Carleson Measures and $ H^{P} $ Interpolating Sequences in The

$Carleson Measures and $ H^{P} $ Interpolating Sequences in The$

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p n p n p Deﬁnition 1.2. We say that S is H (D ) interpolating if Rp(H (D )) ⊇ ℓ (S). We can state, for 1 ≤ p< ∞:

p n Deﬁnition 1.3. We say that the sequence S is Carleson if the operator Rp is bounded from H (D ) to ℓp(S), i.e. p Dn ∃C > 0, ∀f ∈ H ( ), kRpfkℓp(S) ≤ CkfkHp(Dn). Deﬁnition 1.4. Let µ be a Borel measure in Dn. We shall say that µ is a Carleson measure if ∃p, 1 ≤ p< ∞, such that: p Dn p ∃C > 0, ∀f ∈ H ( ), f ∈ L (µ) and kfkLp(µ) ≤ CkfkHp(Dn).

n n For z ∈ D define the rectangle Rz := {ζ ∈ T :: |ζj − zj/ |zj|| ≤ 1 −|zj| , j =1, ..., n}. n n Now, for any open set Ω in T , the (generalised) Carleson region is ΓΩ := {z ∈ D :: Rz ⊂ Ω}. Definition 1.5. The measure µ is a rectangular Carleson measure in Dn if: Dn ∃C > 0 :: ∀z ∈ , |µ| (ΓRz ) ≤ C |Rz| . The "natural" generalisation of the Carleson embedding Theorem from the disc to the polydisc would be that: µ is a Carleson measure iff it is a rectangular Carleson measure. But Car- leson [Carleson, 1974] gave a counter example to this and A. Chang [Chang, 1979] gave the following characterisation.

Theorem 1.6. The measure µ is a Carleson measure in Dn iff, for any open set Ω: ∃C > 0 :: ∀Ω, |µ| (ΓΩ) ≤ C |Ω| . Because this characterisation does not depend on p, if µ is p-Carleson for a 1 ≤ p < ∞ then it is q-Carleson for any q. This justifies the absence of p in the definition of Carleson measure.

For S a sequence of points in Dn, deﬁne the measure: 2 χS := ((1 −|a| ))δa. a∈S X Then we can see easily that the sequence S is Carleson iﬀ the measure χS is a Carleson measure.

2 In one variable the interpolating sequences were characterized by L. Carleson [Carleson, 1958] for H∞(D) and by H. Shapiro and A. Shieds [Shapiro and Shields, 1961] for Hp(D) by the same condition: ∀b ∈ S, dG(a, b) ≥ δ > 0, a∈S, a6=b Ya−b where dG(a, b) := 1−ab¯ is the Gleason distance. In several variables for the unit ball Ω= B or for the unit polydisc Ω= Dn, this characterisation is still an open question, even for H2(Ω). Nevertheless we have already some necessary conditions.

Theorem 1.7. ( [Varopoulos, 1972]) If the sequence S is H∞(Dn) interpolating then the measure χS is rectangular Carleson. We shall need

Deﬁnition 1.8. We say that the Hp(Ω) interpolating sequence S has the linear extension property, LEP, if there is a bounded linear operator E : ℓp(S) → Hp(Ωn) such that kEk < ∞ and for any λ ∈ ℓp(S), E(λ)(z) interpolates the sequence λ in Hp(Ω) on S.

∞ n Theorem 1.9. ( [Amar, 1980]) If the sequence S is H (D ) interpolating then the measure χS is Carleson and for any p ≥ 1 the sequence S is Hp(Dn) interpolating with the LEP.

For the ball we have a better result by P. Thomas [Thomas, 1987]. See also [Amar, 2008a].

1 Theorem 1.10. If the sequence S is H (B) interpolating then the measure χS is Carleson. The ﬁrst main result of this work is an analogous result for the polydisc:

Theorem 1.11. Let p> 2 and suppose that the sequence S of points in Dn is Hp(Dn) interpolating. Then the sequence S is Carleson.

We also have some suﬃcient conditions. Let Ω be the ball or the polydisc.

Theorem 1.12. ( [Berndtsson, 1985] for the ball; [Berndtsson et al., 1987] for the polydisc) Let S be a sequence of points in Ω. Let (a), (b) and (c) denote the following statements: (a) There is a constant δ > 0 such that

∀b ∈ S, dG(a, b) ≥ δ > 0. a∈S, a6=b (b) S is an interpolatingY sequence for H∞(Ω). (c) The sequence S is separated and is a Carleson sequence. Then (a) implies (b), (b) implies (c). However, the converse for each direction is false for n ≥ 2.

