SOME APPLICATIONS OF THE FUNK-HECKE THEOREMicm-sch.math.snu.ac.kr/icm-talk/BezICMKorea.pdf ·...

SOME APPLICATIONS OF THE FUNK-HECKE

THEOREM

Neal Bez

Saitama University, Japan

ICM 2014 Satellite Conference in Harmonic Analysis

Chosun University, Gwangju, Korea

4 August 2014

Theorem (Funk–Hecke)

Let d ≥ 2, k ∈ N0 and Yk be a spherical harmonic of degree k.Then ∫

Sd−1

F (ω · θ)Yk(ω) dω = λkYk(θ)

for any θ ∈ Sd−1 and any function F ∈ L1([−1, 1], (1− t2)d−3

Here, the constant λk is given by

λk = |Sd−2|∫ 1

−1F (t)Pk,d(t)(1− t2)

d−32 dt

where Pk,d is the Legendre polynomial of degree k in d dimensions.

Theorem (Funk–Hecke)

Let d ≥ 2, k ∈ N0 and Yk be a spherical harmonic of degree k.Then ∫

Sd−1

F (ω · θ)Yk(ω) dω = λkYk(θ)

for any θ ∈ Sd−1 and any function F ∈ L1([−1, 1], (1− t2)d−3

2 ).Here, the constant λk is given by

λk = |Sd−2|∫ 1

−1F (t)Pk,d(t)(1− t2)

d−32 dt

where Pk,d is the Legendre polynomial of degree k in d dimensions.

Convex geometry (a brief detour)

Corollary

Let F ∈ L1([−1, 1], (1− t2)d−3

2 ) and let λk be as above in theFunk–Hecke theorem.

Suppose that G is a continuous functionsuch that ∫

Sd−1

F (ω · θ)G (θ)dθ = 0

for each ω ∈ Sd−1.

(1) λk 6= 0 for all k ∈ N0 ⇒ G = 0.

(2) λk 6= 0 for all odd k ∈ N0 ⇒ G is even.

(3) λk 6= 0 for all even k ∈ N0 ⇒ G is odd.

Corollary

Let F ∈ L1([−1, 1], (1− t2)d−3

2 ) and let λk be as above in theFunk–Hecke theorem. Suppose that G is a continuous functionsuch that ∫

Sd−1

F (ω · θ)G (θ)dθ = 0

(1) λk 6= 0 for all k ∈ N0 ⇒ G = 0.

Corollary

Let F ∈ L1([−1, 1], (1− t2)d−3

Sd−1

F (ω · θ)G (θ)dθ = 0

(1) λk 6= 0 for all k ∈ N0 ⇒ G = 0.

Corollary

Let F ∈ L1([−1, 1], (1− t2)d−3

Sd−1

F (ω · θ)G (θ)dθ = 0

(1) λk 6= 0 for all k ∈ N0 ⇒ G = 0.

Corollary

Let F ∈ L1([−1, 1], (1− t2)d−3

Sd−1

F (ω · θ)G (θ)dθ = 0

(1) λk 6= 0 for all k ∈ N0 ⇒ G = 0.

Trivially,

Funk–Hecke theorem

⇒∫Sd−1

F (ω · θ)Yk(ω)dω = λkYk(θ)

⇒∫

(Sd−1)2

F (ω · θ)Yk(ω)G (θ)dθdω = λk

∫Sd−1

Yk(θ)G (θ)dθ.

For (F ,G ) as in the corollary, we obtain

∫Sd−1

Yk(θ)G (θ) dθ = 0.

Completeness of spherical harmonics in L2(Sd−1) immediatelyimplies (1).Additionally, Yk has same parity as k. This yields (2) and (3).

Trivially,

⇒∫Sd−1

⇒∫

(Sd−1)2

∫Sd−1

Yk(θ)G (θ)dθ.

∫Sd−1

Yk(θ)G (θ) dθ = 0.

Trivially,

⇒∫Sd−1

⇒∫

(Sd−1)2

∫Sd−1

Yk(θ)G (θ)dθ.

∫Sd−1

Yk(θ)G (θ) dθ = 0.

Completeness of spherical harmonics in L2(Sd−1) immediatelyimplies (1).

Additionally, Yk has same parity as k. This yields (2) and (3).

Trivially,

⇒∫Sd−1

⇒∫

(Sd−1)2

∫Sd−1

Yk(θ)G (θ)dθ.

∫Sd−1

Yk(θ)G (θ) dθ = 0.

A number of geometric quantities (such as volumes,d − 1-dimensional section areas,...) may be expressed in terms of∫

ω·θ≥0G (θ)(θ · ω)` dθ or

∫ω·θ=0

G (θ) dθ

in which case the corollary is often applicable.

Apply to F (t) = t`1[0,1](t) and F is a Dirac mass at the origin,respectively.

Applications of a Result on Spherical Integration to the Theory ofConvex Sets by K. J. Falconer, The American MathematicalMonthly, (1983).

∫ω·θ=0

G (θ) dθ

∫ω·θ=0

G (θ) dθ

The Busemann–Petty problem

Busemann–Petty (1956): If K1 and K2 are origin-symmetric convexbodies in Rd such that

vold−1(K1 ∩ H) ≤ vold−1(K2 ∩ H)

for all hyperplanes H containing the origin, is it true that

vold(K1) ≤ vold(K2)?

No for d ≥ 5: Larman–Rogers (d ≥ 12), Ball (d ≥ 10),Giannopoulos, Bourgain (d ≥ 7), Gardner, Papadimitrakis (d ≥ 5).

