To each functionfonwe associate two auxiliary functions that respectively measure the size off and the behavior of the Poisson integral off. The first
auxiliary function can be defined wheneverf is a measurable function on any measure space (X, μ). This is thedistribution function
m(λ)=μ({x ∈X :|f(x)|> λ}),
defined forλ >0. The distribution functionm(λ) is a decreasing function ofλ, and it determines theLpnorms of f. If f ∈L∞, thenm(λ)=0 forλ≥ f∞, andm(λ)>0 forλ <f∞; and so we have
f∞=sup{λ:m(λ)>0}.
Lemma 4.1. If (X, μ) is a measure space, if f(x) is measurable, and if 0< p<∞, then
|f|pdμ= ∞
0
pλp−1m(λ)dλ.
(4.1)
Proof. We may assume f vanishes except on a set of σ-finite measure, because otherwise both sides of (4.1) are infinite. Then Fubini’s theorem shows that both sides of (4.1) equal the product measure of the ordinate set {(x, λ) : 0< λ <|f(x)|p}. That is,
|f|p dμ= |f|
0
pλp−1dλdμ= ∞
0
pλp−1μ(|f|> λ)dλ
= ∞
0
pλp−1m(λ)dλ.
We shall also need a simple estimate ofm(λ) known as Chebychev’s in- equality. Let f ∈Lp,0< p<∞and let
Eλ= {x ∈X :|f(x)|> λ}, so thatμ(Eλ)=m(λ). Chebychev’s inequality is
m(λ)≤ fpp/λp. It follows from the observation that
λpμ(Eλ)≤
Eλ|f|pdμ≤ fpp. A functionf that satisfies
m(λ)≤A/λp
is called aweak Lp function. Thus Chebychev’s inequality states that every Lpfunction is a weakLpfunction. The function|xlogx|−1on [0,1] is not in L1, but it satisfiesm(λ)=o(1/λ) (λ→ ∞), and so it is weakL1.
The other auxiliary function we shall define only for functions on. Recall Lebesgue’s theorem that if f(x) is locally integrable on, then
limh→0 k→0
1 h+k
x+k
x−h f(t)dt = f(x) (4.2)
for almost everyx ∈. To make Lebesgue’s theorem quantitative we replace the limit in (4.2) by the supremum, and we put the absolute value inside the integral. Write|I| for the length of an intervalI. The Hardy–Littlewood maximal functionof f is
M f(x)=sup
x∈I
1
|I|
I
|f(t)|dt
forf locally integrable on. Now if f ∈ Lp,p≥1, thenM f(x)<∞almost everywhere. This follows from Lebesgue’s theorem, but we shall soon see a different proof in Theorem 4.3 below. The important thing aboutMf is that it majorizes many other functions associated withf.
Theorem 4.2. Forα >0and t ∈, letα(t)be the cone inH with vertex t and angle2 arctanα, as shown in FigureI.3,
α(t)= {(x,y) :|x−t|< αy,0< y<∞}.
Let f ∈L1(dt/(1+t2))and let u(x,y)be the Poisson integral of f(t), u(x,y)=
Py(s)f(x−s)ds.
Then
supα(t)|u(x,y)| ≤AαM f(t), t ∈, (4.3)
where Aαis a constant depending only onα.
Figure I.3.The coneα(t), α=23.
The condition f ∈L1(dt/(1+t2)) merely guarantees that
Py(s)f(x−s)ds converges.
Proof. We may assumet =0. Let us first consider the points (0,y) on the axis of the coneα(0). Then
u(0,y)=
Py(s)f(s)ds,
and the kernelPy(s) is a positive even function which is decreasing for positive s. That meansPy(s) is a convex combination of the box kernels (1/2h)χ(−h,h)(s) that arise in the definition ofMf. Take step functions hn(s), which are also nonnegative, even, and decreasing ons >0, such thathn(s) increases withn toPy(s). Thenhn(s) has the form
N j=1
ajχ(−xj,xj)(s) withaj ≥0, and
hn ds=
j2xjaj ≤1. See Figure I.4. Hence
hn(s)f(s)ds ≤
hn(s)|f(s)|ds≤
N j=1
2xjaj
1 2xj
xj
−xj
|f(s)|ds≤M f(0).
Then by monotone convergence
|u(0,y)| ≤
Py(s)|f(s)|ds ≤M f(0).
