aa r X i v : . [ m a t h . P R ] O c t ON µ -DVORETZKY RANDOM COVERING OF THE CIRCLE. AIHUA FAN AND DAVIT KARAGULYAN
Abstract.
In this paper we study the Dvoretzky covering problem with non-uniformly distributed centers. When the probability law of the centers admitsan absolutely continuous density which satisfies a regular condition on the set ofessential infimum points, we give a necessary and sufficient condition for coveringthe circle. When the lengths of covering intervals are of the form ℓ n = cn , we givea necessary and sufficient condition for covering the circle, without imposing anyregularity on the density function. Main Statement
Let µ be a Borel probability measure on the circle T := R / Z ≡ [0 ,
1) identified withthe interval [0 ,
1) and let ( ℓ n ) n ≥ be a sequence of positive numbers with 0 < ℓ n < ξ ) n ≥ is a sequence of i.i.d. random variables having µ as probabilitylaw. Then, for each n ≥
1, we consider the random interval I n := ( ξ n − ℓ n / , ξ n + ℓ n / ℓ n and centered at ξ n . Sometimes, we say that I n is the ball B ( ξ n , r n )centered at ξ n and of radius r n := ℓ n /
2. Under what condition on µ and on ( ℓ n ) havewe P (cid:0) T = lim sup n →∞ I n (cid:1) = 1?If the answer is affirmative, we say that T is covered for the µ -Dvoretzky covering .We can also ask if a given compact set is covered or not. Without loss of generality,we always assume that ( ℓ n ) is decreasing.When µ is the Lebesgue measure on T , it is the classical Dvoretzky covering problem[4](1956), to which a necessary and sufficient condition for T to be covered is(1.1) ∞ X n =1 n exp( ℓ + · · · + ℓ n ) = ∞ . This is called
Shepp’s condition , which was obtained by L. Shepp [22](1972). Acompact set F ⊂ T is covered if and only if(1.2) Cap Φ ( F ) = 0where Φ( t, s ) = exp ∞ X n =1 ( ℓ n − | t − s | ) + . This is called
Kahane’s condition , which was obtained by J-P. Kahane [17](1987).Recall that Cap Φ ( F ) = 0 means that for any Borel probability measure σ supported by F we have Z T Z T Φ( t, s ) dσ ( t ) dσ ( s ) = + ∞ . We refer to [19] for the theory of capacity. The above cited results due to Sheppand Kahane will be our basic useful facts. Kahane’s book [18] contains referenceson the study of classical Dvoretzky covering before 1985. For later works, let usonly cite [1, 3, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, 20, 21, 24]. A first study on µ -Dvoretzky covering problem for Gibbs measures µ was made in [23], where the authorused the method taken from [12] in order to find the optimal covering exponent t for ℓ n = a/n t , ( a > , t > µ -Dvoretzky covering problem when µ admits adensity. In the following, f will always denote a density function and the measure ofdensity f will be denoted by µ f . The solution to the problem depends on the essentialinfimum of the density function. Definition 1.
For a given Borel function f : T → R , the essential infimum of f isdenoted by m f , i.e. m f := ess inf T f = sup { a ∈ R : a ≤ f ( x ) a.e. } where ”a.e.” refers to the Lebesgue measure.We can define the local essential infima function E f ( x ) and the local essentialsupprema function E f ( x ) (see Section 2 for definitions). The set of essential infimumpoints of f , denoted by K f is defined to be the set of those x such that E f ( x ) = m f .It will be proved that K f is a non-empty compact set (Proposition 2.1).The flatness of the density will also play a role. It is considered as a regularity ofthe density. Definition 2.
A point x ∈ T has the flatness property for the measure µ f and thesequence { ℓ n } n ≥ if(1.3) ∞ X n =1 | µ f ( B ( x, r n )) − m f ℓ n | < ∞ . The set of all flat points will be denoted by F f ( { ℓ n } n ≥ ) or simply by F f .Here is another regularity of the density function f : there exists a sequence ofpoints { x n } n ≥ ⊂ T such that(1.4) lim n →∞ E f ( x n ) = m f . In is easy to see, that if f is continuous at least at one point of the set K f , then thecondition (1.4) is fulfilled.We are now ready to state one of the main theorems in this paper. Theorem 1.1.
Let µ f be a Borel probability measure on T with the density function f and { ℓ n } ∞ n =1 a sequence of positive numbers. Assume that the condition (1.4) isfulfilled. We distinguish two cases. N µ -DVORETZKY RANDOM COVERING OF THE CIRCLE. 3 (1) Assume m f = 0 and K f is countable. Then, the circle is covered for the µ f -Dvoretzky covering if and only if the following two conditions are satisfied: (1.5) ∀ x ∈ K f , ∞ X n =1 µ f ( B ( x, r n )) = ∞ ;(1.6) ∀ a > , ∞ X n =1 n e a ( ℓ + ··· + ℓ n ) = ∞ . (2) Assume m f > and there exists a sequence { a n } n ≥ , with a = 0 , so that theset K f \ ∪ n ≥ ( a n + F f ∩ K f ) is at most countable. Then, the circle is covered for the µ f -Dvoretzky covering if and only if the following two conditions are satisfied: (1.7) ∀ a > m f , ∞ X n =1 n e a ( ℓ + ··· + ℓ n ) = ∞ ; Cap Φ ( mf ) ( F f ∩ K f ) = 0 , where the capacity refers to the kernel Φ ( m f ) ( t, s ) = exp ∞ X n =1 m f ( ℓ n − | t − s | ) + . We have the following corollaries of Theorem 1.1.
Corollary 1.1.
