MAXIMAL AVERAGE ALONG VARIABLE LINES. 1. Introduction

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1 MAXIMAL AVERAGE ALONG VARIABLE LINES JOONIL KIM Abstract. We prove the L p boundedness of the maximal operator associated with a family of lines l x = {(x, x 2) t(, a(x )) : t [0, )} when a is a positive increasing function.. Introduction Let v : R 2 R 2 be a vector field. We define a maximal function along a family of lines {x tv(x) : t R} by (.) Mf(x) = sup r>0 r r 0 f(x tv(x)) dt, where x = (x, x 2 ). We assume that our vector field v depends on only one variable x given by v(x, x 2 ) = (, a(x )), where a is a real valued function defined on [x 0, ). Consider the maximal operator: (.2) M a f(x) = χ [x0, )(x ) sup r>0 r r 0 f(x t, x 2 a(x )t) dt. In [2], Carbery, Seeger, Wainger and Wright study the maximal operator M a and the corresponding Hilbert transform. They obtain the L p boundedness of those operators under the convexity assumption of a and the additional size conditions of a and a such that (.3) 0 < c a (t) I l a(t) C for t I l = {u : 2 l a (u) 2 l+ }, where C and c are constants independent of l. In this paper we prove the L p - boundedness of the maximal operator M a without the assumption of (.3) Mathematics Subect Classification. Primary42B20,42B25.

2 2 JOONIL KIM Theorem. Suppose that a : [t 0, ) [0, ) is a monotonic increasing function such that a (t 0 ) = 0. Then M a is bounded on L p (R 2 ) for < p. Remark. () Let v in (.) be a Lipschitz vector field from R 2 into the unit circle S. By A. Zygmund, it is conectured that the average of the L 2 -function along the line segment {x tv(x) : t ( r, r)} converges to itself almost every x as r approaches to zero. It is still an open problem and seems to be unknown even if the Lipschitz condition is replaced by the smoothness condition on v as proposed in [6]. (2) If v is real analytic or satisfies a certain finite type condition, then the local version of M is bounded in L p with p >, see [] and [3]. (3) In [4], N. Katz has made an interesting weak type (2,2) bound of M when v is appropriately restricted to a finite number of points in S. (4) In [5], M. Lacey and X. Li study the Hilbert transform H v along the line {x tv(x) : t }. They proved that H v extends to be a bounded map from L 2 (R 2 ) to itself under the assumption that v has + ɛ derivatives. 2. Sketch of Proof Let t = sup{t : a (t) = 0}. Then for x [t 0, t ], M a f(x) is controlled by the one dimensional Hardy-Littlewood maximal function. Thus we may assume that a (x ) > 0. Let ϕ (y) = ϕ(y/2 )/2 with ϕ C0 (/2, ). By the diadic decomposition of the interval [0, r], we define a diadic version of M a by M a f(x) = sup Z M f(x), where M f(x) = f(x t, x 2 a(x )t)ϕ (t)dt. Then it suffices to prove the L p (R 2 ) boundedness of M a in proving Theorem. By using δ(x 2 y 2 a(x )(x y )) = e iλ[x 2 y 2 a(x )(x y )] dλ, we rewrite M f(x) = e iλ[x 2 y 2 a(x )(x y )] ϕ (x y )f(y, y 2 )dy dy 2 dλ.

