Operators related to the Jacobi setting, for all admissible - - PowerPoint PPT Presentation

Operators related to the Jacobi setting, for all admissible parameter values

Peter Sjögren University of Gothenburg Joint work with A. Nowak and T. Szarek

Alba, June 2013

() 1 / 18

Let Pα,β

n

be the classical Jacobi polynomials, seen via the transformation x = cos θ as functions of θ ∈ [0, π]. Here α, β > −1.

() 2 / 18

Let Pα,β

n

be the classical Jacobi polynomials, seen via the transformation x = cos θ as functions of θ ∈ [0, π]. Here α, β > −1. The Pα,β

n

are orthogonal with respect to the measure dµα,β(θ) =

• sin θ

2 2α+1 cos θ 2 2β+1 dθ in [0, π], and we take them to be normalized in L2(dµα,β).

() 2 / 18

Let Pα,β

n

be the classical Jacobi polynomials, seen via the transformation x = cos θ as functions of θ ∈ [0, π]. Here α, β > −1. The Pα,β

n

are orthogonal with respect to the measure dµα,β(θ) =

• sin θ

2 2α+1 cos θ 2 2β+1 dθ in [0, π], and we take them to be normalized in L2(dµα,β). They are eigenfunctions of the Jacobi operator J α,β = − d2 dθ2 − α − β + (α + β + 1) cos θ sin θ d dθ + α + β + 1 2 2 , with eigenvalues

• n + α+β+1

2

2 .

() 2 / 18

The corresponding Poisson operator exp(−t √ J α,β), t > 0, can be defined spectrally and has the integral kernel Hα,β

t

(θ, ϕ) =

∞

• n=0

e−t|n+ α+β+1

2

|Pα,β

n

(θ)Pα,β

n

(ϕ), called the Jacobi-Poisson kernel.

() 3 / 18

The corresponding Poisson operator exp(−t √ J α,β), t > 0, can be defined spectrally and has the integral kernel Hα,β

t

(θ, ϕ) =

∞

• n=0

e−t|n+ α+β+1

2

|Pα,β

n

(θ)Pα,β

n

(ϕ), called the Jacobi-Poisson kernel. By means of Hα,β

t

, one can express the kernels of many operators associated with the Jacobi setting.

() 3 / 18

The corresponding Poisson operator exp(−t √ J α,β), t > 0, can be defined spectrally and has the integral kernel Hα,β

t

(θ, ϕ) =

∞

• n=0

e−t|n+ α+β+1

2

|Pα,β

n

(θ)Pα,β

n

(ϕ), called the Jacobi-Poisson kernel. By means of Hα,β

t

, one can express the kernels of many operators associated with the Jacobi setting. We list some of these operators and their kernels.

() 3 / 18

Riesz transforms DN(J α,β)−N/2 of order N ≥ 1, with kernels cN ∞ ∂N

θ Hα,β t

(θ, ϕ)tN−1 dt.

() 4 / 18

Riesz transforms DN(J α,β)−N/2 of order N ≥ 1, with kernels cN ∞ ∂N

θ Hα,β t

(θ, ϕ)tN−1 dt. Laplace transform type multipliers m √ J α,β

• with m

given by m(z) = zLφ(z), where L denotes the Laplace transform and φ ∈ L∞(R+).

() 4 / 18

Riesz transforms DN(J α,β)−N/2 of order N ≥ 1, with kernels cN ∞ ∂N

θ Hα,β t

(θ, ϕ)tN−1 dt. Laplace transform type multipliers m √ J α,β

• with m

given by m(z) = zLφ(z), where L denotes the Laplace transform and φ ∈ L∞(R+). The kernel is − ∞ φ(t) ∂tHα,β

t

(θ, ϕ) dt.

() 4 / 18

Riesz transforms DN(J α,β)−N/2 of order N ≥ 1, with kernels cN ∞ ∂N

θ Hα,β t

(θ, ϕ)tN−1 dt. Laplace transform type multipliers m √ J α,β

• with m

given by m(z) = zLφ(z), where L denotes the Laplace transform and φ ∈ L∞(R+). The kernel is − ∞ φ(t) ∂tHα,β

t

(θ, ϕ) dt. With φ(t) = const. t−2iγ, γ = 0, this gives the imaginary power (J α,β)iγ of the Jacobi operator.

