AA New Inequality For The Hilbert Transform
Sakin DemirOctober 13, 2020
Abstract
Suppose that { a j } ∈ l . Then we prove that there is a constant C such that ∞ (cid:88) n =1 (cid:93) (cid:40) k ∈ Z : (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n (cid:88) i = − n (cid:48) a k + i i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) > λ (cid:41) ≤ Cλ ∞ (cid:88) i = −∞ | a i | for all λ > Mathematics Subject Classifications:
Key Words:
Hilbert Transform, Inequality.Let ( X, B , µ ) be a measure space, τ : X → X an invertible measure-preserving transformation. The ergodic Hilbert transform of a measurablefunction f , is defined as Hf ( x ) = lim n →∞ n (cid:88) k = − n (cid:48) f ( τ k x ) k . The prime denotes that the term with zero denominator is omitted in thesummation.It is well known that Hf is of weak type ( p, p ) for 1 ≤ p < ∞ , and ofstrong type ( p, p ) for 1 < p < ∞ . There are several different methods in theliterature to see these facts. The most immediate one is to transfer the sameinequalities for the Hilbert transform on R by Calder´on transfer principleas in the relation between the Hardy-Littlewood maximal function and the1 a r X i v : . [ m a t h . C A ] O c t rgodic maximal function.For { a j } ∈ l the Hilbert transform on Z is defined by H a ( k ) = lim n →∞ n (cid:88) i = − n (cid:48) a k + i i . Our main goal of this research is to prove the following:Suppose that { a j } ∈ l has finite support. Then we prove that there is aconstant C such that ∞ (cid:88) n =1 (cid:93) (cid:40) k ∈ Z : (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n (cid:88) i = − n (cid:48) a k + i i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) > λ (cid:41) ≤ Cλ ∞ (cid:88) i = −∞ | a i | for all λ >
0. Then it will be clear by means of a transference argument thatthe same type of inequality for the ergodic Hilbert transform also remainstrue.The following lemmas are due to L. H. Loomis [3], who rediscovered anidea that essentially goes back to G. Boole [2]. We give the proofs of themfor completeness:
Lemma 1.
Let a , a , . . . , a n ≥ and g ( s ) = (cid:80) ni =1 a i s − t i . Then m { s : g ( s ) > λ } = m { s : g ( s ) < − λ } = 1 λ n (cid:88) i =1 a i , where m denotes the Lebesgue measure on R .Proof. Since g ( t i − ) = −∞ , g ( t i +) = ∞ and g (cid:48) ( s ) < s , there areprecisely n points m i such that g ( m i ) = λ , and t i < m − i < t i +1 , i =1 , , . . . , n − , t n , m n . The set where g ( s ) > λ thus consists of the intervals( t i , m i ) and has total length n (cid:88) i =1 ( m i − t i ) = n (cid:88) i =1 m i − n (cid:88) i =1 t i . (1)But the numbers m i are the roots of the equation n (cid:88) i =1 a i s − t i = λ, n (cid:88) i =1 a i (cid:34)(cid:89) j (cid:54) = i ( s − t i ) (cid:35) = λ n (cid:89) i =1 ( s − t i ) , or λs n − (cid:104) λ (cid:88) t j + (cid:88) a i (cid:105) s n − + · · · = 0 , so that n (cid:88) i =1 m i = n (cid:88) i =1 t i + 1 λ n (cid:88) i =1 a i . (2)The first part of the lemma follows from (1) and (2); the proof for g ( s ) < − λ is almost identical. Lemma 2.
There is a constant C such that if { a k } ∈ l and λ > , then (cid:93) (cid:40) k ∈ Z : (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∞ (cid:88) i = −∞(cid:48) a k + i i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) > λ (cid:41) ≤ Cλ ∞ (cid:88) i = −∞ | a i | . Proof.
By treating the positive and negative ones separately, we may assumethat all the a i are positive. We will count A λ = (cid:40) k : ∞ (cid:88) i = −∞(cid:48) a k + i i > λ (cid:41) ;a similar method will apply to A (cid:48) λ = (cid:40) k : ∞ (cid:88) i = −∞(cid:48) a k + i i < − λ (cid:41) . Choose a finite set A ⊂ A λ , and choose N so large that A ⊂ [ N, N ] and, foreach k ∈ A , N (cid:88) i = − N (cid:48) a i i − k > λ. Then g k ( s ) = N (cid:88) i = − N (cid:48) a i i − s > λ s = k ∈ A , and hence g k ( s ) > λ for s ∈ [ k, k + 1), because g (cid:48) k ( s ) >
0. Ifwe let g ( s ) = N (cid:88) i = − N (cid:48) a i i − s > λ and h k ( s ) = a k k − s , then g = g k + h k , so that for each k ∈ A ( k, k + 1) ⊂ { s : g k ( s ) > λ } ⊂ (cid:26) s : g ( s ) > λ (cid:27) ∪ (cid:26) s : h k ( s ) < − λ (cid:27) . Therefore, we get (cid:93)A = m (cid:32) (cid:91) k ∈ A ( k, k + 1) (cid:33) ≤ m (cid:26) s : g ( s ) > λ (cid:27) + (cid:88) k ∈ A m (cid:26) s : h k ( s ) < − λ (cid:27) ≤ Cλ N (cid:88) i = − N a i + (cid:88) k ∈ A Cλ a k ≤ Cλ (cid:107) a (cid:107) as desired. Lemma 3.
