FUNCTIONS OF THE LAPLACIAN ON MANIFOLDS WITH LOWER RICCI BOUND
G. Mauceri1 S. Meda2 M. Vallarino2
1Dipartimento di Matematica Università di Genova
2Dipartimento di Matematica Università di Milano Bicocca
Bardonecchia, June 2009
FUNCTIONS OF THE LAPLACIAN
M noncompact, complete Riemannian manifold, L = −div grad Laplace-Beltrami operator (L ≥ 0)
L = Z ∞
b λ dEλ, b = inf σ2(L) ≥ 0.
Define
f (L) = Z ∞
b f (λ ) dEλ If f ∈ L∞(E ) then f (L) is bounded on L2(M) .
For which f is f (L) bounded on Lp(M) for p 6= 2 ?
Applications: PDE’s, summability of eigenfunction expansions, potential theory . . .
FUNCTIONS OF THE LAPLACIAN
M noncompact, complete Riemannian manifold, L = −div grad Laplace-Beltrami operator (L ≥ 0)
L = Z ∞
b λ dEλ, b = inf σ2(L) ≥ 0.
Define
f (L) = Z ∞
b f (λ ) dEλ If f ∈ L∞(E ) then f (L) is bounded on L2(M) .
For which f is f (L) bounded on Lp(M) for p 6= 2 ?
Applications: PDE’s, summability of eigenfunction expansions, potential theory . . .
FUNCTIONS OF THE LAPLACIAN
M noncompact, complete Riemannian manifold, L = −div grad Laplace-Beltrami operator (L ≥ 0)
L = Z ∞
b λ dEλ, b = inf σ2(L) ≥ 0.
Define
f (L) = Z ∞
b f (λ ) dEλ If f ∈ L∞(E ) then f (L) is bounded on L2(M) .
For which f is f (L) bounded on Lp(M) for p 6= 2 ?
Applications: PDE’s, summability of eigenfunction expansions, potential theory . . .
FUNCTIONS OF THE LAPLACIAN
M noncompact, complete Riemannian manifold, L = −div grad Laplace-Beltrami operator (L ≥ 0)
L = Z ∞
b λ dEλ, b = inf σ2(L) ≥ 0.
Define
f (L) = Z ∞
b f (λ ) dEλ If f ∈ L∞(E ) then f (L) is bounded on L2(M) .
For which f is f (L) bounded on Lp(M) for p 6= 2 ?
Applications: PDE’s, summability of eigenfunction expansions, potential theory . . .
FUNCTIONS OF THE LAPLACIAN
M noncompact, complete Riemannian manifold, L = −div grad Laplace-Beltrami operator (L ≥ 0)
L = Z ∞
b λ dEλ, b = inf σ2(L) ≥ 0.
Define
f (L) = Z ∞
b f (λ ) dEλ If f ∈ L∞(E ) then f (L) is bounded on L2(M) .
For which f is f (L) bounded on Lp(M) for p 6= 2 ?
Applications: PDE’s, summability of eigenfunction expansions, potential theory . . .
Manifolds of C
∞bounded geometry
M has C∞ bounded geometry if
I rinj(M) > 0
I
∇kR(M)
≤ Ck k = 0, 1, 2, . . . uniformly in all charts
Then
V Br(p) ≤ C(1 + r )αeβr β is thegrowth exponentof M
Manifolds of C
∞bounded geometry
M has C∞ bounded geometry if
I rinj(M) > 0
I
∇kR(M)
≤ Ck k = 0, 1, 2, . . . uniformly in all charts
Then
V Br(p) ≤ C(1 + r )αeβr β is thegrowth exponentof M
Manifolds of C
∞bounded geometry
M has C∞ bounded geometry if
I rinj(M) > 0
I
∇kR(M)
≤ Ck k = 0, 1, 2, . . . uniformly in all charts
Then
V Br(p) ≤ C(1 + r )αeβr β is thegrowth exponentof M
Taylor’s theorem
Replace L by H =√
L − b . Then
f (L) = m(H) where m(λ ) = f (λ2+b).
Note that m is even.
The symbol class S
wΣw |Im z| ≤ w
We say that m ∈Sw iff
I m is holomorphic and even in Σw;
I
m(j)(z)
≤ C 1 + |z|−j
∀z ∈ Σw, j ∈ N
Taylor’s theorem
Replace L by H =√
L − b . Then
f (L) = m(H) where m(λ ) = f (λ2+b).