() 4 / 18

Riesz transforms DN(J α,β)−N/2 of order N ≥ 1, with kernels cN ∞ ∂N

θ Hα,β t

(θ, ϕ)tN−1 dt. Laplace transform type multipliers m √ J α,β

• with m

given by m(z) = zLφ(z), where L denotes the Laplace transform and φ ∈ L∞(R+). The kernel is − ∞ φ(t) ∂tHα,β

t

(θ, ϕ) dt. With φ(t) = const. t−2iγ, γ = 0, this gives the imaginary power (J α,β)iγ of the Jacobi operator. Laplace-Stieltjes transform type multipliers m √ J α,β

• ,

where m = Lν and ν is a signed or complex Borel measure on (0, ∞) satisfying

• e−t|α+β+1|/2 d|ν|(t) < ∞. The kernel is now
• (0,∞)

Hα,β

t

(θ, ϕ) dν(t).

() 4 / 18

These kernels are scalar-valued. But allowing kernels taking values in a Banach space B, one can write more operator kernels.

() 5 / 18

These kernels are scalar-valued. But allowing kernels taking values in a Banach space B, one can write more operator kernels. Examples: The Poisson maximal operator, with kernel

• Hα,β

t

(θ, ϕ)

• t>0

and B = L∞(R+; dt).

() 5 / 18

These kernels are scalar-valued. But allowing kernels taking values in a Banach space B, one can write more operator kernels. Examples: The Poisson maximal operator, with kernel

• Hα,β

t

(θ, ϕ)

• t>0

and B = L∞(R+; dt). The mixed square functions, with kernels

• ∂N

θ ∂M t Hα,β t

(θ, ϕ)

• t>0

and B = L2(R+; t2M+2N−1dt). Here M + N > 0.

() 5 / 18

Old result:

Theorem (Nowak and Sjögren, J.F.A.A. 2012)

For α, β ≥ −1/2, the following among the operators listed above are bounded on Lp(dµα,β), 1 < p < ∞, and of weak type (1,1) for µα,β:

() 6 / 18

Old result:

Theorem (Nowak and Sjögren, J.F.A.A. 2012)

For α, β ≥ −1/2, the following among the operators listed above are bounded on Lp(dµα,β), 1 < p < ∞, and of weak type (1,1) for µα,β: The Riesz transforms, the imaginary powers of the Jacobi operator, the maximal operator and the mixed square functions.

() 6 / 18

Old result:

Theorem (Nowak and Sjögren, J.F.A.A. 2012)

For α, β ≥ −1/2, the following among the operators listed above are bounded on Lp(dµα,β), 1 < p < ∞, and of weak type (1,1) for µα,β: The Riesz transforms, the imaginary powers of the Jacobi operator, the maximal operator and the mixed square functions. These operators are also bounded between the corresponding weighted spaces with a weight in the Muckenhoupt class Aα,β

p

, defined with respect to the measure µα,β.

() 6 / 18

Old result:

Theorem (Nowak and Sjögren, J.F.A.A. 2012)

For α, β ≥ −1/2, the following among the operators listed above are bounded on Lp(dµα,β), 1 < p < ∞, and of weak type (1,1) for µα,β: The Riesz transforms, the imaginary powers of the Jacobi operator, the maximal operator and the mixed square functions. These operators are also bounded between the corresponding weighted spaces with a weight in the Muckenhoupt class Aα,β

p

, defined with respect to the measure µα,β. New result:

Theorem (Nowak, Sjögren and Szarek, 2013)

The preceding theorem holds for all α, β > −1 and all the operators listed above.

() 6 / 18

To prove both theorems, one shows that the operators are well-behaved Calderón-Zygmund operators

() 7 / 18

To prove both theorems, one shows that the operators are well-behaved Calderón-Zygmund operators

• n the space of

homogeneous type defined by the interval [0, π] with the measure µα,β and the ordinary distance.

() 7 / 18

To prove both theorems, one shows that the operators are well-behaved Calderón-Zygmund operators

• n the space of

homogeneous type defined by the interval [0, π] with the measure µα,β and the ordinary distance. The operators can be seen to be bounded on L2(dµα,β) (or L∞(dµα,β) in the case of the maximal operator).

