Embed Size (px)
Citation preview
Banach Spaces of analytic functions(MAGIC 082)
Jonathan R. PartingtonUniversity of Leeds, School of Mathematics
March 25, 2014
1
LECTURE 1
Books:
K. Hoffman, Banach spaces of analytic functions.P. Koosis, Introduction to Hp spaces.W. Rudin, Real and complex analysis.N. Nikolski, Operators, functions and systems, an easy reading, Vol. 1.
1 Introduction
Recall that a Banach space is a complete normed space, and a Hilbert spaceis a special kind of Banach space, where the norm is given by an inner prod-uct. You should have met Lp and `p for 1 ≤ p ≤ ∞ and C(K) (the space ofcontinuous functions on K).
We are going to work with complex Banach and Hilbert spaces whose ele-ments are functions.
1.1 Examples (treated informally for the moment)
1. The Hardy spaces Hp (1 ≤ p ≤ ∞) are Banach spaces consisting ofanalytic functions in the unit disc
D = z ∈ C : |z| < 1,
whose boundary values are in Lp(T), where
T = z ∈ C : |z| = 1
is the unit circle. Thus
f(z) =∞∑n=0
anzn
for |z| < 1, and on |z| = 1, f corresponds to an Lp function (pointwise radiallimits almost everywhere, in fact).Recall that functions of period 2π have Fourier series, that is,
g(t) ∼∞∑
n=−∞
cneint,
2
where
cn =1
2π
∫ 2π
0
g(t)e−int dt.
If g ∈ Lp(0, 2π) for some 1 ≤ p ≤ ∞, then g ∈ L1(0, 2π), and
|cn| ≤1
2π
∫ 2π
0
|g(t)||eint| dt = ‖g‖L1 ,
so the sequence (cn) is bounded. Now if cn = 0 for all n < 0, we can define
f(z) =∞∑n=0
cnzn,
and it converges for |z| < 1.So we think of Hp either as a space of functions defined on the disc, or asfunctions on the circle, putting f(eit) = g(t), or indeed as functions of period2π.
In the special case p =∞, H∞ consists of all bounded analytic functions onD, and it forms a Banach algebra.
2. The harmonic Hardy spaces. We may proceed similarly for the set of all Lp
functions on the circle. Now the natural extension to the disc of∑∞−∞ cne
int
is going to be harmonic rather than analytic, i.e., we take
∞∑n=0
cnzn +
∞∑n=1
c−nzn.
(Recall that harmonic functions satisfy
∂2f
∂x2+∂2f
∂y2= 0
and include analytic functions such as zn and anti-analytic ones such as zn.)We define hp to be the space of harmonic functions in the disc with Lp bound-ary values.
3. The disc algebra A(D). These are the functions continuous on the closedunit disc D, and analytic on the open unit disc. We use the supremum norm.
3
This is a closed subalgebra of H∞.
4. The Bergman spaces. These are again spaces of analytic functions in thedisc, but now in Lp on the disc, i.e., f ∈ Ap for 1 ≤ p < ∞ if and only if fis analytic in D and
‖f‖p =
(1
π
∫D|f(z)|p dA(z)
)1/p
<∞,
where dA(z) is two-dimensional Lebesgue measure, i.e.,
‖f‖pp =1
π
∫ 1
r=0
∫ 2π
θ=0
|f(reiθ)|pr dr dθ.
In the case p = 2, if f has a Taylor series
f(z) =∞∑n=0
anzn,
then
‖f‖2 =∞∑n=0
|an|2
n+ 1.
5. The Wiener algebra, W . Take all functions of period 2π that can bewritten as
g(t) =∞∑
n=−∞
cneint,
with ‖g‖W =∑∞−∞ |cn| < ∞. Here the series converges uniformly so g is
actually continuous. Transferring to the circle we can write
f(z) =∞∑−∞
cnzn,
with∑∞−∞ |cn| <∞.
The positive or analytic Wiener algebra W+ consists of all analytic functionsf(z) =
∑∞n=0 cnz
n in the disc with
‖f‖W =∞∑0
|cn| <∞.
4
These functions all lie in A(D).As the name suggests, both W and W+ are Banach algebras.
Elementary inclusions:
W+ ⊂ A(D) ⊂ H∞ ⊂ H2 ⊂ H1
(and many others).
6. The Paley–Wiener space PW (K), where K ⊂ R is compact. Let f be afunction on R which can be written as
f(t) =1
2π
∫ ∞−∞
f(w)eiwt dw,
where f(w) = 0 for w 6∈ K, and f|K ∈ L2(K). Here f is the Fourier transformof f , and can be defined (for f ∈ L1(R)) by
f(w) =
∫ ∞−∞
f(t)e−iwtdt.
It turns out that such f can be extended to be analytic on the whole of C,and that
‖f‖2 =
∫ ∞−∞|f(t)|2 dt =
1
2π
∫ ∞−∞|f(w)|2 dw,
makes PW (K) into a Banach space, indeed even a Hilbert space.An important special case is K = [−a, a], where a > 0. Then PW (K) is thespace of band-limited signals (no high-frequency components).
7. The Segal–Bargmann (or Fock) space SB2 consisting of all functions fanalytic on the whole of C such that
‖f‖2 =1
π
∫C|f(z)|2e−|z|2 dA(z) <∞.
This has applications in quantum mechanics.
5
2 Hardy spaces on the disc
We begin with H2, which is a Hilbert space. Write
T = z ∈ C : |z| = 1 = eiθ : 0 ≤ θ ≤ 2π.
Normalize it to have measure 1, so all integrals will be with respect todθ
2π.
2.1 Proposition
The set of functions
zn : n ∈ Z = einθ : n ∈ Z
forms an orthonormal basis for L2(T), and its linear span is dense in Lp(T)for all 1 ≤ p <∞.Here we use the inner product on L2(T) given by
〈f, g〉 =1
2π
∫ 2π
0
f(eiθ)g(eiθ) dθ.
Proof: We see that
〈zn, zm〉 =1
2π
∫ 2π
0
einθe−imθ dθ =
0 if m 6= n,
1 if m = n,
and hence we have the orthonormality.
Recall now that if f ∈ Lp(0, 2π) and ε > 0, then there is a continuous periodicfunction g with ‖f − g‖Lp < ε/2. Next, g has a Fourier series
∑∞n=−∞ cne
inθ,whose Fejer sums
Gm(θ) =1
m+ 1
m∑k=0
k∑n=−k
cneinθ
converge uniformly to g on [0, 2π]. Hence there is a trigonometric polynomialGm with
‖g −Gm‖Lp < ε/2,
since uniform convergence implies Lp convergence. Now
‖f −Gm‖ ≤ ‖f − g‖+ ‖g −Gm‖ < ε.
6
Translating this to the circle with z = eiθ we see that for f ∈ Lp(T) we have
‖f −N∑−N
anzn‖p < ε
for suitable a−N , . . . , aN ∈ C.
7
LECTURE 2
2.2 Definition
The Hardy space H2 consists of all functions
f(z) =∞∑n=0
anzn
analytic in the unit disc D such that
‖f‖2 =∞∑n=0
|an|2 <∞.
It can naturally be identified with the closed subspace of L2(T) consisting ofall functions
∑∞n=0 ane
inθ (convergence in the L2 sense) with
∞∑n=0
|an|2 <∞,
i.e., the closed linear span of einθ : n ≥ 0.Note that if |z| < 1 then∑
|anzn| ≤(∑
|an|2)1/2 (∑
|z2n|)1/2
<∞,
by Cauchy–Schwarz, so all these power series have radius of convergence atleast 1, and define analytic functions in D.
2.3 Proposition
H2 is a Hilbert space with inner product⟨∞∑0
anzn,∞∑0
bnzn
⟩=∑
anbn,
or equivalently
〈f, g〉 =1
2π
∫ 2π
0
f(eiθ)g(eiθ) dθ.
8
Proof: This is clear, since it is a closed subspace of L2(T).Alternatively, it’s clear since the correspondence
(an)∞n=0 ←→∞∑n=0
anzn
identifies H2 with `2(Z+).
Analytic and harmonic extensions
Recall the Dirichlet problem: given f continuous on the unit circle, find gharmonic on the interior which extends it to a continuous function on theclosed disc. These can be real or complex functions.Recall also that if g is analytic, then its real and imaginary parts are bothharmonic, so g itself is. but if g = u+ iv, then g = u− iv is also harmonic.So if f(eiθ) = einθ for some n ∈ Z, then a solution is
g(reiθ) =
rneinθ = zn if n ≥ 0,
rme−imθ = zm if n = −m < 0,
so that if f(einθ) ∼∑∞
n=−∞ aneinθ, then
g(reiθ) =∞∑
n=−∞
anr|n|einθ
is the ‘natural’ extension to D.
2.4 Definition
The Poisson kernel P (z, w), defined for 0 ≤ |w| < |z| ≤ 1, is the function
P (z, w) =|z|2 − |w|2
|z − w|2.
When |z| = 1, this equals 1−|w|2|1−wz|2 , since |z −w| = |zz −wz| = |1−wz|. This
equals
1− ww(1− wz)(1− wz)
=1
1− wz+
1
1− wz− 1
=∞∑n=0
wnzn +∞∑n=1
wnzn.
9
Thus
P (eit, reiθ) =∞∑
n=−∞
r|n|ein(t−θ) = Pr(t− θ),
say.Clearly the Poisson kernel is a positive continuous function of z and w for0 ≤ |w| < |z| ≤ 1.Note that
∞∑n=−∞
r|n|ein(t−θ) =∞∑
n=−∞
r|n|ein(θ−t),
so Pr(t− θ) = Pr(θ − t).
2.5 Theorem
If f ∈ L1(T), then the function F defined on D by
F (reiθ) =1
2π
∫ 2π
0
Pr(θ − t)f(eit) dt,
is a harmonic function, the harmonic extension of f .
Proof: If f is real, then since for z = eit and w = reiθ we have
Pr(θ − t) =1
1− wz+
1
1− zw− 1
= 2 Re
(1
1− zw− 1
2
)= Re
(1 + zw
1− zw
),
we see that
F (w) = F (reiθ) =1
2πRe
∫ 2π
0
(1 + e−itw
1− e−itw
)f(t) dt,
which is the real part of an anlytic function of w, hence harmonic.
Now, a general f is g + ih with f, g real, and F = G + iH, where G,H arethe corresponding harmonic functions. Hence F is harmonic.
10
2.6 Proposition
If f(eint) = eint and 0 ≤ r < 1, then its harmonic extension to D satisfies
F (reiθ) = r|n|einθ,
i.e., F (w) = wn for n ≥ 0 and w−n for n ≤ 0.
Proof: We have
1
2π
∫ 2π
0
∞∑k=−∞
r|k|eik(θ−t)eint dt = r|n|einθ,
since we can integrate term-by-term as we have uniform convergence forr < 1, and
1
2π
∫ 2π
0
r|k|eik(θ−t)eint dt =
0 if n 6= k
r|n|einθ if n = k.