3 MAXIMAL AVERAGE ALONG VARIABLE LINES 3 Let us choose an even function ψ C0 (, ) such that ψ on [ /2, /2]. Let χ(ξ) = ψ(ξ/2) ψ(ξ). Then for each fixed and λ 0, = n= χ(λa (x )2 2+n ) = ψ(λa (x )2 2 ) + According to the size λa (x )2 2, we decompsoe M f(x) = M loc f(x) + n=0 n= n=0 n= M n f(x) χ(λa (x )2 2+n ). where M loc f(x) = e iλ[x 2 y 2 a(x )(x y )] ϕ (x y )ψ(λa (x )2 2 )f(y)dydλ, M n f(x) = e iλ[x 2 y 2 a(x )(x y )] ϕ (x y )χ(λa (x )2 2+n )f(y)dydλ. Thus our maximal function M a f is maorized by M loc f + M glo f where (2.) (2.2) M loc f(x) = sup Z M glo f(x) = n=0 n= loc M f(x), sup Z M n f(x). For the local part estimate M loc, we give a weak type (,) estimate by making an appropriate Vitali-type covering lemma for a family of variable parallelograms. For this, we need only the condition that a is increasing function. This will be established in Section 3. For the global part M glo, we obtain the L 2 bound by estimating M n [M n ] L 2 L 2 = O(2 c n ), M n [M n 2 ] L 2 L 2 = O(2 c 2 ), for some c > 0. To derive the L p theory, we use the bootstrap argument combined with the Littlewood-Paley decomposition on the second coordinate of the frequency variable. This will be done in Section 4. In [2], they handled each piece M by comparing the sizes of 2 and I l where I l = {u : 2 l a (u) 2 l+ }. The case 2 < I l was reduced to the case having nonvanishing rotational curvature by localization argument. The comparability condition (.3) was used for making the uniform lower bound of

4 4 JOONIL KIM the rotational curvature. For the case 2 > I l, the condition (.3) is used for controlling the size of the curvature λa (x )2 2 in the almost orthogonality estimates. However we note that our main estimate is independent of the size I l. 3. Weak type (,) estimates for M loc For each x R 2 and Z, we set H x () = {y : 2 < x y < 2, x 2 y 2 a(x )(x y ) < 2 2 a (x )}. Define the maximal operator associated with the class of the above parallelograms by (3.) N f(x) = sup f(y) dy. Z H x () H x() Let us rewrite M loc f(x) as ( ) ϕ (x y ) 2 2 a (x ) ψ x2 y 2 a(x )(x y ) (3.2) 2 2 a f(y, y 2 )dy. (x ) Since ψ in (3.2) is a rapidly decreasing function, we observe that M loc f(x) CN f(x). Thus the weak type (,)-boundedness of M loc will be proved by the following estimates: Proposition. Suppose that a is a positive increasing function, then N is weak type (, ). For the proof, we make a variant of Vitali-covering lemma corresponding to the parallelogram H x () s. Let us define two sets B x () and B x() associated with H x () such that B x () = {y : x y < 2, x 2 y 2 a(x )(x y ) < 2 2 a (x )}, B x() = {y : x y < 2 +3, x 2 y 2 a(x )(x y ) < 2 2(+5)) a (x )}.

5 MAXIMAL AVERAGE ALONG VARIABLE LINES 5 Then note that H x () B x () B x() and the size of each set is comparable with the others: (3.3) H x () = a (x )2 3, B x () = a (x )2 3+2, B x() = a (x )2 3(+5). With the above size information we need an engulfing property which is crucial in the Vitali-type covering lemma. Lemma. Let be an integer valued function on R 2 and a be a positive increasing function. If H z ((z)) H x ((x)) and H z ((z)) H x ((x)), then B z ((z)) B x((x)). Proof of Lemma. Suppose that x z. Then a (x ) a (z ), since a is a increasing function. Thus we obtain that (z) (x) in view of the inequality 2 3(z) a (z ) = H z ((z)) H x ((x)) = 2 3(x) a (x ). Since there exists y H z ((z)) H x ((x)), 2 (z) > z y = (z x ) + (x y ) > (x). Thus it follows that (z) = (x). From this and H z ((z)) H x ((x)), it also follows that a (z ) a (x ). So we have (3.4) a (z ) = a (x ) and (z) = (x) if x z. Suppose that z x. For y H z ((z)) H x ((x)), we observe that Therefore, 2 (x) > x y = (x z ) + (z y ) z y > 2 (z). (3.5) a (z ) a (x ) and (z) (x) if z x. Now we show that B z ((z)) B x((x)). Let us take w B z ((z)). By the hypothesis, there exists y H z ((z)) H x ((x)). Thus we use the definition of B z () and H z () to obtain that (3.6) 2 (z) < z w < 2 (z) and S(z, w) < 2 2(z) a (z ), 2 (z) < z y < 2 (z) and S(z, y) < 2 2(z) a (z ), 2 (x) < x y < 2 (x) and S(x, y) < 2 2(x) a (x ),

