The Hilbert matrix operator

on spaces of analytic functions

A. G. Siskakis

Madrid, May 2012

First part

1. Introduction.

2. The Hilbert matrix on Hardy spaces.

3. The Hilbert matrix on Bergman spaces.

4. Some representations of the Hilbert matrix.

Second part

1. The point spectrum on Hardy spaces.

2. Generalized Hilbert operators.

1. The Hilbert matrix.

The (one-sided) Hilbert matrix is the one-sided

infinite matrix

H =

(1

i+ j +1

)∞i,j=0

=

1 12

13

14 .

12

13

14

15 .

13

14

15

16 .

14

15

16

17 .

. . . . .

.

Observe that H is symmetric. Its (i, j) entry is

1

i+ j +1= λi+j

where (λk) is the moment sequence

λk =∫ 1

0xk dx, k = 0,1, · · ·

of the Lebesgue measure on [0,1].

2. Some History.

Hilbert’s double series theorem (∼1900):

If∞∑k=0

a2k <∞ then∞∑i=0

∞∑j=0

aiaji+j+1 <∞.

The proof was published by H. Weyl (1908).

This was generalized by Hardy and Riesz (1925)

as follows:

If∞∑k=0

|ak|p <∞ and∞∑k=0

|bk|q <∞, 1p+ 1

q= 1, then

∞∑i=0

∞∑j=0

|ai||bj|i+ j +1

≤π

sin(πp)

( ∞∑k=0

|ak|p)1/p( ∞∑

k=0

|bk|q)1/q

.

Equivalently, by duality,

∞∑i=0

∣∣∣∣∣∣∞∑j=0

aj

i+ j +1

∣∣∣∣∣∣p1/p

≤π

sin(πp)

( ∞∑k=0

|ak|p)1/p

,

with the constant πsin(πp)

best possible.

Thus on the space lp of p-summable sequences

a = (ak) with norm,

∥a∥p =

∞∑k=0

|ak|p1/p

,

the operator

H : lp → lp, (ak) −→ (Ak),

where

Ak =∞∑j=0

aj

k+ j +1,

is bounded and

∥H∥lp→lp =π

sin(πp), 1 < p <∞.

The matrix of H with respect to the usual Schauder

basis {ej},

ej = (0, · · · ,0,1,0, · · · ), (1 in jth place),

of lp is clearly the Hilbert matrix:

(A0, A1, A2, ...) =

1 1

213

.12

13

14

.13

14

15

.. . . .

a0a1a2.

.

3. The Hilbert matrix on analytic functions.

Let D = {z : |z| < 1}, the unit disc,

A(D) = space of all analytic functions on D.

Let f(z) =∑∞n=0 anz

n ∈ A(D), and let the Hilbertmatrix H multiply the Taylor coefficients (an).We obtain formally the power series

H(f)(z) =∞∑n=0

Anzn =

∞∑n=0

∞∑k=0

akn+ k+1

zn.This series is not always defined: If

f(z) =1

1− z=

∞∑n=0

zn

then An =∞∑

k = 0

1n+k+1 = ∞ for each n. Thus

H cannot be defined on the whole space A(D).On the other hand if f(z) is a polynomial then

An =N∑k=0

akn+ k+1

, N = degree of f ,

so the power series H(f)(z) is well defined andconverges for |z| < 1, thus H(f) is analytic onD.

Integral representation. If f is a polynomial

then

H(f)(z) =∫ 1

0f(t)

1

1− tzdt.

Indeed in this case

N∑k=0

akn+ k+1

=∫ 1

0tnf(t) dt,

and

H(f)(z) =∞∑n=0

(∫ 1

0tnf(t) dt

)zn

=∫ 1

0f(t)

∞∑n=0

tnzn dt

=∫ 1

0f(t)

1

1− tzdt.

Some other representations will be given later.

Restricting the domain. We need to restrict

the domain of H to linear subspaces X ⊂ A(D)on which it is well defined.

Two such classes of spaces are the Hardy spaces

Hp and the Bergman spaces Ap. We will con-

centrate on these two cases.

4. The Hilbert matrix on Hardy spaces.

Hardy spaces. Let f : D → C analytic. For0 < p ≤ ∞ we say that f ∈ Hp if

∥f∥Hp = supr<1

(∫ 2π

0|f(reiθ)|p

dθ

2π

)1/p<∞,

or

∥f∥∞ = supz∈D

|f(z)| <∞, (for p = ∞).

We use ∥ ∥p instead of ∥ ∥Hp. Each Hp is a linearspace. For 1 ≤ p ≤ ∞, the function ∥ ∥p is acomplete norm, making Hp a Banach space. Forp = 2 the norm of f(z) =

∑∞k=0 akz

k ∈ H2 turnsout to be

∥f∥22 = supr

∫ 2π

0|f(reiθ)|2

dθ

2π

= supr

∫ 2π

0f(reiθ)f(reiθ)

dθ

2π

= supr

∞∑k=0

|ak|2r2k

=∞∑k=0

|ak|2

and this identifies H2 with l2. In particular H2

is a Hilbert space, with inner product

⟨f, g⟩ = limr→1

∫ 2π

0f(reiθ)g(reiθ)

dθ

2π

=∞∑k=0

akbk, (f ∼∑akz

k, g ∼∑bkz

k)

Properties of Hp:

• It is easy to see that if 1 < p < q <∞ then

H1 ⊃ Hp ⊃ Hq ⊃ H∞,

with strict containment in each case.

• If f ∈ Hp then the limit

f∗(eiθ) =: limr→1

f(reiθ)

exists for almost all θ ∈ [0,2π], the resulting

function f∗(eiθ) is p-integrable on the circle T,and

∥f∗∥Lp(T) = ∥f∥Hp.

• If f, g ∈ Hp and f∗(eiθ) = g∗(eiθ) on a set of

positive measure on T then f ≡ g. (a form of

the identity principle).

