Exercises and Further Results
1. Preliminaries
Let f(θ)∈L1(T),T denoting the unit circle∂D. Supposing for a moment that f(θ) is real valued, we letu(z) be the Poisson integral of f(θ) and let ˜u(z) denote the harmonic conjugate function ofu(z), normalized so that ˜u(0)=0.
Since
Pz(ϕ)=Reeiϕ+z eiϕ−z, we have
(u+iu)(z)˜ = 1 2π
eiϕ+z
eiϕ−z f(ϕ)dϕ and
u(z)˜ = −1 2π
Qz(ϕ)f(ϕ)dϕ, where
−Qz(ϕ)=Im
eiϕ+z eiϕ−z
= 2rsin(θ−ϕ)
1−2rcos(θ−ϕ)+r2, z =r eiθ. These formulae define theconjugate functionu(z) even when˜ f(θ) is complex valued.
98
The kernel−Qz(ϕ)= Qr(θ−ϕ) is theconjugate Poisson kernel.†The ker- nelsQr do not behave at all like an approximate identity, becauseQr(θ) is an odd function and becauseQr1 ∼log 1(1−r). Nevertheless, the conjugate function ˜u(z) has nontangential limits almost everywhere onT.
Lemma 1.1. If f ∈L1(T), then u(z)˜ has nontangential limit f˜(θ) almost everywhere on T.
Proof. We may suppose f(θ)≥0. The analytic functiong(z)=u(z)+iu(z)˜ then has nonnegative real part, andG(z)=g(z)/(1+g(z)) is bounded. The functionG(z) has nontangential limitG(θ) almost everywhere, andG(θ)=1 on at most a set of measure zero. Consequentlyg=G/(1−G) has a finite nontangential limit almost everywhere, and so does its imaginary part ˜u(z).
The linear mapping
f(θ)→ f˜(θ)
is called theconjugation operator. The conjugate function ˜f can also be cal- culated using a principal value integral.
Lemma 1.2. Let f(θ)∈L1(T). For almost everyθ f˜(θ)=lim
ε→0
1 2π
|θ−ϕ|>εcot
θ−ϕ 2
f(ϕ)dϕ.
(1.1)
In particular, the limit in(1.1)exists almost everywhere. Moreover, u(r e˜ iθ)− 1
2π
|θ−ϕ|>1−rcot
θ−ϕ 2
f(ϕ)dϕ
≤C M f(θ), (1.2)
where C is an absolute constant and where Mf is the Hardy–Littlewood maximal function of f.
Proof. Notice that forθ =0,
r→1limQr(θ)=lim
r→1
2rsinθ
1−2rcosθ+r2 = sinθ
1−cosθ =cotθ
2 =Q1(θ), which is the kernel in (1.1). Setε=1−r. Forε < θ < π, we have
Q1(θ)−Qr(θ)= (1−r)2sinθ
(1−cosθ)(1−2rcosθ+r2)
= 1−r
1+rQ1(θ)Pr(θ)≤ 1−r
1+rQ1(1−r)Pr(θ).
†The minus sign in−Qz(ϕ) is to ensure thatQz(ϕ)=Qr(ϕ) whenz=r. With this notation, Pz−i Qz. is analytic in|z|<1 andeiθ→Pr(θ)+i Qr(θ) extends to be analytic in|z|<1.
Since (1−r)Q1(1−r)≤2, this gives
|Q1(θ)−Qr(θ)| ≤ 2 1+rPr(θ) (1.3)
forε <|θ|< π, On the other hand, for|θ| ≤ε=1−r, we have
|Qr(θ)| ≤2/ε.
(1.4)
Now
u(r e˜ iθ)− 1 2π
|θ−ϕ|>εcot
θ−ϕ 2
f(ϕ)dϕ
≤ 1 2π
|ϕ|≤ε|Qr(ϕ)||f(θ−ϕ)|dϕ + 1
2π
|ϕ|>ε|Q1(ϕ)−Qr(ϕ)||f(θ−ϕ)|dϕ.
