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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. Definition and basic properties2. Harmonic versus holomorphic functions3. Poisson integrals4. Functions in hp-spaces and their limits to the boundary5. Boundary limits of conjugate harmonic functions6. The Cauchy projection7. Blaschke products and the F. and M. Riesz theorem8. Dual spaces
Chapter II Hardy spaces on the half-plane
1. Definitions and basic facts2. Poisson integrals3. The Fourier transform and the Paley-Wiener theorem4. The Cauchy projection and the Hilbert transform5. Transference of Hp-functions and applications
Chapter III Pointwise convergence to the boundary and conjugate harmonic func-tions in hp
1. The Hardy-Littlewood maximal function2. Poisson integrals and maximal function3. Pointwise convergence to the boundary4. Poisson integrals of singular measures5. 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 .
We shall usually prefer to denote by T, rather than ∂D or similar, the boundaryof D. So
(1.1) T = z ∈ C : |z| = 1 = eit : t ∈ R/2πZ .
The reason is that emphasis will be put on the group structure of T. The naturalidentification between T and R/2πZ (both algebraic and topological) will be alwaysassumed. Hence functions defined on T will be identified with functions on R/2πZ,i.e. with functions on the line, periodic of period 2π.
Integrals on T will be understood with respect to the normalized Lebesgue mea-sure 1
2πdt. We shall use as alternative notation for an integral on T any of the
following1:
∫
T
f(eit) dt ,1
2π
∫ π
−πf(eit) dt ,
∫
T
f(t) dt ,1
2π
∫ π
−πf(t) dt
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 = r dr dt .
This may cause some confusion in the occasions where we shall use line integralsin C, ∫
γ
f(z) dz or
∮
γ
f(z) dz ,
in which case the symbol dz denotes a linear differential form. However, the meaningof the symbol dz will be revealed in each case by the domain of integration.
1To be precise, in the first two integrals T is identified with the unit circle, in the last two with
R/2πZ. We may switch from one notation to the other, omitting explicit reference to composition
with the map t 7→ eit.
Typeset by AMS-TEX
4 CHAPTER I
Let f(z) be a holomorphic function on D. Given r ∈ [0, 1) and p ≥ 1, define
(1.2) Mp(f, r) =
( ∫
T
|f(reit)|p dt)1/p
,
and also
(1.3) M∞(f, r) = maxeit∈T
|f(reit)| .
If we setfr(e
it) = f(reit)
for 0 ≤ r < 1, we can then say that
Mp(f, r) = ‖fr‖p .
Definition2. Let 1 ≤ p ≤ ∞. We denote by Hp(D) the space of holomorphicfunctions f on D such that
(1.4) sup0≤r<1
Mp(f, r) = ‖f‖Hp <∞ .
It is easy to check that (1.4) defines a norm. For the implication ‖f‖Hp =0 ⇒ f = 0, one has to observe that any f ∈ Hp(D) is continuous on D, so that‖fr‖p = 0 ⇒ fr = 0, if r < 1.
It is quite obvious that H∞(D) consists of the bounded holomorphic functionson D.
Proposition 1.1. If 1 ≤ p < q ≤ ∞, then Hq(D) ⊂ Hp(D), and, for f ∈ Hq(D),
‖f‖Hp ≤ ‖f‖Hq .
Proof. The inequality ‖fr‖p ≤ ‖fr‖q follows easily from Holder’s inequality if q <∞, and from the trivial majorization if q = ∞. Taking suprema in r, the inequalityis preserved.
All these inclusions are proper. Interesting examples in this respect are thefunctions
fα(z) =1
(1 − z)α,
with3 α > 0. Given p, we want to determine the values of α for which fα is inHp(D). It is obvious that fα 6∈ H∞(D) for α > 0, so we take p <∞.
The answer is based on the following lemma4.
2We define Hardy spaces only for p ≥ 1. The definition makes perfect good sense also for
p < 1, except that in this case (1.4) does not define a norm. Hp-spaces with p < 1 have manyintereseting features, that we will not discuss in this course.
3It is possible to choose a determination of the α-power of 1 − z on D for every α ∈ C,bacause 1− z does not vanish on D and D is simply connected. The “principal” determination is
(1 − z)α = eα log(1−z), with arg(1 − z) ∈ (−π/2, π/2) (observe that <e(1 − z) > 0 for z ∈ D).4We write f g for r → 1 to denote that the ratio |f/g| 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 fixed, consider the integral
Is(r) =
∫ π
−π
1
|1 − reit|s dt ,
as a function of r ∈ [0, 1). Then, for r → 1,
(1) if s < 1, Is(r) 1 ;(2) if s = 1, I1(r) | log(1 − r)| ;(3) if s > 1, Is(r) (1 − r)−(s−1) .
Proof. Take r > 12 . Considering that the triangle with vertices 1, r and reit is obtuse
in r, and that | sin θ| ≥ 2π|θ| for θ ∈ [−π/2, π/2], we have that, for t ∈ [−π, π],
(1.5)
|1 − reit| > maxr|1 − eit|, 1 − r
≥ 1
2
(1
2|1 − eit| + 1 − r
)
=1
2
(∣∣∣ sint
2
∣∣∣ + 1 − r)
≥ 1
2
( 1
π|t| + 1 − r
)
≥ 1
2π(|t| + 1 − r) .
We also have
|1 − reit| ≤ |1 − eit| + |eit − reit| = |1 − eit| + 1 − r ≤ |t| + 1 − r ,
so that
Is(r) ∫ π
−π
1
(|t| + 1 − r)sdt ,
for every s > 0. Then,
∫ π
−π
1
(|t| + 1 − r)sdt = 2
∫ π
0
1
(t+ 1 − r)sdt
=
2(log(1 − r + π) − log(1 − r)
)if s = 1 ,
21−s
((1 − r + π)1−s − (1 − r)1−s
)if s 6= 1 .
The conclusion follows easily.
Proposition 1.3. fα ∈ Hp(D) if and only if αp < 1.
Proof. Observe that
Mp(fα, r) =( 1
2πIαp(r)
)1/p
.
It follows from Lemma 1.2 that Mp(fα, r) is bounded in r if αp < 1, and
limr→1
Mp(fα, r) = +∞
if αp ≥ 1.
We show now that Hp(D) is complete. A couple of preliminary facts have theirown independent importance.
6 CHAPTER I
Lemma 1.4. Let f ∈ Hp(D). Then, for every z ∈ D,
|f(z)| ≤ Cp‖f‖Hp
(1 − |z|)1/p .
Proof. The statement is obvious for p = ∞, so we assume that p is finite.Take r with |z| < r < 1, and let γr be the circle centered at the origin and radius
r, oriented counterclockwise. Then
f(z) =1
2πi
∮
γr
f(w)
w − zdw .
Setting w = reit, we have dw = ireit dt, so that
f(z) =1
2π
∫ π
−π
f(reit)
1 − zr e
−it dt .
If z = |z|eiθ, this becomes
f(z) =1
2π
∫ π
−π
f(reit)
1 − |z|r e
i(θ−t)dt .
By Holder’s inequality,
|f(z)| ≤Mp(f, r)
(1
2π
∫ π
−π
1∣∣1 − |z|
r ei(θ−t)
∣∣p′ dt)1/p′
,
where p′ is the dual exponent of p.Making the change of variable θ−t = u and using the periodicity of the integrand,
we find that∫ π
−π
1∣∣1 − |z|
r ei(θ−t)
∣∣p′ dt =
∫ π
−π
1∣∣1 − |z|
r eiu
∣∣p′ du = Ip′( |z|r
).
By Lemma 1.2, since p′ > 1,
|f(z)| ≤ ‖f‖Hp
( 1
2πIp′
( |z|r
))1/p′
≤ Cp‖f‖Hp
(1 − |z|
r
)− p′−1
p′
= Cp‖f‖Hp
(1 − |z|
r
)−1/p
.
Letting r → 1, we conclude the proof.
Corollary 1.5. Convergence in Hp(D) implies uniform convergence on compactsubsets of D.
Proof. Let K ⊂ D be compact. Then there is r < 1 such that K ⊂ Dr, the closeddisc of radius r centered at the origin. By Lemma 1.4, if fn → f in Hp(D),
‖fn − f‖∞,K = maxz∈K
|fn(z) − f(z)| ≤ Cp‖fn − f‖Hp
(1 − r)1/p,
also tends to zero.
THE UNIT DISC 7
Theorem 1.6. Hp(D) is a Banach space.
Proof. Let fn be a Cauchy sequence in Hp(D). By Lemma 1.4, for every r < 1fn is a Cauchy sequence in C(Dr), with the uniform norm. By the completenessof C(Dr), there is gr ∈ C(Dr) such that fn → gr uniformly on Dr.
Obviously, if r1 < r2 < 1, gr1 and gr2 coincide on Dr1 . It follows that the variousgr, with r < 1, are all restrictions of a unique function g continuous on D.
We now prove that g is holomorphic in D. By Morera’s theorem, this is true ifand only if for every closed arc γ in D,
∮
γ
g(z) dz = 0 .
Let γ be such an arc. Since γ is contained in Dr for some r < 1, fn → guniformly on γ. Therefore
∮
γ
g(z) dz = limn→∞
∮
γ
fn(z) dz ,
and each of these integrals is zero because the fn are holomorphic.We finally prove that fn → g in Hp(D). Given ε > 0, let n be such that
‖fn − fm‖Hp < ε for n,m ≥ n. Take r < 1. Since fm → g uniformly on the circle|z| = r, if n ≥ n,
Mp(fn − g, r) = limm→∞
Mp(fn − fm, r) ≤ limm→∞
‖fn − fm‖Hp ≤ ε .
Since this holds for every r < 1, ‖fn − g‖Hp ≤ ε.
2. Harmonic versus holomorphic functions
A C2-function u defined on an open set Ω ⊆ Rn is called harmonic on Ω if itsLaplacian ∆u, defined as
∆u =n∑
j=1
∂2xju
is identically zero on Ω. A holomorphic function f is harmonic on its domain inR2. This follows from the fact that holomorphic functions are C2 (in fact analytic)and from the Cauchy-Riemann equation
∂zf =1
2
(∂xf + i∂yf) = 0 .
Differentiating in x, we find that
∂2xf = −i∂x∂yf = −∂2
yf .
Since ∆ is real, if u is harmonic, so are u, <eu and =mu. In particular, anti-holomorphic functions are also harmonic.
Harmonic functions are characterized by the mean value property. Let Sn−1 bethe unit sphere in Rn, and dσ the surface measure on it.
8 CHAPTER I
Definition. A continuous function u satisfies the mean value property on Ω if, forany x ∈ Ω and any r > 0 such that the closed ball B(x, r) is contained in Ω,
(2.1) u(x) =1
σ(Sn−1)
∫
Sn−1
u(x+ ry) dσ(y) .
In two dimensions, (2.1) becomes
u(z) =
∫
T
u(z + reit) dt .
Theorem 2.1. Let u be a continuous function on Ω. Then u is harmonic if andonly if the mean value property holds.
Proof. Suppose u is C2 in Ω. Given x ∈ Ω, consider the function
ϕ(r) =1
σ(Sn−1)
∫
Sn−1
u(x+ ry) dσ(y) ,
defined for r ∈ [0, r0), where r0 is the radius of the largest ball centered at x andcontained in Ω. It is easy to verify that ϕ is continuous, ϕ(0) = u(x) and that
ϕ′(r) =1
σ(Sn−1)
∫
Sn−1
∂
∂ru(x+ ry) dσ(y) .
For any r, the last integral can be written as an integral on the boundary of theball B(x, r), in terms of the surface measure dσr on it:
∫
Sn−1
∂
∂ru(x+ ry) dσ(y) =
1
rn−1
∫
∂B(x,r)
ν · ∇u(y) dσr(y) ,
where ν denotes the outer normal to ∂B(x, r).By Green’s formula, we obtain that
ϕ′(r) =1
σ(Sn−1)rn−1
∫
B(x,r)
∆u(y) dy .
It follows that, if u is harmonic, then ϕ′ = 0, hence ϕ is identically equal to u(x),which proves the mean value property.
Conversely, if u is C2 and satisfies the mean value property, the same argumentshows that
∫B(x,r)
∆u(y) dy = 0 for any ball B(x, r) ⊂ Ω. This implies that u is
harmonic.It remains to prove that if u is only continuous on Ω and satisfies the mean value
property, then it is C2. Take a ball B(x0, r) ⊂ Ω, and choose a C2-function ψ(x) onRn, which is radial, non-negative, non-identically zero, and supported on B(0, r/2).Define
(2.2) v(x) =
∫
B(0,r/2)
u(x+ y)ψ(y) dy ,
THE UNIT DISC 9
which is well-defined for x ∈ B(x0, r/2). Since ψ is radial, we can consider its“profile” ψ0, defined on the positive half-line so that ψ(x) = ψ0(|x|). We thenhave, integrating in polar coordinates:
v(x) =
∫ r2
0
∫
Sn−1
u(x+ ρw)ψ0(ρ)ρn−1 dσ(w) dρ
=
∫ r2
0
ψ0(ρ)ρn−1
( ∫
Sn−1
u(x+ ρw) dσ(w)
)dρ
=
(σ(Sn−1)
∫ r2
0
ψ0(ρ)ρn−1 dρ
)u(x) .
Since the expression in parentheses is positive, we conclude that v is a non-zeroconstant multiple of u on B(x0, r/2). On the other hand, changing variable ofintegration in (2.2), we find that, for x ∈ B(x0, r/2),
v(x) =
∫
B(x,r/2)
u(y)ψ(y − x) dy =
∫
B(x0,r)
u(y)ψ(y− x) dy
(the last equality takes into account the fact that ψ = 0 in the extra part ofthe domain of integration). This last expression shows, by differentiating underintegral sign, that v is C2 on B(x0, r/2). The same is then true for u, and, sincex0 is arbitrary, u is C2 on Ω.
Corollary 2.2. Harmonic functions satisfy the maximum modulus principle: if uis harmonic on a connected open set Ω, and it attains its maximum modulus at apoint in Ω, then it is constant.
Proof. Suppose that x0 ∈ Ω is such that |u(x0)| = maxx∈Ω |u(x)|. By replacing uby eiθu for an appropriate θ, we can assume that u(x0) is real and non-negative.
Let B(x0, r) ⊂ Ω. Since <eu is harmonic,
u(x0) = <eu(x0)
=1
σ(Sn−1)
∫
Sn−1
<eu(x0 + ry) dσ(y) .
Therefore
1
σ(Sn−1)
∫
Sn−1
(u(x0) − <eu(x0 + ry)
)dσ(y) = 0 ,
with a continuous integrand u(x0)−<eu(x0 +ry) ≥ 0. Therefore u(x0) = <eu(x0 +ry) for every y ∈ Sn−1. Since |u(x0 + ry)| ≤ u(x0), we conclude that u(x0 + ry) =u(x0).
This proves that u is constant on a neighborhood of x0. It follows that the setof points x ∈ Ω where u(x) = u(x0) is both closed and open. Since Ω is connected,this set is all of Ω.
We shall now restrict ourselves to n = 2, and focus our attention on the relationsbetween harmonic and holomorphic functions. We shall need the following lemmaof Fourier analysis.
10 CHAPTER I
Lemma 2.3. Let f be a C2-function on T, and denote by f(n), with n ∈ Z itsFourier coefficients. Then
n2|f(n)| ≤ ‖f‖C2 ,
and the Fourier series of f converges to f uniformly on T.
Proof. Integrating by parts twice,
n2f(n) =
∫
T
n2f(t)e−int dt = −∫
T
f ′′(t)e−int dt .
Therefore,
n2|f(n)| ≤ ‖f ′′‖∞ ≤ ‖f‖C2 .
By the Weierstrass test5, it follows that the series
∑
n∈Z
f(n)eint
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 .
Theorem 2.4. Let u be harmonic on a neighborhood of the closed disc B(z0, R) ⊂C. Then u admits a power series expansion
(2.3) u(z) = a0 +
∞∑
n=1
an(z − z0)n +
∞∑
n=1
a−n(z − z0)n ,
uniformly convergent on B(z0, R). In particular, u is real-analytic, and it is thesum of a holomorphic and an anti-holomorphic function.
Proof. We can assume for simplicity that z0 = 0 and R = 1, so that B(0, R) = D.The restriction of u to T is C∞, hence its Fourier coefficients an form an absolutelysummable sequence. The function
v(reit) =
∞∑
n=0
anzn +
∞∑
n=1
a−nzn
is uniformly convergent on D and harmonic in the interior. Since v = u on theboundary, the maximum principle implies that v = u in D.
Corollary 2.5. Let u be harmonic on a connected and simply connected domainΩ. Then there are holomorphic functions f and g on Ω, unique up to additiveconstants, such that u = f + g.
Proof. By Theorem 2.4, any z ∈ Ω has a spherical neighborhood Uz where u = fz+gz, with fz, gz holomorphic. If Uz ∩Uz′ 6= ∅, on the intersection fz− fz′ = gz′ − gz.Since the left-hand side is holomorphic and the right-hand side anti-holomorphic,this implies that
fz − fz′ = gz′ − gz = cz,z′
5The Weierstrass test says that if the functions fn satisfy inequalities |fn(x)| ≤ Mn on a setE, with Mn ≥ 0 and
P
Mn <∞, then the seriesP
fn converges uniformly on E.
THE UNIT DISC 11
is constant. Therefore f ′z = f ′
z′ on Uz ∩ Uz′ , so that all these derivatives can beunified into a unique holomorphic function h on Ω. Since Ω is simply connected,h admits a primitive f , holomorphic on all of Ω. Then fz − f = bz is constant oneach Uz. On each Uz,
u− f = gz + fz − f = gz + bz ,
showing that u − f is anti-holomorphic on Ω. Setting g = u− f we have thatu = f + g. If f1 and g1 are holomorphic and u = f1 + g1 on Ω, we repeat theargument given before: f − f1 = g1 − g must be constant.
Corollary 2.6. Let u be a real-valued harmonic function on a connected and sim-ply connected domain Ω. There is a unique, up to additive constants, real-valuedharmonic function u on Ω such that f = u+ iu is holomorphic.
Proof. According to Corollary 2.5, we can write u = 12(f+g), with f, g holomorphic.
Since u is real, =mf = =mg. Then f−g is a real-valued holomorphic function, henceit is constant. Adjusting, if necessary, f and g by adding appropriate constants, wecan assume that f = g. So u = <ef , and we then take u = =mf .
If v is another real-valued harmonic function such that u + iv is holomorphic,v − u = −i(u + iv) + i(u + iu) is a real-valued holomorphic function. Then it isconstant.
Definition. The function u is called a harmonic conjugate of u on Ω. If u = u1 +iu2 is harmonic and complex-valued, the harmonic conjugates of u are the functionsu = u1 + iu2, with u1 and u2 harmonic conjugates of u1 and u2 respectively.
If Ω = D, one usually normalizes “the” conjugate harmonic function of u byimposing that u(0) = 0. Consider the power series expansion (2.3) of u,
u(z) = a0 +
∞∑
n=1
anzn +
∞∑
n=1
a−nzn
The fact that u is real is reflected by the conditions
a0 ∈ R , a−n = an .
Then
u(z) = a0 + 2<e
( ∞∑
n=1
anzn
).
We then take
u(z) = 2=m
( ∞∑
n=1
anzn
),
in order to have
(2.6) u+ iu = a0 + 2
∞∑
n=1
anzn .
The power series expansion of u is therefore
(2.7) u(z) = −i∞∑
n=1
anzn + i
∞∑
n=1
a−nzn .
By linearity, formulas (2.6) and (2.7) remain valid if u is complex-valued.
12 CHAPTER I
3. Poisson integrals
The Dirichlet problem on D consists in assigning a continuous function f on Tand seeking for a function u continuous on D and harmonic in D, which coincideswith f on T. In other words, we want to solve
(3.1)
∆u = 0 in Du = f on T
in u ∈ C(D). The maximum modulus principle implies that a solution, if it exists,is unique. In fact, one just has to observe that the difference of two solutions wouldbe continuous on D, harmonic in D, and identically zero on T.
We shall prove the existence of the solution and give its expression.
If f(eit) = eint with n ∈ Z, the solution of (3.1) is easily found as
u(z) = zn if n ≥ 0 ,u(z) = z|n| if n < 0 .
Suppose next that f ∈ C2(T), and let
f(eit) =∑
n∈Z
f(n)eint
be its Fourier series. It is natural to construct
(3.2) u(z) =∞∑
n=0
f(n)zn +∞∑
n=1
f(−n)zn .
It follows from Lemma 2.3 that this series converges uniformly on D. Since eachsummand is harmonic in D, so is the sum (verify the mean value property). Thenthis is the solution of (3.1).
We use some Fourier analysis to derive an integral formula giving u. By (3.2),the Fourier series of ur(e
it) = u(reit) is, for r < 1,
ur(eit) =
∑
n∈Z
f(n)r|n|eint .
We set
(3.3) Pr(eit) =
∑
n∈Z
r|n|eint ,
where the series converges uniformly on T. Then6
ur(eit) = f ∗ Pr(eit) =
∫
T
f(ei(t−u))Pr(eiu) du .
The functions Pr in (3.3) form the Poisson kernel on T.
6The following identity can be verified directly, expanding one of the two factors in Fourier
series and integrating term by term. It is however a consequence of the general identity f ∗ g(n) =
f(n)g(n) and of the uniqueness theorem for Fourier series expansions.
THE UNIT DISC 13
Lemma 3.1. The Poisson kernel equals
Pr(eit) =
1 − r2
1 + r2 − 2r cos t, r < 1 .
The function P (reit) = Pr(eit) is harmonic in D.
Proof. Consider the geometric series∞∑
n=0
rneint =
∞∑
n=0
(reit)n =1
1 − reit.