Assume that for all x ∈ K f , there is an open neighborhood U of x and a Lipschitz function g x ∈ Lip( U ) , such that f ( t ) ≤ g x ( t ) for almost every t ∈ U and f ( t ) = g x ( t ) on K f ∩ U . (1) If m f = 0 and P ∞ n =1 ℓ n < ∞ , there will be no µ f -Dvoretzky covering. (2) If m f > , the circle is covered if and only if the following two conditions aresatisfied (1.8) ∀ a > m f , ∞ X n =1 n e a ( ℓ + ··· + ℓ n ) = ∞ ; Cap Φ ( mf ) ( K f ) = 0 , Corollary 1.2.
Assume there exists U ⊂ T , so that f ( x ) = m f , for a.e. x ∈ U .Then the circle is covered for the µ f -Dvoretzky covering if and only if ∞ X n =1 n e m f ( ℓ + ··· + ℓ n ) = ∞ . In the special case of ℓ n = cn ( c > f . Theorem 1.2.
Suppose ℓ n = cn ( c > ). Let µ f be an arbitrary Borel probabilitymeasure with density f . A necessary and sufficient condition for covering the circlein the µ f -Dvoretzky covering is cm f ≥ . The µ -Dvoretzky covering problem is subtle and is not treated in this paper when µ is singular. But we have the following local comparison principle. This principlewill serve us as a tool in our present study and it has its own interests. Theorem 1.3.
Consider two Dvoretzky covering respectively defined by two Borelprobability measures µ and ν . Assume that µ | U ≤ ν | U for some non-empty open set U ⊂ T . Let K ⊂ U be a compact set in U . If K is covered for the µ -Dvoretzkycovering, then it is covered for the ν -Dvoretzky covering. AIHUA FAN AND DAVIT KARAGULYAN
We organize the rest of the paper as follows. In Section 2, we prove that the set ofessential infimum is a non-empty compact set. Section 3 contains two basic resultswhich are respectively qualified local Billard criterion and local Kahane criterion. InSection 4, we prove the comparison principle (Theorem 1.3) which is our third basicresult. After these preparations, we find sufficient conditions for covering T in Section5, and necessary conditions for covering T in Section 6. Then Theorem 1.1 togetherwith its corollaries and Theorem 1.2 are proved in Section 7.2. Set of essential infimum
We leave the Dvoretzky covering problem for a while. In this section, we study theset of points where a measurable function attains its ”minimal” value.Let f : T → R be a Borel measurable function and I ⊂ T an interval. The essentialinfimum of the function f on the interval I is defined as followsess inf I f := sup { a ∈ R : a ≤ f ( x ) for almost all x ∈ I } . We will denote ess inf T f by m f . Let x ∈ T be fixed. The essential infimum at x of f is defined to be the following limit E f ( x ) := lim n →∞ ess inf B ( x , n ) f. The set of essential infimum points , denoted K f , is defined by K f := { x ∈ T : E f ( x ) = m f } . Similarly we define the essential suprema ess sup I f and E f ( x ). Clearly E f ( x ) ≤ E f ( x ) . It is also cleat that E f ( x ) = E f ( x ) if f is continuous at x . Proposition 2.1.
For every Borel measurable function f : T → R , K f is a non-emptycompact set.Proof. We prove the non-emptyness of K f by a dissection argument using the factess inf I ∪ J f = min(ess inf I f, ess inf J f ) . Indeed, cut T into T = I ∪ J = [0 , / ∪ [1 / , T f = min(ess inf I f, ess inf J f ) . We continue this process and construct a sequence of nested intervals I ⊃ I ⊃ . . . I n ⊃ . . . , so that I n +1 is one of the two halves of the interval I n and such thatess inf I n f = ess inf I n +1 f = ess inf T f. Since | I n +1 | = | I n | /
2, then T ∞ n =1 I n is a single point, say { x } . We claim that x ∈ K .Indeed, for arbitrary n ∈ N there is m ∈ N such that B ( x , /n ) ⊃ I m , which implies ess inf B ( x , n ) f ≤ ess inf I m f = ess inf T f. N µ -DVORETZKY RANDOM COVERING OF THE CIRCLE. 5 Let n → ∞ . Thus we get E f ( x ) = ess inf T f .Now show that K f is closed. Assume { x n } n ≥ ⊂ K f and lim n →∞ x n = x . For any n ∈ N there exists m ≥ x m ∈ B ( x , n ) , which implies that ess inf B ( x , n ) f ≤ E f ( x m ) = m f . It follows that E f ( x ) = m f , i.e. x ∈ K f . (cid:3) We now show that the regularity condition (1.4) is not always fulfilled.Let A ⊂ [0 ,
1] be a set, which is constructed in the same way as the classical Cantorset, but at each step, we remove the central interval of shorter length. This resultsin a compact set with positive Lebesgue measure, which contains no intervals and isnowhere dense. Consider the characteristic function of the complementary of A : f A ( x ) = 1 − χ A ( x ) , x ∈ [0 , . Observe that(2.1) ∀ x ∈ [0 , , E f A ( x ) = 1; ∀ x ∈ A, E f A ( x ) = 0 . Then, we define the following function f (0) = 0; ∀ x ∈ [2 − n − , − n ) , f ( x ) = f A (2 n +1 x −
1) + 1 n + 1 ( n ≥ . From (2.1), we get that for each n ≥ [ n +1 , n ] f = 1 n + 1 ; ∀ x ∈ [2 − n − , − n ) , E f ( x ) = 1 + 1 n + 1 . Finally we get K f = { } ; E f (0) = 0; ∀ x ∈ [0 , , E f ( x ) ≥ . Thus, the Condition 1.4 is not fulfilled by this function f . However, recall that thecondition (1.4) is satisfied for a function f which is continuous at one point of K f .3. Two basic results due to Billard and Kahane
The proof of Theorem 1.1 will be based on Theorem 1.3 and on the following twocriteria, which have their own interests.3.1.
