SLIDE 1 Generalized BMO spaces
- n Riemannian manifolds
- G. Mauceri 1
- S. Meda 2
- M. Vallarino 3
1 Dipartimento di Matematica
Università di Genova
2 Dipartimento di Matematica
Università di Milano Bicocca
3 Dipartimento di Matematica
Politecnico di Torino
XXXIII Convegno Nazionale di Analisi Armonica Alba, 17-20 Giugno 2013
SLIDE 2 The classical H1 −BMO theory
(X,d,µ) doubling metric measure space H1(X) = f =
λjaj : aj CW-atoms,
a is a CW-atom iff
◮
suppa ⊂ B
◮
a2 ≤ µ(B)−1/2
◮
SLIDE 3 The classical H1 −BMO theory
(X,d,µ) doubling metric measure space H1(X) = f =
λjaj : aj CW-atoms,
a is a CW-atom iff
◮
suppa ⊂ B
◮
a2 ≤ µ(B)−1/2
◮
fH1 = inf
f =
λjaj.
SLIDE 4 The classical H1 −BMO theory
BMO(X) is the space of functions f such that fBMO = sup
B
inf
c∈C
µ(B)
|f(x)−c|2 dµ(x) 1/2
SLIDE 5 The classical H1 −BMO theory
BMO(X) is the space of functions f such that fBMO = sup
B
inf
c∈C
µ(B)
|f(x)−c|2 dµ(x) 1/2 = sup
B
µ(B)−1/2πB(f)L2(B) πB : ⊥ projection on the orthogonal of constants in L2(B) .
SLIDE 6 The classical H1 −BMO theory
◮
H1(X)∗ = BMO(X)
◮
CZO′s : H1(X) → L1(X), L∞(X) → BMO(X)
◮
[H1(X),BMO(X)]θ = Lp(X) , θ = 1−1/p
SLIDE 7 The classical H1 −BMO theory
◮
H1(X)∗ = BMO(X)
◮
CZO′s : H1(X) → L1(X), L∞(X) → BMO(X)
◮
[H1(X),BMO(X)]θ = Lp(X) , θ = 1−1/p
◮
Y Banach space, T linear operator. If TaY ≤ C ∀a CW-atoms then TH1→Y ≤ C . [Meda, Sjögren,Vallarino]
SLIDE 8 Riemannian manifolds
(M,g) noncompact, complete Riemannian manifold dµ(x) =
- detg(x)dx Riemannian measure
L = −divgrad Laplace-Beltrami operator
SLIDE 9 Riemannian manifolds
(M,g) noncompact, complete Riemannian manifold dµ(x) =
- detg(x)dx Riemannian measure
L = −divgrad Laplace-Beltrami operator Natural singular integrals on M
◮ Riesz transforms: ∇L−1/2 , ∇kL−k/2 ◮ spectral multipliers: m(L) , Liu,u ∈ R
SLIDE 10 Riemannian manifolds
(M,g) noncompact, complete Riemannian manifold dµ(x) =
- detg(x)dx Riemannian measure
L = −divgrad Laplace-Beltrami operator Natural singular integrals on M
◮ Riesz transforms: ∇L−1/2 , ∇kL−k/2 ◮ spectral multipliers: m(L) , Liu,u ∈ R
Develop an analogue of the H1 −BMO theory on M providing end-point estimates for ∇kL−k/2 , m(L) .
SLIDE 11 Previous results
(M,g,µ) doubling
◮ [Russ], [Marias, Russ] ◮ [Auscher, McIntosh, Russ]
SLIDE 12 Previous results
(M,g,µ) doubling
◮ [Russ], [Marias, Russ] ◮ [Auscher, McIntosh, Russ]
(M,g,µ) non-doubling: local Hardy spaces
◮ [Taylor] ◮ [Carbonaro, McIntosh, Morris]
Endpoint estimates for s. i. operator that have only local singularities. But ∇kL−k/2 , Liu are singular also at ∞ .