Then
Pr(eit) = 2<e
( ∞∑
n=0
rneint)− 1
= 2<e1 − re−it
|1 − reit|2 − 1
= 21 − rcost
1 + r2 − 2r cos t− 1
=1 − r2
1 + r2 − 2r cos t.
Moreover, the identity P (z) = 2<e(1 − z)−1 − 1 shows that P is harmonicin D.
We have so proved that, for f ∈ C2(T) and r < 1,
u(reit) = f ∗ Pr(eit)
=
∫
T
f(ei(t−u))1 − r2
1 + r2 − 2r cosudu
=
∫
T
f(eiu)1 − r2
1 + r2 − 2r cos(t− u)du .
Another way of writing the same identity is
(3.4) u(z) =
∫
T
f(eiu)P (e−iuz) du .
Observe now that (3.4) makes sense for f continuous and defines a harmonicfunction u in D (again, verify the mean value property using the fact that P isharmonic). It takes some work to verify that u is continuous on D. We do thisnow, introducing some preliminary more general notions.
Definition. Let ϕrr<1 be a family of integrable functions on T. We say thatthey form an approximate identity for r → 1 if 7
(1)∫
T|ϕr(t)| dt ≤ C for some constant C and every r;
(2)∫
Tϕr(t) dt = 1 for every r;
(3) for every δ > 0,
limr→1
∫
δ≤|t|≤π|ϕr(t)| dt = 0 .
7To simplify the notation, here we think of the ϕr as periodic functions on the line. We thenwrite ϕr(t) instead of ϕr(eit). The same for f in the next proposition.
14 CHAPTER I
The notion of “approximate identity” is quite general. We have chosen hereto give the definition with a parameter r tending to 1. The same definition canbe given, with the obvious modifications, with parameters varying on a differentset, for instance for a sequence of functions. Later on, we will find approximateidentities depending on a positive real parameter ε tending to 0.
The name “approximate identity” is justified by the following property.
Proposition 3.2. Let ϕr be an approximate identity on T. If f ∈ C(T),
(3.5) limr→1
f ∗ ϕr = f
uniformly on T. If f ∈ Lp(T), with 1 ≤ p <∞, the limit (3.5) holds in the Lp-norm.
Proof. Suppose f is continuous. By uniform continuity, given ε > 0, there existsδ > 0 such that |f(t− u) − f(t)| < ε for |u| < δ. By condition (3), there is r0 < 1such that for r0 < r < 1, ∫
δ≤|t|≤π|ϕr(t)| dt < ε .
Observe that, by condition (2), we can write
f(t) =
∫
T
f(t)ϕr(u) du .
If r0 < r < 1, we have
∣∣f ∗ ϕr(t) − f(t)∣∣ =
∣∣∣∣∫
T
(f(t− u) − f(t)
)ϕr(u) du
∣∣∣∣
≤ 1
2π
∫
|u|<δ
∣∣f(t− u) − f(t)∣∣∣∣ϕr(u)
∣∣ du
+1
2π
∫
δ≤|u|≤π
∣∣f(t− u) − f(t)∣∣∣∣ϕr(u)
∣∣ du
<1
2πε‖ϕr‖1 +
1
π‖f‖∞ε
≤ 1
2π
(C + 2‖f‖∞
)ε .
This proves the uniform convergence of f ∗ ϕr to f for f continuous.Suppose now that f ∈ Lp(T), with 1 ≤ p < ∞. Given ε > 0, there is g ∈ C(T)
such that ‖f − g‖p < ε. Then
‖f ∗ ϕr − f‖p ≤ ‖(f − g) ∗ ϕr‖p + ‖g ∗ ϕr − g‖p + ‖g − f‖p .
We apply Young’s inequality8 to the first summand, to obtain
‖(f − g) ∗ ϕr‖p ≤ ‖f − g‖p‖ϕr‖1 < Cε .
8More generally, Young’s inequality says that if f ∈ Lp(T), g ∈ Lq(T), and 1p
+ 1q≥ 1, then
f ∗ g ∈ Lr(T), where 1r
= 1p
+ 1q− 1. This property is not specific of convolution on T, but it
holds for convolution on any locally compact group (for instance on R). Another general propertyis that, when p and q are conjugate exponents (hence r = ∞), f ∗ g is continuous.
THE UNIT DISC 15
We then observe that
‖g ∗ ϕr − g‖p ≤ ‖g ∗ ϕr − g‖∞ ,
so that, by the previous part of the proof, there is r0 such that, for r0 < r < 1,‖g ∗ ϕr − g‖p < ε. Then,for r0 < r < 1,
‖f ∗ ϕr − f‖p < (C + 2)ε .
This proves that f ∗ ϕr tends to f in Lp(T).
Observe that the assumption f ∈ L∞(T) does not imply that f ∗ ϕr → f in theL∞-norm. Since each f ∗ ϕr is continuous, this would imply that f is continuoustoo.
With minor modifications to the proof, one can verify that the following moregeneral statement holds.
Corollary 3.3. Assume that the family ϕrr<1 satisfies (1), (3) and
(2’) limr→1
∫
T
ϕr(t) dt = c .
Thenlimr→1
f ∗ ϕr = cf
uniformly on T if f ∈ C(T), and in the Lp-norm if f ∈ Lp(T), with 1 ≤ p <∞.
Proposition 3.4. The Poisson kernel Pr is an approximate identity for r → 1.
Proof. Since the Pr are positive functions, (1) will be a consequence of (2). Usingthe power series expansion (3.3), we have, since the convergence is uniform,
∫
T
Pr(t) dt =∑
n∈Z
r|n|∫
T
eint dt = 1 ,
because all the terms with n 6= 0 give integral zero.We then prove (3). Using (1.5), we have, for t ∈ [−π, π],
(3.6) Pr(t) =1 − r2
|1 − reit|2 ≤ π2 1 − r2
(|t| + 1 − r)2< 2π2 1 − r
(|t| + 1 − r)2.
So, given δ > 0,∫
δ≤|t|≤πPr(t) dt < 2π2
∫
|t|≥δ
1 − r
(|t| + 1 − r)2dt
= 2π2
∫
|u|≥ δ1−r
1
(|u| + 1)2du ,
where we have made the change of variable t = (1 − r)u. Since (|u| + 1)−2 isintegrable on the line,
limr→1
∫
|u|≥ δ1−r
1
(|u| + 1)2du = 0 ,
and this proves (3).
16 CHAPTER I
Theorem 3.5. For f ∈ C(T) the unique solution to the Dirichlet problem (3.1) isu(reit) = f ∗ Pr(eit).Proof. It follows from the uniform convergence of ur to f and the maximum prin-ciple.
The same approach that we have used to obtain the Poisson integral u of acontinuous function f can be used to give a formula for its conjugate harmonicfunction u. Using the Fourier series expansion of f , (3.2) and (2.7), we find that
u(z) = −i∞∑
n=1
f(n)zn + i
∞∑
n=1
f(−n)zn .
Then
(3.7) ur(eit) = −1
∑
n6=0
f(n)sgnn r|n|eint = f ∗ Pr(eit) ,
where
(3.8)
Pr(eit) = −i
∑
n6=0
sgnn r|n|eint
= 2=m
∞∑
n=0
rneint
= 2=m1
1 − reit
=2r sin t
1 + r2 − 2r cos t.
The functions Pr form the conjugate Poisson kernel.
4. Functions in hp-spaces and their limits to the boundary
If u is harmonic in D, we define the means Mp(u, r) according to (1.2) and (1.3).We also define hp(D) as the space of harmonic functions u such that
‖u‖hp = sup0≤r<1
Mp(u, r) <∞ .
The statements of Proposition 1.1, Lemma 1.4, Corollary 1.5 and Theorem 1.6remain valid with Hp replaced by hp. Only Lemma 1.4 requires a modification inthe proof, which goes as follows.
Lemma 4.1. Let u ∈ hp(D). Then, for every z ∈ D,
|u(z)| ≤ Cp‖u‖hp
(1 − |z|)1/p .
THE UNIT DISC 17
Proof. Take ρ so that |z| = r < ρ < 1, and set v(z) = u(ρz). Then v is harmonicin D and continuous on D. By Theorem 3.5,
(4.1) ur = vr/ρ = v1 ∗ Pr/ρ = ur ∗ Pr/ρ .
By Young’s inequality,
|u(z)| ≤ ‖ur‖∞ ≤ ‖uρ‖p‖Pr/ρ‖p′ ≤ ‖u‖hp‖Pr/ρ‖p′ ,
where p′ is the dual exponent of p. We then estimate the Lq-norm of Pr for any q.We know that ‖Pr‖1 = 1. Moreover
‖Pr‖∞ = max|t|≤π
1 − r2
1 + r2 − 2r cos t=
1 − r2
(1 − r)2=
1 + r
1 − r.
If 1 < q <∞,
‖Pr‖q =
( ∫
T
Pr(eit)q dt
) 1q
≤ ‖Pr‖q−1
q∞ ‖Pr‖
1q
1
≤(1 + r
1 − r
) q−1
q
.
Majorizing 1 + r with 2, we then have
‖Pr‖q ≤ 21/q′(1 − r)−1/q′ .
Therefore
|u(z)| ≤ Cp‖u‖hp
(1 − r
ρ
)1/p,
for every ρ ∈ (r, 1). Letting ρ tend to 1, we have the conclusion.
Lemma 4.2. If u is harmonic in D, Mp(u, r) is a non-decreasing function of r.In particular, u ∈ hp(D) if and only if
limr→1
Mp(u, r) <∞ ,
and ‖u‖hp equals this limit.
Proof. If r′ < r < 1,
ur′ = vr′/r = v1 ∗ Pr′/r = ur ∗ Pr′/r ,
by (4.1). By Young’s inequality9, for any p ∈ [1,∞],
‖ur′‖p ≤ ‖ur‖p‖Pr′/r‖1 = ‖ur‖p .
9For p = ∞, one can simply invoke the maximum principle.
18 CHAPTER I
Lemma 4.2 applies in particular to holomorphic functions and Hp-spaces.A simple way to construct hp-functions, for any p ∈ [1,∞], consists in taking
f ∈ Lp(T) and constructing the Poisson integral
u(reit) = f ∗ Pr(eit) .
It follows from Young’s inequality that ‖u‖hp ≤ ‖f‖p.For p = 1, the above construction can be extended to more general objects
defined on the boundary. Take a regular Borel measure µ on T – we write µ ∈M(T) – and define
u(reit) = µ ∗ Pr(eit) =
∫
T
Pr(ei(t−t′)) dµ(t′) .
Since the convolution of a finite measure and an L1-function is in L1, ur ∈ L1(T)for r < 1, and
(4.2) ‖ur‖1 ≤ ‖µ‖M‖Pr‖1 ≤ ‖µ‖M .
The usual verification of the mean value property shows that u is harmonic inD, so that u ∈ h1(D) and ‖u‖h1 ≤ ‖µ‖M .
The question we will discuss now is if every hp-function can be obtained in thisway, i.e. if every hp-function is the Poisson integral of an Lp-function on T if1 < p ≤ ∞, and of a regular Borel measure10 if p = 1.
Theorem 4.3. Consider the operator P mapping f (understood as either a func-tion or a Borel measure on T) into the harmonic function Pf on D given by
(4.3) (Pf)r = f ∗ Pr .
Then P maps Lp(T) isometrically onto hp(D) for 1 < p ≤ ∞, and it maps M(T)isometrically onto h1(D).
The limit limr→1(Pf)r exists in Lp if and only if one of the following holds
(1) f ∈ Lp(T) and 1 ≤ p < ∞;(2) p = ∞ and f ∈ C(T).
In each of these cases, (Pf)r → f in the Lp-norm.For general elements f of M(T) or L∞(T), (Pf)r tends to f in the corresponding
weak*-topology11.
Proof. Suppose first that 1 < p < ∞. Take f ∈ Lp(T) and set u = Pf . ByPropositions 3.2 and 3.4, ur → f in the Lp-norm. In particular,
(4.4) limr→1
Mp(u, r) = ‖f‖p ,
so that ‖Pf‖hp = ‖f‖p. We prove next that P : Lp(T) → hp(D) is onto. Takeu ∈ hp(D) and let rjj∈N be a sequence of radii tending to 1. Since ‖urj
‖ ≤ ‖u‖hp ,
10We regard L1(T) as a subspace of M(T), identifying the function f with the measure µ suchthat dµ(t) = f(t)dt.
11We recall that M(T) is the dual space of C(T) and L∞(T) is the dual space of L1(T). The
weak*-topology on the dual space X ′ of a Banach space X is the weakest topology induced bythe elements of X as linear functionals on X ′.
THE UNIT DISC 19
it follows from the Banach-Alaoglu theorem12 that some subsequence urjk has a
limit f ∈ Lp(T) in the weak* topology of Lp(T). Set v = Pf and take r < 1. Since
Pr ∈ Lp′
(T), and by (4.3),
(4.5)
v(reit) = f ∗ Pr(eit)
=
∫
T
f(eit′
)Pr(t− t′) dt′
= limk→∞
∫
T
urjk(eit
′
)Pr(t− t′) dt′
= limk→∞
urjk∗ Pr(eit)
= limk→∞
u(rrjkeit)
= u(reit)
For p = 1, we know by (4.2) that P : M(T) → h1(D) is continuous with normat most 1. In order to prove that this map is onto, we can repeat the argumentgiven above, using the fact that M(T) is the dual space of C(T), according to theRiesz representation theorem. As before, we take u ∈ h1(D), a sequence rj ofradii tending to 1, and we regard urj
as a bounded sequence in M(T). From it,we can extract a subsequence urjk
having a weak* limit µ ∈ M(T). The proof
that u = Pµ follows the same lines as in (4.5). Moreover, If g ∈ C(T),
∣∣∣∣∫
T
g(t) dµ(t)
∣∣∣∣ = limk→∞
∣∣∣∣∫
T
g(t)urjk(t) dt
∣∣∣∣
≤ lim infk→∞
∫
T
|g(t)||urjk(t)| dt
≤ ‖g‖∞ lim infk→∞
‖urjk‖1
= ‖g‖∞‖u‖h1 .
Therefore
‖µ‖M = sup‖g‖∞=1
∣∣∣∣∫
T
g(t) dµ(t)
∣∣∣∣ ≤ ‖u‖h1 .
Putting this together with (4.2), we have that ‖Pµ‖h1 = ‖µ‖M .
Consider now p = ∞, and let f ∈ L∞(T). Then u = Pf satisfies ‖ur‖∞ ≤‖f‖∞‖Pr‖1, so that u ∈ h∞(D) and ‖u‖h∞ ≤ ‖f‖∞. Using the fact that L∞(T)is the dual space of L1(T), the same proof given above for p = 1, shows thatP : L∞(T) → h∞(D) is onto and isometric.
We pass now to the second part of the statement.
By Propositions 3.2 and 3.4, limr→1(Pf)r exists in the Lp-norm and is equal tof in cases (1) and (2). Consider p = 1 and take µ ∈ M(T). In order to prove that(Pµ)r → µ as r → 1 in the weak* topology, we take g ∈ C(T). Using Fubini’s
12The Banach-Alaoglu theorem says that, if X is a separable Banach space and X ′ is its dualspace, the closed balls in X ′ are sequentially compact in the weak* topology.
20 CHAPTER I
theorem and the parity of Pr (see Lemma 3.1),
∫
T
g(t)(Pµ)r(t) dt =
∫
T
g(t)(µ ∗ Pr)(t) dt
=
∫
T
∫
T
g(t)Pr(t− t′) dµ(t′) dt
=
∫
T
( ∫
T
g(t)Pr(t− t′) dt
)dµ(t′)
=
∫
T
(g ∗ Pr)(t′) dµ(t′) .
Since g is continuous, g ∗ Pr → g uniformly. Hence
limr→1
∫
T
g(t)(Pµ)r(t) dt =
∫
T
g(t′) dµ(t′) ,
showing that µ is the weak* limit of (Pµ)r. If the (Pµ)r have a strong limitas r → 1, this limit is in L1(T), and at the same time it must coincide with µ,because the norm topology is stronger than the weak* topology. This implies thatµ ∈ L1(T).
The same argument, replacing g ∈ C(T) with g ∈ L1(T), shows that, if f ∈L∞(T), then (Pf)r → f in the weak* topology, and that the convergence is innorm if and only if f is continuous.
Observe that the inclusions hp(D) ⊂ hq(D) if q < p also give other consequencesfor convergence to the boundary. For instance, if u is bounded on D, i.e. u ∈h∞(D), and f is its boundary function, then ur → f in any Lq-norm for q <∞.
Corollary 4.4. For r < 1, let u(r)(z) = u(rz). If u ∈ hp(D), with 1 < p < ∞,
u(r) → u in the hp-norm as r → 1. The same is true for p = 1, provided u] ∈ L1(T)
and for p = ∞, provided u] is continuous. For 1 < p < ∞, harmonic polynomialsare dense in hp(D).
Proof. The first part is a direct consequence of Theorem 4.3. Given ε > 0, taker < 1 so that ‖u] − ur‖p < ε. Since each half of the Taylor series of u,
u(z) =
∞∑
n=0
anzn +
∞∑
n=1
a−nzn ,
converges to u uniformly on compact sets, there is N such that the sum sN (z) ofthe two partial sums of order N satisfies |u(z) − sN (z)| < ε for |z| = r. Then‖u] − (uN )r‖p < 2ε.
5. Boundary limits of conjugate harmonic functions
A natural question concerns conjugate harmonic functions: given u ∈ hp(D),is its conjugate harmonic function u also in hp(D)? At this stage we only givethe answer for p = 2 (which is positive) and for p = 1,∞ (which is negative).
THE UNIT DISC 21
The (positive) answer for the other values of p will be given in Chapter III. Inthe last part of this section we discuss under which assumptions the congiugateharmonic function of a function u continuous up to the boundary can be extendedcontinuously to the boundary.
The case p = 2 is special in the sense that h2(D), being isometric with L2(T), isa Hilbert space. To be precise, given u, v ∈ h2(D), there are f, g ∈ L2(T) such thatu = Pf , v = Pg. If we set
〈u, v〉h2 = 〈f, g〉L2 ,
this inner product induces the h2-norm13. It is also clear that P transforms or-thonormal bases of L2(T) into orthonormal bases of h2(D). In particular, from thebasis eintn∈Z of L2(T) we derive the orthonormal basis znn∈N ∪ znn≥1 ofh2(D).
Therefore, if
u =∑
n≥0
anzn +
∑
n≥1
a−nzn ,
then
‖u‖2h2 =
∑
n∈Z
|an|2 .
It follows from (2.7) that
(5.1) ‖u‖2h2 =
∑
n6=0
|an|2 ≤ ‖u‖2h2 .
Then u is also in h2(D). We have so proved the following statement.
Proposition 5.1. The conjugate function operator T : u 7→ u maps h2(D) con-tinuously into itself.
Consider now p = 1, and take
P (reit) =1 − r2
1 + r2 − 2r cos t∈ h1(D) .
Its conjugate harmonic function is
P (reit) =2r sin t
1 + r2 − 2r cos t.
13It follows from Lemma 4.2 and the polarization identity
〈x, y〉 =1
4
`
‖x+ y‖2 + ‖x+ iy‖2 − ‖x− y‖2 − ‖x− iy‖2´
holding on any complex Hilbert space, that
〈u, v〉h2 = limr→1
〈ur, vr〉L2 .
22 CHAPTER I
In fact P and P are real, P (0) = 0, and
F (z) = P (z) + iP (z) =2
1 − z− 1
is holomorphic. If P were in h1(D), it would follow that F ∈ H1(D), but this is incontrast with Proposition 1.3.
This example is interesting because it emphasizes one negative property of theconjugate Poisson kernel: the functions Pr do not form an approximate identity.More precisely, the fact that P 6∈ h1(D) means that
(5.2) limr→1
‖Pr‖1 = +∞ .
Passing to the case p = ∞, the following example shows that the harmonicconjugate of a bounded function need not be bounded.
Consider an open vertical strip S = z : a < <ez < b in the complex plane.By the Riemann mapping theorem, there is a conformal map ϕ from D onto S.Adding, if necessary, an imaginary constant to ϕ, we can assume that ϕ(0) ∈ R.If u = <eϕ, it follows that a < u(z) < b, so that u ∈ h∞(D). But its harmonicconjugate u is =mϕ, which is not bounded14.
One may ask if the conclusion u ∈ h∞(D) holds if u is assumed to be continuouson D. The answer is negative again, but instead of giving a counterexample, weuse a functional analytic argument.
Lemma 5.2. Let X and Y be two Banach spaces of harmonic functions in D.Assume that convergence in each of these spaces implies uniform convergence oncompact subsets of D, and that the conjugate function operator T maps functionsin X into functions in Y . Then T : X → Y is continuous.
Proof. By the closed graph theorem, it is sufficient to prove that if a sequence(un, Tun)
converges to (u, v) ∈ X × Y , then v = Tu. Since T maps real-valued
functions into real-valued functions, we can assume that the un are all real-valued.The same is therefore true for u and v.
On every compact set, un + iTun converges uniformly to u + iv. Hence u + ivis holomorphic. Since vn(0) = 0 for every n, also v(0) = 0. We conclude thatv = u.
Proposition 5.3. There exist functions u, harmonic in D and continuous on D,whose harmonic conjugate u is not bounded in D.
Proof. Denote by h∞c (D) the closed subspace of h∞(D) consisting of those harmonicfunctions in D which admit a continuous extension to D. Assume that T mapsh∞c (D) into h∞(D). By Lemma 5.2, this map would be continuous, hence therewould exists a constant C > 0 such that
‖u‖h∞ ≤ C‖u‖h∞
14Another example is u(z) = =mlog(1− z), which is bounded, but whose harmonic conjugate,−<e log(1 − z) = − log |1 − z|, is not.
