Operaors relaed o he Jacobi seing, for all admissible parameer values Peer Sjögren Universiy of Gohenburg Join work wih A. Nowak and T. Szarek Alba, June 2013 () 1 / 18
Le Pn α,β be he classical Jacobi polynomials, seen via he ransformaion x = cos θ as funcions of θ [0, π]. Here α,β > 1. () 2 / 18
Le Pn α,β be he classical Jacobi polynomials, seen via he ransformaion x = cos θ as funcions of θ [0, π]. Here α,β > 1. The P α,β n are orhogonal wih respec o he measure dµ α,β (θ) = ( sin θ ) 2α+1 ( cos θ 2β+1 dθ in [0,π], 2 2) and we ake hem o be normalized in L 2 (dµ α,β ). () 2 / 18
Le Pn α,β be he classical Jacobi polynomials, seen via he ransformaion x = cos θ as funcions of θ [0, π]. Here α,β > 1. The P α,β n are orhogonal wih respec o he measure dµ α,β (θ) = ( sin θ ) 2α+1 ( cos θ 2β+1 dθ in [0,π], 2 2) and we ake hem o be normalized in L 2 (dµ α,β ). They are eigenfuncions of he Jacobi operaor J α,β = d 2 wih eigenvalues α β + (α + β + 1) cos θ dθ2 sin θ ( ) 2. n + α+β+1 2 d ( α + β + 1 ) 2, dθ + 2 () 2 / 18
The corresponding Poisson operaor exp( J α,β ), > 0, can be defined specrally and has he inegral kernel H α,β (θ,ϕ) = n=0 called he Jacobi-Poisson kernel. α+β+1 n+ e 2 P α,β n (θ)pn α,β (ϕ), () 3 / 18
The corresponding Poisson operaor exp( J α,β ), > 0, can be defined specrally and has he inegral kernel H α,β (θ,ϕ) = n=0 called he Jacobi-Poisson kernel. α+β+1 n+ e 2 P α,β n (θ)pn α,β (ϕ), By means of H α,β, one can express he kernels of many operaors associaed wih he Jacobi seing. () 3 / 18
The corresponding Poisson operaor exp( J α,β ), > 0, can be defined specrally and has he inegral kernel H α,β (θ,ϕ) = n=0 called he Jacobi-Poisson kernel. α+β+1 n+ e 2 P α,β n (θ)pn α,β (ϕ), By means of H α,β, one can express he kernels of many operaors associaed wih he Jacobi seing. We lis some of hese operaors and heir kernels. () 3 / 18
Riesz ransforms D N (J α,β ) N/2 c N 0 of order N 1, wih kernels θ N Hα,β (θ,ϕ) N 1 d. () 4 / 18
Riesz ransforms D N (J α,β ) N/2 c N 0 of order N 1, wih kernels θ N Hα,β (θ,ϕ) N 1 d. ( J ) Laplace ransform ype mulipliers m α,β wih m given by m(z) = zlφ(z), where L denoes he Laplace ransform and φ L (R + ). () 4 / 18
Riesz ransforms D N (J α,β ) N/2 c N 0 of order N 1, wih kernels θ N Hα,β (θ,ϕ) N 1 d. ( J ) Laplace ransform ype mulipliers m α,β wih m given by m(z) = zlφ(z), where L denoes he Laplace ransform and φ L (R + ). The kernel is 0 φ() H α,β (θ,ϕ) d. () 4 / 18
Riesz ransforms D N (J α,β ) N/2 c N 0 of order N 1, wih kernels θ N Hα,β (θ,ϕ) N 1 d. ( J ) Laplace ransform ype mulipliers m α,β wih m given by m(z) = zlφ(z), where L denoes he Laplace ransform and φ L (R + ). The kernel is 0 φ() H α,β (θ,ϕ) d. Wih φ() = cons. 2iγ, γ 0, his gives he imaginary power (J α,β ) iγ of he Jacobi operaor. () 4 / 18