Local Billard necessary condition.
The following theorem gives us a neces-sary condition for covering the circle. The idea of second moment used in the proofcame from P. Billard [2]. Therefore the condition will be refered to as (local) Billardcondition.
Theorem 3.1 (local Billard criterion) . Let F ⊂ T be a non-empty compact set.Suppose that µ is a probability measure on T such that (3.1) sup t ∈ F ∞ X n =1 µ ( B ( t, r n )) < ∞ . AIHUA FAN AND DAVIT KARAGULYAN
Then F is not covered for the µ -Dvoretzky covering if there exists a probability measure σ supported by F such that (3.2) Z F Z F exp ∞ X n =1 µ ( B ( t, r n ) ∩ B ( s, r n )) dσ ( t ) dσ ( s ) < ∞ . Proof.
Consider the martingale M n = Z F Q n ( t ) dσ ( t )where Q n ( t ) = n Y j =1 − χ B ( ξ n ,r n ) ( t )1 − µ ( B ( t, r n )) . Notice that Q n ( t ) = 0 means that t is covered by one of the intervals I j (1 ≤ j ≤ n ).If M n doesn’t tend to zero, then some point in F is not covered. By Fubini theoremand simple computations, we get E M n = Z F Z F n Y j =1 − µ ( B ( t, r n ) − µ ( B ( s, r n )) + µ ( B ( t, r n ) ∩ B ( s, r n ))(1 − µ ( B ( t, r n ))(1 − µ ( B ( s, r n ))) dtds. Using 1 − x = e − x + O ( x ) and the conditions (3.1) and (3.2), we get E M n = O (1) whichimplies that the limit of M n is not almost surely zero. (cid:3) The condition (3.2) means that F has a positive capacity. The condition (3.1) canbe relaxed to the pointwise finiteness, because F can be approximated by compactsets on each of which the sum is uniformly bounded.3.2. Local Kahane criterion.
The next result can be considered as a local versionof Kahane’s theorem. For a >
0, define the kernalΦ ( a ) ( t, s ) := exp a ∞ X k =1 ( l k − | t − s | ) + ! . As we see below, this criterion is not perfect because we need the assumption thatthe density is constant around the set to be covered in question. But it will be one ofour basic tools, because we can approximate our density function by functions whichare locally constant.
Theorem 3.2 (local Kahane criterion) . Consider a probability measure µ f having itsdensity function f . Let I ⊂ T be an open interval. Suppose that f ( x ) = a for almostall x ∈ I . Then a compact subset F of I is covered for the µ f -Dvoretzky covering ifand only if Cap Φ ( a ) ( F ) = 0 .Proof. The idea of proof is to compare the µ f -Dvoretzky covering with a classicDvoretzky covering. Let M be the distribution function of µ f , i.e. M ( x ) = µ f ([0 , x ]) = Z x f ( t ) dt (0 ≤ x ≤ . Let X be a random variable which is uniformly distributed on [0 , Y = M − ( X ) N µ -DVORETZKY RANDOM COVERING OF THE CIRCLE. 7 where the inverse function is defined by M − ( x ) = inf { t ∈ [0 ,
1] : M ( t ) ≥ x } . It is well known that the probability law of Y is µ f . So, take a sequence ( ω n ) of i.i.d.random variables uniformly distributed on [0 ,
1) and use ( ξ n ) to model a µ f -Dvoretzkycovering, where ξ n = M − ( ω n ).Observe that the restriction M : I → M ( I ) is affine and invertible. It follows thatfor any interval J ⊆ I , its image M ( J ) is an interval of length | M ( J ) | = Z J f ( t ) dt = | J | a and the center of M ( J ) is the image of the center of J . So, if I n = ( ξ n − ℓ n / , ξ n + ℓ n / ⊂ I , then M ( I n ) is centered at ( ω n ) and of length aℓ n .Now choose a proper subset J ⊂ I such that F ⊂ J . Then choose N ∈ N so largethat ℓ n ≤ dist { J, ∂I } for all n ≥ N . Assume F ⊂ lim sup I n ( ξ n ). Then for any x ∈ F , x ∈ I n ( ξ n ) for an infinite number of n ′ s with n ≥ N (these n ’s depend on x ). These ξ n ’s must fall into the interval J . For such n , M ( x ) ∈ M n ( I n ( ξ n )). It follows that P ( F ⊂ lim sup n →∞ I n ) = 1 = ⇒ P ( M ( F ) ⊂ lim sup n →∞ M ( I n )) = 1 . The converse implication can be similarly proved, because M : I → M ( I ) is invertible.In other words, F ⊂ J is covered for the µ f -Dvoretzky coverring by intervals of length { ℓ n } n ≥ if and only if M ( F ) is covered for the classic Dvoretzky covering by intervalsof length { aℓ n } n ≥ . But, by Kahane’s theorem, the set M ( F ) is covered for theclassic Dvoretzky covering if and only if Cap Φ ( a ) ( M ( F )) = 0, which is equivalent toCap Φ ( a ) ( F ) = 0, because M is affine around F . (cid:3) Two elementary facts.
Recall that we always assume that ( ℓ n ) is decreasing.The following fact is known when F = T . The general case is not trivial. The proofgiven below involves both Shepp’s condition and Kahane’s condition. Lemma 3.1.
Let F be a compact set with | F | > and a > a positive number. Thenthe following two conditions are equivalent: (3.3) Cap Φ ( a ) ( F ) = 0 , (3.4) ∞ X n =1 n exp( a ( ℓ + · · · + ℓ n )) = ∞ . Proof.