THE UNIT DISC 23
for every u ∈ h∞c (D). Given r < 1, consider the linear functional on C(T)
ψr(f) = f ∗ Pr(1) =
∫
T
f(t)Pr(−t) dt .
If u is the Poisson integral of f , we know from Theorem 4.3 that ‖u‖h∞ = ‖f‖∞.Moreover, ψr(f) = u(r). We would then have
|ψr(f)| ≤ ‖u‖h∞ ≤ C‖f‖∞ .
This would imply that
‖Pr‖1 = supf∈C(T) , ‖f‖∞≤1
∣∣∣∣∫
T
f(t)Pr(−t) dt∣∣∣∣
= supf∈C(T) , ‖f‖∞≤1
|ψr(f)|
≤ C ,
in contrast with (5.2).
On the other hand, if we impose a little more regularity on the behaviour of uon the boundary, we obtain a positive result.
Proposition 5.4. Suppose that u is harmonic in D, continuous on D, and α-Lipschitz 15 on T for some α > 0. Then u extends continuously to D.
Proof. We shall prove that ur has a uniform limit u1 for r → 1. It is a simpleexercise to prove that this condition is in fact equivalent to the existence of acontinuous extension of u on D.
Let f be the restriction of u to T, so that ur = f ∗ Pr. Consider
f ∗ Pr(t) =
∫
T
f(t− t′)Pr(t′) dt′ =
∫
T
f(t− t′)2r sin t′
1 + r2 − 2r cos t′dt′ .
Since Pr is odd in t′,∫
TPr(t
′) dt′ = 0. We then have
(5.3) limr→1
f ∗ Pr(t) = limr→1
∫
T
(f(t− t′) − f(t)
)Pr(t
′) dt′ .
We prove that this limit equals the expression formally obtained by taking thepointwise limit of the integrand, i.e.
(5.4)
f(t) =1
2π
∫ π
−π
(f(t− t′) − f(t)
) sin t′
1 − cos t′dt′
=1
2π
∫ π
−π
(f(t− t′) − f(t)
)cot
t′
2dt′ .
15A function f is α-Lipschitz (or α-Holder) on a metric space X if there is a constant C suchthat |f(x) − f(y)| ≤ Cd(x, y)α for every x, y ∈ X.
24 CHAPTER I
Observe that this integral is absolutely convergent, due to the fact that f isα-Lipschitz. In fact,
(5.5)∣∣f(t− t′) − f(t)
∣∣∣∣∣ cot
t′
2
∣∣∣ ≤ C
|t′|1−α ,
which is integrable on [−π, π].Using (1.5), we find that
(5.6) |Pr(t′)| =2r| sin t′||1 − reit′ |2 ≤ π2 2|t′|
(|t′| + 1 − r)2≤ 2π2
|t′| .
Hence
(5.7)∣∣f(t− t′) − f(t)
∣∣∣∣Pr(t′)∣∣ ≤ C
|t′|1−α ,
and dominated convergence can be applied to show that limr→1 f ∗ Pr(t) = f(t)pointwise. Moreover,
‖f − f ∗ Pr‖∞ ≤ maxt∈[−π,π]
1
2π
∫ π
−π
∣∣f(t− t′) − f(t)∣∣∣∣∣ cot
t′
2− Pr(t
′)∣∣∣ dt′
≤ C
2π
∫ π
−π|t′|α
∣∣∣ cott′
2− Pr(t
′)∣∣∣ dt′ .
We can apply dominated convergence again, because the integrand tends to zeropointwise and ∣∣∣ cot
t′
2− Pr(t
′)∣∣∣ ≤
∣∣∣ cott′
2
∣∣∣ +∣∣Pr(t′)
∣∣ ≤ C
|t′| .
Therefore,limr→1
‖f − f ∗ Pr‖∞ = 0 .
This result admits the following local version.
Proposition 5.5. Suppose that f ∈ L1(T) is α-Lipschitz on an open interval I ⊂ T
for some α > 0. Then f in (5.4) is well defined on I and limr→1 f∗Pr = f uniformlyon compact subsets of I. In particular, the harmonic conjugate u of u = Pf admitsa continuous extension to D ∪ I.Proof. For fixed t ∈ I, the integral in (5.4) is absolutely convergent, because the
estimate (5.5) holds for t− t′ ∈ I, and cot t′
2 is bounded for t− t′ 6∈ I.Let J be a compact subinterval of I. Without loss of generality, we can assume
that J = [−δ, δ] with δ < π2
and that the double interval J ′ = [−2δ, 2δ] is alsocontained in I. For t ∈ J , we then have
∣∣f(t) − Pr ∗ f(t)∣∣ =
1
2π
∣∣∣∣∫ π
−π
(f(t− t′) − f(t)
)(cot
t′
2− Pr(t
′))dt′
∣∣∣∣
=1
2π
∣∣∣∣∫ π
−π
(f(s) − f(t)
)(cot
t− s
2− Pr(t− s)
)ds
∣∣∣∣
≤ 1
2π
( ∫
J ′
∣∣f(s) − f(t)∣∣∣∣∣ cot
t− s
2− Pr(t− s)
∣∣∣ ds
+
∫
[−π,π]\J ′
∣∣f(s) − f(t)∣∣∣∣∣ cot
t− s
2− Pr(t− s)
∣∣∣ ds)
=1
2π
(I1(t, r) + I2(t, r)
).
THE UNIT DISC 25
We now want to prove that
limr→1
supt∈J
∣∣f(t) − Pr ∗ f(t)∣∣ = 0 .
In the integral I1(t, r), both t and s are in I, so that
|I1(t, r)| ≤∫
J ′
|t− s|α∣∣∣ cot
t− s
2− Pr(t− s)
∣∣∣ ds
=
∫
t−t′∈J ′
|t′|α∣∣∣ cot
t′
2− Pr(t
′)∣∣∣ dt′
≤∫ 2π
−2π
|t′|α∣∣∣ cot
t′
2− Pr(t
′)∣∣∣ dt′ ,
which is independent of t. By dominated convergence, as in the proof of Proposition5.4, this last quantity tends to zero as r → 1, so that
limr→1
supt∈I′
I1(t, r) = 0 .
Passing to I2(t, r), let M = maxt∈J |f(t)|. Then
I2(t, r) =
∫
[−π,π]\J ′
∣∣f(s) − f(t)∣∣∣∣∣ cot
t− s
2− Pr(t− s)
∣∣∣ ds
≤∫
[−π,π]\J ′
(|f(s)|+M
) ∣∣∣ cott− s
2− Pr(t− s)
∣∣∣ ds .
Observe that, if t ∈ J and s 6∈ J ′, |t− s| > δ. Therefore
(5.8)
∣∣∣ cott− s
2− Pr(t− s)
∣∣∣ =∣∣∣ sin(t− s)
1 − cos(t− s)− 2r sin(t− s)
1 + r2 − 2r cos(t− s)
∣∣∣
=| sin(t− s)|
1 − cos(t− s)
∣∣∣1 − 2r(1 − cos(t− s)
)
1 + r2 − 2r cos(t− s)
∣∣∣
=∣∣∣ cot
t− s
2
∣∣∣ (1 − r)2
1 + r2 − 2r cos(t− s)
The quantity∣∣ cot t−s
2
∣∣/(1 + r2 − 2r cos(t− s)
)remains bounded for |t− s| > δ.
Given ε > 0, we can then find r0 < 1 such that, for r > r0,∣∣∣ cot
t− s
2− Pr(t− s)
∣∣∣ < ε
for every t, s with |t− s| > δ.Therefore, if r > r0,
I2(t, r) ≤ ε2π(‖f‖1 +M
).
This shows thatlimr→1
supt∈I′
I2(t, r) = 0 ,
concluding the proof.
26 CHAPTER I
6. The Cauchy projection
The Hardy space Hp(D) is a closed subspace of hp(D). It is clear from Section 3that the Poisson integral of a function (or measure) f on T is holomorphic in D if
and only if f(n) = 0 for n < 0. If we define
Lp+(T) = f ∈ Lp(T) : f(n) = 0 for n < 0 ,
and similarlyM+(T) = µ ∈M(T) : µ(n) = 0 for n < 0 ,
the statement of Theorem 4.3 can be repeated word by word, replacing hp withHp, Lp with Lp+, and M with M+.
In the case p = 2 we are working with Hilbert spaces, and we want to describethe orthogonal projection from h2(D) to H2(D), that we shall denote by C (forCauchy).
Lemma 6.1. For u ∈ h2(D), let u be its harmonic conjugate. Then
(6.1) Cu =1
2(u+ iu) +
1
2u(0) .
Denoting by u] ∈ L2(T) the boundary function of u, i.e. u] = limr→1 ur, then
(6.2) Cu(z) =
∫
T
u](eit)
1 − ze−itdt .
Proof. Since znn≥0 ∪ znn≥1 is an orthonormal basis of h2(D), and znn≥0
spans H2(D), if we write
u(z) =∞∑
n=0
anzn +
∞∑
n=1
a−nzn ,
it follows that
Cu(z) =∞∑
n=0
anzn .
Then (6.1) follows from (2.6). Consider now the Fourier series of u],
u](eit) =∑
n∈Z
aneint ,
with convergence in the L2(T)-norm. Since
(Cu)r(eit) =
∞∑
n=0
anrneint ,
one obtains (Cu)r from u] by multiplying each Fourier coefficient u](n) = an by rn
if n ≥ 0 and by 0 if n < 0. Consider therefore the function
(6.3) Cr(eit) =
∞∑
n=0
rneint =1
1 − reit,
THE UNIT DISC 27
called the Cauchy kernel. Then (Cu)r(n) = u](n)Cr(n) for every n ∈ Z, so that
Cu(reiθ) = (Cu)r(eiθ)
= u] ∗ Cr(eiθ)
=
∫
T
u](eit)
1 − rei(θ−t)dt ,
giving (6.2).
Observe that (6.2) can be rewritten as a contour integral,
(Cu)(z) =1
2πi
∮
∂D
u](w)
w − zdw ,
an expression resembling the ordinary Cauchy integral formula. This is compatiblewith the fact that if u is already in H2(D) (i.e. it is holomorphic), we then havethe identity, for |z| < r < 1,
u(z) =1
2πi
∮
γr
u(w)
w − zdw
=1
2π
∫ π
−π
ur(eit)
reit − zreit dt
=
∫
T
ur(eit)
1 − zre−it
dt
where γr is the circle of radius r oriented counterclockwise. Letting r → 1, oneobtains (6.2), since we are assuming that Cu = u.
The orthogonal projection of h2(D) onto H2(D) corresponds, passing to bound-ary values, to the orthogonal projection C] of L2(T) onto L2
+(T), via the followingcommutative diagram:
L2(T)C]
−→ L2+(T)
↓ P ↓ Ph2(D)
C−→ H2(D)
SoC]f = lim
r→1(CPf)r ,
where the limit is meant in the L2-norm.We want to give an expression of C] that does not involve the harmonic extension
to the interior. Because of Lemma 6.1, this problem can be reduced to that offinding a more direct formula for the operator
H : L2(T) −→ L2(T) ,
mapping f intoHf = lim
r→1Pr ∗ f ,
again in the L2-norm. This operator is bounded, by Proposition 5.1, and
(6.4)
Hf(eit) = limr→1
f ∗ Pr(eit)
= limr→1
∫
T
f(ei(t−t′))
2r sin t′
1 + r2 − 2r cos t′dt′ .
28 CHAPTER I
Proposition 6.2. If f ∈ L2(T),
(6.5) Hf(eit) = limε→0
1
2π
∫
ε<|t′|<πf(ei(t−t
′)) cott′
2dt′ ,
where the limit is in the L2-norm.
Proof. For 0 < ε < 1, define
Hεf(eit) =1
2π
∫
ε<|t′|<πf(ei(t−t
′)) cott′
2dt′ .
If we prove that
(6.6) limε→0
‖f ∗ P1−ε −Hεf‖2 = 0 ,
this will imply the assertion, by (6.4).We can write
f ∗ P1−ε(t) −Hεf(t) =1
2π
( ∫
|t′|<εf(ei(t−t
′))P1−ε(t′) dt′
+
∫
ε<|t′|<πf(ei(t−t
′))(P1−ε(t
′) − cott′
2
)dt′
)
= f ∗ ϕε(t) ,
where
ϕε(t′) =
P1−ε(t′) if |t′| < ε
P1−ε(t′) − cot t′
2if ε ≤ |t′| ≤ π .
By (5.8),
∣∣∣P1−ε(t′) − cot
t′
2
∣∣∣ =∣∣∣ cot
t′
2
∣∣∣ ε2
1 + (1 − ε)2 − 2(1 − ε) cos t′
=∣∣∣ cot
t′
2
∣∣∣ ε2
ε2 + 4(1 − ε) sin2 t′
2
≤∣∣∣ cot
t′
2
∣∣∣ ε2
ε2 + 4π2 (1 − ε)t′2
.
We are interested in this quantity for ε ≤ |t′| ≤ π, where we can use the inequality
∣∣∣ cott′
2
∣∣∣ ≤ 1
| sin t′
2 |≤ π
|t′| ≤π
ε.
We can also assume that ε is small enough so that 4π2 (1 − ε) > 1. For ε and t′
subject to these restrictions,
∣∣∣ cott′
2− P1−ε(t
′)∣∣∣ ≤ πε
ε2 + t′2.
THE UNIT DISC 29
On the other hand, if |t′| < ε,
∣∣P1−ε(t′)∣∣ =
2(1 − ε)| sin t′|1 + (1 − ε)2 − 2(1 − ε) cos t′
≤ 2ε
ε2 + t′2.
Putting these two inequalities together, we have
|ϕε(t′)| ≤πε
ε2 + t′2,
for t′ ∈ [−π, π]. We can then apply Corollary 3.3 with c = 0. We leave theverification of conditions (1) and (3) to the reader, and observe that ϕε has integral0 because it is odd on [−π, π]. This gives (6.6).
We cannot eliminate the limit in (6.5) and write the integral over all of T,because the integrand would not be absolutely convergent in general. For the samereason, we cannot replace the integrals outside of the simmetric intervals [−ε, ε]with integrals outside of arbitrary intervals containing 0, such as, for instance,[−ε, 2ε], because the result would not be the same.
This matter can be nicely formalized in the language of distribution theory.Observe that
(6.7) limε→0
∫
ε<|t|<πf(t) cot
t
2dt
does not exist, or is infinite, for a generic continuous function f on T (take forinstance f(t) = sgn t
log |t| near t = 0).
If, however, f is a C1-function16, we have, using the oddness of cot t2 ,
limε→0
∫
ε<|t|<πf(t) cot
t
2dt = lim
ε→0
∫
ε<|t|<π
(f(t) − f(0)
)cot
t
2dt
=
∫ π
−π
(f(t) − f(0)
)cot
t
2dt ,
where this last integral is absolutely convergent. Moreover the linear functional
(6.8) f 7−→ Φ(f) = limε→0
∫
ε<|t|<πf(t) cot
t
2dt
is continuous w.r. to the norm
‖f‖C1 = ‖f‖∞ + ‖f ′‖∞
16f α-Lipschitz would be enough, but distribution theory is focused on spaces of Ck- or C∞-functions.
30 CHAPTER I
on C1(T). In fact
∣∣Φ(f)∣∣ =
∣∣∣∣∫ π
−π
(f(t) − f(0)
)cot
t
2dt
∣∣∣∣
≤∫ π
−π
∣∣f(t) − f(0)∣∣∣∣∣ cot
t
2
∣∣∣ dt
≤ ‖f ′‖∞∫ π
−π|t|
∣∣∣ cott
2
∣∣∣ dt
≤ C‖f‖C1 .
This property is expressed by saying that Φ is a distribution of order 1 on T.According to a standard terminology, a distribution such as Φ, involving limits ofintegrals that would not be abolutely convergent otherwise, is called a principalvalue distribution. One writes
limε→0
1
2π
∫
ε<|t|<πf(t) cot
t
2dt = p.v.
∫
T
f(t) cott
2dt .
The convolution Φ ∗ f is defined for f ∈ C1(T) as
Φ ∗ f(t) = Φ(f(t− ·)
)= p.v.
∫
T
f(t− t′) cott′
2dt′ ,
and one can easily prove that Φ ∗ f ∈ C(T).
One also writes
p.v. cott
2∗ f , or
(p.v. cot
t
2
)∗ f
for Φ ∗ f .
As a consequence of Proposition 6.2, we have the following extension result.
Corollary 6.3. For f ∈ C1(T), Hf = Φ ∗ f . Therefore the convolution operatorf 7→ Φ ∗ f extends in a unique way to a continuous operator from L2(T) to itself.
Moreover
C]f =1
2
(f + iHf + f(0)
).
7. Blaschke products and the F. and M. Riesz theorem
Working with holomorphic functions, one comment must be made on productsof Hp-functions17.
17We do not discuss this issue for hp-functions because, in general, the product of two harmonicfunctions is not harmonic.
THE UNIT DISC 31
Lemma 7.1. Suppose f ∈ Hp(D) and g ∈ Hq(D), with 1 ≤ p, q ≤ ∞ and 1p + 1
q =1s≤ 1. Then fg ∈ Hs(D), ‖fg‖Hs ≤ ‖f‖Hp‖g‖Hq . If p, q > 1, denoting by f ]
and g] the boundary functions of f, g respectively, then (fg)] is also a function and(fg)] = f ]g].
Proof. Suppose first that p and q are both finite. By Holder’s inequality,
Ms(fg, r) ≤Mp(f, r)Mq(g, r)
for every r < 1. Therefore fg ∈ Hs(D) and ‖fg‖Hs ≤ ‖f‖Hp‖g‖Hq . Moreover,
‖(fg)r − f ]g]‖s = ‖frgr − f ]g]‖s≤ ‖(fr − f ])gr + (gr − g])f ]‖s≤ ‖(fr − f ])gr‖s + ‖(gr − g])f ]‖s≤ ‖fr − f ]‖p‖gr‖q + ‖gr − g]‖q‖f ]‖p≤ ‖fr − f ]‖p‖g‖Hq + ‖gr − g]‖q‖f‖Hq .
This shows that (fg)r tends to f ]g] in Ls(D), so that (fg)] = f ]g].If one or both exponents are ∞, the identity (fg)] = f ]g] follows from the
above and the inclusion H∞(D) ⊂ Ht(D) for t < ∞. Once this is established, theinequality of norms is obvious.
As we see, this result implies (when q = p′) that an H1-function which is the
product of an Hp- and an Hp′ -function is the Poisson integral of an L1-funtion onthe torus. On the other hand, the picture given by Lemma 7.1 is not complete,because it does not say what happens on the boundary when one of the two ex-ponents, say p, is equal to 1. In that case we are forced to assume that q = ∞,but stil we do not know if the product f ]g] of a Borel measure times a boundedmeasurable function makes sense.
All these problems are washed out by the following theorem.
Theorem 7.2 (F. and M. Riesz). Let µ be a Borel measure in M+(T). Then µis absolutely continuous with respect to Lebesgue measure.
In other words, M+(T) = L1+(T).
Corollary 7.3. If f ∈ H1(D), the limit limr→1 fr exists in the L1-norm. Holo-morphic polynomials are dense in H1(D).
Proof. The first part is an immediate consequence of Theorems 7.2 and 4.3. Forthe second part, proceed as in the proof of Corollary 4.4.
There are several proofs of the brothers Riesz theorem. The one we give involvesan important notion in Hp-theory, the Blaschke products.
For α ∈ D, consider the function
(7.1) ψα =z − α
αz − 1.
One easily verifies that ψα is holomorphic on C\1/α (hence on a neighborhoodof D), injective, and that on T
|ψα(eit)| =|eit − α||α− e−it| = 1 ,
32 CHAPTER I
hence18 |ψα(z)| < 1 for z ∈ D.Suppose now that f ∈ Hp(D) is not identically zero, and that f(α) = 0 for some
α ∈ D. Then the function
g(z) =f(z)
ψα(z)
has a removable singularity at z = α. Hence g is holomorphic in D.
Lemma 7.4. g ∈ Hp(D) and ‖g‖Hp = ‖f‖Hp .
Proof. We clearly have
Mp(f, r) = Mp(gψα, r) ≤ Mp(g, r) .
Given ε > 0, it is possible to find r0 > |α| such that if |z| > r0, then |ψα(z)| >1 − ε. Then, if r > r0,
Mp(f, r) = Mp(gψα, r) ≥ (1 − ε)Mp(g, r) .
Passing to the limit as r → 1 and using the arbitrarity of ε, we have the conclu-sion.
This suggests a way to “remove zeroes” from a function f ∈ Hp(D) which isnot identically 0, without modifying its norm. If the zeroes of f in D are a finitenumber, α0, α1, . . . , αn (counted with their multiplicities), it is sufficient to repeatthe above construction n + 1 times to obtain a factorization of f(z) as g(z)B(z),with
(7.2) B(z) =
n∏
j=0
ψαj(z) =
n∏
j=0
z − αjαjz − 1
.
Then g has no zeroes in D and ‖g‖Hp = ‖f‖Hp .If the zeroes of f in D are infinitely many, they form a countable set without
accumulation points in D itself. Ordering these zeroes by increasing modulus, andcounting them with their multiplicities, we obtain a sequence αj, with |αj| ≤|αj+1|, and limj→∞ |αj | = 1.
This is true for any non-identically zero holomorphic function in D. Somethingmore can be said about distribution of zeroes of Hp-functions.
Lemma 7.5. Let f ∈ Hp(D) be not identically zero, and let f(z) = azkh(z), withk ≥ 0 and h(0) = 1. Then
∑
j :αj 6=0
(1 − |αj |
)≤ log ‖h‖Hp = log ‖f‖Hp − log |a| .
Proof. The last equality being obvious, we can assume that k = 0 and a = 1, i.e.that f = h.