() 7 / 18

To prove both theorems, one shows that the operators are well-behaved Calderón-Zygmund operators

• n the space of

homogeneous type defined by the interval [0, π] with the measure µα,β and the ordinary distance. The operators can be seen to be bounded on L2(dµα,β) (or L∞(dµα,β) in the case of the maximal operator). So the main crux is to verify that their off-diagonal kernels satisfy the standard estimates in this space.

() 7 / 18

To prove both theorems, one shows that the operators are well-behaved Calderón-Zygmund operators

• n the space of

homogeneous type defined by the interval [0, π] with the measure µα,β and the ordinary distance. The operators can be seen to be bounded on L2(dµα,β) (or L∞(dµα,β) in the case of the maximal operator). So the main crux is to verify that their off-diagonal kernels satisfy the standard estimates in this space. It also requires some effort to prove that the kernels are really as described above.

() 7 / 18

The argument for the old theorem is based on the following integral formula for the Poisson kernel Hα,β

t

(θ, ϕ) =

• Ψα,β(t, θ, ϕ, u, v) dΠα(u) dΠβ(v),

valid for α, β > −1/2,

() 8 / 18

The argument for the old theorem is based on the following integral formula for the Poisson kernel Hα,β

t

(θ, ϕ) =

• Ψα,β(t, θ, ϕ, u, v) dΠα(u) dΠβ(v),

valid for α, β > −1/2, where Ψα,β(t, θ, ϕ, u, v) = cα,β sinh t

2

(cosh t

2 − 1 + q(θ, ϕ, u, v))α+β+2 ,

() 8 / 18

The argument for the old theorem is based on the following integral formula for the Poisson kernel Hα,β

t

(θ, ϕ) =

• Ψα,β(t, θ, ϕ, u, v) dΠα(u) dΠβ(v),

valid for α, β > −1/2, where Ψα,β(t, θ, ϕ, u, v) = cα,β sinh t

2

(cosh t

2 − 1 + q(θ, ϕ, u, v))α+β+2 ,

and q(θ, ϕ, u, v) = 1 − u sin θ 2 sin ϕ 2 − v cos θ 2 cos ϕ 2 ≥ 0,

() 8 / 18

The argument for the old theorem is based on the following integral formula for the Poisson kernel Hα,β

t

(θ, ϕ) =

• Ψα,β(t, θ, ϕ, u, v) dΠα(u) dΠβ(v),

valid for α, β > −1/2, where Ψα,β(t, θ, ϕ, u, v) = cα,β sinh t

2

(cosh t

2 − 1 + q(θ, ϕ, u, v))α+β+2 ,

and q(θ, ϕ, u, v) = 1 − u sin θ 2 sin ϕ 2 − v cos θ 2 cos ϕ 2 ≥ 0, and the probability measures Πα and Πβ are defined by dΠα(u) = Γ(α + 1) √π Γ(α + 1/2) (1 − u2)α−1/2 du, −1 < u < 1.

() 8 / 18

We want to extend the integral formula from α, β > −1/2 to α, β > −1, by analytic continuation.

() 9 / 18

We want to extend the integral formula from α, β > −1/2 to α, β > −1, by analytic continuation. The defining expression for the measure dΠα makes sense for all α with ℜα > −1 except α = −1/2 and produces a local, complex measure in (−1, 1) with infinite mass near ±1.

() 9 / 18

We want to extend the integral formula from α, β > −1/2 to α, β > −1, by analytic continuation. The defining expression for the measure dΠα makes sense for all α with ℜα > −1 except α = −1/2 and produces a local, complex measure in (−1, 1) with infinite mass near ±1. At the point α = −1/2, there is a weak limit dΠ−1/2 = 1

2(δ1 + δ−1).

() 9 / 18

We want to extend the integral formula from α, β > −1/2 to α, β > −1, by analytic continuation. The defining expression for the measure dΠα makes sense for all α with ℜα > −1 except α = −1/2 and produces a local, complex measure in (−1, 1) with infinite mass near ±1. At the point α = −1/2, there is a weak limit dΠ−1/2 = 1

2(δ1 + δ−1).

But the integral against Ψα,β(t, θ, ϕ, u, v) will diverge at the endpoints if α or β, or their real parts, go below -1/2.