A simple calculation shows that there exists a constant C a > Z F Z F exp a ∞ X k =1 ( ℓ k − | t − s | ) + ! dsdt ≤ C a Z T Z T exp ∞ X k =1 ( aℓ k − | u − v | ) + ! dudv. (First replace F by T and then make a change of variable). Then (3.4) is implied by(3.3) because the last double integral equals to the infinity if and only if (3.4) holds.Now assume (3.4). It is nothing but Shepp’s condition for the classic Dvoretzkycovering with the sequence of lengths { aℓ n } n ≥ . Then any non-empty compact set AIHUA FAN AND DAVIT KARAGULYAN K ′ ⊂ T is covered. If F ′ has positive Lebesgue measure, it supports the restriction ofLebesgue measure. Thus, by Kahane’s condition, we have Z F ′ Z F ′ exp ∞ X k =1 ( aℓ k − | u − v | ) + ! dudv = ∞ . Take F ′ = aF , where aF is the scaling of F by a coefficient a . Then, by making achange of variable, we get Z F Z F exp ∞ X k =1 a ( ℓ k − | t − s | ) + ! dtds = ∞ . (cid:3) The following fact shows that P ∞ n =1 ℓ n = ∞ is a strong condition for the Dvoretzkycovering problem. Lemma 3.2.
The condition P ∞ n =1 ℓ k = ∞ implies ∀ a > , ∞ X n =1 n exp( a ( ℓ + · · · + ℓ n )) = ∞ . Proof.
This is known, however we will include a proof for completeness. The condition P ∞ n =1 ℓ n = ∞ implies that ℓ n > n / holds for infinitely many n . Then, for such a n ,by the monotonicity of ( ℓ n ) we have n X k =1 ℓ k > nn / > n / . Therefore the general term of the series in question doesn’t tend to zero. So, theseries diverges. (cid:3) A comparison principle: Proof of Theorem 1.3
Intuitively, if µ | U ≤ ν | U , then sets contained in an open set U are easierly coveredfor the ν -Dvoretzky covering than for the µ -Dvoretzky covering, because there aremore chances to get points in U for the ν -Dvoretzky model. We give here a rigourousproof of this intuition which is stated as Theorem 1.3. The proof benefits from ourdiscussion with Meng Wu.4.1. Model of µ -Dvoretzky covering. Let µ = µ + µ be a decomposition of aprobability measure µ on T . Then we set e µ := µ µ ( T ) , e µ := µ µ ( T ) . For simplicity, we write α = µ ( T ) and α = µ ( T ), so that µ = α e µ + α e µ .Let Ω = { , } N × T N . A point in Ω is denoted by ω := ( ǫ ; ξ ) := ( ǫ , · · · , ǫ n , · · · ; ξ , · · · , ξ n , · · · ) ∈ Ω . N µ -DVORETZKY RANDOM COVERING OF THE CIRCLE. 9 Using the decomposition µ = α e µ + α e µ , we define a Borel probability measure P on Ω as follows. For any integer n ≥
1, any ( a , · · · , a n ) ∈ { , } n and any( A , · · · , A n ) ∈ B ( T ) n , define(4.1) P ( ǫ = a , · · · , ǫ n = a n ; ξ ∈ A , · · · , ξ n ∈ A n ) := n Y j =1 α a j e µ a j ( A j ) = n Y j =1 µ a j ( A j ) . Notice that P is not exactly a tensor product. But it has a nice desintegration as weshow below.Let π be the ( α , α )-Bernoulli measure on { , } N . For ǫ fixed, letΛ ( ǫ ) = { n ∈ N : ǫ n = 1 } , and we define two measures respectively on T Λ ( ǫ ) and T N \ Λ ( ǫ ) where P Λ ( ǫ ) = O n ∈ Λ ( ǫ ) e µ , P N \ Λ ( ǫ ) = O n ∈ N \ Λ ( ǫ ) e µ , Let P ( ǫ ) = P Λ ( ǫ ) ⊗ P N \ Λ ( ǫ ) . Notice that P ( ǫ ) is a version of the conditional probability P ( ·| ǫ ). Lemma 4.1.
The measure P has the following desintegration (4.2) P = Z P ǫ dπ ( ǫ ) = Z P Λ ( ǫ ) ⊗ P N \ Λ ( ǫ ) dπ ( ǫ ) . Consequently, the following statements hold: (1)
All the random vectors ( ǫ j , ξ j ) ( j = 1 , , · · · ) are i.i.d.. (2) Each ǫ j obeys the ( α , α ) -Bernoulli law; (3) Each ξ j admits µ as probability law; (4) For ǫ fixed, with respect to P ( ǫ ) , { ξ n } n ∈ Λ ( ǫ ) are i.i.d. random variables withprobability law e µ and { ξ n } n ∈ N \ Λ ( ǫ ) are i.i.d. random variables with probability law e µ .Proof. Using the fact µ + µ = µ , we check that the measure P is well defined.The desintegration (4.2) is nothing but the definition (4.1). It follows the otherproperties. (cid:3) Since ( ξ n ) is a sequence of i.i.d. random variables with µ as probability law, therandom intervals I n ( ξ n ) = ( ξ n − ℓ n / , ξ n + ℓ n /
2) define a model of µ -covering. Recallthat µ = µ + µ Lemma 4.2.
Let U ⊂ T be a non-empty open interval. Suppose µ ( U ) = 0 . Thenfor any compact set K ⊂ U , for P -a.e. ( ǫ, ξ ) ∈ Ω , we have (4.3) K ⊂ lim sup n ∈ N I n ( ξ n ) ⇐⇒ K ⊂ lim sup n ∈ Λ ( ǫ ) I n ( ξ n ) . Proof.