We initially suppose that f is holomorphic on a larger disc DR = z : |z| < Rwith R > 1, and also that f does not vanish for |z| = 1. We then have only finitely
18More precisely, ψα is a conformal map of D onto itself (i.e. a Mobius transformation). It canbe shown that any Mobius transformation has the form eiθψα for some α ∈ D and some eiθ ∈ T.
THE UNIT DISC 33
many zeroes α0, . . . , αn in D, and we can decompose f as f = gB, with B given by(7.2). Observe that B is defined in a larger disc D′ = D1+δ and non-zero in D′ \D,so that g is holomorphic in D′.
Since g has no zero in D′, log g admits a holomorphic determination19 on D′.Then log |g| = <e log g is harmonic in D′. By the mean value theorem,
log |g(0)| =∫
T
log |g(eit)| dt
=
∫
T
log |f(eit)| dt−∫
T
log |B(eit)| dt
=
∫
T
log |f(eit)| dt .
But
log |g(0)| = log |f(0)| −n∑
j=0
log∣∣ψαj
(0)∣∣
= −n∑
j=0
log |αj | ,
so that
(7.3) −n∑
j=0
log |αj | =
∫
T
log |f(eit)| dt .
Passing to a general f ∈ Hp(D), with f(0) = 1, we apply (7.3) to fr(z) = f(rz),with r < 1 and r 6= |αj | for every j. The zeroes of fr in D are the α′
j = αj/r, with|αj| < r. Therefore,
−∑
j:|αj|<rlog
|αj|r
=
∫
T
log |f(reit)| dt .
Observe now that, for 0 < x < 1, logx < x− 1. Therefore
1 − |αj|r
< − log|αj|r
,
and ∑
j:|αj|<r
(1 − |αj |
r
)<
∫
T
log |f(reit)| dt .
The left-hand side is an increasing function of r, so that, by the Beppo-Levitheorem,
limr→1
∑
j:|αj|<r
(1 − |αj|
r
)=
∞∑
j=0
(1 − |αj|
).
19This is true since D′ is simply connected. In fact g′
gis holomorphic in D, and it admits a
primitive h. Then (eh/g)′ = 0, so that eh = cg. Adding an appropriate constant to h, we obtainc = 1.
34 CHAPTER I
This limit is then controlled by the supremum over r on the right-hand side, i.e.
∞∑
j=0
(1 − |αj |
)≤ sup
r<1
∫
T
log |f(reit)| dt .
It remains to prove that this supremum is finite if f ∈ Hp(D). It is sufficient totake p = 1, by the inclusion properties of Hardy spaces. By Jensen’s inequality20,if r 6= rj for every j,
exp( ∫
T
log |f(reit)| dt)
=
(exp
( ∫
T
log |f(reit)|p dt)) 1
p
≤( ∫
T
|f(reit)|p dt) 1
p
≤ ‖f‖Hp ,
so that ∞∑
j=0
(1 − |αj|
)≤ log ‖f‖Hp .
This concludes the proof.
In principle, we would like to decompose an Hp-function f(z) with infinitelymany zeroes αj as the product g(z)B(z), with the finite product (7.2) replacedby the analogous infinite product. The requirements would be
(1) that the partial products converge uniformly on compact subsets of D,(2) that the limit function vanish only at the zeroes of f .
These requirements already show that some modification to (7.2) must be made.Suppose that an infinite product of complex numbers aj converges to a non-zerovalue, i.e.
∞∏
j=0
aj = limn→∞
n∏
j=0
aj = limn→∞
An = A 6= 0 .
Then necessarily aj 6= 0 for every j and
limj→∞
aj = limj→∞
AjAj−1
= 1 .
Consider therefore a point z ∈ D different from the αj , i.e. such that f(z) 6= 0.In general, we cannot expect that limj→∞ ψαj
(z) = 1. However, this difficulty isbypassed if we by multiply each ψαj
by a contant factor of modulus 1.So, for α ∈ D, define
(7.4) ψα(z) =α
|α|ψα(z) =α
|α|z − α
αz − 1if α 6= 0, and ψ0(z) = z .
20Let m be a probability measure on a set X, and let f : X → R be measurable. If ϕ : R → R
is convex, then
ϕ“
Z
X
f(x)dm(x)”
≤
Z
X
ϕ`
f(x)´
dm(x) .
Here it is applied with ϕ(t) = et.
THE UNIT DISC 35
For αj 6= 0,
ψαj(z) − 1 =
αjz − |αj|2 − |αj|αjz + |αj ||αj|(αjz − 1)
=1 − |αj||αj|
αjz + |αj |αjz − 1
.
Since|αjz − 1| ≥ 1 − |αj||z| > 1 − |z| ,
we have
(7.5)∣∣ψαj
(z) − 1∣∣ < 2
1 − |αj|1 − |z| .
It follows that
(7.6) limj→∞
ψαj(z) = 1 .
We then construct the Blaschke product
(7.7) B(z) =∞∏
j=0
ψαj(z) .
Proposition 7.6. Let αj be a sequence of complex numbers αj ∈ D, such that∑∞j=0
(1−|αj|
)<∞. Then the Blaschke product (7.7) converges unconditionally 21
and uniformly on compact sets to a function B(z) ∈ H∞(D) vanishing only at thepoints αj. Moreover, |B(eit)| = 1 almost everywhere on T.
Proof. Let Bn(z) =∏nj=0 ψαj
(z) be the partial products of (7.4). We discussunconditional convergence first.
If z is one of the αj, then Bn(z) = 0 for n large, and this holds independently ofordering.
If z is not one of the αj, it follows from (7.6) that for j ≥ j0 = j0(z), the points
ψαj(z) are contained in the disc D centered at 1 and radius 1/2. Let log z be the
principal determination of the logarithm in D, i.e. such that log 1 = 0. Then, ifn > j0,
Bn(z) = Bj0−1(z)n∏
j=j0
elog ψαj(z)
= Bj0−1(z) exp
( n∑
j=j0
log ψαj(z)
).
We can then say that the Blaschke product is unconditionally convergent at zto a non-zero limit if and only if the series
∞∑
j=j0
log ψαj(z)
21This means independently of reorderings of the terms.
36 CHAPTER I
is unconditionally (i.e. absolutely) convergent. Since logww−1 is holomorphic in D and
continuous on D, | logw| ≤ C|w − 1| for w ∈ D. By (7.5),
(7.8)
∞∑
j=j0
∣∣ log ψαj(z)
∣∣ ≤ C∞∑
j=j0
∣∣1 − ψαj(z)
∣∣
≤ C
1 − |z|
∞∑
j=j0
(1 − |αj |
).
We prove next uniform convergence on compact sets. Given a compact subsetK of D, there is r < 1 such that |z| ≤ r for z ∈ K. We take now j0 = j0(K) large
enough so that |αj| > r for j ≥ j0, and ψαj(z) ∈ D for j ≥ j0 and z ∈ K. This can
be done by (7.5).We verify the Cauchy condition for the Bn in the uniform norm on K. If z ∈ K
and j0 < n < m,
∣∣Bn(z) − Bm(z)∣∣ =
∣∣Bn(z)∣∣∣∣∣∣1 −
m∏
j=n+1
ψαj(z)
∣∣∣∣
≤∣∣∣∣1 − exp
( m∑
j=n+1
log ψαj(z)
)∣∣∣∣ .
By (7.8),
∣∣∣∣m∑
j=n+1
log ψαj(z)
∣∣∣∣ ≤m∑
j=n+1
∣∣ log ψαj(z)
∣∣
≤ C
r
1 + r
1 − r
m∑
j=n+1
(1 − |αj |
).
Given ε, 0 < ε < 1, we can then take n0 large enough so that, if n0 ≤ n < m,
∣∣∣∣m∑
j=n+1
log ψαj(z)
∣∣∣∣ < ε ,
for all z ∈ K. Let a > 0 be such that |1 − ew| < a|w| for |w| < 1. Then
∣∣Bn(z) −Bm(z)∣∣ ≤
∣∣∣∣1 − exp( m∑
j=n+1
log ψαj(z)
)∣∣∣∣
< a
∣∣∣∣m∑
j=n+1
log ψαj(z)
∣∣∣∣
< aε ,
for all z ∈ K.Finally, we prove that |B(eit)| = 1 a.e. on T. Since |B(z)| < 1 for z ∈ D,
‖B‖H∞ ≤ 1, so that |B(eit)| ≤ 1 a.e. on T.
THE UNIT DISC 37
Let
Rn(z) =B(z)
Bn(z)=
∞∏
j=n+1
ψαj(z) .
Then Rn is also a Blaschke product holomorphic in D, and
(7.9) |B(z)| ≤ |Rn(z)| ≤ 1 .
On the boundary we have the identity
B(eit) = Bn(eit)Rn(e
it) ,
by Lemma 7.1. Moreover, |Bn(eit)| = 1 for every t, so that |B(eit)| = |Rn(eit)|.Observe that
|Rn(0)| = M1(Rn, 0) ≤∫
T
|Rn(eit)| dt =
∫
T
|B(eit)| dt .
But
limn→∞
Rn(0) = limn→∞
∞∏
j=n+1
|αj| = limn→∞
eP
∞
j=n+1log |αj | = 1 .
Therefore, ∫
T
|B(eit)| dt ≥ 1 ,
and, combining this with (7.9), we conclude that |B(eit)| = 1 a.e.
Theorem 7.7. Let f ∈ Hp(D), not identically zero, and let B be its Blaschke
product. Then g = fB ∈ Hp(D) and ‖g‖Hp = ‖f‖Hp .
Proof. With Bn denoting the partial product of B. By Lemma 7.4, ‖f/Bn‖Hp =‖f‖Hp . Fix r < 1. Since the functions |f/Bn| converge monotonically to |f/B|, itfollows that, if 1 > r 6= |αj | for every j, then
Mp(f/B, r) = limn→∞
Mp(f/Bn, r) ≤ ‖f‖Hp .
Therefore, g ∈ Hp(D) and ‖g‖Hp ≤ ‖f‖Hp . On the other hand, since |f(z)| ≤|g(z)| for every z ∈ D, we trivially have ‖f‖Hp ≤ ‖g‖Hp .
An immediate consequence is the following factorization theorem, a kind of in-verse of Lemma 7.1.
Corollary 7.8. Let f ∈ Hs(D), not identically zero, with 1 ≤ s ≤ ∞, and letp, q ∈ [1,∞] be such that 1
p + 1q = 1
s . Then there exist g ∈ Hp(D) and h ∈ Hq(D)
such that f = gh, and ‖g‖pHp = ‖h‖qHq = ‖f‖sHs .
Proof. Let B be the Blaschke product of f and ϕ = fB . Then ϕ ∈ Hs(D) and has
no zeroes in D. Letg(z) = ϕ(z)
sp = e
sp
logϕ(z) .
Then |g(z)| = |ϕ(z)| sp , so that
Mp(g, r)p =
∫
T
|g(reit)|p dt =
∫
T
|ϕ(reit)|s dt = Ms(ϕ, r)s .
38 CHAPTER I
Therefore g ∈ Hp(D) and ‖g‖pHp = ‖ϕ‖sHs = ‖f‖sHs .
If we set h(z) = ϕ(z)sqB(z), then g(z)h(z) = ϕ(z)
sp+ s
qB(z) = f(z), and moreover
Mq(h/B, r)q = Ms(ϕ, r)
s ,
as before. Therefore,
‖h‖qHq = ‖h/B‖qHq = ‖ϕ‖sHs = ‖f‖sHs .
We can prove now the F. and M. Riesz theorem.
Proof of Theorem 7.2. Let f(z) be the Poisson integral of µ. Since µ(n) = 0 forn < 0, then
f(z) =∞∑
n=0
µ(n)zn
is holomorphic. Therefore f ∈ H1(D). We can assume that µ 6= 0, so that f is notidentically zero. We write f = gh, with g, h ∈ H2(D), according to Corollary 7.8.By Lemma 7.1, f has a boundary function
f(eit) = g(eit)h(eit) ∈ L1(T) .
It follows that dµ(t) = f(eit) dt, i.e. µ is absolutely continuous w.r. to the Lebesguemeasure.
8. Dual spaces
The Cauchy projection
C : u(z) =∞∑
n=0
bnzn +
∞∑
n=1
b−nzn 7−→ Cu(z) =
∞∑
n=0
bnzn
is well defined as a linear map from the space of all harmonic functions on D, withvalues in the space of homorphic functions on D. It follows from (6.1) that C mapshp(D) into Hp(D) if and only if the conjugate function operator maps hp(D) intoitself.
Therefore we know from Section 5 that this does not happen if p = 1 or ∞, andwe still have to see that it happens instead if 1 < p <∞. The scope of this sectionis to remark that this issue is relevant for determining the dual space of Hp(D).
Consider the inner product in h2(D):
(8.1) 〈u, v〉h2 =
∫
T
u](eit)v](eit) dt = limr→1
∫
T
u(reit)v(reit) dt =∑
n∈Z
anbn ,
if u], v] are the boundary functions and an, bn their Fourier coefficients (i.e. theTaylor coefficients of u and v respectively). By the general theory of Hilbert spaces,the map
v 7−→ ϕv(u) = 〈u, v〉h2
THE UNIT DISC 39
establishes a conjugate-linear isometric isomorphism between h2(D) and its dualspace h2(D)∗.
In the same way, restricting to H2(D),
g 7−→ ϕg(f) = 〈f, g〉H2
establishes a conjugate-linear isometric isomorphism between H2(D) and its dualspace H2(D)∗.
Let us see first how (8.1) can be extended to u ∈ hp(D) and v ∈ hp′
(D), wherep′ is the conjugate exponent of p.
Lemma 8.1. Let u ∈ hp(D) and v ∈ hp′
(D). If p = 1, assume that either u] ∈L1(T) or v] ∈ C(T) (and symmetrically if p = ∞). Then the limit
B(u, v) = limr→1
∫
T
u(reit)v(reit) dt
exists and it equals∫
Tu](eit)v](eit) dt.
Proof. Suppose first that 1 < p <∞. Then, denoting by ‖ ‖p the norm in Lp(T),∣∣∣∣∫
T
u(reit)v(reit) dt−∫
T
u](eit)v](eit) dt
∣∣∣∣
≤∣∣∣∣∫
T
(u(reit) − u](eit)
)v(reit) dt
∣∣∣∣ +
∣∣∣∣∫
T
u](eit)(v(reit) − v](eit)
)dt
∣∣∣∣≤ ‖ur − u]‖p‖vr‖p′ + ‖u]‖p‖vr − v]‖p′≤ ‖ur − u]‖p‖v‖hp′ + ‖u‖hp‖vr − v]‖p′ .
By Theorem 4.3, ‖ur − u]‖p → 0, and ‖vr − v]‖p′ → 0, and this proves thestatement.
If p = 1 (and p = ∞ is the same), we have∣∣∣∣∫
T
u(reit)v(reit) dt−∫
T
u](eit)v](eit) dt
∣∣∣∣
≤ ‖ur − u]‖1‖vr‖∞ +
∣∣∣∣∫
T
(vr(e
it) − v](eit))u](eit) dt
∣∣∣∣ .
In this case, the conclusion follows from the fact that ‖ur − u]‖1 → 0, and thatvr → v] in the weak* topology of L∞(T), again by Theorem 4.3.
Corollary 8.2. The sesquilinear form B in Lemma 8.1 is continuous on hp(D)×hp
′
(D), precisely, if u ∈ hp(D), v ∈ hp′
(D), then∣∣B(u, v)∣∣ ≤ ‖u‖hp‖v‖hp′ .
If an, bn are their respective Taylor coefficients, then the series∑
n∈Z anbn isAbel summable, and
limr→1
∑
n∈Z
anbnr|n| = B(u, v) .
Proof. The first statement follows from Lemma 8.1. For the second, it is sufficientto observe that
∑
n∈Z
anbnr|n| =
∫
T
u(√reit)v(
√reit) dt .
In terms of linear functionals on Hp(D), the following is an immediate conse-quence.
40 CHAPTER I
Proposition 8.3. If g ∈ Hp′(D), the linear functional
ϕg(f) = B(f, g)
is continuous on Hp(D), and ‖ϕg‖Hp(D)∗ ≤ ‖g‖Hp′ . The map
Φ : g 7→ ϕg
is a norm-decreasing conjugate-linear immersion of Hp′(D) into Hp(D)∗.
Proof. The only fact that does not follows directly from Corollary 8.2 is the injec-tivity of Φ. But if ϕg = 0 for some g(z) =
∑∞n=0 bnz
n ∈ Hp′(D), then
bn = B(zn, g) = 0 ,
for every n. Therefore g = 0.
It is then natural to ask if Φ is surjective onto Hp(D)∗.For p <∞ there is an interesting connection with the boundedness of the Cauchy
projection.
Theorem 8.4. Suppose p < ∞. Then Φ : Hp′(D) → Hp(D)∗ is surjective if and
only if the Cauchy projection is bounded from hp′
(D) to Hp′(D).
Proof. By Lemma 5.2,
(8.3) ‖g‖Hp′ ≤ A‖ϕg‖Hp(D)∗ ≤ A‖g‖Hp′ ,
for some constant A > 0.Take u ∈ hp
′
(D), and denote u] its boundary function. The linear functionalψ(f) = B(f, u) is continuous on Hp(D), because
|ψ(f)| ≤ ‖f ]‖p‖u]‖p′ = ‖u‖hp′‖f‖Hp .
In particular, ‖ψ‖Hp(D)∗ ≤ ‖u‖hp′ . Being Φ surjective, there is g ∈ Hp′(D) suchthat ψ = ϕg, and
‖g‖Hp′ ≤ A‖ψ‖Hp(D)∗ ≤ A‖u‖hp′ ,
by (8.3). We show that g = Cu, by observing that, if
u(z) =
∞∑
n=0
anzn +
∞∑
n=1
a−nzn ,
and
g(z) =
∞∑
n=0
bnzn ,
then ψ(zn) gives u](n) = an for n ≥ 0, but at the same time it equals ϕg(zn) = bn.
Therefore,‖Cu‖Hp′ ≤ A‖u‖hp′ .
THE UNIT DISC 41
Conversely, suppose that the Cauchy projection maps hp′
(D) into Hp′(D), andlet ψ be a continuous linear functional on Hp(D). Define
ψ] : Lp+(T) −→ C
as ψ](f ]) = ψ(f). By the Hahn-Banach theorem, ψ] admits a continuous extension
to all of Lp(T), with the same norm. So there is h ∈ Lp′
(T) such that
‖h‖p′ = ‖ψ]‖Lp+
(T)∗ = ‖ψ‖Hp(D)∗ ,
and
ψ(f) =
∫
T
f ](eit)h(eit) dt ,
for every f ∈ Hp(D).
If u ∈ hp′
(D) is the Poisson integral of h, so that h = u],
ψ(f) = B(f, u) .
Let
u(z) =∞∑
n=0
bnzn +
∞∑
n=1
b−nzn ,
and
f(z) =∞∑
n=0
anzn
be the series expansions of u and f . By Corollary 8.2,
ψ(f) = limr→1
∞∑
n=0
anbnrn = B(f, Cu) .
Hence ψ = Φ(Cu), and Cu ∈ Hp′(D).
As we have anticipated, we shall see in Chapter III that C is bounded from hp(D)to Hp(D) for 1 < p <∞, so that, for these values of p, Φ provides an identification
between Hp(D)∗ and Hp′(D).Since C does not map bounded harmonic functions into bounded holomorphic
functions (see Proposition 5.3), not all the continuous linear functionals on H1(D)can be represented as ϕg for some g ∈ H∞(D).
Consider, for example, v(z) = arg(1 − z) = =m log(1 − z) ∈ h∞(D). Thenϕv is bounded on H1(D), and, for f ∈ H1(D), B(f, v) = B(f, g) with g(z) =12i log(1 − z) 6∈ H∞(D).
More general examples arise from the following Hardy’s inequality.
Theorem 8.5. Let f(z) =∑∞n=0 anz
n be in H1(D). Then
∞∑
n=0
|an|n+ 1
≤ ‖f‖H1 .
42 CHAPTER I
Proof. We assume first that f(0) = a0 = 0, and prove in this case the strongerinequality
∞∑
n=1
|an|n
≤ ‖f‖H1 .
The function
g(t) =
∫ t
0
f ](eis) ds
is continuous on the real line, and, since∫
Tf ](eit) dt = f(0) = 0, it is periodic of
period 2π. Hence it defines a continuous function on T, whose derivative is f ], andwhose Fourier coefficients are
g(0) =1
2π
∫ 2π
0
∫ t
0
f ](eis) ds dt
=1
2π
∫ 2π
0
f ](eis)
∫ 2π
s
dt ds
=1
2π
∫ 2π
0
f ](eis)(2π − s) ds
= − 1
2π
∫ 2π
0
sf ](eis) ds ,
and
g(n) =1
2π
∫ 2π
0
e−int∫ t
0
f ](eis) ds dt
=1
2π
∫ 2π
0
f ](eis)
∫ 2π
s
e−int dt ds
=1
2π
∫ 2π
0
f ](eis)1
in(e−ins − 1) ds
=f ](n)
in,
if n 6= 0. Hence,
g(n) =
anin
if n > 0
0 if n < 0 .
It follows that the function
F (z) = ig(0) +
∞∑
n=1
annzn ,
which is a holomorphic primitive of f(z)/z in D, coincides with the Poisson integralof ig, hence it is continuous on D.
In particular, by Theorem 4.3,
ig(0) + limr→1
∞∑
n=1
annrn = F (1) = 0 .
THE UNIT DISC 43
Assume first that an ≥ 0 for every n > 0. By monotone convergence,
∞∑
n=1
ann
= limr→1
∞∑
n=1
annrn
=i
2π
∫ 2π
0
sf ](eis) ds
≤ 1
2π
∫ 2π
0
s|f ](eis)| ds
≤ ‖f‖H1 .
