The duality between Poincar´ e type inequalities on Hamming cube and square function inequalities on dyadic lattice: How to use tree to climb on hypercube based on works with Paata Ivanisvili Harmonic Analysis and applications to PDE and GMT; on the occasion of 60th birthday of Steve Hofmann May 29, 2018 Alexander Volberg
1. Hamming cube Consider the Hamming cube {− 1 , 1 } n of an arbitrary dimension n ≥ 1. For any f : {− 1 , 1 } n → R define the discrete gradient � f ( x ) − f ( y ) � 2 � |∇ f | 2 ( x ) = , 2 y ∼ x where the summation is over all neighbor vertices of x in {− 1 , 1 } n . Set � E f = 1 f ( x ) . 2 n x ∈{− 1 , 1 } n Alexander Volberg
2. Isoperimetric inequalities and Monge–Amp` ere with drift What follows is a joint work with Paata Ivanisvili. Theorem For 1 < p ≤ 2 , any n ≥ 1 and any f : {− 1 , 1 } n → R we obtain s ( p ) p ( E | f | p − | E f | p ) ≤ �∇ f � p p , where s ( p ) is the smallest positive zero of the confluent hypergeometric function 1 F 1 ( p / 2(1 − p ) , 1 / 2 , x 2 / 2) . Constant s (2) = 1 , s (1+) = 0. The latter is not good. Our approach is based on a certain duality between the classical square function estimates on Euclidean space and the gradient estimates on the Hamming cube. Alexander Volberg
3. As a corollary, we have the following estimate for the constant of Poincar´ e inequality. Let c ( p ) be the largest constant such that any n ≥ 1 and any f : {− 1 , 1 } n → R c poincare ( p ) p E | f − E f | p ≤ �∇ f � p p . Let ˆ s ( p ) be the best (largest) constant in s ( p ) p ( E | f | p − | E f | p ) ≤ �∇ f � p ˆ p . Then we have immediately from the previous slide s ( p ) ≤ ˆ s ( p ) ≤ c poincare ( p ) . Alexander Volberg
4. There is some kind of converse inequality. Notice that if 1 ≤ p ≤ 2 then there exists K ( p ) ≤ 2 : ∀ x ∈ R , p (1 − x )+ | x | p − 1 ≤ K ( p ) | 1 − x | p . f Put x = | E f | , and apply E . Then E | f | p −| E f | p ≤ p E ( | E f | p − f | E f | p − 1 )+ E | f | p −| E f | p ≤ K ( p ) E | f − E f | p . Then s ( p ) ≤ c ( p ) ≤ K ( p ) 1 / p ˆ s ( p ) ≤ ˆ s ( p ) . Alexander Volberg
5. Why p (1 − x ) + | x | p − 1 ≤ C p | 1 − x | p implies p ≤ 2? Consider x = 1 − ε . Then (1 − ε ) p − 1 + p ε = a p ε 2 + o ( ε 2 ) , a p > 0. This can be ≤ C p ε p for ε → 0 only if p ≤ 2. Alexander Volberg
6. The constant s p is sharp when p → 2 − . On the other hand it degenerates to 0 when p → 1+ which should not be the case for the best possible constant by a result of Talagrand. It will be explained later that s p in a “dual” sense coincides with the sharp constants found by B. Davis in L q norm estimates between stopping times and Brownian motion d q � T 1 / 2 � q ≤ � B T � q , q ≥ 2; (1) � B T � p ≤ d p � T 1 / 2 � p , 0 < p ≤ 2 . (2) Here B t is the Brownian motion starting at zero, and T is any stopping time. It was explained in B. Davis [4] that the same sharp estimates (3) and (4) hold with B T replaced by an integrable function g on [0 , 1] with mean zero, and T 1 / 2 replaced by the dyadic square function of g . Alexander Volberg
6a. Analogy with square function: T 1 / 2 =( | t 1 − t 0 | + · · · + | t n − t n − 1 | ) 1 / 2 =( E | B t 1 − B t 0 | 2 + · · · + E | B t n − B t n − 1 | 2 ) 1 / 2 We notice the big difference between d q � T 1 / 2 � q ≤ � B T � q , q ≥ 2; (3) � B T � p ≤ d p � T 1 / 2 � p , 0 < p ≤ 2 . (4) and slide 2 inequality: s ( p ) p ( E | f | p − | E f | p ) ≤ �∇ f � p p that for the given power p , 1 < p ≤ 2, we need “dual” constant s p = d p − 1 in the theorem. Inequality of slide 2 cannot be extended p to the full range of exponents p unlike (3-4). Alexander Volberg
7. Why slide 2 cannot be extended to the full range of p ? Notice that ( E | f | p − | E f | p ) ≤ C ( p ) �∇ f � p p cannot be extended for the range of exponents p > 2 with some finite constant C ( p ), p > 2. Indeed, assume the contrary. If this were true then Gaussian E would also work. Take f ( x ) = 1 + ax . Using Jensen’s inequality we obtain �� � p / 2 � (1 + a 2 ) p / 2 = | 1 + ax | p d γ ≤ C ( p ) | a | p + 1 . | 1 + ax | 2 d γ ≤ R R (5) Therefore taking a → 0 we obtain the contradiction because pa 2 / 2 ≤ C ( p ) | a | p is false for for p > 2. Alexander Volberg
