MEAN AND WEAK CONVERGENCE OF FOURIER-BESSEL SERIES by J. J. GUADALUPE, M. PEREZ, F. J. RUIZ and J. L. VARONA

Transcription

1 MEAN AND WEAK CONVERGENCE OF FOURIER-BESSEL SERIES by J. J. GUADALUPE, M. PEREZ, F. J. RUIZ and J. L. VARONA ABSTRACT: We study the uniform boundedness on some weighted L saces of the artial sum oerators associated to Fourier-Bessel series, obtaining necessary and sufficient conditions for this boundedness in terms of the weights. On the other hand, we study the weak and restricted weak tye of the artial sum oerators in the end oints of the interval of strong boundedness.. Introduction. Let J α be the Bessel function of order α > 1, and j α n (x) = 1/ J α+1 (α n ) 1 J α (α n x) (where {α n } is the increasing sequence of the zeroes of J α ) the Bessel system of order α, which is orthonormal and comlete in L ((, 1); xdx), and therefore the Fourier series of a function f L ((, 1); xdx) with resect to this system converges to f in the L -norm. The next ste is to ask for which (1, ), the above convergence is true. The roblem can be osed, by the Banach-Steinhauss theorem, in terms of the uniform boundedness on L ((, 1); xdx) of the artial sum oerators S α n f(x) = n c k jk α (x), c k = c k (f) = k=1 j α k (y)f(y)ydy. The roblem was solved by Wing [13] (in the case α 1/) and Benedek and Panzone [1] (general case). Moreover, these authors obtained some weighted norm estimates of the form S α n f(x)u(x) C f(x)v (x) (.1) where C is a constant indeendent of n, U(x) = V (x) are functions of radial tye x a and by. we denote the L ((, 1); xdx)-norm. The first objective of this aer is to extend inequality (.1) to a wider class of weights U(x) and V (x), not necessarily the same. We obtain sufficient conditions in terms of A - conditions (roositions 1 and ) and we solve comletely the roblem (theorem 1) when the weights are of the form U(x) = x a (1 x) b m k=1 x x k b k ; < x 1 <... < x m < 1; a, b, b k R. (.) AMS (1991) Subject Classification: 4C1 Keywords and hrases: Convergence of Fourier series, Bessel functions, weighted norm inequalities, A -weights, weak and restricted weak tye. Research suorted by DGICYT under grant PB C-. 1

2 The inequality (.1) is closely related to the roblem of the boundedness of the Hilbert transform with two weights. In this sense, the use of airs of weights in the Muckenhout A classes (see [9], [1]) will be one of the main ingredients in the roofs. We obtain also checkable necessary conditions of integrability on U, V from the inequality (.1) (theorem ) in the sirit of Máté, Nevai and Totik s conditions (see [8]) for systems of orthogonal olynomials on [ 1, 1]. On the other hand, we shall study a roblem related to the end oints of the interval of mean convergence: by interolation, the range of, s such that the uniform boundedness (.1) holds is always an interval. When U(x) = V (x) = 1 (for examle), this interval is oen, say (, 1 ), and then the natural question is if Sn α are uniformly of weak tye (, ) or/and ( 1, 1 ). That is, if we denote g L ((,1);U(x) xdx) = {su λ U(x) xdx} 1/ λ> {x: g(x) >λ} then the question is if S α n f L ((,1);U(x) xdx) C f(x)v (x) (.3) with a constant C indeendent of n, and = or 1. This end oint roblem has been studied by different authors for another oerators (see [], [3], [6], [11]), but in the case of Fourier-Bessel series it seems to be new, even in the unweighted case. In theorem 3, we give necessary conditions on U, V for (.3) to be true. In the articular case α 1/ and U(x) = V (x) of the form (.), we show that (.3) is false for both end oints (theorem 4). Finally, when α 1/ and U(x) = V (x) = 1 we show (theorem 5) that S α n are uniformly of restricted weak tye at the end oints, i.e., (.3) is true, but only for functions f = χ E where E is a measurable set contained in (, 1). The organization of the aer is as follows: in section 1, we exhibit the main results with the notations which are needed to make them understandable. In section, we collect several technical lemmas concerning A weights and the boundedness of some oerators associated to the roblem which we shall use for the roofs of the main results which will be given in section Main results. Let 1 < < and a < b. A (a, b) will stand for the Muckenhout classes (see [4], [9]) consisting of those airs of nonnegative functions (u, v) on (a, b) such that ( I 1 u)( I 1 v / ) / C for every interval I (a, b), I I where + =. We say that (u, v) A δ (a, b) (where δ > 1) if (u δ, v δ ) A (a, b). It is clear from Holder s inequality that A δ (a, b) A (a, b). We say that a sequence {(u n, v n )} n=1 A δ (a, b) uniformly if (u n, v n ) A δ (a, b), n and the constant aearing in the definition is indeendent of n.