Note that m is even.
The symbol class S
wΣw |Im z| ≤ w
We say that m ∈Sw iff
I m is holomorphic and even in Σw;
I
m(j)(z)
≤ C 1 + |z|−j
∀z ∈ Σw, j ∈ N
Taylor’s theorem
Replace L by H =√
L − b . Then
f (L) = m(H) where m(λ ) = f (λ2+b).
Note that m is even.
The symbol class S
wΣw |Im z| ≤ w
We say that m ∈Sw iff
I m is holomorphic and even in Σw;
I
m(j)(z)
≤ C 1 + |z|−j
∀z ∈ Σw, j ∈ N
Taylor’s theorem
Theorem Suppose that M has C∞ bounded geometry and that 1 < p < ∞ . If m ∈Sw with
w > |1/p − 1/2| β then
m(H) : Lp(M) → Lp(M).
If m ∈Sβ /2 then m(H) is of weak type (1, 1) . Remarks
I If β > 0 holomorphy is necessary (Clerc-Stein 1976)
I The width w is related to the growth exponent β .
Taylor’s theorem
Theorem Suppose that M has C∞ bounded geometry and that 1 < p < ∞ . If m ∈Sw with
w > |1/p − 1/2| β then
m(H) : Lp(M) → Lp(M).
If m ∈Sβ /2 then m(H) is of weak type (1, 1) . Remarks
I If β > 0 holomorphy is necessary (Clerc-Stein 1976)
I The width w is related to the growth exponent β .
Ricci versus Riemann
Riemann tensor R(X , Y ) = ∇X∇Y− ∇Y∇X− ∇[X ,Y ] It measures the noncommutativity of ∇
Ricci tensor Ric(X , Y ) = tr : Z 7→ R(Z , Y )X In normal geodesic coordinates
dVRiemann= h
1 −1
6Ricjkxjxk+O(|x |3) i
dVEuclidean It measures the excess or defect of the Riemannian volume with respect to the Euclidean volume on the tangent space.
Ricci versus Riemann
Riemann tensor R(X , Y ) = ∇X∇Y− ∇Y∇X− ∇[X ,Y ] It measures the noncommutativity of ∇
Ricci tensor Ric(X , Y ) = tr : Z 7→ R(Z , Y )X In normal geodesic coordinates
dVRiemann= h
1 −1
6Ricjkxjxk+O(|x |3) i
dVEuclidean It measures the excess or defect of the Riemannian volume with respect to the Euclidean volume on the tangent space.
Ricci versus Riemann
Riemann tensor R(X , Y ) = ∇X∇Y− ∇Y∇X− ∇[X ,Y ] It measures the noncommutativity of ∇
Ricci tensor Ric(X , Y ) = tr : Z 7→ R(Z , Y )X In normal geodesic coordinates
dVRiemann= h
1 −1
6Ricjkxjxk+O(|x |3) i
dVEuclidean It measures the excess or defect of the Riemannian volume with respect to the Euclidean volume on the tangent space.
Lower Ricci bounded vs C
∞bounded geometry.
Def.
M has lower Ricci bounded geometry ifI rinj(M) > 0
I RicM≥ −κ2.
Consequences:
I V Br(p) ≤ C(1 + r )αeβr
I M islocallydoubling
I covering theorem: M =SNj=1S∞m=1B1(xjm) B2(xjm) ∩B2(xj`) = /0 if xjm6= xi`.
Lower Ricci bounded vs C
∞bounded geometry.
Def.
M has lower Ricci bounded geometry ifI rinj(M) > 0
I RicM≥ −κ2.
Consequences:
I V Br(p) ≤ C(1 + r )αeβr
I M islocallydoubling
I covering theorem: M =SNj=1S∞m=1B1(xjm) B2(xjm) ∩B2(xj`) = /0 if xjm6= xi`.
Berger: “Up to the end of the 1980’s, Ricci curvature was believed to be only useful to control volumes,... ”.
Since then many people have obtained several geometric and analytical results (Harnack estimates, isoperimetric
inequalities) assuming only lower Ricci bounded curvature.
Varopoulos, Coulhon, Saloff-Coste, Chavel-Feldman, . . . Theorem Taylor’s theorem holds under the assumption that M has lower Ricci bounded geometry.