Riesz ransforms D N (J α,β ) N/2 c N 0 of order N 1, wih kernels θ N Hα,β (θ,ϕ) N 1 d. ( J ) Laplace ransform ype mulipliers m α,β wih m given by m(z) = zlφ(z), where L denoes he Laplace ransform and φ L (R + ). The kernel is 0 φ() H α,β (θ,ϕ) d. Wih φ() = cons. 2iγ, γ 0, his gives he imaginary power (J α,β ) iγ of he Jacobi operaor. ( J ) Laplace-Sieljes ransform ype mulipliers m α,β, where m = Lν and ν is a signed or complex Borel measure on (0, ) saisfying e α+β+1 /2 d ν () <. The kernel is now H α,β (θ,ϕ) dν(). (0, ) () 4 / 18
These kernels are scalar-valued. Bu allowing kernels aking values in a Banach space B, one can wrie more operaor kernels. () 5 / 18
These kernels are scalar-valued. Bu allowing kernels aking values in a Banach space B, one can wrie more operaor kernels. Examples: The Poisson maximal operaor, wih kernel { H α,β (θ,ϕ) } >0 and B = L (R + ; d). () 5 / 18
These kernels are scalar-valued. Bu allowing kernels aking values in a Banach space B, one can wrie more operaor kernels. Examples: The Poisson maximal operaor, wih kernel { H α,β (θ,ϕ) } >0 and B = L (R + ; d). The mixed square funcions, wih kernels { N θ M H α,β (θ,ϕ) } >0 and B = L 2 (R + ; 2M+2N 1 d). Here M + N > 0. () 5 / 18
Old resul: Theorem (Nowak and Sjögren, J.F.A.A. 2012) For α, β 1/2, he following among he operaors lised above are bounded on L p (dµ α,β ), 1 < p <, and of weak ype (1,1) for µ α,β : () 6 / 18
Old resul: Theorem (Nowak and Sjögren, J.F.A.A. 2012) For α, β 1/2, he following among he operaors lised above are bounded on L p (dµ α,β ), 1 < p <, and of weak ype (1,1) for µ α,β : The Riesz ransforms, he imaginary powers of he Jacobi operaor, he maximal operaor and he mixed square funcions. () 6 / 18
Old resul: Theorem (Nowak and Sjögren, J.F.A.A. 2012) For α, β 1/2, he following among he operaors lised above are bounded on L p (dµ α,β ), 1 < p <, and of weak ype (1,1) for µ α,β : The Riesz ransforms, he imaginary powers of he Jacobi operaor, he maximal operaor and he mixed square funcions. These operaors are also bounded beween he corresponding weighed spaces wih a weigh in he Muckenhoup class A α,β defined wih respec o he measure µ α,β. p, () 6 / 18
Old resul: Theorem (Nowak and Sjögren, J.F.A.A. 2012) For α, β 1/2, he following among he operaors lised above are bounded on L p (dµ α,β ), 1 < p <, and of weak ype (1,1) for µ α,β : The Riesz ransforms, he imaginary powers of he Jacobi operaor, he maximal operaor and he mixed square funcions. These operaors are also bounded beween he corresponding weighed spaces wih a weigh in he Muckenhoup class A α,β defined wih respec o he measure µ α,β. New resul: Theorem (Nowak, Sjögren and Szarek, 2013) The preceding heorem holds for all α, β > 1 and all he operaors lised above. p, () 6 / 18
