Bourgain’s Entropy Estimates Revisited John T. Workman 7 0 0 2 Abstract n a This serves as a near complete set of notes on Bourgain’s well-known paper Almost J sure convergence and bounded entropy [2]. Thetwoentropyresultsaretreated,asisone 5 of the applications. These notes are designed to be independent of Bourgain’s paper 2 and self-contained. There are, at times, differences between Bourgain’s notation and ] my own. The same goes for organization. However, the proofs herein are essentially A his. C . h t a m [ 1 Preliminaries 2 v Our setting will be a probability space (X, ,µ). We are interested in certain sequences 1 F 2 of operators. In particular, given a sequence of operators on L2(µ) we want to make some 6 uniform estimate on their entropy. What kind of conditions must this sequence satisfy? We 1 1 are led to the following definitions. 6 0 Definition. Given a pseudo-metric space (Y,d) (that is, d need not separate points) and / h a subset S Y, define the δ-entropy number of S to be the minimal number of (closed) t ⊂ a δ-balls in the d pseudo-metric needed to cover S. We denote this by N(S,d,δ). m : Notation. Denote by (X) the set of measurable functions f : X R. By Lp(µ), it will v M → i always be meant the subset of (X) which has finite Lp-norm. That is, Lp(µ) consists of X M real-valued functions only. r a Definition. Let T : (X) (X), j N, be a sequence of linear operators. We say j M → M ∈ (T ) is a Bourgain sequence if the following are satisfied: j 1. T : L1(µ) L1(µ) are bounded, j → 2. T : L2(µ) L2(µ) are isometries, j → 3. T are positive, i.e., if f 0 a.s.[µ] then T (f) 0 a.s.[µ], j j ≥ ≥ 4. T (1) = 1 (here, 1 refers to the constant function 1), j 1 5. T satisfy a mean ergodic condition, i.e., j J 1 T f f(x)µ(dx) in L1(µ)-norm for all f L1(µ). j J → ∈ j=1 ZX X The mean ergodic condition will prove useful as is. The first four assumptions lead to the following properties. Lemma 1. Let T : (X) (X) be a linear operator which satisfies assumptions (1) - M → M (4) above. Then, T : L∞(µ) L∞(µ) with Tf L∞(µ) f L∞(µ). Further, T(f2) = T(f)2 → k k ≤ k k a.s.[µ] for all f L2(µ). ∈ Proof. Byassumption4,T(c) = cforallconstantfunctionsc. Byassumption3, T(g) T(h) ≤ a.s.[µ] whenever g h a.s.[µ]. Let f L∞(µ). Then, f f L∞(µ) a.s.[µ], so that T(f) ≤ ∈ ≤ k k ≤ T( f L∞(µ)) = f L∞(µ) a.s.[µ]. Similarly, T(f) = T( f) T( f L∞(µ)) = f L∞(µ) k k k k − − ≤ k k k k a.s.[µ]. Thus, T(f) f L∞(µ) a.s.[µ], giving the first statement. | | ≤ k k We now approach the second statement. First, suppose A . As 0 χ 1, we A ∈ F ≤ ≤ have 0 T(χ ) 1. By assumption 2, µ(A) = χ 2 = T(χ ) 2 = T(χ )2dµ. ≤ A ≤ k AkL2(µ) k A kL2(µ) X A On the other hand, µ(A) = 1 χ 2 = 1 1 χ 2 = 1 T(1 χ ) 2 = − k AckL2(µ) − k − AkL2(µ) − k −R A kL2(µ) 1 1 T(χ ) 2 = 1 (1 T(χ ))2dµ = 2T(χ ) T(χ )2dµ. Setting these two −k − A kL2(µ) − X − A X A − A expressions of µ(A) equal, we have T(χ )dµ = T(χ )2dµ. As 0 T(χ ) 1 a.s.