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A STUDY OF THE REAL HARDY INEQUALITY MOHAMMAD SABABHEH Department of Science and Arts Princess Sumaya University for Technology Amman 11941 Jordan Real Hardy Inequality Mohammad Sababheh vol. 10, iss. 4, art. 104, 2009 EMail: sababheh@psut.edu.jo Title Page Received: 11 November, 2008 Accepted: 20 October, 2009 Communicated by: L. Pick 2000 AMS Sub. Class.: Contents JJ II 42A16, 42A05, 42B30. J I Key words: Hardy’s inequality, inequalities involving Fourier transforms. Page 1 of 18 Abstract: We show that some Hardy-type inequalities on the circle can be proved to be true on the real line. Namely, we discuss the idea of getting Hardy inequalities on the real line by the use of corresponding inequalities on the circle. In the last section, we prove the truth of a certain open problem under some restrictions. Go Back Full Screen Close Contents 1 Introduction 3 2 Setup 5 3 From The Circle To The Line 7 Real Hardy Inequality 4 Another Hardy Inequality 10 Mohammad Sababheh vol. 10, iss. 4, art. 104, 2009 Title Page Contents JJ II J I Page 2 of 18 Go Back Full Screen Close 1. Introduction When McGehee, Pigno and Smith [4] proved the Littlewood conjecture, many questions regarding the best possible generalization of Hardy’s inequality were asked. The longstanding question is: Does there exist a constant c > 0 such that ∞ X |fˆ(n)| (1.1) n=1 n ≤ ckf k1 + c ∞ X |fˆ(−n)| n=1 n 1 for all f ∈ L (T)? The truth of this inequality is an open problem. Many attempts were made to answer this question and many partial results were obtained. We refer the reader to [2], [3], [6] and [7] for some partial results. Almost all articles in the literature treat Hardy-type inequalities on the circle and a very few articles treat them on the real line. In [8] it was proved that a constant c > 0 exists such that for all Z x 1 1 f ∈ g ∈ L (R) : g(t)dt ∈ L (R) −∞ we have Z ∞ ˆ |f (−ξ)|2 |fˆ(ξ)|2 2 dξ ≤ ckf k1 + c dξ. ξ ξ 0 0 Although this is not the first proved Hardy inequality on the real line, its proof is the first proof which uses the construction of a bounded function on R whose Fourier coefficients have some desired decay properties. We have two main goals in this article. The first is to prove Hardy inequalities on the real line using well known inequalities on the circle and the second is to prove a real Hardy inequality on the real line which is related to the open problem (1.1). Z ∞ Real Hardy Inequality Mohammad Sababheh vol. 10, iss. 4, art. 104, 2009 Title Page Contents JJ II J I Page 3 of 18 Go Back Full Screen Close Proving a Hardy inequality usually involves quite a difficult construction and this is because of the way we prove such inequalities. Again we refer the reader to [2], [4], [6], [7] and [8] for more information on how and why we construct bounded functions with desired Fourier coefficients. Real Hardy Inequality Mohammad Sababheh vol. 10, iss. 4, art. 