A generalized Hilbert operator acting on mean Lipschitz...

Notation and definitionsGeneralized Hilbert matrix

Mean Lipschitz spacesHµ acting on mean Lipschitz spaces

A generalized Hilbert operatoracting on mean Lipschitz spaces

Noel MerchánUniversidad de Málaga, Spain

Madrid International Workshop onOperator Theory and Function Spaces

Madrid, Spain. October 16, 2018

Noel Merchán noel@uma.es A generalized Hilbert operator acting on mean Lipschitz spaces

1 Notation and definitions

2 Generalized Hilbert matrixHilbert matrixA generalized Hilbert matrixIntegral operatorCarleson measures

3 Mean Lipschitz spaces

4 Hµ acting on mean Lipschitz spaces

The unit disc and the unit circleD = {z ∈ C : |z| < 1}, the unit disc.T = {ξ ∈ C : |ξ| = 1}, the unit circle.

Spaces of analytic functions in the unit disc

Hol(D) is the space of all analytic functions in D.

Hardy spaces

If 0 < r < 1 and f ∈ Hol(D), we set

Mp(r , f ) =

∫ 2π

0|f (reit )|p dt

, 0 < p <∞,

M∞(r , f ) = sup|z|=r|f (z)|.

If 0 < p ≤ ∞, we consider the Hardy spaces Hp,

{f ∈ Hol(D) : ‖f‖Hp

def= sup

0<r<1Mp(r , f ) <∞

Hardy spaces

Mp(r , f ) =

∫ 2π

0|f (reit )|p dt

, 0 < p <∞,

M∞(r , f ) = sup|z|=r|f (z)|.

def= sup

0<r<1Mp(r , f ) <∞

Hardy spaces

Mp(r , f ) =

∫ 2π

0|f (reit )|p dt

, 0 < p <∞,

M∞(r , f ) = sup|z|=r|f (z)|.

def= sup

0<r<1Mp(r , f ) <∞

BMOA ={

f ∈ H1 : f(

eiθ)∈ BMO

Bloch space

{f ∈ Hol(D) : sup

z∈D(1− |z|2)|f ′(z)| <∞

H∞ ⊂ BMOA ⊂ B.

BMOA ={

f ∈ H1 : f(

eiθ)∈ BMO

Bloch space

{f ∈ Hol(D) : sup

z∈D(1− |z|2)|f ′(z)| <∞

BMOA =

{f ∈ H2 : sup

a∈D‖f ◦ ϕa − f (a)‖H2 <∞

Where ϕa(z) = z−a1−az , a ∈ D.

Bloch space

{f ∈ Hol(D) : sup

z∈D(1− |z|2)|f ′(z)| <∞

BMOA =

{f ∈ H2 : sup

a∈D‖f ◦ ϕa − f (a)‖H2 <∞

Bloch space

{f ∈ Hol(D) : sup

z∈D(1− |z|2)|f ′(z)| <∞

BMOA =

{f ∈ H2 : sup

a∈D‖f ◦ ϕa − f (a)‖H2 <∞

Bloch space

{f ∈ Hol(D) : sup

z∈D(1− |z|2)|f ′(z)| <∞

Hilbert matrixA generalized Hilbert matrixIntegral operatorCarleson measures

Hilbert matrixLet H be the Hilbert matrix,

n + k + 1

)n,k≥0

1 1/2 1/3 1/4 . . .

1/2 1/3 1/4 1/5 . . .1/3 1/4 1/5 1/6 . . .1/4 1/5 1/6 1/7 . . .

......

n + k + 1

)n,k≥0

1 1/2 1/3 1/4 . . .

1/2 1/3 1/4 1/5 . . .1/3 1/4 1/5 1/6 . . .1/4 1/5 1/6 1/7 . . .

......

n + k + 1

)n,k≥0

1 1/2 1/3 1/4 . . .

1/2 1/3 1/4 1/5 . . .1/3 1/4 1/5 1/6 . . .1/4 1/5 1/6 1/7 . . .

......

Hilbert matrixThe Hilbert matrix H can be viewed as an operator betweensequence spaces.