We remove now the assumption that an ≥ 0, and write f = f1f2, with f1, f2 ∈H2(D), and ‖f1‖2
H2 = ‖f2‖2H2 = ‖f‖H1 , according to Corollary 7.8. Since f(0) = 0,
one of the two factors must vanish at 0, say f1(0) = 0.If
f1(z) =
∞∑
n=1
bnzn , f2(z) =
∞∑
n=0
cnzn ,
let
g1(z) =
∞∑
n=1
|bn|zn , g2(z) =
∞∑
n=0
|cn|zn .
Then ‖g1‖H2 = ‖f1‖H2 , ‖g2‖H2 = ‖f2‖H2 . Hence g = g1g2 ∈ H1(D) and‖g‖H1 ≤ ‖g1‖H2‖g2‖H2 = ‖f‖H1 .
Observe that|an| =
∣∣∣∑
j+k=n
bjck
∣∣∣ ≤∑
j+k=n
|bj ||ck| = an ,
which is the n-th Taylor coefficient of g. Therefore,
∞∑
n=1
|an|n
≤∞∑
n=1
ann
≤ ‖g‖H1 ≤ ‖f‖H1 .
If f(0) 6= 0, we just replace f by zf , observing that ‖zf‖H1 = ‖f‖H1 .
Corollario 8.6. Let g(z) =∑∞n=0 bnz
n with bn = O(n). Then ϕg(f) = B(f, g) isa continuous linear functional on H1(D).
Proof. If f(z) =∑∞
n=0 anzn ∈ H1(D), it follows from Hardy’s inequality that
B(f, g) = limr→1
∞∑
n=1
anbnrn =
∞∑
n=1
anbn ,
by dominated convergence. Therefore,
∣∣B(f, g)∣∣ ≤ ‖f‖H1 sup
n∈N
|bn|n+ 1
.
To conclude, the case p = ∞ must be treated separately. As one can reasonablyexpect, the dual space of H∞(D) is strictly bigger than Φ
(H1(D)
). This can be
shown in many ways.
44 CHAPTER I
One argument is based on the following remark. Let HC(D) the space of holo-morphic functions that admit a continuous extension to D. Then HC(D) is aclosed subspace of H∞(D), and proper, because it does not contain the infiniteBlaschke products. By the Hahn-Banach theorem, there is a non-zero continuouslinear functional ϕ on H∞(D) which vanishes identically on HC(D). Suppose thatϕ(f) = B(f, g) for some g ∈ H1(D), and let
g(z) =
∞∑
n=0
anzn .
Since zn ∈ HC(D) for n ≥ 0,
0 = B(zn, g) = an ,
so that g = 0. But this contradicts the fact that ϕ is not identically zero.This argument motivates a new question: can Φ be surjective from H1(D) onto
the dual space of HC(D)? The answer is negative again. It is possible to modifythe first part of the proof of Theorem 8.4 to show that, if this is the case, then theCauchy projection maps h1(D) into H1(D), which we know to be false.
THE HALF-PLANE 45
CHAPTER II
HARDY SPACES ON THE HALF-PLANE
1. Definitions and basic facts
Let now D+ be the upper half-plane in C, D+ = x+ iy : y > 0.
Definition. If f is holomorphic (resp. harmonic) in D+, we say that f ∈ Hp(D+)(resp. f ∈ hp(D+)), for 1 ≤ p ≤ ∞, if for every y > 0 fy(x) = f(x + iy) is inLp(R), and
(1.1) supy>0
‖fy‖p <∞ .
We keep the notation Mp(f, r) for ‖fr‖p, and ‖f‖Hp (resp. ‖f‖hp) for the supre-mum in (1.1).
Lemma 1.1. If f ∈ hp(D+), and z = x+ iy ∈ D+, then
|f(z)| ≤ Cp‖f‖hp
y1p
.
Proof. We assume that p <∞, the other case being trivial.
For every r < y,
f(z) =1
2π
∫ π
−πf(z + reit) dt ,
by the mean value property. Integrating in polar coordinates around z, we thenhave
(1.2)
1
|B(z, y)|
∫
B(z,y)
f(w) dw =1
πy2
∫ y
0
∫ π
−πf(z + reit) dt r dr
=2
y2
∫ y
0
f(z) r dr
= f(z) .
Therefore, using Holder’s inequality and the inclusion B(z, y) ⊂ Sy = u + iv :
Typeset by AMS-TEX
46 CHAPTER II
0 < v < 2y,
|f(z)| ≤ 1
πy2
∫
B(z,y)
|f(w)| dw
≤(
1
πy2
∫
B(z,y)
|f(w)|p dw) 1
p
≤(
1
πy2
∫
Sy
|f(w)|p dw) 1
p
=
(1
πy2
∫
Sy
|f(u+ iv)|p du dv) 1
p
≤(
1
πy2
∫ 2y
0
Mp(f, v)p dv
) 1p
≤ ‖f‖hp
(1
πy2
∫ 2y
0
dv
) 1p
=( 2
πy
) 1p ‖f‖hp .
As for the corresponding spaces on the unit disc, we then have the followingimmediate consequence.
Theorem 1.2. Hp(D+) and hp(D+) are Banach spaces, and convergence in theirnorm implies uniform convergence on compact sets.
The information given by Lemma 1.1 concerns the behaviour of an hp-function fboth for y → 0 and for y → ∞. It does not say anything, however, on the behaviourof f(z) when z tends to infinity within a horizontal strip. The best one can say isthe following.
Lemma 1.3. For 0 < a < b, let Sa,b = x+ iy : a ≤ y ≤ b. If f ∈ hp(D+), withp <∞, , then
limz→∞, z∈Sa,b
f(z) = 0 .
Proof. For z ∈ Sa,b, let Bz the disc centered at z of radius a. By (1.2),
(1.3) |f(z)| ≤ 1
πa2
∫
Bz
|f(w)| dw ≤(
1
πa2
∫
Bz
|f(w)|p dw) 1
p
.
Observe that Bz ⊂ S0,b+a, and that
∫
S0,b+a
|f(w)|p dw =
∫ b+a
0
∫
R
|f(x+ iy)|p dx dy ≤ (b+ a)‖f‖hp <∞ .
Therefore, given ε > 0, there is M > 0 such that
(1.4)
∫ b+a
0
∫
|x|>M|f(x+ iy)|p dx dy < πa2εp .
THE HALF-PLANE 47
If |<ez| > M + a, Bz ⊂ x + iy : |x| > M, 0 < y < b + a. Putting (1.3) and(1.4) together, we obtain that |f(z)| < ε.
In contrast with the corresponding spaces on the unit disc, no Hardy space onD+ is contained in any other, as it can be seen with appropriate examples. Forinstance, if α > 0,
fα(z) =1
(z + i)α
is in Hp(D+) if and only if pα > 1, and
gα(z) =1
zα(z + i)2
is in Hp(D+) if and only if pα < 1.
2. Poisson integrals
Before discussing Poisson integrals in D+, we must recall some facts about con-volution on R on one side, and about conformal mappings on the other side.
The convolution of two continuous functions f and g with compact support isdefined as
(2.1) f ∗ g(x) =
∫
R
f(x− t)g(t) dt .
By a simple change of variables, one verifies that f ∗ g = g ∗ f . It is also easy toverify that
supp (f ∗ g) ⊆ supp f + supp g = x+ x′ : x ∈ supp f, y ∈ supp g .
When it will be necessary to distinguish between convolution on T and convolu-tion on R, we shall write ∗T and ∗R accordingly.
The integral (2.1) may not make sense for more general functions, unless certainintegrability conditions are satisfied22. One can prove that the integral (2.1) isconvergent for almost every x if f ∈ Lp(R), g ∈ Lq(R), with 1 ≤ p, q ≤ ∞ and
(2.2)1
p+
1
q≥ 1 .
In this case f ∗ g ∈ Lr(R), with
1
r=
1
p+
1
q− 1 ,
and the Young inequality
(2.3) ‖f ∗ g‖r ≤ ‖f‖p‖g‖q22Observe that if f(x) = (1 + |x|)−α and g(x) = (1 + |x|)−β, the integral (2.1) is divergent for
every x if α+β ≤ 1. Compare this example with the restrictions on the exponents p and q below.
48 CHAPTER II
holds. When r = ∞ (i.e. p and q are conjugate exponents), f ∗g is also continuous.If, in addition, 1 < p, q <∞, f ∗ g vanishes at infinity.
The convolution of two finite, regular Borel measures µ, ν ∈ M(R) is defined asthe measure µ ∗ ν such that
(2.4)
∫
R
f(x) d(µ ∗ ν)(x) =
∫
R
∫
R
f(s+ t) dµ(s) dν(t) ,
for f ∈ C0(R). One has the inequality
(2.5) ‖µ ∗ ν‖1 ≤ ‖µ‖1‖ν‖1 .
In a more subtle way, the convolution µ ∗ f of a measure µ ∈ M(R) with afunction f ∈ Lp(R), 1 ≤ p ≤ ∞, is also well defined, and
(2.6) ‖µ ∗ f‖p ≤ ‖µ‖1‖f‖p .
Recall now that if A and B are two connected, simply connected, proper opensubsets of C, there exists a conformal mapping23 ϕ from A onto B. For A = D,the unit disc, and B = D+, one can explicitely write such mappings. One of themis
(2.7) ϕ(z) = i1 + z
1 − z,
and it is called the Cayley transform.
Lemma 2.1. The Cayley transform ϕ maps D onto D+, it is invertible, and
(2.8) ϕ−1(w) =w − i
w + i.
Moreover ϕ has a continuous extension to D \ 1, and
(2.9) ϕ(eiθ) = − cotθ
2∈ R = ∂D+ .
Proof. One has
ϕ(z) = i(1 + z)(1 − z)
|1 − z|2 = i1 − |z|2 + 2i=mz
|1 − z|2 ,
so that
=mϕ(z) =1 − |z|2|1 − z|2 > 0 ,
if z ∈ D . Hence ϕ maps D into D+.It is easy to verify that ϕ is injective and that (2.8) gives its inverse function. If
w ∈ D+, |w−i| < |w+i| by a simple geometric consideration, so that |ϕ−1(w)| < 1.This shows that ϕ is onto.
The extension of ϕ to the boundary is easy to derive.
We can use the Cayley transform to trasfer harmonic functions from D to D+
and viceversa.
23i.e. holomorphic and invertible. The fact we are stating is called the Riemann mapping
theorem.
THE HALF-PLANE 49
Lemma 2.2. Let A,B be open subsets of C. If ϕ : A → B is holomorphic, and uis harmonic on B, then u ϕ is harmonic on A.
Proof. We can assume that u is real-valued. Fix z ∈ A, and let U ⊂ B be acircular neighborhood of w = ϕ(z). By Corollary 2.6 od Chapter I, u|B is the realpart of a holomorphic function f . Then f ϕ is holomorphic on the neighborhoodV = ϕ−1(U) of z, and u ϕ = <e(f ϕ) is harmonic on V .
Lemma 2.3. Let u be continuous on D+, harmonic in D and bounded. Then
(2.10) u(x+ iy) =1
π
∫
R
u(t)y
(x− t)2 + y2dt .
Proof. If ϕ is the Cayley transform, the function v(z) = u ϕ(z) is in h∞(D). Inaddition, v is continuous on D \ 1. By dominated convergence,
∫
T
v(eiθ) dθ = limr→1
∫
T
v(reiθ) dθ = v(0) .
Since v(0) = u(i) and v(eiθ) = u(− cot θ2 ) by (2.9), the change of variable − cot θ2 = tgives
u(i) =1
2π
∫ 2π
0
v(eiθ) dθ =1
π
∫ +∞
−∞u(t)
1
t2 + 1dt ,
which is (2.10) for x+ iy = i.For a general point x0 + iy0 ∈ D+, consider the function u(z) = u(x0 + y0z),
which satisfies the same hypotheses as u. Then, with simple changes of variables,
u(x0 + iy0) = u(i)
=1
π
∫ +∞
−∞u(x0 + y0t)
1
t2 + 1dt
=1
π
∫ +∞
−∞u(x0 + t)
y0t2 + y2
0
dt
=1
π
∫ +∞
−∞u(t)
y0(x0 − t)2 + y2
0
dt .
Definition. For y > 0, the function
Py(x) =1
π
y
x2 + y2
is called the Poisson kernel on R.
Formula (2.10) can be written as
(2.11) uy = u0 ∗ Py .
50 CHAPTER II
Corollary 2.4. Let u ∈ hp(D+), with 1 ≤ p ≤ ∞. Given 0 < y1 < y2, then
uy2 = uy1 ∗ Py2−y1 .
In particular, the Poisson kernel has the semigroup property
(2.12) Py+y′ = Py ∗ Py′ .
Proof. Let v(z) = u(z+ iy1). By Lemma 1.1, v satisfies the assumptions of Lemma2.3. Therefore
uy2 = vy2−y1 = v0 ∗ Py2−y1 = uy1 ∗ Py2−y1 .Applying this identity to
u(x+ iy) = Py(x) ∈ h1(D+)
(observe that u is harmonic because it equals − 1π=m 1
x+iy ), we obtain (2.12).
A further important property of the Poisson kernel is that it forms an approx-imate identity on the real line. By definition, an approximate identity on R (forε→ 0) is a family of functions ϕεε>0 satisfying
(1)∫
R|ϕε(t)| dt ≤ C for some constant C and every ε;
(2) limε→0
∫Rϕε(t) dt = 1 for every r;
(3) for every δ > 0,
limε→0
∫
|t|>δ|ϕε(t)| dt = 0 .
Proposition 3.2 of Chapter I extends to approximate identities on R.
Lemma 2.5. Take ϕ ∈ L1(R) such that∫
Rϕ(x) dx = 1. Then the functions
ϕε(x) = 1εϕ(xε
)form an approximate identity for ε→ 0. In particular, the Poisson
kernels Py form an approximate identity for y → 0.
Proof. A simple change of variable shows that ‖ϕε‖1 = ‖ϕ‖1 and∫
Rϕε(t) dt =∫
Rϕ(t) dt = 1. This proves (1) and (2). By the same change of variable,
∫
|t|>δ|ϕε(t)| dt =
∫
|t|> δε
|ϕ(t)| dt ,
which tends to 0 with ε.For the Poisson kernel, observe that
(2.13) Py(x) =1
yP1
(xy
),
and
‖P1‖ =1
π
∫
R
1
x2 + 1dx = 1 .
We are now in a position to extend to the spaces hp(D) the results proved aboutnorm or weak* convergence to the boundary. We summarize them in the followingstatement.
THE HALF-PLANE 51
Theorem 2.6. If u ∈ hp(D+), then Mp(u, y) = ‖uy‖p is decreasing in y, so that
‖u‖hp = limy→0
Mp(u, y) .
The operator P mapping f (understood as either a function or a Borel measureon R) into the harmonic function Pf on D+ given by
(Pf)(x+ iy) = f ∗ Py(x) ,maps Lp(R) isometrically onto hp(D+) for 1 < p ≤ ∞, and it maps M(R) isomet-rically onto h1(D+).
The limit limy→0(Pf)y exists in Lp if and only if one of the following holds
(1) f ∈ Lp(R) and 1 ≤ p <∞;(2) p = ∞ and f ∈ C(R).
In each of these cases, (Pf)y → f in the Lp-norm.For general elements f of M(R) or L∞(R), (Pf)y tends to f in the corresponding
weak*-topology.
Proof. Once we have proven that Pf is harmonic, it is a matter of adapting theproof of Theorem 4.3 in Chapter I.
We verify the mean value property for u = Pf assuming that f is a function(the case f ∈M(R) is left to the reader). If B(z0, r) ⊂ D+, then
(4.6)
∫
T
u(z0 + reiθ) dθ =
∫
T
∫
R
P (z0 + reiθ − t)f(t) dt dθ .
We can apply Fubini’s theorem and change the order of integration for the fol-lowing reason. For fixed θ, write z0 + reiθ = xθ + iyθ. Then
P (z0 + reiθ − t)f(t) = P (xθ + iyθ − t)f(t)
is integrable in t, because it is the product of an Lp′
- and an Lp-function. By (2.13),
‖Py‖q =
( ∫
R
1
yq
∣∣∣P1
(xy
)∣∣∣q
dx
) 1q
=Cq
y1− 1q
.
if 1 ≤ q <∞, and similarly for q = ∞.There is a δ > 0 such that =myθ ≥ δ for all θ. Therefore
∫
R
∣∣P (z0 + reiθ − t)∣∣∣∣f(t)
∣∣ dt ≤ Cp,δ‖f‖p
uniformly in θ. Therefore the double integral in (4.6) is absolutely convergent.We apply Fubini’s theorem, and use the fact that the function P (x + iy) is
harmonic in D+, as observed in the proof of Corollary 2.4. Then∫
T
u(z0 + reiθ) dθ =
∫
R
f(t)
∫
T
P (z0 + reiθ − t) dθ dt
=
∫
R
f(t)P (z0 − t) dt
= u(z0) .
The following corollary can be seen as a replacement for the (non-existing) in-clusion relations among the hp(D+).
52 CHAPTER II
Corollary 2.7. If 1 < p < ∞, and 1 ≤ q ≤ ∞, then hp(D+) ∩ hq(D+) is dense inhp(D+). This does not hold if p = 1,∞ and q 6= p.
Proof. By Theorem 2.6, it is sufficient to prove the corresponding statement forLp(R) (1 < p ≤ ∞) and M(R).
If f ∈ Lp(R) with 1 < p < ∞, and 1 ≤ q < p, then fR = fχ[−R,R] is in bothLp(R) and Lq(R), and fR → f in the Lp-norm as R→ ∞. If p < q ≤ ∞, the sameholds for fR = fχ|f |<R.
For p = 1 6= q, observe that M(R) ∩ Lq(R) = L1(R) ∩ Lq(R), so that its closurein M(R) is only L1(R).
For p = ∞, observe that the constant function 1 cannot be the uniform limit offunctions in Lq(R) if q <∞.
We must also mention the behaviour of Mp(u, y) as y → +∞.
Proposition 2.8. If 1 < p <∞ and u ∈ hp(D+), then limy→+∞Mp(u, y) = 0.
Proof. Given ε > 0, there is a continuous function ϕ on R with compact supportsuch that ‖u] − ϕ‖p < ε. Then, for y > 0,
Mp(u, y) = ‖u] ∗ Py‖p≤ ‖(u] − ϕ) ∗ Py‖p + ‖ϕ ∗ Py‖p< ε+ ‖ϕ‖1‖Py‖p≤ ε+ y
− 1
p′ ‖ϕ‖1 .
So, if y is large enough, Mp(u, y) < 2ε.
3. The Fourier transform and the Paley-Wiener theorem
We begin this section by recalling some of the basic facts about the Fouriertransform on R.
If f ∈ L1(R) and ξ ∈ R, one defines the Fourier transform of f at ξ as
(3.1) f(ξ) =
∫
R
f(x)e−iξx dx .
The inequality |f(ξ)| ≤ ‖f‖1 is an immediate consequence of the definition. The
function f so defined is continuous on R and vanishes at infinity (this last fact isknown as the Riemann-Lebesgue theorem). Then the linear operator F mapping
f into Ff = f is continuous from L1(R) into C0(R).One also defines the Fourier transform of a finite Borel measure µ ∈M(R) as
µ(ξ) =
∫
R
e−iξx dµ(x) .
Then |µ(ξ)| ≤ ‖µ‖1, µ is continuous, but we can no longer say that it vanishesat infinity.
THE HALF-PLANE 53
Further properies of F are the following.
(1) if f(x) = f(−x), then f(ξ) = f(−ξ);(2) f(ξ) = f(−ξ);(3) f ∗ g(ξ) = f(ξ)g(ξ);
(4) if fε(x) = 1εf(xε ), then fε(ξ) = f(εξ);
(5) if f is absolutely continuous, so that also f ′ ∈ L1(R), then f ′(ξ) = iξf(ξ);
(6) if both f and xf are integrable, than f is C1 and (f)′(ξ) = −i (xf)(ξ);
(7) if f ∈ L1(R) ∩ L2(R), then f ∈ L2(R) and the Plancherel formula holds:
(3.2)1
2π
∫
R
∣∣f(ξ)∣∣2 dξ =
∫
R
|f(x)|2 ;
(8) because of (3.2), F extends to every f ∈ L2(R), and ‖f‖2 =√
2π‖f‖2;(9) since Lp(R) ⊂ L1(R) + L2(R), if 1 < p < 2, F is well defined on Lp(R),
f ∈ Lp′
(R), and the Hausdorff-Young inequality holds:
(3.3) ‖f‖p′ ≤ ‖f‖p ;
(10) if f is integrable24, the inversion formula holds:
(3.4) f(x) =1
2π
∫
R
f(ξ)eixξ dξ .
Lemma 3.1. The Fourier transform of Py is
Py(ξ) = e−y|ξ| .
If u ∈ hp(D+), with 1 ≤ p ≤ 2, and u] is its boundary function (or measure),then
(3.5) uy(ξ) = u](ξ)e−y|ξ| .
Proof. By (4) above and (2.13),
Py(ξ) = P1(yξ) ,
and by (3) and (2.12),
Py+y′(ξ) = P1
((y + y′)ξ
)= P1(yξ)P1(y
′ξ) .
Moreover, by (1) and (2), P1 is real and even. Putting these information together
with the fact that P1 ∈ C0(R), we find that P1(ξ) = e−a|ξ| for some a > 0.
24This happens, in particular, if f is differentiable and f ′ ∈ L2(R).