6 6 JOONIL KIM where S(u, v) = u 2 v 2 a(u )(u v ) for u, v R 2. From (3.4)-(3.6), (3.7) x w x y + y z + z w < 2 (x)+3. From (3.4)-(3.6), S(x, w) = S(z, w) S(z, y) + S(x, y) + (a(x ) a(z ))(y w ) (3.8) < a (x )2 2((x)+3) + < a (x )2 2((x)+4). x z a (t)dt(y w ) Therefore from (3.7) and (3.8), w B x((x)). By using Lemma we can prove a variant of the Vitali-covering lemma: Lemma 2. Suppose that the hypothesis of Lemma is true. Let {B xα ((x α ))} N α= be a class of N parallelograms. Then there exists a subsequence {x αk } M k= of {x α } N α= satisfying α=n ν=m (3.9) B xα ((x α )) Bx αk ((x αk )), (3.0) α= k= H xαk ((x αk )) H xαl ((x αl )) = if k l. Proof of Lemma 2. Choose B xα ((x α )) from B = {B xα ((x α ))} N α= so that H xα ((x α )) is one of the largest among H = {H xα ((x α ))} N α=. For m 2, we select B xαm ((x αm )) from the class B m = B m {B xα ((x α )) : H xα ((x α )) H xαm ((x αm )) } so that H xαm ((x αm )) is one of the largest among the class H m = H m {H xα ((x α )) : H xα ((x α )) H xαm ((x αm )) }. Our selection is done in M (which is less then equal to N) steps. Take any B xα ((x α )) with α N. Then H xα ((x α )) meets some H xαm ((x αm )) and H xα ((x α )) H xα m ((x α m )), otherwise it would be a candidate of M +-th step. By Lemma or Lemma 2, B xα ((x α )) B x α m ((x α m )). This proves (3.9). In each step we have chosen H xαm ((x αm )) satisfying (3.0). Proof of Proposition. Let D be a compact set contained in {x : N f(x) > λ}. We show that D C λ f L (R 2 ).

7 MAXIMAL AVERAGE ALONG VARIABLE LINES 7 By (3.), for each x D there exist (x) Z such that (3.) λ < f(y) dy. H x ((x)) H x ((x)) Note that D x D B x((x)) where (x) satisfying (3.). From this and compactness of the set D, there exist x,, x N D for some N such that (3.2) D α=n α= B xα ((x α )). So from (3.3),(3.),(3.2) and Lemma 3, we have α=n D α= ν=m 2 5 B xα ((x α )) k= ν=m k= H xαk ((x αk )) 25 λ B x αk ((x αk )) ν=m k= H xαk ((x αk )) f(y) dy 25 λ f L (R 2 ). The first inequality follows from (3.2), the second inequality from (3.9), the third from (3.3), the fourth from (3.) and the last from (3.0). Remark. Our proof of Lemma is based on the increasing property of a combined with the geometry of H x () {y : 2 > x y > 2 }. If one define a parallelogram as H x() {y : 2 > y x > 2 }, we are not able to use the property such as (3.4) and (3.5) which are crucial factor for the proof of our Vitali-type covering lemma. For this reason, the case including the interval [ r, 0] in (.2) is not directly covered by Theorem. It would be interesting to extend our result to the maximal function defined on the full interval [ r, r] in (.2). 4. L p estimate for M glo In proving the L p boundedness of M glo, we show that there exists a constant C > 0 independent of n such that for n 0: (4.) (4.2) ( Z M n f 2) /2 L (R 2 2 ) C2 n 8 f L (R 2 2 ) sup Z M n f L p (R 2 ) C f L p (R 2 ) for < p 2.