Yes for d ≤ 4: Gardner (d = 3), Zhang (d = 4).

Short proof of d = 4 by Koldobsky using Funk–Hecke theorem.

vold−1(K1 ∩ H) ≤ vold−1(K2 ∩ H)

Applications in analysis and PDE

Suppose T is the integral operator on L2(Sd−1) given by

Tg(θ) =

∫Sd−1

g(ω)K (ω · θ) dω.

Funk–Hecke theorem ⇒ T diagonalises with respect to sphericalharmonics, with explicit expressions for the eigenvalues.

Expanding g =∑

k∈N0Yk ,

Tg(θ) =∑k∈N0

λkYk(θ)

λk = |Sd−2|∫ 1

−1K (t)Pk,d(t)(1− t2)

d−32 dt.

Tg(θ) =

∫Sd−1

Expanding g =∑

k∈N0Yk ,

Tg(θ) =∑k∈N0

λkYk(θ)

λk = |Sd−2|∫ 1

−1K (t)Pk,d(t)(1− t2)

d−32 dt.

Tg(θ) =

∫Sd−1

Expanding g =∑

k∈N0Yk ,

Tg(θ) =∑k∈N0

λkYk(θ)

λk = |Sd−2|∫ 1

−1K (t)Pk,d(t)(1− t2)

d−32 dt.

Lieb’s sharp HLS inequality: On Rd , with λ ∈ (0, d):∣∣∣∣ ∫Rd

f (x)g(y)

|x − y |λdxdy

∣∣∣∣ ≤ π λ2

Γ(d−λ2 )

Γ(d − λ2 )

(Γ(d)

Γ(d2 )

)1−λd

‖f ‖p‖g‖p

where p := 2d2d−λ .

Equality for f = g = (1 + | · |2)−2d−λ

Equivalently (via stereographic projection) on Sd , with λ ∈ (0, d):∣∣∣∣ ∫Sd

F (θ)G (φ)

|θ − φ|λdθdφ

∣∣∣∣ ≤ π λ2

Γ(d−λ2 )

Γ(d − λ2 )

(Γ(d)

Γ(d2 )

)1−λd

‖F‖p‖G‖p

where p := 2d2d−λ . Equality for F = G = 1Sd .

Frank–Lieb: recent rearrangement-free proof using Funk–Hecketheorem. They proved with the same technique the analogoussharp HLS inequality on the Heisenberg group Hn.

For Hn, previously sharp form only known in a special case of λ(Jerison–Lee). Originally considered by Folland–Stein (focus noton sharp constants).

f (x)g(y)

|x − y |λdxdy

∣∣∣∣ ≤ π λ2

Γ(d−λ2 )

Γ(d − λ2 )

(Γ(d)

Γ(d2 )

)1−λd

‖f ‖p‖g‖p

where p := 2d2d−λ . Equality for f = g = (1 + | · |2)−

2d−λ2 .

F (θ)G (φ)

|θ − φ|λdθdφ

∣∣∣∣ ≤ π λ2

Γ(d−λ2 )

Γ(d − λ2 )

(Γ(d)

Γ(d2 )

)1−λd

‖F‖p‖G‖p

f (x)g(y)

|x − y |λdxdy

∣∣∣∣ ≤ π λ2

Γ(d−λ2 )

Γ(d − λ2 )

(Γ(d)

Γ(d2 )

)1−λd

‖f ‖p‖g‖p

2d−λ2 .

F (θ)G (φ)

|θ − φ|λdθdφ

∣∣∣∣ ≤ π λ2

Γ(d−λ2 )

Γ(d − λ2 )

(Γ(d)

Γ(d2 )

)1−λd

‖F‖p‖G‖p

where p := 2d2d−λ .

Equality for F = G = 1Sd .

f (x)g(y)

|x − y |λdxdy

∣∣∣∣ ≤ π λ2

Γ(d−λ2 )

Γ(d − λ2 )

(Γ(d)

Γ(d2 )

)1−λd

‖f ‖p‖g‖p

2d−λ2 .

F (θ)G (φ)

|θ − φ|λdθdφ

∣∣∣∣ ≤ π λ2

Γ(d−λ2 )

Γ(d − λ2 )

(Γ(d)

Γ(d2 )

)1−λd

‖F‖p‖G‖p

f (x)g(y)

|x − y |λdxdy

∣∣∣∣ ≤ π λ2

Γ(d−λ2 )

Γ(d − λ2 )

(Γ(d)

Γ(d2 )

)1−λd

‖f ‖p‖g‖p

2d−λ2 .

F (θ)G (φ)

|θ − φ|λdθdφ

∣∣∣∣ ≤ π λ2

Γ(d−λ2 )

Γ(d − λ2 )

(Γ(d)

Γ(d2 )

)1−λd

‖F‖p‖G‖p

Frank–Lieb: recent rearrangement-free proof using Funk–Hecketheorem.

They proved with the same technique the analogoussharp HLS inequality on the Heisenberg group Hn.

f (x)g(y)

|x − y |λdxdy

∣∣∣∣ ≤ π λ2

Γ(d−λ2 )

Γ(d − λ2 )

(Γ(d)

Γ(d2 )

)1−λd

‖f ‖p‖g‖p

2d−λ2 .

F (θ)G (φ)

|θ − φ|λdθdφ

∣∣∣∣ ≤ π λ2

Γ(d−λ2 )

Γ(d − λ2 )

(Γ(d)

Γ(d2 )

)1−λd

‖F‖p‖G‖p

f (x)g(y)

|x − y |λdxdy

∣∣∣∣ ≤ π λ2

Γ(d−λ2 )

Γ(d − λ2 )

(Γ(d)

Γ(d2 )

)1−λd

‖f ‖p‖g‖p

2d−λ2 .