3 In the inequality S α n f(x)u(x) C f(x)v (x) we shall always mean that C is a constant indeendent of f and n (and the same for the weak L -norm). In our two first results, we distinguish the cases α 1/ and 1 < α < 1/. The reason for that is the different assymtotic behavior of the Bessel functions in each case. Proosition 1. (Case α 1/). Let 1 < <. Let U, V be weights on (, 1). If (U(x) x 1 /, V (x) x 1 / ) A δ (, 1) for some δ > 1 (δ = 1 if U = V ), then S α n f(x)u(x) C f(x)v (x). Proosition. (Case 1 < α < 1/). Let 1 < < and M n = (α n + α n+1 )/. Let U, V be weights on (, 1). If there exists δ > 1 (δ = 1 if U = V ) such that ((M 1 n + x) (α+1/) U(x) x α+1, (M 1 n + x) (α+1/) V (x) x α+1 ) A δ (, 1), (1.1) ((Mn 1 + x) (α+1/) U(x) x α +1, (Mn 1 + x) (α+1/) V (x) x α +1 ) A δ (, 1) (1.) uniformly, then Sn α f(x)u(x) C f(x)v (x). Note. When U = V it suffices δ = 1 because from reverse Holder s inequality it follows that (u, u) A (u, u) A δ for some δ > 1 (see [4]). Alying both roositions to articular weights of the form (.), we have Theorem 1. Let α > 1, 1 < < and the weights U(x) = x a (1 x) b m k=1 m x x k b k, V (x) = x A (1 x) B x x k B k, where < x 1 <... < x m < 1 and a, A, b, B, b k, B k R. If the conditions: B b, B < 1, 1 < b; k=1 B k b k, B k < 1, 1 < b k (1 k m); A a, a 4 + A 4 < a A 4 are fullfilled, then S α n f(x)u(x) C f(x)v (x). (1.3) + min{ 1 4, α + 1 } (1.4) The following general result about necessary conditions will imly, in articular, that (1.3), (1.4) are best ossible. Theorem. If the inequality Sn α f(x)u(x) C f(x)v (x) holds, then U, V must satisfy the conditions U(x) x α+1 dx < +, V (x) x α +1 dx < +, U(x) x 1 / dx < + (1.5) V (x) x 1 / dx < + (1.6) U(x) C V (x) a.e. (1.7) Corollary 1. With the same notation as in theorem 1, the conditions (1.3) and (1.4) are also necessary for the uniform boundedness S α n f(x)u(x) C f(x)v (x). Regarding weak boundedness, we begin with a version of theorem in this case: 3

4 Theorem 3. Let α > 1 and 1 < <. If the inequality S α n f L ((,1);U(x) xdx) C f(x)v (x) holds, then U, V must satisfy conditions (1.6), (1.7) and su λ U(x) xdx < +, λ> {x:x α >λ} su λ U(x) xdx < +. (1.5) λ> {x:x 1/ >λ} From this theorem, straightforward calculations show that the weak boundedness (.3) fails in the lower end oint of the interval of strong boundedness given by (1.3) and (1.4). However, theorem 3 does not work in the uer end oint. In this case, we have obtained that the weak boundedness is also false (in the articular case α 1/ and U(x) = V (x) = x a (1 x) b m k=1 x x k b k ), but the roof is more intricate, being necessary a more detailed analysis of the artial sum oerators. Something very similar occurs when one considers Jacobi-Fourier series (see [], [6]). The result can be stated as follows: Theorem 4. Let α 1/, 1 < < and U(x) = x a (1 x) b m k=1 x x k b k. Then, the following are equivalent: (i) S α n f L ((,1);U(x) xdx) C f(x)u(x) (ii) 1 + a 1 < 1 4 ; 1 < b < 1; 1 < b k < 1 (1 k m). Finally, following the method of [, theorem 1], we establish this (,)-restricted weak tye: Theorem 5. Let α 1/ and = 4 or = 4/3.Then for every measurable set E (, 1).. Technical lemmas. S α n (χ E ) L ((,1);xdx) C χ E We denote by K α n (x, y) = n k=1 jα k (x)jα k (y) the kernel of the oerator Sα n, that is, S α n f(x) = K α n (x, y)f(y)ydy. It is known the following decomosition of the kernel (see [1], [13]): M n Kn α (x, y) = J α (M n x)j α+1 (M n y) (y x) + J α(m n y)j α+1 (M n x) (x y) + M n M n M n + J α (M n x)j α+1 (M n y) (y + x) + J α(m n y)j α+1 (M n x) (x + y) O(1)(xy) α + O(1) (xy) 1/ x y = Kn,i(x, α y). 4 i=1

