Harmonic Analysis and Its Applications In these lectures, we concentrate on the motivations, development and applications of the Calderon-Zygmund operator theory. Lecture 1. The differential operators with constant coefficients and the first generation of Calderon-Zygmund operators Consider the following differential operator with constant coefficients: X @®u Lu(x) = a : (1:1) ® @x® ® By taking the Fourier transform, X d ® (Lu)(») = a®(¡2¼i») uˆ(»): (1:2) ® This suggests one to consider the following more general Fourier multiplier: Definition 1.3: An operator T is said to be the Fourier multiplier if (dT f)(») = m(»)fˆ(»): (1:4) (1.2) shows that any classical differential operator is a Fourier multiplier. 2 2 Example 1: Suppose f 2 L (R) and F is an analytic extension of f on R+ given by Z Z i 1 i (x ¡ t) ¡ iy F (x + iy) = ¡ f(t)dt = ¡ f(t)dt ¼ x + iy ¡ t ¼ (x ¡ t)2 + y2 Z Z 1 y i (x ¡ t) = f(t)dt ¡ f(t)dt: (1:5) ¼ (x ¡ t)2 + y2 ¼ (x ¡ t)2 + y2 R Letting y ! 0; then 1 y f(t)dt ! f(x) for a. e. x, and, in general, the second ¼ (x¡t)2+y2 R 1 term above has no limit. However, one can show p.v x¡t f(t)dt exists for a. e. x. Thus limy!0F (x + iy) = f(x) + iH(f)(x) (1:6) where H is called the Hilbert transform defined by Z 1 f(t) H(f)(x) = ¡ dt: (1:7) ¼ x ¡ t 1 Example 2: Consider the Laplacian Xn @2u 4u = : (1:8) @x2 j=1 j By taking the Fourier transform, Xn d 2 2 (¡ 4 u)(») = (4¼j»jj) uˆ(») = 4¼j»j uˆ(»): j=1 Define the Riesz transforms Rj; 1 · j · n; by » (Rdf)(») = j fˆ(»): (1:9) j j»j Then d2 @ u d ( )(») = ¡4¼»i»juˆ(») = (RiRj 4 u)(»): (1:10) @xi@xj Thus, @2u = RiRj 4 u: (1:11) @xi@xj 2 Since, it is easy to see that H and Rj; 1 · j · n; are bounded on L ; so we obtain klimy!0F (x + iy)k2 · Ckfk2 (1:12) and @2u k k2 = kRiRj 4 uk2 · Ck 4 uk2: (1:13) @xi@xj The Hilbert transform, by taking the Fourier transform, can be written as Hd(f)(») = ¡isign(»)fˆ(») (1:14) and the Riesz transform, by taking the inverse Fourier transform, can be written as Z y R (f)(x) = c p:v: j f(x ¡ y)dy; 1 · j · n: (1:15) j n jyj1+n In 1952, Calderon and Zygmund introduced the following first generation of singular inte- gral operators: Definition 1.16: Z Ω(y) T (f)(x) = p:v: f(x ¡ y)dy; jyjn 2 where Ω satisfies the following conditions: Ω(¸y) = Ω(y) (1:17) for all ¸ > 0; Ω 2 C1(Sn¡1); (1:18) Z Ω(y)dσ(y) = 0: (1:19) First we point out that the first generation of Calderon-Zygmund operators are well defined on S(Rn); the Schwartz test function space. To see this, one can define Z Z Ω(y) Ω(y) p:v: f(y)dy = lim f(y)dy = jyjn ²!0 jyjn jyj>² Z Z Ω(y) Ω(y) [f(y) ¡ f(0)]dy + f(y)dy (1:20) jyjn jyjn jyj<1 jyj¸1 for all f 2 S since Ω has zero average. It is then easy to see that both integrals above converge. R Ω(y) Remark 1.21: A necessary condition for p:v: jyjn f(x ¡ y)dy exists is that Ω has zero average on Sn¡1: In fact, let f 2 S be such that f(x) = 1 for jxj · 2: Then for jxj < 1; Z Z Ω(y) Ω(y) T (f)(x) = lim dy + f(x ¡ y)dy: (1:21) ²!0 jyjn jyjn ²<jyj<1 jyj¸1 R 1 The second integral is convergent but the first equals lim²!0 Ω(y)dσ(y)log( ² ): Thus, Sn¡1 if this limit is finite, then the integral of Ω on Sn¡1 is zero. Theorem 1.22(Calderon and Zygmund): If T is an operator of the first generation of Calderon-Zygmund operator, then T is bounded on Lp; 1 < p < 1: Moreover, there is a constant C such that kT (f)kp · Ckfkp: (1:23) The method of the proof of theorem 1.22 is called the real variable method of Calderon and Zygmund. This method includes the following steps. Step 1: T is bounded on L2: To do this, since T is a convolution operator, by the Plancheral theorem, it suffices to show the Fourier transform of K, the kernel of T , is bounded. In fact, we