7 Hardy Spaces

7 Hardy Spaces.

For 0

Z 1 2π sup |u(reiθ)|pdθ < ∞ (7.1) 0≤r<1 2π 0 From the Poisson representation formula, valid for 1 >r0 >r≥ 0 Z r02 − r2 2π u(r0ei (θ−ϕ)) u(reiθ)= dϕ (7.2) 02 0 2 2π 0 r − 2rr cos ϕ + r R 2π | iθ |p we get the monotonicity of the quantity M(r)= 0 u(re ) dθ,whichis obvious for p = 1 and requires an application of H¨older’s inequality for p>1. Actually M(r) is monotonic in r for p>0.Toseethiswenote that g(reiθ)=log|u(reiθ)| is subharmonic and therefore, using Jensen’s inequality, Z r02 − r2 2π exp[pg(r0ei (θ−ϕ))] dϕ 2π r02 − 2rr0 cos ϕ + r2 0 Z r02 − r2 2π g(r0ei (θ−ϕ)) ≥ exp[p dϕ] 02 0 2 2π 0 r − 2rr cos ϕ + r ≥ exp[pg(reiθ]

If 1

1 1 − r2 2π 1 − 2r cos ϕ + r2

36 and we get the representation (7.2) for u(reiθ)(withr0 =1)intermsofthe boundary function f on S. Z 1 − r2 2π f(ei (θ−ϕ)) u(reiθ)= dϕ (7.3) 2 − 2 2π 0 1 2r cos ϕ + r Now it is clear that actually

lim u(reiθ)=f(θ) r→1 in Lp. Since we can consider the real and imaginary parts seperately, these considerations apply to Hardy functions in Hp as well. The Poisson kernel is harmonic as a function of r, θ and has as its harmonic conjugate the function 1 2R sin θ 2π 1 − R cos θ + R2 r → with R = r0 . Letting R 1, the imaginary part is see to be given by convolution of the real part by 1 2sinθ 1 θ = cot 2π 2(1 − cos θ) 2π 2 which tells us that the real and imaginary parts at any level |z| = r are related through the Hilbert transform in θ. We need to normalize so that Im u(0) = 0. It is clear that any function in the Hardy Spaces is essentially determined by the boundary value of its real (or imaginary part) on S.The conjugate part is then determined through the Hilbert transform and to be in the Hardy class Hp, both the real and imaginary parts should be in Lp(R). For p>1, since the Hilbert transform is bounded on Lp, this is essentially just the condition that the real part be in Lp. However, for p ≤ 1, to be in Hp both the real and imaginary parts should be in Lp, which is stronger than just requiring that the real part be in Lp. We prove a factorization theorem for functions u(z) ∈Hp for p in the range 0

Theorem 7.1. Let u(z) ∈Hp for some p ∈ (0, ∞). Then there exists a factorization u(z)=v(z)F (z) of u into two analytic functions v and F on D with the following properties. |F (z)|≤1 in D and the boundary value ∗ iθ iθ ∗ F (e ) = limr→1 F (re ) that exists in every Lp(S) satisﬁes |F | =1a.e. on S. Moreover F contains all the zeros of u so that v is zero free in D.

37 Proof. Suppose u has just a zero at the origin of order k and no other zeros. Then we take F (z)=zk and we are done. In any case, we can remove the zero if any at 0 and are therefore free to assume that u(z) =6 0. Suppose u has a − ﬁnite number of zeros, z ,... ,z . For each zero z consider f (z)= z zj .A 1 n j zj 1−zz¯j | − | | − | | | | | simple calculation yields z zj = 1 zz¯j for z = 1. Therefore fzj (z) =1 | | n on S and fzj (z) < 1inD.Wecanwriteu(z)=v(z)Πi=1fzj (z). Clearly the factorization u = Fv works with F (z)=Πfzi (z). If u(z)isanalyticinD,we can have a countable number of zeros accumulating near S.Wewanttouse ∈H ∞ the fact that u p for some p>0 to control the inﬁnite product Πi=1fzi (z), that we may now have to deal with. Since log |u(z)| is subharmonic and we can assume that u(0) =06 Z 1 2π −∞

