HARDY SPACES IN ONE COMPLEX VARIABLE prof. Fulvio Ricci A.A. 2004-2005 1 2 Contents Chapter I Hardy spaces on the unit disc 1. De(cid:12)nition and basic properties 2. Harmonic versus holomorphic functions 3. Poisson integrals 4. Functions in hp-spaces and their limits to the boundary 5. Boundary limits of conjugate harmonic functions 6. The Cauchy projection 7. Blaschke products and the F. and M. Riesz theorem 8. Dual spaces Chapter II Hardy spaces on the half-plane 1. De(cid:12)nitions and basic facts 2. Poisson integrals 3. The Fourier transform and the Paley-Wiener theorem 4. The Cauchy projection and the Hilbert transform 5. Transference of Hp-functions and applications Chapter III Pointwise convergence to the boundary and conjugate harmonic functions in hp 1. The Hardy-Littlewood maximal function 2. Poisson integrals and maximal function 3. Pointwise convergence to the boundary 4. Poisson integrals of singular measures 5. Lp-estimates for the conjugate harmonic function THE UNIT DISC 3 CHAPTER I HARDY SPACES ON THE UNIT DISC 1. Definition and basic properties We begin by presenting the main properties of Hardy spaces on the unit disc D = z C : z < 1 : f 2 j j g We shall usually prefer to denote by T, rather than @D or similar, the boundary of D. So (1.1) T = z C : z = 1 = eit : t R=2(cid:25)Z : f 2 j j g f 2 g The reason is that emphasis will be put on the group structure of T. The natural identi(cid:12)cation between T and R=2(cid:25)Z (both algebraic and topological)will be always assumed. Hence functions de(cid:12)ned on T will be identi(cid:12)ed with functions on R=2(cid:25)Z, i.e. with functions on the line, periodic of period 2(cid:25). Integrals on T will be understood with respect to the normalized Lebesgue measure 1 dt. We shall use as alternative notation for an integral on T any of the 2(cid:25) following1: 1 (cid:25) 1 (cid:25) f(eit)dt ; f(eit)dt ; f(t)dt ; f(t)dt ZT 2(cid:25) Z (cid:25) ZT 2(cid:25) Z (cid:25) (cid:0) (cid:0) The spaces Lp(T) must be intended w.r. to the normalized Lebesgue measure. The 2-dimensional Lebesgue measure on C will be denoted by dz. Hence, in the polar coordinates z = reit, dz = rdrdt : This may cause some confusion in the occasions where we shall use line integrals in C, f(z)dz or f(z)dz ; Z I (cid:13) (cid:13) inwhichcasethesymboldz denotesalineardi(cid:11)erentialform. However,themeaning of the symbol dz will be revealed in each case by the domain of integration. 1Tobeprecise,inthe(cid:12)rsttwointegralsTisidenti(cid:12)edwiththeunitcircle,inthelasttwowith R=2(cid:25)Z. Wemayswitchfromonenotationto theother,omittingexplicitreferencetocomposition with the map t7!eit. Typeset by AMS-TEX 4 CHAPTER I Let f(z) be a holomorphic function on D. Given r [0;1) and p 1, de(cid:12)ne 2 (cid:21) 1=p (1.2) M (f;r) = f(reit) pdt ; p (cid:18)ZTj j (cid:19) and also (1.3) M (f;r) = max f(reit) : 1 eit Tj j 2 If we set f (eit) = f(reit) r for 0 r < 1, we can then say that (cid:20) M (f;r) = f : p r p k k De(cid:12)nition2. Let 1 p . We denote by Hp(D) the space of holomorphic (cid:20) (cid:20) 1 functions f on D such that (1.4) sup M (f;r) = f < : p Hp k k 1 0 r<1 (cid:20) It is easy to check that (1.4) de(cid:12)nes a norm. For the implication f = Hp k k 0 f = 0, one has to observe that any f Hp(D) is continuous on D, so that ) 2 f = 0 f = 0, if r < 1. r p r k k ) It is quite obvious that H (D) consists of the bounded holomorphic functions 1 on D. Proposition 1.1. If 1 p < q , then Hq(D) Hp(D), and, for f Hq(D), (cid:20) (cid:20) 1 (cid:26) 2 f f : Hp Hq k k (cid:20) k k Proof. The inequality f f follows easily from Ho(cid:127)lder’s inequality if q < r p r q k k (cid:20) k k , and from the trivial majorization if q = . Taking suprema in r, the inequality i1s preserved. (cid:3) 1 All these inclusions are proper. Interesting examples in this respect are the functions 1 f (z) = ; (cid:11) (1 z)(cid:11) (cid:0) with3 (cid:11) > 0. Given p, we want to determine the values of (cid:11) for which f is in (cid:11) Hp(D). It is obvious that f H (D) for (cid:11) > 0, so we take p < . (cid:11) 1 62 1 The answer is based on the following lemma4. 2We de(cid:12)ne Hardy spaces only for p (cid:21) 1. The de(cid:12)nition makes perfect good sense also for p < 1, except that in this case (1.4) does not de(cid:12)ne a norm. Hp-spaces with p < 1 have many intereseting features, that we will not discuss in this course. 3It is possible to choose a determination of the (cid:11)-power of 1 (cid:0) z on D for every (cid:11) 2 C, bacause 1(cid:0)z does not vanish on D and D is simply connected. The \principal" determination is (1(cid:0)z)(cid:11) =e(cid:11)log(1(cid:0)z), with arg(1(cid:0)z)2((cid:0)(cid:25)=2;(cid:25)=2) (observe that 0 for z 2D). 4We write f (cid:16) g for r ! 1 to denote that the ratio jf=gj is bounded from above and from below by positive constants on a neighborhood of 1. THE UNIT DISC 5 Lemma 1.2. For s > 0 (cid:12)xed, consider the integral (cid:25) 1 I (r) = dt ; s Z 1 reit s (cid:25) (cid:0) j (cid:0) j as a function of r [0;1). Then, for r 1, 2 ! (1) if s < 1, I (r) 1 ; s (cid:16) (2) if s = 1, I (r) log(1 r) ; 1 (cid:16) j (cid:0) j (3) if s > 1, I (r) (1 r) (s 1) . s (cid:0) (cid:0) (cid:16) (cid:0) Proof. Taker > 1. Consideringthatthetrianglewithvertices1,randreit isobtuse 2 in r, and that sin(cid:18) 2 (cid:18) for (cid:18) [ (cid:25)=2;(cid:25)=2], we have that, for t [ (cid:25);(cid:25)], j j (cid:21) (cid:25)j j 2 (cid:0) 2 (cid:0) 1 reit > max r 1 eit ;1 r j (cid:0) j j (cid:0) j (cid:0) 1 1(cid:8) (cid:9) 1 eit +1 r (cid:21) 2 2j (cid:0) j (cid:0) (cid:16) (cid:17) 1 t = sin +1 r (1.5) 2 2 (cid:0) (cid:16)(cid:12) (cid:12) (cid:17) 1 (cid:12)1 (cid:12) (cid:12) t +(cid:12)1 r (cid:21) 2 (cid:25)j j (cid:0) (cid:16) (cid:17) 1 ( t +1 r) : (cid:21) 2(cid:25) j j (cid:0) We also have 1 reit 1 eit + eit reit = 1 eit +1 r t +1 r ; j (cid:0) j (cid:20) j (cid:0) j j (cid:0) j j (cid:0) j (cid:0) (cid:20) j j (cid:0) so that (cid:25) 1 I (r) dt ; s (cid:16) Z ( t +1 r)s (cid:25) (cid:0) j j (cid:0) for every s > 0. Then, (cid:25) 1 (cid:25) 1 dt = 2 dt Z ( t +1 r)s Z (t+1 r)s (cid:25) 0 (cid:0) j j (cid:0) (cid:0) 2 log(1 r+(cid:25)) log(1 r) if s = 1 ; = (cid:0) (cid:0) (cid:0) (cid:26) 2(cid:0) (1 r+(cid:25))1 s (1 r)1 (cid:1)s if s = 1 : 1 s (cid:0) (cid:0) (cid:0) (cid:0) (cid:0) 6 (cid:0) (cid:0) (cid:1) The conclusion follows easily. (cid:3) Proposition 1.3. f Hp(D) if and only if (cid:11)p < 1. (cid:11) 2 Proof. Observe that 1 1=p M (f ;r) = I (r) : p (cid:11) (cid:11)p 2(cid:25) (cid:16) (cid:17) It follows from Lemma 1.2 that M (f ;r) is bounded in r if (cid:11)p < 1, and p (cid:11) lim M (f ;r) = + p (cid:11) r 1 1 ! if (cid:11)p 1. (cid:3) (cid:21) We show now that Hp(D) is complete. A couple of preliminary facts have their own independent importance. 