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Conjugate Functions and Maximal Functions

Dalam dokumen Bounded Analytic Functions (Halaman 122-139)

Exercises and Further Results

3. Conjugate Functions and Maximal Functions

then

μ(λ)≤2m(λ)+ 2 λ2

λ

0

sm(s)ds.

(3.4) Proof. Let

Uλ= {t :u(t)> λ},

so thatm(λ)= |Uλ|, and let Eλ=\Uλ. Form the regionR =

tEλ (t).

The boundaryR consists of two subsets,

Eλ=∩R and = {y>0} ∩R.

The set is the union of some tents whose bases are the component intervals of the open setUλ, as shown in Figure III.1.

Figure III.1. OnR,|u(z)| ≤λ.

SinceF(z)∈ H2, Cauchy’s theorem and the density ofAN in H1yields

R F2(z)dz =0.

Expanding out the real part of the integral gives

Eλ

(u2v2)d x +

(u2v2)d x

2uvd y=0. On ,|d y| =d x and−2uvu2+v2, so that

2uvd y

(u2+v2)d x.

Hence we have

0≤

Eλ

(u2v2)d x+2

u2 d x

and

Eλ

v2d x

Eλ

u2d x +2

u2d x. (3.5)

Along , we have|u(z)| ≤λ, so that

u2d xλ2

d x =λ2|Uλ| =λ2m(λ).

To estimate the other term on the right side of (3.5), write

Eλ

u2d x

Eλ

(u)2d x =

0<s<uu∗≤λ

2s ds d x

= λ

0

{s<u}d x 2s ds= λ

0

(m(s)−m(λ))2s ds

= λ

0

2sm(s)dsλ2m(λ).

With Chebychev’s inequality, (3.5) now yields

|{tEλ:|v(t)|> λ}| ≤(12) λ

0

2sm(s)dsm(λ)+2m(λ).

Then since

μ(λ)≤ |Uλ| + |{tEλ:|v(t)|> λ}|, (3.4) is proved.

Inequality (3.4) is the main step in the proof of Theorem 3.1. The rest of the proof is more routine.

Lemma 3.3. If0< p≤2, and if uLp, then

|u(x+i y)|2d x 1/2

≤2y1/2−1/p

|u|p dt 1/p

, y>0.

Proof. Fixy>0. Then

|u(x+i y)|p ≤ inf

|tx|<y|u(t)|p ≤(1/2y) x+y

xy |u(t)|p dt, and so

sup

x |u(x+i y)| ≤(2y)−1/pup. Sincep≤2, this yields

|u(x+i y)|2d x 1/2

(2y)(p−2)/pu2−pp

|u(x+i y)|pd x 1/2

≤(2y)1/2−1/p

|u|pdt 1/p

, as asserted.

To prove Theorem 3.1 for the line and for 0< p<2, assumeuLpand fixy0 >0. By Lemma 3.3 there is a conjugate functionv(z) defined ony> y0

such that f =u+ivsatisfies

y>ysup0

|f(x+i y)|2d x <.

For everyy0>0 there is only one suchv, and hencevdoes not depend ony0. Let

μ(λ)= |{x :|v(x+i y0)|> λ}|.

Sinceu0(z)=u(z+i y0) hasu0u, Lemma 3.2 implies that μ(λ)≤2m(λ)+ 2

λ2 λ

0

sm(s)ds,

wherem(λ)= |{t :u(t)> λ}|. Integrating this inequality against p−1 now gives

|v(x+i y0)|pd x =

0

p−1μ(λ)

≤2

0

p−1m(λ)+2

0

psm(s)

s λp−3 dλds

=2upp+ 2 2−p

0

psp−1m(s)ds

=2

1+ 1

2− p

upp.

The right side is independent ofy0and we have (3.1) with constant cp =2

1+ 1

2− p

, p<2.

It does not matter that this constant blows up as p→2 because the Riesz theorem gives another proof of (3.1) forp>1.

The casep≥2 of Theorem 3.1 can of course be proved without recourse to the Riesz theorem. The case p>2 follows from the casep<2 by a duality, and an interpolation then gives the remaining case p=2.

On the unit disc the same reasoning can be used once we establish an inequal- ity like (3.4). Let f =u+iu˜ ∈ H2, where ˜u(0)=0. LetUλ= {θ:u(θ)>

λ}, whereuis defined by (3.3), letEλ=T\Uλ, and letm(λ)= |Uλ|. Also let μ(λ)= |{θ:|uθ)|> λ}|. Ifm(λ) is not small we automatically have

μ(λ)≤Cm(λ)+ 1 λ2

λ

0

sm(s)ds, (3.6)

which is the inequality we are after, and so we assumem(λ) is small. As before, form

R =

θEλ

z: |eiθz|

1− |z| <2

.