2.7 Corollary
If f ∈ L1(T) and
cn =1
2π
∫ 2π
0
f(t)e−int dt,
for n ∈ Z, are its Fourier coefficients, then the harmonic extension of f isgiven by
F (w) =∞∑n=0
cnwn +
∞∑n=1
c−nwn,
converging uniformly on |w| ≤ r for 0 ≤ r < 1.
Proof: Now
1
2π
∫ 2π
0
∞∑k=−∞
r|k|eik(θ−t)f(t) dt =1
2π
∞∑k=−∞
r|k|eikθ∫ 2π
0
f(t)e−ikt dt,
as required, since the partial sums converge in L1 and we can integrate term-by-term.
Note that if f ∈ L1, then its Fourier series∑∞
n=−∞ cneint need not converge
anywhere to f(t), so letting |w| → 1 is tricky.
11
2.8 Definition
The Hardy space Hp for 1 ≤ p ≤ ∞ is the subspace of Lp(T) consisting ofall functions f such that
1
2π
∫ 2π
0
f(t)e−int dt = 0
for all n < 0 (i.e., 〈f(z), zn〉 = 0). Such functions have an analytic extensionto D and can be represented by
F (z) =∞∑n=0
cnzn,
with
cn =1
2π
∫ 2π
0
f(t)e−int dt (n ≥ 0).
Also
F (reiθ) =1
2π
∫ 2π
0
Pr(θ − t)f(eit) dt.
Note that Hp equals⋂∞n=1 ϕ
−1n 0, where
ϕn(f) =1
2π
∫ 2π
0
f(t)eint dt.
Since|ϕn(f)| ≤ ‖f‖p‖eint‖q = ‖f‖p,
where1
p+
1
q= 1 (Holder’s inequality), we see that each ϕn is continuous and
so ϕ−1n 0 is closed. Hence Hp is a closed subspace of Lp.
Normally we don’t distinguish between f and F .We normally think of Hp as the class of analytic functions in the disc withLp boundary values (i.e., they are Poisson integrals of Lp functions).
2.9 Corollary
Hp is a Banach space.
Proof: This is immediate since it is a closed subspace of Lp(T), which iscomplete.
12
We have defined H∞ to be the space of analytic functions corresponding toL∞ boundary values – we now want to show that in fact we can define it tobe just the space of bounded analytic functions in D.
2.10 Theorem
If f ∈ L∞(T) with
1
2π
∫ 2π
0
f(t)e−int dt = 0 for n < 0, (1)
then F (z) as given by Theorem 2.5 satisfies ‖F‖∞ = ‖f‖∞. Conversely,any bounded analytic function F in the disc is the harmonic extension of afunction f satisfying (1).
Proof: We have
F (z) =1
2π
∫ 2π
0
Pr(θ − t)f(t) dt,
with z = reiθ, so
|F (z)| ≤(
1
2π
∫ 2π
0
Pr(θ − t) dt)‖f‖∞,
and1
2π
∫ 2π
0
Pr(θ − t) dt =1
2π
∫ 2π
0
∞∑n=−∞
r|n|ein(t−θ) dt = 1, (2)
as only n = 0 gives a non-zero value.
Conversely, if F (z) =∑∞
n=0 anzn is bounded in the disc, then, writing
Fr(eiθ) = F (reiθ) we have that
Fr(eiθ) =
∞∑n=0
anrneinθ.
Also,∞∑n=0
|anrn|2 =1
2π
∫ 2π
0
|Fr(eiθ)|2 dθ ≤ ‖F‖2∞.
13
Letting r → 1 we have∑∞
n=0 |an|2 ≤ ‖F‖2∞, so that the function
f(eiθ) =∞∑n=0
aneinθ
exists in L2(T) and‖Fr − f‖2 → 0
as r → 1. Now Fr → f in measure, and hence a subsequence (Frk) convergespointwise almost everywhere. Then
|f(eiθ)| ≤ lim sup |Frk(eiθ)| a.e.,
i.e., ‖f‖∞ ≤ ‖F‖∞.
Since F (z) =∑∞
n=0 anzn and f has Fourier coefficients (an) we see that F is
the harmonic extension of f and ‖F‖∞ = ‖f‖∞.
Note that ‖Fr‖∞ ‖f‖∞ by the maximum modulus theorem. By workingharder one can prove Fatou’s theorem that Fr → f a.e. as r → 1 (i.e., notjust a subsequence).
14
LECTURE 3
So we now have H2 as a space of power series
f(z) =∞∑n=0
anzn
with
‖f‖2 =∞∑n=0
|an|2 <∞,
and H∞ as all power series with f bounded in D: alternatively, as analyticfunctions that are the Poisson extensions of functions in L2(T) (respectively,L∞(T)), their “boundary values”, whose only non-zero Fourier coefficientsare the (an).
The other important case is p = 1, but we proceed indirectly.
2.11 Proposition
If f ∈ H1, then
sup0<r<1
1
2π
∫ 2π
0
|f(reiθ)| dθ ≤ ‖f‖H1 .
Proof: Recall that
f(reiθ) =1
2π
∫ 2π
0
Pr(θ − t)f(eit) dt,
so
1
2π
∫ 2π
0
|f(reiθ)| dθ =1
2π
∫ 2π
0
∣∣∣∣ 1
2π
∫ 2π
0
Pr(θ − t)f(eit) dt
∣∣∣∣ dθ≤ 1
2π
∫ 2π
0
|f(eit)| 1
2π
∫ 2π
0
Pr(θ − t)dθ dt
= ‖f‖H1 ,
using (2).
We later prove that in fact we have equality above.
15
2.12 Definition
A function of the form
B(z) = eiϕn∏j=1
z − zj1− zjz
,
where ϕ ∈ R and |zj| < 1 for j = 1, . . . , n, is called a finite Blaschke product.
2.13 Proposition
Let B be a finite Blaschke product as above. Then(i) B is analytic in D and continuous in D.(ii) B has zeroes at z1, . . . , zn only, and poles at 1/z1, . . . , 1/zn only.(iii) |B(eiθ)| = 1 for θ ∈ R.
Proof: This is elementary. For the last part, note that∣∣∣∣ eiθ − zj1− zjeiθ
∣∣∣∣ =
∣∣∣∣1− e−iθzj1− eiθzj
∣∣∣∣ = 1
since it is of the form |w/w|.
2.14 Definition
An H∞ function that has unit modulus almost everywhere on T is called aninner function.
Examples are finite Blaschke products and the ‘singular inner function’
exp
(z − 1
z + 1
).
If we have an Hp function, we can use Blaschke products to factor out itszeroes, i.e., we can write f(z) = B(z)g(z), where B is an infinite Blaschkeproduct and g(z) 6= 0 for z ∈ D. We now show this.
2.15 Lemma
16
If f ∈ Hp, f 6≡ 0, then the zeroes (zn) of f in D are at most countable innumber and satisfy ∑
n
(1− |zn|) <∞.
Proof: It is sufficient to prove this for H1 since Hp ⊂ H1 for all p ≥ 1.
Without loss of generality f(0) 6= 0, as we can divide out a power of z, whichdoesn’t change the H1 norm.
Take r < 1 and let z1, . . . , zm be the zeroes in |z| < r. Since they can’taccumulate in D there are only finitely many and we will also assume thatthere are no zeroes on |z| = r.
Write fr(z) = f(rz); this has zeroes at z1/r, . . . , zm/r in |z| ≤ 1.
So
f(rz) =m∏k=1
z − zk/r1− zkz/r
g(z),
with g analytic and nonzero in D. Now log g is analytic, so by Cauchy’sformula
log g(0) =1
2πi
∮|z|=1
log g(z)dz
z
=1
2π
∫ 2π
0
log g(eiθ) dθ.
Take real parts. Then
log |g(0)| =1
2π
∫ 2π
0
log |g(eiθ)| dθ
=1
2π
∫ 2π
0
log |f(reiθ)| dθ.
Now
log |f(0)| =m∑k=1
log |zk/r|+ log |g(0)|,
17
so
log |f(0)|+m∑k=1
log |r/zk| =1
2π
∫ 2π
0
log |f(reiθ)| dθ
≤ log1
2π
∫ 2π
0
|f(reiθ)| dθ
≤ log ‖f‖H1 ,
by Proposition (2.11). Here we have used Jensen’s inequality in the formthat ∫
Ω
log f ≤ log
∫Ω
f,
for a concave function f such as log.Now let r → 1. We see that
log |f(0)|+∞∑k=1
log |1/zk| ≤ ‖f‖H1 <∞,
Now log(1 +x) ∼ x for x small, so log |1/zk| ∼ 1−|zk|, and from this we candeduce that
∞∑k=1
(1− |zk|) <∞.
2.16 Theorem
Let f ∈ Hp. Then the infinite (or finite) Blaschke product
B(z) = zm∏zn 6=0
zn|zn|
zn − z1− znz
,
where (zn) are the zeroes of f , with m of them at 0, converges uniformlyon compact subsets of D to define an H∞ function whose only zeroes areat the (zn), with the correct multiplicities. Moreover, |B(z)| ≤ 1 in D and|B(eiθ)| = 1 a.e. for eiθ ∈ T.
Proof: It’s sufficient to consider f(z)/zm and ignore zeroes at 0. Write
bn(z) =zn|zn|
zn − z1− znz
,
18
noting that bn(0) = |zn| > 0. Also
|bn(eiθ)| = 1,
as in Proposition 2.13.
To get∏bn(z) converging to an analytic function with zeroes at the (zk) only,
we show that∑|1−bn(z)| converges uniformly on discs |z| ≤ R for R < 1.
Again this follows from the fact that log(1 + x) ∼ x when x is small.
Now
|1− bn(z)| =
∣∣∣∣1 +zn|zn|
z − zn1− znz
∣∣∣∣=
∣∣∣∣ |zn| − |zn|znz + znz − znznzn(1− znz)
∣∣∣∣=
(1− |zn|)(znz + |zn|)|zn||1− znz|
≤ (1− |zn|)(
1 + |z|1− |z|
),
giving convergence.So B(z) ∈ H∞ and |B(eiθ)| ≤ 1 a.e.Now write Bn =
∏nk=1 bk. Then B/Bn is another Blaschke product, and so∣∣∣∣ B(0)
Bn(0)
∣∣∣∣ ≤ ∫ 2π
0
∣∣∣∣ B(eiθ)
Bn(eiθ)
∣∣∣∣ dθ=
1
2π
∫ 2π
0
|B(eiθ)| dθ.
Let n→∞ and we see that
1
2π
∫ 2π
0
|B(eiθ)| dθ ≥ 1,
and we conclude that |B(eiθ)| = 1 a.e.
19
Now for f ∈ H∞ we have
‖f‖∞ = limr→1
sup|z|=r|f(z)|,
and for f ∈ H2 we have
‖f‖2 = limr→1
(1
2π
∫ 2π
0
|f(reiθ)|2 dθ)1/2
,
i.e.,
limr→1
(∞∑n=0
|an|2r2n
)1/2
.
We shall now get the same result for H1, i.e., ‖f‖1 = limr→1 ‖fr‖1, wherefr(z) = f(rz). For this we need the Poisson kernel again. Recall that
P (z, w) =|z|2 − |w|2
|z − w|2
for 0 ≤ |w| < |z| ≤ 1.