Denoting C − c by C1, X X (1 −|zj|) ≤ − log |zj|≤C1

−iaj One sees from this that actually the inﬁnite product F (z)=Πjfzj (z)e

38 −iaj converges. with proper phase factors aj. We write −zj = |zj|e .Then

z − z |z | − −iaj j j 1 fzi (z)e =1+ 1 − zz¯j zj z − z|z |2 + z|z |−z |z | = j j j j j zj(1 − zz¯j ) (1 −|z |)(z + z|z |) = j j j zj(1 − zz¯j)

| − −iaj |≤ −| | −| | −1 Therefore 1 fzj (z)e C(1 zj )(1 z ) and if we redeﬁne Fn(z) by

n −iaj Fn(z)=Πj=1fzj (z)e we have the convergence

∞ −iaj lim Fn(z)=F (z)=Πj=1fzj (z)e n→∞ uniformly on compact subsets of D as n →∞. It follows from |Fn(z)|≤1 | |≤ u(z) on D that F (z) 1onD. The functions vn(z)= Fn(z) are analytic in D (as the only zeros of Fn are zeros of u) and are seen easily to converge to u the limit v = F so that u = Fv.MoreoverFn(z) are continuous near S and |Fn(z)|≡1onS. Therefore, Z Z 2π 2π 1 iθ p 1 iθ p sup |vn(re )| dθ = lim sup |vn(re )| dθ 2π → 2π 0

Since vn(z) → v(z) uniformly on compact subsets of D, by Fatou’s lemma, Z Z 1 2π 1 2π sup |v(reiθ)|pdθ ≤ sup |u(reiθ)|pdθ (7.4) 0

39 In other words we have succeeded in writing u = Fv with |F (z)|≤1, re- moving all the zeros of u, but v still satsfying (7.4). In order to complete the proof of the theorem it only remains to prove that |F (z)| = 1 a.e. on S. From (7.4) and the relation u = vF, it is not hard to see that Z 1 2π lim |v(reiθ)|p(1 −|F (reiθ)|p)dθ =0 → r 1 2π 0 Since F (reiθ) is known to have a boundary limit F ∗ to show that |F ∗| =1 a.e. all we need is to get uniform control on the Lebesgue measure of the set {θ : |v(reiθ)|≤δ}. It is clearly suﬃcient to get a bound on Z 1 2π sup | log |v(reiθ)||dθ 0

iθ ∗ iθ Theorem 7.2. Suppose u ∈Hp. Then limr→1 u(re )=u (e ) existsinthe following sense Z 2π lim |u(rei,θ) − u∗(eiθ)|pdθ =0 → r 1 0 Moreover, if p ≥ 1, u has the Poisson kernel representation in terms of u∗.

Proof. If u ∈Hp, according to Theorem 7.1, we can write u = vF with v ∈Hp which is zero free and |F |≤1. Choose an integer k such that kp > 1. k iθ ∗ Since v is zero free v = w for some w ∈Hkp.Noww(re ) has a limit w in ∗ Lkp(S). Since |F |≤1 and has a radial limit F it is clear the u has a limit ∗ ∗ ∗ k ∗ u ∈ Lp(S)givenbyu =(w ) F .If0

40 We can actually prove a better version of Theorem 7.1. Let u ∈Hp for some p>0, be arbitrary but not identically zero. We can start with the inequality Z r2 − r2 2π log |u(rei (θ0−ϕ))| −∞ < log |u(r eiθ0 )|≤ 0 dϕ (7.5) 0 2 − 2 2π 0 r 2rr0 cos ϕ + r0