Now R = Eλ , where is a union of tentlike curves, one over each component ofUλ.By Cauchy’s theorem

(u(0))2= 1 2πi

R f2(z)dz/z.

OnEλ,

1 2πi

dz z =

2π, while on ,

1 2πi

dz z =

2π ± dr 2πir.

Ifm(λ)= |Uλ|is small enough, then 1−r is small on , and a calculation then shows that

dr r dθ

<1

almost everywhere on . Taking the real part of the integral above then gives 0≤ 1

2π

Eλ

(u2−( ˜u)2)+ 1 2π

(u2−( ˜u)2)+ 1 2π

2uu˜dr r . We have the estimate

2|uvdr

r

(u2+( ˜u)2)dθ, so that we obtain

Eλ

( ˜u)2

Eλ

u2+2

u2

whenm(λ) is small. This inequality implies (3.6) just as in the proof of Lemma 3.2 and consequently (3.6) is true for all values ofm(λ). The remainder of the proof of Theorem 3.1 now shows that (3.2) holds for 0< p<2, and the Riesz theorem gives (3.2) for 2≤ p<∞.

Corollary 3.4. If0< p<and if u(z)is a real-valued harmonic function, then u(z)=Re f(z), fHp, if and only if uLp. There are constants c1

and c2, depending only on p, such that

c1upfpc2up.

Proof. The inequalityc1upfpwas proved in Chapter II. The other inequality is immediate from the theorem.

Corollary 3.4, which is of course equivalent to the theorem, is very important because it enables theHPspaces,p<∞, to be defined without any reference to analytic functions. The HP spaces can further be characterized without recourse to harmonic functions (see Fefferman and Stein [1972]).

Corollary 3.5. If0< p<and if u(z) is a harmonic function such that uLp, then there is a conjugate function v such thatvLPand

vppCpupp.

Proof. By Corollary 3.4, f =u+ivis in Hp withfppc2pupp. The maximal theorem then shows that fLp withfppAc2pupp.Since triviallyvf, we conclude thatvppCpupp.

There is another inequality even stronger than (3.1) in which the nontangen- tial maximal function is replaced by the vertical maximal function. Again let u(z) be a harmonic function onH , and write

u+(t)=sup

y>0|u(t+i y)|, t. Obviouslyu+(t)≤u(t). Conversely, we have

Theorem 3.6. If0< p<and if u(z) is harmonic onH , then

|u|p dtcp

|u+|p dt, where cpis a constant depending only on p.

As a corollary, we see that a harmonic functionu(z) is the real part of anHP function if and only ifu+Lp, and consequently that (3.1) holds for cones of any angle.

The proof of Theorem 3.6 rests on a remarkable inequality due to Hardy and Littlewood.

Lemma 3.7. If u(z)is harmonic on the disc (z0,R) and if0< p<, then

|u(z0)| ≤Kp

⎝ 1 πR2

(z0,R)

|u(z)|pd x d y

1/p

, (3.7)

where KPdepends only on p.

Whenp≥1, this lemma is a trivial application of H¨older’s inequality and the mean value property. Whenp<1 the inequality is rather surprising, because

|u(z)|p is not always subharmonic.

Proof. We only have to treat the case p<1. We may change variables and assumez0=0,R=1. Write

mq(r)= 1

2π 2π

0

|u(r e)|q 1/q

. We can assume that

2 1

0

mp(r)pr dr = 1 π

(0,1)

|u(z)|pd x d y=1

and then thatm(r)=sup{|u(z)|:|z| =r}>1 for 0<r <1, because oth- erwise (3.7) is true withKP =1.

Sincep<1 we have m1(r)≤ 1

2π

|u(r e)|pdθm(r)1−p =mp(r)pm(r)1−p. Also, estimating the supremum of the Poisson kernel gives

m(ρ)≤ 2m1(r)

1−ρ/r, 0< ρ <r <1. Setρ=rα, whereα >1 will be determined later. Then

1

1/2logm(rα)dr r

1

1/2log 2

1−ρ/r dr

r + p 1

1/2logmp(r)dr r +(1−ρ)

1

1/2logm(r)dr r .