2.17 Proposition
Let P (z, w) be the Poisson kernel. Then:(i) If f is analytic in D, then we have
f(w) =1
2π
∫ 2π
0
P (reiθ, w)f(reiθ) dθ
for |w| < r < 1;
(ii)1
2π
∫ 2π
0
P (reiθ, w) dθ = 1 for |w| < r;
(iii)1
2π
∫ 2π
0
P (z, seit) dt = 1 for |z| > s.
Proof: (i)∫ 2π
0
P (reiθ, w)f(reiθ) dθ =1
2πi
∮|z|=r
r2 − wwz(z − w)(r2/z − w)
f(z) dz,
20
where z = eiθ and dz = izdθ. This is
1
2πi
∮|z|=r
(r2 − ww)f(z) dz
(z − w)(r2 − zw)= f(w),
since the only pole is at w with residue f(w).
(ii) Put f(z) = 1 in (i).
(iii)
1
2π
∫ 2π
0
P (z, seit)dt =1
2πi
∮|w|=s
(zz − s2) dw
w(z − w)(z − s2/w)
=1
2πi
∮|w|=s
(zz − s2) dw
(z − w)(zw − s2)
=1
2πi
∮|w|=s
(zz − s2) dw
(zz − wz)(w − s2/z)= 1
the only pole inside the contour being at w = s2/z.
2.18 Corollary
If 0 < s < r < 1 and f is analytic in D, then
1
2π
∫ 2π
0
|f(seit)| dt ≤ 1
2π
∫ 2π
0
|f(reiθ)| dθ.
Proof: From Proposition 2.17 we have
1
2π
∫ 2π
0
|f(seit)| dt =1
2π
∫ 2π
t=0
∣∣∣∣ 1
2π
∫ 2π
θ=0
P (reiθ, seit)f(reiθ) dθ
∣∣∣∣ dt≤ 1
2π
∫ 2π
t=0
1
2π
∫ 2π
θ=0
P (reiθ, seit)|f(reiθ)| dθ dt
=1
2π
∫ 2π
0
|f(reiθ)| dθ.
21
2.19 Corollary
For f ∈ H1, the limit
limr→1
1
2π
∫ 2π
0
|f(reiθ)| dθ
exists and equals ‖f‖H1 . Moreover ‖fr − f‖1 → 0, where fr(eiθ) = f(reiθ).
Proof: By Corollary 2.18 the integral increases with r. Also, for anyh ∈ H1 and r < 1 we have
1
2π
∫ 2π
0
|h(reiθ)| dθ ≤ 1
2π
∫ 2π
0
∣∣∣∣ 1
2π
∫ 2π
0
Pr(θ − t)h(eit)
∣∣∣∣ dt≤ 1
2π
∫ 2π
0
|h(eit)| dt.
We adopt the notation hr(z) = h(rz), interpreted for h ∈ L1(T) using thePoisson extension to D.Now, given f ∈ H1, and ε > 0, we may choose a trigonometric polynomial gwith
g(eiθ) =N∑
n=−N
aneinθ,
such that ‖f − g‖1 < ε/3.Now
‖fr − f‖1 ≤ ‖fr − gr‖1 + ‖gr − g‖1 + ‖g − f‖1
≤ ε/3 + ‖gr − g‖1 + ε/3 < ε
for r close to 1, since
(gr − g)(eiθ) =N∑
n=−N
an(r|n| − 1)einθ,
which tends to zero uniformly as r → 1.Thus ‖fr − f‖1 → 0 and ‖fr‖1 ‖f‖1, as required.
22
LECTURE 4
For f ∈ H1 or H2, we have ‖fr − f‖p → 0 (as we have seen), while forf ∈ H∞ we don’t always have ‖f − fr‖∞ → 0, since the fr are continuouson T and f needn’t be. However, ‖fr‖∞ → ‖f‖∞.
2.20 Theorem (Riesz)
Let f ∈ Hp with f 6≡ 0, and let B be the Blaschke product formed using thezeroes (zn) of f . Then f = gB, where ‖f‖p = ‖g‖p.
We prove this in the course just for p = 1, 2 and ∞.
Proof: Let g(z) = f(z)/B(z) and gn(z) = f(z)/Bn(z), where Bn is theBlaschke product corresponding to the first n zeroes of f .If r < 1, then
1
2π
∫ 2π
0
|gn(reiθ)|p dθ ≤ limR→1
1
2π
∫ 2π
0
|f(Reiθ)|p
|Bn(Reiθ)|pdθ
= limR→1
1
2π
∫ 2π
0
|f(Reiθ)|p dθ,
since |Bn(Reiθ)| → 1 uniformly as R→ 1. So
1
2π
∫ 2π
0
|gn(reiθ)|p dθ ≤ ‖f‖pHp .
But |gn| |g| as n → ∞, so by the Monotone Convergence Theorem wehave
1
2π
∫ 2π
0
|g(reiθ)|p dθ ≤ ‖f‖pHp .
But |g(z)| ≥ |f(z)| for z ∈ D, so we have equality.There is a similar (easier) argument for p =∞.
2.21 Theorem (Riesz factorization theorem)
23
A function f lies in H1 if and only if there exist g, h ∈ H2 such that f = gh(pointwise product). We can choose g and h such that ‖f‖1 = ‖g‖2‖h‖2.
Proof: Note first that if g, h ∈ H2, then gh ∈ H1 and
‖gh‖H1 ≤ ‖g‖H2‖h‖H2 ,
by Cauchy–Schwarz.
Conversely, given f ∈ H1, write f(z) = f1(z)B(z), where B is as in Theo-rem 2.20, ‖f‖ = ‖f1‖ and f1 is never 0 in D.
Since f1 is never 0 we can define an analytic square root, say f1(z) = g(z)2
in D. For we write f1(z) = Reit, where t is chosen continuously, and theng(z) =
√Reit/2.
Nowf(z) = g(z)(g(z)B(z)),
and‖f‖H1 = ‖g‖2
H2 = ‖g‖H2‖gB‖H2 ,
since ‖gB‖H2 = ‖g‖H2 .
2.22 Definition
The disc algebra A(D) is the space
f ∈ C(T) :1
2π
∫ 2π
0
f(eit)e−int dt = 0 for all n < 0.
It is a closed subspace of C(T), hence a Banach space. All functions in A(D)lie in H2 so have analytic extensions to D given by
f(eit) =∞∑n=0
cneint
(convergence in L2), and
f(reit) =∞∑n=0
cnrneint
(convergence locally uniformly).
24
2.23 Theorem
The closure of the space of polynomials in Lp(T) is the space Hp for 1 ≤ p <∞ and A(D) for p =∞.
Proof: Since the polynomials lie in A(D) ⊂ H2 ⊂ H1, we see that ineach case the closure is contained in the relevant Banach space.
Given f ∈ H1, H2 or A(D) we write fr(z) = f(rz) as usual and a calculationusing the Poisson kernel shows that ‖fr − f‖ → 0 in the appropriate normas r → 1.
But
fr(eiθ) =
∞∑n=0
cnrneinθ,
where the (cn) are bounded: this can in turn be approximated arbitrarilyclosely in norm by a finite sum
N∑n=0
cnrneinθ
since ∥∥∥∥∥∞∑
n=N+1
cnrnzn
∥∥∥∥∥ ≤ ∑n=N+1
|cn|rn → 0
as N →∞.
2.24 Example
Here is a function in H∞ that does not lie in the subalgebra A(D).Let
f(z) = exp
(z − 1
z + 1
).
This is bounded in modulus by 1 in the unit disc, since the Mobius mapping
M : z 7→ z − 1
z + 1
25
takes D into the left half-plane
C− = x+ iy ∈ C : x < 0,
and then |f(z)| = ex < 1.However, f is discontinuous on the circle, as M maps the circle to the imag-inary axis, and f(eiθ) = es, where s→ ±i∞ as θ → π.
26
3 Operators on H2 and L2(T)We look at three important classes of operators: multiplication (Laurent)operators, Toeplitz operators, and Hankel operators.
Notation: P : L2(T)→ H2 denotes the orthogonal projection,
∞∑n=−∞
aneint 7→
∞∑n=0
aneint.
Clearly ‖Pf‖2 ≤ ‖f‖2.
3.1 Proposition
Let ϕ ∈ L∞(T). Then the transformation Mϕ : L2(T)→ L2(T) defined by
Mϕf(eit) = ϕ(eit)f(eit)
is a bounded operator of norm ‖Mϕ‖ = ‖ϕ‖∞.Moreover,
sup‖Mϕf‖2 : f ∈ H2, ‖f‖2 = 1 = ‖ϕ‖∞.
Proof: Clearly ‖Mϕ‖ ≤ ‖ϕ‖∞, since
‖Mϕf‖22 =
1
2π
∫ 2π
0
|ϕ(eit)f(eit)|2 dt
≤ ‖ϕ‖2∞
1
2π
∫ 2π
0
|f(eit)|2 dt = ‖ϕ‖2∞‖f‖2
2.
Conversely, given ε > 0 there is a set Aε ⊂ T of positive measure with
|ϕ(eit)| > ‖ϕ‖∞ − ε on Aε.
We consider the function
χAε =
1 on Aε,
0 on T \ Aε.
Now χAε ∈ L2(T), with
‖χAε‖2 =1
2π
∫ 2π
0
χAε(eit)2 dt = µ(Aε),
27
where µ is the normalized Lebesgue measure on the circle. Also
‖MϕχAε‖2 =1
2π
∫ 2π
0
|ϕ(eit)|2χAε(eit)2 dt
> (‖ϕ‖∞ − ε)2µ(Aε).
Hence‖MϕχAε‖/‖χAε‖ > ‖ϕ‖∞ − ε,
and so ‖Mϕ‖ ≥ ‖ϕ‖∞ − ε. Since this is true for all ε > 0 we conclude that
‖Mϕ‖ = ‖ϕ‖∞.
The last part is harder, but will be needed. Now χAε will not lie in H2, butif its Fourier series is
χAε(eit) =
∞∑n=−∞
cneint,
then the functions fm defined by
fm(eit) = eimt∞∑
n=−∞
cneint =
∞∑n=−∞
cnei(n+m)t
also satisfy‖fm‖2 = ‖χAε‖2 = µ(Aε)
1/2
and‖Mϕfm‖ = ‖eimtMϕχAε‖ > (‖ϕ‖∞ − ε)µ(Aε)
1/2,
since multiplying by an inner function such as eimt doesn’t change the L2
norm.Now
‖Pfm − fm‖ =
∥∥∥∥∥∞∑
n=−m
cnei(m+n)t −
∞∑n=−∞
cnei(m+n)t
∥∥∥∥∥=
(−m−1∑n=−∞
|cn|2)1/2
,
which tends to 0 as m→∞.