6 CHAPTER I Lemma 1.4. Let f Hp(D). Then, for every z D, 2 2 f f(z) C k kHp : p j j (cid:20) (1 z )1=p (cid:0)j j Proof. The statement is obvious for p = , so we assume that p is (cid:12)nite. 1 Take r with z < r < 1, and let (cid:13) be the circle centered at the origin and radius r j j r, oriented counterclockwise. Then 1 f(w) f(z) = dw : 2(cid:25)i I w z (cid:13)r (cid:0) Setting w = reit, we have dw = ireitdt, so that 1 (cid:25) f(reit) f(z) = dt : 2(cid:25) Z 1 ze it (cid:0)(cid:25) (cid:0) r (cid:0) If z = z ei(cid:18), this becomes j j 1 (cid:25) f(reit) f(z) = dt : 2(cid:25) Z(cid:0)(cid:25) 1(cid:0) jzrjei((cid:18)(cid:0)t) By Ho(cid:127)lder’s inequality, 0 1 (cid:25) 1 1=p f(z) M (f;r) dt ; j j (cid:20) p (cid:18)2(cid:25) Z(cid:0)(cid:25) 1(cid:0) jzrjei((cid:18)(cid:0)t) p0 (cid:19) (cid:12) (cid:12) where p is the dual exponent of p. (cid:12) (cid:12) 0 Makingthechangeofvariable(cid:18) t = uandusingtheperiodicityoftheintegrand, (cid:0) we (cid:12)nd that (cid:25) 1 (cid:25) 1 z Z(cid:0)(cid:25) 1(cid:0) jzrjei((cid:18)(cid:0)t) p0 dt = Z(cid:0)(cid:25) 1(cid:0) jzrjeiu p0 du = Ip0(cid:16)jrj(cid:17) : (cid:12) (cid:12) (cid:12) (cid:12) By Lemma 1.(cid:12)2, since p > 1(cid:12), (cid:12) (cid:12) 0 0 1 z 1=p f(z) f Hp Ip0 j j j j (cid:20) k k 2(cid:25) r (cid:16) (cid:16) (cid:17)(cid:17) z p0(cid:0)01 C f 1 j j (cid:0) p p Hp (cid:20) k k (cid:0) r (cid:16) (cid:17) z 1=p = C f 1 j j (cid:0) : p Hp k k (cid:0) r (cid:16) (cid:17) Letting r 1, we conclude the proof. (cid:3) ! Corollary 1.5. Convergence in Hp(D) implies uniform convergence on compact subsets of D. Proof. Let K D be compact. Then there is r < 1 such that K D , the closed r (cid:26) (cid:26) disc of radius r centered at the origin. By Lemma 1.4, if f f in Hp(D), n ! f f n Hp f f = max f (z) f(z) C k (cid:0) k ; n ;K n p k (cid:0) k1 z K j (cid:0) j (cid:20) (1 r)1=p 2 (cid:0) also tends to zero. (cid:3) THE UNIT DISC 7 Theorem 1.6. Hp(D) is a Banach space. Proof. Let f be a Cauchy sequence in Hp(D). By Lemma 1.4, for every r < 1 n f g f is a Cauchy sequence in C(D ), with the uniform norm. By the completeness n r f g of C(D ), there is g C(D ) such that f g uniformly on D . r r r n r r 2 ! Obviously, if r < r < 1, g and g coincide on D . It follows that the various 1 2 r1 r2 r1 g , with r < 1, are all restrictions of a unique function g continuous on D. r We now prove that g is holomorphic in D. By Morera’s theorem, this is true if and only if for every closed arc (cid:13) in D, g(z)dz = 0 : I (cid:13) Let (cid:13) be such an arc. Since (cid:13) is contained in D for some r < 1, f g r n ! uniformly on (cid:13). Therefore g(z)dz = lim f (z)dz ; n I n I (cid:13) !1 (cid:13) and each of these integrals is zero because the f are holomorphic. n We (cid:12)nally prove that f g in Hp(D). Given " > 0, let n(cid:22) be such that n ! f f < " for n;m n(cid:22). Take r < 1. Since f g uniformly on the circle n m Hp m k (cid:0) k (cid:21) ! z = r, if n n(cid:22), j j (cid:21) M (f g;r)= lim M (f f ;r) lim f f " : p n p n m n m Hp (cid:0) m (cid:0) (cid:20) m k (cid:0) k (cid:20) !1 !1 Since this holds for every r < 1, f g ". (cid:3) n Hp k (cid:0) k (cid:20) 2. Harmonic versus holomorphic functions A C2-function u de(cid:12)ned on an open set (cid:10) Rn is called harmonic on (cid:10) if its (cid:18) Laplacian (cid:1)u, de(cid:12)ned as n (cid:1)u = @2 u xj Xj=1 is identically zero on (cid:10). A holomorphic function f is harmonic on its domain in R2. This follows from the fact that holomorphic functions are C2 (in fact analytic) and from the Cauchy-Riemann equation 