Maximal Functions and Boundary Values

Exercises and Further Results

3. Maximal Functions and Boundary Values

Corollary 3.2. Let1≤ p≤ ∞and let f(t)∈ L^p(). Then f(t) is almost everywhere the nontangential limit of an H^p(dt) function if and only if its Poisson integral

f(z)= Py∗ f(x)

is analytic onH . The Poisson integral f(z) is then the corresponding H^p function.

Proof. If f ∈L^p() and if f(z)= Py∗ f(x) is analytic, then by Chapter I, f(z)∈H^p(dt) and f(z) converges nontangentially to f(t).

Conversely, suppose f(z) is some H^p function. If p>1, then by Theorem 5.3, Chapter I, f(z) has nontangential limit f(t) and f(z)= Py∗ f(x). The case p=1 requires Theorem 3.1. When f(z)∈H¹, (1.3) and Lemma I.3.4 yield

f(z+iε)=

Pz(t)f(t+iε)dt, ε >0.

By (3.3), this means f(z) is the Poisson integral of its boundary function f(t).

See Exercise 2 for some other characterizations of the boundary values of H^pfunctions. Because of (3.2), we often identify f(z)∈ H^pwith its boundary function f(t).^†However, we prescribe no method of regaining f(z) from f(t) whenp<1.

Theorem 3.1 is of course also true on the unit disc. Equality (3.2) for the correspondence f(z)→ f(e^iθ) then shows thatH^p(D) is isometric to a closed subspace of the Lebesgue spaceL^p(dθ/2π). And ifp≥1, f(z) is the Poisson integral of f(e^iθ). The analog of (3.3),

rlim→1

|f(r e^iθ)− f(e^iθ)|^pdθ 2π =0,

coupled with the fact that f(r z),r <1, is trivially the uniform sum of its Taylor series, tells us this: For p<∞, the space of boundary functions of H^p(D)coincides with the closure in L^p(dθ/2π)of the analytic polynomials.

Moreover, if a sequencepn(e^iθ) of polynomials converges to f(e^iθ) inL^p, then by (3.2) for the disc,pn− f^p_H^p →0. Since H^p(D)⊂ H¹(D),p≥1, we see that for p≥1, the boundary function of an H^p(D) function has Fourier series

f(e^iθ)∼^∞

n=0

ane^inθ

†And we often writefpfor the equalfH^p.

supported on the nonnegative integers, and its Fourier coefficients an = 1

2π

f(e^iθ)e^−inθdθ = lim

r→1

1 2πi

|z|=r

f(z)dz zⁿ⁺¹ are the Taylor coefficients of theH^pfunction f(z)=

anzⁿ. ThusH^ptheory is a natural bridge between Fourier analysis and complex analysis.

In the upper half plane we see from (3.3) that the uniformly continuous functions inH^∞∩H^p are dense inH^p, because for f ∈H^p,

|f(x+i y)| ≤ 2 πy

_1/p

fH^p, y>0, and

|f(x+i y)| ≤ 2 y

4 πy

_1/p

fH^p,

by the preceding inequality and by Schwarz’s lemma, scaled to the disc (z,y/2). We will need some very smooth classes of analytic functions that are dense inH^p(dt) and that will play the role of the polynomials in the disc case. LetNbe a positive integer, and letANbe the family ofH^∞(dt) functions satisfying

(i) f(z) is continuous onH and f(t) is infinitely differentiable, f ∈C^∞. (ii) lim_|z|→∞|z|^N|f(z)| =0,z∈H .

Corollary 3.3. Let N be a positive integer. For0< p<∞, the classAN is dense in H^p(dt). For f(z)∈H^∞, there are functions fn(z)inAN such that fn∞ ≤ f∞and such that|fn(t)| ≤ |f(t)|almost everywhere.

Proof. Were it not for the decay condition (ii), we could approximatef by the smooth functions f(z+i/n), which converge inH^pnorm to f(t) if p<∞ and which converge boundedly pointwise almost everywhere tof(t) ifp= ∞.

Now there are some special functionsgk(z) such that (a) gk(z)∈AN,

(b) |gk(z)| ≤1,z∈H , (c) gk(z)→1,z ∈H .