Berger: “Up to the end of the 1980’s, Ricci curvature was believed to be only useful to control volumes,... ”.
Since then many people have obtained several geometric and analytical results (Harnack estimates, isoperimetric
inequalities) assuming only lower Ricci bounded curvature.
Varopoulos, Coulhon, Saloff-Coste, Chavel-Feldman, . . . Theorem Taylor’s theorem holds under the assumption that M has lower Ricci bounded geometry.
Berger: “Up to the end of the 1980’s, Ricci curvature was believed to be only useful to control volumes,... ”.
Since then many people have obtained several geometric and analytical results (Harnack estimates, isoperimetric
inequalities) assuming only lower Ricci bounded curvature.
Varopoulos, Coulhon, Saloff-Coste, Chavel-Feldman, . . . Theorem Taylor’s theorem holds under the assumption that M has lower Ricci bounded geometry.
Outline of the proof
Decompose
m(H) = m[(H) + m](H) where
I m[(H) is alocalCZ operator:
supp f ⊂ B1(x ) ⇒ supp m[(H) f ⊂ B2(x )
I m](H) has kernel k](x , y ) ∈ C∞(M × M) such that sup
y ∈M
Z
M
k](x , y )
dx < ∞, sup
x ∈M
Z
M
k](x , y )
dy < ∞
Return
Outline of the proof
Decompose
m(H) = m[(H) + m](H) where
I m[(H) is alocalCZ operator:
supp f ⊂ B1(x ) ⇒ supp m[(H) f ⊂ B2(x )
I m](H) has kernel k](x , y ) ∈ C∞(M × M) such that sup
y ∈M
Z
M
k](x , y )
dx < ∞, sup
x ∈M
Z
M
k](x , y )
dy < ∞
Return
Outline of the proof
Decompose
m(H) = m[(H) + m](H) where
I m[(H) is alocalCZ operator:
supp f ⊂ B1(x ) ⇒ supp m[(H) f ⊂ B2(x )
I m](H) has kernel k](x , y ) ∈ C∞(M × M) such that sup
y ∈M
Z
M
k](x , y )
dx < ∞, sup
x ∈M
Z
M
k](x , y )
dy < ∞
Return
The wave equation method
Write m(λ ) =
Z ∞
−∞m(t) cos(tλ ) dtb
Remark u(t, x ) = cos(tH)f (x ) is the solution of (
∂tt2u = Lu + b u,
u(0, x ) = f (x ), ∂tu(0, x ) = 0, Properties of cos(tH) :
I kcos(tH)k2,2=1
I Finite speed of propagation:
kcos(tH)(x , y ) = 0 ∀(x, y ) ∈ M × M, d (x, y ) ≥ t. Thus mc1(t) =mc2(t) for |t| ≥ R implies
km1(x , y ) = km2(x , y ) for d (x , y ) ≥ R
The wave equation method
Write m(H) =
Z ∞
−∞m(t) cos(tb H)dt
Remark u(t, x ) = cos(tH)f (x ) is the solution of (
∂tt2u = Lu + b u,
u(0, x ) = f (x ), ∂tu(0, x ) = 0, Properties of cos(tH) :
I kcos(tH)k2,2=1
I Finite speed of propagation:
kcos(tH)(x , y ) = 0 ∀(x, y ) ∈ M × M, d (x, y ) ≥ t. Thus mc1(t) =mc2(t) for |t| ≥ R implies
km1(x , y ) = km2(x , y ) for d (x , y ) ≥ R
The wave equation method
Write m(H) =
Z ∞
−∞m(t) cos(tb H)dt Remark u(t, x ) = cos(tH)f (x ) is the solution of
(
∂tt2u = Lu + b u,
u(0, x ) = f (x ), ∂tu(0, x ) = 0,
Properties of cos(tH) :
I kcos(tH)k2,2=1
I Finite speed of propagation:
kcos(tH)(x , y ) = 0 ∀(x, y ) ∈ M × M, d (x, y ) ≥ t. Thus mc1(t) =mc2(t) for |t| ≥ R implies
km1(x , y ) = km2(x , y ) for d (x , y ) ≥ R
The wave equation method
Write m(H) =
Z ∞
−∞m(t) cos(tb H)dt Remark u(t, x ) = cos(tH)f (x ) is the solution of
(
∂tt2u = Lu + b u,
u(0, x ) = f (x ), ∂tu(0, x ) = 0, Properties of cos(tH) :
I kcos(tH)k2,2=1
I Finite speed of propagation:
kcos(tH)(x , y ) = 0 ∀(x, y ) ∈ M × M, d (x, y ) ≥ t.