• We have

{f∗ : f ∈ Hp} ={p(eiθ) : p polynomial

},

closure in Lp(T). Thus Hp can be identified,

isometrically, with this closed subspace of Lp(T).

Further, for 1 ≤ p ≤ ∞, an f ∈ Lp(T) is the

boundary function of some Hp function if and

only if the negative part of the Fourier series of

f is zero.

• The Riesz Projection P+ : Lp(T) → Hp is

P+(f)(z) =1

2π

∫ 2π

0

f(θ)

1− e−iθzdθ,

or, in terms of Fourier series,

P+ :∞∑−∞

f(n)einθ −→∞∑n=0

f(n)einθ.

P+ is a bounded operator for 1 < p < ∞ (more

on this later).

• If f ∈ Hp then |f(z)| ≤ Cp∥f∥p(1−|z|)1/p

, z ∈ D.

• Hardy’s inequality: If f(z)=∞∑n=0

anzn ∈ H1 then

∞∑n=0

|an|n+1

≤ π∥f∥H1.

• Fejer-Riesz inequality: If f ∈ Hp then∫ 1

−1|f(t)|pdt ≤

1

2∥f∥pHp, (0 < p <∞).

• If φ : D → D is analytic then the composition

operator Cφ(f) = f ◦ φ is bounded on Hp and

∥Cφ∥Hp→Hp ≤(1+ |φ(0)|1− |φ(0)|

)1/p

• We will also need the following:

1. 11−z ∈ Hp for p < 1, but f /∈ H1. This implies

that for r ∈ R,

fr(z) =1

(1− z)r∈ Hp

if and only if r < 1p, and lim

r→1p∥fr∥p = ∞.

2. For s ∈ R the function

gs(z) =1

(1− z)(1z log1

1−z)s

is in H1 if and only if s > 1.

Hardy spaces of the half-plane.

Let Π = {z : Re(z) > 0}, the right half-plane.

For 0 < p <∞, Hp(Π) contains all analyticf : Π → C such that

∥f∥pHp(Π) = sup

0<x<∞

∫ ∞

−∞|f(x+ iy)|p dy <∞.

Hp(Π) are Banach spaces for 1 ≤ p <∞. We willonly need the fact that Hp(Π) are isometricallyisomorphic to Hp. Indeed let

µ : D → Π, µ(z) =1+ z

1− z,

the conformal map and let

V : Hp(Π) → Hp,

be the linear map

V (f)(z) = (4π)1/p(1− z)−2/pf(µ(z)).

It can be checked that V is 1-1, onto, ∥V (f)∥Hp =∥f∥Hp(Π) for each f ∈ Hp(Π), and has inverse

V −1 : Hp → Hp(Π),

V −1(g)(z) = π−1/p(1 + z)−2/pg(µ−1(z)).

The Hilbert matrix on Hp.

Let f(z) =∑∞n=0 anz

n ∈ H1. Then∣∣∣∣∣∞∑n=0

an

n+ k+1

∣∣∣∣∣ ≤∞∑n=0

|an|n+ k+1

≤∞∑n=0

|an|n+1

≤ π∥f∥H1,

by Hardy’s inequality. Thus the series∞∑n=0

( ∞∑k=0

ak

n+ k+1

)zn,

is well defined and converges on D.

In other words the matrix H transforms eachf ∈ H1 (and therefore each f ∈ Hp for p > 1)into a function H(f) ∈ A(D).

But H(H1) * H1. To see this let 1 < s ≤ 2 andconsider the test functions

fs(z) =1

(1− z)(1zlog 1

1−z)s∈ H1.

Assuming that

H(fs)(z) =∞∑n=0

(∫ 1

0tnfs(t)dt

)zn ∈ H1

we can apply Hardy’s inequality to obtain

∞∑n=0

∫ 10 t

nfs(t)dt

n+1dt <∞.

But∞∑n=0

1

n+1

∫ 1

0tnfs(t)dt =

∫ 1

0

( ∞∑n=0

tn

n+1

)fs(t)dt

=

∫ 1

0(1

tlog

1

1− t)fs(t)dt

=

∫ 1

0

1

(1− t)(1tlog 1

1−t)s−1

dt

= ∞,

because 0 < s − 1 ≤ 1, a contradiction. Thus

H(H1) * H1.

Remark. In a recent paper of B. Lanucha, M.

Nowak and M. Pavlovic it was proved:

If f(z) =∑∞n=0 anz

n ∈ H1 with an ≥ 0 then

H(f) ∈ H1 ⇔∞∑n=0

an log(n+1)

n+1<∞.

We also see that H(H∞) * H∞. In fact for f ≡ 1we find

H(1)(z) =1

zlog

1

1− z/∈ H∞.

On the other hand since H2 ≃ l2,

H : H2 → H2,

is bounded and ∥H∥H2→H2 = π.

For other p > 1, observe that the integral repre-

sentation

H(f)(z) =∫ 1

0f(t)

1

1− tzdt

is valid for all f ∈ H1. This is verified by using

the Fejer-Riesz inequality and the absolute con-

vergence of Taylor series to justify the exchange

of sums and integrals.

We can view this as an “improper line integral”.

The change of the path of integration

t = t(s) =s

(s− 1)z+1, 0 ≤ s ≤ 1,

gives

H(f)(z) =∫ 1

0

1

(s− 1)z+1f

(s

(s− 1)z+1

)ds.

This says that

H =∫ 1

0Ts ds, Ts(f) = wsf ◦ φs,

i.e. H is an average of weighted composition

operators with

ws(z) =1

(s− 1)z+1, φs(z) =

s

(s− 1)z+1.

Observe that φs maps D into D.

We would like to estimate ∥Ts∥Hp→Hp.