To prove boh heorems, one shows ha he operaors are well-behaved Calderón-Zygmund operaors () 7 / 18
To prove boh heorems, one shows ha he operaors are well-behaved Calderón-Zygmund operaors on he space of homogeneous ype defined by he inerval [0, π] wih he measure µ α,β and he ordinary disance. () 7 / 18
To prove boh heorems, one shows ha he operaors are well-behaved Calderón-Zygmund operaors on he space of homogeneous ype defined by he inerval [0, π] wih he measure µ α,β and he ordinary disance. The operaors can be seen o be bounded on L 2 (dµ α,β ) (or L (dµ α,β ) in he case of he maximal operaor). () 7 / 18
To prove boh heorems, one shows ha he operaors are well-behaved Calderón-Zygmund operaors on he space of homogeneous ype defined by he inerval [0, π] wih he measure µ α,β and he ordinary disance. The operaors can be seen o be bounded on L 2 (dµ α,β ) (or L (dµ α,β ) in he case of he maximal operaor). So he main crux is o verify ha heir off-diagonal kernels saisfy he sandard esimaes in his space. () 7 / 18
To prove boh heorems, one shows ha he operaors are well-behaved Calderón-Zygmund operaors on he space of homogeneous ype defined by he inerval [0, π] wih he measure µ α,β and he ordinary disance. The operaors can be seen o be bounded on L 2 (dµ α,β ) (or L (dµ α,β ) in he case of he maximal operaor). So he main crux is o verify ha heir off-diagonal kernels saisfy he sandard esimaes in his space. I also requires some effor o prove ha he kernels are really as described above. () 7 / 18
The argumen for he old heorem is based on he following inegral formula for he Poisson kernel H α,β (θ,ϕ) = Ψ α,β (,θ,ϕ, u, v) dπ α (u) dπ β (v), valid for α, β > 1/2, () 8 / 18
The argumen for he old heorem is based on he following inegral formula for he Poisson kernel H α,β (θ,ϕ) = Ψ α,β (,θ,ϕ, u, v) dπ α (u) dπ β (v), valid for α, β > 1/2, where Ψ α,β (,θ,ϕ, u, v) = c α,β sinh 2 (cosh, 2 1 + q(θ,ϕ, u, v))α+β+2 () 8 / 18
The argumen for he old heorem is based on he following inegral formula for he Poisson kernel H α,β (θ,ϕ) = Ψ α,β (,θ,ϕ, u, v) dπ α (u) dπ β (v), valid for α, β > 1/2, where Ψ α,β (,θ,ϕ, u, v) = c α,β sinh 2 (cosh, 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 argumen for he old heorem is based on he following inegral formula for he Poisson kernel H α,β (θ,ϕ) = Ψ α,β (,θ,ϕ, u, v) dπ α (u) dπ β (v), valid for α, β > 1/2, where Ψ α,β (,θ,ϕ, u, v) = c α,β sinh 2 (cosh, 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 he probabiliy measures Π α and Π β are defined by dπ α (u) = Γ(α + 1) π Γ(α + 1/2) (1 u 2 ) α 1/2 du, 1 < u < 1. () 8 / 18
We wan o exend he inegral formula from α, β > 1/2 o α, β > 1, by analyic coninuaion. () 9 / 18