Before we construct the functionsgk, we note that the functions fn(z)=gn(z)f(z+i/n)

inAN then give the desired approximation.

The heart of the proof, constructing the functions gk(z), will be done in the unit disc, wherew= −1 corresponds toz= ∞. Letαk <1, αk →1. The function

hk(w)= w+αk

1+αkw _N+1

has an (N +1)-fold zero at−αk. Thesehk(w) are bounded by 1, and withN fixed they converge to 1 uniformly on compact subsets of ¯D\{−1}. Then the functions

gk(z)=hk(αk, w), w= i−z i+z, satisfy (a)–(c).

Now we can clarify the relation betweenH^p(D) andH^p(dt).

Corollary 3.4. Let0< p<∞, let f(z)be an analytic function in the upper half plane and let

F(z)= π^−1/p (z+i)^2/p f(z).

Then|f(z)|^phas a harmonic majorant if and only if F(z)∈H^p. In that case F_H^pp =u(i),

(3.4)

where u(z)is the least harmonic majorant of|f(z)|^p.

Proof. Let g(w)= f(z),z=i(1−w)/(1+w). The corollary asserts that g∈H^p(D) if and only if F∈ H^p(dt), and that gp= Fp. If N >2/p and if F∈AN, then the corresponding function g(w) is bounded onDand, becausedθ/2π corresponds todt/π(1+t²),

|g(θ)|^pdθ 2π =

|F(t)|^pdt.

By the density ofAN inH^p, it follows from (3.2) thatg∈ H^p(D) whenever F∈H^p(dt) and that

gHp = FHp.

Since Lemma 1.2 shows F ∈H^p(dt) if g(w)∈H^p(D), that concludes the proof.

Incidentally, the fact that we have the equality (3.4) instead of the inequality (1.1) means that in the formula (1.2) for the least harmonic majorant of|f(z)|^p, the constant term isc=0. We shall see in Section 4 that the least harmonic majorant is the Poisson integral of theL¹function|f(t)|^p.

The next corollary is noteworthy because of the recent discovery, forp≤1, of its converse, which will be proved in Chapter III.

Corollary 3.5. Let0< p<∞and let u(z)be a real-valued harmonic func- tion on the upper half planeH . If u(z)is the real part of a function f(z)∈H^p, then

u^∗(t)= sup

α(t)|u(z)| is in L^p().

The proof from Theorem 3.1 is trivial.

Let us reexamine the boundary values in the case p=1. If the harmonic functionu(z) satisfies

sup

y

|u(x+i y)|d x <∞,

thenu(z) need not be the Poisson integral of its nontangential limitu(t). All we can say is thatu(z) is the Poisson integral of a finite measure. However, ifu(z) is also an analytic function, then the measure is absolutely continuous, and its density is the boundary valueu(t). The reason is that the maximal function u^∗(t) is integrable.

Theorem 3.6. If f(z)∈H¹(dt), then f(z)is the Poisson integral of its bound- ary values:

f(z)=

Py(x−t)f(t)dt. (3.5)

Conversely, ifμis a finite complex measure onsuch that the Poisson integral f(z)= Py∗μ(x)is an analytic function onH , thenμis absolutely continuous and

dμ= f(t)=dt,

where f(t)is the boundary function of the Poisson integral f(z)ofμ.

Proof. If f(z)∈H¹, then (3.5) was already obtained in the proof of Corol- lary 3.2. Conversely, if f(z)= Py,∗μ(x) is an analytic function, then by Minkowski’s integral inequality it is an H¹ function and hence it is the Poisson integral of its boundary value f(t). The difference measuredv(t)= dμ(t)− f(t)dthas Poisson integral zero, and sov=0 by Theorem I.3.1.

Lemma 3.7. Let f(z)∈H¹. Then the Fourier transform fˆ(s)=

_∞

−∞ f(t)e^−2πistdt =0 for all s≤0.

Proof. By the continuity of f → fˆ, we may suppose

∈AN. Then for s≤0,F(z)= f(z)e^−2πiszis also inAN. The result now follows from Cauchy’s theorem because

_π

0

|F(Re^iθ)|R dθ →0 (R→ ∞).

Notice that

Pz(t)= 1 2πi

1

t−z − 1 t−z¯

. (3.6)

Also notice that for f ∈H¹, Lemma 3.7 applied to (t−z)¯ ⁻¹f(t) yields f(t)

t−z¯dt =0, Imz >0.