To do this use the isometry V : Hp(Π) → Hp and

equivalently estimate ∥Ts∥Hp(Π)→Hp(Π) where

Ts = V −1TsV : Hp(Π) → Hp(Π).

For f ∈ Hp(Π) we find,

Ts(f)(z) = (1− s)−2p

(s+

2

z − 1

)2p−1

f(ψs(z))

where ψs(z) = s1−sz+

11−s maps Π into Π.

Then integrate for the norm in Hp(Π) to ob-

tain the desired estimate for ∥Ts∥Hp(Π)→Hp(Π) =

∥Ts∥Hp→Hp as in the following Lemma.

Lemma. Suppose 0 < s < 1. We have:

(i) If 2 ≤ p <∞ then

∥Ts∥Hp→Hp ≤ s1p−1

(1− s)−1p .

(ii) If 1 < p < 2 then

∥Ts(f)∥Hp ≤ s1p−1

(1− s)−1p∥f∥Hp.

for all f ∈ Hp with f(0) = 0.

We then have for 2 ≤ p <∞,

∥H∥Hp→Hp ≤∫ 1

0∥Ts∥Hp→Hp ds

≤∫ 1

0s1p−1

(1− s)−1p ds

= B(1

p,1−

1

p)

= Γ(1

p)Γ(1−

1

p)

=π

sin(πp)

where we have used the standard identities for

the classical Beta and Gamma functions B and

Γ. A variation of this argument gives an analo-

gous result for 1 < p < 2, so we have,

Theorem [Diamantopoulos, S. , Studia Math. (2000)]

(i) If 2 ≤ p <∞ then

∥H∥Hp→Hp ≤π

sin(πp).

(ii) If 1 < p < 2 then

∥H(f)∥Hp ≤π

sin(πp)∥f∥Hp

for each f ∈ Hp with f(0) = 0.

H as a Hankel operator.

Consider the Riesz projection

P+ : Lp(T) −→ Hp, f ∼∞∑−∞

aneinθ −→

∞∑n=0

anzn,

and recall that P+ is a bounded operator when1 < p <∞.

Consider also the multiplication operator

Mϕ : Lp(T) −→ Lp(T), f −→ ϕf,

where the function

ϕ(t) = i(π − t)e−it ∈ L∞(T),has ∥ϕ∥L∞(T) = π and has Fourier coefficients

ϕ(n) =1

n+1, n ≥ 0.

Let J be the isometric “conjugation”(or “flip”)operator

J : Hp −→ Lp(T), f(eit) −→ f(e−it)

A calculation gives

H = P+ ◦Mϕ ◦ J,Lp(T)

Mϕ−−→ Lp(T)xJ yP+

Hp H−−→ Hp

It follows immediately that H : Hp → Hp isbounded for 1 < p <∞ and

∥H∥Hp→Hp ≤ ∥P+∥Lp(T)→Hp∥Mϕ∥Lp(T)→Lp(T)= ∥ϕ∥∞∥P+∥Lp(T)→Hp

= π∥P+∥Lp(T)→Hp.

Thus if we knew the norm of P+ we would havean estimate for ∥H∥Hp→Hp.

The norm of the Riesz projection.

I. Gohberg and N. Krupnik (1968) proved:

∥P+∥Lp(T)→Hp ≥1

sin(πp), 1 < p <∞,

and conjectured that equality holds for the norm.

B. Hollenbeck and I. Verbitsky (J. Funct. Anal.2000) proved the conjecture. Thus

∥P+∥Lp(T)→Hp =1

sin(πp), 1 < p <∞.

It follows that for H = P+MϕJ we have the uni-form upper estimate,

∥H∥Hp→Hp ≤π

sin(πp), 1 < p <∞.

Lower estimate of ∥H∥Hp→Hp.

Theorem [M. Dostanic, M. Jevtic and D. Vukotic

(J. Funct. Anal. 2008)]. Let 1 < p <∞, then

∥H∥Hp→Hp ≥π

sin(πp).

Sketch of proof:

Step 1. Fix ε ∈ (0,1), let γ ∈ (ε,1) and consider

the functions

fγ(z) =1

(1− z)γ/p.

Then fγ ∈ Hp and

∥fγ∥Hp → ∞, as γ → 1.

We can write

H(fγ)(z) =∫ 1

0(1− t)−γ/p

1

1− tzdt

=∫ 1

0

x−γ/p

1− z+ xzdx, (set 1− t = x)

=∫ ∞

0

x−γ/p

1− z+ xzdx−

∫ ∞

1

x−γ/p

1− z+ xzdx

= gγ(z)−Rγ(z)

Step 2. The function z1−γ/pgγ(z) is analytic on

C \ [−∞,0] ∪ [1,∞], (z =1

1− x).

If 0 < z < 1 we have

z1−γ/pgγ(z) =z1−γ/p

1− z

∫ ∞

0

x−γ/p

1+ z1−zx

dx

= (1− z)−γ/p∫ ∞

0

u−γ/p

1+ udu,

(set u = z1−zx)

= Γ(γ/p)Γ(1− γ/p)(1− z)−γ/p

=π

sin(γπp )(1− z)−γ/p,

and by the identity principle, for z ∈ D \ (−1,0],

z1−γ/pgγ(z) =π

sin(γπp )(1− z)−γ/p.

It follows that

∥gγ(z)∥Lp(T) = ∥z1−γ/pgγ(z)∥Lp(T)=

π

sin(γπp )∥fγ∥Hp.

Next Rγ(z) is analytic on C \ [−∞,0], thus in

particular well defined a.e. on T, and we have

∥H∥Hp→Hp · ∥fγ∥Hp ≥ ∥H(fγ)∥Hp

≥∣∣∣∥gγ∥Lp(T) − ∥Rγ∥Lp(T)

∣∣∣=

∣∣∣∣∣∣ π

sin(γπp )∥fγ∥Hp − ∥Rγ∥Lp(T)

∣∣∣∣∣∣ .Thus

∥H∥Hp→Hp ≥

∣∣∣∣∣∣ π

sin(γπp )−

∥Rγ∥Lp(T)∥fγ∥Hp

∣∣∣∣∣∣Step 3. Show that

∥Rγ∥Lp(T) ≤ C = C(ε)

where C does not depend on γ ∈ (ε,1).