Thus mc1(t) =mc2(t) for |t| ≥ R implies km1(x , y ) = km2(x , y ) for d (x , y ) ≥ R
The wave equation method
Write m(H) =
Z ∞
−∞m(t) cos(tb H)dt Remark u(t, x ) = cos(tH)f (x ) is the solution of
(
∂tt2u = Lu + b u,
u(0, x ) = f (x ), ∂tu(0, x ) = 0, Properties of cos(tH) :
I kcos(tH)k2,2=1
I Finite speed of propagation:
kcos(tH)(x , y ) = 0 ∀(x, y ) ∈ M × M, d (x, y ) ≥ t.
Thus mc1(t) =mc2(t) for |t| ≥ R implies km1(x , y ) = km2(x , y ) for d (x , y ) ≥ R
The decomposition of m(H)
Choose a cut-off φ ∈ Cc∞(R)
and write m(H) = m[(H) + m](H) , where m[(H) =
Z ∞
−∞m(t) φ (t) cos(tH) dtb m](H) =
Z ∞
−∞m(t) 1 − φ (t) cos(tH) dtb Then, by the finite speed of propagation
k[(x , y ) = 0 for d (x , y ) ≥ 1 so that m[(H) is local.
The decomposition of m(H)
Choose a cut-off φ ∈ Cc∞(R)
and write m(H) = m[(H) + m](H) , where m[(H) =
Z ∞
−∞m(t) φ (t) cos(tH) dtb m](H) =
Z ∞
−∞m(t) 1 − φ (t) cos(tH) dtb Then, by the finite speed of propagation
k[(x , y ) = 0 for d (x , y ) ≥ 1 so that m[(H) is local.
The role of C
∞bounded geometry in Taylor’s thm
If M has C∞ bounded geometry
∀y ∈ M ∃φy, ψy ∈ L2 Br(y )
such that if k > n/4 + 1
I δy =Hkφy+ ψy
I kφyk
L2 Br(y )+ kψyk
L2 Br(y )≤ C
If M has only lower Ricci bounded geometry we use ultracontractivityof theheat semigroupto obtain aSobolev inequality.
The role of C
∞bounded geometry in Taylor’s thm
If M has C∞ bounded geometry
∀y ∈ M ∃φy, ψy ∈ L2 Br(y )
such that if k > n/4 + 1
I δy =Hkφy+ ψy
I kφyk
L2 Br(y )+ kψyk
L2 Br(y )≤ C
If M has only lower Ricci bounded geometry we use ultracontractivityof theheat semigroupto obtain aSobolev inequality.
The role of ultracontractivity
Let n = dim(M) . The heat semigroup satisfies the u.c. estimate ke−tLk1,2≤ C (1 + t−n/4) e−bt ∀t > 0.
Now
(I + H2)−σ = 1 Γ(k )
Z ∞
0
sσ −1 e−s(I+H2)ds and, since H2=L − b , we get theSobolevestimate
k(I + H2)−σk1,2 ≤ C provided σ > n/4.
The role of ultracontractivity
Let n = dim(M) . The heat semigroup satisfies the u.c. estimate ke−tLk1,2≤ C (1 + t−n/4) e−bt ∀t > 0.
Now
(I + H2)−σ = 1 Γ(k )
Z ∞
0
sσ −1 e−s(I+H2)ds and, since H2=L − b , we get theSobolevestimate
k(I + H2)−σk1,2 ≤ C provided σ > n/4.
The role of ultracontractivity
Let n = dim(M) . The heat semigroup satisfies the u.c. estimate ke−tLk1,2≤ C (1 + t−n/4) e−bt ∀t > 0.
Now
(I + H2)−σ = 1 Γ(k )
Z ∞
0
sσ −1 e−s(I+H2)ds and, since H2=L − b , we get theSobolevestimate
k(I + H2)−σk1,2 ≤ C provided σ > n/4.