[µ], R X A R X A ≤ A ≤ it must be that T(χ ) = 0,1 a.s.[µ]. Namely, T takes indicator functions to a.s. indicator A R R functions. Now suppose A,B are disjoint. Then, χ χ = 0. So, 0 = χ χ dµ = χ ,χ = ∈ F A B X A B h A Bi T(χ ),T(χ ) = T(χ )T(χ )dµ. As the integrand is necessarily nonnegative a.s.[µ], h A B i X A B R we have T(χ )T(χ ) = 0 a.s.[µ]. A B R Let s = n c χ be a simple function, where A are pairwise disjoint. Then, s2 = i=1 i Ai j c2χ . Now, T(s) = c T(χ ), which gives T(s)2 = c c T(χ )T(χ ) = i Ai P i Ai i k i k Ai Ak c2T(χ )2 = c2T(χ ) = T(s2) a.s.[µ]. Pi i Ai i i APi P P Let f L2(µ) and ǫ > 0. Denote by T the operator norm of T on L1(µ). Choose a P ∈ P k k simplefunctionssothat s f , s f < ǫ/(4 f )and s2 f2 < ǫ/(2 T ). L2(µ) L2(µ) L1(µ) | | ≤ | | k − k k k k − k k k Then, T(f2) T(f)2 L1(µ) k − k T(f2) T(s2) + T(s2) T(s)2 + T(s)2 T(f)2 L1(µ) L1(µ) L1(µ) ≤ k − k k − k k − k = T(f2 s2) + (T(f) T(s))(T(f)+T(s)) L1(µ) L1(µ) k − k k − k T f2 s2 + T(f +s) T(f s) L1(µ) L2(µ) L2(µ) ≤ k kk − k k k k − k < ǫ/2+2 f f s L2(µ) L2(µ) k k k − k < ǫ. As ǫ is arbitrary, we have T(f2) = T(f)2 a.s.[µ]. Our results will focus on a more general sequence. Let S : (X) (X), n N, be n M → M ∈ a sequence of linear operators where each S : L2(µ) L2(µ) is bounded (not necessarily n → uniformly). We have the following Banach principle-type statements. 2 Theorem 2. Let S : L2(µ) L2(µ) be bounded. Suppose that for some 2 p < , n → ≤ ∞ sup S f is finite a.s.[µ] for all f Lp(µ). Then, there is a finite-valued function θ(ǫ) so n| n | ∈ that µ x X : sup S f(x) > θ(ǫ) < ǫ n ∈ | | n n o for every f 1. Lp(µ) k k ≤ Theorem 3. Let S : L2(µ) L2(µ) be bounded. Suppose S f converges a.s.[µ] for all n n → f L (µ). Then, for each ǫ > 0 and η > 0, there exists ρ(ǫ,η) > 0 such that ∞ ∈ µ x X : sup S f(x) > η < ǫ n ∈ | | n n o for all f L∞(µ) 1 and f L1(µ) ρ(ǫ,η). k k ≤ k k ≤ Theorem 2 is a classical result, and Theorem 3 is due to Bellow and Jones [1]. We postpone the proofs of these two theorems until Section 5. The principal assumption we make on the sequence (S ) is that it commutes with a Bourgain sequence (T ), that is, n j S T = T S for all n,j. More on this later. n j j n 2 Normal Random Variables The proofs of the two entropy results rely heavily on the theory of Gaussian (or normal) random variables and Gaussian processes. It is advantageous at this point to review a few fundamental results. Let (Ω, ,P) be another probability space. B Definition. We say a random variable g : Ω R is normal (or Gaussian) with mean m and → variance σ2 if it has the density function 1 (x m)2 f(x) = exp − , √2πσ2 − 2σ2 (cid:18) (cid:19) i.e., P(g A) = f(x)dx for all Borel sets A. A normal random variable with mean 0 and ∈ A variance 1 is called a standard normal random variable. R Recall, for a random variable g with mean 0, the variance is given by σ2 = E(g2) = g 2 , where E( ) is expectation. Also, if g is a normal random variable with mean 0, k kL2(P) · then it is centered, that is, P(g > 0) = P(g < 0) = 1/2. The following results arewell-known in probability theory, and we present them without proof. Lemma 4. If g is a random variable on Ω, then g(ω) P(dω)= ∞P( g > t)dt and g(ω) 2P(dω) = 2 ∞tP( g > t)dt. | | | | | | | | ZΩ Z0 ZΩ Z0 3 Lemma 5. Let g be a normal random variable with mean 0. Then, the moment generating function is given by eλg(ω)P(dω) = eλ2σ2/2, λ R. ∈ ZΩ Lemma 6. For each 1 p < , there exists a constant C > 0 so that g p Lp(P) ≤ ∞ k k ≤ C g for all normal random variables g with mean 0. p L2(P) k k Lemma 7. Let g ,g ,...,g be independent standard normal random variables. Then, for 1 2 m any constants a , a g is a normal random variable with mean 0 and variance a2. i i i i P P Lemma4isprovenbytwosimpleapplicationsofFubini’sTheorem. TheproofofLemma5 is a standard result and is found in most probability texts. Lemma 7 follows immediately from independence. Only Lemma 6 is a somewhat deep result. In fact, something stronger is true; the Lp and Lq norms of a normal random variable are uniformly equivalent for any 1 p,q < . We need only the case q = 2. A proof of the general result can be found in [7] ≤ ∞ (Corollary 3.2). We now state an important estimate for normal random variables [3]. The proof is postponed until Section 6. Theorem 8. Let G ,...,G be normal random variables each with mean 0. If for some 1 N constant s we have P ω Ω : sup G (ω) s 1/2, then sup G 6s. { ∈ n| n | ≤ } ≥ k n| n|kL1(P) ≤ We now turn our attention to Gaussian processes. Definition. Let T be a countable indexing set. We say a collection of random variables (G : t T) is a Gaussian process if each G has mean 0 and all finite linear combinations t t ∈ a G are normal random variables. t t t P Note that this definition is not entirely standard, in particular the requirement that each G have mean 0. It is added here because throughout all Gaussian processes we deal with t have this property, and it makes life simpler later on. If each G is itself a finite linear combination of mean 0 normal random variables, then t (G : t T) is trivially a Gaussian process. Define a pseudo-metric on T by d (s,t) = t G ∈ G G . Denote the entropy number of T by N(T,d ,δ). The following fundamental s t L2(P) G k − k result is Sudakov’s inequality. Theorem 9. There exists a universal constant R such that if (G : t T) is a Gaussian t ∈ process, then sup δ logN(T,d ,δ) R sup G . G t ≤ | | δ>0 (cid:13)t T (cid:13)L1(P) ∈ p (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) For the remainder of the paper, the notation R is fixed on this constant. A proof of Sudakov’s inequality can be found in [7] (Theorem 3.18). 