardness for Max-Cut. For the ratio result, we can take ɛ = π 4 in place of ɛ0 and obtain hardness o... |

237 | On the power of unique 2-prover 1-round games
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- 2002
(Show Context)
Citation Context ...ce parameter”. In the proof we present, the main aim is obtaining an analogue of Borell’s Theorem; i.e., NSf (δ) ≥ Ω(Var[f ]) · � δ. (In fact, the inapproximability applications of Bourgain’s Theorem =-=[11, 14]-=- only need such noise sensitivity bounds, not Fourier tail bounds.) Unfortunately the subadditivity of Gaussian rotation sensitivity — the source of the “square-root gain” for noise sensitivity — does... |

181 | Optimal inapproximability results for Max-Cut and other 2-variable CSPs
- Khot, Kindler, et al.
(Show Context)
Citation Context ... 2 sets by taking ℓ → ∞ in our Theorem 1.2. Taking ɛ = ɛ0 ≈ π in Borell’s Theorem and combining with the Invariance Principle [19] yields the opti3.876 mal .8786-factor UG-hardness result for Max-Cut =-=[12]-=-. Theorem 1.2 gives a simple proof of Borell’s Theorem for ɛ = π 4 , and using this in place of ɛ0 gives .8787-factor hardness. Theorem 1.1 is also the key to a very simple proof of the following “Her... |

129 | The unique games conjecture, integrability gap for cut problems and embeddings of negative type metrics into `1
- Khot, Vishnoi
- 2005
(Show Context)
Citation Context ...ce parameter”. In the proof we present, the main aim is obtaining an analogue of Borell’s Theorem; i.e., NSf (δ) ≥ Ω(Var[f ]) · � δ. (In fact, the inapproximability applications of Bourgain’s Theorem =-=[11, 14]-=- only need such noise sensitivity bounds, not Fourier tail bounds.) Unfortunately the subadditivity of Gaussian rotation sensitivity — the source of the “square-root gain” for noise sensitivity — does... |

95 |
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(Show Context)
Citation Context ...tion some applications of Borell’s Theorem which our proof is sufficient to obtain. The first is the Gaussian Isoperimetric Inequality (GII) for sets of volume 1 2 . Gaussian Isoperimetric Inequality =-=[21, 3]-=-. Fix α ∈ [0,1]. Then for any A ⊆ Rd satisfying vol(A) = α it holds that surf(A) ≥ surf(H), where H is a halfspace of volume α. Here surf(A) denotes Gaussian surface area. In particular, for α = 1 1 2... |

43 |
Krzysztof Oleszkiewicz. Noise stability of functions with low influences: invariance and optimality
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(Show Context)
Citation Context ...metric Inequality (GII); we obtain a simple proof of the GII for volume- 1 2 sets by taking ℓ → ∞ in our Theorem 1.2. Taking ɛ = ɛ0 ≈ π in Borell’s Theorem and combining with the Invariance Principle =-=[19]-=- yields the opti3.876 mal .8786-factor UG-hardness result for Max-Cut [12]. Theorem 1.2 gives a simple proof of Borell’s Theorem for ɛ = π 4 , and using this in place of ɛ0 gives .8787-factor hardness... |

38 | Isoperimetry and Gaussian Analysis
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(Show Context)
Citation Context ...definition (for any measurable A ⊆ R d ): √ π surf(A) = 2 · limsup ɛ→0 + RSf (ɛ) ∈ [0,∞]. (2) ɛ In Appendix A we discuss why the definitions coincide for nice enough sets. We also mention that Ledoux =-=[17]-=- has shown surf(A) ≤ γ +(A) always and hence using surf(A) in the GII is formally stronger. ℓ . 3Accepting surf(A) as the definition of Gaussian surface area, it is immediate that the GII follows fro... |

38 |
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(Show Context)
Citation Context ...tion some applications of Borell’s Theorem which our proof is sufficient to obtain. The first is the Gaussian Isoperimetric Inequality (GII) for sets of volume 1 2 . Gaussian Isoperimetric Inequality =-=[21, 3]-=-. Fix α ∈ [0,1]. Then for any A ⊆ Rd satisfying vol(A) = α it holds that surf(A) ≥ surf(H), where H is a halfspace of volume α. Here surf(A) denotes Gaussian surface area. In particular, for α = 1 1 2... |