H ({an}∞n=0) =

1 1/2 1/3 1/4 . . .

1/2 1/3 1/4 1/5 . . .1/3 1/4 1/5 1/6 . . .1/4 1/5 1/6 1/7 . . .

......

a0a1a2a3...

{an}∞n=0 7→

{ ∞∑k=0

n + k + 1

}∞n=0

H ({an}∞n=0) =

1 1/2 1/3 1/4 . . .

1/2 1/3 1/4 1/5 . . .1/3 1/4 1/5 1/6 . . .1/4 1/5 1/6 1/7 . . .

......

a0a1a2a3...

{an}∞n=0 7→

{ ∞∑k=0

n + k + 1

}∞n=0

H ({an}∞n=0) =

1 1/2 1/3 1/4 . . .

1/2 1/3 1/4 1/5 . . .1/3 1/4 1/5 1/6 . . .1/4 1/5 1/6 1/7 . . .

......

a0a1a2a3...

{an}∞n=0 7→

{ ∞∑k=0

n + k + 1

}∞n=0

H ({an}∞n=0) =

1 1/2 1/3 1/4 . . .

1/2 1/3 1/4 1/5 . . .1/3 1/4 1/5 1/6 . . .1/4 1/5 1/6 1/7 . . .

......

a0a1a2a3...

{an}∞n=0 7→

{ ∞∑k=0

n + k + 1

}∞n=0

In the same way we can consider H as an operator in Hol(D)multiplicating the matrix by the sequence of Taylor coefficients

of a function f (z) =∞∑

n=0anzn ∈ Hol(D).

We define formally the operator in Hol(D)

H(f )(z) =∞∑

( ∞∑k=0

n + k + 1

n=0anzn ∈ Hol(D).

H(f )(z) =∞∑

( ∞∑k=0

n + k + 1

n=0anzn ∈ Hol(D).

H(f )(z) =∞∑

( ∞∑k=0

n + k + 1

Hilbert matrix as an operator

The operator H is well defined on H1.The operator H : Hp → Hp is bounded if 1 < p <∞,(Diamantopoulos & Siskakis, 2000).Dostanic, Jevtic & Vukotic (2008) found the exact norm of H asan operator from Hp to Hp (1 < p <∞).However, H is not bounded on H1 and neither on H∞.

A generalized Hilbert matrix

Let µ be a finite positive Borel measure on [0,1).Let Hµ = (µn,k )n,k≥0 be the Hankel matrix with entries

µn,k =

∫[0,1)

tn+k dµ(t).

µ0 µ1 µ2 µ3 · · ·µ1 µ2 µ3 µ4 · · ·µ2 µ3 µ4 µ5 · · ·µ3 µ4 µ5 µ6 · · ·...

......

.... . .

If µ is the Lebesgue measure on the interval [0,1) we get theclassical Hilbert matrix.

µn,k =

∫[0,1)

tn+k dµ(t).

......

.... . .

µn,k =

∫[0,1)

tn+k dµ(t).

......

.... . .

µn,k =

∫[0,1)

tn+k dµ(t).

......

.... . .

µn,k =

∫[0,1)

tn+k dµ(t).

......

.... . .

The matrix Hµ induces formally an operator on Hol(D) in thesame way than H:

Hµ(f )(z) =∞∑

( ∞∑k=0

µn,kak

The matrix Hµ induces formally an operator on Hol(D) in thesame way than H:

Hµ(f )(z) =∞∑

( ∞∑k=0

µn,kak

Integral operator

For a finite positive Borel measure on [0,1) µ we also definethe integral operator

Iµ(f )(z) =

∫[0,1)

f (t)1− tz

dµ(t),

when the right side has sense and it defines an analyticfunction.Hµ and Iµ are closely related. If f ∈ Hol(D) is good enoughthen Hµ(f ) = Iµ(f ).