4 3 The First Entropy Result Recall, we consider a sequence S : L2(µ) L2(µ) of bounded operators. For f L2(µ), n → ∈ define a pseudo-metric on N by df(n,n′) = Snf Sn′f L2(µ). For δ > 0, let Nf(δ) := k − k N(N,d ,δ), that is, the δ-entropy number of the set S f : n N in L2(µ). We now state f n { ∈ } and prove the first of Bourgain’s entropy results. Proposition 1. Let S : L2(µ) L2(µ) be bounded (not necessarily uniformly), and as- n → sume (S ) commutes with a Bourgain sequence (T ). Suppose that for some 1 p < , n j ≤ ∞ sup S f < a.s.[µ] for all f Lp(µ). Then, there exists a constant C > 0 such that n| n | ∞ ∈ δ(logN (δ))1/2 C f for all δ > 0 and f L2(µ). f L2(µ) ≤ k k ∈ Proof. As (X,µ) is a probability space, Lp(µ) Lq(µ) when p < q. So, if p < 2, then ⊃ sup S f < a.s.[µ] for all f Lp(µ) Lq(µ) for any q 2. Therefore, assume without n| n | ∞ ∈ ⊃ ≥ loss of generality that p 2. ≥ For M N, let M = 1,2,...,M . Note, sup N(M,d ,δ) = N (δ). Therefore, it ∈ { } M f f suffices to find C, independent of M, such that (logN(M,d ,δ))1/2 C f for all f,δ. f L2(µ) ≤ k k Fix M N. ∈ Suppose we could show δlog(N(M,d ,δ))1/2 C f for all f L (µ) and all f L2(µ) ∞ ≤ k k ∈ δ > 0. Let f L2(µ) and δ > 0. Let D = max( S ,..., S ) be the maximum of the 1 M ∈ k k k k L2(µ) operator norms. Choose f L (µ) with f f and f f < δ/(2D). 1 ∞ 1 1 L2(µ) ∈ | | ≤ | | k − k Then, S f S f < δ/2 for all n M and N(M,d ,δ) N(M,d ,δ/2). Hence, k n − n 1kL2(µ) ∈ f ≤ f1 δ(logN(M,d ,δ))1/2 2C f 2C f , and we have the desired estimate with f 1 L2(µ) L2(µ) ≤ k k ≤ k k 2C. Therefore, it suffices to prove the result for all L (µ) functions. Fix f L (µ). ∞ ∞ ∈ We will fix J at some large integer. By the mean ergodic condition on T , we have j J J 1 T (f2) f2 = f 2 − j → k kL1(µ) k kL2(µ) j=1 X in L1(µ)-norm. But, J 1 T (f2) f 2 a.s.[µ] by Lemma 1. Of course, if a sequence | − j | ≤ k kL∞(µ) of functions h converges to a constant c in L1(µ)-norm and h B a.s.[µ] for all n, it n P | n| ≤ follows h c in Lq(µ)-norm for all 1 q < . Hence, as p 2, choose J so large that n → ≤ ∞ ≥ J 1 T (f2) 2 f 2 . (1) J j ≤ k kL2(µ) (cid:13) (cid:13) (cid:13) Xj=1 (cid:13)Lp/2(µ) (cid:13) (cid:13) Similarly, for each pair n(cid:13)(cid:13),n′ ∈ M, J(cid:13)(cid:13)−1 Tj(Snf − Sn′f)2 → kSnf − Sn′fk2L2(µ) = d (n,n)2 in L1(µ)-norm, and thus in probability. So, for J large enough, f ′ P J 1 1 µ(Cn,n′) := µ x ∈ X : J−1 Tj(Snf −Sn′f)2(x) ≥ 4df(n,n′)2 > 1− 16M2. ( ) j=1 X Choose J big enough so that this holds for each pair n,n. ′ Let g ,...,g be a sequence of