Theorem 3.8. Let dμ(t)be a finite complex measure onsuch that either dμ(t)

t−z =0 on Imz<0, (a)

or

μ(s)ˆ =

e^−2πistdμ(t)=0 on s<0.

(b)

Then dμis absolutely continuous and dμ= f(t)dt, where f(t)∈H¹. Proof. If (a) holds, then by (3.6) f(z)= Py∗μ(x) is analytic and the result follows from Theorem 3.6.

Assume (b) holds. The Poisson kernelPy(t) has Fourier transform

e^−2πistPy(t)dt =e^−2π|s|y,

because ˆPy(−s)= Pˆy(s) since Pyis real, and because ifs≤0,e^−2πiszis the bounded harmonic function with boundary valuese^−2πist. Let fy(x)= Py∗ μ(x). By Fubini’s theorem, fy has Fourier transform

fˆy(s)=

e^{−2πx y}μˆ(s), s≥0, 0, s<0. Since ˆfy∈L¹, Fourier inversion implies

fy(x)=

e^{2πi xs}fˆy(s)ds = _∞

0

e2πi(x+i y)sμ(s)ˆ ds.

Differentiating under the integral sign then shows that f(z)= fy(x) is analytic, and Theorem 3.6 now implies that f(z)∈ H¹.

The disc version of Theorem 3.6, or equivalently Theorem 3.8, is one half of the famous F. and M. Riesz theorem. The other half asserts that if f(z)∈ H¹, f ≡0, then |f(t)|>0 almost everywhere. This is a consequence of a stronger result proved in the next section.

The theorem on Carleson measures, Theorem I.5.6, also extends to theH^p spaces, 0< p≤1, because the key estimate in its proof was the maximal theorem.

Sect. 4 (1/π) (log|f(t)|/(1+t ))dt>−∞ 61 Theorem 3.9(Carleson). Letσbe a positive measure in the upper half plane.

Then the following are equivalent:

(a) σ is a Carleson measure: for some constant N(σ), σ(Q)≤N(σ)h

for all squares

Q = {x0 <x<x0+h,0< y<h}.

(b) For0< p<∞,

|f|^pdσ ≤ Af^p_Hp, f ∈H^p. (c) For some p,0< p<∞, f ∈ L^p(σ)for all f ∈ H^p.

Proof. That (a) implies (b) follows from (3.1) and Lemma I.5.5 just as in the proof of Theorem I.5.6.

Trivially, (b) implies (c). On the other hand, if (c) holds for some fixed p<∞, then (b) holds for the same valuep. This follows from the closed graph theorem, which is valid here even when p<1 (see Dunford and Schwartz [1958, p. 57]). One can also see directly that if there are{fn} in H^p with fnp =1 but

|fn|^pdσ → ∞, then the sumαnfn, when theαn, are chosen adroitly, will give anH^p function for which (c) fails.

Now suppose (b) holds for some p>0. Let Q be the square {x0 <x <

x0+y0,0< y<y0}and let f(z)= 1

π y0

(z−z¯0)² _1/p

, where z0 =x0+i y0. Then f ∈H^p and f^pp =

Pz0(t)dt =1. Since

|f(z)|^p ≥(5πy0)⁻¹,z ∈Q, we have

σ(Q)≤σ({z:|f(z)|>(5πy0)^−1/p})≤5πAy0, so that (a) holds.

4. (1/π)

(log | f (t ) | /(1 + t

²

)) dt > − ∞

A fundamental result of H^p theory is that the condition of this section’s title characterizes the moduli of H^p functions|f(t)|among the positive L^p functions. In the disc, this result is due to Szeg¨o for p=2 and to F. Riesz for the otherp. For functions analytic across∂D, the inequality (4.1) below was first noticed by Jensen [1899] and for this reason the inequality is sometimes called Jensen’s inequality. We prefer to use that name for the inequality about averages and convex functions given in Theorem I.6.1.

In this section the important thing about anH^pfunction will be the fact that the subharmonic function

log|f(r e^iθ)| ≤(1/p)|f(r e^iθ)|^p

is majorized by a positiveL¹function ofθ. It will be simpler to work at first on the disc.