F (θ)G (φ)

|θ − φ|λdθdφ

∣∣∣∣ ≤ π λ2

Γ(d−λ2 )

Γ(d − λ2 )

(Γ(d)

Γ(d2 )

)1−λd

‖F‖p‖G‖p

For Hn, previously sharp form only known in a special case of λ(Jerison–Lee).

Originally considered by Folland–Stein (focus noton sharp constants).

f (x)g(y)

|x − y |λdxdy

∣∣∣∣ ≤ π λ2

Γ(d−λ2 )

Γ(d − λ2 )

(Γ(d)

Γ(d2 )

)1−λd

‖f ‖p‖g‖p

2d−λ2 .

F (θ)G (φ)

|θ − φ|λdθdφ

∣∣∣∣ ≤ π λ2

Γ(d−λ2 )

Γ(d − λ2 )

(Γ(d)

Γ(d2 )

)1−λd

‖F‖p‖G‖p

Christ–Liu–Zhang (in a pair of papers about 3 weeks ago):established sharp HLS inequalities on quaternionic Heisenberggroups, and the octonionic Heisenberg group.

Used the Frank-Liebstrategy and versions of the Funk–Hecke theorem adapted to thesegroups.

Beckner: solved conjecture of Stein on contraction properties ofPoisson semigroup on Sd using Lieb’s sharp HLS and Funk–Hecketheorem.

Christ–Liu–Zhang (in a pair of papers about 3 weeks ago):established sharp HLS inequalities on quaternionic Heisenberggroups, and the octonionic Heisenberg group. Used the Frank-Liebstrategy and versions of the Funk–Hecke theorem adapted to thesegroups.

Sharp weighted Fourier extension/Kato-smoothing

For (x , t) ∈ Rd × R, consider

Sf (x , t) = |x |−τ2

e i(x ·ξ+ 12t|ξ|2)|ξ|1−

τ2 f (ξ) dξ

where τ ∈ (1, d).

Interpretation as Fourier extension operator associated with theparaboloid {(ξ, 1

2 |ξ|2) : ξ ∈ Rd} ⊂ Rd+1,

or as a solution operator through

Sf (x , t) = (2π)d |x |−τ2 (−∆)

12− τ

4 u(x , t)

where i∂tu + 12 ∆u = 0 with initial data f .

Sf (x , t) = |x |−τ2

e i(x ·ξ+ 12t|ξ|2)|ξ|1−

τ2 f (ξ) dξ

2 |ξ|2) : ξ ∈ Rd} ⊂ Rd+1,

Sf (x , t) = (2π)d |x |−τ2 (−∆)

12− τ

4 u(x , t)

Sf (x , t) = |x |−τ2

e i(x ·ξ+ 12t|ξ|2)|ξ|1−

τ2 f (ξ) dξ

2 |ξ|2) : ξ ∈ Rd} ⊂ Rd+1,

Sf (x , t) = (2π)d |x |−τ2 (−∆)

12− τ

4 u(x , t)

Sf (x , t) = |x |−τ2

e i(x ·ξ+ 12t|ξ|2)|ξ|1−

τ2 f (ξ) dξ

2 |ξ|2) : ξ ∈ Rd} ⊂ Rd+1,

Sf (x , t) = (2π)d |x |−τ2 (−∆)

12− τ

4 u(x , t)

Recall

Sf (x , t) = |x |−τ2

e i(x ·ξ+ 12t|ξ|2)|ξ|1−

τ2 f (ξ) dξ.

Theorem (B–Sugimoto)

For each k ∈ N0,S∗Sf = λk f

wheref (η) = Yk( η

|η|)f0(|η|)|η|−d−1

Yk is any spherical harmonic of degree k, f0 ∈ L2(0,∞), and

λk = (2π)d+121−τ Γ(τ − 1)Γ(k + d−τ2 )

Γ( τ2 )2Γ(k + d+τ2 − 1)

Moreover, (λk)k∈N0 is a strictly decreasing sequence.

Hence, ‖S‖L2(Rd )→L2(Rd+1) =√λ0 which is attained if and only if f

is radially symmetric.

Recall

Sf (x , t) = |x |−τ2

e i(x ·ξ+ 12t|ξ|2)|ξ|1−

τ2 f (ξ) dξ.

|η|)f0(|η|)|η|−d−1

λk = (2π)d+121−τ Γ(τ − 1)Γ(k + d−τ2 )

Γ( τ2 )2Γ(k + d+τ2 − 1)

Recall

Sf (x , t) = |x |−τ2

e i(x ·ξ+ 12t|ξ|2)|ξ|1−

τ2 f (ξ) dξ.

|η|)f0(|η|)|η|−d−1

λk = (2π)d+121−τ Γ(τ − 1)Γ(k + d−τ2 )

Γ( τ2 )2Γ(k + d+τ2 − 1)

Recall

Sf (x , t) = |x |−τ2

e i(x ·ξ+ 12t|ξ|2)|ξ|1−

τ2 f (ξ) dξ.

|η|)f0(|η|)|η|−d−1

λk = (2π)d+121−τ Γ(τ − 1)Γ(k + d−τ2 )

Γ( τ2 )2Γ(k + d+τ2 − 1)

Relabelling s = τ2 − 1 (τ ∈ (1, d)⇔ s ∈ (−1

2 ,d2 − 1)).