independent standard normal random variables on a 1 J probability space (Ω, ,P). Define the functions F,F on the product space X Ω by ∗ B × 5 J F(x,ω) = J 1/2 g (ω)T f(x) and F (x,ω) = sup S F(x,ω) . − j j ∗ n | | j=1 n M X ∈ Bythecommutativity assumption, S F(x,ω) = J 1/2 g (ω)T S f(x). Note,foreachfixed n − j j n x such that T S f(x) is finite for all j,n, (S F(x, ) : n M) is a Gaussian process. The j n n · P ∈ focus of the proof will be finding the “correct” x to fix. We define four sets , , , X. First, let A B C D ⊂ = x X : T S f(x) < for all 1 j J, n M . j n A ∈ | | ∞ ≤ ≤ ∈ n o As each T S f L2(µ), it is clear that µ( ) = 1. Set j n ∈ A = x X : Tj(Snf Sn′f)2(x) = Tj(Snf Sn′f)(x)2 for all 1 j J, n,n′ M B ∈ − − ≤ ≤ ∈ n o By Lemma 1, µ(B) = 1. Let C = n,n′Cn,n′, i.e., T 1/2 J 1 = x X : J−1 Tj(Snf Sn′f)2(x) df(n,n′) for all n,n′ M . C  ∈ − ≥ 2 ∈  !  Xj=1  Now,   1 1 µ c µ( c ) < = . Cn,n′ ≤ Cn,n′ 16M2 16 (cid:18)n,n[′ M (cid:19) Xn,n′ Xn,n′ ∈ Equivalently, µ( ) > 15/16. C Finally, let R = 5√2C θ(1/4), where where θ is the finite-valued function from Theo- ′ p rem 2 and C is the constant from Lemma 6. Note, R does not depend on f or M. Set p ′ = x X : P ω Ω : F (x,ω) R f 1/2 . ∗ ′ L2(µ) D ∈ ∈ ≤ k k ≥ n o Now, by Lemmas 1, 6, and 7,(cid:8)and (1), we have (cid:9) 6 1/p F( ,ω) P(dω) = F(x,ω) pµ(dx) P(dω) Lp(µ) k · k | | ZΩ ZΩ(cid:18)ZX (cid:19) 1/p F(x,ω) pµ(dx)P(dω) ≤ | | (cid:18)ZΩZX (cid:19) p 1/p J = J 1/2g (ω)T f(x) P(dω)µ(dx) − j j ZX ZΩ(cid:12)(cid:12)Xj=1 (cid:12)(cid:12) ! (cid:12) (cid:12) (cid:12) p (cid:12) 1/p J (cid:12) (cid:12) = J 1/2g ( )T f(x) µ(dx) − j j  ·  ZX (cid:13)(cid:13)Xj=1 (cid:13)(cid:13)Lp(P) (cid:13) (cid:13)  (cid:13) (cid:13) p  1/p (cid:13) J (cid:13) Cp J 1/2g ( )T f(x) µ(dx) ≤  p − j · j  ZX (cid:13)(cid:13)Xj=1 (cid:13)(cid:13)L2(P) (cid:13) (cid:13)  (cid:13) p/2 (cid:13) 1/p  (cid:13)1 J (cid:13) = C (T f)(x)2 µ(dx) p j  J  ZX (cid:12)(cid:12) Xj=1 (cid:12)(cid:12) (cid:12) (cid:12)  J(cid:12) 1/2 (cid:12)  1 (cid:12) (cid:12) = C T (f2) p j J (cid:13) (cid:13) (cid:13) Xj=1 (cid:13)Lp/2(µ) (cid:13) (cid:13) C √(cid:13)2 f . (cid:13) ≤ p(cid:13) k kL2(µ) (cid:13) By Chebyshev’s inequality, P ω : F( ,ω) > 5√2C f 1/5 < 1/4, or equiva- Lp(µ) p L2(µ) { k · k k k } ≤ lently, P ω Ω : F( ,ω) 5√2C f > 3/4. Lp(µ) p L2(µ) ∈ k · k ≤ k k n o Fix an ω in the above set. Then, F( ,ω) Lp(µ). By Theorem 2, · ∈ µ x X : sup S F(x,ω) F( ,ω) θ(1/4) > 3/4. n Lp(µ) ∈ n N| | ≤ k · k n ∈ o Of course, F (x,ω) sup S F(x,ω) . Further, we have a bound on F( ,ω) by the ∗ ≤ n| n | k · kLp(µ) choice of ω. Thus, µ x X : F (x,ω) 5√2C θ(1/4) f > 3/4. ∗ p L2(µ) ∈ ≤ k k n o As this holds for all such ω, we have P ω Ω : µ x X : F (x,ω) R f > 3/4 > 3/4. ∗ ′ L2(µ) ∈ ∈ ≤ k k n o We now apply Fubini’s th(cid:8)eorem. In