Theorem 4.1. If0< p≤ ∞and if f(z)∈H^p(D), f ≡0, then 1

2π

log|f(e^iθ)|dθ >−∞.

If f(0) ≡0, then

log|f(0)| ≤ 1 2π

log|f(e^iθ)|dθ, (4.1)

and more generally, if f(z0) ≡0 log|f(z0)| ≤ 1

2π

log|f(e^iθ)|Pz0(θ)dθ.

(4.2)

Proof. By Theorem I.6.7 and by the subharmonicity of log|f|, log|f(z)| ≤lim

r→1

1 2π

log|f(r e^iθ)|Pz(θ)dθ.

Since log|f(r e^iθ)| →log|f(e^iθ) almost everywhere, and since these func- tions are bounded above by the integrable function (1/p)|f^∗(θ)|^p, where f^∗ is the maximal function, we have

log⁺|f(r e^iθ)|Pz₀(θ)dθ →

log⁺|f(e^iθ)|Pz₀(θ)dθ.

Fatou’s lemma can now be applied to the negative parts to give us

r→1lim 1 2π

log|f(r e^iθ)|Pz0(θ)dθ ≤ 1 2π

log|f(e^iθ)|Pz0(θ)dθ.

This proves (4.2) and (4.1). The remaining inequality follows by removing any zero at the origin.

Note that the same result in the upper half plane log|f(z0)| ≤

log|f(t)|Pz0(t)dt, f ∈H^p, (4.3)

follows from Theorem 4.1 and from Lemma 1.1 upon a change of variables.

Corollary 4.2. If f(z)∈ H^p and if f(t)=0on a set of positive measure, then f =0.

Corollary 4.2 gives the other half of the F. and M. Riesz theorem. Ifdμ(t) is a finite measure such thatpy∗μ(x) is analytic, then not only isdμabsolutely continuous todt, but alsodtis absolutely continuous todμ.

Corollary 4.3. Let0< p,r ≤ ∞. If f(z)∈H^pand if the boundary function f(t)∈L^r, then f(z)∈H^r.

Sect. 4 (1/π) (log|f(t)|/(1+t ))dt>−∞ 63 This corollary is often written

H^p ∩L^r ⊂H^r.

Proof. Applying Jensen’s inequality, with the convex function ϕ(s)= exp(r s) and with the probability measurePy(x−t)dt, to (4.3) gives

|f(z)|^r ≤

|f(t)|^rPy(x−t)dt. Integration inxthen yields f ∈ H^r.

Theorem 4.4. Let h(t)be a nonzero nonnegative function in L^p(). Then there is f(z)∈H^p(dt)such that|f(t)| =h(t)almost everywhere if and only

if

logh(t)

1+t² dt >− ∞.

(4.4)

Proof. It has already been proved that (4.4) is a necessary condition. To show (4.4) is sufficient, note that since logh≤(1/p)|h|^p, (4.4) holds if and only if logh∈ L¹(dt/(1+t²)). Letu(z) be the Poisson integral of logh(t) and letv(z) be any harmonic conjugate ofu(z). (SinceH is simply connected, there exists a harmonic conjugate functionv(z) such thatu+ivis analytic. The conjugate functionv(z) is unique except for an additive constant.) The function we are after is

f(z)=e^u(z)+iv(z),

which is an analytic function onH . When p<∞Jensen’s inequality again gives

|f(z)|^p ≤

Py(x−t)|h(t)|^pdt

and thereforef ∈ H^p. Ifp= ∞thenuis bounded above and so f ∈ H^∞. When p= ∞, (4.4) is especially important. Leth≥0,h∈L^∞, and sup- pose 1/h∈ L^∞. Then f =e^u+iv ∈ H^∞and also

1/f =e^−(u+iv)∈H^∞.

In other words,f is aninvertible function in H^∞. We write f ∈(H^∞)⁻¹. For emphasis, we state this fact separately, writingg=logh, so thatg∈L^∞ if h∈L^∞and 1/h∈L^∞.

Theorem 4.5. Every real-valued function g(t)in L^∞has the formlog|f(t)|, where f ∈(H^∞)⁻¹is an invertible function in H^∞.