We wan o exend he inegral formula from α, β > 1/2 o α, β > 1, by analyic coninuaion. The defining expression for he measure dπ α makes sense for all α wih Rα > 1 excep α = 1/2 and produces a local, complex measure in ( 1, 1) wih infinie mass near ±1. () 9 / 18
We wan o exend he inegral formula from α, β > 1/2 o α, β > 1, by analyic coninuaion. The defining expression for he measure dπ α makes sense for all α wih Rα > 1 excep α = 1/2 and produces a local, complex measure in ( 1, 1) wih infinie mass near ±1. A he poin α = 1/2, here is a weak limi dπ 1/2 = 1 2 (δ 1 + δ 1 ). () 9 / 18
We wan o exend he inegral formula from α, β > 1/2 o α, β > 1, by analyic coninuaion. The defining expression for he measure dπ α makes sense for all α wih Rα > 1 excep α = 1/2 and produces a local, complex measure in ( 1, 1) wih infinie mass near ±1. A he poin α = 1/2, here is a weak limi dπ 1/2 = 1 2 (δ 1 + δ 1 ). Bu he inegral agains Ψ α,β (,θ,ϕ, u, v) will diverge a he endpoins if α or β, or heir real pars, go below -1/2. () 9 / 18
We wan o exend he inegral formula from α, β > 1/2 o α, β > 1, by analyic coninuaion. The defining expression for he measure dπ α makes sense for all α wih Rα > 1 excep α = 1/2 and produces a local, complex measure in ( 1, 1) wih infinie mass near ±1. A he poin α = 1/2, here is a weak limi dπ 1/2 = 1 2 (δ 1 + δ 1 ). Bu he inegral agains Ψ α,β (,θ,ϕ, u, v) will diverge a he endpoins if α or β, or heir real pars, go below -1/2. In order o ge convergence, we firs resric he inegral o u, v > 0, using he even par of he inegrand. Ψ α,β E (,θ,ϕ, u, v) = 1 4 Ψ α,β (,θ,ϕ, ±u, ±v) ± ± () 9 / 18
Since he measures dπ α are even, he formula can be wrien H α,β (θ,ϕ) = 4 (,θ,ϕ, u, v) dπ α(u) dπ β (v). Ψ α,β (0,1] 2 E () 10 / 18
Since he measures dπ α are even, he formula can be wrien H α,β (θ,ϕ) = 4 (,θ,ϕ, u, v) dπ α(u) dπ β (v). Ψ α,β (0,1] 2 E By subracing an adding he values of Ψ α,β E (,θ,ϕ, u, v) a he endpoins u, v = 1, we can rewrie his as () 10 / 18
Since he measures dπ α are even, he formula can be wrien H α,β (θ,ϕ) = 4 (,θ,ϕ, u, v) dπ α(u) dπ β (v). Ψ α,β (0,1] 2 E By subracing an adding he values of Ψ α,β E (,θ,ϕ, u, v) a he endpoins u, v = 1, we can rewrie his as ( H α,β (θ, ϕ) = 4 (,θ,ϕ, u, v) Ψα,β (,θ,ϕ, u, 1) + 2 + 2 (0,1] 2 Ψ α,β E (0,1] (0,1] Ψ α,β E E ) (,θ,ϕ, 1, v) + Ψα,β E (,θ,ϕ, 1, 1) dπ α (u) dπ β (v) ( ) Ψ α,β E (,θ,ϕ, u, 1) Ψα,β E (,θ,ϕ, 1, 1) dπ α (u) ( ) (,θ,ϕ, 1, v) Ψα,β E (,θ,ϕ, 1, 1) dπ β (v) Ψ α,β E + Ψ α,β E (,θ,ϕ, 1, 1). () 10 / 18
The righ-hand side here can be coninued analyically from α,β > 1/2 o Rα, Rβ > 1. () 11 / 18