By (1.3), (1.4), and Theorem I.4.2, the last two integrals are dominated by CM f(θ) and (1.2) is proved.
To prove (1.1) we use the fact that the odd functionsQr(θ) andQ1(θ)χ|θ|>ε
are orthogonal to constants. Thus u(r e˜ iθ)− 1
2π
ε<|ϕ|<πcot ϕ
2
f(θ−ϕ)dϕ
= 1 2π
|ϕ|≤εQr(ϕ)(f(θ−ϕ)− f(θ))dϕ + 1
2π
ε<|ϕ|<π
1−r
1+rQ1(ϕ)Pr(ϕ)(f(θ−ϕ)− f(θ))dϕ, and so by (1.3) and (1.4),
u(r e˜ iθ)− 1 2π
ε<|ϕ|<πcot ϕ
2
f(θ−ϕ)dϕ
≤ 1 πε
|ϕ|≤ε|f(θ−ϕ)− f(θ)|dϕ + 1
2π 2
1+rPr(ϕ)|f(θ−ϕ)− f(θ)|dϕ.
By Chapter I, Exercise 11, the last two integrals tend to zero on the Lebesgue set off.
It is not difficult to prove the nontangential analog of (1.2), u(z)˜ − 1
2π
|θ−ϕ|>εcot
θ−ϕ 2
f(ϕ)dϕ
≤CαM f(θ) whenz∈ α(θ) andε=1− |z|. The details are left to the reader.
The inequality
2
θ −cotθ 2
≤ 2 π
shows that the principal value in (1.1) has the same behavior as theHilbert transform,
Hf(θ)= lim
ε→0
1 π
ε<|ϕ−θ|<π
f(ϕ)
θ−ϕdϕ= lim
ε→0Hεf(θ).
(1.5)
Although Hf(θ)= f˜(θ), the difference arises by convolving f with the bounded function (2/θ −cotθ/2), so thatH f(θ) exists almost everywhere and
|f˜(θ)−H f(θ)| ≤(2/π)f1. (1.6)
There exist continuous f(θ) for which ˜f(θ) is not even bounded. For exam- ple, letF=u+iu˜be the conformal mapping ofDonto{z:|x|<1/(1+y2)}
withF(0)=0,F(0)>0. Then Carath´eodory’s theorem on the continuity of conformal mappings implies thatu(θ)=limr→1u(r eiθ) is continous and that u(θ) is unbounded. A more elementary example can be given using (1.5). Let˜ f(θ) be an odd function, so that no cancellation can occur in (1.5) atθ =0.
Then limr→1u(r˜ ) behaves like
ε→0lim 2 π
π
ε
f(θ)
θ dθ,
and this integral can diverge even thoughf is continuous.
When f(θ) is a continuous function onT, we write ω(δ)=ωf(δ)= sup
|θ−ϕ|<δ|f(θ)− f(ϕ)|
for themodulus of continuityoff. The modulus of continuity is a nondecreasing function satisfying
δ→0limω(δ)=0 and ω(δ1+δ2)≤ω(δ1)+ω(δ2).
A function is calledDini continuousif a
0
ω(t)
t dt <∞ for somea>0.
Theorem 1.3. If f(θ)is a Dini continuous function on T,then f exists at˜ every point of T, f is a continuous function,˜ and
ωf˜(δ)≤C δ
0
ω(t) t dt+δ
π
δ
ω(t) t2 dt
, (1.7)
where C is a constant not depending on f orω.
Observe that ifδ0is small and if 0< δ < δ0, then δ
π
δ
ω(t) t2 dt ≤
δ0
δ
ω(t)
t dt+ δ δ0
π
δ0
ω(t) t dt,
so that when f is Dini continuous the continuity of ˜f follows from (1.7). For 0< α <1, we say f ∈αifω(δ)=O(δα). Then (1.7) shows that conjugation preserves the Lipschitz classesα.