iθ0 where z0 = r0e is such that r0 = |z0| < 1and|u(z0)| > 0. We can use the uniform integrability of log+ |u(reiθ)| as r → 1, and conclude from Fatou’s lemma that Z 2π | log |u(ei (θ0−ϕ))|| dϕ < ∞ − 2 0 1 2r0 cos ϕ + r0 Since the Poisson kernel is bounded above as well as below (away from zero) we conclude that the boundary function u(eiθ)satisfies Z 2π | log |u(eiθ)||dθ < ∞ 0 We define f(reiθ) by the Poisson integral Z 1 − r2 2π log |u(ei (θ−ϕ))| f(reiθ)= dϕ − 2 4π 0 1 2r cos ϕ + r to be Harmonic with boundary value log |u(eiθ)|. From the inequality (7.5) it follows that f(reiθ) ≥ log |u(reiθ)| We then take the conjugate harmonic function g so that w(·)givenbyw(reiθ)=f(reiθ)+ig(reiθ)isanalytic.We define v(z)=ew(z) so that log |v| = f.Wecanwriteu = Fv that produces a factorization of u with a zero free v and F with |F (z)|≤1onD.Sincethe boundaru values of log |u| and log |v| match on S, the boundary values of F which exist must satisfy |F | = 1 a.e. on S.Wehavethereforeproved

Theorem 7.3. Any u in Hp,withp>0, can be factored as u = Fv with the following properties: |F |≤1 on D, |F | =1on S, v is zero free in D and log |v|, which is harmonic in D, is given by the Poisson formula in terms iθ iθ of its boundary value log |v(e )| =log|u(e )| which is in L1(S).Sucha factorization is essentially unique, the only ambiguity being a mutiplicatve constant of absolute value 1.

41 Remark. The improvement over Theorem 7.1 is that we have made sure that log |v| is not only Harmonic in D but actually takes on its boundary value in the sense L1(S). This provides the uniqueness that was missing before. As an example consider the Poisson kernel itself. z+1 u(z)=e z−1

θ |u(z)| < 1onD, u(reiθ) → ei cot 2 as r → 1. Such a factor is without zeros and would be left alone in Theorem 7.1, but removed now. There are characterizations of the factor F that occurs in u = vF.Let us suppose that u ∈H2 is not identically zero.. If we denote by H∞,the space of all bounded analytic functions in D, clearly if H ∈H∞ and u ∈H2, then Hu ∈H2.WedenotebyK the closure in H2 of Hu as H varies over H∞. It is clear that K = H2 if and only if K contains any and therefore all of the units i.e. invertible elements in H2. In any case since u ≡ 0isruled out, let us pick a ∈ D, a =0suchthat6 |u(a)| > 0andtakeka ∈ K to be the 1 K orthogonal projection of fa(z)= 1−az¯ in . Note that by Cauchy’s formula for any v ∈H2, Z Z 1 2π 1 2π 1 f (eiθ)v(eiθ)dθ = v(eiθ)deiθ = v(a) (7.6) a iθ − 2π 0 2πi 0 e a

Then (fa − ka) ⊥K. Writing the orthogonality relations in terms of the n boundary values, and noting that z ka ∈Kfor n ≥ 0, Z 2π iθ iθ inθ iθ n [fa(e ) − ka(e )]e ka(e )dθ ==0 (7.7) 0 n On the other hand for n ≥ 0, since z ka ∈H2, by (7.6) Z 2π iθ inθ iθ n fa(e )e ka(e )dθ =2πa ka(a) 0 Combining with equation (7.7) we get for n ≥ 0, Z 2π inθ iθ 2 n e |k(e )| dθ =2πka(a)a 0 But |k|2 is real and therefore k (a) must be real and a  Z  n 2π 2πka(a)a if n>0 inθ| iθ |2 e k(e ) dθ = 2πka(a)ifn =0 0  n 2πka(a)¯a if n<0

42 iθ 2 iθ This implies that |ka(e )| ≡ cPa(e )onS where Pa is the Poisson kernel. If c = 0, it follows that fa ⊥K, which in turn implies by (7.6) that