There is a constant C such that if { a k } ∈ l and λ > , then (cid:93) (cid:40) k ∈ Z : sup n ≥ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n (cid:88) i = − n (cid:48) a k + i i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) > λ (cid:41) ≤ Cλ ∞ (cid:88) i = −∞ | a i | . Proof.
We assume as before that all the a i are positive and drop the absolutevalue signs. Let A ⊂ (cid:40) k : sup n ≥ n (cid:88) i = − n (cid:48) a k + i i > λ (cid:41) be closed and bounded. For each k ∈ A there is an interval of integers I k = [ k − n − k, k + n k ] such that (cid:88) i ∈ I k (cid:48) a i i − k > λ. g k ( s ) = (cid:88) i ∈ I k (cid:48) a i i − s , g ( s ) = ∞ (cid:88) i = −∞(cid:48) a i i − s , h k ( s ) = (cid:88) i/ ∈ I k (cid:48) a i i − s . If k ∈ A , then g k ( k ) > λ , so that either g ( k ) > λ or h k ( k ) < − λ . In thefirst case ( k ∈ A ), by Lemma 2, k falls into a single (independent of k ) setof measure no more than Cλ (cid:107) a (cid:107) . To deal with the left over k ’s ( k ∈ A ),replace { I k } by a disjoint subfamily which still covers at least of A , byat each stage selecting an interval of maximal disjoint from the previouslychosen ones. Find N such that (cid:91) k ∈ A I k ⊂ [ − N, N ]and ˜ h k ( k ) ≤ − λ k ∈ A , where ˜ h k ( s ) = (cid:88) i ∈{− N,...,N }− I k a i i − s . Then also ˜ h k ( s ) < − λ on ( k − n k , k ), so that we find (cid:93)A = (cid:93)A + (cid:93)A ≤ Cλ (cid:107) a (cid:107) + 6 (cid:88) k ∈ A n k ≤ Cλ (cid:107) a (cid:107) + 6 m (cid:32) (cid:91) k ∈ A (cid:26) s : ˜ h k ( s ) < − λ (cid:27)(cid:33) ≤ Cλ (cid:107) a (cid:107) + 6 m (cid:32) (cid:91) k ∈ A (cid:32)(cid:40) s : N (cid:88) i = − N (cid:48) a i i − s < − λ (cid:41) ∪ (cid:26) s : g k ( s ) > λ (cid:27)(cid:33)(cid:33) ≤ Cλ (cid:107) a (cid:107) + 6 m (cid:40) s : N (cid:88) i = − N (cid:48) a i i − s < − λ (cid:41) ∪ (cid:26) s : g k ( s ) > λ (cid:27) + 6 (cid:88) k ∈ A m (cid:26) s : g k ( s ) > λ (cid:27) ≤ Cλ (cid:107) a (cid:107) + 24 Cλ (cid:107) a (cid:107) + 6 (cid:88) k ∈ A Cλ (cid:88) i ∈ I k a i ≤ Cλ (cid:107) a (cid:107) .
5e can now state and prove our main result:
Theorem 1.