104, 2009 Title Page Contents JJ II J I Page 4 of 18 Go Back Full Screen Close 2. Setup Let L1 denote the space of all integrable functions (equivalent classes) defined on R. For f ∈ L1 , we define the Fourier transform of f to be Z ˆ f (ξ) = f (x)e−ixξ dξ. R If fˆ ∈ L1 then the inversion formula for f holds: Z 1 f (x) = fˆ(ξ)eiξx dξ. 2π R If f ∈ L2 then the Fourier transform of f is defined to be the L2 − limit: Z n ˆ f (ξ) = lim f (x)e−iξx dx. n→∞ −n In this case fˆ ∈ L2 and 1 f (x) = lim n→∞ 2π Z n fˆ(ξ)eixξ dξ, in the L2 sense. −n The Plancheral theorem then says: 1 kf k2 = √ kfˆk2 . 2π Lemma 2.1. For f, g ∈ L2 we have Z Z 1 ˆ f (x)g(x)dx = fˆ(ξ)g(ξ)dξ. 2π R R Real Hardy Inequality Mohammad Sababheh vol. 10, iss. 4, art. 104, 2009 Title Page Contents JJ II J I Page 5 of 18 Go Back Full Screen Close We refer the reader to any standard book in analysis for more on these concepts, see for example [5]. Observe that if f ∈ L1 is such that fˆ is compactly supported, then fˆ ∈ L2 and hence f ∈ L2 . Now, given f ∈ L1 (R), define (2.1) ϕN (t) = 2π ∞ X fN (t + 2πj), where fN (x) = N f (N x). j=−∞ Mohammad Sababheh vol. 10, iss. 4, art. 104, 2009 Then we have (2.2) n ϕN ∈ L1 (T), ϕ̂N (n) = fˆ N Real Hardy Inequality and lim kϕN kL1 (T) = kf kL1 (R) . N →∞ For a discussion of this idea we refer the reader to [1, p. 160-162]. Thus, the equations in (2.1) and (2.2) enable us to move from a function integrable on the line to a function integrable on the circle. This idea will be used efficiently in the next section to prove some Hardy-type inequalities on the real line using only a corresponding inequality on the circle. Title Page Contents JJ II J I Page 6 of 18 Go Back Full Screen Close 3. From The Circle To The Line In this section we discuss how one can use a Hardy inequality on the circle to obtain an inequality on the real line. Recall that Hardy’s inequality on the circle states that a constant C > 0 exists such that for all f ∈ L1 (T) with fˆ(n) = 0, n < 0 we have ∞ X |fˆ(n)| n=1 n Real Hardy Inequality ≤ Ckf k1 . Mohammad Sababheh vol. 10, iss. 4, art. 104, 2009 As a matter of notation, let H 1 (R) be defined by n o H 1 (R) = f ∈ L1 (R) : fˆ(ξ) = 0, ∀ξ < 0 . Title Page Contents In the following theorem we use the above stated Hardy’s inequality to get the well known Hardy inequality on the line. We should remark that proving such an inequality (on the line) is a difficult task, but transforming the problem from the circle to the line greatly simplifies the proof. Theorem 3.1. There exists an absolute constant C > 0 such that Z ∞ ˆ |f (ξ)| ≤ Ckf k1 ξ 0 II J I Page 7 of 18 Go Back Full Screen Close for all f ∈ H 1 (R). Proof. Let f ∈ H 1 (R) be arbitrary and let, for N ∈ N, ϕN (as in (2.1) and (2.2)) be such that: ϕN ∈ L1 (T), JJ lim kϕN kL1 (T) = kf kL1 (R) N →∞ and ϕ̂N (n) = fˆ(n/N ). Clearly ϕ̂N (n) = 0, ∀n < 0, hence Hardy’s inequality applies and we have ∞ X |ϕ̂N (n)| n n=1 ≤ CkϕN kL1 (T) . However, this implies that N X 1 |fˆ(n/N )| ≤ CkϕN kL1 (T) . N n/N n=1 We take the limit as N → ∞ to get Z 1 ˆ |f (ξ)| (3.1) ≤ Ckf k1 . ξ 0 Thus, we have shown (3.1) for any function in H 1 (R). Now we proceed to prove the inequality stated in the theorem. That is, we would like to replace the upper limit of the above integral by ∞. For this, let M ∈ N be arbitrary and put h(x) = f (x/M ). Then khk1 = M kf k1 and ĥ(ξ) = M fˆ(M ξ). Now apply (3.1) on h to obtain Z 1 ˆ Z 1 |ĥ(ξ)| |f (M ξ)| ≤ Ckhk1 ⇒ dξ ≤ Ckf k1 . ξ ξ 0 0 Put M ξ = t to get Z (3.2) 0 M |fˆ(ξ)| dξ ≤ Ckf k1 . ξ Now, letting M tend to ∞, we obtain the result. Real Hardy Inequality Mohammad Sababheh vol. 10, iss. 4, art. 104, 2009 Title Page Contents JJ II J I Page 8 of 18 Go Back Full Screen Close In fact, the idea of this proof can be used to prove many Hardy inequalities on the real line! It is proved that, see for example [3], for all functions f ∈ L1 (T) we have ∞ X |fˆ(n)|2 n=1 n ≤ Ckf k21 +C ∞ X |fˆ(−n)|2 n=1 n . Using an argument similar to that of Theorem 3.1 we can prove: Real Hardy Inequality Mohammad Sababheh Theorem 3.2. There exists a constant C > 0 such that for all functions f ∈ L1 (R) we have Z ∞ ˆ 2 Z ∞ ˆ |f (ξ)| |f (−ξ)|2 2 dξ ≤ Ckf k1 + C dξ. ξ ξ 0 0 vol. 10, iss. 4, art. 104, 2009 Title Page Contents This modifies the result proved in [8]. JJ II J I Page 9 of 18 Go Back Full Screen Close 4. Another Hardy Inequality In this section we prove (1.1) on the real line under a certain condition on the signs of the Fourier coefficients. We should remark here that the method used to prove this inequality is standard and all recent articles use this idea; we need a bounded function whose Fourier coefficients obey some desired decay conditions. One last remark before proceeding, although the given proof is for a real Hardy inequality, we can imitate the given steps to prove a similar inequality on the circle. For j ≥ 1 put Z j 1 4 ixξ fj (x) = j e dξ, x ∈ R. 4 4j−1 Then we have our first lemma: Lemma 4.1. Let fj be as above, then ( 2π , 4j−1 < ξ < 4j , 4j ˆ fj (ξ) = 0, otherwise. Real Hardy Inequality Mohammad Sababheh vol. 10, iss. 4, art. 104, 2009 Title Page Contents JJ II J I Page 10 of 18 Proof. Let Go Back ( gj (ξ) = 2 2π , 4j 4 0, otherwise, j−1 j <ξ<4 , Full Screen Close 2 then gj ∈ L . Therefore, gj is the Fourier transform of some function hj ∈ L and Z 1 g(ξ)eixξ dξ hj (x) = 2π R = f (x). However, ĥj = gj , which implies fˆj = gj as claimed. Observe that the above proof implies that fj ∈ L2 and Lemma 4.2. For fj as above, we have √ kfj k2 = 6π 4j/2 . Real Hardy Inequality 2 Proof. Since fj ∈ L we have Mohammad Sababheh vol. 10, iss. 4, art. 104, 2009 1 kfj k2 = √ kfˆj k2 2π 1 2π =√ 2π 4j √ 6π = j/2 . 4 Z ! 