28
Hence‖Pfm‖ → µ(Aε)
1/2
and
‖MϕPfm‖ = ‖Mϕfm +Mϕ(Pfm − fm)‖≥ ‖Mϕfm‖ − ‖Mϕ(Pfm − fm)‖> (‖ϕ‖∞ − ε)µ(Aε)
1/2
for m sufficiently large. Thus
sup‖Mϕf‖2 : f ∈ H2, ‖f‖2 = 1 = ‖ϕ‖∞.
3.2 Corollary
If ϕ ∈ H∞ then Mϕ : H2 → H2 satisfies ‖Mϕ‖ = ‖ϕ‖∞.
Proof: This follows from the previous result, with the exception of show-ing that ϕf ∈ H2 if ϕ ∈ H∞ and f ∈ H2.We know it is in L2 and the analyticity in D is easily seen to follow bymultiplying the Taylor series together.
Matrix notation.
Write (en)∞n=−∞ for the orthonormal basis of H2 given by en(z) = zn forz ∈ T. Then (en)∞n=0 is an orthonormal basis of H2.If ϕ ∈ H∞ with ϕ(z) =
∑∞k=0 dkz
k, then calculating Mϕen we get the infinitematrix
d0 0 0 0 . . .d1 d0 0 0 . . .d2 d1 d0 0 . . .d3 d2 d1 d0 . . ....
......
.... . .
,
representing the operator Mϕ acting on H2. It is constant on the maindiagonals, i.e., a Toeplitz matrix. It is also lower triangular.
29
3.3 Definition
If ϕ ∈ L∞(T), then the Toeplitz operator with symbol ϕ, Tϕ, is the operatorTϕ : H2 → H2 defined by Tϕf = P (Mϕf), for f ∈ H2.
Clearly, ‖Tϕ‖ ≤ ‖ϕ‖∞, and we have equality if ϕ ∈ H∞. We’ll see that infact we have equality always.
Matrix of Tϕ.
Let ϕ(eit) =∑∞
k=−∞ dkeikt. Then
Tϕen = P
∞∑k=−∞
dkeikteint =
∞∑p=0
dp−neipt,
with p = n+ k.We then have the (full) Toeplitz matrix.
d0 d−1 d−2 d−3 . . .d1 d0 d−1 d−2 . . .d2 d1 d0 d−1 . . .d3 d2 d1 d0 . . ....
......
.... . .
,
Special cases: if ϕ ≡ 1, then Tϕ is the identity. If ϕ(z) = z, then Tϕ is theshift, zn 7→ zn+1.
30
LECTURE 5
3.4 Theorem
For ϕ ∈ L∞(T), ‖Tϕ‖ = ‖ϕ‖∞, that is,
supf∈H2,‖f‖=1
‖Tϕf‖ = ‖ϕ‖∞.
Proof: Clearly,
‖Tϕ‖ = ‖PMϕ‖ ≤ ‖Mϕ‖ = ‖ϕ‖∞,
so we need to show “≥”.Given ε > 0 there is an f ∈ H2 of norm 1 with ‖Mϕf‖ > ‖ϕ‖∞ − ε, byProposition 3.1.Without loss of generality, f is a polynomial, say, p(z) =
∑Nk=0 ckz
k, since if
f(z) =∑∞
k=0 akzk, then truncating it to fN(z) =
∑Nk=0 akz
k we have fN → fand ‖fN‖ → ‖f‖ = 1, so
fN‖fN‖
→ f and TϕfN‖fN‖
→ Tϕf.
So take p = fN/‖fN‖ for suitably large N .Now we claim that as m→∞,
‖Tϕzmp‖ → ‖Mϕzmp‖ = ‖zmϕp‖ > ‖ϕ‖∞ − ε.
It is enough to show that ‖Tϕzmzk −Mϕzmzk‖ → 0 for each k ≥ 0 and use
the triangle inequality to deduce that
‖Tϕzmp−Mϕzmp‖ → 0.
But with z = eit we have
‖Tϕeimteikt −Mϕeimteikt‖ =
∥∥∥∥∥(I − P )∞∑
r=−∞
dreirteimteikt
∥∥∥∥∥=
∥∥∥∥∥−1−m−k∑r=−∞
drei(r+m+k)t
∥∥∥∥∥=
(−1−m−k∑r=−∞
|dr|2)1/2
→ 0,
31
as m→∞, and the result follows.
Convolution operators
Let (dk)∞k=−∞ be a complex sequence, and consider the convolution operator
on `2(Z+) defined by
(Tdu)k =k∑
j=−∞
djuk−j (k = 0, 1, 2, . . .).
Such operators are sometimes called discrete-time linear systems, with u be-ing the input and y = Tdu the output.
They are called causal if dj = 0 for j < 0, in which case yk only depends onthe “past”, i.e., u0, . . . , uk.
If we consider the standard basis for `2(Z+), namely e0 = (1, 0, 0, . . .), e1 =(0, 1, 0, 0, . . .), e2 = (0, 0, 1, 0, 0, . . .), etc., then Td is represented by theToeplitz matrix
d0 d−1 d−2 d−3 . . .d1 d0 d−1 d−2 . . .d2 d1 d0 d−1 . . .d3 d2 d1 d0 . . ....
......
.... . .
,
and so we know that, when Td is bounded, its norm given by
sup‖u‖2≤1
‖Tdu‖2 = ‖G‖∞,
where G is the L∞(T) function with Fourier coefficients (dk).
For Td to be causal, that is the same as saying that G is in H∞, and Tdis equivalent to an analytic Toeplitz operator (or Laurent operator), repre-sented by a lower-triangular matrix.
32
If instead of working with `2(Z+) we work with `∞(Z+) (“bounded-input tobounded-output”), then
|(Tdu)k| ≤k∑
j=−∞
|dj||uk−j| ≤∞∑
j=−∞
|dj|‖u‖∞,
so that, taking the sup over k, we have ‖Td‖ ≤ ‖d‖`1(Z). By choosing suitableu one can show that we get equality.
Thus Td is a bounded operator on `∞(Z+) if and only if G lies in the Wieneralgebra, that is,
G(eit) =∞∑
j=−∞
djeijt, with
∞∑j=−∞
|dj| <∞.
The causal operators correspond to the analytic Wiener algebra (i.e., dj = 0for j < 0), and these are the ones with an analytic extension to D.
We now look at another class of operators in some detail.
3.5 Definition
We write (H2)⊥ for the orthogonal complement of H2 in L2(T), i.e., theclosed subspace spanned by the negative basis vectors en(z) = zn, for z ∈ T,with n < 0.
If ϕ ∈ L∞(T) then the Hankel operator Γϕ : H2 → (H2)⊥ is defined byΓϕf = (I − P )Mϕf , i.e., Mϕ = Tϕ + Γϕ. There are several equivalentdefinitions, but we shall use this one.
3.6 Proposition
We have ‖Γϕ‖ ≤ ‖ϕ‖∞, and if ϕ(eit) =∑∞
n=−∞ dneint, then the matrix of Γϕ
with respect to the bases e0, e1, e2, . . . of H2 and e−1, e−2, . . . of (H2)⊥ isd−1 d−2 d−3 . . .d−2 d−3 d−4 . . .d−3 d−4 d−5 . . .d−4 d−5 d−6 . . .
......
......
,
33
which is constant on ‘minor’ diagonals. Also Γϕ = 0 if and only if ϕ ∈ H∞(whereas Tϕ = 0 if and only if ϕ = 0).
Proof:
‖Γϕf‖2 = ‖(I − P )Mϕf‖2 ≤ ‖Mϕf‖2 ≤ ‖ϕ‖∞‖f‖2.
Also
Γϕek = (I − P )∞∑
n=−∞
dneinteikt
=−1−k∑n=−∞
dnei(n+k)t =
∞∑r=1
d−r−ke−irt,
letting n+ k = −r.Clearly, Γϕ = 0 if and only if dn = 0 for each n < 0, which is the same assaying that ϕ ∈ H∞.Similarly, looking at the matrix, we see that Tϕ = 0 if and only if ϕ = 0 (andwe have already seen that ‖Tϕ‖ = ‖ϕ‖∞).
3.7 Theorem (Nehari)
If Γϕ : H2 → (H2)⊥ is a bounded Hankel operator, then there exists a symbolψ ∈ L∞(T) such that Γϕ = Γψ and ‖Γψ‖ = ‖ψ‖∞. Hence
‖Γϕ‖ = inf‖ϕ+ F‖ : F ∈ H∞ = dist(ϕ,H∞).
Proof: Observe that if ϕ(eit) =∑∞
k=−∞ dkeikt, then
〈Γeint, e−imte−it〉 = d−n−m−1 (n,m ≥ 0)
= 〈Γ(einteimt), e−it〉.
Hence⟨Γ
(N∑n=0
bneint
),
M∑m=0
cme−imte−it
⟩=
⟨Γ
(N∑n=0
bneint
M∑m=0
cmeimt
), e−it
⟩.
34
Thus we can define a linear functional on products of polynomials, by
α(f1f2) = 〈Γ(f1f2), e−it〉 = 〈Γf1, f2e−it〉,
We see that|α(f1f2)| ≤ ‖Γ‖‖f1‖2‖f2‖2.
But we have seen that products of polynomials are dense in H1, and so αhas a unique extension by
α(f1f2) = 〈Γf1, f2e−it〉, (f1, f2 ∈ H2),
with|α(f1f2)| ≤ ‖Γ‖‖f1‖2‖f2‖2.
and hence, by Theorem 2.21 we have defined α as a linear functional on H1
with|α(g)| ≤ ‖Γ‖‖g‖1 (g ∈ H1).
We may now use the Hahn–Banach theorem to extend α, so that we haveα : L1(T)→ C with the same norm.This implies that we have a representation
α(g) =1
2π
∫ 2π
0
g(eit)h(eit) dt
for some h ∈ L∞(T) with
‖h‖∞ = ‖α‖ ≤ ‖Γ‖.
More calculations:
〈Γeikt, e−it〉 = d−k−1 (k ≥ 0)
=1
2π
∫ 2π
0
eikth(eit) dt = 〈h, e−ikt〉.
Hence if
h(eit) =∞∑
l=−∞
h`ei`t,
then we have h−k = d−k−1.
35
We write ψ(eit) = e−ith(eit) and so ψ−k−1 = d−k−1, and
‖ψ‖∞ = ‖h‖∞ ≤ ‖Γ‖,
as claimed.Finally, since one has Γϕ = Γψ if and only if F := ψ−ϕ ∈ H∞, we have that
‖Γϕ‖ ≤ inf‖ϕ+ F‖∞ : F ∈ H∞,
and there exists a ψ such that ψ = ϕ+ F and ‖ψ‖∞ ≤ ‖Γϕ‖ ≤ ‖ψ‖∞.Hence ‖ψ‖∞ = ‖Γ‖ψ, and we have the result.
3.8 Example
Hilbert’s Hankel matrix
Γ =
1 1
213
. . .12
13
14
. . .13
14
15
. . ....
......
. . .
has norm π as an operator on H2 (take as symbol ψ(eiθ) = i(θ − π) for0 ≤ θ < 2π). We thus have Hilbert’s inequality
|〈Γx, y〉| ≤ π‖x‖ ‖y‖,
or ∣∣∣∣∣∞∑n=0
∞∑m=0
xnymn+m+ 1
∣∣∣∣∣ ≤ π
(∞∑n=0
|xn|2)1/2( ∞∑
m=0
|ym|2)1/2
.