SLIDE 13 Previous results
(M,g,µ) doubling
◮ [Russ], [Marias, Russ] ◮ [Auscher, McIntosh, Russ]
(M,g,µ) non-doubling: local Hardy spaces
◮ [Taylor] ◮ [Carbonaro, McIntosh, Morris]
Endpoint estimates for s. i. operator that have only local singularities. But ∇kL−k/2 , Liu are singular also at ∞ . (M,g,µ) non-doubling: global Hardy spaces
◮ [GM, Meda, Vallarino]
SLIDE 14 A class of nondoubling manifolds
We assume
◮
M has bounded geometry inj M > 0, Ric M bounded below
◮
M has spectral gap b = infσ2(L) > 0 (M,g,µ) is locally doubling but not globally doubling.
SLIDE 15 A class of nondoubling manifolds
We assume
◮
M has bounded geometry inj M > 0, Ric M bounded below
◮
M has spectral gap b = infσ2(L) > 0 (M,g,µ) is locally doubling but not globally doubling. Examples
◮ noncompact semisimple Lie groups with finite centre and any
invariant metric
◮ noncompact symmetric spaces with Killing metric ◮ Damek-Ricci spaces ◮ Cartan-Hadamard manifolds with spectral gap.
SLIDE 16 The space H1
A function a ∈ As (atoms at scale s ) if (1) suppa ⊂ B, rB < s (2) a2 ≤ µ(B)−1/2 (3)
H1
s (M) =
j λj a,
aj ∈ As,
s = inf
λj a, aj ∈ As [Russ], [Carbonaro, M, Meda]
SLIDE 17 Properties of H1
Main properties [Carbonaro, M, Meda]
◮
H1
s1(M) = H1 s2(M) ,
fH1
s1 ≈ fH1 s2
Henceforth H1(M) = H1
s0(M) ,
s0 = 1
2 inj M ◮
[H1(M),L2(M)]θ = Lp(M) , θ = 2(1−1/p)
SLIDE 18 Properties of H1
Main properties [Carbonaro, M, Meda]
◮
H1
s1(M) = H1 s2(M) ,
fH1
s1 ≈ fH1 s2
Henceforth H1(M) = H1
s0(M) ,
s0 = 1
2 inj M ◮
[H1(M),L2(M)]θ = Lp(M) , θ = 2(1−1/p) However Liu and ∇L−1/2 do not map H1(M) → L1(M) More cancellation is needed
SLIDE 19 Special atoms and the space X 1(M)
Quasiharmonic functions q(M) = {u ∈ C∞(M) : Lu = const in M} q2(B) =
- u ∈ L2(B) : Lu = const in B
- q2(B) =
- u : Lu = const in a ngbhd of B
- q(M) ⊂ q2(B) ⊂ q2(B)
SLIDE 20 Special atoms and the space X 1(M)
Quasiharmonic functions q(M) = {u ∈ C∞(M) : Lu = const in M} q2(B) =
- u ∈ L2(B) : Lu = const in B
- q2(B) =
- u : Lu = const in a ngbhd of B
- q(M) ⊂ q2(B) ⊂ q2(B)
A function A ∈ SAs (special atoms at scale s ) if (1) suppA ⊂ B, rB < s (2) A2 ≤ µ(B)−1/2 (3)
∀u ∈ q2(B).
SLIDE 21 Special atoms and the space X 1(M)
Quasiharmonic functions q(M) = {u ∈ C∞(M) : Lu = const in M} q2(B) =
- u ∈ L2(B) : Lu = const in B
- q2(B) =
- u : Lu = const in a ngbhd of B
- q(M) ⊂ q2(B) ⊂ q2(B)
A function A ∈ SAs (special atoms at scale s ) if (1) suppA ⊂ B, rB < s (2) A2 ≤ µ(B)−1/2 (3)
∀u ∈ q2(B). If B has no “holes" and ∂B is smooth then (3) is equivalent to (3’)
∀u ∈ q(M). (3”)
∀u ∈ q2(B) True if s ≤ s0 = 1
2 inj M .
SLIDE 22 The space X 1
X 1
s =
j λj A,
Aj ∈ SAs,
s = inf
λj A, Aj ∈ SAs
SLIDE 23 The space X 1
X 1
s =
j λj A,
Aj ∈ SAs,
s = inf
λj A, Aj ∈ SAs Main properties [M, Meda, Vallarino]
◮
U = L(σI +L)−1 : H1 → X 1
s
is an isomorphism for all s > 0 and σ large enough (hard work!).