Integral operator

Iµ(f )(z) =

∫[0,1)

f (t)1− tz

dµ(t),

Integral operator

Iµ(f )(z) =

∫[0,1)

f (t)1− tz

dµ(t),

DefinitionLet I be an interval of T. We define the Carleson squareassociated to I asS(I) = {reiθ : eiθ ∈ I, 1− |I|2π ≤ r < 1}.

0 S( I ) I

DefinitionLet µ be a finite measure on D. µ is a Carleson measure ifthere is a constant C > 0 such that

µ (S(I)) ≤ C|I| for every I ⊂ T interval.

Theorem (Carleson, 1962)Let µ be a finite measure on D. Then µ is a Carleson measureif and only if there exist a constant C > 0 such that∫

D|f (z)|dµ(z) ≤ C‖f‖H1 for all f ∈ H1.

Logarithmic Carleson measuresLet µ be a positive Borel measure on D, 0 ≤ α <∞, and0 < s <∞ we say that µ is an α-logarithmic s-Carlesonmeasure if there exists a positive constant C such that

µ (S(I))

2π|I|

)α≤ C|I|s, for any interval I ⊂ T.

Logarithmic Carleson measuresLet µ be a positive Borel measure on D, 0 ≤ α <∞, and0 < s <∞ we say that µ is an α-logarithmic s-Carlesonmeasure if there exists a positive constant C such that

µ (S(I))

2π|I|

)α≤ C|I|s, for any interval I ⊂ T.

Widom (1966) and Power (1980) characterized those positiveBorel measures on [0,1) such that Hµ is bounded (or compact)from H2 into itself.

Galanopoulos and Peláez (2010) characterized those positiveBorel measures on [0,1) such that Hµ is bounded (or compact)in the Hardy space H1.

Chatzifountas, Girela and Peláez (2013) described thosepositive Borel measures on [0,1) such that Hµ is bounded (orcompact) from Hp into Hq with 0 < p,q <∞.

Girela and M. (2017) described those positive Borel measureson [0,1) such that Hµ is bounded on conformally invariantspaces.

Theorem (Girela, M.)

Let µ be a positive Borel measure on [0,1) such that∫[0,1) log 2

1−t dµ(t) < ∞. Then the following three conditionsare equivalent:

(i) The operator Iµ is bounded from B into BMOA.(ii) The operator Iµ is bounded from BMOA into itself.(iii) The measure µ is a 1-logarithmic 1-Carleson measure.Moreover, if (i) holds, then the operator Hµ is also well definedon the Bloch space and

Hµ(f ) = Iµ(f ), for all f ∈ B,

and hence the operator Hµ is bounded from B into BMOA.

Qs spacesFor 0 ≤ s <∞ we define the space Qs as

{f ∈ Hol(D) : sup

∫D|f ′(z)|2(1− |ϕa(z)|2)s dA(z) <∞

D ( Qs1 ( Qs2 ( BMOA = Q1 ( B = Qs, 0 < s1 < s2 < 1 < s.

{f ∈ Hol(D) : sup

∫D|f ′(z)|2(1− |ϕa(z)|2)s dA(z) <∞

{f ∈ Hol(D) : sup

∫D|f ′(z)|2(1− |ϕa(z)|2)s dA(z) <∞

Let µ be a positive Borel measure on [0,1) and let0 < s1, s2 <∞. Then the following conditions are equivalent.

(i) The operator Iµ is well defined in Qs1 and, furthermore, it isa bounded operator from Qs1 into Qs2 .

(ii) The operator Hµ is well defined in Qs1 and, furthermore, itis a bounded operator from Qs1 into Qs2 .

(iii) The measure µ is a 1-logarithmic 1-Carleson measure.

Let µ be a positive Borel measure on [0,1) and let0 < s1, s2 <∞. Then the following conditions are equivalent.

(i) The operator Iµ is well defined in Qs1 and, furthermore, it isa bounded operator from Qs1 into Qs2 .

(ii) The operator Hµ is well defined in Qs1 and, furthermore, itis a bounded operator from Qs1 into Qs2 .

(iii) The measure µ is a 1-logarithmic 1-Carleson measure.