36
LECTURE 6
To get a more explicit solution to the Nehari problem we need a technicallemma on H2 functions.
We writelog x = log+ x+ log− x,
where log+ x = max(log x, 0) and log− x = min(0, log x).
3.9 Lemma
If f ∈ H2 and f 6≡ 0, then
1
2π
∫ 2π
0
log− |f(eiθ)| dθ > −∞,
and hence f 6= 0 a.e. on T.
The result can be extended to H1, since every H1 function is the product oftwo H2 functions.
(Of course, arbitrary Lp functions can vanish on any set of positive measurethat we like.)
Proof: Without loss of generality f has no zeroes in D, as we can divideout a Blaschke product, whose values on the circle have modulus 1 almosteverywhere.As in Lemma 2.15 we have the inequality
log |f(0)| ≤ 1
2π
∫ 2π
0
log |f(eiθ)| dθ
Now log+ x ≤ x2 for x > 0, because for y ≥ 0 we have
y ≤ ey2
= 1 + y2 + . . . ,
since 1 + y2 − y = (1 + y/2)2 + 3y2/4 ≥ 0.
37
So1
2π
∫ 2π
0
log+ |f(eiθ)| dθ ≤ 1
2π
∫ 2π
0
|f(eiθ)|2 dθ = ‖f‖22.
Hence
1
2π
∫ 2π
0
log− |f(eiθ)| dθ =1
2π
∫ 2π
0
log |f(eiθ)| dθ − 1
2π
∫ 2π
0
log+ |f(eiθ)| dθ
≥ log |f(0)| − ‖f‖22 > −∞,
giving the result.
3.10 Corollary
If f ∈ A(D), and f 6≡ 0, then, regarding f as a function on D, its zero setZ = z ∈ D : f(z) = 0 satisfies:
(i) Z ∩ D is countable and∑
zn∈Z∩D(1− |zn|) <∞;(ii) Z ∩ T has measure 0;(iii) Z is closed.
Proof: This follows (via Lemmas 2.15 and 3.9) from the fact thatA(D) ⊂ H2 and the observation that f is continuous on D.
Solution of the Nehari problem
3.11 Theorem (Sarason)
If Γ : H2 → (H2)⊥ is a bounded Hankel operator and if f ∈ H2 with f 6≡ 0satisfies ‖Γf‖ = ‖Γ‖ ‖f‖, then there is a unique symbol ψ for Γ = Γψ ofminimal norm, i.e., ‖ψ‖∞ = ‖Γψ‖, and it is given by ψ = Γf
f, i.e.,
ψ(eiθ) =(Γf)(eiθ)
f(eiθ)a.e.
Moreover |ψ(eiθ)| is constant almost everywhere.
38
Proof: We have that Γψf = (I − P )Mψf , so that
‖Γψ‖‖f‖2 = ‖Γψf‖2 ≤ ‖ψf‖2 ≤ ‖ψ‖∞‖f‖2.
But if we have equality throughout then Γf = ψf and so
ψ =Γf
f,
since f 6= 0 a.e. on T. Moreover, since ‖ψf‖2 = ‖ψ‖∞‖f‖2 we must have|ψ(eiθ)| = ‖ψ‖∞ a.e.
3.12 Example
Take a rank-1 Hankel operator with matrix1 α α2 . . .α α2 α3 . . .α2 α3 α4 . . ....
......
. . .
,
where |α| < 1. One can check that
‖Γ‖ = 1/(1− |α|2).
Let f(z) = 1/(1− αz), so that
(Γf)(z) =1
1− |α|2
(1
z+α
z2+α2
z3+ . . .
),
and thus
ψ(z) =(Γf)(z)
f(z)=
1
1− |α|21− αzz − α
.
Note that (z−α)/(1−αz) is a Blaschke product, so ψ has modulus 1/(1−|α|2)a.e.
Best approximation problems (Nehari problems)
3.13 Corollary
39
If ϕ ∈ L∞(T) is such that Γϕ attains its norm (i.e., ‖Γϕf‖ = ‖Γϕ‖ ‖f‖ forsome f ∈ H2 with f 6≡ 0), then ϕ has a unique best approximant h in H∞,i.e., a closest element, so that
‖ϕ− h‖∞ = dist(ϕ,H∞)),
given by
h = ϕ− Γf
f, and ‖ϕ− h‖∞ = ‖Γϕ‖.
Moreover |(ϕ− h)(eiθ)| = ‖Γϕ‖ a.e.
Proof: if g ∈ H∞ then Γϕ−g = Γϕ, and hence
‖ϕ− g‖∞ ≥ ‖Γϕ−g‖ = ‖Γϕ‖.
But there is a ψ ∈ L∞(T), namely ψ = (Γf)/f , such that Γϕ = Γψ and
‖ψ‖∞ = ‖Γψ‖ = ‖Γϕ‖.
Hence h = ϕ− ψ ∈ H∞, because Γh = Γϕ − Γψ = 0. Also,
‖ϕ− h‖∞ = ‖ψ‖∞ = ‖Γϕ‖,
so h is a best approximant. Since ψ is unique, so is h. Since ψ has constantmodulus a.e., so has ϕ− h.
Note. If ϕ ∈ C(T), then Γϕ is compact and hence attains its norm (seelater); so there is always a best approximant in H∞, but it does not alwayslie in A(D).
3.14 Example
Take
ϕ(z) =3
z+
2
z2,
which lies in C(T). Note that the closest element in H2 (in the L2(T) norm)is just the orthogonal projection onto H2, which is 0. To get the closest
40
element in H∞ (in the L∞(T) norm), form the rank-two Hankel matrix3 2 0 . . .2 0 0 . . .0 0 0 . . ....
......
. . .
.
This is symmetric, eigenvalues 4 and −1, and orthogonal eigenvectors
(2, 1, 0, 0, . . .)T and (1,−2, 0, 0, . . .)T .
Hence a maximizing vector is f(z) = 2 + z, mapped into
(Γf)(z) =8
z+
4
z2.
Then h = ϕ− Γff
, and h(z) =3
2 + z, and h ∈ H∞.
Indeed ‖ϕ‖∞ = 5, so 0 is not the best uniform approximant, whereas
(ϕ− h)(z) =4
z2
1 + 2z
z + 2,
which has constant modulus 4.
Two classical problems
1. The Caratheodory–Fejer extension problem.
Given a polynomial a0+a1z+a2z2+. . .+anz
n, choose coefficients an+1, an+2, . . .to minimize ∥∥∥∥∥
∞∑k=0
akzk
∥∥∥∥∥∞
;
i.e., minimize ∥∥a0 + a1z + a2z2 + . . .+ anz
n + zn+1h(z)∥∥∞
over all h analytic and bounded in D.
41
2. The Nevanlinna–Pick interpolation problem.
Given z1, . . . , zn ∈ D and w1, . . . , wn ∈ C, find a function g analytic in Dsuch that g(zk) = wk for k = 1, . . . , n, and ‖g‖∞ is minimized.
3.15 Theorem
The Caratheodory–Fejer problem can be solved using the solution to the Ne-hari problem.
Proof: Letg(z) = a0 + a1z + . . .+ anz
n.
We are trying to minimize‖g − zn+1h‖∞
over h ∈ H∞. This is equivalent to
minh∈H∞
∥∥∥ g
zn+1− h∥∥∥∞,
We have seen that the solution is
h = ϕ− Γϕf
f,
where ϕ(z) =g(z)
zn+1.
It is easily seen that Γϕ has finite rank, so a maximizing vector f exists.
Example. Let g(z) = 2 + 3z. Then ‖g‖∞ = 5.The recipe tells us to look at
ϕ(z) =2 + 3z
z2=
2
z2+
3
z,
which is the ϕ we looked at earlier. Then
h(z) =3
2 + z.
42
So the minimal-norm extension is
f(z) = 2 + 3z − 3z2
2 + z=
4 + 8z
2 + z= 4
z + 1/2
1 + z/2,
i.e., a constant times an inner function, with ‖f‖∞ = 4.
3.16 Theorem
The Nevanlinna–Pick problem can be solved using the solution to the Nehariproblem.
Proof: There certainly do exist functions f ∈ H∞ with f(zk) = wk fork = 1, . . . , n; for example, a suitable polynomial.
If g is another such function then f − g vanishes at z1, . . . , zn, we we canwrite
f(z)− g(z) = B(z)h(z),
where
B(z) =n∏k=1
z − zk1− zkz
is the appropriate Blaschke product, and h ∈ H∞ with ‖h‖∞ = ‖f − g‖∞,as in Proposition 2.13.
Moreover, any h ∈ H∞ gives an interpolant g = f −Bh.
Our problem therefore is to find
inf‖f −Bh‖∞ : h ∈ H∞.
This isinf‖B−1f − h‖∞ : h ∈ H∞,
since B is inner. And this is dist(B−1f,H∞), which is the Nehari problemagain.
43
Example. Find a minimal-norm interpolant g such that g(0) = 1 andg(1
2) = 0.
We start with an interpolant f(z) = 1− 2z of H∞ norm 3.Now
(f −Bh)(z) = (1− 2z)−z(z − 1
2)
1− 12zh(z),
so we minimize
‖B−1f − h‖ =
∥∥∥∥z − 2
z− h∥∥∥∥∞.
Writing h = 1 + h, we have to find
infh
∥∥∥∥z − 2
z− (1 + h)
∥∥∥∥∞
= infh
∥∥∥∥−2
z− h∥∥∥∥∞.
Now the best H∞ approximant to ϕ(z) = −2/z turns out to be the 0 function(just as in H2 norm), as the corresponding Hankel operator is
Γ =
−2 0 . . .0 0 . . .. . . . . . . . .
which attains its norm at f = 1, and then
ϕ(z)− (Γf(z))/f(z) = −2/z − (−2/z) = 0.
Now we work out g = f −Bh for this choice (h = 1), and and then the bestinterpolant is
g(z) = −2z − 1
2
1− 12z,
of norm 2. Since ‖f‖∞ = 3, we did better.
Finite-rank Toeplitz and Hankel operators.
3.17 Proposition
The only finite-rank Toeplitz operator is T0 = 0.
44
We shall later use a similar proof to show that there are no non-zero compactToeplitz operators.
Proof: Let (en)n≥0 be the usual orthonormal basis of H2.Any finite rank operator T on a Banach space has the form
Tx =r∑
k=1
fk(x)xk,
where x1, . . . , xr are a basis for the range of T and f1, . . . , fr ∈ X∗.In our case fk(x) = 〈x, yk〉 for some yk ∈ H2. So
‖Ten‖ =
∥∥∥∥∥r∑
k=1
〈en, yk〉xk
∥∥∥∥∥≤
r∑k=1
|〈en, yk〉| ‖xk‖ → 0,
noting that |〈en, yk〉| is the absolute value of the nth Taylor coefficient of ykand these tend to zero as n → ∞, since they are square-summable. Recallthat the matrix of T is
d0 d−1 d−2 d−3 . . .d1 d0 d−1 d−2 . . .d2 d1 d0 d−1 . . .d3 d2 d1 d0 . . ....