SLIDE 24 The space X 1
X 1
s =
j λj A,
Aj ∈ SAs,
s = inf
λj A, Aj ∈ SAs Main properties [M, Meda, Vallarino]
◮
U = L(σI +L)−1 : H1 → X 1
s
is an isomorphism for all s > 0 and σ large enough (hard work!).
◮
X 1
s1 = X 1 s2 ,
fX 1
s1 ≈ fX 1 s2
Henceforth X 1 = X 1
s0 ,
s0 = 1
2 inj M
SLIDE 25 The space X 1
X 1
s =
j λj A,
Aj ∈ SAs,
s = inf
λj A, Aj ∈ SAs Main properties [M, Meda, Vallarino]
◮
U = L(σI +L)−1 : H1 → X 1
s
is an isomorphism for all s > 0 and σ large enough (hard work!).
◮
X 1
s1 = X 1 s2 ,
fX 1
s1 ≈ fX 1 s2
Henceforth X 1 = X 1
s0 ,
s0 = 1
2 inj M ◮
[X 1,L2(M)]θ = Lp(M) , θ = 2(1−1/p)
SLIDE 26 The space X 1
X 1
s =
j λj A,
Aj ∈ SAs,
s = inf
λj A, Aj ∈ SAs Main properties [M, Meda, Vallarino]
◮
U = L(σI +L)−1 : H1 → X 1
s
is an isomorphism for all s > 0 and σ large enough (hard work!).
◮
X 1
s1 = X 1 s2 ,
fX 1
s1 ≈ fX 1 s2
Henceforth X 1 = X 1
s0 ,
s0 = 1
2 inj M ◮
[X 1,L2(M)]θ = Lp(M) , θ = 2(1−1/p)
◮ If T = ∇L−1/2,Liu then TA1 ≤ C
∀A ∈ SAs.
SLIDE 27
The space X 1
fin(M)
QUESTION: Suppose Y Banach, T linear s. t. (UBA) T AY ≤ C ∀A ∈ SAs. Does T : X 1(M) → Y boundedly? NOT OBVIOUS!
SLIDE 28 The space X 1
fin(M)
QUESTION: Suppose Y Banach, T linear s. t. (UBA) T AY ≤ C ∀A ∈ SAs. Does T : X 1(M) → Y boundedly? NOT OBVIOUS! X 1 ∋ f =
λj Aj, fX 1 ≈
SLIDE 29 The space X 1
fin(M)
QUESTION: Suppose Y Banach, T linear s. t. (UBA) T AY ≤ C ∀A ∈ SAs. Does T : X 1(M) → Y boundedly? NOT OBVIOUS! X 1 ∋ f =
λj Aj, fX 1 ≈
λj Aj =
λj TAj
SLIDE 30 The space X 1
fin(M)
QUESTION: Suppose Y Banach, T linear s. t. (UBA) T AY ≤ C ∀A ∈ SAs. Does T : X 1(M) → Y boundedly? NOT OBVIOUS! X 1 ∋ f =
λj Aj, fX 1 ≈
λj Aj =
λj TAj TfY ≤
- j
- λj
- TAjY ≤ C
- j
- λj
- ≈ C fX 1.
SLIDE 31 The space X 1
fin(M)
QUESTION: Suppose Y Banach, T linear s. t. (UBA) T AY ≤ C ∀A ∈ SAs. Does T : X 1(M) → Y boundedly? NOT OBVIOUS! (UBA) implies TfY ≤ CfX 1
fin,
where X 1
fin(M) := finite linear span of SAs
fX 1
fin = inf
f =
λj Aj, Aj ∈ SAs0
SLIDE 32 The space X 1
fin(M)
QUESTION: Suppose Y Banach, T linear s. t. (UBA) T AY ≤ C ∀A ∈ SAs. Does T : X 1(M) → Y boundedly? NOT OBVIOUS! (UBA) implies TfY ≤ CfX 1
fin,
where X 1
fin(M) := finite linear span of SAs
fX 1
fin = inf
f =
λj Aj, Aj ∈ SAs0
Answer to QUESTION is affirmative if fX 1
fin ≈ fX 1
Note : fX 1
fin ≈ fX 1 ⇐
⇒ (X 1)∗ = (X 1
fin)∗
SLIDE 33
The dual of X 1
fin
πB : L2
loc(M) → q2(B)⊥
“ ⊥ projection"
SLIDE 34
The dual of X 1
fin
πB : L2
loc(M) → q2(B)⊥
“ ⊥ projection" Y := space of all G = (GB)B∈B s.t.