Suppose that { a j } ∈ l . Then there is a constant C such that ∞ (cid:88) n =1 (cid:93) (cid:40) k ∈ Z : (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n (cid:88) i = − n (cid:48) a k + i i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) > λ (cid:41) ≤ Cλ ∞ (cid:88) i = −∞ | a i | for all λ > .Proof. Let us first define the integer block B n = {− n, − ( n − , − ( n − , . . . , n − , n − , n } for each n ∈ Z . Let A n = (cid:40) k ∈ Z : (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n (cid:88) i = − n (cid:48) a k + i i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) > λ (cid:41) and A = (cid:40) k ∈ Z : sup n ≥ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n (cid:88) i = − n (cid:48) a k + i i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) > λ (cid:41) . Then we have A n ⊂ A for all n ≥ . This imples that (cid:93) A n ≤ (cid:93) A for all n ≥ (cid:93) A < ∞ by Lemma 3we see that (cid:93) A n < ∞ for all n ≥
1. This shows that A n has finitely manyelements for all n ≥ (cid:93) is the counting measure on Z , and thus A n isa bounded set for each n ≥
1. Therefore, we can select a sequence { t n } oftranslates so that ( A n − t n ) ∩ ( A n (cid:48) − t n (cid:48) ) = φ if n (cid:54) = n (cid:48) and sup n ≥ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:88) i ∈B n − t n (cid:48) a k + i i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ sup n ≥ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:88) i ∈B n (cid:48) a k + i i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . Since (cid:93) ( A n − t n ) = (cid:93) A n we only need to prove that 6 (cid:88) n =1 (cid:93) ( A n − t n ) ≤ Cλ ∞ (cid:88) i = −∞ | a i | for some constant C .We now have ∞ (cid:88) n =1 (cid:93) ( A n − t n ) = ∞ (cid:88) n =1 (cid:93) (cid:40) k ∈ Z : (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:88) i ∈B n − t n (cid:48) a k + i i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) > λ (cid:41) = (cid:93) ∞ (cid:91) n =1 (cid:40) k ∈ Z : (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:88) i ∈B n − t n (cid:48) a k + i i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) > λ (cid:41) ≤ (cid:93) (cid:40) k ∈ Z : sup n ≥ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:88) i ∈B n − t n (cid:48) a k + i i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) > λ (cid:41) ≤ (cid:93) (cid:40) k ∈ Z : sup n ≥ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:88) i ∈B n (cid:48) a k + i i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) > λ (cid:41) ≤ Cλ ∞ (cid:88) i = −∞ | a i | (by Lemma 3)as desired. Corollary 2.
Let ( X, B , µ ) be a measure space, τ : X → X an invertiblemeasure-preserving transformation. Then there exists a constant C > suchthat ∞ (cid:88) n =1 µ (cid:40) x : (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n (cid:88) i = − n (cid:48) f ( τ i x ) i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) > λ (cid:41) ≤ Cλ (cid:107) f (cid:107) , for all f ∈ L ( X ) and λ > .Proof. The transference argument we are about use to proof our Corollaryis the modification of the proof of Lemma 1 in K. Petersen [4] to our case.One can also directly apply a well known variant of the transfer principle ofA. P. Calder´on [1] to Theorem 1 to get the desired result.7y considering f + and f − separately, we may assume that f ≥
0. Wewill show that ∞ (cid:88) n =1 µ (cid:40) x : (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n (cid:88) i = − n (cid:48) f ( τ i x ) i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) > λ (cid:41) ≤ Cλ (cid:107) f (cid:107) , where C is a constant independent of f and λ .For fixed x and K , let a k = f ( τ k x ) and a Kk = (cid:26) a k if | k | ≤ K, | k | > K, so that { a Kk } ∈ l . For each j ∈ Z , let G j ( x ) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n (cid:88) k = − n (cid:48) a k + j k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) , and G Kj ( x ) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n (cid:88) k = − n (cid:48) a Kk + j k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . Then G j ( x ) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n (cid:88) k = − n (cid:48) a Kk + j k + a k + j − a Kk + j k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ G Kj ( x ) + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n (cid:88) k = − n (cid:48) a k + j − a Kk + j k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) , so that G j ( x ) ≤ G Kj ( x ) for | j | ≤ K .Now let E = { x : G ( x ) > λ } , so that { x : G j ( x ) > λ } = τ − j E .Let ¯ E = (cid:8) ( x, j ) : G Kj ( x ) > λ (cid:9) . Then, if (cid:93) continues to denote the countingmeasure on Z , ∞ (cid:88) n =1 µ × (cid:93) ( ¯ E ) = (cid:90) X ∞ (cid:88) n =1 (cid:93) (cid:8) j : G Kj ( x ) > λ (cid:9) dµ ( x ) ≤ (cid:90) X Cλ ∞ (cid:88) j = −∞ (cid:12)(cid:12) a Kj (cid:12)(cid:12) dµ ≤ (cid:90) X Cλ K (cid:88) − K | a j | dµ ≤ Cλ [2 K + 1] (cid:107) f (cid:107) , µ × (cid:93) ( ¯ E ) ≥ K (cid:88) j = − K µ (cid:8) x : G Kj ( x ) > λ (cid:9) ≥ K (cid:88) j = − K µ { x : G j ( x ) > λ } = K (cid:88) j = − K µ (cid:0) τ − j E (cid:1) = (2 K + 1) µ ( E ) . Thus, we have ∞ (cid:88) n =1 µ ( E ) ≤ Cλ (cid:107) f (cid:107) and this completes our proof. References [1] A. P. Calder´on,
Ergodic theory and translation-invariant operators ,Proc. Nat. Acad. Sci. USA 59 (1968) 349-353.[2] G. Boole,
On the comparision of trancendents with certain applicationsto the theory of definite integrals , Philos. Trans. Roy. Soc. London 147(1957) 745-803.[3] L. H. Loomis,
A note on the Hilbert transform , Bull. AMS 52 (1946)1082-1086.[4] K. Petersen,