Step 4. Let γ → 1 to obtain

∥H∥Hp→Hp ≥π

sin(γπp ).

As a conclusion

∥H∥Hp→Hp =π

sin(πp), 1 < p <∞.

5. H on Bergman spaces.

Bergman spaces. Let f : D → C analytic. For0 < p <∞ we say that f ∈ Ap if

∥f∥pAp =∫D|f(z)|p dA(z) <∞,

where dA(z) = 1πdxdy denotes the normalized

area measure on D.

For 1 ≤ p <∞ these are Banach spaces, and forp = 2,∫D|f(z)|2 dA(z) =

∫Df(z)f(z) dA(z)

=∫ 1

0

∫ 2π

0

∞∑n=0

anrneinθ

∞∑k=0

akrke−ikθ

1

πrdθdr

= 2∫ 1

0

∞∑n=0

|an|2r2n+1

=∞∑n=0

|an|2

n+1.

A2 is Hilbert space with inner product

⟨f, g⟩ =∫Df(z)g(z)dA(z)

=∞∑n=0

anbn

n+1, (f ∼

∑akz

k, g ∼∑bkz

k)

Properties of Ap:

• If 1 < p < q <∞ then

A1 ⊃ Ap ⊃ Aq ⊃ H∞,

with strict containment in each case. In additionfor each p we have

Hp ⊂ A2p.

• In contrast to Hardy spaces, functions f ∈ Ap

need not have boundary values on T.

• If 2 < p <∞ and f(z)=∞∑n=0

anzn ∈ Ap then

∞∑n=0

|an|n+1

≤ Cp∥f∥Ap.

(substitute of Hardy’s inequality)

• If f ∈ Ap then

|f(z)| ≤Cp∥f∥Ap

(1− |z|)2/p, z ∈ D.

As a consequence, when f ∈ Ap with p > 2,∫ 1

0|f(t)| dt ≤ C′

p∥f∥Ap

• If φ : D → D is analytic then the composition

operator Cφ(f) = f ◦ φ is bounded on Ap and

∥Cφ∥Hp→Hp ≤(1+ |φ(0)|1− |φ(0)|

)2/p

• For r ∈ R,

fr(z) =1

(1− z)r∈ Ap

if and only if r < 2p, and lim

r→2p∥fr∥p = ∞.

H on Bergman spaces.

On A2, H is not defined. Indeed

f(z) =∞∑n=0

1

log(n+1)zn ∈ A2

since∞∑n=0

1

(n+1) log2(n+1)<∞,

while

H(f)(0) =∞∑n=0

1

(n+1) log(n+1)is divergent.

If 2 < p <∞ the substitute of Hardy’s inequality

for f ∈ Ap implies that the series

H(f)(z) =∞∑n=0

∞∑k=0

akn+ k+1

znis a well defined analytic function on D.

The growth estimate for Ap implies that if f ∈ Ap

with p > 2 then the integral∫ 1

0f(t)

1

1− tzdt

is absolutely convergent for z ∈ D and defines an

analytic functions on D. A further calculation

gives

H(f)(z) =∫ 1

0f(t)

1

1− tzdt,

the integral representation for Bergman func-

tions.

We can change the variable again to obtain

H =∫ 1

0Ts ds, Ts(f) = wsf ◦ φs,

where ws and ϕs are the same as for Hardy

spaces.

We want to estimate ∥Ts∥Ap. For this we needthe following identities

1. ws(z)2 = 1s(1−s)ϕ

′t(z),

2. ws(ϕ−1s (z)) = z

s,

which are valid for all 0 ≤ s < 1, z ∈ D. Thus

∥Ts(f)∥pAp =

∫D|ws(z)|p|f(ϕs(z))|p dA(z)

=

∫D|ws(z)|p−4|ws(z)|4|f(ϕs(z))|p dA(z)

=1

s2(1− s)2

∫D|ws(z)|p−4|f(ϕs(z))|p|ϕ′

s(z)|2 dA(z)

=1

s2(1− s)2

∫ϕs(D)

|ws(ϕ−1s (u))|p−4|f(u)|p dA(u)

=1

sp−2(1− s)2

∫ϕs(D)

|u|p−4|f(u)|p dA(u)

(assumimg p ≥ 4)

≤1

sp−2(1− s)2

∫D|f(u)|p dA(u)

=1

sp−2(1− s)2∥f∥pAp.

For 2 < p < 4 a little more work gives a similarresult with an additional constant C in the laststep. This gives the following

Lemma For 0 < s < 1 we have:

(i) If 4 ≤ p <∞ then ∥Ts∥Ap ≤ s2p−1

(1− s)−2p .

(ii) If 2 < p < 4 then ∥Ts∥Ap ≤ Cs2p−1

(1− s)−2p

for some constant C = Cp.

The usual calculation then gives

∥H∥Ap ≤∫ 1

0t2/p−1(1− t)−2/p dt

= Γ(2

p)Γ(1−

2

p)

=π

sin(2πp ),

so we have the following.

Theorem [Diamantopoulos, Illinois J. Math. 2004]

1. ∥H∥Ap ≤ πsin(2πp )

for 4 ≤ p <∞.

2. ∥H∥Ap ≤ Cpπ

sin(2πp )for 2 < p < 4.

Remark. Cp → ∞ as p→ 2. This was improved

to a uniform C valid for all 2 < p < 4 by M.

Dostanic, M. Jevtic and D. Vukotic.