54 CHAPTER II
In particular, P1 ∈ L1(R), so that we can use the inversion formula in order todetermine a. We are so led to compute
1
2π
∫
R
e−a|ξ|eixξ dξ =1
π
∫ ∞
0
e−aξ cos(xξ) dξ
=1
π<e
( ∫ ∞
0
e−(a−ix)ξ dξ
)
=1
π<e
1
a− ix
=1
π
a
x2 + a2.
This must be equal to P1(x), hence a = 1. The rest of the proof is obvious.
The Fourier transform allows to describe for which functions f ∈ Lp(R), thePoisson integral u(x + iy) = f ∗ Py(x) is holomorphic. The main result concernsp = 2, and is called the Paley-Wiener theorem.
Theorem 3.2. A function u ∈ h2(D+) is holomorphic if and only if the Fourier
transform u] of its boundary function u] ∈ L2(R) is zero on the negative half-line. Hence a holomorphic function f on D+ is in H2(D+) if and only if f is theFourier-Laplace transform,
(3.6) f(z) =1
2π
∫ ∞
0
g(ξ)eizξ dξ ,
of some g ∈ L2(R+), and in this case ‖f‖H2 = 1√2π
‖g‖2.
Proof. Suppose that u is holomorphic. Calling γa the horizontal curve γa(t) = t+ia,consider the line integral
Ia =
∫
γa
u(z)e−iξz dz =
∫ ∞
−∞u(x+ ia)e−iξ(x+ia) dx = eaξ ua(ξ) .
We claim that Ia = Ib if a, b > 0. Suppose a < b, fix M > 0 and let RM bethe rectangle with vertices −M + ia, M + ia, M + ib, −M + ib. Denote by γa,M ,
γb,M the arcs of γa and γb describing the horizontal edges of RM , and by γ+M , γ−M
the arcs describing the vertical edges, oriented upwards. By the Cauchy integralformula applied to the boundary of RM , we have
(3.7)
0 =
∫
γa,M
u(z)e−iξz dz +
∫
γ+
M
u(z)e−iξz dz
−∫
γb,M
u(z)e−iξz dz −∫
γ−
M
u(z)e−iξz dz .
Now, ∫
γ+
M
u(z)e−iξz dz = i
∫ b
a
u(M + it)e−iξ(M+it) dt ,
THE HALF-PLANE 55
so that
∣∣∣∣∫
γ+
M
u(z)e−iξz dz
∣∣∣∣ ≤∫ b
a
|u(M + it)|eξt dt
≤ eb|ξ|(b− a) maxt∈[a,b]
|u(M + it)| ,
which tends to 0 as M → ∞ by Lemma 1.3. Then, passing to the limit as M → ∞in (3.7), we obtain that Ia = Ib.
So, by (3.5),
eaξua(ξ) = e−a(|ξ|−ξ)u](ξ)
is independent of a. This forces u](ξ) to be a.e. zero for ξ < 0.
Since u] ∈ L2(R), then uy = e−y|ξ|u] ∈ L1(R) for y > 0, and the inversionformula (3.6) gives
u(x+ iy) =1
2π
∫ ∞
0
u](ξ)e−yξeixξ dξ =1
2π
∫ ∞
0
u](ξ)ei(x+iy)ξ dξ .
Conversely, take g ∈ L2(R+) and define f by (3.6). The integral is absolutelyconvergent, because |eizξ| = e−ξ=mz ∈ L2(R+). If γ is a closed arc in D+, then
∫
γ
f(z) dz =1
2π
∫
γ
∫ ∞
0
g(ξ)eizξ dξ dz .
If the support of γ is contained in the half-plane =mz ≥ a > 0, then |eizξ| ≤ e−aξ ,and g(ξ)eizξ is integrable on γ × R+. We can then change order of integration andobtain that ∫
γ
f(z) dz =1
2π
∫ ∞
0
g(ξ)
∫
γ
eizξ dz dξ = 0 ,
because eizξ is holomorphic in z. Then f is holomorphic by Morera’s theorem.Observe that fy is the inverse Fourier transform of g(ξ)e−yξχR+
. Therefore, byPlancherel’s formula (7),
‖fy‖2 =1√2π
‖e−y·g‖2 ,
and
‖f‖H2 =1√2π
limy→0
‖e−y·g‖2 =1√2π
‖g‖2 .
Corollary 3.3. Let 1 ≤ p ≤ 2, u ∈ hp(D+) and u] be its boundary function (or
measure). Then u is holomorphic if and only if u] is zero on the negative half-line. If
this is the case, then u is the Fourier-Laplace transform (3.6) of g = u] ∈ Lp′
(R+).
Proof. Let v(ε)(z) = u(z + iε). Then v(ε) ∈ hp(D+) ∩ h∞(D+), hence in h2(D+).
If we assume that u ∈ Hp(D+), then v(ε) ∈ H2(D+). Therefore,
v](ε)(ξ) = uε(ξ) = e−ε|ξ|u](ξ)
56 CHAPTER II
is zero on the negative half-line. By Holder’s inequality, uy = e−y|·|u] ∈ L1(R) fory > 0, so that the Fourier inversion formula gives (3.6) as before.
Conversely, if u ∈ hp(D+) and supp u] ⊂ [0,+∞), the same is true for v(ε), which
belongs to h2(D+). Therefore, v(ε) is holomorphic in D+, i.e. u is holomorphic inD+ + iε. By the arbitrarity of ε, u is holomorphic in D+.
4. The Cauchy projection and the Hilbert transform
By the Paley-Wiener theorem and Theorem 2.6, the map which assigns to a
harmonic function u ∈ h2(D+) the Fourier transform u] is, up to a constant, anisometry of h2(D+) onto L2(R), and the image of H2(D+) is
L2+ = g ∈ L2(R) : supp g ⊆ [0,+∞) ∼ L2(R+) .
Since the orthogonal projection of L2(R) onto L2+ is obviously the map g 7→
gχR+, we derive the following recipe for constructing the orthogonal projection
from h2(D+) onto H2(D+): starting with u ∈ h2(D+), take u], multiply it by χR+,
and apply (3.6). This gives the Cauchy projection on D+,
(4.1)
Cu(x+ iy) =1
2π
∫ ∞
0
u](ξ)ei(x+iy)ξ dξ
= F−1(u]e−yξχR+
)(x)
= u] ∗ F−1(e−yξχR+)(x) .
Define the Cauchy kernel Cy(x) = F−1(e−yξχR+)(x). Explicitely,
(4.2)
Cy(x) =1
2π
∫ ∞
0
e−yξeixξ dξ
= − 1
2πi
1
x+ iy.
We can then write (4.1) in the following way.
Theorem 4.1. The orthogonal projection C from h2(D+) onto H2(D+) is givenby
(4.3) Cu(z) = u] ∗ Cy(x) =1
2πi
∫ ∞
−∞
u](t)
t− zdt ,
with z = x+ iy.
If u is already in H2(D+), then Cu = u, and (4.3) formally coincides with theordinary Cauchy integral formula
u(z) =1
2πi
∫
γ0
u](w)
w − zdw ,
THE HALF-PLANE 57
where we denote by γ0 the curve describing the real axis with its natural orientation.Notice however that the curve is not bounded25 and it is not contained in the interiorof the domain where u is holomorphic.
Write now
Cy(x) = − 1
2πi
x− iy
x2 + y2=
1
2π
y
x2 + y2+
i
2π
x
x2 + y2=
1
2
(Py(x) + iPy(x)
),
noticing that the real part of 2Cy is the Poisson kernel Py and setting
(4.4) Py(x) =1
π
x
x2 + y2.
Therefore (4.3) takes the form
(4.5) Cu(x+ iy) =1
2u(x+ iy) +
i
2u] ∗ Py(x) .
Observe however that Py is not integrable for any y > 0, so that the convolution
u] ∗ Py is not defined for u ∈ h∞(D+). On the other hand, Py is in Lq(R) for anyq > 1 and in C0(R), so that the convolution is well defined for u ∈ hp(D+) withp <∞.
Proposition 4.2. Let u ∈ hp(D+) with p <∞. Then
u(x+ iy) = u] ∗ Py(x)
is harmonic, u+ iu is holomorphic, and
limz→∞ ,=mz≥a
u(z) = 0 ,
for every a > 0. If u is real-valued, the u is the only real-valued function satisfyingthe above properties.
Proof. We verify the mean value property for u. If B(z0, r) ⊂ D+, then
(4.6)
∫
T
u(z0 + reiθ) dθ =
∫
T
∫
R
P (z0 + reiθ − t)u](t) dt dθ .
We can apply Fubini’s theorem and change the order of integration for the fol-lowing reason. For fixed θ, write z0 + reiθ = xθ + iyθ. Then
P (z0 + reiθ − t)u](t) = P (xθ + iyθ − t)u](t)
is integrable in t, because it is the product of an Lp′
- and an Lp-function. Since
Py(x) =1
yP1
(xy
),
25It becomes a nice closed curve if embedded in the Riemann sphere.
58 CHAPTER II
and P1 ∈ Lq(R) for q > 1, we have, as in the proof of Theorem 2.6,
‖Py‖p′ ≤ Cpy− 1
p ,
for p <∞.The proof continues as for Theorem 2.6.
Observe that in the proof we have implicitely obtained the inequality
(4.7) |u(x+ iy)| ≤ Cpy− 1
p ‖u‖hp .
for u ∈ hp(D+) and p <∞.By (4.5),
u = iu− 2iCu ,
so that, if u ∈ h2(D+), also u ∈ h2(D+). If u] is the boundary function, we have
(4.8) u](ξ) = iu](ξ) − 2iχR+(ξ)u](ξ) = −i(sgn ξ)u](ξ) ,
so that, by Plancherel’s inequality,
(4.9) ‖u‖h2 = ‖u‖h2 .
The operations described in (4.8) take place on the real line, and define a linearoperator H : L2(R) → L2(R),
Hf = F−1(− i sgn ξf(ξ)
).
This operator is called the Hilbert transform. Proposition 6.2 in Chapter I hasthe following analogue (with a very similar proof, that we omit).
Proposition 4.3. For f ∈ L2(R),
Hf(x) = limε→0
1
π
∫
|t|>εf(x− t)
1
tdt ,
where the limit is in the L2-norm.
Much of the content of Sections 5 and 6 of Chapter I can be repeated here, withthe unit disc replaced by the upper half-plane.
The Cauchy projection is bounded on hp(D+) for p 6= 2 if and only if the sameis true for the conjugate function operator mapping u to u. As for the unit disc(and with the same arguments) this turns out to be false if p = 1. Also for p = ∞,when the formulas above do not hold, it is not true in general that the conjugatefunctions of a bounded harmonic function are bounded26. This is not even true ifu is bounded and continuous on D+.
We shall see later that C is bounded from hp(D+) onto Hp(D+) for 1 < p <∞.
26Take u(x + iy) = arccot xy
= arg z = =mlog z. It is bounded, but its harmonic conjugates,
− log |z| + constant, are unbounded.
THE HALF-PLANE 59
5. Transference of Hp-functions and applications
We shall now establish simple relations between Hp-functions on the unit discand the upper half-plane. This will allow us to transfer toD+ many results obtainedin Chapter I for D.
The simplest situation occurs for p = ∞: if ϕ denotes the Cayley transform(2.7), the map
(5.1) f 7−→ f ϕ−1 = T∞f
is an isometry of H∞(D) onto H∞(D+), for the simple reason that f is boundedif and only if f ϕ−1 is bounded, and the two ∞-norms are equal27.
Clearly the same does not hold for p < ∞ (take f = 1). In order to see whatneeds to be modified, let us consider the simple case where f is continuous on Dand holomorphic in the interior.
Then g = f ϕ−1 is continuous on D+, and let us try to compute the Lp-normof g|R . By (2.9), with the change of variable x = − cot θ
2we have
‖g|R‖p =
( ∫ ∞
−∞|f ϕ−1(x)|p dx
) 1p
=
(1
2
∫ 2π
0
|f(eiθ)|p dθ
sin2 θ2
) 1p
,
and this differs from the Lp-norm of g by the factor sin−2 θ2
coming from the change
of variable. This suggest that we must multiply g by a power of (ϕ−1)′(z) = −2i(z+i)2 .
Precisely, for p <∞ and f ∈ Hp(D), we set
(5.2) Tpf(z) =1
π1p (z + i)
2p
f ϕ−1(z) .
Proposition 5.1. If 1 < p ≤ ∞, Tp is an isometry of Hp(D) onto Hp(D+). Ifp = 1, T1 is an isometry of H1(D) onto the subspace H1
0 (D+) of those f ∈ H1(D+)such that f ] ∈ L1(R).
At the end of this section we shall prove that in fact H10 (D+) = H1(D+), as a
consequence of the F. and M. Riesz theorem for R.
Proof. The case p = ∞ having been discussed already, we take p < ∞. Assumefirst that f is continuous on D, bounded and holomorphic in the interior, and letg = Tpf . Then g is continuous on D+ and
Mp(g, y)p ≤ 1
π‖f‖p∞
∫
R
1
x2 + (y + 1)2dx ≤ ‖f‖p∞ ,
so that g ∈ Hp(D+). Let g] be its boundary function (or measure) on R, in thesense of Theorem 2.6. If 1 < p < ∞, gy → g] in the Lp-norm as y → 0, there is asequence yn → 0 such that gyn
→ g] almost everywhere. Therefore g] = g|R .
27This is true also for bounded harmonic functions, but what we shall do next for Hp-functionswith p <∞ has no analogue for harmonic functions.
60 CHAPTER II
If p = 1 we get to the same conclusion by a different argument. Since gy → g]
in the weak* topology, for any ϕ ∈ Cc(R),∫
R
ϕ(x) dg](x) = limy→0
∫
R
ϕ(x)g(x+ iy) dx =
∫
R
ϕ(x)g(x) dx .
Hence dg](x) = g(x) dx.We now have
‖g|R‖p =
(1
π
∫ ∞
−∞|f ϕ−1(x)|p dx
|x+ i|2) 1
p
.
If x = ϕ(eiθ), then
x+ i = i1 + eiθ
1 − eiθ+ i =
2i
1 − eiθ,
so that1
|x+ i|2 = sin2 θ
2.
Therefore the change of variable x = − cot θ2
gives dx|x+i|2 = 1
2dθ and then
‖g|R‖p = ‖f|T‖p .We have so proved that Tp is an isometry from a dense subspace of Hp(D) (see
Corollaries 4.3 and 7.3 in Chapter I) onto its image, call it V , in Hp(D+). ThereforeTp extends to an isometry (also denoted by Tp) of all Hp(D) onto the closure of Vin Hp(D+). The proof will be finished if we prove that Tp has a dense image inHp(D+) if p > 1 or in H1
0 (D+) if p = 1.Take g ∈ Hp(D+), resp. in H1
0 (D+), ε > 0, and define
hε(z) =g(z + iε)
(1 − iεz)2.
Observe that for z = x+iy ∈ D+, |1−iεz| ≥ 1+εy ≥ 1, so that the denominatordoes not vanish, and moreover
|hε(z)| ≤ |g(z + iε)| .This implies that hε ∈ Hp(D+). We prove that limε→0 hε = g in Hp(D+). In
fact
‖hε − g‖Hp =
( ∫
R
∣∣∣ gε(x)
(1 − iεx)2− g](x)
∣∣∣p
dx
) 1p
≤( ∫
R
|gε(x) − g](x)|p|1 − iεx|2p dx
) 1p
+
( ∫
R
|g](x)|p∣∣∣ 1
(1 − iεx)2− 1
∣∣∣p
dx
) 1p
≤( ∫
R
|gε(x) − g](x)|p dx) 1
p
+
( ∫
R
|g](x)|p∣∣∣ 1
(1 − iεx)2− 1
∣∣∣p
dx
) 1p
.
THE HALF-PLANE 61
As ε→ 0, the first term tends to 0 by Theorem 2.6, and the second by dominatedconvergence, since
∣∣ 1(1−iεx)2 − 1
∣∣ ≤ 2.
We show that hε is the image under Tp of a function in Hp(D). We computetherefore
T−1p hε(z) = π
1p
(ϕ(z) + i
) 2phε ϕ(z)
= π1p
( 2i
1 − z
) 2p 1(1 − iεϕ(z)
)2 g(ϕ(z) + iε
)
= π1p
( 2i
1 − z
) 2p (1 − z)2(1 + ε− (1 − ε)z
)2 g(ϕ(z) + iε
)
= π1p (2i)
2p
(1 − z)2
p′
(1 + ε− (1 − ε)z
)2 g(ϕ(z) + iε
).
The factor g(ϕ(z) + iε
)is bounded because =m
(ϕ(z) + iε
)> ε, and the denom-
inator does not vanish on D. therefore T−1p hε is bounded, hence in Hp(D).
This fact will allow us to transfer the factorization results proved in the unit discto D+.
Lemma 5.2. Let 1 ≤ p ≤ ∞. If f ∈ Hp(D+) is not identically zero and αjj∈N
is the sequence of its zeroes (repeated according to their multiplicities), then
(5.3)∞∑
j=0
=mαj|αj|2 + 1
<∞ .
Proof. For p > 1, we take g = T−1p f ∈ Hp(D), and apply Lemma 7.5 in Chapter I.
The zeroes of g are the points
βj = ϕ−1(αj) =αj − i
αj + i.
Then ∞∑
j=0
(1 − |βj |2) ∞∑
j=0
(1 − |βj |) <∞ ,
and, if αj = aj + ibj ,
1 − |βj |2 =|αj + i|2 − |αj − i|2
|αj + i|2
=4=mαj|αj + i|2
=4bj
a2j + (bj + 1)2
bja2j + b2j + 1
==mαj
|αj|2 + 1.
62 CHAPTER II
Take now f ∈ H1(D+). Replacing, if necessary, f(z) by f((1 + δ)z
), we can
assume that f(i) 6= 0. Notice that, in doing so, the zeroes change from αj to(1 + δ)−1αj , but this change does not affect the convergence of the series (5.3).
For ε > 0, let f ε(z) = f(z + iε). Then f ε ∈ H1(D+), ‖fε‖H1 ≤ ‖f‖H1 , and(fε)](x) = f(x + iε) = fε(x) ∈ L1(R). Therefore, T−1
1 fε ∈ H1(D). If ε is smallenough, say ε < ε0, T
−11 fε(0) = −4πf
((1 + ε)i
)6= 0. We then set
gε =1
T−11 fε(0)
T−11 fε .
If Eε = j : =mαj > ε, the zeroes of f ε in D+ are the αεj = αj− iε with j ∈ Eε,
and the zeroes of gε in D are βεj = ϕ−1(αεj), j ∈ Eε.Since gε(0) = 1, we can apply Lemma 7.5 in Chapter I to obtain that
∑
j∈Eε
(1 − |βεj |
)≤ sup
r<1
∫
T
log |gε(reit)| dt
≤ log ‖gε‖H1(D)
= log ‖f ε‖H1(D+) − log∣∣4πf
((1 + ε)i
)∣∣
Setting αj = aj + ibj and taking ε < ε0, we have
1 − |βεj | =mαεj
|αεj + i|2 =bj − ε
a2j + (bj − ε+ 1)2
bj − ε
a2j + b2j + 1
.
Therefore,
∑
j∈Eε
bj − ε
a2j + b2j + 1
≤ C(
log ‖fε‖H1(D+) − log∣∣4πf
((1 + ε)i
)∣∣).
If ε decreases, the left-hand side increases for each j ∈ N (taking into accountthat the sets Eε also increase). Passing to the limit by monotone convergence,
∑
j∈N
bja2j + b2j + 1
≤ C(log ‖f‖H1(D+) − log |4πf(i)|
)<∞ .
Blaschke products onD+ are defined as compositions Bϕ−1 = T∞B of Blaschkeproducts on D with the inverse Cayley transform. If w = ϕ−1(z) and β = ϕ−1(α),then, according to (7.4) in Chapter I, we define
σα(z) = ψβ(w) =β
|β|w − β
βw − 1=
|α2 + 1|α2 + 1
z − α
z − α
if α 6= i, and
σi(z) = ϕ−1(z) =z − i
z + i.
Once we are at this stage, we can simply state the analogues of the results ofSection 7 in Chapter I. They can be transfered directly from D via the maps Tp forp > 1, or reproved in exactly the same way for p = 1.
THE HALF-PLANE 63
Proposition 5.3. Let αj be a sequence of points in D+ satisfying (5.3). TheBlaschke product
B(z) =∞∏
j=0
σαj(z)
converges unconditionally and uniformly on compact subsets of D+ to a functionB ∈ H∞(D+) vanishing in D+ only at the points αj. Moreover, |B](x)| = 1 foralmost every x ∈ R.
Theorem 5.4. Let f ∈ Hp(D+), 1 ≤ p ≤ ∞, be not identically zero, and let B be
its Blaschke product. Then g = fB
∈ Hp(D+) and ‖g‖Hp = ‖f‖Hp.
Corollary 5.5. Let f ∈ Hs(D+), not identically zero, with 1 ≤ s ≤ ∞, and letp, q ∈ [1,∞] be such that 1
p+ 1q
= 1s. Then there exist g ∈ Hp(D+) and h ∈ Hq(D+)
such that f = gh, and ‖g‖pHp = ‖h‖qHq = ‖f‖sHs .
Finally, we can state the F. and M. Riesz theorem for the line.
Theorem 5.6 (F. and M. Riesz). Let µ ∈M(R) be a measure such that µ(ξ) = 0for ξ ≤ 0. Then µ is absolutely continuous w.r. to Lebesgue measure.
The proof follows from the factorization theorem 5.5 like in the disc.
Corollary 5.7. If f ∈ H1(D+), then f ] ∈ L1(R), and therefore limy→0 fy = f ]
in the L1-norm. The map T1 of Proposition 5.1 is an isometry of H1(D) ontoH1(D+).