Recall that conditioned on ǫ ∈ { , } N , the sequence { ξ n } n ∈ Λ ( ǫ ) are i.i.d. ran-dom variables with probability law e µ and { ξ n } n ∈ N \ Λ ( ǫ ) are i.i.d. with probability law e µ . Since e µ ( U c ) = 1, for P ( ǫ ) -a.e. ξ we have ξ n ∈ U c for all n ∈ N \ Λ ( ǫ ) . Since K iscompact, it has positive distance from the boundary of U , we must have P ( ǫ ) ( K ∩ lim sup n ∈ N \ Λ ( ǫ ) I n ( ξ n ) = ∅ ) = 0 . From this we deduce (4.3), with the aid of the desintegration (4.2). (cid:3)
However, notice that ǫ j and x j are not independent.4.2. Proof of Theorem 1.3.
We can model the covering as above, according to thefollowing decompositions µ = µ + µ with µ = µ | U , µ = µ | U c ν = ν + ν with ν = µ | U , ν = ( ν | U − µ | U ) + ν | U c . The corresponding measures will be denoted respectively by P µ and P ν .Suppose K is a.s. covered in the µ -covering. Then by Lemma 4.2, we have π − a.e. ǫ, P ( ǫ ) µ (cid:18) K ⊂ lim sup n ∈ Λ ( ǫ ) I n ( ξ n ) (cid:19) = 1 . But ( ξ n ) n ∈ Λ ( ǫ ) has the same probability law under P ( ǫ ) µ and under P ( ǫ ) ν , because of µ = ν (see Lemma 4.1 (4)). Thus π − a.e. ǫ, P ( ǫ ) ν (cid:18) K ⊂ lim sup n ∈ Λ ( ǫ ) I n ( ξ n ) (cid:19) = 1 . It follows obviously that π − a.e. ǫ, P ( ǫ ) ν (cid:18) K ⊂ lim sup n ∈ N I n ( ξ n ) (cid:19) = 1 . Finally using once more the desintegration we get P ν (cid:18) K ⊂ lim sup n ∈ N I n ( ξ n ) (cid:19) = Z P ( ǫ ) ν (cid:18) K ⊂ lim sup n ∈ N I n ( ξ n ) (cid:19) dπ ( ǫ ) = 1 . Comparison with the classic Dvoretzky covering.Corollary 4.1.
Let us consider a probability measure µ f on T with density f . Let F be a non-empty compact set contained in an open set U .1) Assume f ( x ) ≥ a for almost all x ∈ U . The set F is covered for the µ f -Dvoretzkycovering if Cap Φ ( a ) ( F ) = 0 .2) Assume f ( x ) ≤ a for almost all x ∈ U . The set K is not covered for the µ f -Dvoretzky covering if Cap Φ ( a ) ( F ) > .Proof. Let h be a density function on T so that h ( x ) = a for all x ∈ U .1) Assume Cap Φ ( a ) ( F ) = 0. Then, according to the local Kahane criterion (Theo-rem 3.2), the set F is covered for the µ h -Dvoretzky covering. Then, by Theorem 1.3,the set F is covered for the µ f -Dvoretzky covering for f | U ≥ h | U .2) Assume Cap Φ ( a ) ( F ) >
0. Then, according to the local Kahane criterion (Theo-rem 3.2), the set F is not covered for the µ h -Dvoretzky covering. Then, by Theorem1.3, the set F is not covered for the µ f -Dvoretzky covering for f | U ≤ h | U .. (cid:3) N µ -DVORETZKY RANDOM COVERING OF THE CIRCLE. 11 Sufficient conditions for covering T In this section we discuss several sufficient conditions for covering the circle, whichis decomposed into K f and T \ K f . Proposition 5.1.
The set T \ K f is covered for the µ f -Dvoretzky covering under thecondition (5.1) ∀ a > m f , ∞ X n =1 n exp( a ( ℓ + · · · + ℓ n )) = ∞ . If, additionally, f satisfies the property (1.4) , then the condition (5.1) will also benecessary for covering the set T \ K f .Proof. Let K cf = T \ K f . Let us assume (5.1). Notice that K cf is an open set, byProposition 2.1. By definition, E f ( x ) > m f for all x ∈ K cf . Therefore, for every x ∈ K cf there is an integer n x ∈ N and a number a x > m f such that f ( t ) ≥ a x , for almost all t ∈ B ( x, n x ) . Thus, by Corollary 4.1, the interval ( x − n x , x + n x ) is covered for the µ f -Dvoretzkycovering. However, the open set K cf is a union of countably many such intervals B ( x, n x ). So, K cf is covered.Now assume the condition (1.4), which implies that for any ε >
0, there is N ∈ N such that E f ( x N ) < m f + ε, which implies that there is k ∈ N such that f ( x ) < m f + ε, for almost all x ∈ B ( x , /k ) . Let a = m f + ε . Notice that ε is arbitrary. By Corollary 4.1, the condition (5.1) mustbe necessary for covering K cf . (cid:3) Corollary 5.1. If f is continuous at some point of K f , then the condition (5.1) isnecessary and sufficient for covering the set T \ K f .Proof. Assume f is continuous at x ∈ K f . Then E f ( x ) = m f , which clearly implies(1.4) if we take x n = x for all n ≥ (cid:3) Sufficient conditions for covering T .Proposition 5.2. Assume m f > . The circle is covered for the µ f -Dvoretzky cov-ering if (5.2) ∀ a > m f , ∞ X n =1 n e a ( ℓ + ··· + ℓ n ) = ∞ ; Cap Φ ( mf ) ( K f ) = 0 . Proof.