The first integral converges with valueCα, and the second integral is bounded by

1

1/2mpp(r)dr r ≤4

1

0

mpp(r)r dr ≤2 by our assumption, and so we have

1

1/2logm(rα)dr

r ≤(Cα+2)+(1−p) 1

1/2logm(r)dr r . Since we have assumed logm(r)≥0, a change of variables gives

1

1/2logm(r)dr r

1

(1/2)αlogm(r)dr r =α

1

1/2logm(rα)dr r

α(Cα+2)+α(1− p) 1

1/2logm(r)dr r .

Takingα(1−p)<1 now yields

1/2<r<1inf logm(r)< α(Cα+2) 1−α(1− p)

1

log 2 =logKp, and (3.7) follows from the maximum principle.

Lemma 3.7 provides an alternate proof of Lemma 3.3 but with a different constant.

Proof of Theorem 3.6. Fix q with 0<q< p. Let z=x+i y∈ (t)= {|xt|<y}. Thenuis harmonic on(z,y/2), and Lemma 3.7 gives

|u(z)|q ≤ 4Kqq

πy2

(z,y/2)|u(ξ+)|qdξdηCq

y

|ξx|<y/2|u+(ξ)|q

Cq

y

|ξt|<3y/2|u+(ξ)|qdξ,

since|xt|<y. The last integral is dominated byCqMg(t) whereg=(u+)q and whereMgis the Hardy–Littlewood maximal function ofg. We have proved that

|u(t)|pCq (Mg(t))p/q.

Sincep/q>1 andgLp/q, the maximal theorem gives us

|u(t)|p dtC

|Mg(t)|p/q dtC

|u+(t)|pdt, and Theorem 3.6 is proved.

Notes

For the classesα, Theorem 1.3 dates back to Privalov [1916]. The weak- type estimate

|{θ :|f˜(θ)|> λ}| ≤(A/λ)f, (N.1)

and its corollary

f˜pApf1, 0< p<1, (N.2)

which of course follow from Theorem 2.1, are due to Kolmogoroff [1925].

The weak-type estimate for the Hilbert transform had been published earlier by Besicovitch [1923]. M. Riesz announced his theorem in [1924], but he delayed publishing the proof until [1927]. Corollary 2.6 is due to Zygmund [1929] and Theorem 2.7 is from Stein and Weiss [1959].

There are numerous proofs of the Riesz theorem. The one in the text was chosen because it is short and because it stresses harmonic estimates. The proof of Theorem 2.1 copies the Carleson proof of Kolmogoroff’s inequality in Katznelson [1968]. I learned this proof of Theorem 2.7 from Brian Cole, while driving in Los Angeles.

The real-variables proofs of the Riesz theorem, which are more amenable to higher-dimensional generalizations, lead to the theory of singular in- tegrals (see Calder´on and Zygmund [1952], and Stein [1970]). The Calder´on-Zygmund approach to Theorem 2.1 is outlined in Exercise 11, and some of its ideas will appear at the end of Chapter VI. Another real vari- ables proof, relying on a beautiful lemma due to Loomis [1946], is elegantly presented in Garsia [1970].

Three other proofs deserve mention: the original proof in Riesz [1927] or in Zygmund [1955]; Calder´on’s refinement in Calder´on [1950a] or Zygmund [1968]; and the Green’s theorem proof, due to P. Stein [1933], which is given in Duren [1970] and also in Zygmund [1968].

The best possible constants Ap in Theorem 2.3 were determined by Pichorides [1972] and independently by B. Cole (unpublished). Using Brownian motion, B. Davis [1974, 1976] found the sharp constants in Kolmogoroff’s inequalities (N.1) and (N.2). Later Baernstein [1979] gave a classical proof of the Davis results. The idea is to reduce norm inequalities about conjugate functions to a problem about subharmonic functions on the entire plane. By starting with nonnegative functions, we could work with har- monic functions on a half plane instead. The cost is higher constants and some loss in generality. Gamelin’s recent monograph [1979] gives a nice exposi- tion of Cole’s beautiful theory of conjugate functions in uniform algebras and derives the sharp constants for several theorems.

The Burkholder–Gundy–Silverstein theorem, Theorem 3.1, is now a funda- mental result. On the one hand it explains why conjugate functions obey the same inequalities that maximal functions do. On the other hand, the casep≤1 of the theorem shows that the nontangential maximal function is more pow- erful than the Hardy–Littlewood maximal function. It is not the harmonicity but the smoothness of the Poisson kernel that is decisive here. For example, let ϕ(t) be a positive, Dini continuous, compactly supported function on. Let

fL1() be real valued, and, in analogy with the Poisson formula, define U(x,y)= 1

y

ϕ xt

y

f(t)dt, y>0. Then for 0< p≤1, f =ReF,FHp, if and only if

supy>0|U(x,y)| ∈Lp().