Corollary (Simon, Watanabe)

Let d ≥ 2 and s ∈ (−12 ,

d2 − 1). If i∂tu + 1

2 ∆u = 0 then∫R

|u(x , t)|2 dxdt

|x |2(s+1)≤ C(d , s)‖u(0)‖2

Hs(Rd ),

C(d , s) = π2−2s Γ(2s + 1)Γ(d2 − s − 1)

Γ(s + 1)2Γ(d2 + s).

The constant is sharp and equality holds if and only ifu(0) ∈ Hs(Rd) is radially symmetric.

Relabelling s = τ2 − 1 (τ ∈ (1, d)⇔ s ∈ (−1

2 ,d2 − 1)).

Corollary (Simon, Watanabe)

Let d ≥ 2 and s ∈ (−12 ,

d2 − 1). If i∂tu + 1

2 ∆u = 0 then∫R

|u(x , t)|2 dxdt

|x |2(s+1)≤ C(d , s)‖u(0)‖2

Hs(Rd ),

C(d , s) = π2−2s Γ(2s + 1)Γ(d2 − s − 1)

Γ(s + 1)2Γ(d2 + s).

The constant is sharp and equality holds if and only ifu(0) ∈ Hs(Rd) is radially symmetric.

A calculation shows that

S∗Sf (η) = C(d , τ)

∫Sd−1

f (|η|ω)

|ω − η′|d−τdω

for some explicitly computable constant C(d , τ). Here, η′ = η|η| .

Applied to

f (η) = Yk( η|η|)f0(|η|)|η|−

d−12

the angular and radial variables split, and the result follows fromthe Funk–Hecke theorem.

A calculation shows that

S∗Sf (η) = C(d , τ)

∫Sd−1

f (|η|ω)

|ω − η′|d−τdω

for some explicitly computable constant C(d , τ). Here, η′ = η|η| .

Applied to

f (η) = Yk( η|η|)f0(|η|)|η|−

d−12

the angular and radial variables split, and the result follows fromthe Funk–Hecke theorem.

The 12-derivative endpoint: localisation

Recall that for d ≥ 2 and s ∈ (−12 ,

d2 − 1),∫

|u(x , t)|2 dxdt

|x |2(s+1)≤ C(d , s)‖u(0)‖2

Hs(Rd ),

whenever i∂tu + 12 ∆u = 0, and this fails for s = −1

One recovers a full half-derivative gain by spatial localising

supR>0

∫|x |≤R

|∇u(x , t)|2 dxdt ≤ C‖u(0)‖2

H12 (Rd )

Proved separately by Constantin–Saut, Sjolin and Vega.

Vega–Visciglia proved there is a reverse form

supR>0

∫|x |≤R

|∇u(x , t)|2 dxdt ≥ c‖u(0)‖2

H12 (Rd )

d2 − 1),∫

|u(x , t)|2 dxdt

|x |2(s+1)≤ C(d , s)‖u(0)‖2

Hs(Rd ),

supR>0

∫|x |≤R

|∇u(x , t)|2 dxdt ≤ C‖u(0)‖2

H12 (Rd )

supR>0

∫|x |≤R

|∇u(x , t)|2 dxdt ≥ c‖u(0)‖2

H12 (Rd )

d2 − 1),∫

|u(x , t)|2 dxdt

|x |2(s+1)≤ C(d , s)‖u(0)‖2

Hs(Rd ),

supR>0

∫|x |≤R

|∇u(x , t)|2 dxdt ≤ C‖u(0)‖2

H12 (Rd )

supR>0

∫|x |≤R

|∇u(x , t)|2 dxdt ≥ c‖u(0)‖2

H12 (Rd )

The 12-derivative endpoint: angular regularity

Theorem (≤ Hoshiro, Sugimoto; ≥ Fang–Wang)

Let d ≥ 2 and s ∈ (−12 ,

d2 − 1). Then there exist constants

0 < c(d , s) ≤ C(d , s) <∞

such that whenever i∂tu + 12 ∆u = 0 we have

|(1− Λ)1+2s

4 u(x , t)|2 dxdt

|x |2(1+s)≤ C(d , s)‖u(0)‖2

Hs(Rd )

and∫R

|(1− Λ)1+2s

4 u(x , t)|2 dxdt

|x |2(1+s)≥ c(d , s)‖u(0)‖2

Hs(Rd ).

Here, Λ is the Laplace–Beltrami operator on Sd−1. (Heuristically,(−Λ)σ ∼ |x |2σD2σ.)

Let d ≥ 2 and s ∈ (−12 ,

0 < c(d , s) ≤ C(d , s) <∞

such that whenever i∂tu + 12 ∆u = 0 we have∫

|(1− Λ)1+2s

4 u(x , t)|2 dxdt

|x |2(1+s)≤ C(d , s)‖u(0)‖2

Hs(Rd )

and∫R

|(1− Λ)1+2s

4 u(x , t)|2 dxdt

|x |2(1+s)≥ c(d , s)‖u(0)‖2

Hs(Rd ).

Here, Λ is the Laplace–Beltrami operator on Sd−1.

(Heuristically,(−Λ)σ ∼ |x |2σD2σ.)

Let d ≥ 2 and s ∈ (−12 ,

0 < c(d , s) ≤ C(d , s) <∞

such that whenever i∂tu + 12 ∆u = 0 we have∫

|(1− Λ)1+2s

4 u(x , t)|2 dxdt

|x |2(1+s)≤ C(d , s)‖u(0)‖2

Hs(Rd )

and∫R

|(1− Λ)1+2s

4 u(x , t)|2 dxdt

|x |2(1+s)≥ c(d , s)‖u(0)‖2

Hs(Rd ).