34 |
Geometric bounds on the Ornstein-Uhlenbeck velocity process
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- 1985
(Show Context)
Citation Context ...ny ɛ of the form π 2ℓ Theorem 1.2. Let A ⊆ Rd have Gaussian measure 1 2 . Then RSA(ɛ) ≥ ɛ π for any ɛ = , ℓ ∈ N. π 2ℓ Theorem 1.2 is an isoperimetric bound for Gaussian space; it was proved by Borell =-=[5]-=- for all ɛ ∈ [0, π 2 ]. Taking ɛ → 0 in Borell’s Theorem yields the Gaussian Isoperimetric Inequality (GII); we obtain a simple proof of the GII for volume- 1 2 sets by taking ℓ → ∞ in our Theorem 1.2... |

29 |
Extremals of functionals with competing symmetries
- Carlen, Loss
- 1990
(Show Context)
Citation Context ... Ehrhard symmetrization and differential equations; in fact it gives a much stronger statement than the above. For another proof using two-point symmetrization on the sphere and Poincaré’s lemma, see =-=[1, 7]-=-. When α = 1 2 , the minimizing halfspaces in Borell’s Theorem pass through the origin and there is a closed form for their rotation sensitivity. This easy result is known as Sheppard’s Formula [20]: ... |

20 |
Sobolev inequalities, the Poisson semigroup and analysis on the sphere
- Beckner
- 1992
(Show Context)
Citation Context ...r −A = A c ). The rotation sensitivity of a set A is a kind of measure of its boundary size. isoperimetric problem was solved by Borell [5]: The associated Borell’s Theorem. Fix ɛ ∈ [0, π 2 ] and α ∈ =-=[0,1]-=-. Then for any A ⊆ Rd satisfying vol(A) = α it holds that RSA(ɛ) ≥ RSH(ɛ), where H is a halfspace of volume α. Borell’s proof uses Ehrhard symmetrization and differential equations; in fact it gives a... |

18 | Some connections between isoperimetric and Sobolev-type inequalities - Bobkov, Houdré - 1997 |

16 |
On the application of the theory of error to cases of normal distributions and normal correlation
- Sheppard
(Show Context)
Citation Context ...[1, 7]. When α = 1 2 , the minimizing halfspaces in Borell’s Theorem pass through the origin and there is a closed form for their rotation sensitivity. This easy result is known as Sheppard’s Formula =-=[20]-=-: Sheppard’s Formula. Let H ⊆ R d be a halfspace through the origin. Then RSH(ɛ) = ɛ π . 2.2 Subadditivity of rotation sensitivity We now prove that rotation sensitivity is subadditive. Theorem 2.4. L... |

15 | Learning geometric concepts via gaussian surface area
- Klivans, O’Donnell, et al.
- 2008
(Show Context)
Citation Context ... 2 , the Hermite weight of f at degree k, and W >k [f ] = ∑ j>k W j [f ], the tail weight of f beyond degree k. Hermite tail weights are of interest in, e.g., approximation theory and learning theory =-=[15]-=-, since W >k [f ] = min{E[(f − p) 2 ] : p is a polynomial of degree at most k}. In particular, Var[f ] = W >0[f ]. We have the following well-known connection between rotation sensitivity and Hermite ... |

9 | Hypercontraction principle and random multilinear forms. Probability Theory and Related Fields - Krakowiak, Szulga - 1988 |

8 | On polynomial chaos and integrability - Borell - 1984 |

5 |
The Gaussian surface area and noise sensitivity of degree-d polynomial threshold functions
- Kane
- 2010
(Show Context)
Citation Context ...ensional Gaussian space. For functions on Gaussian space we define “rotation sensitivity”, RSf (ɛ), a slightly different parametrization of noise sensitivity. With a very simple proof (reminiscent of =-=[9]-=-) we show that rotation sensitivity is subadditive. In particular: Theorem 1.1. Let f : R d → {−1,1}. Then RSf (ɛ) ≥ 1 ℓ RSf (ℓɛ) for any ɛ ∈ R, ℓ ∈ N. As a direct corollary we obtain that the ɛ-rotat... |

1 | On Elliptic Curves, the ABC Conjecture, and Polynomial Threshold Functions - Kane - 2011 |