Exercises and Further Results
1. Dual Extremal Problems
Theorem 1.2. Let1≤ p<∞, and let f ∈Lp, /∈Hp. Then the distance from f to Hp is
dist(f,Hp)= inf
g∈Hpf −gp
(1.3)
=sup
f Fdθ 2π
: F∈ H0q,Fq ≤1
.
There exists unique g∈Hpsuch thatdist(f,Hp)= f −gpand there exists unique F ∈H0q,Fq =1such that
f Fdθ
2π =dist(f,Hp). (1.4)
Proof. The identity (1.3) follows from (1.2). Letgn ∈ Hpbe such thatf − gnp →dist(f,Hp). The Poisson integrals of f −gn are bounded on any compact subset of the disc, so by normal families there is an analytic function gonDsuch thatgn(z)→g(z),z∈ D, if we replace{gn}by a subsequence.
Taking means over circles of radiusr <1, we see thatg∈ Hp and thatf − gp≤limf −gnp. Thusf −gp =dist(f,Hp) and there exists at least one best approximationg∈Hp.
LetFn ∈H0q,Fnq ≤1, be such that
f Fndθ/2π →dist(f,Hp). Since 1<q ≤ ∞, the Banach–Alaoglu theorem can be used to obtain a weak- star limit point F of {Fn}. Then Fq ≤1,F∈H0q, and
f Fdθ/2π = dist(f,Hp). Since
g F dθ =0, we have dist(f,Hp)=
(f −g)F dθ/2π ≤ f −gpFq ≤ f −gp. (1.5)
Equality must hold throughout this chain of inequalities. This meansFq =1, and there exists a dual extremal functionFfor which (1.4) holds.
Now letg∈ Hpbe any best approximation off and letF∈ H0qbe any dual extremal function. Because equality holds in (1.5), the conditions for equality in H¨older’s inequality give us
f −g
f −gp = F|F|¯ q−2 and F =(f −g) |f −g|p−2 f −gp−1p
whenp>1. When p=1 we get
(f −g)F= |f −g|
instead.
Let p>1. Since|F|>0 almost everywhere, the first equation shows that gis unique. Since F∈ H0q is determined by its values on any set of positive measure, the second equation shows thatFis unique.
Similarly, whenp=1, the third equation shows thatFis unique. The third equation then shows that Im(g F) is unique. Sinceg F ∈H01, this determines g F uniquely, and thengis determined almost everywhere because |F|>0 almost everywhere.
Note that the best approximationg∈Hp and the dual extremal function F∈H0q,Fq =1, are characterized by the relation
(f −g)Fdθ
2π = f −gpFq = f −gp, (1.6)
which is (1.5) with equality. Sometimes it is possible to compute dist(f,Hp) by finding solutionsFandgof (1.6) (see Exercise 3).
When p= ∞some of the conclusions of the theorem can be rescued if we use the bottom row of the table instead of the second row, and use (1.1) instead of (1.2).
Theorem 1.3. If f ∈L∞, then the distance from f to H∞is dist(f,H∞)= inf
g∈H∞f −g∞=sup
f Fdθ 2π
: F∈ H01,F1 ≤1
. There exists g∈H∞such thatf −g∞ =dist(f,H∞). If there exists F ∈ H01,F1≤1, such that
dist(f,H∞)=
f Fdθ 2π, then the best approximation g∈H∞is unique and
|f −g| =dist(f,H∞) almost everywhere.
Proof. The dual expression for the distance follows from (1.1). Just as in the proof of Theorem 1.2, a normal families argument shows there is a best approximating functiong∈H∞. If a dual extremal function F∈ H01exists, then
dist(f,H∞)=
(f −g)Fdθ
2π = f −g∞F1, so that
(f −g)
dist(f,H∞) = (f −g) f −g∞ = F¯
|F|
(1.7)
almost everywhere, and there is a unique best approximationg.
If a dual extremal functionFexists, (1.7) does not imply thatFis unique.
But it does, of course, imply that the argument ofFis unique.
Example 1.4. Let f(0)=e−2iθ. Taking F0(θ)=e2iθ we see that 1
2π
f F0dθ=1.