Now we shall need the following important deﬁnition.

Deﬁnition 1.13. Let S be a sequence of points in Ω. We say that S is a dual bounded sequence p p in H (Ω) if there is a sequence {ρa}a∈S ⊂ H (Ω) such that, with kb,p′ the normalised reproducing ′ kernel in Hp (Ω) for the point b ∈ Ω: ′ ∃ C > 0, ∀a, b ∈ S, hρa,kb,p i = δa,b and kρakHp(Ω) ≤ C.

3 Using this deﬁnition we have, with Ω the ball B or the polydisc Dn:

Theorem 1.14. ( [Amar, 2008b]) Let S be a sequence of points in Ω. Suppose S is Hp(Ω) dual bounded with either p = ∞ or p ≤ 2. Moreover suppose that S is Carleson. Then S is Hq(Ω) interpolating with the LEP for any 1 ≤ q < p.

For the ball we have a better result.

Theorem 1.15. ( [Amar, 2009]) Let S be a sequence of points in B. Suppose S is Hp(B) dual bounded. Then S is Hq(B) interpolating with the LEP for any 1 ≤ q < p.

The second main new result here is the extension of Theorem 1.15 to the bidisc.

p 2 Theorem 1.16. Let S be a dual bounded sequence in H (D ), such that the associated measure χS is Carleson. Then S is Hs(D2) interpolating with the LEP, for any s ∈ [1,p[.

2 First main result.

We have the easy lemma.

Lemma 2.1. Let S be a dual bounded sequence in Hp(Dn), then for any 1 ≤ q ≤ p, S is dual bounded in Hq(Dn).

Proof. p n To see this, just take ρa := γaka,r where γa is the dual sequence in H (D ). Then, −1 ′ ′ hρa,kb,q i = hγaka,r,kb,q i = γa(b)ka,r(b)kkbkp′ . But −1 ′ hγa,kb,p i = γa(b)kkbkp′ = δab, using −1 ′ δa,b = hγa,kb,p i = kkbkp′ γa(b). 2 −1/p′ −2 2 −1 Recall that kkakp = ((1 −|a| )) hence defining χa := kkak2 = ((1 −|a| )) we get kkakp′ = 2 1/p 1/p ((1 −|a| )) = χa .So we get 1 1 1 1 1 1/q q − p q − p + r′ ′ hρa,kb,q i = γa(b)ka,r(b)χb = δabχa ka,r(a)= δa,bχa ka(a). 2 −1 But ka(a)= kkak2 = χa so finally: 1 1 1 q − p + r′ −1 hρa,kb,q′ i = δa,bχa = δab 1 1 1 provided that r = q − p which is possible because 1 ≤ q ≤ p. p Dn It remains to prove that ∃C > 0 :: ∀a ∈ S, kρakHq (Dn) ≤ C. But because γa ∈ H ( ) and r Dn 1 1 1 ka,r ∈ H ( ) and q = r + p we get kρakq ≤kγakpkka,rkr ≤ C using that ′ ∀a ∈ S, kγakp ≤ C, kka,rkr ≤ C . This finishes the proof of the lemma.

Now we are in position to prove our ﬁrst main result.

4 Theorem 2.2. Let p> 2 and suppose that the sequence S of points in Dn is Hp(Dn) interpolating. Then the sequence S is Carleson.

Proof. Suppose that the sequence S ⊂ Dn is Hp(Dn) interpolating. 2 n Because p ≥ 2 we have that S is dual bounded in H (D ) by Lemma 2.1. Call {ρa}a∈S the dual sequence to the normalised reproducing kernels ka,2. We have ∃C > 0 :: ∀a ∈ S, kρak2 ≤ C. Now we set SN the truncation of S to its N ﬁrst terms and set ES := Span{ka, a ∈ S} and N ES := Span{ka, a ∈ SN }. 2 Dn N Let also PN be the orthogonal projection from H ( ) onto ES . We have kPN ρak2 ≤kρak2.