8 8 JOONIL KIM 4.. Proof of (4.). Fix λ R and n 0, Z. We define an operator T,n λ on L2 (R ) by T,nf(x) λ = e iλa(x)(x y) χ(λa (x)2 2+n )ϕ (x y)f(y)dy, where x, y R. Then by dy 2 integration in the definition of M n f(x, x 2 ), we can observe that M n f(x, x 2 ) = [T λ,n f λ 2 (x )] (x 2 ). where is the inverse Fourier transform with respect to λ variable, and 2 f λ (x ) = e iλx 2 f(x, x 2 )dx 2. So by applying the Plancherel theorem on the second variable we have T,n λ f 2 λ 2 L (R 2 ) dλ. M n f 2 L 2 (R 2 ) = Thus by repetition of the Plancherel Theorem on the second variable, it suffices to show that for each fixed λ ( Z T,nf λ 2) /2 L (R 2 ) C2 n 8 f L (R 2 ). By using the Cotlar-Stein Lemma combined with the duality, we have only to check the following two estimates: (4.3) (4.4) T λ,n[t λ,n] L 2 (R ) L 2 (R ) C2 n 2, T λ,n[t λ 2,n] L (R 2 ) L (R 2 ) C2 2 4 where C > 0 is independent of λ. We write T λ,n[t λ 2,n] f(x) = K, 2 (x, z)f(z)dz where K, 2 (x, z) is (4.5) e iψ(x,z,y,λ) χ(λa (x)2 2+n )χ(λa (z)2 22+n )ϕ (x y)ϕ 2 (z y)dy, where Ψ(x, z, y, λ) = λ(a(x)(x y) a(z)(z y)). Note that we shall use x, y and z as real numbers in the proof of (4.3) and (4.4).

9 MAXIMAL AVERAGE ALONG VARIABLE LINES 9 Proof of (4.3). Let us now split our kernel K, (x, z) = K (x, z) + K 2 (x, z) so that K (x, z) = χ A (x, z)k, (x, z) K 2 (x, z) = χ A c(x, z)k, (x, z) where χ A is a characteristic function supported on the set A = {(x, z) : x z 2 n 2 2 }. On the support of A c = R 2 A, the derivative of the phase function Ψ with respect to y is bounded below such that (4.6) λ(a(x) a(z)) = λ x z a (u)du 2 n /2 2. Thus the integration by parts yields that K 2 (x, z) dx C2 n 2, K 2 (x, z) dz C2 n 2. By measuring the support of A we obtain that K (x, z) dx C2 n 2, K (x, z) dz C2 n 2. Hence the above four estimates prove (4.3). Proof of (4.4). Assume without loss of generality 2. Let us define a set B: B = {(x, z) : a(x) a(z) a (x) ( 2 ) }. Let us now split our kernel K, 2 (x, z) = L (x, z) + L 2 (x, z) so that L (x, z) = χ B (x, z)k, 2 (x, z) L 2 (x, z) = χ B c(x, z)k, 2 (x, z). On the support of B c, the derivative of our phase function Ψ with respect to y is λ(a(x) a(z)) > λa (x) ( 2 ) 2 2 λa (x) ,

10 0 JOONIL KIM which is bigger then n So the integration by parts yields that L 2 (x, z) dx C2 2 2, (4.7) L 2 (x, z) dz C From the fact that λa (x)2 2+n λa (z)2 22+n, we have a (z) a (x)2 2( 2 ). Therefore for each fixed x, we measure the support of z so that {z : (x, z) B} C a (x) ( 2 ) a (x)2 2( = 2 2 ( 2 ) 2. 2 ) Using this we can estimate L (x, z) dx C, (4.8) L (x, z) dz C Hence (4.4) follows from (4.7) and (4.8) Proof of (4.2). Let I k = {t : 2 k < a (t) 2 k } and define P k f(x, x 2 ) = χ Ik (x )f(x, x 2 ) where χ Ik is the characteristic function supported on the set I k. Then (4.9) M n f(x) = χ Ik 2 (x )P k 2 M n f(x) k= where P k 2 M n f(x) is χ Ik 2 (x ) ϕ (x y )e iλ[x 2 y 2 a(x )(x y )] χ(2 2+n a (x )λ)dλf(y)dy. On the support of the kernel, 2 k < a (x )2 2 2 k and 2 n < λa (x )2 2 2 n. So 2 n k < λ < 2 n k+. This implies that (4.0) P k 2 M n f(x) = P k 2 M n L 2 n+k f(x), where L 2 n+k f(ξ, ξ 2 ) = ( χ(2 n+k ξ 2 ) + χ(2 n+k ξ 2 ) + χ(2 n+k+ ξ 2 ) ) f(ξ, ξ 2 ). Thus from (4.9) and (4.0) combined with the fact k χ I k 2 (x ) 2 2, sup Z M n f(x) ( k Z sup P k 2 M n (L 2 n+k f)(x) 2) /2 (4.).

