POCEEDINGS OF THE AMEICAN MATHEMATICAL SOCIETY Volume 24, Number 9, September 996 OSCILLATOY SINGULA INTEGALS ON L p AND HADY SPACES YIBIAO PAN (Communicated by J. Marshall Ash) Abstract. We consider boundedness properties of oscillatory singular integrals on L p and Hardy spaces. By constructing a phase function, we prove that H boundedness may fail while L p boundedness holds for all p (, ). This shows that the L p theory and H theory for such operators are fundamentally different.. Introduction Let Φ be a real-valued C function on [, ], Φ() = Φ () =, λ, and define T λ by () iλφ(x y) f(y) T λ f(x) =p.v. e x y x y f(y)dy. Such operators are called oscillatory singular integral operators and have been studied extensively ([2], [3], [6], [9], [], [2]). If Φ is sufficiently smooth and Φ (k) () for some k>, then the L p and H boundedness of T λ is well-known ([3], [6], [7]). Theorem A. Suppose Φ is sufficiently smooth and Φ (k) () for some k>; then T λ are uniformly bounded on L p () (<p< )and H (). The space H () is the usual Hardy space H. Both the L p and H uniform boundedness of {T λ } λ may fail if Φ (n) () = for all n ([5], [7]). On the other hand, the condition that Φ (k) () forsome k> is not a necessary condition. By using results of Nagel et al. ([4]) on Hilbert transforms along curves, one can obtain the following L 2 result: Theorem B. Suppose Φ is even, Φ() = Φ () = and Φ >. Then the operators T λ are uniformly bounded on L 2 () if and only if there is a C>such that (2) Φ (Ct) > 2Φ (t), for every t>. More recent results due to Carlsson et al. ([]) imply that the uniform L p boundedness of T λ holds under exactly the same condition. eceived by the editors November 5, 994 and, in revised form, March 25, 995. 99 Mathematics Subject Classification. Primary 42B2. 282 c 996 American Mathematical Society
2822 YIBIAO PAN Theorem C. Suppose Φ is given as in Theorem B and p (, ). Then T λ are uniformly bounded on L p if and only if Φ satisfies (2). esults concerning odd Φ can be found in [4] and []. For the uniform H boundedness of T λ we have the following ([8]): Theorem D. Suppose Φ() = Φ () = and (3) Φ (t), for t>. Then T λ are uniformly bounded on H (). To give an example of Φ which has vanishing derivatives at t = but satisfies (2) and (3), we cite the function Φ(t) =e 4/t2. We point out that condition (3) is strictly stronger than condition (2). By using interpolation and Theorem B, it is easy to see that (2) is a necessary condition for the uniform H boundedness of T λ (when Φ is even and convex). In light of Theorem C, it seems reasonable to speculate that (2) may also be a sufficient condition for the uniform H boundedness of T λ. This turns out to be false. The purpose of this paper is to construct an even, convex function Φ which satisfies (2) such that the corresponding T λ s are not uniformly bounded on H (). Theorem E. There exists a function Φ which is even and satisfies (i) Φ() = Φ () = and (ii) Φ > such that (4) sup T λ Lp L p <, λ for <p<,and (5) sup T λ H H =. λ 2. Construction of Φ For k, we choose a function P k (t) on[2 k+ 2 k /8, 2 k+ ] such that (i) P k (2 k+ 2 k /8) = 2 2k +7 2 5k 2,P k (2 k+ )=2 2k+2 ; (ii) P k (2k+ 2 k /8) = 2 4k+, P k (2k+ )=2 4k+5 ; (iii) <P k (t)<c2k,fort [2 k+ 2 k /8, 2 k+ ], where C is some positive constant. Define g on [, ] by g() = and g(t) =2 2k +2 4k+ (t 2 k )ift [2 k, 2 k+ 2 k /8); g(t) =P k (t)ift [2 k+ 2 k /8, 2 k+ ]. Clearly we have g C ([, ]). Define Φ(t) for t by Φ(t) = t g(s)ds. For t weletφ(t)=φ( t). Therefore we have Φ C 2 ([, ]), Φ() = Φ () =, and Φ (t) > fort. LetT λ be given by (). The following result can be found in [4]. Lemma. Let {n j } be a sequence of positive integers such that n j+ /n j σ> for j =,2,..., and let {ξ j,γ j } be a sequence of pairs of real numbers. If j= (ξ2 j + η2 j ) <, then there is a function f C([, 2π]) such that 2π f(t)cosn j tdt = ξ j,
OSCILLATOY SINGULA INTEGALS ON L p AND HADY SPACES 2823 and for j =,2,... (6) and (7) 2π f(t)sinn j tdt = γ j, By Lemma, there exists a function η C([, 2π]) such that 2π 2π cos(2 j t)η(t)dt =; sin(2 j t)η(t)dt =/j, for j =,2,... Define b(t) on by b(t) = η(t) if t [, 2π], b(t) = η( t) if t [ 2π, ), and b(t) ift [ 2π, 2π]. We have b H () and b H =c >. Proposition. Let Φ be defined as above. Then for p (, ), thereexistsc p > such that (8) sup T λ Lp L p C p. λ Proof. Since 2 2k Φ (t) 2 2k+2 for t [2 k, 2 k+ ], we have Φ (8t) > 2Φ (t) for t. Therefore (8) follows from Theorem C. Proposition 2. Let a(x) beafunctiondefinedon[, δ] and satisfying a (2δ) and δ a(x)dx =. Then δ e iλφ(x y) (9) a(y)dy x >2δ x dx T λa +C, for some C which is independent of a and λ. Proof. δ x >2δ T λ a T λ a(x) dx x 2δ x x y a(y) dydx + x >2δ = I I 2 + I 3. By Proposition and Hölder s inequality, we have I Cδ /2 T λ a 2 C. x δ e iλφ(x y) a(y)dy dx
2824 YIBIAO PAN For I 2 we have Therefore, (9) holds. I 2 2δ δ 2 a x 2 dx C. We now prove Theorem E. Let N be a large integer and λ =2 N+5 π. Define a N (x) by a N (x)=(2 N/3+4 π)b(2 N/3+4 πx). Therefore we have supp(a N ) [ 2 N/3 3, 2 N/3 3 ], and a N H some c independent of N. For k, t [2 k, 2 k+ 2 k /8] we have c > for Therefore, for N/3 k, Φ(t) =Φ(2 k )+2 2k (t 2 k )+2 4k (t 2 k ) 2. 2 k +2 k /8 x 2 k+ 2 k /4 2 k +2 k /8 x 2 k+ 2 k /4 x =ln(4/9) x e iλφ(x y) a N (y)dy dx e iλ(φ(2k )+2 2k (x y 2 k )) a N (y)dy dx Cλ 26k e iλ 22k y a N (y)dy Cλ 26k where we used (7). Hence we obtain x 2 x N/3 2ln(4/9) =2ln(4/9)(2N/3+2k+) C 2 N+6k, N/3 k N/4 e iλφ(x y) a N (y)dy dx (2N/3+2k+) C k N/4 2 N+6k c ln N, where c is some absolute constant. By Proposition 2, we have for λ =2 N+5 π.thuswehave T λ a N H T λ a N L c ln N a N H, sup T λ H H =. λ
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