Moreover the sequence {PN ρa} is still a dual sequence to {ka,2}a∈SN because ∀a, b ∈ S, hPN ρa,kb,2i = hρa,PN kb,2i = hρa,kb,2i = δa,b. N Abusing the notation, we still denote by ρa the dual sequence to {ka,2}a∈SN in ES , i.e. noting ρa instead of PN ρa. 1 1 1 Let = + and f ∈ Hp(Dn), g ∈ Hr(Dn). Because fg ∈ H2(Dn), kP (fg)k ≤kfgk , we have 2 p r N 2 2 that:

1/p 1/r PN (fg)= hfg,ka,2iρa = f(a)χa g(a)χa ρa. (2.1) aX∈SN aX∈SN Let {ǫ } be a random Bernouilli i.i.d.. Consider the ﬁnite sum ǫ µ ρ . We have, ﬁxing a a∈S a∈SN a a a ζ ∈ Tn, with X(ζ) := ǫ µ ρ (ζ): a∈SN a a a 2 P P 2 2 Var(X(ζ)) = E( ǫaµaρa(ζ) )= |µa| |ρa(ζ)|

a∈SN a∈SN because the ǫ are independent. X X a By the Theorem of Fubini-Tonelli we get 2 2

E ǫaµaρa = E( ǫaµaρa(ζ) )dζ =   T n a∈SN H2 Z a∈SN X X   2 2 2 2 = |µa| |ρa(ζ)| dζ = |µa| kρak2. Tn Z aX∈SN aX∈SN We always have that kρak2 ≥ 1 because 1= hρa,ka,2i≤kρak2kka,2k2 = kρak2 and because S hence 2 SN is dual bounded in H , we have that kρak2 ≤ C. So we get 2 2 2 |µa| ≤ E ǫaµaρa ≤ C |µa| .   a∈SN a∈SN H2 a∈SN X X X We notice that the constants here are independent of N.

Hence we can ﬁnd a sequence {ǫa}a∈SN such that 2 2 ǫaµaρa ≥ |µa|

aX∈SN 2 aX∈SN which means, see [Amar, 1977, p. 18] for another way to get it,

5 2 2 sup ǫaµaρa ≥ |µa| (2.2) {ǫa} a∈SN 2 a∈SN X X

p p n Now take λ ∈ ℓ (S) and for {ǫa}a∈S a random Bernouilli i.i.d., we can choose a f ∈ H (D ) such 1/p p Dn that hf,ka,pi = χa f(a)= ǫaλa and kfkHp(Dn) ≤ Ckλkℓp(S) because S is H ( ) interpolating. Applying (2.2) to PN (fg) we get 1/p 1/r 1/r µa := f(a)χa g(a)χa = ǫaλag(a)χa , and using (2.1), 2 1/r 1/r 2 sup ǫaλag(a)χa ρa ≥ λag(a)χa . ǫ a a∈S a∈S XN 2 XN 1/r 1 1 1 So, setting νa := g(a)χa , and ν := {νa}a∈SN , we get by Hölder inequalities, using 2 = p + r , 2 2 2 kλkℓp kνkℓr ≥ |νaλa| a∈SN and X

2 2 sup |νaλa| = kνkℓr (2.3) p λ∈ℓ ,kλkℓp ≤1 aX∈SN 1/p So we choose f :: hf,ka,pi = χa f(a) = ǫaλa with λ making (2.3) and {ǫa}a∈SN making (2.2) (recall that SN is ﬁnite). We get for this choice, with kfkp ≤ C, because kλkℓp ≤ 1, 2 2 1/r 1/r 2 kPN (fg)k2 = ǫaλag(a)χa ρa ≥ λag(a)χa . a∈S a∈S XN 2 XN Hence kνk r ≤kP N (fg)k ≤kfgk . ℓ (SN ) 2 2 1 1 1 But kfgk2 ≤kfkpkgkr using 2 = p + r , hence kνkℓr(SN ) ≤kfgk2 ≤kfkpkgkr ≤ Ckgkr. r r 1/r r Now kνk r = g(a)χ ≤ Ckgk so we get ℓ (SN ) a∈SN a r r r |g(a)|Pχa ≤ C kgkr. a∈SN BecauseX none of the constants depends on N, we get: r r |ga| χa ≤ Ckgkr, a∈S which meansX that S is a Carleson sequence.

3 Second main result.

3.1 Interpolation of Hp spaces. We set the assumption: 1 Tn Tn p Tn 1 (AS) We have (L ( ),BMO( ))θ = L ( ), where p =1 − θ.