11 MAXIMAL AVERAGE ALONG VARIABLE LINES By using 2 2 a (x ) 2 k we can see that P k 2 M n f(x) is maorized by χ Ik 2 (x ) ϕ (x y ) 2 ψ ( x 2 y 2 a(x )(x y )) f(y k+n 2 k+n, y 2 )dy. For each fixed k and n, we define the maximal function where and N n,k f(x) = sup N n,k f(x) N n,k f(x) = χ I k 2 (x ) f(y, y 2 ) dy R x () R x() R x () = {y : 2 < x y 2, x 2 y 2 a(x )(x y ) < 2 k+n }. Observe from the fact that ψ is rapidly decreasing function, sup P k 2 M n f(x) CN n,k f(x). k Thus by the Littlewood-Paley inequality for the square sum of L 2 n+kf in (4.), we see that it suffices to show that there exist C independent of n such that for < p 2: (4.2) ( k Z N n,k ( f k ) 2) /2 L p (R 2 ) C ( k Z f k 2) /2 L p (R 2 ). We shall show that for < p <, there is a constant C independent of n, k such that:, (4.3) N n,k f L p (R 2 ) C f L p (R 2 ). Now we show that (4.3) implies (4.2). If we have (4.3), then (4.2) follows for p = 2. This and the positivity of the kernel of the operator N n,k imply that {N n,k f k } L 2 (l ) C {f k } L 2 (l ). From (4.3) we have {N n,k f k } L p (l p ) C {f k } L p (l p ) for any p >. By interpolation of above two vector valued norm space, we have (4.2) for 4 3 < p 2. Repeat this process until we obtain (4.2) for any < p 2.

12 2 JOONIL KIM Now let us turn to show (4.3). We have from the support condition for each fixed n, k, (4.4) N n,k f(x) N n,k f R (x) where f R (y) = χ R (y)f(y) and R = x I 2k R x (). From (4.4), n,k (4.5) N f R (x) p dx dx 2 N n,k f p L p (R 2 ) Z Z f R (x) p dx dx 2. By (4.5), we see that (4.3) is proved if we show that (4.6) R R 2 = if 2 > 0. Proof of (4.6). Without loss of generality > It suffices to show that for 2 k 2 < a (x ) 2 k 2 and 2 k 2 2 < a (z ) 2 k 2 2 : R x ( ) R z ( 2 ) =. Since a is increasing, it follows that x z. Assume that y R x ( ) R z ( 2 ), then 2 < x y 2 and 2 2 < z y 2 2. This means that y is on the left to x with distance 2 and y is on the left to z with distance 2 2. This is a contradiction with the fact that x z and 2 > Hence (4.6) is proved. References [] J. Bourgain, A remark on the maximal function associated to an analytic vector field, Analysis at Urbana, edited by E. Berkson, T. Peck and J. Uhl, Cambridge University Press, 989, pp [2] A. Carbery, A. Seeger, S. Wainger, J. Wright, Classes of singular integral operators along variable lines. J. Geom. Anal. 9 (999), no. 4, [3] M. Christ, A. Nagel, E.M. Stein, S. Wainger, Singular and maximal Radon transforms: Analysis and geometry, Ann of Math. 50 (2000) [4] N. Katz, A partial result on Lipschitz differentiation. Harmonic analysis at Mount Holyoke (South Hadley, MA, 200), , Contemp. Math., 320, Amer. Math. Soc., Providence, RI, [5] M. Lacey, X. Li, On the Hilbert transform and C +ɛ families of lines, Ann of Math, to appear.

13 MAXIMAL AVERAGE ALONG VARIABLE LINES 3 [6] E.M. Stein, S. Wainger, Problems in harmonic analysis related to curvature, Bull. Amer. Math. Soc. 84 (978), Chung-Ang University Math Department, Huksuk-dong 22, Dongak-ku, Seoul , Republic of Korea address: ikim@cau.ac.kr