12 4j Title Page dξ 4j−1 Contents JJ II J I Page 11 of 18 Lemma 4.3. For fj as above and for M ∈ N, let M X FM (x) = fj (x) − fj (x) , j=1 then kFM k∞ ≤ C where C is some absolute constant (independent of M ). Go Back Full Screen Close Proof. Observe first that Z j 4 2 sin(xξ)dξ fj (x) − fj (x) = j 4 4j−1 2 cos(4j x) − cos(4j−1 x) = j 4 x 2 2 sin2 (4j−1 x/2) [−1 + 16 cos2 (4j−1 x) cos2 (4j−1 x/2)] = j 4 x 60 sin2 (4j−1 x/2) ≤ j . 4 |x| Real Hardy Inequality Mohammad Sababheh vol. 10, iss. 4, art. 104, 2009 Title Page Note that FM is an even function, so it suffices to consider only the case x > 0. Now, fix x > 0 and observe that ∞ X 1 sin2 (4j−1 x/2) FM (x) ≤ 60 4j x j=1 2 2 j−1 j−1 X 1 sin (4 x/2) X 1 sin (4 x/2) , = 60 + j j 4 x 4 x −j −j 4 ≤x 4 Contents JJ II J I Page 12 of 18 Go Back >x Full Screen where, by convention, the second sum is zero if 4−j ≤ x for all j ≥ 1. Now ∞ X 1 sin2 (4j−1 x/2) 1X 1 ≤ , 4j x x j=k 4j −j 4 ≤x x −j where kx is the smallest positive integer such that 4 ≤ x for all j ≥ kx . Close Consequently X 1 sin2 (4j−1 x/2) 14 1 ≤ j 4 x x 3 4kx −j 4 ≤x 14 4 x= . x3 3 ≤ On the other hand, if P 4−j >x 1 sin2 (4j−1 x/2) 4j x Real Hardy Inequality is not zero, we get Mohammad Sababheh 4 (1/x) X 1 sin2 (4j−1 x/2) logX 1 sin2 (4j−1 x/2) = 4j x 4j x −j j=1 4 >x log4 (1/x) ≤ X j=1 1 4j x 1 j−1 4 x 2 log4 (1/x) = X 1 x 4j 64 j=1 log4 (1/x) 1 4 x 64 1 ≤ . 192 ≤ −4 FM (x) ≤ 60 4 1 + 3 192 Title Page 2 Contents JJ II J I Page 13 of 18 Go Back 3 Full Screen Close Thus, we have vol. 10, iss. 4, art. 104, 2009 := C. Now let X be the set of all f ∈ L1 such that the sign of fˆ(ξ) is constant in the block (4j−1 , 4j ). Then we have our main result: Theorem 4.4. There is an absolute constant K > 0 such that Z ∞ ˆ Z ∞ ˆ |f (ξ)| |f (−ξ)| (4.1) dξ ≤ Kkf k1 + K dξ ξ ξ 1 1 Real Hardy Inequality for all f ∈ X. Mohammad Sababheh Proof. Let f ∈ X be such that fˆ is compactly supported, let fj be as above and let M ∈ N be such that the support of fˆ is contained in [−M, M ]. Denote the sign of fˆ in the block (4j−1 , 4j ) by σj and put F (x) = M X σj fj (x) − fj (x) . j=1 Then kF k∞ ≤ C, where C is the constant of Lemma 4.3. Moreover, F̂ (ξ) = 2π σ, 4j j where j is the unique index such that ξ ∈ [4j−1 , 4j ]∪[−4j , −4j−1 ] if −4M ≤ ξ ≤ 4M and F̂ (ξ) = 0 otherwise. Moreover, since fˆ is of compact support, we have f ∈ L2 . Now we apply a standard duality argument: Z Ckf k1 ≥ f (x)F (x)dx R Z 1 ˆ = f (ξ)F̂ (ξ)dξ (see Lemma 2.1) 2π R vol. 10, iss. 4, art. 104, 2009 Title Page Contents JJ II J I Page 14 of 18 Go Back Full Screen Close 1 ≥ 2π Z 1 ∞ Z 4j X 1 fˆ(ξ)F̂ (ξ)dξ − fˆ(ξ)F̂ (ξ)dξ 2π −1 j−1 j=1 4 ∞ Z j 1 X 4 ˆ f (−ξ)F̂ (−ξ)dξ . − 2π 4j−1 j=1 That is, 1 (4.2) 2π Real Hardy Inequality ∞ Z j Z 1 X 4 1 ˆ(ξ)F̂ (ξ)dξ f fˆ(ξ)F̂ (ξ)dξ ≤ Ckf k1 + 2π −1 j−1 j=1 4 ∞ Z j 1 X 4 ˆ + f (−ξ)F̂ (−ξ)dξ . 