It is interesting that asp→1,Ap≤ A/(p−1) for some constantA, and if p1 = ∞, then limp→∞ Ap ≤A1. For the maximal function we obtain
App = p2p+1/(p−1).
Other splittings of f = f0+ f1give more accurate estimates of the dependen- cies of AponA0and A1 (see Zygmund [1968, Chapter XII]).
If p=1, we know only thatu(z) is the Poisson integral of a finite measure μonand
|dμ| ≤sup
y
|u(x+i y)|d x,
becauseμis a weak-star limit of the measuresu(x+i y)d x,y→0. Define M(dμ)(t)=sup
t∈I(|μ|(I)/|I|).
The proof of Theorem 4.2 shows thatu∗(t)≤AαM(dμ)(t). And the proof of part (a) of Theorem 4.3 shows thatM(dμ)(t) is weakL1and
|{t :M(dμ)(t)> λ} ≤ 2 λ
d|μ|.
Therefore (5.2) holds in the casep=1.
The nontangential maximal functionu∗will be more important to us than the Hardy–Littlewood maximal functionMf. The next corollary is stated to emphasize the strength of Theorem 5.1.
Corollary 5.2. If u(z)is harmonic onH and if p>1, then
sup
y |u(x+i y)|p d x ≤Bpsup
y
|u(x+i y)|pd x.
Note that Corollary 5.2 is false at p=1. Take u(x,y)= Py∗ f(x), f ∈ L1, f >0. Then supy|u(x,y)| ≥M f(x) andM f ∈L1.
Theorem 5.3 (Fatou). Let u(z) be harmonic on H , let 1≤ p≤ ∞ and assume
sup
y u(x+i y)Lp(d x)<∞.
Then for almost all t the nontangential limit
α(t)z→tlim u(z)= f(t) exists.
If p>1,u(z)is the Poisson integral of the boundary value function f(t), and if1< p<∞,
u(x+i y)− f(x)p→0 (y→0).
If p=1, then u(z)is the Poisson integral of a finite measureμon, andμ is related to the boundary value function f(t)by
dμ= f(t)dt+dν, where dν is singular to Lebesgue measure.
Proof. First, let 1≤ p<∞ and assume u(z) is the Poisson integral of a function f(t)∈Lp. We will showu(z) has nontangential limit f(t) for almost allt. We can assumef is real valued. Let
f(t)=lim
z→tu(z)−lim
z→tu(z),
where z is constrained to α(t). Then by the maximal theorem f(t)≤ 2u∗(t)≤2AαM f(t), so thatf, as well as the limes superior and the limes inferior, is finite almost everywhere. The functionf(t) represents the non- tangential oscillation ofuatt, anduhas a nontangential limit attif and only iff(t)=0.
By Theorem 5.1, and by Chebychev’s inequality if p>1, we have
|{t :f(t)> ε}| ≤ Bp
2 εfp
p . (5.3)
Now ifg∈Lpand if in additiong∈C0(), then by Theorem 3.1,g =0 for allt, and sof =f+g. Takeg∈C0() so thatf +gp ≤ε2. Then
|{t :f(t)> ε}| = |{t :f+g(t)> ε}| ≤ Bp
2
εf +gp
p
≤cpεp. Consequentlyf(t)=0 almost everywhere anduhas a nontangential limit almost everywhere. The limit coincides with f(t) almost everywhere because u(x,y) converges in Lp norm to f(x). That proves the theorem in the case 1< p<∞and, provided thatu(z) is the Poisson integral of anL1 function, in the case p=1.
Let p= ∞, and letu(z)=(Py∗ f)(x), with f(t)∈L∞. Let A>0 and write f(t)= f1(t)+ f2(t) where f2 =0 on (−A,A) and f1 ∈L1. Then u(z)=u1(z)+u2(z), where uj(z)=(Py∗ fj)(x), j =1,2. It was proved above that u1(z) has nontangential limit f1(t) almost everywhere, and by Lemma 3.3 uz(z) has limit f2(t)=0 everywhere on (−A,A). Hence u(z) converges to f(t) nontangentially almost everywhere on (−A,A). Letting A→ ∞we have the result for p= ∞.
Now letp=1 and assume sup
y u(x+i y)L1(d x)<∞.
Then u(z) is the Poisson integral of a finite measure μon . Write dμ= f(t)dt+dν, wheredνis singular tod x, and letu1(z)=(Py∗ f)(x),u2(z)= (Py∗ν)(x). Then u(z)=u1(z)+u2(z). It was shown above that u1(z) has nontangential limit f(t) almost everywhere. Becauseν is singular, the next lemma showsu2(z) has nontangential limit zero almost everywhere, and that concludes the proof.
Lemma 5.4. Ifνis a finite singular measure on, then(Py∗ν)(x)converges nontangentially to zero almost everywhere.