5 So, we can decomose S α n f(x) = 6 Sn,if(x) α = i=1 We shall use the following notations Hf(x) =.v. f(y) dy; Jf(x) = x y 6 i=1 The main tools for estimating the oerators S α n,i f are: K α n,i(x, y)f(y)ydy. f(y) 1 x y dy; J f(x) = f(y) x + y dy. J α (z) = z α α Γ(α + 1) 1 + O(z α+ ), z (.1) J α (z) = 1/ (πz) 1/ cos(z απ π 4 ) + O(z 3/ ) z (.) for α > 1 (see [1]) and the following results: J α (α n x) x dx 1 πα n, n (.3) Lemma 1. Let 1 < <. If (u, v) A δ (, 1) for some δ > 1, then where T = H, J or J. T f(x) L ((,1);u(x)dx) C f(x) L ((,1);v(x)dx). Proof. The result for H is due to Neugebauer [1]. In fact, Neugebauer showed that under the hyothesis there exists a w A (i.e. (w, w) A ) and constants C 1, C such that C 1 u(x) w(x) C v(x) a.e., and then we can aly that H is bounded on L ((, 1); w(x)dx) if and only if w A (see [7]). The last fact is also true for Mf, being M the Hardy-Littlewood maximal oerator (see [9]). Let now T 1 f(x) = 1 x x f(y)dy the Hardy function and T f(x) = x Then, it is easy to see that f(y) y dy its adjoint. J f(x) T 1 ( f )(x) + T ( f )(x); T 1 f(x) J ( f )(x); T f(x) J ( f )(x). So, the weighted boundedness for J is equivalent to the one for T 1 and T. The oerator T 1 is controlled by M; and for its adjoint T we use the duality roerty for A -classes: (u, v) A if and only if (v /, u / ) A. So, J is controlled by oerators which are bounded under the hyothesis. Finally, we can deal with the oerator J in the same way as J by making the change of variables X = 1 x, Y = 1 y which does not change A conditions. Note. The constants C aearing in the weighted norm inequalities deend only on the constants aearing in the definition of the A classes. Thus, for sequences of weights belonging to A δ uniformly, the corresonding weighted norm inequalities are also uniform. For the roof of theorem 1, we need to analyze when certain sequences of weights of radial tye belong to A classes uniformly. For that, we give the 5