64 CHAPTER II
POINTWISE CONVERGENCE 65
CHAPTER III
POINTWISE CONVERGENCE TO THE BOUNDARY
AND CONJUGATE HARMONIC FUNCTIONS IN hp
1. The Hardy-Littlewood maximal function
We attack now a question that has been in the background so far, i.e. thepointwise behaviour of hp- or Hp-functions. In particular, in the cases where f ] isa function, we wnt to know (e.g. in the disc, to fix the notation) if
limr→1
f(reit) = f ](eit)
for almost every eit ∈ T. If f ∈ H∞(D) and f ] is continuous, the answer is positive,since we know that the functions fr tend to f ] uniformly. Partial answers can begiven easily also in other cases. If f ∈ Hp(D) with 1 < p < ∞, then fr → f ]
in norm, and therefore there is a sequence rn → 1 such that frn→ f almost
everywhere. But this is much less than what we are looking for.Sharp answers to our question follow from considerations about maximal func-
tions. This notion makes sense and is useful in many different situations, and it isworth therefore to discuss their properties in a general context.
Let X be a topological space. A quasi-distance on X is a function d from X×Xto R such that
(1) d(x, y) ≥ 0 for all x, y ∈ X;(2) d(x, y) = 0 if and only if x = y;(3) d(x, y) = d(y, x) for all x, y ∈ X;(4) there is a constant c ≥ 1 such that
(1.1) d(x, z) ≤ c(d(x, y) + d(y, z)
)
for all x, y, z ∈ X;(5) the balls B(x, r) = y : d(x, y) < r form a fundamental neighborhood
system at each x ∈ X.
Let m be a positive regular Borel measure on X. One says that m is doubling(more precisely, d-doubling) if there is a constant c′ such that
(1.2) m(B(x, 2r)
)≤ c′m
(B(x, r)
)
for all x ∈ X and r > 0.
Typeset by AMS-TEX
66 CHAPTER III
Definition. A triple (X, d,m), where d is a quasi-distance on X and m is a d-doubling measure, is called a space of homogeneous type28.
Examples.
The following are spaces of homogeneous type29:
(1) The unit circle T, with the distance d(eit, eit′
) = |eit − eit′ | (or equivalently
with the quotient distance of R/2πZ) and the normalized Lebesgue measure.(2) Rn, with Euclidean distance and the Lebesgue measure.(3) R, with the Euclidean distance and the measure dm(x) = |x|αdx, with
α > −n.(4) Rn, with the Lebesgue measure and the distance
d(x, y) = max|x1 − y1|1/d1 , . . . , |xn − yn|1/dn
,
where d1, . . . , dn > 0.(5) Z, with the distance d(n,m) = |n−m| and the counting measure m(E) =
cardE.(6) The unit sphere Sn−1 ⊂ Rn, with the induced distance from Rn and the
Hausdorff measure.(7) The unit sphere S2n−1 ⊂ Cn, with the Hausdorff measure and the distance
d(ζ, ζ ′) = |1 − 〈ζ, ζ ′〉|, where 〈ζ, ζ ′〉 is the Hermitean inner product in Cn.
Let X be a space of homogeneous type.
Definition. Let f be locally integrable w.r. to m. The function
(1.3) Mf(x) = supx∈B
1
m(B)
∫
B
|f(y)| dy ,
is called the Hardy-Littlewood maximal function of f and the operator M : f 7−→Mf is called the Hardy-Littlewood maximal operator.
ClearlyM is not linear (observe thatMf(x) ≥ 0 for every f), but only sub-linear,in the sense that
(1.4) M(f + g) ≤Mf +Mg , M(λf) = |λ|Mf .
Lemma 1.1. The function Mf is lower-semicontinuous, hence measurable.
Proof. Let M(x0) > α. Then there is a ball B containing x such that
1
m(B)
∫
B
|f(y)| dy > α .
Then Mf(x) > α for every x ∈ B.
28Sometimes the definition of space of homogeneous type is given by requiring that X be justa set, and d a function satisfying conditions (1)-(4) only. A topology is then introduced on X by
stating that a set A is open if for every x ∈ A there is a ball B(x, r) ⊂ A. However, this does notguarantee that the balls are open, not even that they are Borel sets. Therefore, one must impose
that the doubling measure m be defined on a σ-algebra containing both the open sets and the
balls.29Some of the following statements would require a proof, that we omit.
POINTWISE CONVERGENCE 67
Remark. The classical definition of Hardy-Littlewood maximal function (for X =Rn, with Euclidean distance and Lebesgue measure) is the following:
(1.5) M ′f(x) = supr>0
1
m(B(x, r)
)∫
B(x,r)
|f(y)| dy ,
i.e. limited to the averages of |f | over balls centered at x. ClearlyM ′f(x) ≤Mf(x);however M ′f is not necessarily lower-semicontinuous. The measurability of M ′ffollows in this case from the fact that the map
F (x, r) =1
m(B(x, r)
)∫
B(x,r)
|f(y)| dy
is continuous in r, so that the sup in (1.5) can be limited to r ∈ Q.In a general space of homogeneous type, F (x, r) need not be continuous in r,
hence the definition (1.3) is preferable.
We want to discuss boundedness of M on the spaces Lp(X), i.e. inequalities ofthe form
‖Mf‖p ≤ C‖f‖p .
Notice that it follows from (1.4) that |Mf−Mg| ≤M(f−g), hence boundednesson Lp is equivalent to continuity, as for linear operators.
Obviously M is bounded on L∞(X). At the other extreme, p = 1, M is notbounded in general. For instance, in the classical siuation (X = Rn, etc.), takingf the characteristic function of the unit ball, one has
Mf(x) ≥ C
1 + |x|n ,
so that Mf 6∈ L1.Nevertheless, the starting point of our proof is that, for p = 1, M satisfies a
weaker form of boundedness. We first recall some properties of the distributionfunction, defined for α > 0,
(1.6) δf (α) = m(x : |f(x)| > α
)
of an m-measurable function f on X.
Lemma 1.2. Let (X,m) be a measure space. If f ∈ Lp(X,m), 1 ≤ p < ∞, thenthe following hold:
(1) the Chebishev inequality
δf (α) ≤(‖f‖p
α
)p;
(2) the identity
‖f‖pp = p
∫ ∞
0
δf (α)αp−1 dα .
68 CHAPTER III
Proof. Let Eα = x : f(x) > α. Then
m(Eα) ≤∫
Eα
|f(x)|pαp
dm(x) ≤(‖f‖p
α
)p.
This gives (1). Suppose now that f ∈ Lp(X,m) assumes a finite number ofvalues. Then |f | =
∑nj=1 cjχAj
, with 0 < c1 < · · · < cn and the Aj pairwisedisjoint. Then
m(Eα) =∑
j:cj>α
m(Aj) ,
and, setting c0 = 0,
p
∫ ∞
0
m(Eα)αp−1 dα = pn∑
j=1
∫ cj
cj−1
αp−1n∑
k=j
m(Ak) dα
=n∑
j=1
(cpj − cpj−1)n∑
k=j
m(Ak)
=
n∑
j=1
cpjm(Ak)
= ‖f‖pp .
For general f , one approximates |f | from below by finite-valued functions.
Definition. Let T be a linear, or sub-linear, operator defined on Lp(X), p < ∞,and taking values in the space of measurable functions on X. One says that T isweak-type (p, p) if, for every α > 0
(1.7) δTf (α) ≤ C(‖f‖p
α
)p.
The expression “weak-type” comes from the fact that (1.7) is a weaker conditionthan boundedness on Lp, as a consequence of Chebishev’s inequality. In fact, if Tis bounded on Lp(X),
δTf (α) ≤(‖Tf‖p
α
)p≤ ‖T‖pLp→Lp
(‖f‖pα
)p.
One also says that an operator T is strong type (p, p) if it is bounded on Lp(X).
Going back to the maximal operator M , we shall see that it is weak-type (1, 1).The proof is based on the following Vitali covering lemma.
Lemma 1.3. Inside a space X of homogeneous type, let Bjj∈J be a finite familyof balls covering a measurable set E. There exists a sub-family Bjj∈J ′ such thatBj ∩Bk = ∅ for j, k ∈ J ′, j 6= k, and
m
( ⋃
j∈J ′
Bj
)≥ κm(E) ,
POINTWISE CONVERGENCE 69
where κ dipends only on the constants c, c′ in (1.1) and (1.2).
Proof. Start with a ball Bj1 of maximum radius. Inductively, take Bjk+1with
maximum radius among the balls disjoint from Bj1∪· · ·∪Bjk . This procedure stopsafter a finite number of steps, precisely when there are no more balls left which aredisjoint from Bj1 ∪ · · · ∪ Bjk . Then set J ′ = j1, . . . , jk and call Bj1 , . . .Bjk the“selected” balls, and the remaining ones the “excluded” balls.
If B is a ball of radius r, denote by B∗ the ball with the same center and radius3c2r, where c is the constant in (1.1). Observe that if two balls B,B ′ have non-empty intersection and the radius of B′ is not larger than the radius of B, thenB′ ⊆ B∗ by (1.1).
Let B′ one of the excluded balls. It necessarily intersects one of the selectedones. Let ¯ be the smallest integer ` such that B′ ∩ Bj` 6= ∅. Then the radius Bj¯is greater than or equal to the radius of B′, so that
B′ ⊆ B∗j¯ .
ThereforeE ⊆
⋃
j∈JBj ⊆
⋃
j∈J ′
B∗j ,
Take ν so that 2ν ≥ 2c. Then
m(B∗j ) ≤ m
(B(x, 2kr)
)≤ c′
νm(B) ,
so that, with κ = 1c′ν ,
m(E) ≤∑
j∈J ′
m(B∗j ) = κ−1
∑
j∈J ′
m(Bj) = κ−1m( ⋃
j∈J ′
Bj
).
Theorem 1.4. The operator M is weak-type (1, 1).
Proof. Given f ∈ L1(X) and α > 0, let Eα = x : Mf(x) > α. By Lemma 1.1,Eα is open and its measure is the supremum of the measures of its compact subsets.
Let E be a compact subset of Eα. Given x ∈ E, Mf(x) > α, so that there is aball Bx containing x such that
1
m(Bx)
∫
Bx
|f(y)| dy > α ,
i.e.
(1.8) m(Bx) ≤1
α
∫
Bx
|f(y)| dy .
Since E is compact, we can extract a finite subcovering Bxjj∈J of E from
Bxx∈E . By Lemma 1.2, we can further extract a finite family Bxjj∈J ′ of
mutually disjoint balls such that
∑
j∈J ′
m(Bxj) ≥ κm
( ⋃
j∈JBxj
)≥ κm(E) .
70 CHAPTER III
Combining this with (1.8), we have
m(E) ≤ κ−1∑
j
m(Bxj) ≤ κ−1
α
∑
j
∫
Bxj
|f(y)| dy ≤ κ−1 ‖f‖1
α.
Taking the supremum over E ⊂ Eα, we obtain that m(Eα) ≤ κ−1‖f‖1/α.
Combining together the weak-type (1,1) property of M and its boundedness onL∞, we can prove that it is bounded on Lp for 1 < p < ∞. What follows is aspecial case of the Marcinkiewicz interpolation theorem30.
Theorem 1.5. Let T be a linear, or sub-linear, operator which is weak-type (1, 1)and bounded on L∞(X,m). Then T is bounded on Lp(X,m) for 1 < p < ∞.
Proof. Given f ∈ Lp(X,m) and α > 0, define
fα(x) =
f(x) if |f(x)| ≤ α
0 if |f(x)| > α ,fα(x) =
f(x) if |f(x)| > α
0 if |f(x)| ≤ α .
Then fα ∈ L∞(X,m) with ‖fα‖∞ ≤ α, and fα ∈ L1(X,m). In fact,
‖fα‖1 =
∫
|f(x)|>α|f(x)| dm(x)
≤∫
|f(x)|>α|f(x)| |f(x)|p−1
αp−1dm(x)
≤ 1
αp−1‖f‖pp .
If C∞ = ‖T‖L∞→L∞ , then ‖Tfα‖∞ ≤ C∞α. Since f = fα + fα,
|Tf(x)| ≤ |Tfα(x)| + |Tfα(x)| ,so that
|Tf(x)| > 2C∞α =⇒ |Tfα(x)| > C∞α ,
in other words,
x : |Tf(x)| > 2C∞α ⊆ x : |Tfα(x)| > C∞α .We use now (2) in Lemma 1.2 and the weak-type (1,1) of T to obtain
‖Tf‖pp = p
∫ ∞
0
m(x : |Tf(x)| > α
)αp−1 dα
= p(2C∞)p∫ ∞
0
m(x : |Tf(x)| > 2C∞α
)αp−1 dα
≤ p(2C∞)p∫ ∞
0
m(x : |Tfα(x)| > C∞α
)αp−1 dα
≤ p2pCp−1∞ C1
∫ ∞
0
‖fα‖1αp−2 dα
= p2pCp−1∞ C1
∫ ∞
0
αp−2
∫
x:|f(x)|>α|f(x)| dx dα
= p2pCp−1∞ C1
∫
X
|f(x)|∫ |f(x)|
0
αp−2 dα dx
= C ′‖f‖pp ,30For its general formulation, see E.M. Stein, G. Weiss, An introduction to Fourier analysis
on Euclidean spaces
POINTWISE CONVERGENCE 71
having denoted by C1 the weak-type (1,1) constant for T .
Corollary 1.6. If 1 < p ≤ ∞, M is bounded on Lp(X).
2. Poisson integrals and maximal function
The reason for introducing the Hardy-Littlewood maximal function is that itcontrols other quantities that intervene in the analysis of the boundary behaviourof hp- (or Hp-) functions. We begin with the upper half-plane D+, where thegeometric picture is more clear.
For every point x0 ∈ R = ∂D+, we define different “approach regions” from theinterior of D+, and for each of them we introduce a maximal operator. The mostnatural approach region is the vertical line x = x0. Correspondingly, for a functionf in some Lp(R), we call vertical maximal function of f the function
(2.1) Mvertf(x) = supy>0
∣∣Pf(x+ iy)∣∣ ,
where P denotes the operator assigning to f its Poisson integral.
Other approach regions are the non-tangential angles. Given a point x0 ∈ R,consider the open infinite angle Γα(x0) inside D+ with vertex in x0, symmetric w.r.to the vertical line x = x0, and with semi-aperture α, 0 < α < π
2. Explicitely,
Γα(x0) = x+ iy : |x− x0| < y tanα .
For a fixed α ∈ (0, π/2), we define the non-tangential maximal function of f ∈Lp(R) as
Mnt,αf(x) = supz∈Γα(x)
|Pf(z)| .
Obviously,
Mvertf(x) ≤Mnt,αf(x) ≤Mnt,α′f(x) ,
if α < α′.
Lemma 2.1. For every α ∈ (0, π/2) there is a constant Cα > 0 such that, iff ∈ Lp(R),
Mnt,αf(x) ≤ CαMf(x) .
In particular, each Mnt,α and Mvert are weak-type (1, 1) and bounded on Lp for1 < p < ∞.
Proof. We can assume that α > π/4, so that a = tanα > 1. Consider the dyadicintervals Iaj = [−a2j , a2j] in the real line, with j ≥ 0. We set Eaj = Iaj \ Iaj−1 for
72 CHAPTER III
j > 0, and Ea0 = Ia0 . Then
(2.2)
P1(x) =1
π
1
1 + x2
≤ 1
πχEa
0(x) +
1
π
∞∑
j=1
1
1 + a222(j−1)χEa
j(x)
≤ 1
πχEa
0(x) +
4
πa2
∞∑
j=1
2−2jχEaj(x)
≤ 4
πχIa
0(x) +
4
π
∞∑
j=1
2−2j(χIa
j(x) − χIa
j−1(x)
)
=4
π
∞∑
j=0
(2−2j − 2−2(j+1))χIaj(x)
= C
∞∑
j=0
2−2jχIaj(x) .
For a general y > 0, (2.2) and the identity
Py(x) =1
yP1
(xy
),
give
(2.3) Py(x) ≤ C∞∑
j=0
2−2j
yχIa
j
(xy
)= C
∞∑
j=0
2−2j
yχIay
j(x) .
It follows that
(2.4)
∣∣f ∗ Py(x)∣∣ ≤
∫
R
|f(x− t)|Py(t) dt
≤ C∞∑
j=0
2−2j
y
∫
Iayj
|f(x− t)| dt
= Ca
∞∑
j=0
2−2j
y
∫ x+ay2j
x−ay2j
|f(t)| dt .
If x+ iy ∈ Γα(x0), then x0 ∈ [x− ay2j, x+ ay2j], so that
1
2ay2j
∫
[x−ay2j ,x+ay2j ]
|f(t)| dt ≤Mf(x0) .
Therefore∣∣f ∗ Py(x)
∣∣ ≤ Ca2∞∑
j=0
2−jMf(x0) ,
andMnt,αf(x0) ≤ Ca2Mf(x0) .
POINTWISE CONVERGENCE 73
Theorem 2.2. Suppose that u is in hp(D+), 1 < p ≤ ∞, or in Hp(D+), 1 ≤ p ≤∞. Then the functions
u∗α(x)def= sup
z∈Γα(x)
∣∣u(z)∣∣ ,
are in Lp(R) and ‖u∗α‖p ≤ Cp,α‖u‖hp. The same is true for
u∗(x)def= sup
y>0
∣∣u(x+ iy)∣∣ .
Proof. For p > 1, u∗α ∈ Lp(R) because u] ∈ Lp(R) and by Lemma 2.1.It remains to discuss the case u ∈ H1(D+). By Corollary 5.5 in Chapter II, we
can factorize u as u = vw, with v, w ∈ H2(D+) and ‖v‖H2 = ‖w‖H2 = ‖u‖12
H1 .Then
u∗α(x) = supz∈Γα(x)
∣∣v(z)∣∣∣∣w(z)
∣∣ ≤ v∗α(x)w∗α(x) .
Therefore,
‖u∗α‖1 ≤ ‖v∗α‖2‖w∗α‖2 ≤ C2
2,α‖v‖H2‖w‖H2 = C22,α‖u‖H1 .
We pass now to the unit disc. The substitute for the vertical maximal functionis the radial maximal function,
Mradf(eit) = supr<1
∣∣Pf(reit)∣∣ .
defined for f ∈ L1(T).As non-tangential access regions, we take the Stolz regions Sρ(e
it), with ρ ∈(0, 1), defines as the open convex envelop of the point eit and the disc with center0 and radius ρ. Near eit, the Stolz region Sρ(e
it) is the angle pointing towards theinterior of the disc, symmetric w.r. to the radius, and of semi-aperture α = arcsin ρ.
We then define the non-tangential maximal function of f ∈ L1(T) as
Mnt,ρf(eit) = supz∈Sρ(eit)
|Pf(z)| .
Lemma 2.3. For every ρ ∈ (0, 1) there is a constant Cρ > 0 such that, if f ∈L1(T),
Mnt,ρf(eit) ≤ CρMf(eit) .
In particular, each Mnt,ρ and Mrad are weak-type (1, 1) and bounded on Lp for1 < p < ∞.
Proof. We use (3.6) in Chapter I,
Pr(eit) ≤ C
1 − r
t2 + (1 − r)2
with t ∈ [−π, π], and proceed as in the proof of Lemma 1.1, just replacing y by
1 − r, and stopping the sums in j as soon as the intervals Ia(1−r)j do not intersect
[−π, π].
74 CHAPTER III
Theorem 2.4. Suppose that u is in hp(D), 1 < p ≤ ∞, or in Hp(D), 1 ≤ p ≤ ∞.Then the functions
u∗ρ(eit)
def= sup
z∈Sρ(eit)
∣∣u(z)∣∣ ,
are in Lp(T) and ‖u∗α‖p ≤ Cp,ρ‖u‖hp. The same is true for
u∗(x)def= sup
r<1
∣∣u(reit)∣∣ .
3. Pointwise convergence to the boundary
In this section we discuss pointwise convergence to the boundary of harmonicand holomorphic functions on D or on D+. In this case we will begin with the unitdisc, where the situation is simplified by the inclusion of all Lp-spaces into L1.
Our initial problem is the radial convergence of a function u to the boundary,i.e. the existence of
limr→1
u(reit) = u](eit) ,
almost everywhere. The results of the previous section, however, induce us toconsider the stronger problem of non-tangential convergence, i.e. the existence of
limz→eit , z∈Sρ(eit)
u(z) = u](eit) ,
for every ρ < 1.
Theorem 3.1. If u = Pf , with f ∈ L1(T), then u converges to f non-tangent-ially a.e.
Proof. We can assume that f is real-valued. Consider then the quantity
δu(eit) = lim supz→eit , z∈Sρ(eit)
u(z) − lim infz→eit , z∈Sρ(eit)
u(z) ,
which is non-negative. Obviously,
δu(eit) ≤ 2Mnt,ρf(eit) .
Therefore, using Lemma 2.3 and denoting by |E| the normalized Lebesgue mea-sure of E ⊂ T, we have, for α > 0,
(3.1)∣∣eit : δu(eit) > α
∣∣ ≤ Cρ‖f‖1
α.
Given ε > 0, there is g ∈ C(T) such that ‖f − g‖1 < ε. By Theorem 4.3 inChapter I, v = Pg is continuous on D, so that
lim supz→eit , z∈Sρ(eit)
(u− v)(z) = lim supz→eit , z∈Sρ(eit)
u(z) − limz→eit , z∈Sρ(eit)
v(z)
= lim supz→eit , z∈Sρ(eit)
u(z) − g(eit) .
POINTWISE CONVERGENCE 75
In the same way,
lim infz→eit , z∈Sρ(eit)
(u− v)(z) = lim infz→eit , z∈Sρ(eit)
u(z) − g(eit) ,
and thereforeδ(u− v)(eit) = δu(eit) .