The estimates of k
]est
Divide M into shells:
Aj(y ) = {x ∈ M : j ≤ d (x , y ) < j + 1}
M = [
j≥0
Aj(y )
The estimates of k
]Then Z
M
k](x , y )
dx =
∑
j≥o
Z
Aj(y )
k](x , y ) dx
≤
∑
j≥0
Aj(y )
1/2kk](·,y )k
L2 Aj(y )
≤ C
∑
j≥0
(1 + j)α /2ejβ /2 kk](·,y )k
L2 Aj(y ).
Lemma
If m ∈Sw then supy
kk](·,y )k
L2 Aj(y )≤ C`(1 + j)−`e−jw ∀j ≥ 0, ` ∈ N.
The estimates of k
]Then Z
M
k](x , y )
dx =
∑
j≥o
Z
Aj(y )
k](x , y ) dx
≤
∑
j≥0
Aj(y )
1/2kk](·,y )k
L2 Aj(y )
≤ C
∑
j≥0
(1 + j)α /2ejβ /2 kk](·,y )k
L2 Aj(y ).
Lemma
If m ∈Sw then supy
kk](·,y )k
L2 Aj(y )≤ C`(1 + j)−`e−jw ∀j ≥ 0, ` ∈ N.
The estimates of k
]Then Z
M
k](x , y )
dx =
∑
j≥o
Z
Aj(y )
k](x , y ) dx
≤
∑
j≥0
Aj(y )
1/2kk](·,y )k
L2 Aj(y )
≤ C
∑
j≥0
(1 + j)α /2ejβ /2 kk](·,y )k
L2 Aj(y ).
Lemma
If m ∈Sw then supy
kk](·,y )k
L2 Aj(y )≤ C`(1 + j)−`e−jw ∀j ≥ 0, ` ∈ N.
The estimates of k
]Then Z
M
k](x , y )
dx =
∑
j≥o
Z
Aj(y )
k](x , y ) dx
≤
∑
j≥0
Aj(y )
1/2kk](·,y )k
L2 Aj(y )
≤ C
∑
j≥0
(1 + j)α /2ejβ /2 kk](·,y )k
L2 Aj(y ).
Lemma
If m ∈Sw then supy
kk](·,y )k
L2 Aj(y )≤ C`(1 + j)−`e−jw ∀j ≥ 0, ` ∈ N.
Proof of the Lemma
Chose a cutoff ψj: ψj(t) = 1 for |t| ≥ j ,
Proof of the Lemma
Chose a cutoff ψj: ψj(t) = 1 for |t| ≥ j , Define
mj](H) = Z ∞
−∞m(t) ψb j(t) cos(tH) dt Then
kj](x , y ) = k](x , y ) if d (x , y ) ≥ j.
because m(t) ψb j(t) =m(t)b for |t| ≥ j . So
sup
y
kk](·,y )k
L2 Aj(y )=sup
y
kkj](·,y )k
L2 Aj(y )
≤ sup
y
kkj](·,y )kL2(M)
= kmj](H)k1,2
Proof of the Lemma
Chose a cutoff ψj: ψj(t) = 1 for |t| ≥ j , Define
mj](H) = Z ∞
−∞m(t) ψb j(t) cos(tH) dt Then
kj](x , y ) = k](x , y ) if d (x , y ) ≥ j.
because m(t) ψb j(t) =m(t)b for |t| ≥ j . So
sup
y
kk](·,y )k
L2 Aj(y )=sup
y
kkj](·,y )k
L2 Aj(y )
≤ sup
y
kkj](·,y )kL2(M)
= kmj](H)k1,2
Proof of the Lemma
Chose a cutoff ψj: ψj(t) = 1 for |t| ≥ j , Define
mj](H) = Z ∞
−∞m(t) ψb j(t) cos(tH) dt Then
kj](x , y ) = k](x , y ) if d (x , y ) ≥ j.
because m(t) ψb j(t) =m(t)b for |t| ≥ j . So
sup
y
kk](·,y )k
L2 Aj(y )=sup
y
kkj](·,y )k
L2 Aj(y )
≤ sup
y
kkj](·,y )kL2(M)
= kmj](H)k1,2
Proof of the Lemma
Chose a cutoff ψj: ψj(t) = 1 for |t| ≥ j , Define
mj](H) = Z ∞
−∞m(t) ψb j(t) cos(tH) dt Then
kj](x , y ) = k](x , y ) if d (x , y ) ≥ j.