() 9 / 18

We want to extend the integral formula from α, β > −1/2 to α, β > −1, by analytic continuation. The defining expression for the measure dΠα makes sense for all α with ℜα > −1 except α = −1/2 and produces a local, complex measure in (−1, 1) with infinite mass near ±1. At the point α = −1/2, there is a weak limit dΠ−1/2 = 1

2(δ1 + δ−1).

But the integral against Ψα,β(t, θ, ϕ, u, v) will diverge at the endpoints if α or β, or their real parts, go below -1/2. In order to get convergence, we first restrict the integral to u, v > 0, using the even part Ψα,β

E (t, θ, ϕ, u, v) = 1

4

• ±
• ±

Ψα,β(t, θ, ϕ, ±u, ±v)

• f the integrand.

() 9 / 18

Since the measures dΠα are even, the formula can be written Hα,β

t

(θ, ϕ) = 4

• (0,1]2 Ψα,β

E (t, θ, ϕ, u, v) dΠα(u) dΠβ(v).

() 10 / 18

Since the measures dΠα are even, the formula can be written Hα,β

t

(θ, ϕ) = 4

• (0,1]2 Ψα,β

E (t, θ, ϕ, u, v) dΠα(u) dΠβ(v).

By subtracting an adding the values of Ψα,β

E (t, θ, ϕ, u, v) at the

endpoints u, v = 1, we can rewrite this as

() 10 / 18

Since the measures dΠα are even, the formula can be written Hα,β

t

(θ, ϕ) = 4

• (0,1]2 Ψα,β

E (t, θ, ϕ, u, v) dΠα(u) dΠβ(v).

By subtracting an adding the values of Ψα,β

E (t, θ, ϕ, u, v) at the

endpoints u, v = 1, we can rewrite this as Hα,β

t

(θ, ϕ) = 4

• (0,1]2
• Ψα,β

E (t, θ, ϕ, u, v) − Ψα,β E (t, θ, ϕ, u, 1)

−Ψα,β

E (t, θ, ϕ, 1, v) + Ψα,β E (t, θ, ϕ, 1, 1)

• dΠα(u) dΠβ(v)

+ 2

• (0,1]
• Ψα,β

E (t, θ, ϕ, u, 1) − Ψα,β E (t, θ, ϕ, 1, 1)

• dΠα(u)

+ 2

• (0,1]
• Ψα,β

E (t, θ, ϕ, 1, v) − Ψα,β E (t, θ, ϕ, 1, 1)

• dΠβ(v)

+ Ψα,β

E (t, θ, ϕ, 1, 1).

() 10 / 18

The right-hand side here can be continued analytically from α, β > −1/2 to ℜα, ℜβ > −1.

() 11 / 18

The right-hand side here can be continued analytically from α, β > −1/2 to ℜα, ℜβ > −1. This will produce an analytic continuation of the left-hand side Hα,β

t

(θ, ϕ) =

∞

• n=0

e−t|n+ α+β+1

2

|Pα,β

n

(θ)Pα,β

n

(ϕ).

() 11 / 18

The right-hand side here can be continued analytically from α, β > −1/2 to ℜα, ℜβ > −1. This will produce an analytic continuation of the left-hand side Hα,β

t

(θ, ϕ) =

∞

• n=0

e−t|n+ α+β+1

2

|Pα,β

n

(θ)Pα,β

n

(ϕ). When α, β > −1/2, one can delete all the absolute value symbols here, and the analytic extension of Hα,β

t

will be Hα,β

t

(θ, ϕ) =

∞

• n=0

e−t(n+ α+β+1

2

)Pα,β

n

(θ)Pα,β

n

(ϕ), ℜα, ℜβ > −1.

() 11 / 18

The right-hand side here can be continued analytically from α, β > −1/2 to ℜα, ℜβ > −1. This will produce an analytic continuation of the left-hand side Hα,β

t

(θ, ϕ) =

∞

• n=0

e−t|n+ α+β+1

2

|Pα,β

n

(θ)Pα,β

n

(ϕ). When α, β > −1/2, one can delete all the absolute value symbols here, and the analytic extension of Hα,β

t

will be Hα,β

t

(θ, ϕ) =

∞

• n=0

e−t(n+ α+β+1

2

)Pα,β

n

(θ)Pα,β

n

(ϕ), ℜα, ℜβ > −1. For real values of α, β > −1, the term with n = 0 will be different in these two sums when α + β < −1.

