THE JOHN-NIRENBERG INEQUALITY WITH SHARP CONSTANTS

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1 THE JOHN-NIRENBERG INEQUALITY WITH SHARP CONSTANTS ANDREI K. LERNER Abstract. We consider the one-dimensional John-Nirenberg inequality: {x I 0 : fx f I0 > α} C 1 I 0 exp C 2 α. A. Korenovskii found that the sharp C 2 here is C 2 = 2/e. It is shown in this paper that if C 2 = 2/e, then the best possible C 1 is C 1 = 1 2 e4/e. 1. Introduction Let I 0 R be an interval and let f LI 0. Given a subinterval I I 0, set f I = 1 1 f and Ωf; I = fx f I I I I I dx. We say that f BMOI 0 if sup Ωf; I <. The classical I I 0 John-Nirenberg inequality [1] says that there are C 1, C 2 > 0 such that for any f BMOI 0, {x I 0 : fx f I0 > α} C 1 I 0 exp C 2 α α > 0. A. Korenovskii [4] see also [5, p. 77] found the best possible constant C 2 in this inequality, namely, he showed that C 2 = 2/e: 1.1 {x I 0 : fx f I0 > α} C 1 I 0 exp 2/e α α > 0, and in general the constant 2/e here cannot be increased. A question about the sharp C 1 in 1.1 remained open. In [4], 1.1 was proved with C 1 = e 1+2/e = The method of the proof in [4] was based on the Riesz sunrise lemma and on the use of nonincreasing rearrangements. In this paper we give a different proof of 1.1 yielding the sharp constant C 1 = 1 2 e4/e = Mathematics Subject Classification. 42A05,42B35. Key words and phrases. BM O, John-Nirenberg inequality, sharp constants. 1

2 2 ANDREI K. LERNER Theorem 1.1. Inequality 1.1 holds with C 1 = 1 2 e4/e, and this constant is best possible. We also use as the main tool the Riesz sunrise lemma. But instead of the rearrangement inequalities, we obtain a direct pointwise estimate for any BM O-function see Theorem 2.2 below. The proof of this result is inspired and close in spirit by a recent decomposition of an arbitrary measurable function in terms of mean oscillations see [2, 6]. We mention several recent papers [7, 8] where sharp constants in some different John-Nirenberg type estimates were found by means of the Bellman function method. 2. Proof of Theorem 1.1 We shall use the following version of the Riesz sunrise lemma [3]. Lemma 2.1. Let g be an integrable function on some interval I 0 R, and suppose g I0 α. Then there is at most countable family of pairwise disjoint subintervals I j I 0 such that g Ij = α, and gx α for almost all x I 0 \ j I j. Observe that the family {I j } in Lemma 2.1 may be empty if gx < α a.e. on I 0. Theorem 2.2. Let f BMOI 0, and let 0 < γ < 1. Then there is at most countable decreasing sequence of measurable sets G k I 0 such that G k min k, 1 I 0 and for a.e. x I 0, 2.1 fx f I0 χ Gk x. Proof. Given an interval I I 0, set EI = {x I : fx > f I }. Let us show that there is at most countable family of pairwise disjoint subintervals I j I 0 such that j I j γ I 0 and for a.e. x I 0, 2.2 f f I0 χ EI0 χ EI 0 + j f f Ij χ EIj. We apply Lemma 2.1 with g = f f I0 and α = f. One can assume that α > 0 and the family of intervals {I j } from Lemma 2.1 is non-empty since otherwise 2.2 holds trivially only with the first term on the right-hand side. Since g Ij = α, we obtain I j = 1 f f I0 dx 1 f f I0 dx α j I j α {x I 0 :fx>f I0 } j = 1 2α Ωf; I 0 I 0 γ I 0.

3 THE JOHN-NIRENBERG INEQUALITY 3 Further, f Ij = f I0 + α, and hence f f I0 = f f I0 χ I0 \ j I j + αχ j I j + j f f Ij χ Ij. This proves 2.2 since f f I0 α a.e. on I 0 \ j I j. The sum on the right-hand side of 2.2 consists of the terms of the same form as the left-hand side. Therefore, one can proceed iterating 2.2. Denote Ij 1 = I j, and let Ij k be the intervals obtained after the k-th step of the process. Iterating 2.2 m times yields f f I0 χ EI0 m χ EI k j x + j i f f I m+1χ i EI m+1 i where Ij 0 = I 0. If there is m such that for any i each term of the second sum is bounded trivially by f χ EI m+1 i, we stop the process, and we would obtain the finite sum with respect to k. Otherwise, let m. Using that i I m+1 i γ l I m l γ m+1 I 0, we get that the support of the second term will tend to a null set. Hence, setting E k = j EI k j, for a.e x EI 0 we obtain 2.3 f f I0 χ EI0 f χ EI0 x + k=1 χ Ek x. Observe that EI j = {x I j : fx > f I0 +α} EI 0. From this and from the above process we easily get that E k+1 E k. Also, E k j I k j, and hence E k γ k I 0. Setting now F I = {x I : fx f I }, and applying the same argument to f I0 fχ F I, we obtain 2.4 f I0 fχ F I0 f χ F I0 x + k=1 χ Fk x, where F k+1 F k and F k γ k I 0. Also, F k E k =. Therefore, summing 2.3 and 2.4 and setting G 0 = I 0 and G k = E k F k, k 1, we get 2.1. Proof of Theorem 1.1. Let us show first that the best possible C 1 in 1.1 satisfies C e4/e. It suffices to give an example of f on I 0 such that for any ε > 0, 2.5 {x I 0 : fx f I0 > 21 ε } = I 0 /2.