6 We shall use the following theorem of Lin [Lin, 1986] (see also [Chang and Feﬀerman, 1985], p. 37) proved for the bidisc:

Theorem 3.1. If 1 ≤ p

1 2 p 1 1−θ By duality, because B.M.O. is the dual of H we get (L ,BMO)θ = L , for p = 2 . Now by use of a Theorem of Wolﬀ we get: 1 T2 T2 p 1 Corollary 3.2. We have (L ( ),BMO( ))θ = L , where p =1 − θ. Hence the assumption (AS) is true for the bidisc. Theorem 3.3. Let S be a dual bounded sequence in Hp(Dn), which is also Carleson. Then S is Hs(Dn) interpolating with the LEP, for any s ∈ [1,p[ provided that the assumption (AS) is true. In order to prove Theorem 3.3, we shall need some tools.

3.2 BMO functions in the polydisc. We shall work in the polydisc Dn and we shall use the family of kernels, for p ∈]1, ∞[: 2 p−1 2 p−1 (1 −|a1| ) (1 −|an| ) P (a, z) := p ×···× p . |1 − a¯1z1| |1 − a¯nzn| D R2 To see that these kernels are "good" ones, we use the Cayley transform → U := + to read them in U n: 1 y1 yn P (rα,z) → ϕt(y) := ϕt1 (y1) ··· ϕtn (yn) := ϕ( ) ··· ϕ( ) t1 ··· tn t1 tn 1 p ∞ 1 R with ϕ(x) := ( x2+1 ) . Because ϕ is C smooth and in L ( ) for p > 1, ϕt(y) is a good kernel in the product of the upper half planes U n. See [Chang and Fefferman, 1985], [Lin, 1986], for further details. In particular we shall use the fact that the non tangential maximal function f → MP f, associated to these kernels sends the real H1(Tn) in L1(Tn) and Lp(Tn) in Lp(Tn) boundedly. We have the Corollary to the proof of the characterisation by A. Chang [Chang, 1979] of the Carleson iθ iθ measures for the bidisc. Let Pr(e ,ζ)= P (re ,ζ) be a kernel from the family above and set: iθ1 iθ2 iθ u(r1e ,r2e ) := P f(re ) := Pr2 ∗ (Pr1 ∗ f)(θ1, θ2). Let U ⊂ T2 be an open connected set and define the region S(U) in D2 to be: D2 S(U) := {(z1, z2) ∈ :: Iz1 ×Iz2 ⊂ U} iθ iθj,0 where Izj := {e :: |θ − θj,0| < 1 − rj} with zj = rj . Define as usual for p ≥ 1 and f ∈ Lp(T2), p 2 p 2 H (D ) := {P f :: MP f ∈ L (T )}. Corollary 3.4. Let µ be a positive measure on D2, let p ≥ 1, then µ is bounded on Hp(D2) iff: µ(S(U)) ≤ C |U| for all connected, open sets U ⊂ T2. Proof. As A. Chang [Chang, 1979] said, it is the same proof as in Stein [Stein, 1970], page 236. These measures are the Carleson measures of the bidisc.

We shall use the following lemma.

7 Lemma 3.5. Let P (a, ζ) be a kernel in the above family. Let ω be a Carleson measure in D2. Then its balayage by P (a, ζ) is a B.M.O.(T2) function.

Proof. Let P ∗ω(ζ) := P (a, ζ)dω(a) D2 be the balayage ofZω by P. We know that the dual space of H1(Tn) is B.M.O.(Tn) by [Chang and Fefferman, 1980], [Chang and Fefferman, 1985], so it suffices to test P ∗ω(ζ) against a smooth function h in H1(D2). We have h(ζ)P ∗ω(ζ)dζ = h(ζ)P (a, ζ)dω(a)dζ. T2 T2×D2 Set Z Z H(a) := h(ζ)P (a, ζ)dζ, T2 then we get by theZ Theorem of Fubini h(ζ)P ∗ω(ζ)dζ = H(a)dω(a). T2 D2 Then,Z because ω is a CarlesonZ measure and H(a) is a function in H1(Dn), ∗ h(ζ)P ω(ζ)dζ ≤ |H(a)| d |ω| (a) ≤ CkωkC khkH1 . T2 D2 Z ∗ Z n So we get that P ω is in B.M.O.(T ) with its norm bounded by the Carleson norm of ω, Ckωk . C

Now we are in position to prove Theorem 3.3, i.e.