......
.... . .
.
We see that ‖Ten‖ ≥ |dm| when n + m ≥ 0, since dm appears in the nthcolumn. Hence for T finite-rank, on letting n→∞, we have that dm = 0 foreach m.
3.18 Theorem (Kronecker)
The Hankel operator
Γ =
d−1 d−2 d−3 . . .d−2 d−3 d−4 . . .d−3 d−4 d−5 . . .d−4 d−5 d−6 . . .
......
......
45
has finite rank if and only if f(z) :=∑−1
k=−∞ dkzk is a rational function of z,
and its rank is the number of poles of f (which must lie in D).
Proof: If rank(Γ) = r, then the first r+1 columns are linearly dependent,so that there are scalars λ1, . . . , λm such that
r+1∑i=1
λid−i−m = 0 (m = 0, 1, 2, . . .)
This implies that
(λ1 + λ2z + · · ·+ λr+1zr)
(d1
z+d2
z2+ . . .
)is a polynomial of degree at most r−1 in z, since the coefficient of a negativepower z−m−1 is
λ1d−m−1 + λ2d−m−2 + · · ·+ λr+1d−m−r−1 = 0.
Thus f , as above, is rational of degree at most r.
Conversely, if
P (z)
(d1
z+d2
z2+ . . .
)= Q(z)
with P , Q polynomial and degP ≤ r, then working backwards we see thatthe rank of Γ is at most r.
The poles of f are the roots of
λ1 + λ2z + · · ·+ λr+1zr = 0.
They must lie in D since otherwise we could split f into partial fractionsf = g + h, where deg g < deg f and h ∈ H∞, so Γf = Γg and rank(Γf ) =deg g < deg f , a contradiction.
3.19 Example
46
Rank 1 Hankel matrices giving bounded operators are of the form
c
1 α α2 . . .α α2 α3 . . .α2 α3 α4 . . .. . . . . . . . . . . .
,
where c ∈ C and α ∈ D. The corresponding symbol isc
z − α.
Now we look at compactness of Toeplitz and Hankel operators.
3.20 Theorem
There are no compact Toeplitz operators except 0.
Proof: Recall that if (en) is a sequence tending weakly to zero, and T iscompact, then (Ten) tends in norm to zero.
For if ‖Ten‖ 6→ 0, then for some ε > 0 and a suitable subsequence we have‖Tenk
‖ ≥ ε and ‖Tenk− w‖ → 0, so w 6= 0. But
〈Tenk, w〉 = 〈enk
, T ∗w〉 → 0,
implying that 〈w,w〉 = 0, which is a contradiction.
We apply this with en(z) = zn, for n ≥ 0.
As in Proposition 3.17, recall the matrix of T isd0 d−1 d−2 d−3 . . .d1 d0 d−1 d−2 . . .d2 d1 d0 d−1 . . .d3 d2 d1 d0 . . ....
......
.... . .
.
We see that ‖Ten‖ ≥ |dm| when n + m ≥ 0, since dm appears in the nthcolumn. Hence for T compact, on letting n → ∞, we have that dm = 0 foreach m.
47
LECTURE 7
3.21 Theorem
The Hankel operator Γϕ is compact if and only if ϕ ∈ H∞ + C(T), which isa closed subalgebra of L∞(T).
Proof: We prove just the easy part of this. It is enough to show that Γϕis compact for ϕ ∈ C(T), since Γϕ = 0 for ϕ ∈ H∞.
But if ϕ ∈ C(T), there is a sequence (ϕn) of rational functions tending uni-formly to ϕ on T.
Now‖Γϕ − Γϕn‖ ≤ ‖ϕ− ϕn‖∞ → 0.
Since each Γϕn has finite rank we deduce that Γϕ is compact.
The converse (due to Sarason) is harder, and we omit details. The basic ideais that a compact Hankel operator is the limit of a sequence of finite-rankHankel operators, whose symbols have the form hn + rn where hn ∈ H∞ andrn is rational (by Kronecker’s theorem). Then it is a matter of choosing thehn so that we get a uniformly convergent sequence of symbols, with limit inH∞ + C(T).
3.22 Corollary
Suppose that the sequence (dn)−1n=−∞ lies in `1, i.e.,
−1∑n=−∞
|dn| < ∞. Then
this Hankel matrix defines a compact operator:d−1 d−2 d−3 . . .d−2 d−3 d−4 . . .d−3 d−4 d−5 . . .d−4 d−5 d−6 . . .
......
......
.
Proof: This follows since the function∑−1
n=−∞ dnzn lies in C(T).
48
Recall that an operator T : H → K, where H and K are Hilbert spaces,
is said to be Hilbert–Schmidt if∞∑n=0
‖Ten‖2 < ∞ for one (and hence all) or-
thonormal bases (en) of H. It is well known that Hilbert–Schmidt operatorsare always compact.
3.23 Proposition
The Hankel matrix d−1 d−2 d−3 . . .d−2 d−3 d−4 . . .d−3 d−4 d−5 . . .d−4 d−5 d−6 . . .
......
......
defines a Hilbert–Schmidt operator if and only if
∞∑n=1
n|d−n|2 <∞.
Proof: Taking en(z) = zn for n ≥ 0, an orthonormal basis of H2, we seethat the square of the Hilbert–Schmidt norm is just the sum of the squaresof the absolute values of the entries of the matrix.
The Hilbert transform
3.24 Definition
The Hilbert transform is the operator defined on L2(T) by
(Hf)(eit) = − 1
πPV
∫ ∞−∞
f(eiθ)
θ − tdθ,
where PV denotes the Cauchy principal value, i.e., limδ→0
∫∞t+δ
+∫ t−δ−∞ .
Strictly speaking,we define it on the orthonormal basis (en)∞n=−∞, whereen(eit) = eint, and extend by linearity and continuity.
49
So H maps the function en with en(eiθ) = einθ to
(Hen)(eit) = − 1
πPV
∫ ∞−∞
einθ
θ − tdθ = − 1
πPV
∫ ∞−∞
einteinϕ
ϕdϕ,
where θ = t+ ϕ. Now
PV
∫ ∞−∞
einϕ
ϕdϕ = PV
∫ ∞−∞
cosnϕ+ i sinnϕ
ϕdϕ = 0 + iπ
for n ≥ 0, and we see easily that
Hen =
−ien if n > 0,
0 if n = 0,
ien if n < 0.
Then H is an operator of norm 1 with iH self-adjoint and
H4(∑
anen
)=∑n6=0
anen.
Note cosnθ 7→ sinnθ and sinnθ 7→ − cosnθ for n 6= 0.
Thus f + iHf is analytic in D, that is, if u is real harmonic in D, then Huis the harmonic conjugate v such that u+ iv is analytic in D, and v(0) = 0;this is unique.
Indeed,
∞∑0
anrn cosnθ +
∞∑1
bnrn sinnθ 7→
∞∑0
anrn sinnθ −
∞∑1
bnrn cosnθ.
The corresponding analytic function is
a0 +∞∑n=1
(an − ibn)zn.
50
4 The shift operator
The shift operator on H2 is defined by
(Sf)(z) = zf(z), (f ∈ H2),
and is thus an analytic Toeplitz operator. When considered on the space`2(Z+) of coefficients, it acts equivalently by
(a0, a1, a2, . . .) 7→ (0, a0, a1, a2, . . .),
the unilateral shift.
The invariant subspaces for S are the closed subspaces M such that SM ⊆M , i.e.,
f ∈M =⇒ Sf ∈M.
We exclude the trivial case M = 0 in what follows.
4.1 Theorem (Beurling)
Let M be a nontrivial (closed) invariant subspace for S. Then there is aninner function Θ ∈ H∞ such that
M = ΘH2 = Θf : f ∈ H2.
Also, Θ is unique to within a constant of modulus 1.
Proof: It is clear that ΘH2 is a closed invariant subspace (closed, sincethe mapping f 7→ Θf is an isometry on H2).
Also if Θ1H2 = Θ2H
2, then Θ1 = fΘ2 and Θ2 = gΘ1 for f, g ∈ H2. f and gmust themselves be inner, and so |f | ≤ 1 and |g| ≤ 1 on D. But fg = 1 so|f | = 1 on D and by the maximum principle f and g are constant.
Note that SM 6= M , as otherwise every function in M would be divisible byan arbitrarily large power of z and M = 0.
So let Θ ∈M SM , and without loss of generality ‖Θ‖2 = 1.
51
Now
0 = 〈SnΘ,Θ〉 =1
2π
∫ 2π
0
|Θ(eit)|2eint dt
for each n ≥ 1. By conjugation it is true for n ≤ −1 as well so |Θ|2 is aconstant, which must be 1.
Thus Θ is inner and ΘH2 ⊆M . Also if ϕ ∈M with ϕ ⊥ ΘH2, then
0 = 〈ϕ, SnΘ〉 =1
2π
∫ 2π
0
ϕ(eit)Θ(eit)eint dt (3)
for all n ≥ 0. But Θ ⊥ SM so
0 = 〈Θ, Snϕ〉 =1
2π
∫ 2π
0
Θ(eit)eintϕ(eit) dt
for all n ≥ 1, which is equivalent to (3) for all n < 0.Thus all the Fourier coefficients of ϕΘ vanish, and ϕ = 0. So M = ΘH2.
4.2 Definition
For u ∈ H2 we write [u] = spanu, Su, S2u, . . . (the notation span meansclosed linear span). This is the smallest (closed) invariant subspace contain-ing u.
We say that u is cyclic, or outer, if [u] = H2.
Obviously non-zero constant functions are cyclic. This implies easily thatinvertible functions in H2, such as u(z) = z + 2, are outer.
Note that outer functions u cannot vanish at any point of the disc, since allfunctions in [u] would vanish at the same point.
4.3 Theorem
Every nontrivial function in H2 has a factorization f = Θu, where Θ is innerand u is outer. This is unique to within unimodular constants.
52
Proof: Consider [f ] which is ΘH2 for some inner function Θ, by Beurling’stheorem. So f = Θu, with u ∈ H2. But then u is outer, as
Θ[u] = [Θu] = [f ] = ΘH2,
and so [u] = H2. The uniqueness of Θ also leads easily to the uniqueness ofthe factorization.
In fact, outer functions have a representation
u(z) = α exp
(1
2π
∫ 2π
0
eit + z
eit − zk(eit) dt
)(z ∈ D),
where |α| = 1 and k ∈ L1(T) is real. Moreover, k = log |u| on T. We do notprove this here.
53
5 Hardy spaces on the half-plane
5.1 Proposition
The Mobius map M : z 7→ 1−z1+z
is a self-inverse bijection from the disc D tothe right half-plane C+ = x+ iy : x > 0. Let 1 ≤ p <∞. Then a functiong defined on T is in Lp(T) if and only if the function G : iR→ C defined by
G(s) = π−1/p(1 + s)−2/pg(Ms)
is in Lp(iR), and moreover ‖g‖p = ‖G‖p.