6 Lemma. Let r, s, R, S R, 1 < < and let {x n } n=1 be a sequence of ositive numbers such that lim x n =. Then (x r (x + x n ) s, x R (x + x n ) S ) A δ (, 1) for some δ > 1 1 < r; 1 < r + s; R < 1; R + S < 1; R r; R + S r + s (.4) Proof. Straightforward calculations show that for a < b 1, c 1, γ > 1, γ + µ > 1, there exists a constant K, deending only on γ, µ, such that b { x γ (x + c) µ Kb dx γ+µ (b a), if c b; Kb γ c µ (b a), if b c. and then (.4) imlies ( b a a x r (x + c) s dx)( b a (x R (x + c) S ) / dx) / K(b a), that is, (x r (x + c) s, x R (x + c) S ) A (, 1) uniformly on c. Now, these airs of weights belong to A δ (, 1), for some δ > 1 because the inequalities 1 < r; 1 < r + s; R < 1; R + S < 1 are strict. Recirocally, the A condition imlies that x r (x + x n ) s and (x R (x + x n ) S ) 1/ 1 are integrables on (, 1) and then 1 < r, R < 1. Taking limits on n we get the integrability of x r+s and (x R+S ) 1/ 1 (even more, the air (x r+s, x R+S ) A (, 1)).This gives 1 < r + s and R + S < 1. Finally, by using that if (u, v) A (, 1) then u v a.e., we have R r, R + S r + s. On the other hand, we also need to deal with roducts of weights in A classes. In this sense it is not difficult to rove Lemma 3. Let {u n (x)} n=, {v n (x)} n=, {U n (x)} n=, {V n (x)} n= be sequences of weights on a finite interval (a, b). Let c (a, b) and ɛ >. Assume that there exist ositive constants C i, (i = 1, ) such that C 1 U n (x) C, C 1 V n (x) C on (a, c + ɛ) and C 1 u n (x) C, C 1 v n (x) C on (c ɛ, b), n. If {(u n, v n )} A (a, c) uniformly and {(U n, V n )} A (c, b) uniformly then {(u n U n, v n V n )} A (a, b) uniformly. In articular, the iterated use of lemma 3 and the fact that ( x c α, x c β ) A δ (a, b) for some δ > 1 (where a c b) if and only if 1 < α, β < 1 and β α gives Lemma 4. Let α > 1 and the weights U(x) = x a (1 x) b m k=1 m x x k b k, V (x) = x A (1 x) B x x k B k where < x 1 <... < x m < 1 and a, A, b, B, b k, B k R. Then, (U, V ) A δ (, 1) if and only if A a, 1 < a, A < 1; B b, 1 < b, B < 1; k=1 B k b k, 1 < b k, B k < 1 (1 k m). The following lemma will be crucial in the study of necessary conditions for the mean and weak convergence of Fourier Bessel series. It is arallel to the one given by Máté, Nevai and Totik [8, theorem ] in the context of orthogonal olynomials. 6

7 Lemma 5. Let α > 1. Let h(x) be a Lebesgue measurable nonnegative function on [ 1, 1], {r n } a sequence such that lim n r n = + and 1 <. Then lim n rn 1/ J α (r n x) h(x)dx M h(x) x / dx where M is a ositive constant indeendent of h(x) and {r n }. In articular, (with r n = α n ) lim jn α (x) h(x)dx M n h(x) x / dx. (.5) Proof. It can be seen in [5]. There, the roof is given only in the case r n = α n but the arguments are the same for any sequence {r n } such that lim n r n = Proofs of the main results. Proof of roosition 1. We use the estimate J α (z) C z 1/, z (, ), α 1/ (3.1) which can be deduced from (.1) and (.). We are going to rove that all the oerators Sn,i α (i = 1 to 6) are uniformly bounded. = C C S α n,1f(x)u(x) xdx = f(y)y 1/ (ym n ) 1/ J α+1 (ym n ) dy (xm n ) 1/ J α (xm n ) U(x) x 1 / dx (y x) f(y)y 1/ (ym n ) 1/ J α+1 (ym n ) dy U(x) x 1 / dx (y x) f(x)x 1/ (xm n ) 1/ J α+1 (xm n ) V (x) x 1 / dx C f(x)v (x) xdx, where we have used (3.1), the hyothesis on U, V and lemma 1 for the Hilbert transform. The boundedness of S α n, is comletely similar and the oerators S α n,3, S α n,4 can be treated as the revious oerators but using the oerator J instead of the Hilbert transform. For the boundedness of S α n,5 we only need Holder s inequality and the fact that u and v 1/( 1) are integrable when (u, v) A (, 1). Finally, it is immediate that S α n,6f(x)u(x) C f(x)v (x) 7

8 if and only if Jf(x) L ((,1);U(x) x 1 / dx) C f(x) L ((,1);V (x) x 1 / dx) and then we can aly lemma 1 for J. Proof of roosition. Here, we shall use that there exists a constant C such that z (, ) z 1/ J α (z) C (1 + z α+1/ ), z 1/ J α+1 (z) C (1 + z α+1/ ) 1 (3.) which follows from (.1) and (.). For α < 1/, we have and so, condition (1.1) is equivalent to 1 + (xm n ) α+1/ (xm n ) α+1/ (1 + xm n ) α 1/ ((1 + (xm n ) α+1/ ) U(x) x 1 /, (1 + (xm n ) α+1/ ) V (x) x 1 / ) A δ (, 1) uniformly. Now, alying (3.) and lemma 1 we get = C C S α n,1f(x)u(x) xdx = f(y)y 1/ (ym n ) 1/ J α+1 (ym n ) dy (xm n ) 1/ J α (xm n ) U(x) x 1 / dx (y x) f(y)y 1/ (ym n ) 1/ J α+1 (ym n ) dy (1 + (xm n ) α+1/ ) U(x) x 1 / dx (y x) f(x)x 1/ (xm n ) 1/ J α+1 (xm n ) (1 + (xm n ) α+1/ ) V (x) x 1 / dx C f(x)v (x) xdx. The boundedness of S α n, is similar since (1.) is equivalent to ((1 + (xm n ) α+1/ ) U(x) x 1 /, (1 + (xm n ) α+1/ ) V (x) x 1 / ) A δ (, 1). The boundedness of S α n,3 and S α n,4 is analogous, taking the oerator J instead of H. From conditions (1.1) and (1.) (taking n = 1) it follows that U(x) x α+1 dx < +, V (x) x α +1 dx < + and this fact together with Holder s inequality leads to the boundedness of S α n,5. 8