Theorem 3.6. Let S be a dual bounded sequence in Hp(Dn), which is also Carleson. Then S is Hs(Dn) interpolating with the LEP, for any s ∈ [1,p[ provided that the assumption (AS) is true.

Proof. First it is known that all the results above are true for Dn, n ≥ 2, except, up to my knowledge, the result on the interpolation between L1(T2) and BMO(T2). p n p n Suppose that S is a dual bounded sequence in H (D ), i.e. there is a sequence {ρa}a∈S ⊂ H (D ) such that: ′ ∃C > 0, ∀a, b ∈ S, kρakp ≤ C, hρa,kb,p i = δa,b. Now ﬁx s

Let, with {γa}a∈S to be precised later,

h := γaνaρaka,q = γaλaρaµaka,q. a∈S a∈S Then we getX X

∀b ∈ S, h(b)= γaνaρa(b)ka,q(b)= γbνbρb(b)kb,q(b). Xa∈S Now we choose {γa}a∈S such that γbνbρb(b)kb,q(b)= νbkkbks′ . We get:

8 2 kkbk2 γbνbρb(b)kb,q(b)= γbkkbkp′ = νbkkbks′ kkbkq by the choice of γb. Hence the function h interpolates the sequence ν. Moreover h depends linearly on ν. It remains to estimate its Hs(Dn) norm. First we estimate γb. We have: kkbks′ γbνbρb(b)kb,q(b)= νbkkbks′ ⇒ γb = . ρb(b)kb,q(b) 2 kb ρb(b) kb(b) kkbk2 But 1= ρb, = ⇒ ρb(b)= kkbk ′ . Also kb,q(b)= = . Hence kkbkp′ kkbkp′ p kkbkq kkbkq D kkbks′ kEkbkq γb = 2 . kkbkp′ kkbk2 Now we know that the Structural Hypotheses are true for the polydisc, see [Amar, 2007] and [Amar, 2008b]. This means that Dn 2 ∃α = αq > 0, ∀a ∈ , kkak2 ≥ αkkakqkkakq′ and, for p,q,s such that 1/s =1/p +1/q, Dn ∃β = βp,q > 0, ∀a ∈ , kkaks′ ≤ βkkakp′ kkakq′ Hence, taking a = b, we get γb ≤ αβ. Now, we have, using Hölder inequalities 1/p 1/p′ p p p′ p′ |h| ≤ Cαβ |λa| |ρa| |µa| |ka,q| a∈S ! a∈S ! with p′ the conjugateX exponent of p. X p p 1/p Let g := a∈S |λa| |ρa| , we have p p p p p p kgkp=P |λa| |ρa| dσ = |λa| kρakp ≤ Ckλkp, T2 Z a∈S a∈S hence g ∈ Lp(Tn).XNow we replace CXby Cαβ, to ease notation. ′ ′ ′ 1/p p p Let f := a∈S |µa| |ka,q| , we have |h| ≤ Cfg hence sP s s s s khks = |h| dσ ≤ C f g dσ. Tn Tn Z Z 1 1 1 Applying Hölder inequalities with exponents p/s, q/s because s = p + q , we get s s s s khks ≤ C kgkpkfkq. To have h ∈ Ls(Tn), it remains to prove that 1/p′ ′ ′ p p q n f := |µa| |ka,q| ∈ L (T ). a∈S ! ′ ′ pX′ p p q/p′ Tn ′ ′ So let F := f = a∈S |µa| |ka,q| , we shall show that F ∈ L ( ), (q>p ⇒ q/p > 1). From 2 1/q′ 2 1/q′ (1P−|a1| ) (1 −|an| ) ka,q(z)= ×···× . (1 − a¯1z1) (1 − a¯nzn) we get 2 p′/q′ 2 p′/q′ p′ (1 −|a1| ) (1 −|an| ) |ka,q| = p′ ×···× p′ = |1 − a¯1z1| |1 − a¯nzn| 2 −p′/q n −p′/q = (1 −|a1| ) ··· (1 −|an| ) ×

9 2 p′−1 2 p′−1 (1 −|a1| ) (1 −|an| ) 2 n × p′ ×···× p′ ×(1 −|a1| ) ··· (1 −|an| ). |1 − a¯1z1| |1 − a¯nzn| We make the hypothesis that the measure 2 2 dω(z) := (1 −|a1| ) ··· (1 −|an| )δa a∈S is a Carleson measureX in Dn. Then we shall apply the Lemma 3.5 to get that the balayage of dω(z) by 2 p′−1 2 p′−1 (1 −|a1| ) (1 −|an| ) P (a, z) := p′ ×···× p′ |1 − a¯1z1| |1 − a¯nzn| is in B.M.O.(Tn), because this kernel veriﬁes the conditions of Lemma 3.5. Now if µ is a bounded measure in Dn then its balayage by P (a, z) is in L1(Tn) because P (a, z) is of norm 1 in L1(Tn) for any a ﬁxed in Dn.