Here, Λ is the Laplace–Beltrami operator on Sd−1. (Heuristically,(−Λ)σ ∼ |x |2σD2σ.)

Let d ≥ 2, s ∈ (−12 ,

d2 − 1), and θ a function on [0,∞).

For eachk ∈ N0, define

βk = π2−2s Γ(2s + 1)Γ(k + d2 − s − 1)

Γ(s + 1)2Γ(k + d2 + s)

|θ(k(k + d − 2))|2.

Setb = inf

`∈N0

β`, B = sup`∈N0

k = {k ∈ N0 : b = βk} and K = {k ∈ N0 : B = βk}.

Write Hk for the space of all linear combinations of functions

η 7→ Yk( η|η|)f0(|η|)|η|−

d−12

where Yk is a spherical harmonic of order k and f0 ∈ L2(0,∞).

Let d ≥ 2, s ∈ (−12 ,

d2 − 1), and θ a function on [0,∞). For each

k ∈ N0, define

βk = π2−2s Γ(2s + 1)Γ(k + d2 − s − 1)

Γ(s + 1)2Γ(k + d2 + s)

|θ(k(k + d − 2))|2.

Setb = inf

`∈N0

β`, B = sup`∈N0

η 7→ Yk( η|η|)f0(|η|)|η|−

d−12

Let d ≥ 2, s ∈ (−12 ,

k ∈ N0, define

βk = π2−2s Γ(2s + 1)Γ(k + d2 − s − 1)

Γ(s + 1)2Γ(k + d2 + s)

|θ(k(k + d − 2))|2.

Setb = inf

`∈N0

β`, B = sup`∈N0

η 7→ Yk( η|η|)f0(|η|)|η|−

d−12

Let d ≥ 2, s ∈ (−12 ,

k ∈ N0, define

βk = π2−2s Γ(2s + 1)Γ(k + d2 − s − 1)

Γ(s + 1)2Γ(k + d2 + s)

|θ(k(k + d − 2))|2.

Setb = inf

`∈N0

β`, B = sup`∈N0

η 7→ Yk( η|η|)f0(|η|)|η|−

d−12

If i∂tu + 12 ∆u = 0 then

b ‖u(0)‖2Hs(Rd )

≤∫R

|θ(−Λ) u(x , t)|2 dxdt

|x |2(1+s)≤ B ‖u(0)‖2

Hs(Rd )

and the constants are sharp. Also, u(0) is an extremiser for thelower bound if and only if Dsu(0) ∈

⊕k∈kHk , and an extremiser

for the upper bound if and only if Dsu(0) ∈⊕

k∈KHk .

Hoshiro, Sugimoto and Fang–Wang equivalences correspond to

θ(ρ) = (1 + ρ)1+2s

Stirling’s formula easily gives b > 0 and B <∞ in this case.

b ‖u(0)‖2Hs(Rd )

≤∫R

|x |2(1+s)≤ B ‖u(0)‖2

Hs(Rd )

k∈KHk .

θ(ρ) = (1 + ρ)1+2s

b ‖u(0)‖2Hs(Rd )

≤∫R

|x |2(1+s)≤ B ‖u(0)‖2

Hs(Rd )

k∈KHk .

θ(ρ) = (1 + ρ)1+2s

b ‖u(0)‖2Hs(Rd )

≤∫R

|x |2(1+s)≤ B ‖u(0)‖2

Hs(Rd )

k∈KHk .

θ(ρ) = (1 + ρ)1+2s

(Re-)Relabelling τ = 2(1 + s) (so that τ ∈ (1, d)).

Seek b = infk∈N0 βk and B = supk∈N0βk , where

βk = π22−τ Γ(τ − 1)Γ(k + d−τ2 )

Γ( τ2 )2Γ(k + d+τ2 − 1)

(1 + k(k + d − 2))τ−1

We also want to identify

k = {k ∈ N0 : b = βk}, K = {k ∈ N0 : B = βk}.

For d ≥ 5, define τ∗, τ∗ ∈ (1, d), τ∗ ≤ τ∗, by

dτ∗−1

(d − τ∗

d + τ∗2− 1 and Γ(d−τ

2 ) = Γ(d+τ∗

2 − 1).

βk = π22−τ Γ(τ − 1)Γ(k + d−τ2 )

Γ( τ2 )2Γ(k + d+τ2 − 1)

(1 + k(k + d − 2))τ−1

k = {k ∈ N0 : b = βk}, K = {k ∈ N0 : B = βk}.

dτ∗−1

(d − τ∗

2 ) = Γ(d+τ∗

2 − 1).

βk = π22−τ Γ(τ − 1)Γ(k + d−τ2 )

Γ( τ2 )2Γ(k + d+τ2 − 1)

(1 + k(k + d − 2))τ−1

k = {k ∈ N0 : b = βk}, K = {k ∈ N0 : B = βk}.

dτ∗−1

(d − τ∗

2 ) = Γ(d+τ∗

2 − 1).

βk = π22−τ Γ(τ − 1)Γ(k + d−τ2 )

Γ( τ2 )2Γ(k + d+τ2 − 1)

(1 + k(k + d − 2))τ−1

k = {k ∈ N0 : b = βk}, K = {k ∈ N0 : B = βk}.

dτ∗−1

(d − τ∗

2 ) = Γ(d+τ∗

2 − 1).

For d = 5 and τ ∈ [τ∗, 5), let k(τ) ∈ [0,∞) satisfy

2k(τ) + 5− τ2k(τ) + 3 + τ

(1 + (k(τ) + 1)(k(τ) + 4)

1 + k(τ)(k(τ) + 3)

) τ−12

and let k∗(τ) = ceiling(k(τ)).