Hence dist(f,H∞)=1 and F0is a dual extremal function. However Fα(z)= z(z+α)(1+αz)¯
1+ |α|2 , |α|<1,
is another dual extremal function. In this problem the best approximating function isg=0.
Example 1.5. Let
f(θ)=
⎧⎨
⎩
1, 0< θ < π/2, 0, π/2< θ≤3π/2,
−1, 3π/2< θ≤2π.
If there wereg∈H∞such thatf −g∞ <1, then Reg> δ >0 on (0, π/2) and Reg<−δ <0 on (−π/2,0). Hence
ε→0lim 1 π
|θ|>ε
Reg(θ)
θ dθ = +∞.
Except for bounded error terms, this integral represents−limr→1 Im g(r).
Thus there is no such boundedg, and so dist(f,H∞)=1 andg=0 is one best approximation. Now letgbe the conformal mapping ofDonto the half discD∩ {Imz>0}. We can arrange thatg(1)=0,g(i)=1, andg(−i)= −1.
Then by checking the values ofgalong the three arcs on whichf is constant, we see thatf −g∞=1. Hence there is not a unique best approximation to
f ∈ H∞. By Theorem 1.3 there is no dual extremal functionF∈ H01. It is interesting to notice where the proof of Theorem 1.2 breaks down in Example 1.5. By the bottom row of the table there are Fn ∈H01,Fn1 ≤1, such that
f Fndθ/2π →1. As linear functionals onL∞, the Fn have some weak-star limit pointσ ∈(L∞)∗, andσ is orthogonal to H∞. In Chapter V we shall see thatσ is a complex measure on a compact Hausdorff space, the maximal ideal space ofL∞. However,σ is not weakly continuous onL∞; that is,σ cannot be represented asσ(h)=
h Fdθ with F∈L1. Otherwise we could conclude that|F| =1 almost everywhere. In fact, in a sense to be made precise in Chapter V,σ is singular todθ, which means, in classical language, that Fn(z)→0,z∈D. The absence of a dual extremal function F∈ H01 is often the central difficulty with an H∞ extremal problem. We confront this difficulty again in Section 4.
If the function f ∈L∞ is continuous, there is a dual extremal functionF and the best approximationgis unique.
Lemma 1.6. If f ∈C, then
dist(f,H∞)=dist(f,Ao)= inf
g∈Aof −g∞.
Proof. There existsg∈H∞ such thatf −g∞=dist(f,H∞). Let fr = f ∗Pr be the Poisson integral off and letgr =g∗Pr. SincePr1 =1 we
have
fr−gr∞= (f −g)∗Pr∞≤ f −g∞. Butgr ∈Ao, andf − fr∞ < εif 1−r is small. Thus
dist(f,Ao)≤lim
r f −gr∞ ≤ f −g∞=dist(f,H∞).
The reverse inequality is clear sinceH∞ ⊃ Ao.
We writeH∞+Cfor the set of functionsg+h,g∈H∞,h∈C.
Theorem 1.7. If f ∈H∞+C, then there exists F∈H01,F1 =1, such that
1 2π
f F dθ =dist(f,H∞), (1.8)
and there exists unique g∈H∞such thatf −g∞=dist(f,H∞).
Proof. Write f =g+h,g∈ H∞,h∈C. Then dist(f,H∞)=dist(h,H∞) and we can assume that f is continuous. By Theorem 1.3 there are Fn ∈ H01,Fn1 ≤1, such that
1 2π
f Fndθ →dist(f,H∞).
Taking a subsequence we can assume Fn(z)→ F(z),z∈D, where F∈ H01,F1≤1. This gives convergence of the Fourier coefficients, so that for all trigonometric polynomialsp(θ),
Fnpdθ 2π →
F pdθ
2π. Takingf − p∞small we see that
dist(f,H∞)=
F f dθ 2π.
Hence (1.8) holds and Theorem 1.3 now implies that f has a unique best approximation inH∞.
As an application we reprove some results from Chapter I, Section 2.