The righ-hand side here can be coninued analyically from α,β > 1/2 o Rα, Rβ > 1. This will produce an analyic coninuaion of he lef-hand side H α,β (θ,ϕ) = n=0 α+β+1 n+ e 2 P α,β n (θ)pn α,β (ϕ). () 11 / 18
The righ-hand side here can be coninued analyically from α,β > 1/2 o Rα, Rβ > 1. This will produce an analyic coninuaion of he lef-hand side H α,β (θ,ϕ) = n=0 α+β+1 n+ e 2 P α,β n (θ)pn α,β (ϕ). When α, β > 1/2, one can delee all he absolue value symbols here, and he analyic exension of H α,β will be H α,β (θ,ϕ) = n=0 α+β+1 (n+ e 2 ) P α,β n (θ)p α,β n (ϕ), Rα, Rβ > 1. () 11 / 18
The righ-hand side here can be coninued analyically from α,β > 1/2 o Rα, Rβ > 1. This will produce an analyic coninuaion of he lef-hand side H α,β (θ,ϕ) = n=0 α+β+1 n+ e 2 P α,β n (θ)pn α,β (ϕ). When α, β > 1/2, one can delee all he absolue value symbols here, and he analyic exension of H α,β will be H α,β (θ,ϕ) = n=0 α+β+1 (n+ e 2 ) P α,β n (θ)p α,β n (ϕ), Rα, Rβ > 1. For real values of α,β > 1, he erm wih n = 0 will be differen in hese wo sums when α + β < 1. () 11 / 18
The righ-hand side here can be coninued analyically from α,β > 1/2 o Rα, Rβ > 1. This will produce an analyic coninuaion of he lef-hand side H α,β (θ,ϕ) = n=0 α+β+1 n+ e 2 P α,β n (θ)pn α,β (ϕ). When α, β > 1/2, one can delee all he absolue value symbols here, and he analyic exension of H α,β will be H α,β (θ,ϕ) = n=0 α+β+1 (n+ e 2 ) P α,β n (θ)p α,β n (ϕ), Rα, Rβ > 1. For real values of α,β > 1, he erm wih n = 0 will be differen in hese wo sums when α + β < 1. Evaluaing hose erms, one finds H α,β (θ,ϕ) = H α,β (θ,ϕ) + χ {α+β< 1} c α,β sinh (α + β + 1). 2 () 11 / 18
Nex we rewrie he inegral formula for H α,β (θ, ϕ), inegraing by pars and going back from Ψ α,β E o Ψ α,β. () 12 / 18
Nex we rewrie he inegral formula for H α,β (θ, ϕ), inegraing by pars and going back from Ψ α,β E o Ψ α,β. In he case when 1 < α,β < 1/2, he resul is 1 1 H α,β (θ,ϕ) = 1 1 1 2 1 2 ± ± 1 1 1 1 u v Ψ α,β (,θ,ϕ, u, v) Π α (u)π β (v) dudv u Ψ α,β (,θ,ϕ, u, ±1)Π α (u) du v Ψ α,β (,θ,ϕ, ±1, v)π β (v) dv + 1 Ψ α,β (,θ,ϕ, ±1, ±1). 4 ± ± () 12 / 18
Nex we rewrie he inegral formula for H α,β (θ, ϕ), inegraing by pars and going back from Ψ α,β E o Ψ α,β. In he case when 1 < α,β < 1/2, he resul is 1 1 H α,β (θ,ϕ) = 1 1 1 2 1 2 ± ± 1 1 1 1 u v Ψ α,β (,θ,ϕ, u, v) Π α (u)π β (v) dudv u Ψ α,β (,θ,ϕ, u, ±1)Π α (u) du v Ψ α,β (,θ,ϕ, ±1, v)π β (v) dv + 1 Ψ α,β (,θ,ϕ, ±1, ±1). 4 ± Here Π α (u) = u 0 dπ α(u ), he (odd) primiive. ± () 12 / 18
Nex we rewrie he inegral formula for H α,β (θ, ϕ), inegraing by pars and going back from Ψ α,β E o Ψ α,β. In he case when 1 < α,β < 1/2, he resul is 1 1 H α,β (θ,ϕ) = 1 1 1 2 1 2 ± ± 1 1 1 1 u v Ψ α,β (,θ,ϕ, u, v) Π α (u)π β (v) dudv u Ψ α,β (,θ,ϕ, u, ±1)Π α (u) du v Ψ α,β (,θ,ϕ, ±1, v)π β (v) dv + 1 Ψ α,β (,θ,ϕ, ±1, ±1). 4 ± Here Π α (u) = u 0 dπ α(u ), he (odd) primiive. This will allow esimaes of derivaives of H α,β ± (θ,ϕ) and H α,β (θ,ϕ). () 12 / 18