Lower bound for ∥H∥Ap.

Theorem [D-J-V]. Let 2 < p <∞, then

∥H∥Ap→Ap ≥π

sin(2πp ).

Sketch of proof:

Consider the test functions

fγ(z) =1

(1− z)γ/p, γ < 2 < p,

then fγ ∈ Ap. A calculation shows that for the

weighted composition operators Ts we have

Ts(fγ)(z) =((s− 1)z+1)γ/p−1

(1− s)γ/pfγ(z),

thus

H(fγ)(z) =

(∫ 1

0

((s− 1)z+1)γ/p−1

(1− s)γ/pds

)fγ(z)

(change of variable u = 1− s)

=

(∫ 1

0

(1− uz)γ/p−1

uγ/pdu

)fγ(z)

= ϕγ(z)fγ(z)

We check that the function

ϕγ(z) =∫ 1

0

(1− uz)γ/p−1

uγ/pdu

is in fact defined and continuous on |z| ≤ 1 forall γ ≤ 2. Since γ/p− 1 < 0 and

|1− uz| ≥ 1− u|z| ≥ 1− u, z ∈ D,

we have

|ϕγ(z)| ≤ |ϕγ(1)|

=∫ 1

0

(1− u)γ/p−1

uγ/pdu

=π

sin(γπp ).

Now let gγ(z) =fγ(z)∥fγ∥Ap

and consider the family

{|gγ(z)|p : 0 ≤ γ ≤ 2, z ∈ D}.

We can check that it satisfies all properties ofan approximate identity:

1. |gγ(z)|p ≥ 02.

∫D |gγ(z)|p dA(z) = 1 for each γ

3. |gγ(z)|p tends uniformly to 0, as γ → 2, oncompact subsets of D \ {1}.

Further the function ϕγ(z) : [0,2] × D → C isuniformly continuous and uniformly bounded inany set of the form

{z ∈ D : |z − 1| < ε} × [0,2].

In addition∫D|H(gγ)(z)|pdA(z)− |ϕ2(1)|p

=∫D(|ϕγ(z)|p − |ϕ2(1)|p)dA(z)

→ 0, as γ → 2

and it follows that

limγ→2

∥ϕγgγ∥Ap = ∥ϕ2∥∞ = ϕ2(1) =π

sin(2πp ).

Since ∥H∥Ap→Ap ≥ ∥H(gγ)∥Ap = ∥ϕγgγ∥Ap foreach γ < 2 we conclude

∥H∥Ap→Ap ≥π

sin(2πp ).

As a consequence,

∥H∥Ap→Ap =π

sin(2πp ), 4 ≤ p <∞.

The value ∥H∥Ap→Ap is not known for 2 < p < 4.

6. Some other representations of H.

a. As an area integral. ([D-J-V]) If 2 < p <∞and f ∈ Ap then

H(f)(z) =∫D

f(w)

(1− w)(1− wz)dA(w).

Indeed∫DwmwndA(w) =

1

n+1, when m = n,

0 when m = n

so∫D

1

1− wwndA(w) =

∞∑m=0

∫DwmwndA(w) =

1

n+1.

If f(z) =∑Nn=0 anz

n is a polynomial then

N∑k=0

ak

n+ k+1=

N∑k=0

ak

∫D

1

1− wwn+kdA(w)

=

∫D

1

1− w

(N∑k=0

akwk

)wndA(w)

=

∫D

f(w)

1− wwndA(w),

and

H(f)(z) =∞∑n=1

(∫D

f(w)

1− wwndA(w)

)zn

=

∫D

f(w)

1− w

( ∞∑n=0

(wz)n

)dA(w)

=

∫D

f(w)

(1− w)(1− wz)dA(w).

The validity of this for all functions in Ap is

verified by appealing to the uniform convergence

of the Taylor series of f .

b. As average of composition operators.

We can change the path in the integral

H(f)(z) =∫ 1

0f(t)

1

1− tzdt

to t = t(s) = ϕs(z) where {ϕs(z) : 0 ≤ s ≤ 1}is a family of functions with ϕs : D → D analytic

for each 0 ≤ s < 1, ϕ0(z) = 0 and ϕ1(z) = 1 for

each z ∈ D, and ∂ϕs(z)∂s exists and is a bounded

analytic function of z for each s, to obtain

H(f)(z) =∫ 1

0

∂ϕs(z)∂s

1− ϕs(z)zf(ϕs(z)) dt.

Families of this form can be created as follows:

Let h : D → C be a starlike univalent function(i.e. f(0) = 0 and if w ∈ h(D) then [0, w] ⊂h(D)), and put

ψs(z) = h−1(sh(z)), 0 ≤ s ≤ 1.

Then ψs : D → D are analytic, ψs(0) = 0 and weset

ϕs(z) =ψs(z)

z,

which are also analytic self-maps of D by Schwarz’sLemma. The path

t(s) = ϕs(z), 0 ≤ s ≤ 1,

joins 0 to 1, and we evaluate the integral alongthis path to find

H(f)(z) =h(z)

z

∫ 1

0

1

(1− zϕs(z))h′(zϕs(z))f(ϕs(z)) ds

or

H(f)(z) =h(z)

z

∫ 1

0w(zϕs(z))f(ϕs(z)) ds

where w(z) = 1(1−z)h′(z).

The previous representation H =∫ 10 Ts ds can be

recovered in this way with the starlike functionh(z) = z

1−z.

As a further example, let

h(z) = log1

1− z,

then we find ψs(z) = 1− (1− z)s, and w(z) ≡ 1.

Thus

H(f)(z) =1

zlog

1

1− z

∫ 1

0f

(1− (1− z)s

z

)ds.