Proof. Ifb(θ) is a bounded function, then the convolution b∗ f(θ)= 1
2π
b(ϕ)f(θ−ϕ)dϕ satisfies
ωb∗f(δ)≤ b∞ωf(δ)≤Cb∞δ π
δ
ω(t) t2 dt.
It is therefore enough to show thatH f(θ) exists almost everywhere and that H f has continuity (1.7).
Since
Hf(θ)= lim
ε→0
1 π
|ϕ−θ|>ε
f(ϕ)− f(θ)
θ−ϕ dϕ,
the Dini continuity ensures thatH f exists at every point.
Let|θ1−θ2| =δ and take θ3=(θ1+θ2)/2. Because a constant function has Hilbert transform zero we can assume f(θ3)=0. We assume θ1< θ2. Then
|Hf(θ1)−Hf(θ2)| ≤ 1 π
θ1+δ
θ1−δ
|f(ϕ)− f(θ1)|
|ϕ−θ1| dϕ +1
π θ2+δ
θ2−δ
|f(ϕ)− f(θ2)|
|ϕ−θ2| dϕ +1
π θ1
θ1−δ
|f(ϕ)|
|ϕ−θ2| dϕ+ 1 π
θ2+δ
θ2
|f(ϕ)|
|ϕ−θ1| dϕ +1
π
3δ/2<|ϕ−θ3|<π|f(ϕ)|
1
θ1−ϕ − 1 θ2−ϕ
dϕ.
The first two integrals are dominated byCδ
0(ω(t)/t)dt. Since f(θ3)=0, the second two integrals are each bounded by
ω 3δ
2
2δ δ
dt
t ≤Cω(δ)≤c δ
δ/2
ω(t) t dt.
Again using f(θ3)=0, we can bound the fifth integral by C|θ2−θ1|
π
3δ/2
ω(t)
t2 dt ≤Cδ π
δ
ω(t) t2 dt. Together, these estimates give us (1.7).
Corollary 1.4. Let I be an open arc on T. Let f(θ)∈L1and assume f is Dini continuous on the arc I. Then f˜(θ)is continuous at each point of I.
Proof. First note that if f(θ)=0 onI, then by (1.1), ˜f is real analytic onI.
IfJis any compact subarc ofI, there is a functiong(θ) Dini continuous onT such thatg= f on a neighborhood ofJ. By Theorem 1.3 and the preceding remark,
f˜=g˜+(f −g)∼ is continuous onJ.
There is a close connection between conjugate functions and partial sums of Fourier series. Let f(θ)∈L1 have Fourier series
∞
−∞
aneinθ.
Suppose for convenience that f(θ) is real, so thata−n =an. Since Pr(θ)+i Qr(θ)= 1+r eiθ
1−r eiθ =1+2
∞ n=1
rneinθ
and sincePr andQr are real, we have Pr(θ)= ∞
−∞
r|n|einθ, Qr(θ)=
n=0
(−i) sgn(n)r|n|einθ.
Hence
u(r eiθ)= Pr ∗ f(θ)= ∞
−∞
anr|n|einθ and
u(r e˜ iθ)= Qr∗ f(θ)= −i
n=0
sgn(n)anr|n|einθ.
In particular, if f(θ) is a trigonometric polynomialN
−Naneinθ, then ˜f(θ) is a trigonometric polynomial of the same degree
f˜(θ)=
N
−N
m(n)aneinθ, (1.8)
where
m(n)=
⎧⎨
⎩
−i, n>0, 0, n=0, i, n<0,
is theFourier multiplierassociated with the conjugation operator. Parseval’s theorem now gives
Theorem 1.5. If f ∈L2, then f˜∈L2and f˜22 = f22− |a0|2, where a0 =(1/2π)
f(θ)dθ. Now consider the operator
P(f)= 12(f +i f˜)+ 12a0, which sends∞
−∞aneinθ into∞
0 aneinθ. The operatorPdiscards theanfor n<0, and so Pis the orthogonal projection of L2 onto H2. In any norm under which the linear functional f →a0(f) is continuous, the operatorPis bounded if and only if the conjugation operator is bounded.