because m(t) ψb j(t) =m(t)b for |t| ≥ j . So
sup
y
kk](·,y )k
L2 Aj(y )=sup
y
kkj](·,y )k
L2 Aj(y )
≤ sup
y
kkj](·,y )kL2(M)
= kmj](H)k1,2
Proof of the Lemma
Chose a cutoff ψj: ψj(t) = 1 for |t| ≥ j , Define
mj](H) = Z ∞
−∞m(t) ψb j(t) cos(tH) dt Then
kj](x , y ) = k](x , y ) if d (x , y ) ≥ j.
because m(t) ψb j(t) =m(t)b for |t| ≥ j . So
sup
y
kk](·,y )k
L2 Aj(y )=sup
y
kkj](·,y )k
L2 Aj(y )
≤ sup
y
kkj](·,y )kL2(M)
= kmj](H)k1,2
Proof of the Lemma
Chose a cutoff ψj: ψj(t) = 1 for |t| ≥ j , Define
mj](H) = Z ∞
−∞m(t) ψb j(t) cos(tH) dt Then
kj](x , y ) = k](x , y ) if d (x , y ) ≥ j.
because m(t) ψb j(t) =m(t)b for |t| ≥ j . So
sup
y
kk](·,y )k
L2 Aj(y )=sup
y
kkj](·,y )k
L2 Aj(y )
≤ sup
y
kkj](·,y )kL2(M)
= kmj](H)k1,2
Estimate of
m
j](H)
1,2 Choose aninteger σ > n/4 .mj](H) = (I + H2)−σ (I + H2)σ mj](H)
(by Sobolev ) ≤C
(I + H2)σmj](H) 2,2
(I + H2)σmj](H) = Z ∞
−∞m(t)ψb j(t)(I + H2)σ cos(tH)dt
= Z ∞
−∞m(t)ψb j(t)(D2t +1)σ cos(tH)dt
= Z
|t|≥j−1
(Dt2+1)σ m(t)ψb j(t)
cos(tH) dt The conclusion follows because kcos(tH)k2,2≤ 1 and
Dt`m(t)b
≤ C`h |t|−h e−wt ∀`, h ∈ N, |t| ≥ 1.
Estimate of
m
j](H)
1,2 Choose aninteger σ > n/4 .mj](H)
1,2 ≤
(I + H2)−σ 1,2
(I + H2)σ mj](H) 2,2
(by Sobolev ) ≤C
(I + H2)σmj](H) 2,2
(I + H2)σmj](H) = Z ∞
−∞m(t)ψb j(t)(I + H2)σ cos(tH)dt
= Z ∞
−∞m(t)ψb j(t)(D2t +1)σ cos(tH)dt
= Z
|t|≥j−1
(Dt2+1)σ m(t)ψb j(t)
cos(tH) dt The conclusion follows because kcos(tH)k2,2≤ 1 and
Dt`m(t)b
≤ C`h |t|−h e−wt ∀`, h ∈ N, |t| ≥ 1.
Estimate of
m
j](H)
1,2 Choose aninteger σ > n/4 .mj](H)
1,2 ≤
(I + H2)−σ 1,2
(I + H2)σ mj](H) 2,2
(by Sobolev ) ≤C
(I + H2)σmj](H) 2,2
(I + H2)σmj](H) = Z ∞
−∞m(t)ψb j(t)(I + H2)σ cos(tH)dt
= Z ∞
−∞m(t)ψb j(t)(D2t +1)σ cos(tH)dt
= Z
|t|≥j−1
(Dt2+1)σ m(t)ψb j(t)
cos(tH) dt The conclusion follows because kcos(tH)k2,2≤ 1 and
Dt`m(t)b
≤ C`h |t|−h e−wt ∀`, h ∈ N, |t| ≥ 1.
Estimate of
m
j](H)
1,2 Choose aninteger σ > n/4 .mj](H)
1,2 ≤
(I + H2)−σ 1,2
(I + H2)σ mj](H) 2,2
(by Sobolev ) ≤C
(I + H2)σmj](H) 2,2
(I + H2)σmj](H) = Z ∞
−∞m(t)ψb j(t)(I + H2)σ cos(tH)dt
= Z ∞
−∞m(t)ψb j(t)(D2t +1)σ cos(tH)dt
= Z
|t|≥j−1
(Dt2+1)σ m(t)ψb j(t)
cos(tH) dt
The conclusion follows because kcos(tH)k2,2≤ 1 and Dt`m(t)b
≤ C`h |t|−h e−wt ∀`, h ∈ N, |t| ≥ 1.