Hence if γ(z) ∈ Lt(dω) by interpolation using the assumption (AS), valid for the bidisc by Lin’s Theorem 3.1, we get that P ∗(γdω)(ζ) ∈ Lt(Tn).

So it remains to′ see that p 2 −p′/q 2 −p′/q q/p′ γ := |µa| (1 −|a1| ) ··· (1 −|an| ) ∈ L (dω). But q/p′ q 2 −1 n −1 2 n γ dω = |µa| (1 −|a1| ) ··· (1 −|an| ) ×(1 −|a1| ) ··· (1 −|an| ), Dn Z a∈S hence X q/p′ q q γ dω = |µa| = kµkℓ(S) < ∞. Dn a∈S Z ′ X ′ ′ 1/p q/p′ Tn p p q Tn So we prove that F ∈ L ( ), hence f := a∈S |µa| |ka,q| ∈ L ( ) and we are done. P 4 Appendix. Assumption (AS)

We shall see that the assumption (AS) would be a corollary of a weak factorisation. We shall suppose that p Dn 2p Dn ∃C > 0, ∀1 ≤ p ≤ 2, ∀f ∈ H ( ), ∃gj, hj ∈ H ( ) :: f = j∈N gjhj and P j∈N kgjk2pkhjk2p ≤ Ckfkp, i.e. we shall suppose that Hp(Dn) has the extended weak factorisation property (EWF) for 1 ≤ p ≤ 2 with aP uniform constant C. This true for p =1 by M. T. Lacey and E. Terwilleger [Lacey and Terwilleger, 2009] and for p> 1 but the constant is a priori not uniform.

Because the Szegö projection P : L2 → H2 is uniformly bounded on Lp(Tn), for 2 ≤ p ≤ 4 we get:

Lemma 4.1. We have: 1 1 − θ θ (H2(Dn), H4(Dn)) = Hp(Dn) with = + . θ p 2 4

10 Proof. p n p n 2 n 4 n First take f ∈ H (D ). Then f ∈ L (T ) and we have P f = f. We already know that (L (T ), L (T ))θ = Lp(Tn) so it exists a holomorphic function F (z,ζ) for z in the strip 0 < ℜz < 1 and in L2(Tn) in ζ on ℜz =0 and in L4(Tn) in ζ on ℜz =1 such that: R F (θ, ·)= f, ∃C > 0, ∀y ∈ , kF (iy, ·)kL2 ≤ CkfkLp , kF (1 + iy, ·)kL4 ≤ CkfkLp . Set G(z,ζ) := Pζ F (z, ·)(ζ), then G(z,ζ) is holomorphic as a function of ζ and, because 2 ≤ p ≤ 4, there is a uniform C > 0 such that: R G(θ,ζ)= PζF (θ)(ζ)= P f(ζ)= f(ζ), ∀y ∈ , kG(iy, ·)kL2 ≤ CkfkLp , kG(1 + iy, ·)kL4 ≤ CkfkLp . Because P is linear, we keep the fact that G(z,ζ) is also holomorphic in z in the strip 0 < ℜz < 1. 2 n 4 n 2 n 4 n p n So we get that f ∈ (H (D ),H (D ))θ hence (H (D ),H (D ))θ ⊃ H (D ). 2 n 4 n 2 n 4 n p n To prove the converse, we take f ∈ (H (D ),H (D ))θ ⊂ (L (T ), L (T ))θ = L (T ) hence it remains to see that f is holomorphic. We use the fact that: iψ iψ iψ iψ f(e )= F (θ, e )= Pθ(iy)F (iy, e )dy + Pθ(1 + iy)F (1+ iy, e )dy R R Z Z where Pz(·) is the Poisson kernel of the strip 0 < ℜz < 1. To see that f extends holomorphically in n iψ iψ1 iψn D , we compute its Fourier coeﬃcients, with ψ =(ψ1, ..., ψn), e := e ×···×e , dψ := dψ1···dψn: fˆ(k) := f(eiψ)eikψdψ Tn Z iψ iψ ikψ = Pθ(iy)F (iy, e )dy + Pθ(1 + iy)F (1 + iy, e )dy e dψ. Tn R R Using theZ TheoremZ of Fubini we get Z