SLIDE 35 The dual of X 1
fin
πB : L2
loc(M) → q2(B)⊥
“ ⊥ projection" Y := space of all G = (GB)B∈B s.t.
◮
GB ∈ q2(B)⊥
SLIDE 36 The dual of X 1
fin
πB : L2
loc(M) → q2(B)⊥
“ ⊥ projection" Y := space of all G = (GB)B∈B s.t.
◮
GB ∈ q2(B)⊥
◮
B ⊂ B′ ⇒ πB(GB′) = GB (compatibility condition)
SLIDE 37 The dual of X 1
fin
πB : L2
loc(M) → q2(B)⊥
“ ⊥ projection" Y := space of all G = (GB)B∈B s.t.
◮
GB ∈ q2(B)⊥
◮
B ⊂ B′ ⇒ πB(GB′) = GB (compatibility condition)
◮
GY = sup
B∈Bs0
µ(B)
|GB|2 dµ 1/2 < ∞
SLIDE 38 The dual of X 1
fin
πB : L2
loc(M) → q2(B)⊥
“ ⊥ projection" Y := space of all G = (GB)B∈B s.t.
◮
GB ∈ q2(B)⊥
◮
B ⊂ B′ ⇒ πB(GB′) = GB (compatibility condition)
◮
GY = sup
B∈Bs0
µ(B)
|GB|2 dµ 1/2 < ∞ THEOREM (X 1
fin)∗ = Y .
More precisely: ∀λ ∈ (X 1
fin)∗, ∃! G ∈ Y s. t. λ(X 1
fin)∗ ≈ GY and
λ(F) =
F GB dµ ∀F ∈ X 1
fin,
suppF ⊂ B.
SLIDE 39 Sketch of proof 1
LEMMA For all B , q2(B)⊥ ⊂ X 1
fin
and fX 1
fin ≤ C(1+rB)µ(B)1/2 f2.
SLIDE 40 Sketch of proof 1
LEMMA For all B , q2(B)⊥ ⊂ X 1
fin
and fX 1
fin ≤ C(1+rB)µ(B)1/2 f2.
Pick λ ∈ (X 1
fin)∗, f ∈ q2(B)⊥ . Then
|λ(f)| ≤ λfX 1
fin
≤ C (1+rB) µ(B)1/2 f2
SLIDE 41 Sketch of proof 1
LEMMA For all B , q2(B)⊥ ⊂ X 1
fin
and fX 1
fin ≤ C(1+rB)µ(B)1/2 f2.
Pick λ ∈ (X 1
fin)∗, f ∈ q2(B)⊥ . Then
|λ(f)| ≤ λfX 1
fin
≤ C (1+rB) µ(B)1/2 f2 By Riesz’s rep. thm ∃GB ∈ q2(B)⊥ s. t. λ(f) = (f,GB) and GB2 ≤ C(1+rB)µ(B)1/2.
SLIDE 42
Sketch of proof 2
By Riesz’s rep. thm ∃GB ∈ q2(B)⊥ s. t. λ(f) = (f,GB) and GB2 ≤ C(1+rB)µ(B)1/2. Thus sup
B∈Bs0
µ(B)−1/2GB2 ≤ C(1+s0) < ∞.
SLIDE 43
Sketch of proof 2
By Riesz’s rep. thm ∃GB ∈ q2(B)⊥ s. t. λ(f) = (f,GB) and GB2 ≤ C(1+rB)µ(B)1/2. Thus sup
B∈Bs0
µ(B)−1/2GB2 ≤ C(1+s0) < ∞. Compatibility condition:
SLIDE 44
Sketch of proof 2
By Riesz’s rep. thm ∃GB ∈ q2(B)⊥ s. t. λ(f) = (f,GB) and GB2 ≤ C(1+rB)µ(B)1/2. Thus sup
B∈Bs0
µ(B)−1/2GB2 ≤ C(1+s0) < ∞. Compatibility condition: If B ⊂ B′ then q2(B)⊥ ⊂ q2(B′)⊥ . Thus (f,GB) = λ(f) = (f,GB′) ∀f ∈ q2(B)⊥.