4 4 ANDREI K. LERNER Let I 0 = [0, 1] and take f = χ [0,1/4] χ [3/4,1]. Then f I0 = 0. Hence, 2.5 would follow from = 1/2. To show the latter fact, take an arbitrary I I 0. It is easy to see that computations reduce to the following cases: I contains only 1/4 and I contains both 1/4 and 3/4. Assume that I = a, b, 1/4 I, and b < 3/4. Let α = 1 a and 4 β = b 1. Then f 4 I = α/α + β and Ωf; I = 2 f f I = 2αβ α + β α + β 1/2 2 {x I:f>f I } with Ωf; I = 1/2 if α = β. Consider the second case. Let I = a, b, a < 1/4 and b > 3/4. Let α be as above and β = b 3. Then 4 Since Ωf; I = 2 α + β + 1/2 sup 0 α,β 1/4 {x I:f>f I } f f I = 4α4β + 1 2α + 2β = 1/2, 4α4β + 1 2α + 2β this proves that = 1/2. Therefore, C e4/e. Let us show now the converse inequality. Let f BMOI 0. Setting ψx = χ G k x, where G k are from Theorem 2.2, we have {x I 0 : ψx > α} = G k χ [k,k+1 α Hence, by 2.1, I 0 min1, k χ [k,k+1 α. {x I 0 : fx f I0 > α} {x I 0 : ψx > α/ } I 0 min k, 1χ [k,k+1 α/. This estimate holds for any 0 < γ < 1. Therefore, taking here the infimum over 0 < γ < 1, we obtain 2/e {x I 0 : fx f I0 > α} ϕ α I 0, where ϕξ = inf 0<γ<1 min k, 1χ [k,k+1 γeξ.

5 THE JOHN-NIRENBERG INEQUALITY 5 Thus, the theorem would follow from the following estimate: 2.6 ϕξ 1 2 e 4 e ξ ξ > 0. It is easy to see that ϕξ = 1 for 0 < ξ 2/e, and in this case 2.6 holds trivially. Next, ϕξ = 2 for 2/e ξ 4/e. Using that the eξ function e ξ /ξ is increasing on 1, and decreasing on 0, 1, we get max 2e ξ /eξ = 1 ξ [2/e,4/e] 2 e4/e, verifying 2.6 for 2/e ξ 4/e. For ξ 1 we estimate ϕξ as follows. Let ξ [m, m + 1, m N. Taking γ i = i/eξ for i = m and i = m + 1, we get m, m + 1 m+1 ϕξ 2 min 2.7 m eξ eξ m = 2 mχ[m,ξm]ξ + eξ m + 1 eξ m+1χ[ξm,m+1ξ, where ξ m = 1 m+1 m+1. Using that the function e ξ /ξ m is increasing e m m on m, and decreasing on 0, m, by 2.7 we obtain that for ξ [m, m + 1, m m+1 me ϕξe ξ ξ 2 m e 1 e 1+1/mm = 2 c eξ m 1 + 1/m m m. Let us show now that the sequence {c m } is decreasing. This would finish the proof since c 1 = 1 2 e4/e. Let ηx = 1 + 1/x x and νx = e ηx/e /ηx x+1. Then c m = 2νm and hence it suffices to show that ν x < 0 for x 1. We have ν x = νx log e ηx 1 ηx/e log1 + 1/x1+x. ηx e ηx Since ηx1 + 1/x > e, we get µx = > x. From this and from the fact that the function 1 + 1/x 1+x is decreasing we obtain 1 e/ηx 1 ηx/e = 1 + 1/µx 1+µx < 1 + 1/x 1+x, which is equivalent to that ν x < 0. References [1] F. John and L. Nirenberg, On functions of bounded mean oscillation, Comm. Pure Appl. Math , [2] T. Hytönen, The A 2 theorem: remarks and complements, preprint. Available at [3] I. Klemes, A mean oscilation inequality, Proc. Amer. Math. Soc. 93, no. 3, 1985,

6 6 ANDREI K. LERNER [4] A.A. Korenovskii, The connection between mean oscillations and exact exponents of summability of functions, Mat. Sb , no. 12, in Russian; translation in Math. USSR-Sb , no. 2, [5] A.A. Korenovskii, Mean oscillations and equimeasurable rearrangements of functions. Lect. Notes Unione Mat. Ital., vol. 4, Springer/UMI, Berlin/Bologna, [6] A.K. Lerner, A pointwise estimate for local sharp maximal function with applications to singular integrals, Bull. London Math. Soc., , no. 5, [7] L. Slavin and V. Vasyunin, Sharp results in the integral-form John-Nirenberg inequality, Trans. Amer. Math. Soc , no. 8, [8] V. Vasyunin and A. Volberg, Sharp constants in the classical weak form of the John Nirenberg inequality, preprint. Available at Department of Mathematics, Bar-Ilan University, Ramat Gan, Israel address: aklerner@netvision.net.il