The first condition in (5.2) implies that T \ K f is covered, Proposition 5.1. Thecovering of K f follows from the second condition in (5.2), Corollary 4.1 and Kahane’sresult. (cid:3) Proposition 5.3.
If we have m f > , then the condition ∞ X n =1 n exp( m f ( ℓ + · · · + ℓ n )) = ∞ , is sufficient for covering the circle.Proof. It follows from Corollary 4.1 and Shepp’s result, because f ( x ) ≥ m f for almostall x ∈ T . (cid:3) We now show that each of the conditions in (1.7) does not imply the other.Let us recall the two conditions in (1.7) :(C1) ∀ a > m f , ∞ X n =1 n e a ( ℓ + ··· + ℓ n ) = ∞ ; (C2) Cap Φ ( mf ) ( K f ) = 0 . Consider the following density function f ( x ) = | x | + 34 , for x ∈ (cid:20) − , (cid:19) . Observe that m f = , K f = { } and f is Lipschitz at x = 0. By Corollary 1.1 (2), theconditions (C1) and (C2) are necessary and sufficient for covering the circle. Since K f = { } , the condition (C2) is equivalent to P ∞ n =1 ℓ n = ∞ , which does not imply(C1). Indeed, when ℓ n = n , we have P ∞ n =1 ℓ n = ∞ , but ∀ a < / , ∞ X n =1 n e a ( ℓ + ··· + ℓ n ) ≈ ∞ X n =1 n (2 − a ) < ∞ . Observe however that in this case (C1) implies (C2).We now show that (C1) does not imply (C2). To see this, let us look at the sequence ℓ n = n − n ln n and the density function f ( x ) = 12 1 [0 , / ( x ) + 32 1 [1 / , ( x ) . We have that m f = 1 / K f = [0 , / a > / ∞ X n =1 n e a ( ℓ + ··· + ℓ n ) ≈ ∞ X n =1 n e (2 a ln n − a ln ln n ) = ∞ X n =1 n (2 − a ) ln a n = ∞ . Hence, (C1) is satisfied. Since | K f | >
0, according to Lemma 3.1 the condition (C2)is equivalent to (3.4), which reads as ∞ X n =1 n e ( ℓ + ··· + ℓ n ) = ∞ . However ∞ X n =1 n ln n < ∞ . Thus, the condition (C1) does not imply (C2) for all density functions f and allsequences { ℓ n } n ≥ . N µ -DVORETZKY RANDOM COVERING OF THE CIRCLE. 13 Necessary conditions for covering T In this section we discuss necessary conditions for covering the circle. The proofof the next proposition is based on Lebesgue’s differentiation theorem and Billard’scriterion.
Proposition 6.1.
Suppose that the sequence { ℓ n } ∞ n =1 satisfies the condition (6.1) lim sup n →∞ nℓ n ℓ + · · · + ℓ n < . If T is covered for the µ f -Dvoretzky covering, then (6.2) ∀ a > m f , ∞ X n =1 n e a ( ℓ + ··· + ℓ n ) = ∞ . Proof.
For ǫ >
0, let A = n x ∈ T : f ( x ) < m f + ǫ o . By the definition of m f , A has positive Lebesgue measure. Consider S n ( x ) := µ f ( B ( x, r n )) ℓ n = 1 ℓ n Z x + r n x − r n f ( t ) dt. By Lebesgue’s differentiation theorem, S n ( x ) converges to f ( x ) almost everywhere.Then, by Egoroff’s theorem, there is a compact set A ′ ⊂ A with | A ′ | > S n ( x ) converges to f ( x ) uniformly in A ′ . Therefore, for all large enough n , we have ∀ x ∈ A ′ , | S n ( x ) − f ( x ) | < ǫ . It follows that for large n and for x ∈ A ′ we have(6.3) µ f ( B ( x, r n )) ≤ (cid:16) f ( x ) + ǫ (cid:17) ℓ n ≤ ( m f + ǫ ) ℓ n . Notice that B ( t, r n ) ∩ B ( s, r n ) ⊂ B ( s, r n ) and f ( x ) ≥ m f for almost all x ∈ T . Itfollows that for all s, t ∈ T (6.4) µ f ( B ( t, r n ) ∩ B ( s, r n )) ≤ m f ( ℓ n − | t − s | ) + + ( µ f ( B ( s, r n ) − m f ℓ n ) . Indeed, this is trivial if | t − s | > ℓ n . Otherwise the inequality reads as µ f ( B ( t, r n ) ∩ B ( s, r n )) ≤ µ f ( B ( s, r n )) − m f | t − s | , which is true, because B ( s, r n ) is the union of B ( s, r n ) ∩ B ( t, r n ) and an interval oflength | t − s | . Assume now s ∈ A ′ . From (6.4) and (6.3) we get(6.5) µ ( B ( t, r n ) ∩ B ( s, r n )) ≤ m f ( ℓ n − | t − s | ) + + ǫℓ n . We claim that there exists 0 < δ < t, s ) ∈ A ′ × A ′ with | t − s | small enough we have(6.6) ∞ X n =1 µ f ( B ( t, r n ) ∩ B ( s, r n ) ≤ (cid:16) m f + ǫ − δ (cid:17) ∞ X n =1 ( ℓ n − | t − s | ) + . Indeed, assume ℓ N +1 < | t − s | ≤ ℓ N we have ∞ X n =1 µ f ( B ( t, r n ) ∩ B ( s, r n )) = N X n =1 µ f ( B ( t, r n ) ∩ B ( s, r n )) . By the estimation (6.5), for ( t, s ) ∈ A ′ × A ′ we have(6.7) ∞ X n =1 µ f ( B ( t, r n ) ∩ B ( s, r n )) ≤ N X n =1 ( ℓ n − | t − s | ) + + ǫ N X n =1 ℓ n . However N X n =1 ( ℓ n − | t − s | ) + = N X n =1 ℓ n − N | t − s | ≥ N X n =1 ℓ n − N ℓ N . By the assumption (6.1), for some 0 < δ <
1, when N is large enough (i.e. | t − s | issmall enough), we have(6.8) N X n =1 ( ℓ n − | t − s | ) + ≥ (1 − δ ) N X n =1 ℓ n . Therefore, by (6.7) and (6.8), we get ∞ X n =1 µ ( B ( t, r n ) ∩ B ( s, r n )) ≤ (cid:18) m f + ǫ − δ (cid:19) N X n =1 ( ℓ n − | t − s | ) + . This is what we have claimed for (6.6) because ( ℓ n − | t − s | ) + = 0 for n > N .We are ready to check (6.2). By Lemma 3.2, we can assume that P ∞ n =1 ℓ n < ∞ . Then, by (6.3), we havesup t ∈ A ′ ∞ X n =1 µ ( B ( t, r n )) = ( m f + ε ) ∞ X n =1 ℓ n < ∞ . Since the compact set A ′ is covered and it has positive Lebesgue measure, we canapply Billard’s local criterion (Theorem 3.1) to confirm that Z A ′ Z A ′ exp ∞ X n =1 µ ( B ( t, r n ) ∩ B ( s, r n )) dtds = ∞ . Then, by the estimation (6.6) we get Z A ′ Z A ′ exp ( m f + ε − δ ) ∞ X n =1 ( ℓ k − | t − s | ) + ! dtds = ∞ . Notice that 0 < δ < ǫ > (cid:3)
Proposition 6.2.