9 Finally, taking limits on n in condition (1.1) we get (U(x) x 1 /, V (x) x 1 / ) A δ (, 1) and then ( see the roof of roosition 1), S α n,6 is bounded. Proof of theorem 1. For α 1/, the result follows from roosition 1 and lemma 4. Let 1 < α < 1/; we must check conditions (1.1) and (1.). If we ut U(x) = x a U 1 (x), V (x) = x A V 1 (x), then U 1 and V 1 are constant-like near. Thus, lemma 3 tells us that we just need to verify the conditions (U 1 (x), V 1 (x) ) A δ (, 1), ((x + 1 M n ) (α+1/) x α +1+a, (x + 1 M n ) (α+1/) x α +1+A ) A δ (, 1), ((x + 1 M n ) (α+1/) x α+1+a, (x + 1 M n ) (α+1/) x α+1+a ) A δ (, 1). With the hel of lemma 4 we obtain the first condition and the two last ones are an easy consequence of lemma. Proof of theorem. Let T n f = S n f S n 1 f = c n (f)j α n. From the hyothesis, we have c n (f) j α n (x)u(x) C f(x)v (x) with constant indeendent of n. Moreover, by alying duality to the oerator f L ((, 1); V (x) xdx) c n (f) we obtain j α n (x)u(x) j α n (x)v (x) 1 C (3.3) uniformly on n. Now, by using (.5) in this inequality we get the two second conditions in (1.5) and (1.6). On the other hand, taking n = 1 in (3.3) and using that J α (α 1 x) Cx α when x (, 1/) (which follows from (.1)), we find the necessary conditions / U(x) x α+1 dx < +, / V (x) x α +1 dx < + which combined with the two already obtained conditions gives (1.5) and (1.6). Finally, we take the function f(x) = min{u(x), V (x) 1, U(x)V (x) 1 }j (x). By using Holder s inequality and (3.3) it is easy to see that f L ((, 1); V (x) xdx) L ((, 1); xdx). For every measurable set E (, 1), let g(x) = f(x)χ E (x). Since S n g g in L (xdx), there exists a subsequence such that S nj g g a.e.. Then, by using Fatou s lemma and the uniform boundedness of {S nj } from L ((, 1); V (x) xdx) into L ((, 1); U(x) xdx) we obtain f(x) U(x) xdx = g(x) U(x) xdx lim inf S nj g(x) U(x) xdx E 9

10 E (, 1), which imlies (1.7) easily. C g(x) V (x) xdx = C f(x) V (x) xdx E Proof of theorem 3. As in the roof of theorem, by alying the uniform boundedness of T n we get j α n (x) L ((,1);U(x) xdx) j α n (x)v (x) 1 C (3.4) with C indeendent of n. We shall use the following Kolmogorov s inequality : g L (h(x)dx) su gχ E Lr (h(x)dx) χ E 1 L E s (h(x)dx) ( r )1/r g L (h(x)dx) where h(x)dx is a Borel measure, < r < <, 1/s = 1/r 1/ and the suremum is taken over all measurable sets E of finite, ositive Lebesgue measure (see [4, lemma V..8,. 485]). This result and (.5) imly the existence of a ositive constant C indeendent of h and n such that lim inf n jα n (x) L (h(x)dx) C x 1/ L (h(x)dx). By alying this condition and (.5) again in (3.4) we obtain the second conditions in (1.5) and (1.6). In order to obtain the two other conditions in (1.5) and (1.6), we observe firstly that it suffices to rove x α χ (,1/) L ((,1);U(x) xdx) <, / V (x) x α +1 dx < + (3.5) since the conditions already obtained imly that we do not have any roblems of integrability at 1. Now, the inequalities (3.5) are easily obtained from (3.4) (with n = 1) and the estimate (.1) in a very similar way as it was done in the roof of theorem. In order to verify (1.7), we take the function f λ (x) = λ 1/ χ {W (x)>λ 1/ }(x)λ 1/ χ { j (x) >λ 1/ }(x) for each λ >, where W (x) = min{u(x), V (x) 1, U(x)V (x) 1 }. By Holder s inequality and (3.4), f λ belongs to L ((, 1); V (x) xdx) L ((, 1); xdx). The function g(x) = f λ (x)χ E (x), (E measurable (, 1)) satisfies g(x) = λ χ { g(x) >λ/} (x) and then, in a similar way as in the revious roof, we get E f λ (x) U(x) xdx C E f λ (x) V (x) xdx, E (, 1), λ >. 1