=2πu(a)=0 which is not possible because of the choice of a. We claim that {kaH} as H varies over H2 is all of K.Ifnot,letv ∈Kbe such that v ⊥ kaH for all n H ∈H2. We have then, for n ≥ 0, taking H = z , Z 2π iθ −inθ iθ n ka(e )e v(e )dθ ==0 0 For n = −m<0, zmv ∈Kand Z 2π iθ −inθ iθ m m m ka(e )e v(e )dθ ===2πa v(a) 0 Now Fourier inversion gives

∞ X ae−iθ k (eiθ)v(eiθ)=v(a) ame−imθ = v(a) a 1 − ae−iθ m=1 1 = c (a) = c (a)P (eiθ)(e−iθ − a¯) 1 eiθ − a 2 a

2 iθ Multiplying by ka and remembering that |ka| = cPa,weobtain(kav)(e )= −iθ c3(a)(e − a¯) This leads to

k (eiθ) k (eiθ)eiθ v(eiθ)= a = a e−iθ − a¯ 1 − ae¯ iθ z ∈H ⊥ Therefore v = kaH with H(z)= 1−az¯ 2 contradicting v Hka for all H ∈H2 and forcing v to be 0. We are nowready to prove the following theorem.

Theorem 7.4. Let u ∈H2 be arbitrary and nontrivial. Then 1 belongs to the span of {znu : n ≥ 0} if and only if Z 1 2π log |u(0)| = log |u(eiθ)| dθ (7.8) 2π 0

43 k − k → · Proof. Let pn(z)u(z) 1 H2 0 for some polynomials pn( ). Then Z 2π 1 iθ log |pn(0)|≤ log |pn(e )| dθ 2π 0

iθ iθ Since log |pn(e )u(e )|→0asn →∞in measure on S and + iθ iθ log |pn(e )u(e )| is uniformly integrable, Z 2π 1 iθ iθ lim sup log |pn(e )u(e )| dθ ≤ 0 n→∞ 2π 0 This implies Z 1 2π log |u(0)|≥ log |u(eiθ)| dθ 2π 0 The reverse inequality is always valid and we are done with one half. As for n the converse, If the span of {z u : n ≥ 0isK⊂H2 is a proper subspace, 2 iθ iθ there is k such that u = kv for some v ∈H2 with |k| (e )=cPa(e ), the Poisson kernel for some a ∈ D. For the Poisson kernel it is easy to verify that Z 2π 1 iθ log |Pa(0)| < log |Pa(e )| dθ 2π 0 for any a ∈ D. Therefore we cannot have (7.8) satisﬁed.

iθ Suppose f(e ) ≥ 0 is a weight that is in L1(S). We consider the Hilbert Space H = L2(S, f) of functionsR u that are square integrable with respect to 2π | iθ |2 iθ ∞ the weight f, i.e. g such that 0 g(e ) f(e )dθ < . The trigonometric functions {einθ : −∞

44 Theorem 7.5. Let us suppose that Z 2π log f(eiθ)dθ > −∞ (7.9) 0

Then H∞ = {0} and the residual error is given by Z 1 2π k iθ − iθk2 iθ e0(e ) e 2 =2π exp[ log f(e )dθ] > 0 (7.10) 2π 0 Proof. We will split the proof into several steps. Step 1. We write f(eiθ)=|ueiθ)|2,whereu is the boundary value of a iθ function u(re )) in H2. Note that, if this were possible. according to The- orem 7.1 one can assume with out loss of generality that u(0) =6 0 and for 0

46 Therfore the representation of f(eiθ)=|u(eiθ)|2,withu(eiθ) the boundary value of u ∈H2 that satisﬁes condition (7.8) is unique to within a multiplica- tive constant of absolute value 1. The signiﬁcance of making the choice of u so that the condition (7.8) is valid, is that we can conclude that {zju(z)} spans all of H2.