In the language of uniform algebra theory, Theorem 4.5 asserts thatH^∞is astrongly logmodular subalgebraofL^∞. It is also alogmodular subalgebra

ofL^∞, which means that the set

log|(H^∞)⁻¹| = {log|f(t)|: f ∈H^∞,1/f ∈H^∞}

is dense inL^∞, the space of realL^∞ functions. The Banach algebra aspects ofH^∞will be discussed in some detail later; for the present we only want to say that Theorem 4.4 is a powerful tool for constructingH^p functions.

Leth(t)≥0 satisfy

|logh(t)|dt 1+t² <∞.

The function

f(z)=e^u(z)+iv(z), where

u(z)= Py∗(logh)(x) (4.5)

and wherev(z) is a harmonic conjugate function ofu(z), is called anouter function. The outer function f(z) is determined by h(t) except for the uni- modular constant factor arising from the choice of the conjugate functionv(z).

The function|f(z)|has boundary valuesh(t) almost everywhere, and Jensen’s inequality with (4.5) shows that f(z)∈H^p if and only if h(t)∈L^p. Outer functions inH^pcan be characterized in several ways.

Theorem 4.6. Let0< p≤ ∞and let f(z)∈ H^p, f ≡0. Then the following are equivalent.

(a) f(z)is an outer function.

(b) For each z∈H , equality holds in(4.3); that is, log|f(z)| =

log|f(t)|Py(x−t)dt. (4.6)

(c) For some point z₀ ∈H ,(4.6)holds.

(d) If g(z)∈H^p and if|g(t)= |f(t)|almost everywhere, then

|g(z)| ≤ |f(z)|, z∈H .

(e) f(z)has no zeros inH and the harmonic functionlog|f(z)|is the Poisson integral of a function k(t)such that

|k(t)|dt

1+t² <∞, e^k(t)∈L^p.

Proof. First, (e) is merely a rewording of the definition of an outer function in H^p, because any functionf(z) without zeros is an exponential, f =e^u+iv,u= log|f|. Thus (a) and (e) are equivalent.

By definition, (a) implies (b). If (b) holds, then we see (d) holds by applying (4.3) to the function g(z) in (d). Moreover, if (d) holds, and if g(z) is an

Sect. 5 (1/π) (log|f(t)|/(1+t ))dt>−∞ 65 outer function determined by log|f(z)|, then the analytic function f(z)/g(z) satisfies

|f(z)/g(z)| =1,

so that f =λg,|λ| =1, andf is an outer function. Hence (a), (b), (d), and (e) are equivalent.

Trivially, (b) implies (c). Now assume (c) and again letg(z) be an outer function determined by log|f(t)|. Then|f(z)/g(z)| ≤1, and if (c) holds, the maximum principle shows that|f/g| =1, and so f(z) is an outer function.

The functionS(z)=e^{i z}has no zeros in the upper half plane, andS(z)∈H^∞, but S(z) is not an outer function, because log|S(z)| = −y is not a Poisson integral.

Corollary 4.7. If f(z)∈ H^p and if for some r>0,1/f(z)∈ H^r, then f(z) is an outer function.

This holds becausef satisfies (4.6).

Corollary 4.8. Let f(z)∈H^p. Either of the following two conditions imply that f(z)is an outer function.

(a) Re f(z)≥0,z∈H .

(b) There exists a C¹are terminating at0such that f(H )⊂\ .

Proof. If (a) holds then f +ε is an outer function for any ε >0, because (f +ε)⁻¹∈ H^∞. Now, since Re f ≤0,

log|f(t)+ε|Py(x−t)dt →

log|f(t)|Py(x−t)dt

as ε→0, by dominated convergence on {t :|f(t)| ≥ ¹₂} and by monotone convergence on {t :|f(t)|< ¹₂}. Hence (4.6) holds for f and f is an outer function.

The above argument shows that f(z) is an outer function if Re f(z)>0 on the set{|f(z)|<1}. Now assume (b). Replacing f(z) byλf(z),|λ| =1, we can also assume that has tangent vector (1, 0) atz=0. This means that ifδ is sufficiently small, the analytic function

g(z)=(f(z)/δ)^1/5

satisfies Reg(z)>0 if|g(z)|<1. Hencegis an outer function, and f =δ⁵g⁵ is an outer function.