will show that for » 2 Sn¡1; Z 1 ¼ m(») = Kˆ (») = Ω(t)[log( ) ¡ i sign(t ¢ »)]dσ(t): (1:24) jt ¢ »j 2 Sn¡1 3 Indeed, since K is homogeneous of degree ¡n, so m(») is homogeneous of degree 0. There- fore, we may assume that » 2 Sn¡1: Since Ω has zero average, Z Ω(y) m(») = lim e¡2¼iy¢»dy ²!0 jyjn 1 ²<jyj< ² 1 Z Z1 Z² dt dt = lim Ω(y)[ (e¡2¼ity¢» ¡ 1) + e¡2¼ity¢» ]dσ(y) ²!0 t t Sn¡1 ² 1 1 Z Z1 Z² dt dt = lim Ω(y)[ (cos(2¼ty ¢ ») ¡ 1) + cos(2¼ty ¢ » ]dσ(y) ²!0 t t Sn¡1 ² 1 1 Z Z² dt ¡ilim Ω(y) sin(2¼ty ¢ ») dσ(y) ²!0 t Sn¡1 ² Making the change of variables s = 2¼ty ¢ » and assume y ¢ » 6= 0; the second term above will be 1 Z Z² Z Z1 dt sin(s) lim Ω(y) sign(2¼ty ¢ ») dσ(y) = Ω(y) sign(y ¢ ») dsdσ(y) ²!0 t s Sn¡1 ² Sn¡1 0 Z ¼ = Ω(t) sign(t ¢ »)dσ(t): 2 Sn¡1 The first, after the change of variables, will be 1 Z Z1 Z² dt dt lim Ω(y)[ (cos(2¼y ¢ ») ¡ 1) + cos(2¼y ¢ » ]dσ(y) = ²!0 t t Sn¡1 ² 1 2¼jy¢»j 1 Z Z1 Z ² 2¼Zjy¢»j ds ds ds = Ω(y)[ (cos(s) ¡ 1) + cos(s) ¡ ]dσ(y) s s s Sn¡1 2¼jy¢»j² 1 1 Z 1 = Ω(t)[log( )dσ(t) jt ¢ »j Sn¡1 since Ω has zero average. Now applying the Plancheral yields d ˆ ˆ kT (f)k2 = k(T (f)k2 = kKˆ fk2 · Ckfk2 = Ckfk2: 4 Step 2: We show T is of week type (1, 1): There is a constant C such that kfk jfx 2 Rn : jT (f)(x) > ¸gj · C 1 (1:25) ¸ for any ¸ > 0 and f 2 L1 \ L2: To do this, we need the following Calderon-Zygmund decomposition. Calderon-Zygmund decomposition 1.26: Given f 2 L1 and non-negative, and given a positive ¸; there exists a sequence fQjg of disjoint cubes such that f(x) · ¸ (1:27) for x2 = [Qj; 1 j [ Q j · kfk ; (1:28) j ¸ 1 Z 1 ¸ < f(x)dx · 2n¸: (1:29) jQjj Qj The proof of this decompositionR is to use the so-calledR stopping time argument. First, 1 1 n choose a large cube Q so that jQj f(x)dx · jQj f(x)dx · ¸: Then divide Q to 2 equal Q R 0 1 subcubes Q : Now we use the stopping time argument as follows: if jQ0j f(x)dx > ¸; we Q0 keep this subcube Q0: Otherwise, divide this subcube as above and keep this procedure. Now we get a sequence fQjg. If x2 = [Qj; this means thatR there is a sequence fQng 1 withjQnj ! 0 as n ! 1; so that x2 Qn for all n and f(x)dx · ¸ which shows jQnj R Qn R 1 1 (1.27). To see (1.29), notice that if ¸ < f(x)dx then ¸ ¸ n f(x)dx which jQj j j2 Qj j n R Qj R 2 Qj 1 1 yields (1.29). Finally, from ¸ < f(x)dx we obtain jQjj < f(x)dx. Summing up jQj j ¸ Qj Qj shows Z 1 X 1 j [ Q j · f(x)dx · kfk : j ¸ ¸ 1 j Qj Now we apply the Calderon-Zygmund decomposition toR show T is of the week type (1, 1 1). Define g(x) = f(x) for x2 = [Qj and g(x) = f(x)dx for x 2 Qj; and b(x) = jQj j Qj f(x) ¡ g(x). Since T is bounded on L2; so Z ¸ kT (g)k2 kgk2 1 jfx 2 Rn : T (g)(x) > gj · C 2 · C 2 · C [ jf(x)j2dx 2 ¸2 ¸2 ¸2 c ([Qj ) X Z Z Z 1 1 n 1 + f(y)dydx] · C [ jf(x)jdx + 2 j [ Qjj] · C kfk1: jQjj ¸ ¸ j c Qj Qj ([Qj ) 5 To estimate T (b)(x), it is easy to see ¸ ¸ jfx 2 Rn : jT (b)(x)j > gj · j [ 2Q j + jfx 2 ([2Q )c : jT (b)(x)j > gj: 2 j j 2 c ¸ 1 By (1.28), it suffices to show jfx 2 (R[2Qj) : jT (b)(x)j > 2 gj · C ¸ kfk1: Rewrite b(x) = P 1 2 bj(x); where bj(x) = [f(x) ¡ f(x)dx]ÂQ (x) and the sum converges in L ; as well jQj j j j Qj c as pointwise. Ifx 2 ([2Qj) ; then, by the fact that bj has zero average, Z Z T (bj)(x) = K(x ¡ y)bj(y)dy = [K(x ¡ y) ¡ K(x ¡ xQj )]bj(y)dy c where xQj is the center of Qj: Thus, by the fact thatx 2 ([2Qj) and the smoothness of K, Z (Qj) jT (bj)(x)j · C n+1 jbj(y)jdy jx ¡ xQj j Qj We now estimate the integral Z X Z Z (Qj) jT (b)(x)jdx · C n+1 jbj(y)jdydx: jx ¡ xQj j c j c ([2Qj ) ([2Qj ) Qj We apply Fubini’sP theorem to change the order of integration in each of the double integrals and it gives C kbjk1: This, in turn, can be estimated by using (1.28) and (1.29).