11 Letting λ and using the dominated convergence theorem, we obtain χ {W (x)>} (x)u(x) xdx C χ {W (x)>} (x)v (x) xdx, E (, 1) E which imlies (1.7). E Proof of theorem 4. (ii) (i). Observing that min{ 1 4, α+1 } = 1/4 if α 1/, it is a articular case of theorem 1. In fact, (ii) imlies strong boundedness (that is, boundedness on L ). (i) (ii). By theorem 3, (i) imlies (1.5) and (1.6) and easy calculations show that the conditions 1/4 1 + a 1 < 1/4; 1 < b < 1 ; 1 < b k < 1 (1 k m) (3.6) follow from (1.5) and (1.6). The strict inequality in a 1 < 1 4 cannot be deduced from conditions (1.5) and (1.6). In what follows, we shall use the following decomosition of the kernel where K α n (x, y) = K α n,13(x, y) + K α n,4(x, y) + K α n,5(x, y) + K α n,6(x, y) K α n,13(x, y) = K α n,1(x, y) + K α n,3(x, y), K α n,4(x, y) = K α n,(x, y) + K α n,4(x, y). So, we can decomose where and S α n f(x) = S α n,13f(x) + S α n,4f(x) + S α n,5f(x) + S α n,6f(x) S α n,13f(x) = S α n,4f(x) = M n J α (M n x)j α+1 (M n y)y y x f(y)ydy M n J α (M n y)j α+1 (M n x)x x y f(y)ydy. The lan is to rove that under conditions (3.6), S α n,4, S α n,5, S α n,6 are bounded from L ((, 1); U(x) xdx) into L ((, 1); U(x) xdx) (in fact, S α n,4 will be of strong tye), while S α n,13 is not bounded in the case 1 4 = 1 + a 1. Weak boundedness of S α n,5. It is simly to observe that S α n,5f(x) C and use Holder s inequality. (xy) α f(y)ydy Cx 1/ y 1/ f(y) ydy 11

12 Weak boundedness of Sn,6. α It would be enough to rove (as in the last art of the roof of roosition 1) that U(x) x 1 / A (, 1), but this is not true, since the condition 1 4 < 1 + a 1 is needed as lemma 4 establishes. We will roceed as follows: we set the oerator Sn,6 α in terms of the oerator J and we only need to show x 1/ χ (,1/) (x)jf(x) L ((,1);U(x) xdx) C x 1/ f(x) L ((,1);U(x) xdx) (3.7) x 1/ χ (1/,1) (x)jf(x) L ((,1);U(x) xdx) C x 1/ f(x) L ((,1);U(x) xdx). (3.8) The inequality (3.8) holds (even with strong norm) because the air of weights (U(x) x 1 / χ (1/,1) (x), U(x) x 1 / ) A δ (, 1) since only the inequality 1 + a 1 < 1 4 is needed in x =. For (3.7) notice that if < x < 1/ then Jf(x) C x 1/ χ (,1/) (x)jf(x) L ((,1);U(x) xdx) C C C f(y) dy x 1/ χ (,1/) (x) L ((,1);U(x) xdx) f(y) dy x 1/ L ((,1);U(x) xdx) f(y) dy and f(y) dy C x 1/ U(x)f(x) L (xdx) x 1/ U(x) 1 f(x) L (xdx) C x 1/ f(x) L ((,1);U(x) xdx), where we have used Holder s inequality and that x 1/ L ((, 1); U(x) xdx) L ((, 1); U(x) xdx) which is immediate from (3.6). The crucial fact is that x 1/ L even though we have the equality in a 1 (this is not true for strong L ). Strong boundedness of Sn,4. α Making the change of variables t = y, z = x and taking the function g(z) = Mn 1/ J α (M n z 1/ )f(z 1/ ), the boundedness of Sn,4 α on L ((, 1); U(x) xdx) turns out to be equivalent to Mn 1/ J α+1 (M n z 1/ )z 1/ Hg(z) L (U(z 1/ ) dz) C Mn 1/ J α (M n z 1/ ) 1 g(z) L (U(z 1/ ) dz). Now, by using the estimate (3.1), we only need to see that Hg(z) L (U(z 1/ ) z /4 dz) C g(z) L (U(z 1/ ) z /4 dz) 1