This theorem, and another proof of Theorem 3.1 forn, are in Fefferman and Stein [1972]. Incidentally, the exact necessary and sufficient conditions for a kernelϕto characterizeHpremains an unsolved problem (see Weiss [1979]).

Another route to Theorem 3.1 is through the atomic theory ofHp spaces.

See Coifman [1974] for the case of1, Latter [1978] for the generalization to n, and Garnett and Latter [1978] for the case of the ball ofn. The theory of atoms is briefly discussed in Exercise 11 of Chapter VI.

The elementary proof of Theorem 3.1 given in the text is from Koosis [1978].

It makes this book 10 pages shorter.

Exercises and Further Results

1. (Peak sets for the disc algebra). Let ET be a compact set of measure zero.

(a) (Fatou [1906]) There exists uL1(T) such that u:T →[−∞,0]

continuously,u−1(−∞)= E, anduisC1onT\E. Then g=u+iu˜

is continuous onD\Eandghas range in the left half plane. The function f0 =eg

is in the disc algebra Ao = HC(T) and f0(z)=0 if and only ifzE.

Thus any closed set of measure zero is azero setfor Ao. Conversely, ifET is the zero set of some nonzero fAo, then|E| =0.(

log|f|dθ >− ∞.) (b) (F. and M. Riesz [1916]) Letgbe the function from part (a) and let

f1 =g/(g−1).

Then f1Ao, f1 ≡1 onE, and|f1(z)|<1,zD\E. This meansEis apeak setfor Ao. Conversely, any peak set for Ao has measure zero. (IfEis a peak set and if f ≡1 on E,|f|<1 on D\E,then 1− f has zero setE.)

(c) Use part (b) to prove the F. and M. Riesz theorem on the disc: Ifμis a finite complex Borel measure onTsuch that

einθ(θ)=0,n=1,2, . . . , thenμis absolutely continuous. (If not, there is compactE,|E| =0 such that

fEeiθ=0.But

n→∞lim

f1ne=0.) This is the original proof of the theorem.

(d) (Rudin [1956], Carleson [1957]) IfETis a compact set of measure zero, thenEis apeak interpolationset for Ao : GivenhC(E) there exists gAo such that the restriction toEishand such that

g= h =sup

E |h(z)|.

By Runge’s theorem there aregnAosuch thatgnhuniformly onE.Take nj so that|gnj+1gnj|<2j on some open neighborhoodVj ofE, and take

kj so that

|gnj+1gnjf1kj|<2j onT\Vj, where f1is the function from part (b). Then

g=gn1+

j=1

f1kj(gnj+1gnj)

is in Ao andg=honE. By being a little more careful, one can also obtain g= h.

(e) The result in (d) can also be derived using a duality. The restriction map S: AoC(E) has adjointS: M(E)→ M(T)/Ao, whereM(E)=C(E) is the space of finite complex Borel measures onEand whereAoM(T) is the subspace of measures orthogonal toAo. The F. and M. Riesz theorem from Chapter II shows thatSis an isometry. IfIis the ideal{fAo :S(f)=0}, then by a theorem in functional analysis the induced map ˜S: Ao/IC(E) is also an isometry and ˜Smaps ontoC(E). Thus ifhC(E) and ifε >0, there existsgAosuch thatg=honEandg≤(1+ε)h. Withmkchosen correctly,

g0= 2kf1mkg

then satisfiesg0 = handS(g0)=h(see Bishop [1962] and Glicksberg [1962]).

(f) (Gamelin [1964]) Letp(z) be any positive continuous function on D. IfhC(E) and if|h(z)| ≤ p(z),z∈E, then there existsgAosuch that g=honEand|g(z)| ≤ p(z),zD.

2. We shall later make use of the following application of 1(b) and 1(c) above. Letμbe a finite complex Borel measure onT. Assumeis singular to. Then there are analytic polynomials pn(z) such that

(i) |pn| ≤1,|z| ≤1, (ii) pnμ, and

(iii) pn →0 almost everywhere.

Condition (iii) can also be replaced bypn → −1 almost everywhere. (Hint:

There are En compact, EnEn+1,|En| =0 such that μ is supported on En.LethnC(En),hn<1,

Enhnμ; letgnAo interpolate hn,gn<1; and let pn be a polynomial approximation ofgn.)