We mus verify he sandard esimaes for he (B-valued) kernels K(θ,ϕ) of he singular operaors of he heorem, in he space of homogeneous ype. () 13 / 18
We mus verify he sandard esimaes for he (B-valued) kernels K(θ,ϕ) of he singular operaors of he heorem, in he space of homogeneous ype. This means K(θ,ϕ) B 1 µ α,β (B(θ, θ ϕ )), where B(θ, r) is he inerval of cener θ and half-lengh r, () 13 / 18
We mus verify he sandard esimaes for he (B-valued) kernels K(θ,ϕ) of he singular operaors of he heorem, in he space of homogeneous ype. This means K(θ,ϕ) B 1 µ α,β (B(θ, θ ϕ )), where B(θ, r) is he inerval of cener θ and half-lengh r, and θ K(θ,ϕ) B + ϕ K(θ,ϕ) B 1 θ ϕ 1 µ α,β (B(θ, θ ϕ )). () 13 / 18
We mus verify he sandard esimaes for he (B-valued) kernels K(θ,ϕ) of he singular operaors of he heorem, in he space of homogeneous ype. This means K(θ,ϕ) B 1 µ α,β (B(θ, θ ϕ )), where B(θ, r) is he inerval of cener θ and half-lengh r, and θ K(θ,ϕ) B + ϕ K(θ,ϕ) B 1 θ ϕ 1 µ α,β (B(θ, θ ϕ )). In he vecor-valued case B C, hese derivaives of K he weak sense. are aken in () 13 / 18
Depending on he operaor, he kernel K(θ, ϕ) will be an expression conaining H α,β (θ,ϕ) or some of is derivaives. () 14 / 18
Depending on he operaor, he kernel K(θ, ϕ) will be an expression conaining H α,β (θ,ϕ) or some of is derivaives. The sraegy is now o use he link beween H α,β (θ,ϕ) and H α,β (θ,ϕ) and he inegral formula for H α,β (θ, ϕ), o esimae he kernel and is derivaives. () 14 / 18
Depending on he operaor, he kernel K(θ, ϕ) will be an expression conaining H α,β (θ,ϕ) or some of is derivaives. The sraegy is now o use he link beween H α,β (θ,ϕ) and H α,β (θ,ϕ) and he inegral formula for H α,β (θ, ϕ), o esimae he kernel and is derivaives. Examples: () 14 / 18
Depending on he operaor, he kernel K(θ, ϕ) will be an expression conaining H α,β (θ,ϕ) or some of is derivaives. The sraegy is now o use he link beween H α,β (θ,ϕ) and H α,β (θ,ϕ) and he inegral formula for H α,β (θ, ϕ), o esimae he kernel and is derivaives. Examples: The Riesz ransforms have kernels and B = C. K(θ,ϕ) = c 0 θ N Hα,β (θ,ϕ) N 1 d, () 14 / 18
Depending on he operaor, he kernel K(θ, ϕ) will be an expression conaining H α,β (θ,ϕ) or some of is derivaives. The sraegy is now o use he link beween H α,β (θ,ϕ) and H α,β (θ,ϕ) and he inegral formula for H α,β (θ, ϕ), o esimae he kernel and is derivaives. Examples: The Riesz ransforms have kernels and B = C. K(θ,ϕ) = c For he square funcions, 0 θ N Hα,β (θ,ϕ) N 1 d, K(θ,ϕ) = { N θ M H α,β (θ,ϕ) } >0, and B = L 2 (R + ; 2M+2N 1 d). () 14 / 18
