The Brunn-Minkowski Theorem and Influences of Boolean Variables

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1 Lecture 7 The Bru-Mikowski Theorem ad Iflueces of Boolea Variables Friday 5, 005 Lecturer: Nati Liial Notes: Mukud Narasimha Theorem 7.1 Bru-Mikowski). If A, B R satisfy some mild assumptios i particular, covexity suffices), the A + B) A) + B) where A + B = {a + b : a A ad b B}. Proof. First, suppose that A ad B are axis aliged boxes, say A = j=1 I j ad B = i=1 J i, where each I j ad J i is a iterval with I j = x j ad J i = y i. We may assume WLOG that I j = 0, x j ad J i = 0, y i ad hece A + B = i=1 0, x i + y i. For this case, the BM iequality asserts that x i + y i ) 1 x 1 i i=1 i=1 i=1 y 1 i ) 1 xi yi 1 x i + y i x i + y i Now, sice the arithmetic mea of umbers is bouded above by their harmoic mea, we have α i ) 1 P αi ad 1 α i )) 1 P 1 αi ). Takig α i = x i x i +y i ad hece 1 α i = y i x i +y i, we see that the above iequality always holds. Hece the BM iequality holds wheever A ad B are axis aliged boxes. Now, suppose that A ad B are the disjoit uio of axis aliged boxes. Suppose that A = α A A α ad B = β B B β. We proceed by iductio o A + B. We may assume WLOG that A > 1. Sice the boxes are disjoit, there is a hyperplae separatig two boxes i A. We may assume WLOG that this hyperplae is x 1 = 0. ) 1 4

2 A A + A Let A + = {x A : x 1 0} ad A = {x A : x 1 0} as show i the figure above. It is clear that both A + ad A are the disjoit uio of axis aliged boxes. I fact, we may let A + = α A + A α ad A = α A A α where A + < A ad A < A. Suppose that A+ ) A) = α. Pick a λ so that {x B : x 1 λ}) B) is cotiu- We ca always do this by the mea value theorem because the fuctio fλ) = { x B :x 1 λ}) B) ous, ad fλ) 0 as λ ad ad fλ) 1 as λ. = α Let B + = {x B : x 1 λ} ad B = {x B : x 1 λ}. By iductio, we may apply BM to both A +, B + ) ad A, B ), obtaiig A + + B +) 1 A +) 1 + B +) 1 A + B ) 1 A ) 1 + B ) 1 Now, A + ) 1 + B +) 1 = α 1 A) B) A ) 1 + B ) 1 = 1 α) 1 A) B) Hece A + + B +) 1 + A + B ) 1 A) B) The geeral case follows by a limitig argumet without the aalysis for the case where equality holds). Suppose that f : S 1 R is a mappig havig a Lipshitz costat 1. Hece fx) fy) x y Let µ be the media of f, so µ = prob {x S : fx) < µ} = 1 43

3 We assume that the probability distributio always admits such a µ at least approximately). The followig iequality holds for every ɛ > 0 as a simple cosequece of the isoperimetric iequality o the sphere. {x S : f µ > ɛ} < e ɛ/ For A S ad for ɛ > 0, let A ɛ = {x S : dist x, A < ɛ} Questio 7.1. Fid a set A S with A = a for which A ɛ is the smallest. The probability used here is the ormalized) Haar measure. The aswer is always a spherical cap, ad i particular if a = 1, the the best A is the hemisphere ad so A ɛ = {x S : x 1 < ɛ}). We will show that for A S with A = 1, A ɛ 1 e ɛ /4. If A is the hemisphere, the A ɛ = 1 Θe ɛ / ), ad so the hemisphere is the best possible set. But first, a small variatio o BM : ) A + B A) B) This follows from BM because ) 1 ) 1 ) 1 A + B A B + = 1 A) B) A) B) For A S, let à = {λa : a A, 1 λ 0}. The A = µ +1Ã). Let B = S \ A ɛ. Lemma 7.. If x à ad ỹ B, the x + ỹ 1 ɛ 8 Ã+ B It follows that is cotaied i a ball of radius at most 1 ɛ 8. Hece ) +1 ) à 1 ɛ + B 8 ) ) à B ) B ) Therefore, e ɛ/4 B. 44