3. Let Am denote the algebra of functionsf such thatf and its first m derivatives belong toAo, and let A= ∩Am. LetAα,0< α≤1 be the space of functions fAosuch that

|f(z)− f(w)| ≤K|zw|α.

(a) LetEbe a closed subset of the unit circle having measure zero, and let {lj}be the lengths of the arcs complementary toE.Necessary and sufficient for Eto be the zero set of a function inAorAαis the conditionjljlog(1/lj)<

∞(see Carleson [1952]).

(b) On the other hand, there is fAm such that f(z)=1,zE, and

|f(z)|<1,zD\E,if and only ifEis a finite set (see Taylor and Williams [1970]; the papers of Alexander, Taylor, and Williams [1971] and Taylor and Williams [1971] have further information on related questions).

4. Theorem 1.3 is sharp in the following sense. Letω(δ) be increasing and continuous on [0,2π], ω(δ1+δ2)≤ω(δ1)+ω(δ2), ω(0)=0.Suppose

1

0

ω(δ)

δ = ∞.

(a) There exists fC(T), f real, such that ωf(δ)≤ω(δ), δ < δ0,

but such that f is not continuous. (Let f(t)=ω(t),0<t < δ0, f(t)= 0,−1<t <0.)

(b) There isgC(T),greal, such that ωg(δ)≥ω(δ)

but such that ˜g is continuous. Here, in outline, is one approach. IfK is the Cantor ternary set, then{xy,xK,yk} =[−1,1].(Try drawing a pic- ture ofK ×K.) LethC(K) be real,ωh(δ)≥ω(δ), and findgHwith continuous boundary values such thatg=hon{e :θK}.

5. If fC(T) is Dini continuous, then Snff uniformly. There is a sharper result: Snff uniformly if ω(δ) log(1)→0(δ→0) (see Zygmund [1968]).

6. If 1< p<∞, and if fLp, then Snffp →0. 7. Let fL1(T). If

|f|log(2+ |f|)dθ <∞, then ˜fL1. If f ≥0 and ˜fL1, then

|f|log(2+ |f|)dθ <.

The first assertion is due to Zygmund [1929], the second is due to M. Riesz;

see Zygmund [1968].

8. Prove the two Kolmogoroff estimates

|{|f˜|> λ}| ≤(A/λ)f1 and f˜pApf1, 0< p<1.

The first is valid on both the line and the circle, the second on the circle.

9. LetEbe a subset ofof finite measure and define E(x)= lim

ε→0

1 π

|xt|

χE(t) xt dt.

Then the distribution ofE depends only on|E|. More precisely,

|{|E(x)|> λ}| = 2|E|

sinhπλ (see Stein and Weiss [1959]).

10. (More jump theorem). Let fL1(T) and define F(z)= 1

2πi

|ζ|=1

f(ζ)

ζz ,|z| =1.

From insideD,F(z) has nontangential limit calledf1(ζ) at almost everyζT, and from|z|>1,F(z) has nontangential limit f2(ζ) at almost everyζT. Moreover,

f1(ζ)− f2(ζ)= f(ζ) almost everywhere.

11. Let fLp(),1≤ p<∞. Write Hεf(x)= 1

π

|xt|

f(t)

xt dt, ε >0. and

Hf(x)=sup

ε>0|Hεf(x)|.

Hf is themaximal Hilbert transform.

(a) Show

Hfc1M(H f)+c2M f,

whereMis the maximal function, andH f =limε→0Hεf. Then conclude by means of (1.10) thatHf2cf2.

(b) Assuming that H is weak-type 1–1, show by interpolation that Hfpcpfp,1< p<2.

(c) Withf fixed, there is a measurable functionε(x),0< ε(x)<∞, such that

|Hε(x)f(x)| ≥ 12Hf(x).

(This process is called linearization.) Suppose ε(x) is a simple function N

j=1εjχEj,EjE=∅, then for fLp,gLq,2< p<∞,

Hε(x)f(x)g(x)d(x)

=

g(x)

j

χEj(x)

|tx|≥εj

f(t) xtdtd x

j

f(t)

|tx|≥εj

χEj(x)g(x) tx d xdt

fp

j

H(χEjg)q.

(d) From (b) and (c) it is almost trivial that H f =limε→0Hεf satisfies H fpCpfp.