Λ21/2 =

{f ∈ Hol(D) : M2(r , f ′) = O

(1− r)1/2

Let µ be a positive Borel measure on [0,1) and let X be aBanach space of analytic functions in D with Λ2

1/2 ⊂ X ⊂ B.Then the following conditions are equivalent.

(i) The operator Hµ is well defined in X and, furthermore, it isa bounded operator from X into the Bloch space B.

(ii) The operator Hµ is well defined in X and, furthermore, it isa bounded operator from X into Λ2

(iii) The measure µ is a 1-logarithmic 1-Carleson measure.(iv)

∫[0,1) tn log 1

1−t dµ(t) = O(1

Λ21/2 =

{f ∈ Hol(D) : M2(r , f ′) = O

(1− r)1/2

∫[0,1) tn log 1

1−t dµ(t) = O(1

Λ21/2 =

{f ∈ Hol(D) : M2(r , f ′) = O

(1− r)1/2

∫[0,1) tn log 1

1−t dµ(t) = O(1

Integral modulus of continuity

If f ∈ Hol(D) has a non-tangential limit f (eiθ) at almost everyeiθ ∈ T and δ > 0, we define for 1 ≤ p <∞

ωp(δ, f ) = sup0<|t |≤δ

∫ π

∣∣∣f (ei(θ+t))− f (eiθ)∣∣∣p dθ

and for p =∞ we define

ω∞(δ, f ) = sup0<|t |≤δ

(ess. supθ∈[−π,π]

|f (ei(θ+t))− f (eiθ)|

ωp(δ, f ) = sup0<|t |≤δ

∫ π

ω∞(δ, f ) = sup0<|t |≤δ

ωp(δ, f ) = sup0<|t |≤δ

∫ π

ω∞(δ, f ) = sup0<|t |≤δ

Mean Lipschitz spacesGiven 1 ≤ p ≤ ∞ and 0 < α ≤ 1, we define the mean Lipschitzspace Λp

Λpα =

{f ∈ Hol(D) : ∃ f (eiθ) a.e. θ, ωp(δ, f ) = O(δα), as δ → 0

Theorem (Hardy & Littlewood, 1932)

If 1 ≤ p ≤ ∞ and 0 < α ≤ 1 then we have that Λpα ⊂ Hp and

Λpα =

{f ∈ Hol(D) : Mp(r , f ′) = O

(1− r)1−α

Λpα =

{f ∈ Hol(D) : Mp(r , f ′) = O

(1− r)1−α

Λpα =

{f ∈ Hol(D) : Mp(r , f ′) = O

(1− r)1−α

Λpα =

{f ∈ Hol(D) : Mp(r , f ′) = O

(1− r)1−α

If 1 < p <∞ and α > 1p then Λp

α ⊂ A ⊂ H∞.

If 1 < p <∞ and α = 1p then f (z) = log

)∈ Λp

Theorem (Bourdon, Shapiro, Sledd, 1989)

Λp1/p ⊂ Λq

1/q ⊂ BMOA ⊂ B, 1 ≤ p < q <∞.

This result is sharp in a very strong sense.

α ⊂ A ⊂ H∞.

)∈ Λp

Λp1/p ⊂ Λq

1/q ⊂ BMOA ⊂ B, 1 ≤ p < q <∞.

α ⊂ A ⊂ H∞.

)∈ Λp

Λp1/p ⊂ Λq

1/q ⊂ BMOA ⊂ B, 1 ≤ p < q <∞.

Generalization of Λpα spaces

Let ω : [0, π]→ [0,∞) be a continuous and increasing functionwith ω(0) = 0 and ω(t) > 0 if t > 0.Then, for 1 ≤ p ≤ ∞, the mean Lipschitz space Λ(p, ω) isdefined as

Λ(p, ω) = {f ∈ Hp : ωp(δ, f ) = O(ω(δ)), as δ → 0} .

With this notation Λpα = Λ(p, δα).

Dini conditionWe say that ω satisfies the Dini condition if there exists apositive constant C such that∫ δ

ω(t)t

dt ≤ Cω(δ), 0 < δ < 1.