The operator f →e−inθP(einθf) discards the coefficientsak fork<−n and leaves the other coefficients unchanged. The operator f →ei(n+1)θ P(e−i(n+1)θf) similarly removesak fork<n+1. Consequently
e−inθP(einθf)−ei(n+1)θP(e−i(n+1)θf)=Sn(f), thenth partial sumn
−nakeikθ of the Fourier series. Similar reasoning shows that
P(f)= lim
n→∞einθSn(e−inθf)
for f ∈L2. This means that the famous Marcel Riesz theorem f˜p ≤ Apfp, 1< p<∞, is equivalent to either of the inequalities
P fp ≤ Bpfp, 1< p<∞ or
sup
n Snfp ≤Cpfp, 1< p<∞.
It is not hard to see that the last inequality holds if and only if Snf − fp →0 (n→ ∞), 1< p<∞.
(For one implication use the uniform boundedness principle; for the other use theLpdensity of the trigonometric polynomials.)
In the upper half plane letu(z) be the Poisson integral of f(t)∈Lp,1≤ p<∞. The conjugate function ˜u(z) is now defined by
u(z)˜ = 1 π
x−t
(x−t)2+y2 f(t)dt = Qy∗ f(x), z=x+i y, where
Qy(t)= 1 π
t t2+y2
is the conjugate kernelfor the upper half plane. The integral defining ˜u(z) converges becauseQy∈Lq for allq>1. Since
Py(x−t)+i Qy(x−t)= 1 πi
1 t−z,
the functionu+iu˜is analytic in the upper half plane. This choice of ˜uinvolves a normalization different from the one used in the disc. Instead of ˜u(i)=0 we require limy→∞u(x˜ +i y)=0, because only with this normalization is it possible for ˜u(z) to be the Poisson integral of anLpfunction,p<∞. Because Qy∈/ L1, forp= ∞we revert to the normalization used on the disc and write
u(z)˜ = Qy(x−t)+ 1 π
t 1+t2
f(t)dt.
Then ˜u(i)=0 and the integral is absolutely convergent when f ∈ L∞. The results obtained above for conjugate functions on the disc can be proved in a similar way for the upper half plane, and we shall not carry out the detailed arguments. We shall, however, point out some minor differences between the two cases.
Whenp<∞, the limit of the conjugate kernelsQy(t), asy→0, coincides with the Hilbert transform kernel 1/πt. Thus
u(x˜ +i y)− 1 π
|x−t|>y
f(t) x−t dt
≤CMf(x),
and asy→0 both quantities in this expression converge almost everywhere to the same function ˜f(x)= Hf(x).
The Hilbert transform of f ∈L∞is defined almost everywhere by Hf(x)= lim
y→0u(x˜ +i y)= lim
ε→0
1 π
|x−t|>ε
1
x−t + t 1+t2
f(t)dt.
The normalization ˜u(i)=0 conveniently makes these integrals converge for larget.
When y>0 is fixed, the functionKy = Py+i Qy = −1/πi(t+i y) is in L2, and its Fourier transform
Kˆy(s)= lim
N→∞
−1 πi
N
−N
e−2πist t +i y dt can be evaluated by Cauchy’s theorem,
Kˆy(s)=
2e−2πsy, s>0, 0, s<0.
Since ˆPy(s)=e−2π|s|y, this gives Qˆy(s)=
−ie−2π|s|y, s >0, ie−2π|s|y, s <0.
It now follows from Plancherel’s theorem thatQy∗ f converges inL2 norm asy→0. Since Qy∗ f → Hf almost everywhere, we see thatHf ∈L2and Qy∗ f −H f2 →0, and we have the identities
Hf(s)=( ˜f)ˆ(s)= −i s
|s| fˆ(s), (1.9)
Hf2 = f˜2 = f2. (1.10)