We knowC1

(5− τ)1/4≤ k(τ) ≤ C2

(5− τ)1/2

for some positive constants C1 and C2.

For d = 5 and τ ∈ [τ∗, 5), let k(τ) ∈ [0,∞) satisfy

2k(τ) + 5− τ2k(τ) + 3 + τ

(1 + (k(τ) + 1)(k(τ) + 4)

1 + k(τ)(k(τ) + 3)

) τ−12

and let k∗(τ) = ceiling(k(τ)).

We knowC1

(5− τ)1/4≤ k(τ) ≤ C2

(5− τ)1/2

for some positive constants C1 and C2.

The case θ(ρ) = (1 + ρ)τ−1

4 , τ ∈ (1, d)

Let d ≥ 2, τ ∈ (1, d) and θ(ρ) = (1 + ρ)τ−1

4 . Then

(d , τ) b B

d = 2, 3 limk→∞ βk β0

d = 4, τ ∈ (1, 2) β0 limk→∞ βkd = 4, τ = 2 π π

d = 4, τ ∈ (2, 4) limk→∞ βk β0

d = 5, τ ∈ (1, τ∗) β0 limk→∞ βkd = 5, τ ∈ [τ∗, τ

∗) βk∗(τ) limk→∞ βkd = 5, τ ∈ [τ∗, 5) βk∗(τ) β0

d ≥ 6, τ ∈ (1, τ∗) β0 limk→∞ βkd ≥ 6, τ ∈ [τ∗, τ

∗) β1 limk→∞ βkd ≥ 6, τ ∈ [τ∗, d) β1 β0

4 , τ ∈ (1, d)

Let d ≥ 2, τ ∈ (1, d) and θ(ρ) = (1 + ρ)τ−1

4 . Then

(d , τ) b B

d = 2, 3 limk→∞ βk β0

d = 4, τ ∈ (1, 2) β0 limk→∞ βkd = 4, τ = 2 π π

d = 4, τ ∈ (2, 4) limk→∞ βk β0

d = 5, τ ∈ (1, τ∗) β0 limk→∞ βkd = 5, τ ∈ [τ∗, τ

∗) βk∗(τ) limk→∞ βkd = 5, τ ∈ [τ∗, 5) βk∗(τ) β0

d ≥ 6, τ ∈ (1, τ∗) β0 limk→∞ βkd ≥ 6, τ ∈ [τ∗, τ

∗) β1 limk→∞ βkd ≥ 6, τ ∈ [τ∗, d) β1 β0

4 , τ ∈ (1, d)

(d , τ) k K

d = 2, 3 ∅ {0}d = 4, τ ∈ (1, 2) {0} ∅

d = 4, τ = 2 N0 N0

d = 4, τ ∈ (2, 4) ∅ {0}d = 5, τ ∈ (1, τ∗) {0} ∅

d = 5, τ ∈ [τ∗, τ∗), k(τ) /∈ N0 {k∗(τ)} ∅

d = 5, τ ∈ [τ∗, τ∗), k(τ) ∈ N0 {k∗(τ), k∗(τ) + 1} ∅

d = 5, τ ∈ [τ∗, 5), k(τ) /∈ N0 {k∗(τ)} {0}d = 5, τ ∈ [τ∗, 5), k(τ) ∈ N0 {k∗(τ), k∗(τ) + 1} {0}

d ≥ 6, τ ∈ (1, τ∗) {0} ∅d ≥ 6, τ = τ∗ {0,1} ∅

d ≥ 6, τ ∈ (τ∗, τ∗) {1} ∅

d ≥ 6, τ ∈ [τ∗, d) {1} {0}

Sharp trace theorem on Sd−1

Let Rf := f |Sd−1 and

Tθg := R θ(−Λ)D−sg . Also

λk = 21−2s Γ(2s − 1)Γ(k + d2 − s)

Γ(s)2Γ(k + d2 − 1 + s)

|θ(k(k + d − 2))|2.

and K = {k ∈ N0 : λk = sup`∈N0λ`}.

Theorem (B–Machihara–Sugimoto)

Let d ≥ 2 and s ∈ ( 12 ,

d2 ). Then

‖R θ(−Λ)f ‖2L2(Sd−1) ≤ sup

k∈N0

λk‖f ‖2Hs(Rd )

where the constant is optimal and the space of extremisers isprecisely the nonzero elements of D−sT ∗θ (

⊕k∈KHk).

Trace theorems with angular regularity considered by Fang–Wang.

Let Rf := f |Sd−1 and Tθg := R θ(−Λ)D−sg . Also

λk = 21−2s Γ(2s − 1)Γ(k + d2 − s)

Γ(s)2Γ(k + d2 − 1 + s)

|θ(k(k + d − 2))|2.

and K = {k ∈ N0 : λk = sup`∈N0λ`}.

Let d ≥ 2 and s ∈ ( 12 ,

d2 ). Then

‖R θ(−Λ)f ‖2L2(Sd−1) ≤ sup

k∈N0

λk‖f ‖2Hs(Rd )

⊕k∈KHk).

λk = 21−2s Γ(2s − 1)Γ(k + d2 − s)

Γ(s)2Γ(k + d2 − 1 + s)

|θ(k(k + d − 2))|2.

and K = {k ∈ N0 : λk = sup`∈N0λ`}.

Let d ≥ 2 and s ∈ ( 12 ,

d2 ). Then

‖R θ(−Λ)f ‖2L2(Sd−1) ≤ sup

k∈N0

λk‖f ‖2Hs(Rd )

⊕k∈KHk).