1 (cid:22) @ f = @ f +i@ f) = 0 : z x y 2 (cid:0) Di(cid:11)erentiating in x, we (cid:12)nd that @2f = i@ @ f = @2f : x (cid:0) x y (cid:0) y Since (cid:1) is real, if u is harmonic, so are u(cid:22), eu and mu. In particular, anti- < = holomorphic functions are also harmonic. Harmonic functions are characterized by the mean value property. Let Sn 1 be (cid:0) the unit sphere in Rn, and d(cid:27) the surface measure on it. 8 CHAPTER I De(cid:12)nition. A continuous function u satis(cid:12)es the mean value property on (cid:10) if, for any x (cid:10) and any r > 0 such that the closed ball B(x;r) is contained in (cid:10), 2 1 (2.1) u(x) = u(x+ry)d(cid:27)(y) : (cid:27)(Sn(cid:0)1) ZSn(cid:0)1 In two dimensions, (2.1) becomes u(z) = u(z +reit)dt : ZT Theorem 2.1. Let u be a continuous function on (cid:10). Then u is harmonic if and only if the mean value property holds. Proof. Suppose u is C2 in (cid:10). Given x (cid:10), consider the function 2 1 ’(r) = u(x+ry)d(cid:27)(y) ; (cid:27)(Sn(cid:0)1) ZSn(cid:0)1 de(cid:12)ned for r [0;r ), where r is the radius of the largest ball centered at x and 0 0 2 contained in (cid:10). It is easy to verify that ’ is continuous, ’(0) = u(x) and that 1 @ ’ (r) = u(x+ry)d(cid:27)(y) : 0 (cid:27)(Sn(cid:0)1) ZSn(cid:0)1 @r For any r, the last integral can be written as an integral on the boundary of the ball B(x;r), in terms of the surface measure d(cid:27) on it: r @ 1 u(x+ry)d(cid:27)(y)= (cid:23) u(y)d(cid:27) (y) ; ZSn(cid:0)1 @r rn(cid:0)1 Z@B(x;r) (cid:1)r r where (cid:23) denotes the outer normal to @B(x;r). By Green’s formula, we obtain that 1 ’ (r) = (cid:1)u(y)dy : 0 (cid:27)(Sn 1)rn 1 Z (cid:0) (cid:0) B(x;r) It follows that, if u is harmonic, then ’ = 0, hence ’ is identically equal to u(x), 0 which proves the mean value property. Conversely, if u is C2 and satis(cid:12)es the mean value property, the same argument shows that (cid:1)u(y)dy = 0 for any ball B(x;r) (cid:10). This implies that u is B(x;r) (cid:26) harmonic. R It remains to prove that if u is only continuous on (cid:10) and satis(cid:12)es the mean value property, then it is C2. Take a ball B(x ;r) (cid:10), and choose a C2-function (x) on 0 (cid:26) Rn, which is radial, non-negative, non-identically zero, and supported on B(0;r=2). De(cid:12)ne (2.2) v(x) = u(x+y) (y)dy ; Z B(0;r=2) THE UNIT DISC 9 which is well-de(cid:12)ned for x B(x ;r=2). Since is radial, we can consider its 0 2 \pro(cid:12)le" , de(cid:12)ned on the positive half-line so that (x) = ( x ). We then 0 0 j j have, integrating in polar coordinates: r 2 v(x) = u(x+(cid:26)w) ((cid:26))(cid:26)n 1d(cid:27)(w)d(cid:26) 0 (cid:0) Z0 ZSn(cid:0)1 r 2 = ((cid:26))(cid:26)n 1 u(x+(cid:26)w)d(cid:27)(w) d(cid:26) 0 (cid:0) Z0 (cid:18)ZSn(cid:0)1 (cid:19) r 2 = (cid:27)(Sn 1) ((cid:26))(cid:26)n 1d(cid:26) u(x) : (cid:0) 0 (cid:0) (cid:18) Z (cid:19) 0 Since the expression in parentheses is positive, we conclude that v is a non-zero constant multiple of u on B(x ;r=2). On the other hand, changing variable of 0 integration in (2.2), we (cid:12)nd that, for x B(x ;r=2), 0 2 v(x) = u(y) (y x)dy = u(y) (y x)dy Z (cid:0) Z (cid:0) B(x;r=2) B(x0;r) (the last equality takes into account the fact that = 0 in the extra part of the domain of integration). This last expression shows, by di(cid:11)erentiating under integral sign, that v is C2 on B(x ;r=2). The same is then true for u, and, since 0 x is arbitrary, u is C2 on (cid:10). (cid:3) 0 Corollary 2.2. Harmonic functions satisfy the maximum modulus principle: if u is harmonic on a connected open set (cid:10), and it attains its maximum modulus at a point in (cid:10), then it is constant. Proof. Suppose that x (cid:10) is such that u(x ) = max u(x) . By replacing u 0 0 x (cid:10) by ei(cid:18)u for an appropria2te (cid:18), we can assujme thjat u(x )2is rjeal ajnd non-negative. 