Estimate of
m
j](H)
1,2 Choose aninteger σ > n/4 .mj](H)
1,2 ≤
(I + H2)−σ 1,2
(I + H2)σ mj](H) 2,2
(by Sobolev ) ≤C
(I + H2)σmj](H) 2,2
(I + H2)σmj](H) = Z ∞
−∞m(t)ψb j(t)(I + H2)σ cos(tH)dt
= Z ∞
−∞m(t)ψb j(t)(D2t +1)σ cos(tH)dt
= Z
|t|≥j−1
(Dt2+1)σ m(t)ψb j(t)
cos(tH) dt The conclusion follows because kcos(tH)k2,2≤ 1 and
Dt`m(t)b
≤ C`h |t|−h e−wt ∀`, h ∈ N, |t| ≥ 1.
Estimate of the kernel of m
[(H)
To show that m[(H) is a CZ operator we need to prove that sup
y
k dyk[(·,y )k
L2 Ar(y )<C r−1, 0 < r < 1 where
Ar(y ) = B1(y ) \ Br(y ),
Remark. dyk[(x , y ) is the kernel of
m[(H) d∗:Cc∞(M; Λ1) →Cc∞(M) mapping1-formsto functions.
Estimate of the kernel of m
[(H)
To show that m[(H) is a CZ operator we need to prove that sup
y
k dyk[(·,y )k
L2 Ar(y )<C r−1, 0 < r < 1 where
Ar(y ) = B1(y ) \ Br(y ),
Remark. dyk[(x , y ) is the kernel of
m[(H) d∗:Cc∞(M; Λ1) →Cc∞(M) mapping1-formsto functions.
Recall that
L = d∗d and L = d d∗+d∗d is the Hodge Laplacianon 1 -forms.
We introduce two variants of Taylor’s method. In the formula m[(H) =
Z ∞
−∞
mc[(t)cos(tH) dt
I replace H =√
L − b by H =√
L − b + k2
I replace cos by aBessel function.
Recall that
L = d∗d and L = d d∗+d∗d is the Hodge Laplacianon 1 -forms.
We introduce two variants of Taylor’s method. In the formula m[(H) =
Z ∞
−∞
mc[(t)cos(tH) dt
I replace H =√
L − b by H =√
L − b + k2
I replace cos by aBessel function.
Recall that
L = d∗d and L = d d∗+d∗d is the Hodge Laplacianon 1 -forms.
We introduce two variants of Taylor’s method. In the formula m[(H) =
Z ∞
−∞
mc[(t)cos(tH) dt
I replace H =√
L − b by H =√
L − b + k2
I replace cos by aBessel function.
Recall that
L = d∗d and L = d d∗+d∗d is the Hodge Laplacianon 1 -forms.
We introduce two variants of Taylor’s method. In the formula m[(H) =
Z ∞
−∞
mc[(t)cos(tH) dt
I replace H =√
L − b by H =√
L − b + k2
I replace cos by aBessel function.
The Bessel function formula
Write H1=√
L − b + k2. Then m[(H) = f (H1) and m[(H) =
Z ∞
−∞
bf (t) cos(tH1)dt with supp bf ⊂ [−1, 1].
Define the Bessel function BN(t) = 21/2−N
√ π Γ(N )
Z 1 0
(1 − s2)N−1 cos(ts) ds.
There exists a differential operator P = P t dtd
such that cos(tH1)=P BN(tH1) ∀t > 0
Thus
The Bessel function formula
Write H1=√
L − b + k2. Then m[(H) = f (H1) and m[(H) =
Z ∞
−∞
bf (t) cos(tH1)dt with supp bf ⊂ [−1, 1].
Define the Bessel function BN(t) = 21/2−N
√ π Γ(N )
Z 1 0
(1 − s2)N−1 cos(ts) ds.
There exists a differential operator P = P t dtd
such that cos(tH1)=P BN(tH1) ∀t > 0
Thus
The Bessel function formula
Write H1=√
L − b + k2. Then m[(H) = f (H1) and m[(H) =
Z ∞
−∞
bf (t) cos(tH1)dt with supp bf ⊂ [−1, 1].