Thus we will have o esimae many inegrals of various derivaives of he funcion Ψ α,β (,θ,ϕ, u, v) = c α,β sinh 2 (cosh 2 1 + q)α+β+2, agains Π α (u) du or Π β (v) dv or boh. () 15 / 18
Thus we will have o esimae many inegrals of various derivaives of he funcion Ψ α,β (,θ,ϕ, u, v) = c α,β sinh 2 (cosh 2 1 + q)α+β+2, agains Π α (u) du or Π β (v) dv Here or boh. 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 o esimae many inegrals of various derivaives of he funcion Ψ α,β (,θ,ϕ, u, v) = c α,β sinh 2 (cosh 2 1 + q)α+β+2, agains Π α (u) du or Π β (v) dv Here or boh. q = q(θ,ϕ, u, v) = 1 u sin θ 2 sin ϕ 2 v cos θ 2 cos ϕ 2. These esimaes are raher echnical, and we give only a few samples. () 15 / 18
We assume ha 1 < α,β < 1/2. () 16 / 18
We assume ha 1 < α,β < 1/2. Firs example: Uniformly in (0, 1] and θ,ϕ [0,π], one has L ϕ θ N M H α,β (θ,ϕ) 1 + K,R=0,1 k,r=0,1,2 ( sin θ 2 + sin ϕ ) Kk ( cos θ 2 2 + cos ϕ ) Rr 2 dπ α,k dπ β,r ( 2 + q) α+β+3/2+(l+n+m+kk+rr)/2. () 16 / 18
We assume ha 1 < α,β < 1/2. Firs example: Uniformly in (0, 1] and θ,ϕ [0,π], one has L ϕ θ N M H α,β (θ,ϕ) 1 + K,R=0,1 k,r=0,1,2 ( sin θ 2 + sin ϕ ) Kk ( cos θ 2 2 + cos ϕ ) Rr 2 dπ α,k dπ β,r ( 2 + q) α+β+3/2+(l+n+m+kk+rr)/2. Here dπ α,k is dπ 1/2 when K = 0 and dπ α+1 when K = 1, and similarly for dπ β,r. () 16 / 18
Second example: For each θ,ϕ [0,π] and s {0, 1}, he norm of ( sin θ 2 + sin ϕ ) Kk ( cos θ 2 2 + cos ϕ 2 in he space L p ((0, 1), W 1 d) ) Rr dπ α,k dπ β,r ( 2 + q) α+β+3/2+w/(2p)+kk/2+rr/2+s/2 () 17 / 18
Second example: For each θ,ϕ [0,π] and s {0, 1}, he norm of ( sin θ 2 + sin ϕ ) Kk ( cos θ 2 2 + cos ϕ 2 ) Rr dπ α,k dπ β,r ( 2 + q) α+β+3/2+w/(2p)+kk/2+rr/2+s/2 in he space L p ((0, 1), W 1 d) is conrolled by 1 1 cons. θ ϕ s µ α,β (B(θ, θ ϕ )). () 17 / 18
Second example: For each θ,ϕ [0,π] and s {0, 1}, he norm of ( sin θ 2 + sin ϕ ) Kk ( cos θ 2 2 + cos ϕ 2 ) Rr dπ α,k dπ β,r ( 2 + q) α+β+3/2+w/(2p)+kk/2+rr/2+s/2 in he space L p ((0, 1), W 1 d) is conrolled by 1 1 cons. θ ϕ s µ α,β (B(θ, θ ϕ )). Here 1 p and W 1. () 17 / 18
Third example: For 1 < <, one has insead sup L ϕ θ N M H α,β (θ,ϕ) <, θ,ϕ [0,π] L p ((1, ), W 1 d) () 18 / 18
Third example: For 1 < <, one has insead sup L ϕ θ N M H α,β (θ,ϕ) <, θ,ϕ [0,π] L p ((1, ), W 1 d) which is proved from he oscillaing series for H α,β, no he inegral formula. () 18 / 18
Third example: For 1 < <, one has insead sup L ϕ θ N M H α,β (θ,ϕ) <, θ,ϕ [0,π] L p ((1, ), W 1 d) which is proved from he oscillaing series for H α,β, no he inegral formula. This leads o he sandard esimaes for his par. () 18 / 18
Third example: For 1 < <, one has insead sup L ϕ θ N M H α,β (θ,ϕ) <, θ,ϕ [0,π] L p ((1, ), W 1 d) which is proved from he oscillaing series for H α,β, no he inegral formula. This leads o he sandard esimaes for his par. THE END () 18 / 18