() 11 / 18

The right-hand side here can be continued analytically from α, β > −1/2 to ℜα, ℜβ > −1. This will produce an analytic continuation of the left-hand side Hα,β

t

(θ, ϕ) =

∞

• n=0

e−t|n+ α+β+1

2

|Pα,β

n

(θ)Pα,β

n

(ϕ). When α, β > −1/2, one can delete all the absolute value symbols here, and the analytic extension of Hα,β

t

will be Hα,β

t

(θ, ϕ) =

∞

• n=0

e−t(n+ α+β+1

2

)Pα,β

n

(θ)Pα,β

n

(ϕ), ℜα, ℜβ > −1. For real values of α, β > −1, the term with n = 0 will be different in these two sums when α + β < −1. Evaluating those terms, one finds Hα,β

t

(θ, ϕ) = Hα,β

t

(θ, ϕ) + χ{α+β<−1} cα,β sinh (α + β + 1)t 2 .

() 11 / 18

Next we rewrite the integral formula for Hα,β

t

(θ, ϕ), integrating by parts and going back from Ψα,β

E

to Ψα,β.

() 12 / 18

Next we rewrite the integral formula for Hα,β

t

(θ, ϕ), integrating by parts and going back from Ψα,β

E

to Ψα,β. In the case when −1 < α, β < −1/2, the result is Hα,β

t

(θ, ϕ) = 1

−1

1

−1

∂u∂vΨα,β(t, θ, ϕ, u, v) Πα(u) Πβ(v) dudv −1 2

• ±

1

−1

∂uΨα,β(t, θ, ϕ, u, ±1) Πα(u) du −1 2

• ±

1

−1

∂vΨα,β(t, θ, ϕ, ±1, v) Πβ(v) dv +1 4

• ±
• ±

Ψα,β(t, θ, ϕ, ±1, ±1).

() 12 / 18

Next we rewrite the integral formula for Hα,β

t

(θ, ϕ), integrating by parts and going back from Ψα,β

E

to Ψα,β. In the case when −1 < α, β < −1/2, the result is Hα,β

t

(θ, ϕ) = 1

−1

1

−1

∂u∂vΨα,β(t, θ, ϕ, u, v) Πα(u) Πβ(v) dudv −1 2

• ±

1

−1

∂uΨα,β(t, θ, ϕ, u, ±1) Πα(u) du −1 2

• ±

1

−1

∂vΨα,β(t, θ, ϕ, ±1, v) Πβ(v) dv +1 4

• ±
• ±

Ψα,β(t, θ, ϕ, ±1, ±1). Here Πα(u) = u

0 dΠα(u′), the (odd) primitive.

() 12 / 18

Next we rewrite the integral formula for Hα,β

t

(θ, ϕ), integrating by parts and going back from Ψα,β

E

to Ψα,β. In the case when −1 < α, β < −1/2, the result is Hα,β

t

(θ, ϕ) = 1

−1

1

−1

∂u∂vΨα,β(t, θ, ϕ, u, v) Πα(u) Πβ(v) dudv −1 2

• ±

1

−1

∂uΨα,β(t, θ, ϕ, u, ±1) Πα(u) du −1 2

• ±

1

−1

∂vΨα,β(t, θ, ϕ, ±1, v) Πβ(v) dv +1 4

• ±
• ±

Ψα,β(t, θ, ϕ, ±1, ±1). Here Πα(u) = u

0 dΠα(u′), the (odd) primitive.

This will allow estimates of derivatives of Hα,β

t

(θ, ϕ) and Hα,β

t

(θ, ϕ).

() 12 / 18

We must verify the standard estimates for the (B-valued) kernels K(θ, ϕ) of the singular operators of the theorem, in the space of homogeneous type.