13 and this is equivalent to check that (U(z 1/ ) z /4 ) A (, 1). Since 1 x 1/ 1 x, x 1/ x k x x k, the A -condition aears as U(x) = x a/+/4 (1 x) b m k=1 which is true from lemma 4 under the hyothesis (3.6). x x k b k A (, 1) The last ste is to rove that, when 1 4 = 1 + a 1 there exists no constant C such that S α n,13f(x) L ((,1);U(x) xdx) C f(x)u(x) (3.9) uniformly on n. The roof is by contradiction. Assume that (3.9) holds, take < d < 1 and f(x) = χ (d,1) (x) sgn(j α+1 (M n x)) g(x) x 1/, where g(x) is a suitable function that we will choose below. It is clear that for x (, d), Then, (3.9) imlies Sn,13f(x) α Mn 1/ J α (M n x) d Mn 1/ J α+1 (M n y) y 1/ g(y) dy. ( Mn 1/ J α+1 (M n y) y 1/ g(y) dy) Mn 1/ J α (M n x)χ (,d) (x) L (U(x) xdx) d C χ (d,1) (x)g(x)x 1/ U(x) with C indeendent of d, n and g. As lim M n = + we can aly lemma 5 and its weak version (as it was done in the roof of theorem 3) to the two factors in the left hand side of the inequality, and we obtain ( y 1 g(y) dy) x 1/ χ (,d) (x) L (U(x) xdx) C x 1/ χ (d,1) (x)g(x)u(x). (3.1) d Now, let us take < d < r < x 1 (therefore U(x) Cx a on (, r)) and g = χ (d,r). Then, it is easy to rove that d y 1 g(y) dy = log r d, x 1/ χ (d,1) (x)g(x)u(x) C(log r d )1/, and also that if 1 4 = 1 + a 1 then x 1/ χ (,d) (x) L (U(x) xdx) C x 1/ χ (,d) (x) L (x a+1 dx) C. Thus, (3.1) imlies the existence of a constant C such that log r d C (log r d )1/ 13

14 which is false. This concludes the roof of theorem 4. Proof of theorem 5. Let α 1/ and = 4. According to the revious roof, we just need to show the (4, 4)-restricted weak tye of S α n,13. Making the change of variables t = y, z = x, this boundedness is equivalent to {z (, 1) : Mn 1/ J α (M n z 1/ )H(Mn 1/ J α+1 (M n t 1/ )t 1/ f(t 1/ ))(z) > λ} C λ 4 f(x) xdx where f is a characteristic function. Now, denoting g(t) = f(t 1/ ) and using that J α (M n z 1/ ) C z 1/4 (see (3.1)), it will be enough to rove that M 1/ n where Observe that A λ,e C λ 4 E, λ >, E (, 1) (3.11) A λ,e = {z (, 1) : z 1/4 H(Mn 1/ J α+1 (M n t 1/ )t 1/ χ E (t))(z) > λ}. A λ,e (1/4, 1) C λ 4 C λ 4 H(Mn 1/ J α+1 (M n t 1/ )t 1/ χ E (t))(z) 4 dz Mn 1/ J α+1 (M n t 1/ )t 1/ χ E (t) 4 dt C λ 4 t 1/4 χ E (t) 4 C λ 4 E due to the boundedness of the Hilbert transform on L 4 ((, 1); dx) and to the estimate (3.1). On the other hand, if E [1/, 1) and z (, 1/4) Hence, H(Mn 1/ J α+1 (M n t 1/ )t 1/ χ E (t))(z) C t 1/4 χ E (t)dt C E. A λ,e (, 1/4) {z (, 1/4) : z 1 C E 4 > λ 4 } C λ 4 E 4 C λ 4 E and (3.11) holds, if E [1/, 1). If E (, 1/), let I k = ( 1 k+1, 1 k ), k =, 3,... We shall use the following notation: 1 I k1 = (, k+ ), I 1 k = ( k+, 1 k 1 ), I 1 k3 = (, 1/), k =, 3,... k 1 14