Proof. We may assumeν ≥0. Becauseν is singular, we have
h→0lim
ν((t−h,t+h))
2h =0
(5.4)
for Lebesgue almost allt. Indeed, if (5.4) were not true, there would be a compact setKsuch that|K|>0, ν(K)=0, and
h→0lim
ν((t−h,t +h))
2h >a>0, all t ∈K.
Cover K by finitely many intervals Ij such that ν(∪Ij)< ε and such that ν(Ij)>a|Ij|. By the covering lemma 4.4, pairwise disjoint intervals{Ji}can be chosen from the{Ij}such that
|K| ≤2 |Ji|< 2
a ν(Ji)< 2ε a, a contradiction forεsufficiently small.
Suppose (5.4) holds att ∈. Letz ∈α(t) and suppose for simplicity that Rez=t. Sinceν ≥0, we have
(Py∗ν)(t)=
|s−t|<AyPy(t−s)dν(s)+
|s−t|≥Ay
Py(t−s)dν(s).
The second integral does not exceed (πA2y)−1
dν. If we approximate Py(s)χ|s|<Ay(s) from below by even step functions, as in the proof of Theorem 4.2, we see that
|s−t|<AyPy(t−s)dν(s)≤ sup
h<Ay
ν((t−h,t+h))
2h .
Choosing A= A(y) so that Ay→0 (y→0) but A2y→ ∞ (y→0), we obtainPy∗ν(t)→0(y→0) if (5.4) holds att. The estimates when|x−t|<
αyare quite similar and we leave them to the reader.
A positive measure σ on H is called a Carleson measure if there is a constantN(σ) such that
σ(Q)≤ N(σ)h (5.5)
for all squares
Q= {x0<x <x0+h,0<y<h}.
The smallest such constantN(σ) is theCarleson normofσ.
Lemma 5.5. Letσ be a positive measure onH , and letα >0. Thenσ is a Carleson measure if and only if there exists A= A(α)such that
σ({|u(z)|> λ})≤ A|{t :u∗(t)> λ}|, λ >0, (5.6)
for every harmonic function u(z) on H , where u∗(t) is the nontangential maximal function of u(z)over the cone{|x−t|< αy}. If A is the least constant such that(5.6)holds, then
c1(α)A≤N(σ)≤c2(α)A.
Proof. We takeα=1. The proof for a differentα is similar. Assumeσ is a Carleson measure. The open set{t :u∗(t)> λ} is the union of a disjoint sequence of open intervals{Ij}, with centersc(Ij). LetTj be the tent
Tj = {z :|x−c(Ij)| +y<|Ij|/2},
an isosceles right triangle with hypotenuseIj. If|u(z)|> λ, thenu∗(t)> λon the interval{|t−x|< y}and this interval is contained in someIj. See Figure I.5. Consequently,
{z:|u(z)|> λ} ⊂∞
j=1
Tj. By (5.5) we therefore have
σ({z:|u(z)|> λ})≤
j
σ(Tj)≤N(σ)
j
|Ij| =N(σ)|{t :u∗(t)> λ}|, and (5.6) holds.
Figure I.5.
Conversely, letIbe an interval{x0 <t <x0+h}and letu(z)= Py∗ f(x) with f(x)=4λχI(x). Thenu(z)> λon the squareQwith baseI, so that by (5.6) and the maximal theorem,
σ(Q)≤A|{t :u∗(t)> λ}| ≤(AC/λ)f1 ≤ ACh, andσ is a Carleson measure.
Theorem 5.6. (Carleson). Let f ∈Lp()and let u(z)denote the Poisson integral of f. If σ is a positive measure on the upper half plane, then the following are equivalent.
(a) σ is a Carleson measure.
(b) For1< p<∞, and for all f ∈Lp(),u(z)∈Lp(σ).
(c) For1< p<∞,
|u(z)|pdσ ≤Cp
|f|p dt, f ∈ Lp. (d) For all f ∈L1(), we have the distribution function inequality
σ({z:|u(z)|> λ})≤(c1/λ)
|f(t)|dt, λ >0.
If(b)or(c)holds for one value of p,1< p<∞, then(a)holds.
The constants Cpdepend only on p and the constant N(σ)in(5.5). In fact, if(a)holds, we can take Cp = N(σ)Bp where Bp is the constant in Theorem 5.1withα=1. If(c)or(d)holds then(5.5)holds with N(σ)≤4pCp. Proof. If (a) holds, then by (5.6) and Theorem 5.1, (c) and (d) hold. Clearly, (c) implies (b), and if (b) holds for somep, then the closed graph theorem for Banach spaces shows that (c) holds for the same value ofp.
Now suppose that (d) holds or that (c) holds for some p,1< p<∞. As in the proof of Lemma 5.5, takeI = {x0 <t <x0+h}, and setu(z)= Py∗ f(x), f(t)=4χI(t). Thenfp =4h1/p, andu(z)>1 on Q =I ×(0,h).
Hence
σ(Q)≤σ({|u(z)|>1})≤Cp
|f|pdt =4pCph, and (5.5) holds.