() 13 / 18

We must verify the standard estimates for the (B-valued) kernels K(θ, ϕ) of the singular operators of the theorem, in the space of homogeneous type. This means K(θ, ϕ)B 1 µα,β(B(θ, |θ − ϕ|)), where B(θ, r) is the interval of center θ and half-length r,

() 13 / 18

We must verify the standard estimates for the (B-valued) kernels K(θ, ϕ) of the singular operators of the theorem, in the space of homogeneous type. This means K(θ, ϕ)B 1 µα,β(B(θ, |θ − ϕ|)), where B(θ, r) is the interval of center θ and half-length r, and ∂θK(θ, ϕ)B + ∂ϕK(θ, ϕ)B 1 |θ − ϕ| 1 µα,β(B(θ, |θ − ϕ|)).

() 13 / 18

We must verify the standard estimates for the (B-valued) kernels K(θ, ϕ) of the singular operators of the theorem, in the space of homogeneous type. This means K(θ, ϕ)B 1 µα,β(B(θ, |θ − ϕ|)), where B(θ, r) is the interval of center θ and half-length r, and ∂θK(θ, ϕ)B + ∂ϕK(θ, ϕ)B 1 |θ − ϕ| 1 µα,β(B(θ, |θ − ϕ|)). In the vector-valued case B = C, these derivatives of K are taken in the weak sense.

() 13 / 18

Depending on the operator, the kernel K(θ, ϕ) will be an expression containing Hα,β

t

(θ, ϕ) or some of its derivatives.

() 14 / 18

Depending on the operator, the kernel K(θ, ϕ) will be an expression containing Hα,β

t

(θ, ϕ) or some of its derivatives. The strategy is now to use the link between Hα,β

t

(θ, ϕ) and Hα,β

t

(θ, ϕ) and the integral formula for Hα,β

t

(θ, ϕ), to estimate the kernel and its derivatives.

() 14 / 18

Depending on the operator, the kernel K(θ, ϕ) will be an expression containing Hα,β

t

(θ, ϕ) or some of its derivatives. The strategy is now to use the link between Hα,β

t

(θ, ϕ) and Hα,β

t

(θ, ϕ) and the integral formula for Hα,β

t

(θ, ϕ), to estimate the kernel and its derivatives. Examples:

() 14 / 18

Depending on the operator, the kernel K(θ, ϕ) will be an expression containing Hα,β

t

(θ, ϕ) or some of its derivatives. The strategy is now to use the link between Hα,β

t

(θ, ϕ) and Hα,β

t

(θ, ϕ) and the integral formula for Hα,β

t

(θ, ϕ), to estimate the kernel and its derivatives. Examples: The Riesz transforms have kernels K(θ, ϕ) = c ∞ ∂N

θ Hα,β t

(θ, ϕ)tN−1 dt, and B = C.

() 14 / 18

Depending on the operator, the kernel K(θ, ϕ) will be an expression containing Hα,β

t

(θ, ϕ) or some of its derivatives. The strategy is now to use the link between Hα,β

t

(θ, ϕ) and Hα,β

t

(θ, ϕ) and the integral formula for Hα,β

t

(θ, ϕ), to estimate the kernel and its derivatives. Examples: The Riesz transforms have kernels K(θ, ϕ) = c ∞ ∂N

θ Hα,β t

(θ, ϕ)tN−1 dt, and B = C. For the square functions, K(θ, ϕ) =

• ∂N

θ ∂M t Hα,β t

(θ, ϕ)

• t>0,

and B = L2(R+; t2M+2N−1dt).

() 14 / 18

Thus we will have to estimate many integrals of various derivatives of the function Ψα,β(t, θ, ϕ, u, v) = cα,β sinh t

2

(cosh t

2 − 1 + q)α+β+2 ,

against Πα(u) du or Πβ(v) dv or both.

() 15 / 18

Thus we will have to estimate many integrals of various derivatives of the function Ψα,β(t, θ, ϕ, u, v) = cα,β sinh t

2

(cosh t

2 − 1 + q)α+β+2 ,

against Πα(u) du or Πβ(v) dv or both. Here q = q(θ, ϕ, u, v) = 1 − u sin θ 2 sin ϕ 2 − v cos θ 2 cos ϕ 2 .

() 15 / 18

Thus we will have to estimate many integrals of various derivatives of the function Ψα,β(t, θ, ϕ, u, v) = cα,β sinh t

2

(cosh t

2 − 1 + q)α+β+2 ,

against Πα(u) du or Πβ(v) dv or both. Here q = q(θ, ϕ, u, v) = 1 − u sin θ 2 sin ϕ 2 − v cos θ 2 cos ϕ 2 . These estimates are rather technical, and we give only a few samples.