Second part

7. The point spectrum of H on Hp.

Let Fn be the nth finite section of the Hilbertmatrix

Fn =

(1

i+ j +1

)n−1

i,j=0,

then

det(Fn) =((n− 1)!!)4

(2n− 1)!!,

where n!! =∏nk=1 k!. For example

det(F3) =1

2160,

det(F6) =1

186313420339200000,

det(F9) ∼1

1042.

In particular the inverse of Fn has very large in-teger entries and the computation of its eigen-values is a very sensitive problem.

In Numerical Analysis, Fn are typical examplesof “ill-conditioned”matrices, difficult to use innumerical computation.

Coming back to the infinite Hilbert matrix H,

Various papers, ∼1950-1960, discuss H and more

general versions of it like

Hλ =

(1

i+ j + λ

), λ ∈ C.

These works concentrate mostly on the spec-

trum on l2 and in particular on “latent roots”.

A latent root is a phony eigenvalue in the sense

that the corresponding eigenvector is not neces-

sarily in the space under consideration. For Hλ,

this will be a complex number c for which there

is a nonzero sequence (xn), not necessarily in l2,

such that∞∑k=0

xkn+ k+ λ

= cxn, n = 0,1, · · ·

In these works one can find results connecting

the eigenvectors to hypergeometric functions.

Sample results of this kind are:

Theorem [Hill, J. London Math. Soc. (1960)]

If xn = xn(λ, µ) is defined by

xn =n∑

k=0

(nk

)(−1)k

Γ(k+ µ)Γ(k+1− µ)

k!Γ(k+ λ)

where 0 < Re(µ) ≤ 12. Then

∞∑k=0

xkn+ k+ λ

=π

sin(πµ)xn, n = 0,1,2, · · ·

i.e πsin(πµ) is a latent root of Hλ.

Theorem. [W.Magnus, Amer, J. Math., (1950]

The spectrum of H = H1 on l2 (and thus on H2)

is the interval [0, π], and there are no eigenval-

ues.

A more complete study of the point spectrum on

Hardy spaces (and on some other spaces defined

by the growth of f) was done recently by A.

Aleman, A. Montes and A. Sarafoleanu [Constr.

Approx., to appear]. They consider matrices

Hλ, λ ∈ C \ Z,

and prove the representation

Hλ(f)(z) =∫ 1

0f(t)

tλ−1

1− tzdt, if Re(λ) > 0,

=1

κ

∫γf(t)

tλ−1

1− tzdt, if Re(λ) ≤ 0,

where κ = e2πiλ − 1 and γ is appropriate closed

curve such as the boundary of a Stolz angle

{z ∈ D : |1− z| ≤ σ(1− |z|)}, σ > 1.

The key observation in this work is that Hλ “al-

most commutes”with two specific second order

linear differential operators with polynomial co-

efficients. More precisely if

D1,λ(f)(z) = (z2 − 1)f ′(z) + λzf(z)

and

D2,λ(f)(z) =

= z(z − 1)2f ′′(z) + (z − 1)[(λ+2)z − λ]f ′(z) + λzf(z)

then

(HλD1,λ −D1,λHλ)(f) =1

κ(λ− 1)

∫γf(t)tλ−2 dt,

and

HλD2,λ = D2,λHλ.

Next, the eigenvalue equation

D1,λ(f) = νf,

has analytic solutions which are constant multi-

ples of

fλ(z) = (1− z2)−λ/2(1+ z

1− z

)−ν/2.

Similarly the eigenvalue equation

D2,λ(g) = νg,

has solutions which are constant multiples of

ga(z) = (1− z)a 2F1(a+1, a+ λ;λ; z),

where a is a root (any of the two) of the quadratic

x2 + x+ λ− ν = 0,

and

2F1(α, β; γ; z) = 1+αβ

1!γz+

α(α+1)β(β+1)

2!γ(γ +1)z2+· · ·

is the classical hypergeometric function.

Observe that these functions ga make sense for

each a ∈ C and that for each a ∈ C we can find

ν such that a is root of the quadratic. In view

of this, consider all a in the strip

{z : −1 < Re(z) ≤ −1

2}.

If a′ is the other root of the quadratic then since

a+a′ = −1 we will have −12 ≤ Re(a′). There are

three cases to consider

1. If a = −12 = a′ then ga ∈ Hp for all p < 2.

2. If a = −12 and λ is such that −(a′+λ) /∈ N∪{0}

then

ga(z) ∼d

(1− z)a, d = 0,

as z → 1 along the radius. In this case ga ∈ Hp

if and only if Re(a) < 1p.

3. If a = −12 and λ is such that −(a′ + λ) = n ∈

N ∪ {0} then

Re(λ) ≤1

2, and ga(z) = (1− z)a

′Q(z),

with Q a polynomial of degree n. In this case

ga ∈ Hp if and only if 1+Re(a) < 1p.

Further for all values of a in the strip, Hλ(ga) is

defined and satisfies

Hλ(ga) = −π

sin(πa)ga.

This last assertion is seen as follows:

By the commutation of Hλ with D2,λ we have

D2,λHλ(ga) = HλD2,λ(ga)

or

D2,λ(Hλ(ga)) = Hλ(νga) = νHλ(ga).

This says that Hλ(ga) is an eigenfunction of

D2,λ, so there is a ξ such that

Hλ(ga) = ξga,

and the value of ξ is found by observing that

ga(0) = 1, thus

ξ = Hλ(ga)(0)

=1

κ

∫γga(t)t

λ−1 dt

= · · ·

= −π

sin(πa).

Thus using D2,λ, and for spaces Hp with 1 < p <2, we have detected a large set of eigenvaluesof the form{

−π

sin(πa): −

1

p< Re(a) ≤ −

1

2

}with corresponding eigenfunctions

ga(z) = (1− z)a 2F1(a+1, a+ λ;λ; z)

(except when possibly a = −λ,−λ − 1). Thesekind of eigenvalues however disappear for Hp

with p ≥ 2.