Figure I.4. Py(s) and its approximationhn(s), which is a positive combination of box kernels (1/2xj)χ(−xj,xj)(s).
Now fix (x,y)∈α(0). Then|x|< αy, and Py(x−s) is majorized by a pos- itive even functionψ(s), which is decreasing ons>0, such that
ψ(s)ds ≤Aα =1+ 2α π .
The function is ψ(s)=sup{Py(x−t) :|t|>s}. Approximatingψ(s) from below by step functionshn(s) just as before, we have
ψ(s)|f(s)|ds ≤AαM f(0) and
|u(x,y)| ≤
ψ(s)|f(s)|ds≤ AαM f(0), which is (4.3).
Theorem 4.2 is, with the same proof, true for Poisson integrals of functions on∂D, where the cone is replaced by the region
α(eiϕ)=
z: |eiϕ−z|
1− |z| < α,|z|<1
, α >1,
which is asymptotic, asz→eiϕ, to an angle with vertexeiϕ. The theorem is quite general. The proof shows it is true withPy(x−s) replaced by any kernel ϕy(x−s) which can be dominated by a positive, even functionψ(s), depending on (x,y), provided thatψ is decreasing ons >0 and that
ψ(s)ds ≤ Aα whenever (x,y)∈α(t) (see Stein, [1970]).
The Hardy–Littlewood maximal theorem is this:
Theorem 4.3. If f ∈Lp(),1≤ p≤ ∞, then M f(t)is finite almost every- where.
(a) If f ∈L1(), then Mf is weak L1,
|{t ∈: M f(t)> λ}| ≤(2/λ)f1, λ >0.
(b) If f ∈Lp(), with1< p≤ ∞, then M f ∈Lp()and M fp ≤Apfp,
where Ap depends only on p.
In (a) we have used|E| to denote the Lebesgue measure of E⊂. That M f <∞almost everywhere follows from (a) or (b). Condition (a) says the operator Mf is weak-type1–1. The weak-type inequality in (a) is the best possible result onMf when f ∈L1. Notice that if f ≡0, thenMf cannot possibly be inL1, because for largex
M f(x)≥ 1
|4x|
3x
−x |f(t)|dt ≥ c
|x|
iff1=0. Iffis supported on a finite intervalI, then
IM f(t)dt <∞if and only if
I|f|log+|f|dt <∞; we leave the proof as an exercise. By letting f(t)=(1/h)χ(0,h)(t), and sending h→0, one can see that the constant in (a) cannot be improved upon.
The proof of Theorem 4.3 will use two additional theorems: a covering lemma of Vitali type for part (a) and the Marcinkiewicz interpolation theorem for part (b).
Lemma 4.4. Letμbe a positive Borel measure onand let{I1, . . . ,In}be a finite family of open intervals in. There is a subfamily{J1, . . . ,Jm}such that the Ji are pairwise disjoint and such that
m i=1
μ(Ji)≥ 1 2μ
n j=1
Ij
.
Proof. By induction{I1, . . . ,In}can be replaced by a subfamily of intervals such that no intervalIj is contained in the union of the others and such that the refined family has the same union as the original family. Write the Ij in the refined family as (αj, βj) and index them so that
α1 ≤α2 ≤ · · ·αn.
Then βj+1 > βj since otherwise Ij+1⊂ Ij, and αj+1> βj−1 since other- wise Ij ⊂ Ij−1∪Ij+1. Therefore the even-numbered intervals and the odd- numbered intervals comprise pairwise disjoint subfamilies. Then
jeven
μ(Ij)+
jodd
μ(Ij)≥μ n
j=1
Ij
,
and for{Ji}we take either the even-numbered intervals or the odd-numbered intervals, which ever gives the larger sum.
Proof of Theorem 4.3(a). Assume f ∈L1 and let λ >0. Then the set Eλ= {t :M f(t)> λ}is open, and therefore measurable. For eacht ∈ Eλwe have an open intervalIcontainingtsuch that
1
|I|
1
|f|ds > λ, which is the same as
|I|< 1 λ
t
|f|ds.
(4.4)
Let K be a compact subset of Eλ and cover K by finitely many intervals I1, . . . ,In that satisfy (4.4). Applying the lemma to {I1, . . . ,In} gives us pairwise disjoint intervalsJ1,J2, . . . ,Jm, that satisfy (4.4) such that
n j=1
Ij
≤2
m j=1
|Jj|.