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We assume that −1 < α, β < −1/2.

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We assume that −1 < α, β < −1/2. First example: Uniformly in t ∈ (0, 1] and θ, ϕ ∈ [0, π], one has

• ∂L

ϕ∂N θ ∂M t Hα,β t

(θ, ϕ)

• 1 +
• K,R=0,1
• k,r=0,1,2
• sin θ

2 + sin ϕ 2 Kk cos θ 2 + cos ϕ 2 Rr ×

• dΠα,K dΠβ,R

(t2 + q)α+β+3/2+(L+N+M+Kk+Rr)/2 .

() 16 / 18

We assume that −1 < α, β < −1/2. First example: Uniformly in t ∈ (0, 1] and θ, ϕ ∈ [0, π], one has

• ∂L

ϕ∂N θ ∂M t Hα,β t

(θ, ϕ)

• 1 +
• K,R=0,1
• k,r=0,1,2
• sin θ

2 + sin ϕ 2 Kk cos θ 2 + cos ϕ 2 Rr ×

• dΠα,K dΠβ,R

(t2 + q)α+β+3/2+(L+N+M+Kk+Rr)/2 . Here dΠα,K is dΠ−1/2 when K = 0 and dΠα+1 when K = 1, and similarly for dΠβ,R.

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Second example: For each θ, ϕ ∈ [0, π] and s ∈ {0, 1}, the norm of

• sin θ

2 + sin ϕ 2 Kk cos θ 2 + cos ϕ 2 Rr ×

• dΠα,K dΠβ,R

(t2 + q)α+β+3/2+W/(2p)+Kk/2+Rr/2+s/2 in the space Lp((0, 1), tW−1dt)

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Second example: For each θ, ϕ ∈ [0, π] and s ∈ {0, 1}, the norm of

• sin θ

2 + sin ϕ 2 Kk cos θ 2 + cos ϕ 2 Rr ×

• dΠα,K dΠβ,R

(t2 + q)α+β+3/2+W/(2p)+Kk/2+Rr/2+s/2 in the space Lp((0, 1), tW−1dt) is controlled by const. 1 |θ − ϕ|s 1 µα,β(B(θ, |θ − ϕ|)).

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Second example: For each θ, ϕ ∈ [0, π] and s ∈ {0, 1}, the norm of

• sin θ

2 + sin ϕ 2 Kk cos θ 2 + cos ϕ 2 Rr ×

• dΠα,K dΠβ,R

(t2 + q)α+β+3/2+W/(2p)+Kk/2+Rr/2+s/2 in the space Lp((0, 1), tW−1dt) is controlled by const. 1 |θ − ϕ|s 1 µα,β(B(θ, |θ − ϕ|)). Here 1 ≤ p ≤ ∞ and W ≥ 1.

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Third example: For 1 < t < ∞, one has instead

• sup

θ,ϕ∈[0,π]

• ∂L

ϕ∂N θ ∂M t Hα,β t

(θ, ϕ)

• Lp((1,∞),tW−1dt)

< ∞,

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Third example: For 1 < t < ∞, one has instead

• sup

θ,ϕ∈[0,π]

• ∂L

ϕ∂N θ ∂M t Hα,β t

(θ, ϕ)

• Lp((1,∞),tW−1dt)

< ∞, which is proved from the oscillating series for Hα,β

t

, not the integral formula.

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Third example: For 1 < t < ∞, one has instead

• sup

θ,ϕ∈[0,π]

• ∂L

ϕ∂N θ ∂M t Hα,β t

(θ, ϕ)

• Lp((1,∞),tW−1dt)

< ∞, which is proved from the oscillating series for Hα,β

t

, not the integral formula. This leads to the standard estimates for this part.

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Third example: For 1 < t < ∞, one has instead

• sup

θ,ϕ∈[0,π]

• ∂L

ϕ∂N θ ∂M t Hα,β t

(θ, ϕ)

• Lp((1,∞),tW−1dt)

< ∞, which is proved from the oscillating series for Hα,β

t

, not the integral formula. This leads to the standard estimates for this part.

THE END

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