A similar analysis for D1,λ detects eigenfunctionsof the form

fn(z) =1

(1− z)λ

(1+ z

1− z

)n, 0 ≤ n ≤ N

where N is the largest integer for which

(1− z)−λ−N ∈ Hp.

The corresponding eigenvalues are (−1)n πsin(πλ).

In all cases the eigenspaces are finite dimen-sional.

The results as written in the article do not coverthe case λ = 1 (but A. Aleman says that he cando that case too, by a variation of the argumentsinvolved).

8. Generalized Hilbert operators

a. Changing the Lebesgue measure.

We have seen that H = (hi,j) where

hi,j =1

i+ j +1= λi+j

with (λk) is the moment sequence

λk =∫ 1

0xk dx, k = 0,1, · · ·

of the Lebesgue measure on [0,1].

For a general finite positive Borel measure µ on[0,1) consider the moments

µk =∫[0,1)

xk dµ(x), k = 0,1, · · ·

and the resulting generalized Hilbert matrix

Hµ =(hi,j

)∞i,j=0

where hi,j = µi+j, that is

Hµ =

µ0 µ1 µ2 .µ1 µ2 µ3 .µ2 µ3 µ4 .. . . .

.

For f(z) =∑∞n=0 anz

n ∈ A(D) we have, by anal-

ogy, the formal transformation

Hµ(f)(z) =∞∑n=0

∞∑k=0

µn+kak

zn.If µ satisfies

(∗) µ((t,1)) = O((1− t)), all t near 1,

then

µn = O(1/(n+1)).

Using Hardy’s inequality then it follows that the

power series Hµ(f) is analytic on D for every

f ∈ H1 and

Hµ(f)(z) =∫[0,1)

f(t)1

1− tzdµ(t).

Condition (∗) is known to be equivalent to that

Hµ : H2 → H2

is bounded.

Recall that the Hilbert matrix H is not bounded

on H1. For general µ we have

Theorem[P. Galanopoulos and J.A. Pelaez, Stu-dia Math. 2010] Suppose µ satisfies (∗), then

1. Hµ : H1 → H1 is bounded if and only if

µ((t,1)) log1

1− t= O((1− t)), all t near 1,

2. Hµ : H1 → H1 is compact if and only if

µ((t,1)) log1

1− t= o((1− t)).

Further, [G-P] proved that Hµ is Hilbert-Schmidton H2 if and only if

(HS)∫[0,1)

µ([t,1])

(1− t)2dµ(t) <∞

(Hilbert-Schmidt means:∑∞k=0 ∥Hµ(ek)∥2H2 < ∞

for any orthonormal basis (ek))

On the Bergman space A2,

Theorem[G-P] If µ satisfies (HS) then for eachf ∈ A2 the power series Hµ(f) represents ananalytic function on D and

Hµ(f)(z) =∫[0,1)

f(t)1

1− tzdµ(t).

But for the boundedness of Hµ on A2 something

more is required. In fact [G-P] proved that if

(HS) holds then Hµ : A2 → A2 is bounded if and

only if∫D|f(z)|2|µ′(z)|2dA(z) ≤ C

∫D|f ′(z)|2dA(z)

for all f for which the right hand-side integral is

finite, where µ(z) is the function

µ(z) =∞∑n=0

µnzn.

To appreciate this, compare with the following

relevant result

Theorem [G-P] For each β ∈ [0,1) there is a µ

such that∫[0,1)

µ([t,1))

(1− t)2

(log

1

1− t

)βdµ(t) <∞,

but Hµ is not bounded on A2.

b. Changing the kernel function.

Observe that

H(f)(z) =∫ 1

0f(t)

1

1− tzdt

=∫ 1

0f(t)g′(tz) dt

where g(z) = log( 11−z). We may consider the

more general transformation

Hg(f)(z) =∫ 1

0f(t)g′(tz) dt,

where g is any analytic function on D.

By Fejer-Riesz’s inequality the integral converges

for each f ∈ H1.

The question arises to describe the symbols g

for which

Hg : Hp → Hp

is a bounded operator (or compact or ...)

Preliminary remarks

• If g is a polynomial of degree n then Hg(f) is

also a polynomial of degree n− 1. Thus Hg is a

finite rank operator when g is a polynomial.

• Since

Hλg1+µg2 = λHg1 + µHg2, λ, µ ∈ C,

the set

V = Vp =: {g : Hg : Hp → Hp is bounded }

is a linear space of analytic functions and con-

tains the polynomials.

• We can define a norm on V ,

∥g∥V =: ∥Hg∥Hp→Hp, g ∈ V,

to make (V, ∥ ∥V ) a normed space.

• If we set

V0 = Vp,0 =: closure of polynomials in V

then Hg : Hp → Hp is a compact operator, for

each g ∈ V0.

• If f(z) =∑∞n=0 anz

n, g(z) =∑∞n=0 bnz

n then

Hg(f)(z) =∞∑n=0

((n+1)bn+1

∫ 1

0tnf(t)dt

)zn

=∞∑n=0

(n+1)bn+1

∞∑k=0

akn+ k+1

zn.In particular if the series of g has gaps, then the

series of Hg(f) has the same gaps.

• We can view Hg as a product of the classical

Hilbert operator H and a coefficient multiplica-

tion operator. For an analytic h(z) ∼∑λnzn

let

Λh(f)(z) =∞∑n=0

λnanzn, f ∼

∑anz

n.

the coefficient multiplier operator. Then we

have

Hg = Λg′ ◦ H.

In particular if g is a function such that

Λg′ : Hp → Hp

is a bounded multiplier, then Hg : Hp → Hp is

bounded. In other words the space V of those g

for which Hg is bounded contains all antideriva-

tives of functions that act as coefficient multi-

pliers of Hp.

It turns out that we can almost completely de-

scribe all g which give bounded Hg in terms of

mean Lipschitz spaces.