Suppose that µ f admits a density f . If K f ∩ F f is covered, then Cap Φ ( mf ) ( K f ∩ F f ) = 0 . N µ -DVORETZKY RANDOM COVERING OF THE CIRCLE. 15 Proof. If K f ∩ F f = ∅ , there is nothing to prove. Suppose then K f ∩ F f = ∅ . Assume m f = 0. Then by the Borel-Cantelli lemma, any fixed point x ∈ K f is only finitelycovered. Since K f ∩ F f is covered, we must have m f >
0. We can assume that P ∞ n =1 ℓ n < ∞ . Otherwise, we get immediately Cap Φ ( mf ) ( K f ) = 0 from Lemma 3.2and Corollary 5.3.Notice that x µ f ( B ( x, r n )) is continuous. So the function P ∞ n =1 | µ f ( B ( x, r n )) − m f ℓ n | is lower semi-continuous. Thus the following sets are compact: K m := K f ∩ ( x ∈ T : ∞ X n =1 | µ f ( B ( x, r n )) − m f ℓ n | ≤ m ) . It is clear that K m is increasing and it tends to K f ∩ F f . Therefore we have only toprove that Cap Φ ( mf ) ( K m ) = 0 for every m ≥ t, s ∈ T we have ∞ X n =1 µ ( B ( t, r n ) ∩ B ( s, r n )) ≤ ∞ X n =1 m f ( ℓ k − | t − s | ) + + ∞ X n =1 | µ ( B ( s, r n )) − m f ℓ n | . Assume s ∈ K m . Then the last sum is bounded by m . Therefore, for all s ∈ K m andall t ∈ T we have(6.9) ∞ X n =1 µ ( B ( t, r n ) ∩ B ( s, r n )) ≤ ∞ X n =1 m f ( ℓ k − | t − s | ) + + m. On the other hand, letting a n = | µ ( B ( s, r n )) − m f ℓ n | , we have(6.10) sup s ∈ K m ∞ X n =1 µ ( B ( s, r n )) ≤ sup s ∈ K m ∞ X n =1 ( a n + 2 a n m f ℓ n + ( m f ℓ n ) ) < ∞ because P ℓ n < ∞ and P a n ≤ m . By the local Billard criterion, for any probabilitymeasure σ on K m we have Z K m Z K m exp ∞ X n =1 µ ( B ( t, r n ) ∩ B ( s, r n )) dσ ( t ) dσ ( s ) = + ∞ , which, together with (6.9), implies(6.11) Z K m Z K m exp ∞ X n =1 m f ( ℓ k − | t − s | ) + dσ ( t ) dσ ( s ) = ∞ . Thus we have proved Cap Φ ( mf ) ( K m ) = 0. (cid:3) We now discuss the condition (6.1) and its relation to (6.2).(A)
The condition (6.1) is not always fulfilled.
We will construct a decreasing sequence ( ℓ n ) such that P ∞ n =1 ℓ n < ∞ and (6.1) isnot satisfied. Let 1 = n < n < · · · < n k < . . . be a sequence of positive integerssuch that for all k ≥ n k n k ≥ ln n k +1 n k +1 . Define ( ℓ n ) as follows ℓ n := ln n k +1 n k +1 , when n + · · · + n k < n ≤ n + · · · + n k + n k +1 . Clearly ( ℓ n ) is decreasing and if n k grows fast enough, then ∞ X n =1 ℓ n < ∞ X k =1 n k · ln n k n k = ∞ X k =1 ln n k n k < ∞ . For n = n + · · · + n k , we have nℓ n ℓ + · · · + ℓ n = ( n + · · · + n k ) ln n k n k ln n + · · · + ln n k . If n k grows fast enough, thenlim n →∞ nℓ n ℓ + · · · + ℓ n ≥ lim k →∞ n n k ln n k + · · · + n k − n k ln n k + ln n k ln n + · · · + ln n k = 1 . (B) (6.1) is not necessary for (6.2).For this we modify the construction above. For all even k s, we alter the constructionabove as follows. For all n ∈ N , with n + · · · + n k ≤ n ≤ n + · · · + n k +1 , we define ℓ n = cn , where c >
1. Note that ln n k n k ≥ cn + · · · + n k . Next we take n k +1 so large that n + ··· + n k +1 X n =1 n e a ( ℓ + ··· + ℓ n ) > k. This is doable, since ac ≥
1. As a result, we will get a sequence ( ℓ n ) n ≥ for which(6.1) fails, but for all a ≥ c the series in (6.2) diverges.7. Proofs of Theorem 1.1 , Corollaries 1.1, 1.2 and Theorem 1.2
Proof of Theorem 1.1.