Condition b1

We say that ω satisfies the b1 condition if there exists a positiveconstant C such that∫ π

ω(t)t2 dt ≤ C

ω(δ)

δ, 0 < δ < 1.

ω(t)t

dt ≤ Cω(δ), 0 < δ < 1.

Condition b1

ω(t)t2 dt ≤ C

ω(δ)

δ, 0 < δ < 1.

ω(t)t

dt ≤ Cω(δ), 0 < δ < 1.

Condition b1

ω(t)t2 dt ≤ C

ω(δ)

δ, 0 < δ < 1.

Admissible weights

AW = Dini ∩ b1.

Theorem (Blasco & de Souza, 1990)If 1 ≤ p ≤ ∞ and ω ∈ AW then,

Λ(p, ω) =

{f ∈ Hol(D) : Mp(r , f ′) = O

(ω(1− r)

1− r

), as r → 1

Admissible weights

AW = Dini ∩ b1.

Λ(p, ω) =

{f ∈ Hol(D) : Mp(r , f ′) = O

(ω(1− r)

1− r

), as r → 1

Admissible weights

AW = Dini ∩ b1.

Λ(p, ω) =

{f ∈ Hol(D) : Mp(r , f ′) = O

(ω(1− r)

1− r

), as r → 1

α ⊂ A ⊂ H∞.

)∈ Λp

Λp1/p ⊂ Λq

1/q ⊂ BMOA ⊂ B, 1 ≤ p < q <∞.

Theorem (Girela, 1997)

If 1 < p <∞ and ω ∈ AW with ω(δ)

δ1/p ↗∞ when δ ↘ 0 then

Λ(p, ω) 6⊂ B.

α ⊂ A ⊂ H∞.

)∈ Λp

Λp1/p ⊂ Λq

1/q ⊂ BMOA ⊂ B, 1 ≤ p < q <∞.

Theorem (Girela, 1997)

If 1 < p <∞ and ω ∈ AW with ω(δ)

δ1/p ↗∞ when δ ↘ 0 then

Λ(p, ω) 6⊂ B.

TheoremLet µ be a positive Borel measure on [0,1) and let X be aBanach space of analytic functions in D with Λ2

∫[0,1) tn log 1

1−t dµ(t) = O(1

X can be BMOA, Qs for s > 0 or Λp1/p for 2 ≤ p <∞.

∫[0,1) tn log 1

1−t dµ(t) = O(1

∫[0,1) tn log 1

1−t dµ(t) = O(1

LemmaLet f ∈ Hol(D) be of the form f (z) =

∑∞n=0 anzn with {an}∞n=0

being a decreasing sequence of nonnegative numbers.If X is a subspace of Hol(D) with

Λ21/2 ⊂ X ⊂ B,

f ∈ X ⇔ an = O(

Lemma (M.)

Let f ∈ Hol(D) be of the form f (z) =∑∞

n=0 anzn with {an}∞n=0being a decreasing sequence of nonnegative numbers.If 1 < p <∞ and X is a subspace of Hol(D) with

Λp1/p ⊂ X ⊂ B

f ∈ X ⇔ an = O(

Theorem (M.)Suppose that 1 < p <∞. Let µ be a positive Borel measure on[0,1) and let X be a Banach space of analytic functions in Dwith Λp

1/p ⊂ X ⊂ B. Then the following conditions areequivalent.

(ii) The operator Hµ is well defined in X and, furthermore, it isa bounded operator from X into Λp

∫[0,1) tn log 1

1−t dµ(t) = O(1

Corollary

Let µ be a positive Borel measure on [0,1) and 1 < p <∞.Then the operator Hµ is well defined in Λp

1/p and, furthermore, itis a bounded operator from Λp

1/p into itself if and only if µ is a1-logarithmic 1-Carleson measure.

Corollary

Let µ be a positive Borel measure on [0,1) and 1 < p <∞.Then the operator Hµ is well defined in Λp

1/p and, furthermore, itis a bounded operator from Λp

1/p into itself if and only if µ is a1-logarithmic 1-Carleson measure.

Lemma (M.)