Then
|K| ≤
n i=1
Ij
≤2
j
1 λ
Jj
|f|ds≤ 2 λ
|f|ds.
Letting|K|increase to|Eλ|gives us part (a).
The proof of part (b) depends on the interpolation theorem of Marcinkiewicz.
Theorem 4.5. Let(X, μ)and(Y, ν)be measure spaces, and let1< p1 ≤ ∞.
Suppose T is a mapping from L1(X, μ)+Lp1(X, μ)to v-measurable functions such that
(i) |T(f +g)(y)| ≤ |T f(y)| + |T g(y)|;
(ii) ν({y:|T f(y)|> λ})≤(A0/λ)f1, f ∈L1; (iii) ν({y:|T f(y)|> λ})≤((A1/λ)fp1)p1, f ∈ Lp1; (when p1= ∞we assume instead that
T f∞≤ A1f∞). Then for1< p< p1,
T fp≤ Apfp, f ∈ Lp, where Ap depends only on A0,A1,p, and p1.
The hypothesis that the domain of T is L1(X, μ)+Lp1(X, μ) is just a device to make sureTf is defined when f ∈Lp,1≤ p≤ p1. For f ∈Lp, write f = fχ|f|>1+ fχ|f|≤1 = f0+ f1. Then|f0| ≤ |f|p ∈L1 and|f1| ≤
|f|p/p1 ∈Lp1. Before proving the theorem, let us use it to prove the remaining part (b) of the Hardy–Littlewood maximal theorem.
Proof of Theorem 4.3(b). In this case the measure spaces are both (, d x).
The operator M clearly satisfies the subadditivity condition (i). Condition (ii) we proved as part (a) of the Hardy–Littlewood theorem. We take p1= ∞ and condition (iii) holds withA1 =1. The Marcinkiewicz theorem then tells us
M fp≤ Apfp, 1< p≤ ∞
which is the assertion in part (b) of Theorem 4.3. It of course follows that M f <∞almost everywhere.
Proof of Theorem 4.5. Fix f ∈Lp,1< p< p1, and,λ >0. Let Eλ = {y:|T f(y)|> λ}.
We are going to get a tight grip onν(Eλ) and then use Lemma 4.1 to estimate T fp. The clever Marcinkiewicz idea is to splitf atλ/2A1. Write
f0= fχ{x:|f(x)|>λ/2A1}, f1 = fχ{x:|f(x)|≤λ/2A1}.
Then|T f(y)| ≤ |T f0(y)| + |T f1(y)|, andEλ⊂ Bλ∪Cλ,where Bλ= {y:|T f0(y)|> λ/2}, Cλ= {y:|T f1(y)|> λ/2}.
Now by (ii) we have
ν(Bλ)≤2A0
λ f01 ≤2A0
λ
|f|>λ/2A1
|f|dμ.
To estimateν(Cλ) we consider two cases. If p1 = ∞, thenf1∞< λ/2A1
and by (iii)Cλ =∅. (This explains the presence ofA1in the definitions of f0
and f1.) Ifp1 <∞, we have by (iii) ν(Cλ)≤
2
λA1f1p1
p1
≤ (2A1)p1 λp1
|f|≤λ/2A1
|f|p1dμ.
We boundν(Eλ) byν(Bλ)+ν(Cλ) and use Lemma 4.1. The case p1 = ∞is easier:
T fpp = ∞
0
pλp−1ν(Eλ)dλ≤ ∞
0
pλp−1 2A0
λ
|f|>λ/2A1
|f|dμ
dλ
≤2A0p
|f|
2A1|f|
0
λp−2dλdμ=2pA0A1p−1p p−1
|f|pdμ, because p−2>−1. Hence T f ∈Lp. If p1<∞, we have the same thing plus an additional term boundingν(Cλ):
T fpp ≤ ∞
0
pλp−1 2A0
λ
|f|>λ/2A1
|f|dμ
dλ +
∞
0
pλp−1(2A1)p1 λp1
|f|≤λ/2A1
|f|p1 dμdλ.
The first integral we just estimated in the proof forp1 = ∞. The second integral is
(2A1)p1p
|f|p1 ∞
2A1|f|λp−p1−1dλdμ= (2A1)pp p1−p
|f|pdμ becausep−p1−1<−1. Altogether this gives
T fp ≤ Apfp
with
App ≤2pA1p−1 A0p
p−1+ A1p p1− p
, which proves the theorem.
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]).