We also haveg(z) = 22/pπ1/p(1 + z)−2/pG(Mz).
Proof: The fact that |z| < 1 if and only if ReMz > 0 is easy, and so ischecking that M M = I.Now
‖G‖pp =
∫ ∞−∞|G(iy)|p dy
= −i∫iR|g(Ms)|p 1
π
1
|1 + s|2ds
= −∫|z|=1
|g(z)|p 1
π
|1 + z|2
4
−2 dz
(1 + z)2,
making the change of variable s =1− z1 + z
withds
dz= − 2
(1 + z)2and 1 + s =
2
1 + z.
Now|1 + z|2
(1 + z)2=
1 + z
1 + z=
1
z
on T and we find that ‖G‖pp = ‖g‖pp.
For p =∞, we take G(s) = g(Ms) and g(z) = G(Mz). Then ‖g‖∞ = ‖G‖∞.
54
5.2 Corollary
The functions En defined by
En(s) =1√π
(1− s)n
(1 + s)n+1, (n ∈ Z),
form an orthonormal basis for L2(iR).
Proof: Recall that (zn)n∈Z is an orthonormal basis of L2(T). Now anorm-preserving map also preserves inner products, so, using the calculationin Proposition 5.1 for p = 2, we see that (En)n∈Z form an o.n.b. for L2(iR).
55
LECTURE 8
5.3 Definition
The space H∞(C+) consists of all functions analytic and bounded in C+,with norm ‖G‖ = sups∈C+
|G(s)|.
Looking at boundary values, we regard H∞(C+) as the closed subspace ofL∞(iR) consisting of all G with G(s) = g
(1−s1+s
), where g ∈ H∞(D).
Likewise H2(C+) consists of those functions analytic in C+ with L2 boundaryvalues, i.e., the closed subspace of L2(iR) consisting of all G with
G(s) =1√π
1
1 + sg
(1− s1 + s
), for g ∈ H2(D).
5.4 Corollary
The functions En defined by
En(s) =1√π
(1− s)n
(1 + s)n+1,
for n ≥ 0 form an orthonormal basis of H2(C+).
Note that En is analytic in C+ if and only if n ≥ 0.
Proof: Immediate, since (zn)n≥0 form an o.n.b. for H2(D).
5.5 Remark
One can define the norm on Hp(C+) more directly via
‖f‖pp = supx>0
∫ ∞−∞|f(x+ iy)|p dy.
Note we do need to take all x > 0, not just a limit as x→ 0; for consider
f(z) =ez
z + 1.
56
This is not in H∞(C+) or H2(C+), even though it is analytic in C+ withboundary values in L∞ ∩ L2(iR).
Laplace and Fourier transforms
5.6 Definition
For f ∈ L1(R) ∩ L2(R) define the 2-sided Laplace transform of f by
(Lf)(s) =
∫ ∞−∞
e−stf(t) dt,
and the Fourier transform of f by
(Ff)(w) =
∫ ∞−∞
e−iwtf(t) dt. (4)
Thus (Ff)(w) = (Lf)(iw). We are most interested in w ∈ R and s ∈ iR,when, since f ∈ L1(R), the integrals converge absolutely.
We regard L2(R) as L2(−∞, 0)⊕ L2(0,∞), decomposing a function into itsvalues on t < 0 and t > 0.
5.7 Theorem (Paley–Wiener–Plancherel)
The Laplace transform (defined initially on (L1 ∩ L2)(R), then extended bycontinuity) determines a linear bijection between L2(R) and L2(iR) such that
‖Lf‖L2(iR) =√
2π‖f‖L2(R).
Also, L2(0,∞) is mapped onto H2(C+).
There is an inverse map defined by
(L−1G)(t) =1
2πi
∫iRG(s)est ds,
for G ∈ (L1 ∩ L2)(iR), extended by continuity to all of L2(iR).
57
Thus,
L2(R) = L2(0,∞)⊕ L2(−∞, 0)
L ↓L2(iR) = H2(C+)⊕H2(C−).
Compare the discrete version:
`2(Z) = `2(Z+)⊕ `2(Z−)
↓L2(T) = H2(D)⊕H2(D)⊥,
where Z+ is the non-negative, and Z− the strictly negative, integers. Herethe mapping takes a sequence (an) to the function eiθ 7→
∑ane
inθ.
We shall sketch the proof of this shortly.
5.8 Corollary
If f ∈ L2(R), then Ff ∈ L2(R) (defined by (4) for f ∈ L1∩L2 and extendedby continuity), and
‖Ff‖2 =√
2π‖f‖2.
Moreover, the inverse map is defined by
(F−1F )(t) =1
2π
∫ ∞−∞
F (w)eiwt dw,
again for F ∈ L1 ∩ L2, then extended by continuity to L2(R).
Notes.
1. If f(t) and f ′(t) are both O((1 + |t|)−2) and f is C2, then f and Ffare in L1(R), so there are no problems about convergence of integrals. Suchfunctions form a dense subspace of L2(R).
2. We write f(w) = (Ff)(w) and F (t) = (F−1F )(t), and note thatˆf(t) =
2πf(−t), and thus F4 = 4π2I.
58
We give a sketch of the proof of Theorem 5.7.
Proof: Consider the dense subspace of L2(−∞,∞) consisting of all func-tions
f(t) =N2∑
j=−N2
ajχ[jM/N,(j+1)M/N ](t),
where M > 0 and N > 1 are integers, and the aj are complex scalars.Then
(Lf)(s) =N2∑
j=−N2
aje−jMs/N 1− e−Ms/N
s,
which lies in L2(iR), and direct calculation shows that its norm is
‖Lf‖L2(iR) =√
2π
√M
N
N2∑j=−N2
|aj|21/2
.
Now there is a unique continuous extension from this subspace to the wholeof L2 (given by an integral for f ∈ L1 ∩ L2).
Note that L carries L2(0,∞) ontoH2(C+), sinceH2(C+) has o.n.b.1√π
(1− s)n
(1 + s)n+1
for n ≥ 0 and
L(tne−tχ(0,∞)) =n!
(1 + s)n+1,
so the orthonormal basis functions in H2(C+) are linear combinations ofthese.
Similarly the functions with n < 0 are the Laplace transforms of functionssupported on (−∞, 0).
An orthonormal basis for L2(0,∞)
A rational orthonormal basis for H2(C+) was given by Corollary 5.4. Thefunctions transform under L−1 to functions t 7→ pn(t)e−t in L2(0,∞), wherepn is a real polynomial of degree n; then
√2πpn(t)e−t form an orthonormal
59
basis for L2(0,∞).
In factpn(t) = ±Ln(2t)/
√π,
where Ln denotes the Laguerre polynomial
Ln(t) =et
n!
dn
dtn(tne−t),
and these satisfy ∫ ∞0
Ln(t)Lm(t)e−t dt = δmn
(Kronecker delta). Similarly√
2πpn(−t)et form an orthonormal basis forL2(−∞, 0).
So, by a series of transformations we have linked an o.n.b. of
`2(Z+), namely (0, . . . , 0, 1, 0, . . .), to an o.n.b. of
H2, namely zn, to an o.n.b. of
H2(C+), namely1√π
(1− s)n
(1 + s)n+1, to an o.n.b. of
L2(0,∞), namely ±√
2Ln(2t)e−t.
60
6 Reproducing kernel Hilbert spaces
6.1 Definition
A reproducing kernel Hilbert space (RKHS) H is a Hilbert space of functionsdefined on a non-empty set S such that for all w ∈ S the evaluation mappingf 7→ f(w) is bounded, and thus there is a function kw (the reproducingkernel) such that
f(w) = 〈f, kw〉 (f ∈ H).
Note that if `w is another such reproducing kernel, then
`w(s) = 〈lw, ks〉 = ks(w)
for all s, w ∈ S, and hence kw is unique and
kw(s) = ks(w).
Another useful formula is that
‖kw‖2 = 〈kw, kw〉 = kw(w).
6.2 Theorem (Aronszajn)
In a RKHS it holds that for all α1, . . . , αn ∈ C and w1, . . . , wn ∈ S we have
n∑i=1
n∑j=1
αikwj(wi)αj ≥ 0. (5)
Moreover, if (w, z) 7→ kw(z) is a function on S × S satisfying (5), then thereis a unique RKHS with kw as the reproducing kernel, namely the completionof the space of all finite linear combinations of kw for w ∈ S, with norm∥∥∥∥∥
n∑j=1
αjkwj
∥∥∥∥∥2
=n∑i=1
n∑j=1
αikwj(wi)αj.
Proof: Evaluating 〈f, f〉 for
f =n∑j=1
αjkwj
61
shows that (5) holds.
Conversely, if (5) holds, then (kwj(wi))
ni,j=1 is a Hermitian matrix and the
formula ⟨n∑j=1
αjkwj,
n∑i=1
βikwi
⟩=
n∑i=1
n∑j=1
βikwj(wi)αj
satisfies the axioms of an inner product on the space of finite linear com-binations of the kwj
. It is now possible to take its completion in the usualway.
Note that reproducing kernels do not have to be linearly independent (forexample, if every function in the space takes the same value at two distinctpoints).
Examples
0. The Hilbert space Cn can be considered as a function space on 1, 2, . . . , n,with reproducing kernel km(j) = δmj, the Kronecker delta.
Clearly,〈(a1, . . . , an), km〉 = am.
1. The Hardy space H2 considered as a function space on D.
Note that if f ∈ H2 with
f(z) =∞∑n=0
anzn,
then
f(w) =∞∑n=0
anwn = 〈f, kw〉,
where
kw(z) =∞∑n=0
wnzn =1
1− wz,
the Cauchy (or Szego) kernel.
62
2. The Bergman space A2 of functionals analytic in D with finite Bergmannorm given by
‖f‖ =
(1
π
∫D|f(z)|2 dA(z)
)1/2
=
(1
π
∫ 1
r=0
∫ 2π
θ=0
|f(reiθ)|2r dr dθ)1/2
.
It can be checked that the norm of a function f represented by a power series
f(z) =∞∑n=0
anzn
is given by
‖f‖ =
(∞∑n=0
|an|2
n+ 1
)1/2
,
and the functions√n+ 1 zn for n ≥ 0 form an orthonormal basis. Hence
f(w) =∞∑n=0
anwn = 〈f, kw〉,
where
kw(z) =∞∑n=0
(n+ 1)wnzn =1
(1− wz)2,
the Bergman kernel.
An analogous space can be defined on an arbitrary open subset U of C, the
norm being(∫
U|f(z)|2 dA(z)
)1/2.
6.3 Proposition
Suppose that H is a reproducing kernel Hilbert space on a set S, with or-thonormal basis (en). Then the reproducing kernel is given by
kw =∑
en(w)en (w ∈ S).
Proof: Clearly there exist constants (an) depending on w such thatkw =
∑anen.
63
Thenan = 〈kw, en〉 = 〈en, kw〉 = en(w),
as claimed.
64
LECTURE 9
More examples.