Let 1 < p <∞, ω ∈ AW and let f (z) =∑∞

n=0 anzn with {an}∞n=0being a decreasing sequence of nonnegative numbers. Then

f ∈ Λ(p, ω) ⇔ an = O(ω(1/n)

n1−1/p

Lemma (M.)

Let 1 < p <∞, ω ∈ AW and let f (z) =∑∞

n=0 anzn with {an}∞n=0being a decreasing sequence of nonnegative numbers. Then

f ∈ Λ(p, ω) ⇔ an = O(ω(1/n)

n1−1/p

Theorem (M.)

Let 1 < p <∞, ω ∈ AW with ω(δ)

δ1/p ↗∞ when δ ↘ 0. Thefollowing conditions are equivalent:

(i) The operator Hµ is well defined in Λ(p, ω) and,furthermore, it is a bounded operator from Λ(p, ω) intoitself.

(ii) The measure µ is a Carleson measure.

Remark

Hµ : Λp1/p → Λp

1/p bounded ⇔ µ is a 1-log 1-Carlesonmeasure.Hµ : Λ(p, ω)→ Λ(p, ω) bounded ⇔ µ is a Carlesonmeasure.

Theorem (M.)

Remark

Theorem (M.)

Remark

Theorem (M.)

Remark

Theorem (M.)

Remark

(i)⇒ (ii) Suppose that Hµ : Λ(p, ω)→ Λ(p, ω) is bounded. Bythe Lemma

f (z) =∞∑

ω(1/n)

n1−1/p zn, f ∈ Λ(p, ω), so Hµ(f ) ∈ Λ(p, ω).

Hµ(f )(z) =∞∑

( ∞∑k=1

ω(1/k)

k1−1/p µn+k

Since∑∞

k=1ω(1/k)k1−1/p µn+k ↘ 0, using again the Lemma we can

prove that µ is a Carleson measure.

f (z) =∞∑

ω(1/n)

Hµ(f )(z) =∞∑

( ∞∑k=1

ω(1/k)

k1−1/p µn+k

Since∑∞

f (z) =∞∑

ω(1/n)

Hµ(f )(z) =∞∑

( ∞∑k=1

ω(1/k)

k1−1/p µn+k

Since∑∞

f (z) =∞∑

ω(1/n)

Hµ(f )(z) =∞∑

( ∞∑k=1

ω(1/k)

k1−1/p µn+k

Since∑∞

f (z) =∞∑

ω(1/n)

Hµ(f )(z) =∞∑

( ∞∑k=1

ω(1/k)

k1−1/p µn+k

Since∑∞

f (z) =∞∑

ω(1/n)

Hµ(f )(z) =∞∑

( ∞∑k=1

ω(1/k)

k1−1/p µn+k

Since∑∞

(ii)⇒ (i) Suppose that µ is a Carleson measure. Using thefollowing Lemma (Girela & González, 2000)

f ∈ Λ(p, ω)⇒ |f (z)| . ω(1− |z|)(1− |z|)1/p , z ∈ D.

We can prove that

f ∈ Λ(p, ω)⇒∫[0,1)

|f (t)||1− tz|

dµ(t) <∞.

So if f ∈ Λ(p, ω) then Iµ(f ) and Hµ(f ) are well defined and

Hµ(f )(z) = Iµ(f )(z), z ∈ D.

f ∈ Λ(p, ω)⇒ |f (z)| . ω(1− |z|)(1− |z|)1/p , z ∈ D.

We can prove that

f ∈ Λ(p, ω)⇒∫[0,1)

|f (t)||1− tz|

dµ(t) <∞.

Hµ(f )(z) = Iµ(f )(z), z ∈ D.

f ∈ Λ(p, ω)⇒ |f (z)| . ω(1− |z|)(1− |z|)1/p , z ∈ D.

We can prove that

f ∈ Λ(p, ω)⇒∫[0,1)

|f (t)||1− tz|

dµ(t) <∞.

Hµ(f )(z) = Iµ(f )(z), z ∈ D.

f ∈ Λ(p, ω)⇒ |f (z)| . ω(1− |z|)(1− |z|)1/p , z ∈ D.