λk = 21−2s Γ(2s − 1)Γ(k + d2 − s)

Γ(s)2Γ(k + d2 − 1 + s)

|θ(k(k + d − 2))|2.

and K = {k ∈ N0 : λk = sup`∈N0λ`}.

Let d ≥ 2 and s ∈ ( 12 ,

d2 ). Then

‖R θ(−Λ)f ‖2L2(Sd−1) ≤ sup

k∈N0

λk‖f ‖2Hs(Rd )

⊕k∈KHk).

When θ = 1 we have

T1T ∗1 G (ω) = 2−2sπ−d2

Γ(d2 − s)

∫Sd−1

G (ϕ)

|ω − ϕ|d−2sdσ(ϕ)

and we may apply the Funk–Hecke theorem.

In this case, f is an extremiser if and only if

f ∈ span(| · |2s−d ∗ dσ) \ {0}.

When d = 3, for s ∈ ( 12 ,

| · |2s−3 ∗ dσ(x) = C (s)(|x |+ 1)2s−1 −

∣∣|x | − 1∣∣2s−1

For (d , s) = (3, 1), extremisers for

‖f |S2‖L2(dσ) ≤ ‖∇f ‖L2(R3)

are precisely nonzero multiples of

f (x) =

{1 if |x | ≤ 11|x | if |x | > 1

When θ = 1 we have

T1T ∗1 G (ω) = 2−2sπ−d2

Γ(d2 − s)

∫Sd−1

G (ϕ)

f ∈ span(| · |2s−d ∗ dσ) \ {0}.

When d = 3, for s ∈ ( 12 ,

| · |2s−3 ∗ dσ(x) = C (s)(|x |+ 1)2s−1 −

∣∣|x | − 1∣∣2s−1

‖f |S2‖L2(dσ) ≤ ‖∇f ‖L2(R3)

f (x) =

{1 if |x | ≤ 11|x | if |x | > 1

When θ = 1 we have

T1T ∗1 G (ω) = 2−2sπ−d2

Γ(d2 − s)

∫Sd−1

G (ϕ)

f ∈ span(| · |2s−d ∗ dσ) \ {0}.

When d = 3, for s ∈ ( 12 ,

| · |2s−3 ∗ dσ(x) = C (s)(|x |+ 1)2s−1 −

∣∣|x | − 1∣∣2s−1

‖f |S2‖L2(dσ) ≤ ‖∇f ‖L2(R3)

f (x) =

{1 if |x | ≤ 11|x | if |x | > 1

When θ = 1 we have

T1T ∗1 G (ω) = 2−2sπ−d2

Γ(d2 − s)

∫Sd−1

G (ϕ)

f ∈ span(| · |2s−d ∗ dσ) \ {0}.

When d = 3, for s ∈ ( 12 ,

| · |2s−3 ∗ dσ(x) = C (s)(|x |+ 1)2s−1 −

∣∣|x | − 1∣∣2s−1

‖f |S2‖L2(dσ) ≤ ‖∇f ‖L2(R3)

f (x) =

{1 if |x | ≤ 11|x | if |x | > 1

Strichartz estimates for the wave equation

For d ≥ 2 and s ∈[

‖e it√−∆f ‖Lp(Rd+1) ≤ C‖f ‖Hs(Rd )

for each f ∈ Hs(Rd) and where

p =2(d + 1)

d − 2s.

The sharp constant and a full characterisation of extremisers isonly known in some rather isolated cases.

Strichartz estimates for the wave equation

For d ≥ 2 and s ∈[

‖e it√−∆f ‖Lp(Rd+1) ≤ C‖f ‖Hs(Rd )

for each f ∈ Hs(Rd) and where

p =2(d + 1)

d − 2s.

The sharp constant and a full characterisation of extremisers isonly known in some rather isolated cases.

It is known that, for all admissible (d , s), an extremiser exists(Bulut, Fanelli–Vega–Visciglia, Ramos).

Finding the exact shape of such extremisers appears to be a ratherdifficult problem; known when (d , s) = (2, 1

2 ), (3, 12 ) (Foschi) and

(d , s) = (5, 1) (B-Rogers).

In each of these cases, the initial datum f? such that

f?(ξ) =e−|ξ|

is extremal.

Finding the exact shape of such extremisers appears to be a ratherdifficult problem;

known when (d , s) = (2, 12 ), (3, 1

2 ) (Foschi) and(d , s) = (5, 1) (B-Rogers).

f?(ξ) =e−|ξ|

is extremal.

2 ), (3, 12 ) (Foschi) and

(d , s) = (5, 1) (B-Rogers).

f?(ξ) =e−|ξ|

is extremal.

2 ), (3, 12 ) (Foschi) and

(d , s) = (5, 1) (B-Rogers).

f?(ξ) =e−|ξ|

is extremal.

Theorem (B-Jeavons)

The one-sided wave propagator satisfies the estimate

‖e it√−∆f ‖L4(R5) ≤

‖f ‖H

34 (R4)

where the constant is sharp and is attained if and only if

f (ξ) =ea|ξ|+ib·ξ+c

where a, c ∈ C such that Re(a) < 0, and b ∈ Rd .

Theorem (B-Rogers)

Let d ≥ 3. Then

‖e it√−∆f ‖4

L4(Rd+1)

≤ C(d)

∫(Rd )2

|f (y1)|2|f (y2)|2|y1|d−1

2 |y2|d−1

2 (1− y ′1 · y ′2)d−3

2 dy1dy2

holds with sharp constant

C(d) = 2−d−1

2 (2π)−3d+1|Sd−1|

which is attained if and only if

f (ξ) =ea|ξ|+b·ξ+c

where a, c ∈ C, b ∈ Cd with |Re(b)| < −Re(a).