fˆ(k)= Pθ(iy)Fˆ(iy,k)dy + Pθ(1 + iy)Fˆ(1 + iy,k)dy R R Z Z hence, if k = (k1, ..., kn) contains a negative entry, then Fˆ(iy,k) = Fˆ(1 + iy,k)=0 because these are holomorphic functions in ζ. So we get that, if k = (k1, ..., kn) contains a negative entry, then fˆ(k)=0 hence f extends holomorphically in Dn. The proof is complete.

Then we have

Theorem 4.2. Suppose that Hp(Dn) has the extended weak factorisation property for 1 ≤ p ≤ 2 1 2 p 1 θ with a uniform constant C. Then we have by the complex method: (H ,H )θ = H with p =1 − 2 . Proof. 1 2 1 θ p Take f ∈ (H ,H )θ with p = 1 − 2 then we get easily that f ∈ H . The point is to prove that if p 1 2 1 θ f ∈ H then f ∈ (H ,H )θ with p =1 − 2 . p 2p So let f ∈ H . By assumption f = j∈N gjhj with gj, hj ∈ H . Because p ≥ 1, we get 2p ≥ 2 hence, by Lemma 4.1, there is a holomorphic function Gj(z) in the strip 0 < ℜz < 1 such that P Gj(θ) = gj and kGj(iy)kH2 ≤ CkgjkH2p and kGj(1 + iy)kH4 ≤ CkgjkH2p . The same there is a holomorphic function Hj(z) in the strip 0 < ℜz < 1 such that Hj(θ) = hj and kHj(iy)kH2 ≤ CkgjkH2p and kHj(1 + iy)kH4 ≤ CkhjkH2p . Let F (z) := j∈N Gj(z)Hj(z). Then F (z) is a holomorphic function in the strip 0 < ℜz < 1 and

kF (iy)kHP1 ≤ kGj(iy)kH2 kHj(iy)kH2 ≤ Ckfkp, j∈N and X

11 kF (1 + iy)kH2 ≤ kGj(1 + iy)kH4 kHj(1 + iy)kH4 ≤ Ckfkp. j∈N Moreover we have X F (θ)= Gj(θ)Hj(θ)= gjhj = f. j∈N j∈N X 1 X2 1 θ Hence we proved that f ∈ (H ,H )θ with p =1 − 2 . The proof is complete.

Tn 1 1 Now we set BMO( ) the real BMO space and HR the real H space. Then

1 p 1 Corollary 4.3. We have (L ,BMO)θ = L , where p =1 − θ. Proof. First Theorem 4.2 gives 1 θ (H1 , L2) = Lp with =1 − R θ p 2 p p 1 2 because if f ∈ L , p> 1, then f ∈ HR hence from Theorem 4.2 we get that f ∈ (HR,HR)θ so, a 1 2 1 2 1 2 p fortiori, f ∈ (HR, L )θ. If f ∈ (HR, L )θ, then f ∈ (L , L )θ = L and we are done. 1 By duality, because B.M.O. is the dual of HR by [Chang, 1979], we get 1 1 − θ (L2,BMO) = Lp with = . θ p 2 Now choose q > 2. We shall use the extrapolation Theorem by T. Wolﬀ [Wolﬀ, 1981] which says that if: 1 q r 2 s (L , L )θ = L and (L ,BMO)θ = L then 1 (L1,BMO) = Lp, where =1 − θ. θ p The proof is complete.

We know that the weak factorisation is true for Hp(Dn), p> 1 and for H1(Dn). An "heuristic" proof of its validity for Hp(Dn), p ≥ 1, with a uniform constant, is the following. We take the proof for H1(Dn) done by M. T. Lacey and E. Terwilleger [Lacey and Terwilleger, 2009] ′ which consists on a careful study of a function b in BMO and we replace BMO by Hp (Dn) the dual space of Hp(Dn). Then we keep careful track of the constants and, hopefully, we get the weak factorisation with uniform constants between 1 ≤ p ≤ 2. Of course this is just heuristic!

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