0 Let B(x ;r) (cid:10). Since eu is harmonic, 0 (cid:26) < u(x ) = eu(x ) 0 0 < 1 = eu(x +ry)d(cid:27)(y) : 0 (cid:27)(Sn(cid:0)1) ZSn(cid:0)1 < Therefore 1 u(x ) eu(x +ry) d(cid:27)(y) = 0 ; 0 0 (cid:27)(Sn(cid:0)1) ZSn(cid:0)1 (cid:0) (cid:0)< (cid:1) with a continuous integrand u(x ) eu(x +ry) 0. Therefore u(x ) = eu(x + 0 0 0 0 (cid:0)< (cid:21) < ry) for every y Sn 1. Since u(x +ry) u(x ), we conclude that u(x +ry) = (cid:0) 0 0 0 2 j j (cid:20) u(x ). 0 This proves that u is constant on a neighborhood of x . It follows that the set 0 of points x (cid:10) where u(x) = u(x ) is both closed and open. Since (cid:10) is connected, 0 this set is a2ll of (cid:10). (cid:3) We shall now restrict ourselves to n = 2, and focus our attention on the relations between harmonic and holomorphic functions. We shall need the following lemma of Fourier analysis. 10 CHAPTER I Lemma 2.3. Let f be a C2-function on T, and denote by f^(n), with n Z its 2 Fourier coe(cid:14)cients. Then n2 f^(n) f ; C2 j j (cid:20) k k and the Fourier series of f converges to f uniformly on T. Proof. Integrating by parts twice, n2f^(n) = n2f(t)e intdt = f (t)e intdt : (cid:0) 00 (cid:0) ZT (cid:0)ZT Therefore, n2 f^(n) f f : 00 C2 j j (cid:20) k k1 (cid:20) k k By the Weierstrass test5, it follows that the series f^(n)eint nXZ 2 is uniformly convergent on T. If g is its sum, integration term by term shows that g^(n) = f^(n) for every n. This implies that g = f. (cid:3) Theorem 2.4. Let u be harmonic on a neighborhood of the closed disc B(z ;R) 0 (cid:26) C. Then u admits a power series expansion 1 1 (2.3) u(z) = a + a (z z )n + a (z z )n ; 0 n 0 n 0 (cid:0) (cid:0) (cid:0) nX=1 nX=1 uniformly convergent on B(z ;R). In particular, u is real-analytic, and it is the 0 sum of a holomorphic and an anti-holomorphic function. Proof. We can assume for simplicity that z = 0 and R = 1, so that B(0;R) = D. 0 The restriction of u to T is C , hence its Fourier coe(cid:14)cients a form an absolutely 1 n summable sequence. The function 1 1 v(reit) = a zn + a z(cid:22)n n n (cid:0) nX=0 nX=1 is uniformly convergent on D and harmonic in the interior. Since v = u on the boundary, the maximum principle implies that v = u in D. (cid:3) Corollary 2.5. Let u be harmonic on a connected and simply connected domain (cid:10). Then there are holomorphic functions f and g on (cid:10), unique up to additive constants, such that u = f +g(cid:22). Proof. By Theorem 2.4, any z (cid:10) has a spherical neighborhood U where u = f + z z 2 gz, with fz;gz holomorphic. If Uz Uz0 = , on the intersection fz fz0 = gz0 gz. \ 6 ; (cid:0) (cid:0) Since the left-hand side is holomorphic and the right-hand side anti-holomorphic, this implies that fz fz0 = gz0 gz = cz;z0 (cid:0) (cid:0) 5The Weierstrass test says that if the functions fn satisfy inequalities jfn(x)j (cid:20) Mn on a set E, with Mn (cid:21)0 and Mn <1, then the series fn converges uniformly on E. P P

HARDY SPACES IN ONE COMPLEX VARIABLE prof. Fulvio Ricci AA 2004-2005 1

Similar HARDY SPACES IN ONE COMPLEX VARIABLE prof. Fulvio Ricci AA 2004-2005 1