particular, (cid:9) 7 4 3/4 < χ{µ(F∗(x,ω)≤R′kfkL2(µ))>3/4}(ω)P(dω) ≤ 3µ{x : F∗(x,ω) ≤ R′kfkL2(µ)}P(dω) ZΩ ZΩ 4 = 3 χ{F∗(x,ω)≤R′kfkL2(µ)}(x,ω)µ(dx)P(dω) ZΩZX 4 = 3 χ{F∗(x,ω)≤R′kfkL2(µ)}(x,ω)P(dω)µ(dx) ZX ZΩ 4 = P ω : F (x,ω) R f µ(dx) ∗ ′ L2(µ) 3 { ≤ k k } ZX 4 = P F (x,ω) R f µ(dx)+ P F (x,ω) R f µ(dx) ∗ ′ L2(µ) ∗ ′ L2(µ) 3 { ≤ k k } { ≤ k k } (cid:20)Z Z c (cid:21) D D 4 (µ( )+1/2). ≤ 3 D The last line follows from the definition of . This gives µ( ) > 1/16. D D The estimates µ( ) = 1, µ( ) = 1, µ( ) > 15/16, and µ( ) > 1/16 together imply A B C D that µ( ) > 0. Fix x . Define G (ω) = S F(x,ω) = n n A ∩ B ∩ C ∩ D ∈ A ∩ B ∩ C ∩ D J 1/2 g (ω)T S f(x). As x , (G : n M) is a Gaussian process. By Sudakov’s − j j n n ∈ A ∈ inequality and Theorem 8, and because x , we see that P ∈ D supδ(logN(M,d ,δ))1/2 R sup G (ω) P(dω) G n ≤ | | δ>0 ZΩn M ∈ = R F (x,ω)P(dω) 6RR f . ∗ ′ L2(µ) ≤ k k ZΩ On the other hand, each Gn Gn′ is a linear combination of independent standard normal − random variables. It follows from Lemma 7 again and because x that ∈ B∩C J 1/2 1 2 dG(n,n′) = Gn Gn′ L2(P) = TjSnf(x) TjSn′f(x) k − k J − (cid:18) Xj=1 (cid:16) (cid:17) (cid:19) 1/2 J 1 = Tj(Snf Sn′f)2(x) J − ! j=1 X 1 d (n,n). f ′ ≥ 2 This implies N(M,d ,δ) N(M,d ,δ/2) for all δ > 0. Hence, δ(logN(M,d ,δ))1/2 f G f ≤ ≤ 12RR f . We note that 12RR is universal, and does not depend on M or f. As ′ L2(µ) ′ k k f L (µ) was arbitrary, this holds all a.s. bounded functions. By our earlier note, ∞ ∈ δ(logN(M,d ,δ))1/2 24RR f =: C f for all f L2(µ) and all δ > 0. Taking f ′ L2(µ) L2(µ) ≤ k k k k ∈ the supremum over M, δ(logN (δ))1/2 C f . f L2(µ) ≤ k k 8 4 The Second Entropy Result It will now be necessary to assume the (S ) are uniformly bounded. Of course, by dividing n out a constant, we may assume each S is an L2(µ)-contraction, i.e., S f f n n L2(µ) L2(µ) k k ≤ k k for all n. Proposition 2. Let S be a sequence of L2(µ) contractions that commute with a Bourgain n sequence (T ). Suppose S f converges a.s.[µ] for all f L (µ). Then, there exists a finite- j n ∞ ∈ valued function C(δ) such that N (δ) C(δ) for all δ > 0 and f 1. f L2(µ) ≤ k k ≤ Proof. Suppose we can show the uniform entropy estimate for all f L (µ), f 1. ∞ L2(µ) ∈ k k ≤ Let f 1 and δ > 0. Choose f L (µ), f f such that f f < δ/2. L2(µ) 1 ∞ 1 1 L2(µ) k k ≤ ∈ | | ≤ | | k − k Then, S f S f < δ/2 for all n and N (δ) N (δ/2) C(δ/2). As δ and f k n − n 1kL2(µ) f ≤ f1 ≤ are arbitrary, we have the uniform estimate with the function C (δ) = C(δ/2). It therefore 0 suffices to prove the result for a.s. bounded functions. We proceed by contradiction. Suppose not, i.e., suppose there is some δ > 0 such that N (δ) is unbounded over all such f. Define the constant R = δ . As per The- f ′ 25R orem 3, pick the constant ρ(1/10,R/10). Choose K N, K > 1 big enough so that ′ ∈ 1(R 1000(logK) 1/2) > R/10 and 2(logK) 1/2 < ρ(1/10,R/10). 