2π j−1 4 Mohammad Sababheh vol. 10, iss. 4, art. 104, 2009 Title Page Contents j=1 σ , where σj is However, when ξ ∈ (4j−1 , 4j ) ∪ (−4j , 4j−1 ), we have F̂ (ξ) = 2π 4j j the sign of fˆ in (4j , 4j−1 ). Thus, when ξ ∈ (4j−1 , 4j ) we have fˆ(ξ)F̂ (ξ) = |fˆ(ξ)|. Moreover Z 1 Z 1 ˆ(ξ)F̂ (ξ)dξ ≤ kf k1 f F̂ (ξ) dξ −1 −1 Z 1 12 Z ≤ kf k1 dξ −1 √ ≤ 2kf k1 kF̂ k2 √ √ = 2kf k1 2πkF k2 1 −1 |F̂ (ξ)|2 dξ 12 JJ II J I Page 15 of 18 Go Back Full Screen Close ∞ X √ ≤ 2 πkf k1 × 2 kfj k2 √ = 4 6πkf k1 , j=1 where we have used the facts F (x) = M X σj fj (x) − fj (x) and kfj k2 = √ 6π4−j/2 . j=1 Real Hardy Inequality Mohammad Sababheh vol. 10, iss. 4, art. 104, 2009 Consequently, (4.2) becomes ∞ Z 4j ∞ Z 4j X X √ |fˆ(−ξ)| |fˆ(ξ)| 6kf k + dξ. (4.3) dξ ≤ Ckf k + 2 1 1 4j 4j j−1 j−1 j=1 4 j=1 4 However, when ξ ∈ [4j−1 , 4j ] we have 1 1 ≤ 4j ξ and 1 1 ≥ . 4j 4ξ Therefore, (4.3) boils down to Z ∞ ˆ Z ∞ ˆ |f (ξ)| |f (−ξ)| ≤ Kkf k1 + K dξ, ξ ξ 1 1 √ where K = 4(C + 2 6) and where C is the constant of Lemma 4.3. This completes the proof for f ∈ X with the property that fˆ is compactly supported. Now for general f ∈ X, consider the convolution f ∗ Kλ where Kλ is the Fejer Kernel of order λ. A standard limiting process yields the result. Title Page Contents JJ II J I Page 16 of 18 Go Back Full Screen Close Remark: 1. We note that the above proof can be adopted to prove that: There is an absolute constant C 0 > 0 such that for any function f ∈ L1 (T) whose Fourier coefficients have the same sign on the block [4j−1 , 4j ) we have ∞ X |fˆ(n)| n=1 n 0 ≤ C kf k1 + C 0 ∞ X |fˆ(−n)| n=1 n . 2. Observe that the condition that fˆ has the same sign on the block (4j−1 , 4j ) is flexible. This is because functions in Lp are in fact equivalent classes. Therefore, even if fˆ obeys our condition but for a set of measure zero, then we may modify our choice by changing the values of fˆ so that the new fˆ satisfies our condition. We should remark that this process does not effect the proof above. Real Hardy Inequality Mohammad Sababheh vol. 10, iss. 4, art. 104, 2009 Title Page Contents JJ II J I Page 17 of 18 Go Back Full Screen Close References [1] Y. KATZNELSON, An Introduction To Harmonic Analysis, John Wiley and Sons, Inc., 1968. [2] I. KLEMES, A note on Hardy’s inequality, Canad. Math. Bull., 36(4) (1993), 442–448. [3] P. KOOSIS, Conference On Harmonic Analysis in Honour of Antoni Zygmund, Volume 2, Wadsworth International Group, Belmont, California, A Division of Wadsworth, Inc., (1981), 740–748. [4] O.C. McGEHEE, L. PIGNO AND B. SMITH, Hardy’s inequality and the L1 norm of exponential sums, Annals of Math., 113 (1981), 613–618. [5] W. RUDIN, Real and Complex Analysis, 3rd edition, McGraw-Hill Book Co., New York, 1987. [6] M. SABABHEH, Two-sided probabilistic versions of Hardy’s inequality, J. Fourier Anal. and Applics., (2007), 577–587. [7] M. SABABHEH, On an argument of Körner and Hardy’s inequality, Analysis Mathematica, 34 (2008), 51–57. [8] M. SABABHEH, Hardy-type inequalities on the real line, J. Inequal. in Pure and Appl. Math., 9(3) (2008), Art. 72. [ONLINE: http://jipam.vu.edu. au/article.php?sid=1009]. Real Hardy Inequality Mohammad Sababheh vol. 10, iss. 4, art. 104, 2009 Title Page Contents JJ II J I Page 18 of 18 Go Back Full Screen Close