We can prove that

f ∈ Λ(p, ω)⇒∫[0,1)

|f (t)||1− tz|

dµ(t) <∞.

Hµ(f )(z) = Iµ(f )(z), z ∈ D.

f ∈ Λ(p, ω)⇒ |f (z)| . ω(1− |z|)(1− |z|)1/p , z ∈ D.

We can prove that

f ∈ Λ(p, ω)⇒∫[0,1)

|f (t)||1− tz|

dµ(t) <∞.

Hµ(f )(z) = Iµ(f )(z), z ∈ D.

ProofUsing again the Lemma, the Minkowski inequality and doingsome work we obtain that

Mp(r , Iµ(f )′

∫ π

∣∣∣∣∣∫[0,1)

tf (t)(1− treiθ)2 dµ(t)

∣∣∣∣∣p

.∫[0,1)|f (t)|

(∫ π

dθ|1− treiθ|2p

dµ(t)

.∫[0,1)

|f (t)|(1− tr)2−1/p dµ(t)

.∫[0,1)

ω(1− t)(1− t)1/p(1− tr)2−1/p dµ(t)

.ω(1− r)

(1− r).

Mp(r , Iµ(f )′

∫ π

∣∣∣∣∣∫[0,1)

∣∣∣∣∣p

.∫[0,1)|f (t)|

(∫ π

dθ|1− treiθ|2p

dµ(t)

.∫[0,1)

|f (t)|(1− tr)2−1/p dµ(t)

.∫[0,1)

ω(1− t)(1− t)1/p(1− tr)2−1/p dµ(t)

.ω(1− r)

(1− r).

Mp(r , Iµ(f )′

∫ π

∣∣∣∣∣∫[0,1)

∣∣∣∣∣p

.∫[0,1)|f (t)|

(∫ π

dθ|1− treiθ|2p

dµ(t)

.∫[0,1)

|f (t)|(1− tr)2−1/p dµ(t)

.∫[0,1)

ω(1− t)(1− t)1/p(1− tr)2−1/p dµ(t)

.ω(1− r)

(1− r).

Mp(r , Iµ(f )′

∫ π

∣∣∣∣∣∫[0,1)

∣∣∣∣∣p

.∫[0,1)|f (t)|

(∫ π

dθ|1− treiθ|2p

dµ(t)

.∫[0,1)

|f (t)|(1− tr)2−1/p dµ(t)

.∫[0,1)

ω(1− t)(1− t)1/p(1− tr)2−1/p dµ(t)

.ω(1− r)

(1− r).

Mp(r , Iµ(f )′

∫ π

∣∣∣∣∣∫[0,1)

∣∣∣∣∣p

.∫[0,1)|f (t)|

(∫ π

dθ|1− treiθ|2p

dµ(t)

.∫[0,1)

|f (t)|(1− tr)2−1/p dµ(t)

.∫[0,1)

ω(1− t)(1− t)1/p(1− tr)2−1/p dµ(t)

.ω(1− r)

(1− r).

Mp(r , Iµ(f )′

∫ π

∣∣∣∣∣∫[0,1)

∣∣∣∣∣p

.∫[0,1)|f (t)|

(∫ π

dθ|1− treiθ|2p

dµ(t)

.∫[0,1)

|f (t)|(1− tr)2−1/p dµ(t)

.∫[0,1)

ω(1− t)(1− t)1/p(1− tr)2−1/p dµ(t)

.ω(1− r)

(1− r).

Mp(r , Iµ(f )′

∫ π

∣∣∣∣∣∫[0,1)

∣∣∣∣∣p

.∫[0,1)|f (t)|

(∫ π

dθ|1− treiθ|2p

dµ(t)

.∫[0,1)

|f (t)|(1− tr)2−1/p dµ(t)

.∫[0,1)

ω(1− t)(1− t)1/p(1− tr)2−1/p dµ(t)

.ω(1− r)

(1− r).

THANK YOU!

A generalized Hilbert operator acting on mean Lipschitz...

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