6 ′ − − ′ − ′ Now, by our assumption, there is some f L , f 1 such that N (δ) > K. In ∞ L2(µ) f ∈ k k ≤ particular, there is a subset I N with I = K (cardinality) and Snf Sn′f L2(µ) > δ for ⊂ | | k − k all n = n I. ′ 6 ∈ As before, we will need to choose an appropriately large J. First, denote B = f L∞(µ). k k Fix a number T > 0 such that T > 3√logK and exp( T2) 7 . Note, the quantity −2B2 ≤ 2KB2 eλ2(2 B2) eλ2(1 B2) is strictly positive for all λ [√logK,T/3]. Let γ > 0 be the minimum − − − ∈ value of this quantity for λ in this interval. As J 1 T (f2) f2 = f 2 in − j → k kL1(µ) k kL2(µ) L1(µ)-norm, it converges in probability. So, pick J large enough so that P J 1 µ(Y) := µ x X : T (f2)(x) f 2 > 1 < γ. (2) ∈ J j −k kL2(µ) ( (cid:12) (cid:12) ) (cid:12) Xj=1 (cid:12) (cid:12) (cid:12) Recall from the proof of Proposit(cid:12)(cid:12)ion 1, J−1 Tj(Snf −Sn′(cid:12)(cid:12)f)2 → kSnf −Sn′fk2L2(µ) in probability for each pair n,n I. So, just as we did before, take J big enough so that if ′ ∈ P 1/2 J 1 Z1 = x ∈ X : J−1 Tj(Snf −Sn′f)2(x) ≥ 2kSnf −Sn′fkL2(µ) for all n,n′ ∈ I !  Xj=1  then µ(Z ) > 4/5. (3)  1  Again, define F(x,ω) = J 1/2 J g (ω)T f(x). Write F(x,ω) = ϕ(x,ω) + H(x,ω) − j=1 j j where P ϕ(x,ω) = F(x,ω)χ and H(x,ω) = F(x,ω)χ . F(x,ω) 6√logK F(x,ω)>6√logK {| |≤ } {| | } Define three subsets of Ω by 9 = ω Ω : sup S H(x,ω) µ(dx) 90 , n A ∈ | | ≤ (cid:26) ZX n∈I (cid:27) = ω Ω : µ x X : sup S F(x,ω) > R(logK)1/2 > 1/5 , n ′ B ∈ ∈ | | (cid:26) n∈I (cid:27) (cid:8) (cid:9) = ω Ω : ϕ(x,ω) µ(dx) 12 . C ∈ | | ≤ (cid:26) ZX (cid:27) Suppose for the moment that we could choose ω . Define ψ(x) = ∈ A ∩ B ∩ C 1(logK) 1/2ϕ(x,ω). Simply from the definition of ϕ, it follows 6 − ψ 1. | | ≤ As ω , ∈ C ψ(x) µ(dx) 2(logK) 1/2 < ρ(1/10,R/10). − ′ | | ≤ ZX By Theorem 3, we have that µ x X : sup S ψ(x) > R/10 < 1/10. (4) n ′ ∈ | | n (cid:26) (cid:27) On the other hand, as ω , we have by Chebyshev that ∈ A µ x X : sup S H(x,ω) > 1000 90/1000 < 1/10, n ∈ | | ≤ (cid:26) n∈I (cid:27) or equivalently µ x X : sup S H(x,ω) 1000 > 9/10. (5) n ∈ | | ≤ (cid:26) n∈I (cid:27) As ω , ∈ B µ x X : sup S F(x,ω) > R(logK)1/2 > 1/5. (6) n ′ ∈ | | (cid:26) n∈I (cid:27) Now, sup S ϕ sup S ϕ sup S F sup S H . So, taking the intersection of the n| n | ≥ I | n | ≥ I | n |− I | n | sets in (5) and (6), we have µ x X : sup S ϕ(x,ω) R(logK)1/2 1000 > 1/10. n ′ ∈ | | ≥ − (cid:26) n (cid:27) Applying the definition of ψ and the choice of K, µ x X : sup S ψ(x) > R/10 > 1/10. n ′ ∈ | | (cid:26) n (cid:27) This clearly contradicts (4). Therefore, it suffices to find such an ω. 10

Bourgain's Entropy Estimates Revisited

Similar Bourgain's Entropy Estimates Revisited