Applying (3.1) to u− v, we obtain that
(3.2)∣∣eit : δu(eit) > α
∣∣ ≤ Cρε
α,
and this holds for every ε > 0. Hence
∣∣eit : δu(eit) > α∣∣ = 0 ,
for every α > 0.Then the set
⋃
n≥1
eit : δu(eit) >
1
n
= eit : δu(eit) > 0
has measure zero, i.e.
lim supz→eit , z∈Sρ(eit)
u(z) = lim infz→eit , z∈Sρ(eit)
u(z)(
= limz→eit , z∈Sρ(eit)
u(z))
almost everywhere. On the other hand, limr→1 ‖ur − f‖1 = 0, so that there is asubsequence rj → 1 such that u(rje
it) → f(eit) a.e. We conclude that
limz→eit , z∈Sρ(eit)
u(z) = f(eit)
almost everywhere.
Corollary 3.2. If ρ < 1 and u ∈ hp(D), 1 < p ≤ ∞, or in Hp(D), 1 ≤ p ≤ ∞,then for almost every eit ∈ T, limz→eit , z∈Sρ(eit) u(z) = u](eit) for every ρ < 1.
In the proof of Theorem 3.1 we have used in a crucial way the density of con-tinuous functions in L1, in order to have (3.2). The same proof can be adapted toprove the analogous result in the upper half-plane. In this case, we approximatef ∈ L1(R) by functions g ∈ C0(R). This also works for f ∈ Lp(R) as long as p <∞,but it breaks down for p = ∞. For this case we use a different argument.
Theorem 3.3. If u = Pf , with f ∈ Lp(R) and 1 ≤ p ≤ ∞, then u converges to fnon-tangentially a.e.
Proof. If f ∈ L1(R), we proceed as described above. Take now f ∈ L∞(R). Thenu ∈ h∞(D+). If ϕ is the Cayley transform in (2.7) of Chapter II, then v = u ϕ ∈h∞(D). We know by Corollary 3.2 that v converges to v](eit) non-tangentially a.e.
Fix now x ∈ R and α > 0, let Γ′α(x) = Γα(x) ∩ z : =mz < 1 and eit =
ϕ−1(x). Since ϕ′(eit) 6= 0, ϕ is a diffeomorphism of a neighborhood of eit onto aneighborhood of x. Therefore ϕ−1
(Γ′α(x)
)is contained in a Stolz angle Sρ(e
it).
76 CHAPTER III
This implies that for a.e. x ∈ R,
(3.3) limz→x , z∈Γα(x)
u(z) = limz→x , z∈Γ′
α(x)u(z) = lim
z→eit , z∈Sρ(eit)v(z) = v]
(ϕ−1(x)
).
In particular,limy→0
u(x+ iy) = v](ϕ−1(x)
)
almost everywhere. Since u is bounded, we can apply dominated convergence toprove that, if g ∈ L1(R),
limy→0
∫
R
uy(x)g(x) dx =
∫
R
v](ϕ−1(x)
)g(x) dx ,
i.e. uy → v] ϕ−1 in the weak* topology. Hence v] ϕ−1 = u] = f , and (3.3) thensays that u converges to f non-tangentially a.e.
At this point, the simplest argument to prove the statement for 1 < p < ∞ is toobserve that any f ∈ Lp(R) decomposes as a sum f = f1 + f∞, with f1 ∈ L1(R)and f∞ ∈ L∞(R). To see this, take
f∞(x) =
f(x) if |f(x)| ≤ 1 ,
0 if |f(x)| > 1 ,
and f1 = f − f∞.
Corollary 3.4. If α < π/2 and u ∈ hp(D+), 1 < p ≤ ∞, or in Hp(D+), 1 ≤ p ≤∞, then limz→x , z∈Γα(x) u(z) = u](x) almost everywhere.
The case p = ∞ in Corollary 3.2 and 3.4 is referred to as “Fatou’s theorem”.
4. Poisson integrals of singular measures
The discussion in the previous section does not say anything about non-tan-gential limits of general h1-functions. We complete the picture here, proving thatCorollary 3.4 can be extended to h1(R) (the same can be done on T, but we omitthe proof). It must be noted however that no maximal function is involved in theproof.
Every h1-function on D+ is the Poisson integral of a measure µ ∈ M(R). Werecall the Lebesgue decomposition of µ as
µ = µa + µs ,
where µa is absolutely continuous with respect to Lebesgue measure m (or µa m), i.e. dµa(x) = h(x) dx with h ∈ L1(R), and µs is singular with respect tothe Lebesgue measure (or µs ⊥ m). This means that there is a set E such thatm(R \E) = 0 and |µs|(E) = 0.
The functionϕ(x) = |µs|(−∞, x)
is non-decreasing, hence differentiable at (Lebesgue-) almost every point, and itsbeing singular implies that ϕ′(x) = 0 a.e. This means that at a.e. x ∈ R,
(4.1) limh→0
|µs|(x− h, x+ h)
h= 0 .
POINTWISE CONVERGENCE 77
Lemma 4.1. Let µ be a singular measure on R, and u = Pµ. Then for almostevery x ∈ R,
limz→x , z∈Γα(x)
u(z) = 0
for every α > 0.
Proof. Take a point where (4.1) holds. We can assume that this point is the origin.Given ε > 0, take δ such that |µs|(−h, h) < εh for h < δ.
With a = tanα, which we can assume to be greater than 1, take x with |x| < ay.Then
|u(x+ iy)| ≤∫ x+ δ
4
x− δ4
Py(x− t) d|µ|(t) +
∫
|t−x|> δ4
Py(x− t) d|µ|(t)
≤∫ x+ δ
4
x− δ4
Py(x− t) d|µ|(t) + Py
(δ4
)‖µ‖1 .
Since Py(δ/4) = 2yδ2+4y2 , the last term tends to 0 as y → 0. We must then show
that
limy→0 , |x|<ay
∫ x+ δ4
x− δ4
Py(x− t) d|µ|(t) = 0 .
Take y < δ4a
, so that (x − δ4, x + δ
4) ⊂ (− δ
2, δ
2). Define Ij = (x − 2jy, x + 2jy)
for j ≥ 0 and 2jy < δ2. Then Ij ⊂ (−δ, δ) for every such j, and (x − δ
4, x + δ
4) is
covered by the Ij .Since supt∈Ij\Ij−1
Py(x− t) = Py(2j−1y) < 4
22jy and (a+ 2j)y < δ,
∫ x+ δ4
x− δ4
Py(x− t) d|µ|(t) ≤ |µ|(I0)y
+∑
j≥1
4
22jy|µ|(Ij \ Ij−1)
≤ C∑
j
|µ|(− (a+ 2j)y, (a+ 2j)y
)
2−2jy
≤ Cε∑
j
(a+ 2j)y
2−2jy
≤ Caε .
Therefore
lim supy→0 , |x|<ay
∫ x+ δ2
x− δ2
Py(x− t) d|µ|(t) ≤ Caε .
The conclusion follows by the arbitrarity of ε.
Theorem 4.2. Let u = Pµ ∈ h1(D+), and let µ = hm + µs be the Lebesguedecomposition of µ, with h ∈ L1(R). Then for a.e. x ∈ R,
limz→x , z∈Γα(x)
u(z) = h(x)
for every α > 0.
The proof follows easily from Theorem 3.3 and Lemma 4.1.
78 CHAPTER III
5. Lp-estimates for the conjugate harmonic function
We discuss now a crucial point in the theory of Hardy spaces, that has been leftaside in the previous chapters: the fact that for 1 < p <∞ the conjugate harmonicfunction of an hp-function is also in hp.
We begin with the unit disc, and recall that if
u(z) =∞∑
n=0
anzn +
∞∑
n=1
a−nzn
is holomorphic in the unit disc, its harmonic conjugate is
(5.1) u(z) = −i∞∑
n=1
anzn + i
∞∑
n=1
a−nzn .
When u is real-valued, u is characterized by the properties that it is real-valued,u+ iu is holomorphic, and u(0) = 0.
Lemma 5.1. If u is harmonic and positive on some open set, and p > 0, then
(5.2) ∆(up) = p(p− 1)up−2|∇u|2 .
If f is holomorphic and non-zero in some open set, and p > 0, then
(5.3) ∆(|f |p
)= p2|f |p−2|f ′|2 .
Proof. We have
∂2x(u
p) = ∂x(pup−1∂xu
)= p(p− 1)up−2(∂xu)
2 + pup−1∂2xu .
summing with the corresponding formula for ∂2y(u
p), we have (5.2).
As to (5.3), setting the standard notation ∂z = 12(∂x − i∂y), ∂z = 1
2(∂x + i∂y),
we have ∆ = 4∂z∂z. We fix a point z0 in the domain of f , and a determinationof log f in a neighborhood of z0. We then set fp/2(z) = e
p2
log f(z). Then, in thisneighbourhood of z0,
∆|f |p = 4∂z∂zfp/2fp/2
= 4(∂zfp/2)(∂zfp/2)
= 4|∂zfp/2|2
= p2|f |p−2|f ′|2 .
Teorema 5.2. If 1 < p < ∞, there is a constant Cp such that ‖u‖hp ≤ Cp‖u‖hp
for every u ∈ hp(D).
Proof. The case p = 2 is already known, with C2 = 1.Consider first the case where 1 < p < 2, and u > 0 in D. Let f = u+ iu. Then
f is holomorphic and non-zero in D.
POINTWISE CONVERGENCE 79
We apply Green’s formula to the two functions up and |f |p on the disc Drcentered at the origin and with radius r < 1. We have
(5.4)
r
∫ 2π
0
∂r(u(reit)p
)dt = p(p− 1)
∫
Dr
u(z)p−2|∇u(z)|2 dz ,
r
∫ 2π
0
∂r|f(reit|p dt = p2
∫
Dr
|f(z)|p−2|f ′(z)|2 dz .
But
(5.5) |∇u|2 = (∂xu)2 + (∂yu)
2 = (∂xu)2 + (∂xu)
2 = |∂xf |2 = |f ′|2 .
Since p < 2 and |u| ≤ |f |, we have |f |p−2 ≤ |u|p−2. Therefore
(5.6)
r
∫ 2π
0
∂r|f(reit|p dt ≤ p2
∫
Dr
|u(z)|p−2|∇u(z)|2 dz
=p
p− 1r
∫ 2π
0
∂r|u(reit|p dt .
In other words,∂r
(Mp(f, r)
p)≤ p′∂r
(Mp(u, r)
p).
Since Mp(u, 0) = Mp(f, 0) = u(0), we conclude that Mp(f, r)p ≤ p′Mp(u, r)
p for
every r < 1, hence ‖f‖Hp ≤ (p′)1/p‖u‖hp .Since |u| ≤ |f |, we also have
‖u‖hp ≤ (p′)1/p‖u‖hp .
If u is a generic function in hp(D), we decompose u] ∈ Lp(T) as the combinationu] = ϕ1 − ϕ2 + iϕ3 − iϕ4 of four non-negative functions, with the supports of ϕ1
and ϕ2 disjoint, and similarly for ϕ3 and ϕ4. Then
u = u1 − u2 + iu3 − iu4 ,
where uj = Pϕj is either identically zero (if so is ϕj), or strictly positive in D, as aconsequence of the fact that the Poisson kernels Pr are strictly positive everywhere.By construction,
∑4j=1 ‖ϕj‖p ≤ C‖u]‖p, so that
4∑
j=1
‖uj‖hp ≤ C‖u‖hp .
Therefore
‖u‖hp ≤4∑
j=1
‖uj‖hp ≤ Cp‖u‖hp ,
which gives the conclusion for 1 < p < 2.Consider now the duality between hp(D) and hp
′
(D) of Section 8 in Chapter I,
B(u, v) = limr→1
∫
T
u(reit)v(reit) dt =
∫
T
u](eit)v](eit) dt = limr→1
∑
n∈Z
anbnr2|n| ,
80 CHAPTER III
if an, bn are the Taylor coefficients of u and v respectively. If 1 < p <∞, we have
|B(u, v)| ≤ ‖u‖hp‖v‖hp′ , ‖u‖hp = sup‖v‖
hp′≤1
|B(u, v)| .
By (5.1), the operator mapping u into u is skew-hermitian w.r. to B, that is
B(u, v) = limr→1
∑
n∈Z
(−i sgn n)anbnr2|n| = −B(u, v) .
Therefore, if u ∈ hp(D) and 2 < p <∞,
‖u‖hp = sup‖v‖
hp′≤1
|B(u, v)|
= sup‖v‖
hp′≤1
|B(u, v)|
≤ ‖u‖hp sup‖v‖
hp′≤1
‖v‖hp′
≤ Cp′‖u‖hp .
This concludes the proof.
We switch now to the upper half-plane. We take a function u ∈ hp(D+) and wesuppose that p < ∞. By Proposition 4.2 in Chapter II, the harmonic conjugate
u(x+ iy) = u] ∗ Py(x) =1
π
∫
R
u](x− t)t
t2 + y2dt
is well defined, and, if u is real-valued, it is characterized as the only real-valuedharmonic function such that u+ iu is holomorphic and tends to 0 at infinity in eachhalf-plane z : =mz ≥ a with a > 0.
We know by (4.9) in Chapter II that, for u ∈ h2(D+), u ∈ h2(D+) and ‖u‖h2 =‖u‖h2 . We prove now the analogue of Theorem 5.2.
Theorem 5.3. If 1 < p < ∞, there is a constant Cp such that ‖u‖hp ≤ Cp‖u‖hp
for every u ∈ hp(D+).
Following the proof of Theorem 5.2, we want to use Green’s formula to provethat, if f = u + iu, then |∂yMp(f, y)
p| ≤ C|∂yMp(u, y)p|. We shall then apply
Green’s formula over rectangles, and use the fact that the integrals over certainedges tend to 0 if the edge is moved to infinity. This requires some preliminaryproof.
Lemma 5.4. Suppose u ∈ hp(D+) with p <∞. Then
limx→∞
|∇u(x+ iy)| = 0 ,
uniformly in y ≥ a, for every a > 0. Moreover,
Mp(∇u, y) ≤1
yMp
(P|u]|, y
);
POINTWISE CONVERGENCE 81
in particular, limy→∞Mp(∇u, y) = 0.
Proof. We have
(5.7) ∇u(x+ iy) = ∇∫
R
u](t)Py(x− t) dt =
∫
R
u](t)∇Py(x− t) dt ,
Since Py(x) = − 1π=m 1
x+iy , using the Cauchy-Riemann equations for (x+ iy)−1,
we have
(5.8) |∇Py(x)| =1
π|x+ iy|2 =1
yPy(x) .
By (5.7),
|∇u(x+ iy)| ≤ 1
y
∫
R
|u](t)|Py(x− t) dt =1
y
(P|u]|
)(x+ iy) .
The first part of the statement then follows from Lemma 1.3 in Chapter II, sinceP|u]| ∈ hp(D+), and the last part is now obvious.
Proof of Theorem 5.3. As in the proof of Theorem 5.2, we consider first the case1 < p < 2. We further suppose that u = Pu], with u] ≥ 0, continuous with compactsupport, and non-identically zero. Then u is strictly positive on D+.
In this hypotheses, we can say right now that u is “not far from” being in hp(D+).In fact
(5.9) Mp(u, y) = ‖u] ∗ Py‖p ≤ ‖u]‖1‖Py‖p ≤ Cpy− 1
p′ ‖u]‖1 ;
(observe that the same estimate also holds for u:
(5.9’) Mp(u, y) ≤ ‖u]‖1‖Py‖p ≤ Cpy− 1
p′ ‖u]‖1 ,
a fact that we shall use later).Therefore
(5.10) v(ε)(z) = u(z + iε) ∈ hp(D+)
for every ε > 0.For a, b, R > 0 with a < b, let Qa,b,R be the rectangle [−R,R]× [a, b]. Applying
Green’s formula, the analogues of (5.4) become
(5.11)
−∫ R
−R∂y
(u(x+ ia)p
)dx+
∫ b
a
∂x(u(R+ iy)p
)dy
+
∫ R
−R∂y
(u(x+ ib)p
)dx−
∫ b
a
∂x(u(−R+ iy)p
)dy
= p(p− 1)
∫
Qa,b,R
u(z)p−2|∇u(z)|2 dz ,
82 CHAPTER III
and
(5.12)
−∫ R
−R∂y|f(x+ ia)|p dx+
∫ b
a
∂x|f(R+ iy)p| dy
+
∫ R
−R∂y|f(x+ ib)|p dx−
∫ b
a
∂x|f(−R+ iy)p| dy
= p2
∫
Qa,b,R
|f(z)|p−2|f ′(z)|2 dz .
We first fix a and b and let R tend to infinity. By Lemma 5.4 and Lemma 1.3 inChapter II,
∂x(u(±R+ iy)p
)= pu(±R+ iy)p−1∂xu(±R+ iy)
tends to zero as R→ +∞, uniformly in y ∈ [a, b].To see that the same holds for ∂x|f(−R+ iy)p|, observe that
∂x|fp| = <e∂zfp2 f
p2 =
p
2<ef
p2−1f
p2 f ′ ,
so that
(5.13)∣∣∂x|fp|
∣∣ ≤ p
2|f |p−1|f ′| .
By (5.5), f ′(x+ iy) tends to zero as x→ ±∞ uniformly in y ∈ [a, b]. The sameis true for |f(z)|, as a consequence of (5.10).
We can then let R→ +∞ in (5.11) and (5.12), to obtain
(5.14)
−∫
R
∂y(u(x+ ia)p
)dx+
∫
R
∂y(u(x+ ib)p
)dx
= p(p− 1)
∫
x∈R , y∈[a,b]
u(z)p−2|∇u(z)|2 dz ,
and
(5.15)
−∫
R
∂y|f(x+ ia)|p dx+
∫
R
∂y|f(x+ ib)|p dx
= p2
∫
x∈R , y∈[a,b]
|f(z)|p−2|f ′(z)|2 dz .
We now make tend b to +∞, and show that the integrals containg b tend to zero.In (5.14) we use Holder’s inequality to obtain
∣∣∣∣∫
R
∂y(u(x+ ib)p
)dx
∣∣∣∣ ≤ p
∫
R
u(x+ ib)p−1|∂yu(x+ ib)| dx
≤ pMp(u, b)p−1Mp(∂yu, b) ,
which tends to 0 by Lemma 5.5. For the second integral in (5.15), we use theanalogue of (5.13) for ∂y(|f |p) to obtain that
∣∣∣∣∫
R
∂y|f(x+ ib)|p dx∣∣∣∣ ≤
p
2Mp(f, b)
p−1Mp(f′, b)
≤ p
2
(Mp(u, b) +Mp(u, b)
)p−1Mp(∇u, b) .
POINTWISE CONVERGENCE 83
This last quantity tends to zero by (5.10) and Lemma 5.4. So (5.14) and (5.15)respectively imply that
(5.16)
−∫
R
∂y(u(x+ ia)p
)dx = p(p− 1)
∫
x∈R , y≥au(z)p−2|∇u(z)|2 dz
−∫
R
∂y|f(x+ ia)|p dx = p2
∫
x∈R , y≥a|f(z)|p−2|f ′(z)|2 dz .
We claim that the two left-hand sides in (5.16) are −∂y(Mp(u, y)
p)|y=b
and
−∂y(Mp(f, y)
p)|y=b
respectively. We postpone this proof, and assume this fact
to be true for the moment.Since Mp(u, y) and Mp(f, y) are decreasing in y, the left-hand sides in (5.16) are
in fact positive quantities. Since
limy→0
Mp(u, y)p = lim
y→0Mp(f, y)
p = 0
by (5.9) and (5.9’), the same argument used in the proof of Theorem 5.2 shows that
Mp(f, y)p ≤ p′Mp(u, y)
p
for every y > 0. The rest of the proof will then proceed as in Theorem 5.2, takinginitially u as the Poisson integral of a continuous, complex-valued function u] on Rwith compact support. Once we have the estimate
‖u‖hp ≤ Cp‖u‖hp
for such functions, we use a density argument to extend it to a generic u ∈ hp(D+).We are so left with the proof of our previous claim, concerning
∂y(Mp(u, y)
p)|y=b
= limh→0
∫
R
u(x+ i(b+ h)
)p − u(x+ ib)p
hdx .
In order to be allowed to move the limit inside the integral, we use dominatedconvergence. We shall use vertical maximal functions for this purpose.
Fix ε > 0 and smaller than b/2, and call u(ε)(z), resp. f(ε)(z), the functionsu(z+iε), resp. f(z+iε). Notice that u(ε), ∂yu(ε) ∈ hp(D+) and f(ε), f
′(ε) ∈ Hp(D+),
by (5.9), (5.9’) and Lemma 5.4.We take h > −(b− ε), as we can, in order to stay above the level ε.By the mean value theorem, for every x there is t = t(x) ∈ (b, b+ h) such that
u(x+ i(b+ h)
)p − u(x+ ib)p
h= ∂y
(u(x+ iy)p
)|y=t
= pu(x+ it)p−1∂yu(x+ it) .
Therefore,
∣∣∣u(x+ i(b+ h)
)p − u(x+ ib)p
h
∣∣∣ ≤ p(Mvertu(ε)(x)
)p−1Mvert(∂yu(ε))(x) .
Applying Holder’s inequality and Theorem 2.2, we can see that this last functionis integrable. The same applies to f .
84 CHAPTER III
We can now complete some statements left incomplete in the previous chapters.We give a list of them.
(i) The Cauchy projection is bounded from hp onto Hp for 1 < p < ∞, bothin D and in D+.
(ii) The conjugate function operator on T (see Propsoition 6.2 in Chapter I) andthe Hilbert transform on R (see Proposition 4.3 in Chapter II) are boundedon Lp for 1 < p <∞.
(iii) Under the sesquilinear map B in Proposition 8.3 in Chapter I, the dual of
Hp(D) is identified with Hp′(D) for 1 < p < ∞. The same can be verifiedon D+.
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