SLIDE 45
Sketch of proof 2
By Riesz’s rep. thm ∃GB ∈ q2(B)⊥ s. t. λ(f) = (f,GB) and GB2 ≤ C(1+rB)µ(B)1/2. Thus sup
B∈Bs0
µ(B)−1/2GB2 ≤ C(1+s0) < ∞. Compatibility condition: If B ⊂ B′ then q2(B)⊥ ⊂ q2(B′)⊥ . Thus (f,GB) = λ(f) = (f,GB′) ∀f ∈ q2(B)⊥. Thus GB′ −GB ∈ q2(B) , i. e. πB(GB′) = GB . Hence G = (GB) ∈ Y .
SLIDE 46
THEOREM. (X 1)∗ = (X 1
fin)∗
SLIDE 47 THEOREM. (X 1)∗ = (X 1
fin)∗
Sketch of proof. X 1
fin ֒
→ X 1 continuous, dense. Thus λ → λ
fin
: (X 1)∗ ֒ → (X 1
fin)∗.
is a continuous embedding.
SLIDE 48 THEOREM. (X 1)∗ = (X 1
fin)∗
Sketch of proof. X 1
fin ֒
→ X 1 continuous, dense. Thus λ → λ
fin
: (X 1)∗ ֒ → (X 1
fin)∗.
is a continuous embedding. To prove that it is onto pick λ ∈ (X 1
fin)∗ .
SLIDE 49 THEOREM. (X 1)∗ = (X 1
fin)∗
Sketch of proof. X 1
fin ֒
→ X 1 continuous, dense. Thus λ → λ
fin
: (X 1)∗ ֒ → (X 1
fin)∗.
is a continuous embedding. To prove that it is onto pick λ ∈ (X 1
fin)∗ .
Recall: U = L(σI +L)−1 : H1 → X 1 is an isomorphism LEMMA. U : H1
fin → X 1 fin
(completion of X 1
fin w.r.t. X 1
fin ).
SLIDE 50 THEOREM. (X 1)∗ = (X 1
fin)∗
Sketch of proof. X 1
fin ֒
→ X 1 continuous, dense. Thus λ → λ
fin
: (X 1)∗ ֒ → (X 1
fin)∗.
is a continuous embedding. To prove that it is onto pick λ ∈ (X 1
fin)∗ .
Recall: U = L(σI +L)−1 : H1 → X 1 is an isomorphism LEMMA. U : H1
fin → X 1 fin
(completion of X 1
fin w.r.t. X 1
fin ).
H1
fin
X 1
fin
C U λ ✲ ✲ Thus λ◦U ∈ (H1
fin)∗
SLIDE 51 THEOREM. (X 1)∗ = (X 1
fin)∗
Sketch of proof. X 1
fin ֒
→ X 1 continuous, dense. Thus λ → λ
fin
: (X 1)∗ ֒ → (X 1
fin)∗.
is a continuous embedding. To prove that it is onto pick λ ∈ (X 1
fin)∗ .
Recall: U = L(σI +L)−1 : H1 → X 1 is an isomorphism LEMMA. U : H1
fin → X 1 fin
(completion of X 1
fin w.r.t. X 1
fin ).
H1
fin
X 1
fin
C U λ ✲ ✲ Thus λ◦U ∈ (H1
fin)∗ = (H1)∗ [M, Meda]
⇒ λ ∈ (X 1)∗
SLIDE 52
A “GLUEING” QUESTION:
If G = (GB) ∈ Y does there exist G ∈ L2
loc(M) such that
GB = πB(G) ?