Step 3. Consider the mapping from L2(S, dθ)intoL2(S, f) that sends iθ iθ → g(e ) g(e ) u(eiθ) . Since the integrability of log f implies that f and therefore u is almost surely nonzero on S, this map is a unitary isomorphism. Whereas any u with |u|2 = f would be enough, our u has a special property. iθ It is the boundary value of a function u(re ) ∈H2, that satisﬁes (7.8). Con- iθ a0 sider g(e )=a0. In the isomorphism it goes over to u . Its inner product with eikθ with k ≥ 1isgivenby Z Z 2π 2π a0 −ikθ iθ iθ iθ −ikθ [ iθ ]e u(e )¯u(e )dθ = a0 u¯(e )e dθ =0 0 u(e ) 0 a0 ⊥ a0 u−a0 This shows that u H1. We claim that the decomposition 1 = u + u is ∩ ⊥ ⊕ the decomposition of 1 into its components in (H0 H1 ) H1. The residual 2 error is given by 2π|a0| which is equal to 2π exp[2u(0)] and agrees with (7.10). We now establish our claim to complete the proof. We need to check a0 ∈ u−a0 ∈ → 1 ≥ that u H0 and u H1. In our isomorphism 1 u and for n 0, n inθ n z u → e . We know that the span of {z u(z):n ≥ 0} in H2 is all of H2. 1 ∈ Therefore u H0. To complete the proof of our claim we need to verify that n u − a0 is in the span of {z u : n ≥ 1}. This is easy because u − a0 = zv(z) ∈H for some v 2. Finally the sameP agument shows that the norm of the ∞ | |2 →∞ projection of 1 onto Hk equals 2π i=k ai which tends to 0 as k . n inθ This proves 1 ⊥ H∞. In fact since U H∞ = H∞, it follows that e ⊥ H∞ for every n. Therefore H∞ = {0}. ∈ ConsiderP the problem of approximating a function f0 L2(µ) by linearR com- inations j≤−1 ajfj. We assume a stationarity in the form ρn = fjfn+jdµ which is independent of j.OfcourseR ρn = ρ−n is a positive deﬁnite function 2π inθ and by Bochner’s theorem ρn = 0 e dF (Rθ) for someP nonnegative mea- | − |2 sure F on S. The object to be minimized is f0 j≤−1 ajfj dµ over all possible a−1,... ,a−k,.... By Bochner’s theorem this is equal to Z 2π X ijθ 2 inf |1 − aje | dF (θ) {aj :j≤−1} 0 j≤−1

47 If dF (θ)=f(θ)dθ,bytherealityofρn, f(θ) is symmetric, and we can replace j ≤−1byj ≥ 1. Then thisR is exactly the problem we considered. 1 2π The minimum is equal to 2π exp[ 2π 0 log f(θ)dθ] and we know how to find the minimizer. R ≥ ∞ ∞ Suppose now f(t) 0isaweightonR with −∞ f(t)dt < .LetHa be the { ibt ≤ } · span of e : b a in L2[R, f].R For a<0, what is the projection ea( )of1 ∞ | − |2 in Ha and what is the value of −∞ 1 ea(t) f(t)dt? R ∞ log f(t) −∞ The natural condition on f is −∞ 1+t2 dt > . Then the Poisson integral Z y ∞ f(t) F x, y dt ( )= 2 2 2π −∞ y +(x − t) defines a harmonic function F in the upper half plane C+ = {(x, y):y>0} 1 { } with boundary value 2 log f on R = (x, y):y =0 and F can be the real part of an analytic function w = F + iG on C+. The function u = ew defines an analytic function on C+ with boundary value u∗ and f(t)=|u∗|2. ∗ Moreover u is the Fourier transform of v in L2(R) that is supported on (−∞, 0). One can again set up an isomorphism between L2[R, 1] and L2[R, f]

→ gb b → by sending g u∗ (g is the Fourier transform of g). This maps v 1and ·− → iat 1 · v( a) eR . The projecton is seen to be the image of v (−∞,a)( )withthe 0 | |2 error being a v (t)dt. Example: 1 | ∗|2 Suppose f(t)= 1+t2 . The factorization f = u is produced by ∗ −1 t u (t)=(i + t) . This produces v(t)=e 1(−∞,0)(t). The projection is the t a iat Fourier transform of va(t)=e 1(−∞,a)(t) which is seen to be e e .