8. Gaussian isoperimetry of Sudakov–Tsirelson and Borell 2 π e − x 2 / 2 denote the density of γ = γ 1 and let Φ − 1 1 Let ϕ ( x ) = √ denote the inverse of the standard Gaussian distribution function Φ( x ) = γ 1 (( −∞ , x ]). The Gaussian isoperimetric inequality due to V.N. Sudakov, B.S. Tsirelson and C. Borell then asserts that, for any measurable set A ⊂ R n , γ + n ( A ) ≥ I ( γ n ( A )) , I ( t ) = ϕ (Φ − 1 ( t )) , t ∈ [0 , 1] . Here equality holds for an arbitrary halfspace A . Remarkable feature is that the function I is independent of the dimension n . γ n ( A h ) − γ n ( A ) , where A h is the h -neighborhood γ + n ( A ) = lim inf h → 0 h of A in Euclidean metric. Alexander Volberg
9. Bobkov’s functional inequality on Hamming cube Sergei Bobkov proved the following generalization of the previous inequality. Let f : → [0 , 1] then � I 2 ( f ) + |∇ f | 2 , I ( E f ) ≤ E � � � 2 � n f ( x ) − f ( s i ( x )) where |∇ f | ( x ) := , and s i ( x ) are all n i =1 2 neighbors of x . Obviously, for f = 1 A , A ⊂ {− 1 , 1 } n , one gets I ( 1 A )( x ) = 0 ∀ x ∈ {− 1 , 1 } n , and we get � ϕ (Φ − 1 ( | A | )) ≤ E |∇ f | = 1 w A ( x ) =: 1 2 surface measure of A , 2 E where w A ( x ) = | neighbors of x from outside | . Alexander Volberg
10. Using Bobkov In particular, two things can be derived: 1) If | A | = 2 n − 1 then � 2 2 E ( w A ( x )) 1 / 2 ≥ √ = π . 2 π 2) For any function f : {− 1 , 1 } n → R one has Talagrand-Poincar´ e inequality E | f − E f | ≤ C 1 E |∇ f | . Also 1 ≤ q < ∞ Talagrand-Poincar´ e inequalities hold: E | f − E f | q ≤ C q q E |∇ f | q . Seems like nobody knows sharp C q on Hamming cube. Alexander Volberg
11. Combinatorics. What we know about g ( p ) := inf n inf A : | A | =2 n − 1 E ( w A ) p / 2 ? 1) g ( p ) = inf n inf A : | A | =2 n − 1 E ( w A ) p / 2 = 0 if p ∈ [0 , 1). Hamming balls are extremizers. � 2 2) g (1) ≥ π , from Bobkov. Sharp? 3) g (2) = 1, discrete Poincar´ e. Sharp. Half-cube is extremal. 4) g ( p ) is monotonically increasing. 5) g ( p ) ≥ s ( p ) p , from our theorem on slide 2. In fact apply theorem to f ( x ) = 1 , x ∈ A ; f ( x ) = − 1 , x ∈ {− 1 , 1 } n \ A . Can be sharp only near p = 2. � 2 � p 2 } . 6) g ( p ) ≥ max { s ( p ) p , π Alexander Volberg
12. Slide 2 for positive functions on Hamming cube It is also interesting to remark that if one considers only nonnegative functions in Theorem 1 then one obtains probably better constant than s p p . For example, it was obtained in [5] that for any smooth f ≥ 0 we have �� �� � p � � ( H ′ 1 / ( p − 1) ( R 1 / ( p − 1) )) p − 1 R n f p d γ − R n |∇ f | p d γ, R n f d γ ≤ where H q is the Hermite function, and R q is the largest zero of H q . Numerical computations show that the constant 1 / ( p − 1) ( R 1 / ( p − 1) )) p − 1 is larger than s p ( H ′ p . √ If p = 3 / 2 , (2 x | x = 1) 1 / 2 = 2, and, therefore, �� � 3 / 2 � �� � R n f 3 / 2 d γ − 1 R n |∇ f | 3 / 2 d γ if f ≥ 0. R n f d γ ≤ √ 2 We do not know whether the same constants work for positive functions on Hamming cube and E replacing � Gaussian measure · d γ . But for p = 3 / 2 we do know that. Alexander Volberg
13. An anonymous Bellman function In this section we want to define a function U : R 2 → R that α satisfies some special properties. Let α ≥ 2 and let β = α − 1 ≤ 2 be the conjugate exponent of α . Let � � ∞ � α � � α � � 2 , x 2 ( − 2 x 2 ) m − α 2 , 1 α N α ( x ) := 1 F 1 = 2 − 1 · · · 2 − m + 1 2 (2 m )! 2 m =0 = 1 − x 2 α 2 + ... be the confluent hypergeometric function. N α ( x ) satisfies the Hermite differential equation N ′′ α ( x ) − xN ′ α ( x ) + α N α ( x ) = 0 for x ∈ R (6) with initial conditions N α (0) = 1 and N ′ α (0) = 0. Let s α be the smallest positive zero of N α ( z ). Alexander Volberg
14. Burgess Davis function For α ≥ 2 set � − α s α − 1 α ( s α ) N α ( x ) , α 0 ≤ | x | ≤ s α ; N ′ u α ( x ) := s α α − | x | α , s α ≤ | x | . Clearly u α ( x ) is C 1 ( R ) ∩ C 2 ( R \ { s α } ) smooth, even, concave function. Concavity follows from matching derivatives and Lemma For any α ≥ 2 we have 0 < s α ≤ 1 . In addition s α is decreasing in α > 0 , and N ′ α ( t ) , N ′′ α ( t ) ≤ 0 on [0 , s α ] for α > 0 . Alexander Volberg
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