SLIDE 53
A “GLUEING” QUESTION:
If G = (GB) ∈ Y does there exist G ∈ L2
loc(M) such that
GB = πB(G) ? In the classical proof of (H1)∗ = BMO this is true
SLIDE 54
A “GLUEING” QUESTION:
If G = (GB) ∈ Y does there exist G ∈ L2
loc(M) such that
GB = πB(G) ? In the classical proof of (H1)∗ = BMO this is true
SLIDE 55
A “GLUEING” QUESTION:
If G = (GB) ∈ Y does there exist G ∈ L2
loc(M) such that
GB = πB(G) ? In the classical proof of (H1)∗ = BMO this is true
SLIDE 56
A “GLUEING” QUESTION:
If G = (GB) ∈ Y does there exist G ∈ L2
loc(M) such that
GB = πB(G) ? DIFFICULTY: if B ⊂ B′ then GB′ −GB ∈ q2(B) may not extend to a function in q2(B′) .
SLIDE 57
The space GBMO
GBMO : space of all G ∈ L2
loc(M) such that
GGBMO = sup
B∈Bs0
µ(B)−1/2 πB(G)2 < ∞.
SLIDE 58
The space GBMO
GBMO : space of all G ∈ L2
loc(M) such that
GGBMO = sup
B∈Bs0
µ(B)−1/2 πB(G)2 < ∞. NOTE: BMO ֒ → GBMO because C ⊂ q2(B)
SLIDE 59
The space GBMO
GBMO : space of all G ∈ L2
loc(M) such that
GGBMO = sup
B∈Bs0
µ(B)−1/2 πB(G)2 < ∞. NOTE: BMO ֒ → GBMO because C ⊂ q2(B) LEMMA GGBMO = 0 iff G ∈ q(M) . Hence GBMO is a norm on GBMOs/q(M) .
SLIDE 60
The space GBMO
GBMO : space of all G ∈ L2
loc(M) such that
GGBMO = sup
B∈Bs0
µ(B)−1/2 πB(G)2 < ∞. NOTE: BMO ֒ → GBMO because C ⊂ q2(B) LEMMA GGBMO = 0 iff G ∈ q(M) . Hence GBMO is a norm on GBMOs/q(M) . The map G → G = (πB(G))B∈B is an isometric embedding of GBMO/q(M) into Y . Its image is the set of “glueable" elements of Y .
SLIDE 61
All G ∈ Y are gluable
THEOREM. GBMO/q(M) ≈ Y
SLIDE 62
All G ∈ Y are gluable
THEOREM. GBMO/q(M) ≈ Y Sketch of the proof. G = (GB) ∈ Y = (X 1)∗ , G → λG ∈ (X 1)∗ ,
SLIDE 63
All G ∈ Y are gluable
THEOREM. GBMO/q(M) ≈ Y Sketch of the proof. G = (GB) ∈ Y = (X 1)∗ , G → λG ∈ (X 1)∗ , U : H1 → X 1 ⇒ (U−1)t : BMO/C → (X 1)∗ isomorphisms
SLIDE 64
All G ∈ Y are gluable
THEOREM. GBMO/q(M) ≈ Y Sketch of the proof. G = (GB) ∈ Y = (X 1)∗ , G → λG ∈ (X 1)∗ , U : H1 → X 1 ⇒ (U−1)t : BMO/C → (X 1)∗ isomorphisms Thus ∃g ∈ BMO s.t. λG = (U−1)t(g +C).
SLIDE 65 All G ∈ Y are gluable
THEOREM. GBMO/q(M) ≈ Y Sketch of the proof. G = (GB) ∈ Y = (X 1)∗ , G → λG ∈ (X 1)∗ , U : H1 → X 1 ⇒ (U−1)t : BMO/C → (X 1)∗ isomorphisms Thus ∃g ∈ BMO s.t. λG = (U−1)t(g +C). Thus, if A ∈ SA , λG(A) = A,(U−1)t(g +C)
X1,(X1)∗
= U−1A,g +C
H1,BMO
=
because U = L(σI +L)−1 ⇒ U−1 = I +σL−1 .
SLIDE 66 Sketch of proof (continued)
λG(A) =
- (I +σL−1)A g dµ =
- A g dµ+σ
- L−1A g dµ
LEMMA. ∀g ∈ BMO ∃G ∈ GBMO such that LG = g in M and GGBMO ≤ CgBMO. Thus σ
- L−1A g dµ = σ
- L−1A LG dµ = σ
- A G dµ
Hence λG(A) =
Now g +σG ∈ GBMO , since BMO ⊂ GBMO . Thus, if supp(A) ⊂ B λG(A) =
GB = πB(g +σG) ∀B .