3. More generally, if H is a Hilbert space of power series in the disc suchthat ∥∥∥∥∥
∞∑n=0
anzn
∥∥∥∥∥2
H
=∞∑n=0
γn|an|2,
i.e., (zn) is an orthogonal basis with norm ‖zn‖2 = γn, then kw if it existsmust be given by
kw(z) =∞∑n=0
wnzn/γn,
and H is a RKHS if and only if kw ∈ H, i.e.,∑∞
n=0 |w|2n/γn < ∞, for allw ∈ D.
For the Hardy space γn = 1 for each n, and for the Bergman space γn =1/(n+ 1).
4. The Hardy–Sobolev space of functions f such that f ′ ∈ H2. It is usual totake
‖f‖2 = ‖f‖2H2 + ‖f ′‖2
H2 ,
which means that γn = 1 + n2.
5. The Dirichlet space of functions f such that f ′ ∈ A2, where we can takeγ0 = 1 and γn = n for n > 0.
The reproducing kernel for the Dirichlet space is given by
kw(z) = 1 +∞∑n=1
wnzn/n = 1− log(1− wz).
6.4 Proposition
Let H be a reproducing kernel Hilbert space of analytic functions in thedisc, and let G ∈ H∞(D) such that MG is a bounded operator on H. ThenM∗
Gkw = G(w)kw, and hence the spectrum of MG contains the closure of theset
G(w) : w ∈ D.
65
In particular, ‖MG‖ ≥ ‖G‖∞.
Proof: For f ∈ H, we have
〈f,M∗Gkw〉 = 〈MGf, kw〉 = G(w)f(w) = 〈f,G(w)kw〉,
and so kw is an eigenvector of M∗G with eigenvalue G(w).
The rest follows from standard results about the spectrum: in particular,that the norm of MG is greater than or equal to the spectral radius of MG.
This is a simple way of proving that the shift S on H2 has spectrum equalto the closed unit disc, although it has no eigenvalues.
Here’s another class of operators that can be studied using reproducing ker-nels.
6.5 Definition
Let ϕ : D→ D be analytic. Then the composition operator Cϕ is defined onH2 by
Cϕf(z) = (f ϕ)(z) = f(ϕ(z)), (f ∈ H2).
6.6 Theorem (Littlewood’s subordination theorem)
The operator Cϕ maps H2 boundedly into H2.
Proof: Suppose first that ϕ(0) = 0: we claim that in this case Cϕ is acontraction.
Let f be a polynomial. We prove that ‖Cϕf‖ ≤ ‖f‖ by induction on thedegree of f . Clearly it is true for deg f = 0. In general,
f(z) = f(0) + zS∗f,
where S∗ is the backward shift, T1/z.
66
ThusCϕf = f(0) + ϕCϕS
∗f
and‖Cϕf‖2 = |f(0)|2 + ‖ϕCϕS∗f‖2 ≤ |f(0)|2 + ‖CϕS∗f‖2,
and by the induction hypothesis ‖CϕS∗f‖ ≤ ‖S∗f‖. So
‖Cϕf‖2 ≤ |f(0)|2 + ‖S∗f‖2 = ‖f‖2.
For general f ∈ H2 we have polynomials fn tending to f in H2 (and locallyuniformly in D), so Cϕfn → Cϕf locally uniformly. Then for each 0 < r < 1we have
1
2π
∫ 2π
0
|Cϕf(reit)|2 dt = limn→∞
1
2π
∫ 2π
0
|Cϕfn(reit)|2 dt
≤ lim sup ‖Cϕfn‖2
≤ lim sup ‖fn‖2 = ‖f‖2.
Then letting r → 1 we see that Cϕ is a contraction when ϕ(0) = 0.
For general ϕ we consider the Mobius map ψa defined by
ψa(z) =a− z1− az
.
Note that ψa ψa is the identity map.
It is easy to see by a change of variable on T (since the derivative of ψa isbounded) that Cψa acts as a bounded operator on L2(T); also, it preservesthe subspace H2. Indeed
‖Cψa‖ ≤(
1 + |a|1− |a|
)1/2
.
Choosing a = ϕ(0) we have (ψa ϕ)(0) = 0. Also
Cψaϕf(z) = f(ψa(ϕ(z))),
soCψaϕ = CϕCψa = CϕC
−1ψa.
67
Thus
‖Cϕ‖ ≤ ‖Cψaϕ‖‖Cψa‖ ≤ ‖Cψa‖ ≤(
1 + |ϕ(0)|1− |ϕ(0)|
)1/2
.
6.7 Theorem
Let ϕ : D → D be analytic, and let kw (for w ∈ D) denote the reproducingkernels for H2. Then C∗ϕkw = kϕ(w). Hence
‖Cϕ‖ ≥ supw∈D
(1− |w|2
1− |ϕ(w)|2
)1/2
.
Proof: Clearly
〈f, C∗ϕkw〉 = 〈Cϕf, kw〉 = f(ϕ(w)) = 〈f, kϕ(w)〉
for f ∈ H2.
Also
‖Cϕ‖ = ‖C∗ϕ‖ ≥ supw∈D
‖C∗ϕkw‖‖kw‖
,
which gives the estimate, recalling that
‖kw‖2 = 〈kw, kw〉 = kw(w) =1
1− |w|2.
In general the norms of many operators (e.g. Toeplitz, Hankel, composition,. . . ) can be estimated very well by “testing” them on reproducing kernels.This is sometimes called the reproducing kernel thesis.
For example a Hankel matrix Γ is bounded if and only if
sup ‖Γkw‖/‖kw‖ <∞
(Bonsall’s theorem, not proved here).
68
6.8 Remark
In a RKHS H the subspace Zw = f ∈ H : f(w) = 0 is just the orthogonalcomplement of the span of kw, and moreover f(wj) = g(wj) for j = 1, . . . , nif and only if (f − g) ⊥ kw1 , . . . , kwn .
For example, in H2, we have seen that f and g take the same values atw1, . . . , wn if and only if the Blaschke product B with zeroes w1, . . . , wn di-vides f − g.
Another way of expressing this is to say that the orthogonal complement ofBH2 is spankw1 , . . . , kwn.
A similar result holds for infinite sequences (wk)∞k=1 if they satisfy the Blaschke
condition∞∑k=1
(1− |zk|) <∞,
so that there is a Blaschke product with the (wk) as its zeroes. See Theo-rem 2.16.
6.9 Definition
Let K ⊂ R be compact and nonempty. Then the Paley–Wiener spacePW (K) consists of all f ∈ L2(R) such that f is supported on K.
So f ∈ L2(K), and so by Cauchy–Schwarz∫K
|f | ≤(∫
K
|f |2)1/2(∫
K
12
)1/2
<∞.
Thus
f(t) =1
2π
∫K
f(w)eiwt dw,
the integral converging pointwise absolutely.
Functions in PW (K) are very well behaved.
69
6.10 Lemma (Riemann–Lebesgue)
If f ∈ L1(R), then f lies in C0(R), i.e., it is continuous and tends to zero at±∞.
Proof: If
f =N2−1∑j=−N2
ajχ[jM/N,(j+1)M/N ],
a step function, then
f(t) =1
2π
N2−1∑j=−N2
ajejMit/N e
Mit/N − 1
it,
which lies in C0(R). For general g ∈ L1(R) there is a sequence (gn) of stepfunctions converging to g in L1 norm.
Now
‖gn − g‖∞ ≤1
2π‖gn − g‖L1 → 0,
so g ∈ C0(R).
6.11 Theorem
PW (K) is a RKHS of continuous functions on R, with
kt(s) = F−1(χK(w)e−iwt)(s) =1
2π
∫K
eiw(s−t) dw.
Proof:
f(t) =1
2π
∫K
f(w)eiwt dw
=1
2π〈f , e−iwtχK(w)〉L2(R)
= 〈f, kt〉L2(R),
where kt(w) = e−iwtχK(w), so kt ∈ PW (K).
70
We used the identity
〈f, g〉 =1
2π〈f , g〉
from Corollary 5.8.
In the special case K = [−b, b] for b > 0, we write PW (b), the space of“band-limited signals”.
The reproducing kernel is
ks(t) =1
2π
∫ b
−beiw(s−t) dw
=1
2πi(s− t)(eib(s−t) − e−ib(s−t)
)=
1
π
sin b(s− t)s− t
(s 6= t),
with ks(s) = b/π.
Note that ks ⊥ kt if ks(t) = 0, i.e., if (s− t) is a nonzero multiple of π/b.
6.12 Proposition
If f ∈ PW (b), then f extends to an entire function (i.e., analytic on thewhole of C) by
f(z) =1
2π
∫ b
−bf(w)eiwz dw,
and
|f(z)| ≤ 1
2π‖f‖L1eb|z|,
that is, f is of exponential type.
Proof: With
f(z) =1
2π
∫ b
−bf(w)eiwz dw,
we have
f (n)(z) =1
2π
∫ b
−bf(w)(iw)neiwz dw,
71
since this integral converges absolutely; thus f is an entire function. Inparticular we have
|f (n)(0)| ≤ 1
2πbn‖f‖1.
and
f(z) =∞∑n=0
znf (n)(0)
n!,
so
|f(z)| ≤∞∑n=0
|z|n
n!
1
2πbn‖f‖1 =
1
2π‖f‖1e
b|z|,
as required.
In fact there is a converse, which we do not prove.
6.13 Theorem
Let b > 0. If f is an entire function such that f|R ∈ L2(R) and there existsC > 0 such that |f(z)| ≤ Ceb|z| for all z ∈ C, then f|R ∈ PW (b).
72
LECTURE 10
We now look at an important orthonormal basis for PW (b).
6.14 Lemma
If F ∈ L2[−b, b], then
F (w) =π
b
∞∑n=−∞
F (nπ/b)e−inπw/b,
converging in L2 norm.
Proof: It is well known that the functions (en)n∈Z, defined by
en(w) =1√2be−inπw/b
form an orthonormal basis for L2[−b, b]. Thus
F =∞∑
n=−∞
〈F, en〉en
which gives
F (w) =1
2b
∞∑n=−∞
∫ b
−bF (y)einπy/b dye−inπw/b
=1
2b
∞∑n=−∞
2πF (nπ/b)e−inπw/b.
6.15 Corollary (Whittaker–Kotel’nikov–Shannon sampling theorem)
If f ∈ PW (b), then
f =π
b
∞∑n=−∞
f(nπ/b)knπ/b,
73
where k denotes the reproducing kernel. The series converges in L2(R) normand uniformly (i.e., in L∞(R) norm). That is, f can be reconstructed fromsamples spaced at intervals π/b.
Proof: By Lemma 6.14, with F = f , we have
f(t) =π
b
∞∑n=−∞
f(nπ/b)F−1(e−inπw/bχ[−b,b])(t),
converging in PW (b) and hence in L2 and uniformly. But
ks = F−1(e−iwsχ[−b,b]),
which gives the result.
Notes. Note that an orthonormal basis for PW (b) is the set√π
bknπ/b : n ∈ Z
.
Since PW (b) ⊂ PW (c) for 0 < b < c, we can also derive a formula for f interms of f(nπ/c), n ∈ Z. This is called “oversampling”.
THE END
74