Corollary 1.8. Let z1,z2, . . . ,zn be distinct points in D, and let w1, w2, . . . , wnbe complex numbers. Among all f ∈H∞such that
f(zj)=wj, 1≤ j ≤n, (1.9)
there is a unique function f of minimal norm. This function has the form cB(z) where B(z)is a Blaschke product of degree at most n−1.
Corollary 1.9. Let zn+1 ∈D be distinct from z1,z2, . . . ,zn. Assume(1.9) has a solution f ∈ H∞withf∞ ≤1. Among such solutions let f0 he one for which |f(zn+1)| is largest. Then f0 is uniquely determined by its value
f0(zn+1)and f0 is a Blaschke product of degree at most n.
Corollary 1.9 is a simple consequence of Corollary 1.8. By normal families there exists an extremal function f0,f0∞≤1. Letwn+1 = f0(zn+1). If f ∈ H∞interpolates (1.9) and if also
f(zn+1)=wn+1,
thenf∞ ≥1, because otherwise for someλ,|λ|small, g= f +λ
n j=1
z−zj
1−z¯jz
is a function satisfying (1.9) such that g ≤1 and |g(zn+1)|>|f0(zn+1)|.
Hence f0has minimal norm among the functions interpolatingw1, . . . , wn+1
atz1, . . . ,zn+1.
Proof of Corollary 1.8. LetB0be the Blaschke product with zerosz1, . . .zn, and let f0be a polynomial that does the interpolation (1.9). The minimal norm of the functions inH∞satisfying (1.9) is
g∈Hinf∞f0−B0g∞= inf
g∈H∞B¯0f0−g∞.
Since ¯B0f0∈C, there is a unique interpolating function f ∈ H∞of minimal norm and there isF∈ H01,F1=1, such that
fB¯0Fdθ
2π = f∞. Hence|f| = |fB¯0| = f∞almost everywhere, and
f F/B0≥0 (1.10)
almost everywhere.
Lemma 1.10. If G∈H1is real almost everywhere on an arc I ⊂T , then G extends analytically across I.
Proof. On|z|>1 defineG(z)=G(1/z). This is anH1 function on|z|>1 with nontangential limitsG(θ) at almost every point ofI. Letζ ∈I and center atζ a discso small that∩T ⊂I. LetV =∩D,W =∩ {|z|>1}. Since we are dealing withH1functions, we have forw∈\T,
1 2πi
∂
G(z)
z−wdz = 1 2πi
∂V
G(z)
z−wdz+ 1 2πi
∂W
G(z)
z−wdz =G(w).
This shows thatGcan be continued acrossI.
To complete the proof of Corollary 1.8, notice thatG = f F/B0 is an H1 function on the annulusr <|z|<1 ifr >|zj|. Using (1.10) and using the lemma locally, we see that G is analytic across T, and thatG is in fact a rational function. Since B0G is analytic acrossT, and since f/f∞ is an inner function, Theorem II.6.3 shows thatf is analytic acrossT. Consequently f/f∞ is a Blaschke product of finite degree, andFis a rational function.
NowB0hasnzeros inDandFhas a zero atz=0. By (1.10) and the argument principle, it follows thatf has at mostn−1 zeros.
In Corollary 1.8 the extremal functionf(z)=cB(z) can have fewer thann− 1 zeros. For example, supposegis the function of minimum norm interpolating
g(zj)=wj, 1≤ j ≤n−1,
and takewn =g(zn). Then f =ghas at mostn−2 zeros. Similarly, the ex- tremal function in Corollary 1.9 can have fewer thannzeros. However, if the in- terpolation (1.9) has two distinct solutions f1, f2withf1∞≤1,f2∞≤1, then the extremal function f0in Corollary 1.9 is a Blaschke product of degree n. For the proof, notice that (1.9) then has a solution f withf∞ <1, by the uniqueness asserted in Corollary 1.8. Then by Rouch´e’s theorem, f0andf0− f have the same number of zeros in|z|<1. Since f0(zj)= f(zj),1≤ j ≤n, it follows that f0has at leastnzeros in|z|<1.
Corollary 2.4 Chapter I, contains more information than we have obtained here, but the duality methods of this section apply to a wider range of problems.