(e) We turn to the weak-type estimate

|{x : Hf(x)>3λ}| ≤(C/λ)f1. (E.1)

Let= {x :M f(x)> λ}and let F=\. Write=

j=1Ij where the closed intervalsIjsatisfy

dist (F,Ij)= |Ij| (see Figure III.2).

Figure III.2. The decomposition= Ij.

Notice that

1

|Ij|

Ij

|f(t)|dt ≤2λ

becauseIjis contained in an interval twice as large that touchesF. Let g(x)= f(x)χF(x)+

j

1

|Ij|

Ij

f(t)dt

χIj(x) and

b(x)= f(x)−g(x)=

j

f(x)− 1

|Ij|

Ij

f(t)dt

χIj(x).

Real analysts callg(x) the good function,b(x) the bad function. The reason g(x) is good is thatg2is not too large,

g22 =

F

|f|2d x+

j

1

|Ij|

Ij

f(t)dt 2

|Ij|

λ

F

|f|d x+2λ

j

Ij

|f(t)|dt ≤2λf1.

Therefore theL2 estimate from part (a) yields

|{x : Hg(x)> λ}| ≤ C

λ2g22 ≤ 2C λ f1

and (E.1) will follow if we can show

|{x : Hb(x)>2λ}| ≤ C λf1.

By the maximal theorem,|| ≤(C/λ)f1, so that we only have to prove

|{xF: Hb(x)>2λ}| ≤ C λf1. (E.2)

Write the bad functionb(x) as

bj(x),where bj(x)=

f(x)− 1

|Ij|

Ij

f(t)dt

χIj(x), and note that bj(x) has the good cancellation property

bj(x)d x =0. Fix xFand letε. Then

Hεb(x)=

j

|xt|

bj(t) xt dt

=

dist(x,Ij)

Ij

bj(t) xt dt+

dist(x,Ij)≤ε

|xt|

bj(t) xt dt.

= Aε(x)+Bε(x).

Lettjbe the center of Ij. When dist (x,Ij)> ε,we have

Ij

bj(t) xt dt =

Ij

bj(t) 1

xt − 1 xtj

dt

=

Ij

bj(t)(ttj) (xt)(xtj) dt

since

bj(t)dt =0.ForxFandtIj,|xt|/|xtj|is bounded above and below, so that

Ij

bj(t) xt dt

c|Ij|

Ij

|bj(t)|

|xt|2 dt. Therefore

supε>0|Aε(x)| ≤ A(x)=

j

c|Ij|

Ij

|bj(t)|

|xt|2 dt.

The trick here was to use the cancellation ofbj(t) to replace the first-order singularity by a second-order singularity. Now we have

F

A(x)d xC

j

c|Ij|

Ij

|bj(t)|

|Ij|

ds s2 dt

C bj1 ≤2Cf1

and

|{xF : sup|Aε(x)|> λ}| ≤ C λf1.

When dist (x,Ij)≤ε, we have|Ij| =dist(F,Ij)≤ε, so that

|Bε(x)| ≤ c ε

ε<|x−1|<2ε| bj(t)|dtcMb(x). Sinceb1 ≤2f1, this yields

|{xF: sup

ε>0|Bε(x)|> λ}| ≤ C λf1.

Now (E.2) follows from the weak-type estimates we have established for supε|Aε(x)| and supε|Bε(x)|(see Calder´on and Zygmund [1952] and Stein [1970]).

IV

Some Extremal Problems

We begin with the basic duality relation,

gHinfpfgp =sup

Ffdθ 2π

: FH0q,Fq =1

. (0.1)

This relation is derived from the Hahn–Banach theorem in Section 1. Then, rather than continuing with a general theory, we use (0.1) to study three im- portant and nontrivial problems.

1. Determining when a continuous function on the circleThas continuous best approximation inH. This problem is discussed in Section 2.

2. Characterizing the positive measuresμonTfor which

|p2 K

|p|2

for all trigonometric polynomialsp(θ). This topic will reappear in Chapter VI, where the main result of Section 3 of this chapter will yield information about the real parts ofHfunctions.

3. Solving the interpolation problem

f(zj)=wj, j =1,2, . . . , (0.2)

fH, with the additional restriction that|f(e)|be constant or thatf

be as small as possible. This problem is treated in some detail, because Theorem 4.1 below will have important and striking applications later and because the rather precise results on this topic require ideas somewhat deeper than the duality relation (0.1).

To compare the duality approach to a more classical method, we conclude the chapter with Nevanlinna’s beautiful solution of (0.2).

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Dalam dokumen Bounded Analytic Functions (Halaman 122-139)