Case m f = 0. By assumption, we have (1.4). Hence,by Proposition 5.1 the condition (1.6) is necessary and sufficient for covering the set T \ K f . Obviously, the condition (1.5) is necessary for covering K f . Otherwise forsome x ∈ T we will have ∞ X n =1 µ f ( B ( x , r n )) < ∞ , a contradiction to the fact that x is covered. Since K f is countable, then the condition(1.5) will also be sufficient for covering the set K f . Case m f >
0. The sufficiency of the first conditions in (1.7) follows from Proposi-tion 5.2. We now show its necessity. As above, the necessity of the first condition in(1.7) follows from the assumptions (1.4). N µ -DVORETZKY RANDOM COVERING OF THE CIRCLE. 17 The second fact in (1.7) implies that K f ∩ F f and K f ∩ ( a n + K f ∩ F f ) are all coveredby Kahane’s result and Corollary 4.1. The points in K f , which are not covered bythe translates of K f ∩ F n are countable, these countable set are covered according tothe condition (1.5). Observe that when m f >
0, the condition (1.5) is automaticallyfulfilled. This is because µ f ( B ( x, r n )) ≥ m f ℓ n and P ℓ n = ∞ . Actually we get more ∀ x ∈ T , ∞ X n =1 µ f ( B ( x, r n )) = ∞ . So, we have only to check the necessity of the second condition in (1.7) which, inturn, follows from Proposition 6.2.7.2.
Proof of Corollary 1.1.
Proof.
By assumption, the set of points where we do not have a Lipschitz dominant g ∈ Lip( U ), is countable, hence the condition (1.5) will be necessary and sufficientfor covering this set. As for all x ∈ K f ∩ F f , there is a set U with x ∈ U and afunction g x = g ∈ Lip( U ), so that f ( t ) ≤ g ( t ) for almost all t ∈ U and f ( t ) = g ( t ), for t ∈ K f ∩ U . Since g is continuous at x and f ( x ) = g ( x ) = m f , then f is continuous at x too, so according to Corollary 5.1 the first condition in 1.7 is necessary and sufficientfor covering the set K cf . To finish the proof, we need to show that for ∀ x ∈ K f wehave the flatness condition.Fix x ∈ K f . Let C be the Lipschitz constant of g . Hence µ ( B ( x, r n )) ≤ Z x + r n x − r n g ( y ) dy ≤ Z x + r n x − r n ( C | x − y | + | g ( x ) | ) dy ≤ Cr n + ℓ n f ( x ) . Since x ∈ K f , then f ( x ) = m f . Hence | µ ( B ( x, r n )) − m f ℓ n | ≤ Cr n . From which ∞ X n =1 | µ ( B ( x, r n )) − m f ℓ n | ≤ C ∞ X n =1 r n . (1) Let m f = 0 and P ∞ n =1 ℓ n < ∞ , then ∞ X n =1 µ ( B ( x, r n )) < ∞ . Hence, there will be no µ f -Dvoretzky covering.(2) If m f >
0, then we can assume ∞ X n =1 ℓ n < ∞ , from which ∞ X n =1 | µ ( B ( x, r n )) − m f ℓ n | < ∞ . Thus, at x we have the flatness property. To finish the proof we will need to applyTheorem 1.1. (cid:3) Proof of Corollary 1.2.
Proof.
Let K = K f ∩ F f . By assumption, K is of positive Lebesgue measure. Suppose T is covered. Then so is K which supports the Lebesgue measure restricted on K ,and by Proposition 6.2, we have(7.1) Z K Z K exp m f ∞ X k =1 ( ℓ k − | t − s | ) + ! dtds = ∞ , which is equivalent to the condition (3.4) according to Lemma 3.1. Thus, the condition(3.4) is necessary for covering the circle. The sufficiency of the condition (3.4) forcovering the circle follows from Proposition 5.3. (cid:3) Proof of Theorem 1.2.
Proof.
Observe that ℓ + · · · + ℓ n = c log n + O (1). Hence(7.2) ∞ X n =1 n e a ( ℓ + ··· + ℓ n ) = ∞ ⇔ ac ≥ . Assume cm f ≥ . By Corollary 5.3 and (7.2) applied to a = m f , the circle iscovered.Conversely assume that the circle is covered. Remark that the condition (6.1) issatisfied. Indeed, nℓ n ℓ + · · · + ℓ n = cc ln n + O (1) = o (1) . Then by Proposition 6.1, we have ∀ ǫ > , ∞ X n =1 n e ( m f + ǫ )( ℓ + ··· + ℓ n ) = ∞ , which, according to (7.2), implies c ( m f + ǫ ) ≥ ǫ >
0. Therefore cm f ≥ . (cid:3) Acknowledgement.
The second author would like to thank the School of Mathemat-ics and Statistics of the Central China Normal University for its hospitality duringhis visit in the Spring semester of 2018, where the work was started.
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Aihua Fan: LAMFA, UMR 7352, CNRS, University of Picardie, 33 rue Saint Leu,80039 Amiens CEDEX 1, France
E-mail address : [email protected] Davit Karagulyan: Department of Mathematics, University of Maryland, CollegePark 20742, USA
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