Mean Lipschitz spaces.

Let 1 ≤ p <∞, 0 < α ≤ 1. The integral modulus

of continuity of order p for an integrable function

f on T is defined to be

ωp(t) = sup0<h≤t

(∫ 2π

0|g(ei(θ+h))− g(eiθ)|pdθ

)1/p

Λ(p, α) contains all f for which

ωp(t) = O(tα), as t→ 0.

If either p or α is kept fixed and the other is

let to increase then Λ (p, α) decreases in size.

If α > 1p then Λ (p, α) consists entirely of con-

tinuous functions. The borderline space Λ(p, 1p

)contains unbounded functions, in fact

log1

1− z∈ Λ

(p,

1

p

), all p > 1,

and these spaces increase with p but they stay

always inside BMOA:

Λ

(q,

1

q

)⊂ Λ

(p,

1

p

)⊂ BMOA, 1 ≤ q < p <∞,

These will be the spaces of interest to us.

Equivalent characterizations:

By taking Poisson integrals of functions on the

boundary we may assume that we are working

with analytic functions on D when dealing with

the above spaces.

Suppose g(z) =∑∞n=0 anz

n is analytic on D. We

write as usual

Mp(r, g) =

(∫ 2π

0|g(reiθ)|p dθ

)1/p

and

∆ng(z) =∑

2n≤k≤2n+1−1

akzk.

the n-th dyadic section of g. Then

Lemma. The following are equivalent:

1. g ∈ Λ(p, α),

2. Mp(r, g′) = O((1− r)α−1

)as r → 1,

3. ∥∆ng∥Hp = O(2−nα

)as n→ ∞,

4. ∥∆ng′∥Hp = O(2n(1−α)

)as n→ ∞.

The following theorems describes the bounded-

ness of Hg on Hp and on Ap.

Theorem 1. [P. Galanopoulos D. Girela, J.A.

Pelaez, and A. S. (2011), preprint]

Suppose g is analytic on D.

1. If 1 < p ≤ 2 then Hg : Hp → Hp is bounded if

and only if g ∈ Λ(p, 1p

).

2. If 2 < p < ∞ and Hg : Hp → Hp is bounded

then g ∈ Λ(p, 1p

).

3. If 2 < p < ∞ and g ∈ Λ(q, 1q

)for some q < p

then Hg : Hp → Hp is is bounded.

Theorem 2.[G-G-P-S]

Suppose 2 < p < ∞. Then Hg : Ap → Ap is

bounded if and only if g ∈ Λ(p, 1p

).

Sketch of proof of:

Hg : Hp → Hp bounded ⇒ g ∈ Λ(p, 1p

),

Write (f∗g)(z) =∑∞n=0 f(n)g(n)z

n, the Hadamardproduct of two power series f and g.

We are going to use the following lemma fromthe general theory of “smooth Cesaro means”.

Lemma Let Φ be a C∞ function with compactsupport in (0,∞), write

AΦ = ∥Φ∥∞ + ∥Φ′′∥∞and let

WΦm(z) =

∑k∈N

Φ(k/m)zk,

Then for each p ∈ (0,∞) there is a Cp such that

∥WΦm ∗ f∥Hp ≤ CpAΦ∥f∥Hp, f ∈ Hp

Next we construct the C∞ function for our pur-pose. Fix 1 < p <∞. Let

aN = 1−1

N, N = 2,3, · · ·

and let ϕN be defined by

ϕN(s) =∫ 1

0

tsN

(1− aNt)2dt, (ϕN ∈ C∞).

Find a C∞ function ΦN : R → C such that:(i) supp(ΦN) ⊂ [12,4],(ii) In the smaller interval [1,2],

ΦN(s) =N

2−1p

ϕN(s), 1 ≤ s ≤ 2.

We can estimate that

AΦN= ∥ΦN∥∞ + ∥Φ′′

N∥∞ ≤ CN1−1

p .

Put

WN(z) =∑k∈N

ΦN(k/N)zk,

a polynomial. Let also

fN(z) =N

1p−2

(1− aNz)2, N = 2,3, · · ·

Then fN ∈ Hp and

supN

∥fN∥Hp = C <∞,

therefore by hypothesis,

supN

∥Hg(fN)∥Hp = C′ <∞.

Applying the Lemma to Hg(fN) ∈ Hp we find

∥WN ∗ Hg(fN)∥Hp ≤ CAΦN∥Hg(fN)∥Hp

≤ C′N1−1

p .

On the other hand, if g(z) =∑∞n=0 bkz

k we find

WN∗Hg(fN)(z)

=∑

N2≤k≤4N

(ΦN(

k

N)(k+1)bk+1

∫ 1

0tkfN(t)dt

)zk

=∑

N2≤k≤N−1

· · ·+∑

N≤k≤2N−1

· · ·+∑

2N≤k≤4N

· · ·

= FN1 (z) + FN2 (z) + FN3 (z),

and we compute, for N ≤ k ≤ 2N ,

ΦN(k/N) =N

2−1p

ϕN(k/N)

=N

2−1p

∫ 10

tkNN

(1−aN t)2dt

=1∫ 1

0 tk N

1p−2

(1−aN t)2dt

=1∫ 1

0 tkfN(t) dt

,

therefore,

FN2 (z) =∑

N≤k≤2N−1

(k+1)bk+1zk.

Thus

∆ng′(z) =

2n+1−1∑k=2n

(k+1)bk+1zk = F2n

2 (z)

Now, from M. Riesz’s projection theorem

∥FN2 ∥Hp ≤ C∥Wn ⋆Hg(fN)∥Hp ≤ C′N1−1

p

Therefore,

∥∆ng′∥Hp = ∥F2n

2 ∥Hp ≤ C2n(1−1

p)

and it follows that g ∈ Λ(p, 1p

), Q.E.D.

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