15 A k λ,e = A λ,e I k, where E km = E I km, m = 1,, 3. Then, A k λ,e Ak1 λ,e Ak λ,e Ak3 λ,e and A λ,e (, 1/4) A k λ,e k= A km λ,e = A k λ/3,e km, m = 1,, 3 { A k1 λ,e + A k λ,e + A k3 λ,e }. (3.1) For m = 3, notice that if z I k and t I k3 then t z t. Hence, using (3.1) k= H(Mn 1/ J α+1 (M n t 1/ )t 1/ χ Ek3 (t))(z) C t 3/4 χ E (t)dt C E 1/4 (the last inequality can be deduced from duality between Lorentz saces). Therefore, The same arguments work in order to rove that A k1 For m =, we have A k3 λ,e {z I k : z < C λ 4 E }. (3.13) λ,e has the estimate (3.13). A k λ,e {z I k : H(M 1/ n J α+1 (M n t 1/ )t 1/ χ Ek (t))(z) > C λ k 4 } C λ 4 k C λ 4 k H(Mn 1/ J α+1 (M n t 1/ )t 1/ χ Ek (t))(z) 4 dz Mn 1/ J α+1 (M n t 1/ )t 1/ χ Ek (t) 4 dt C λ 4 k tχ Ek (t) C λ 4 E k. (3.14) Finally, utting together (3.1), (3.13) and (3.14) we get (3.11) and the roof for = 4 is concluded. The roof for = 4/3 follows from standard arguments of duality between Lorentz saces. Remarks. The decomosition in four terms which we have used to study the weak boundedness, also works to study the strong boundedness, instead of using the decomosition in six terms. However, the results are somehow different. For instance, to obtain the uniform boundedness of the roosition 1, we must change the condition by (U(x 1/ ) x /4, V (x 1/ ) x /4 ) A δ (, 1), (U(x 1/ ) x /4, V (x 1/ ) x /4 ) A δ (, 1). Something similar haens with roosition. If we aly these conditions to articular weights as in theorem 1, the inequalities (1.3), (1.4) are the same. Analogous results can be obtained in a similar way for Fourier-Dini series. References. 15

16 1. A. I. Benedek and R. Panzone, Mean convergence of series of Bessel functions, Rev. Un. Mat. Arg. 6 (197), S. Chanillo, On the weak behaviour of artial sums of Legendre series, Trans. Amer. Math. Soc. 68 (1981), S. Chanillo, The multilier for the ball and radial functions, J. Funct. Anal. 55 (1984), J. García Cuerva and J. L. Rubio de Francia, Weighted norm inequalities and related toics, North-Holland, Amsterdam, J. J. Guadalue, M. Pérez, F. J. Ruiz and J. L. Varona, Two notes on convergence and divergence a.e. of Fourier series with resect to some orthogonal systems, (rerint). 6. J. J. Guadalue, M. Pérez and J. L. Varona, Weak behaviour of Fourier-Jacobi series, J. Arox. Theory 61 (199), R. Hunt, B. Muckenhout and R. Wheeden, Weighted norm inequalities for the conjugate function and Hilbert transform, Trans. Amer. Math. Soc. 176 (1973), A. Máté, P. Nevai and V. Totik, Necessary conditions for weighted mean convergence of Fourier series in orthogonal olynomials, J. Arox. Theory 46 (1986), B. Muckenhout, Weighted norm inequalities for the Hardy maximal function, Trans. Amer. Math. Soc. 165 (197), C. J. Neugebauer, Inserting A weights, Proc. Amer. Math. Soc. 87 (1983), E. Romera and F. Soria, Endoint estimates for the maximal oerator associated to sherical artial sums on radial functions, (rerint). 1. G. N. Watson, A treatise on the theory of Bessel functions, nd. ed., Cambridge Univ. Press, Cambridge, G. M. Wing, The mean convergence of orthogonal series, Amer. J. Math. 7 (195), José J. Guadalue, Mario Pérez and Francisco J. Ruiz. Deartamento de Matemáticas. Facultad de Ciencias. Universidad de Zaragoza, 59 Zaragoza, SPAIN. Juan L. Varona. Deartamento de Matemática Alicada. Colegio Universitario de la Rioja, 61 Logroño, SPAIN. 16