4 7.1 Boolea Iflueces Let f : {0, 1} {0, 1} be a boolea fuctio. For a set S, the ifluece of S o f, I f S) is defied as follows. Whe we pick {x i } i S uiformly at radom, three thigs ca happe. 1. f = 0 regardless of {x i } i S suppose that this happes with probability q 0 ).. f = 1 regardless of {x i } i S suppose that this happes with probability q 1 ). 3. With probability If f S) := 1 q 0 q 1, f is still udetermied. Some examples: Dictatorship) fx 1, x,..., x ) = x 1. I this case If dictatorship S) = { 1 if i S 0 if i S Majority) For = k + 1, fx 1, x,..., x ) is 1 if ad oly if a majority of the x i are 1. For example, if S = {1}, For fairly small sets S, If majority {1}) = prob x 1 is the tie breaker ) k ) ) k 1 = k = Θ k ) S If majority S) = Θ Parity) fx 1, x,..., x ) = 1 if ad oly if a eve umber of the x i s are 1. I this case for every 1 i. If parity {x i }) = 1 Questio 7.. What is the smallest δ = δ) such that there exists a fuctio f : {0, 1} {0, 1} which is balaced i.e., Ef = 1 ) for which If f {x i }) < δ for all x i? Cosider the followig example, called tribes. The set of iputs {x 1, x,..., x } is partitioed ito tribes of size b each. Here, fx 1, x,..., x ) = 1 if ad oly if there is a tribe that uaimously 1. 45

5 Sice we wat Ef = 1, we must have probf = 0) = 1 1 ) b = 1 b. Therefore, b l 1 1 ) = b We use the Taylor series expasio for l1 ɛ) = ɛ ɛ / = ɛ Oɛ ) to get l. b 1 b + O 1 4 b )) = l. This yields = b b l 1 + O1)). Hece b = log log l + Θ1). Hece, If tribes x) = 1 1 ) /b b = ) 1 1 b b 1 1 b = b b = 1 b 1 = Θ b 1 ) log b I this example, each idividual variable has ifluece Θ log /). It was later show that this is lowest possible ifluece. Propositio 7.3. If Ef = 1, the x If f x) 1. This is a special case of the edge isoperimetric iequality for the cube, ad the iequality is tight if f is dictatorship. ) b 1 x = 1 x = 0 The variable x is ifluetial i the cases idicated by the solid lies, ad hece If f x) = Let S = f 1 0). The If f x) = 1 1 es, S c ). # of mixed edges 1 46

6 Oe ca use ˆf to compute iflueces. For example, if f is mootoe so x y fx) fy)), the Therefore, ˆfS) = T 1) S T ˆf{i}) = 1 ft ) 1 ft ) i T = 1 ft ) ft {i})) = 1 # mixed edges i the directio of i = 1 If f x i ) Hece If f x i ) = ˆf{i}). What ca be doe to express If f x) for a geeral f? Defie f i) z) = fz) fz e i ) x = 1 f i) =0 f i) =0 f i) =1 f i) = 1 The x = 0 f i) =0 f i) =0 f i) = 1 f i) =1 If f x i ) = supportf i) ) = f i) w) w 47

7 The last term will be evaluated usig Parseval. For this, we eed to compute the Fourier expressio of f i) expressed i terms of ˆf). f i) S) = 1 f i) T ) 1) S T T = 1 ft ) ft {i}) 1) S T T = 1 ft ) ft {i} 1) S T S T + ft {i}) ft ) 1) {i}) ) = 1 ft ) ft {i} 1) S T 1) { 0 if i S = ˆfS) if i S S T {i}) ) Usig Parseval o f i) alog with the fact that f i) takes o oly values {0, ±1}, we coclude that If f x i ) = 4 i S hatfs) Next time, we will show that if Ef = 1, the there exists a i such that ) i S ˆfS) > Ωl /). Lemma 7.4. For every f : {0, 1} {0, 1}, there is a mootoe g : {0, 1} {0, 1} such that Eg = Ef. For every s, If g S) If f S). Proof. We use a shiftig argumet. x = 1 x = 0 Clearly E f = Ef. We will show that for all S, If f S) If f S). We may keep repeatig the shiftig step util we obtai a mootoe fuctio g. It is clear that the process will termiate by cosiderig the progress measure fx) x which is strictly icreasig. Therefore, we oly eed show that If f ) S) If f S). 48