Theorem (B-Rogers)

Let d ≥ 3. Then

‖e it√−∆f ‖4

L4(Rd+1)

≤ C(d)

∫(Rd )2

|f (y1)|2|f (y2)|2|y1|d−1

2 |y2|d−1

2 (1− y ′1 · y ′2)d−3

2 dy1dy2

holds with sharp constant

C(d) = 2−d−1

2 (2π)−3d+1|Sd−1|

which is attained if and only if

f (ξ) =ea|ξ|+b·ξ+c

where a, c ∈ C, b ∈ Cd with |Re(b)| < −Re(a).

Polar coordinates gives

‖e it√−∆f ‖4

L4(Rd+1) ≤C(d)

2d−3

∫(Sd−1)2

g(ω1)g(ω2)|ω1 − ω2|d−3 dω1dω2

g(ω) =

∫ ∞0|f (rω)|2r

3(d−1)2 dr .

Note that ∫Sd−1

g = (2π)d‖f ‖2

Hd−1

Polar coordinates gives

‖e it√−∆f ‖4

L4(Rd+1) ≤C(d)

2d−3

∫(Sd−1)2

g(ω1)g(ω2)|ω1 − ω2|d−3 dω1dω2

g(ω) =

∫ ∞0|f (rω)|2r

3(d−1)2 dr .

Note that ∫Sd−1

g = (2π)d‖f ‖2

Hd−1

Hρ(g) =

∫(Sd−1)2

g(ω1)g(ω2)|ω1 − ω2|−ρ dω1dω2

and µg = 1|Sd−1|

∫Sd−1 g .

Theorem (B-Jeavons)

Let −2 < ρ < 0, and let g ∈ L1(Sd−1). Then

Hρ(g) ≤ Hρ(µg1) = 2d−2−ρB(d−1−ρ2 , d−1

2 )|Sd−2||Sd−1|

∣∣∣∣∫Sd−1

∣∣∣∣2and equality holds if and only if g is constant.

Hρ(g) =

∫(Sd−1)2

g(ω1)g(ω2)|ω1 − ω2|−ρ dω1dω2

and µg = 1|Sd−1|

∫Sd−1 g .

Theorem (B-Jeavons)

Hρ(g) ≤ Hρ(µg1) = 2d−2−ρB(d−1−ρ2 , d−1

2 )|Sd−2||Sd−1|

∣∣∣∣∫Sd−1

Hρ(g) =

∫(Sd−1)2

g(ω1)g(ω2)|ω1 − ω2|−ρ dω1dω2

and µg = 1|Sd−1|

∫Sd−1 g .

Theorem (B-Jeavons)

Hρ(g) ≤ Hρ(µg1) = 2d−2−ρB(d−1−ρ2 , d−1

2 )|Sd−2||Sd−1|

∣∣∣∣∫Sd−1

Expanding g =∑

k∈N0Yk , the Funk–Hecke theorem gives

Hρ(g) = 2−ρ2

∑k≥0

Ik(d , ρ)

∫Sd−1

|Yk(ω)|2 dω

Ik(d , ρ) = |Sd−2|∫ 1

−1(1− t)−

ρ2 Pk,d(t)(1− t2)

d−32 dt.

LemmaIf −2 < ρ < 0 then

I0(d , ρ) = |Sd−2|2d−2− ρ2 B(d−1−ρ

2 , d−12 ) > 0

and Ik(d , ρ) < 0 for all k ≥ 1.

Hρ(g) ≤ 2−ρ2 I0(d , ρ)

∫Sd−1

|Y0|2 dω = Hρ(µg1).

The lemma fails for ρ < −2 because I2(d , ρ) > 0.

Applying it with (d , ρ) = (4,−1) gives

‖e it√−∆f ‖4

L4(R4+1) ≤C(4)√

∫(S3)2

g(ω1)g(ω2)|ω1 − ω2|dω1dω2

≤ C(4)√2

H−1(1)|µg |2

15π2‖f ‖4

H34 (R4)

Hρ(g) ≤ 2−ρ2 I0(d , ρ)

∫Sd−1

|Y0|2 dω = Hρ(µg1).

‖e it√−∆f ‖4

L4(R4+1) ≤C(4)√

∫(S3)2

≤ C(4)√2

H−1(1)|µg |2

15π2‖f ‖4

H34 (R4)

Hρ(g) ≤ 2−ρ2 I0(d , ρ)

∫Sd−1

|Y0|2 dω = Hρ(µg1).

‖e it√−∆f ‖4

L4(R4+1) ≤C(4)√

∫(S3)2

≤ C(4)√2

H−1(1)|µg |2

15π2‖f ‖4

H34 (R4)

For higher dimensions, we want to use the case (d , ρ) = (d , 3− d).But 3− d < −2 for d ≥ 6.

The case (d , ρ) = (3,−1) was used by Foschi to prove thatconstant functions are extremisers in the Stein–Tomas inequality

‖gdσ‖L4(R3) ≤ C‖g‖L2(dσ).

This was partially extended to higher dimensions byCarneiro–Oliveira e Silva, with a similar obstruction preventing ageneralisation to all dimensions.

Thanks for listening...!

SOME APPLICATIONS OF THE FUNK-HECKE THEOREMicm-sch.math.snu.ac.kr/icm-talk/BezICMKorea.pdf ·...

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