Define the Bessel function BN(t) = 21/2−N
√ π Γ(N )
Z 1 0
(1 − s2)N−1 cos(ts) ds.
There exists a differential operator P = P t dtd
such that cos(tH1)=P BN(tH1) ∀t > 0
Thus
The Bessel function formula
Write H1=√
L − b + k2. Then m[(H) = f (H1) and m[(H) =
Z ∞
−∞
bf (t) cos(tH1)dt with supp bf ⊂ [−1, 1].
Define the Bessel function BN(t) = 21/2−N
√ π Γ(N )
Z 1
0
(1 − s2)N−1 cos(ts) ds.
There exists a differential operator P = P t dtd
such that cos(tH1)=P BN(tH1) ∀t > 0
Thus
m[(H) = Z ∞
−∞
bf (t)PBN(tH1)dt
The Bessel function formula
Write H1=√
L − b + k2. Then m[(H) = f (H1) and m[(H) =
Z ∞
−∞
bf (t) cos(tH1)dt with supp bf ⊂ [−1, 1].
Define the Bessel function BN(t) = 21/2−N
√ π Γ(N )
Z 1
0
(1 − s2)N−1 cos(ts) ds.
There exists a differential operator P = P t dtd
such that cos(tH1)=P BN(tH1) ∀t > 0
Thus
m[(H) = Z ∞
−∞Pt bf (t) BN(tH1)dt
The Bessel function formula
Write H1=√
L − b + k2. Then m[(H) = f (H1) and m[(H) =
Z ∞
−∞
bf (t) cos(tH1)dt with supp bf ⊂ [−1, 1].
Define the Bessel function BN(t) = 21/2−N
√ π Γ(N )
Z 1
0
(1 − s2)N−1 cos(ts) ds.
There exists a differential operator P = P t dtd
such that cos(tH1)=P BN(tH1) ∀t > 0
Thus
m[(H)d∗= Z ∞
−∞
bf (t)PBN(tH1)d∗ dt
m
[(H) d
∗= Z
∞−∞
P
tb f (t) B
N(tH
1) d
∗dt
Pass to kernels, use the finite speed of propagation and apply Schwarz. Matters reduce to proving that for N large enough
BN(tH1)d∗
1,2≤ C t−n/2−1 t ∈ (0, 1).
Lemma
Suppose that for some N > n/2 + 1|F (λ )| ≤ C (1 + λ )−N ∀λ > 0.
Then
F (tH1)d∗
1,2 ≤ C t−n/2−1 t ∈ (0, 1).
Remark:Note that |BN(λ )| ≤ C (1 + λ )−N.
m
[(H) d
∗= Z
∞−∞
P
tb f (t) B
N(tH
1) d
∗dt
Pass to kernels, use the finite speed of propagation and apply Schwarz. Matters reduce to proving that for N large enough
BN(tH1)d∗
1,2≤ C t−n/2−1 t ∈ (0, 1).
Lemma
Suppose that for some N > n/2 + 1|F (λ )| ≤ C (1 + λ )−N ∀λ > 0.
Then
F (tH1)d∗
1,2 ≤ C t−n/2−1 t ∈ (0, 1).
Remark:Note that |BN(λ )| ≤ C (1 + λ )−N.
m
[(H) d
∗= Z
∞−∞
P
tb f (t) B
N(tH
1) d
∗dt
Pass to kernels, use the finite speed of propagation and apply Schwarz. Matters reduce to proving that for N large enough
BN(tH1)d∗
1,2≤ C t−n/2−1 t ∈ (0, 1).
Lemma
Suppose that for some N > n/2 + 1|F (λ )| ≤ C (1 + λ )−N ∀λ > 0.
Then
F (tH1)d∗
1,2 ≤ C t−n/2−1 t ∈ (0, 1).
Remark:Note that |BN(λ )| ≤ C (1 + λ )−N.
m
[(H) d
∗= Z
∞−∞
P
tb f (t) B
N(tH
1) d
∗dt
Pass to kernels, use the finite speed of propagation and apply Schwarz. Matters reduce to proving that for N large enough
BN(tH1)d∗
1,2≤ C t−n/2−1 t ∈ (0, 1).
Lemma
Suppose that for some N > n/2 + 1|F (λ )| ≤ C (1 + λ )−N ∀λ > 0.
Then
F (tH1)d∗
1,2 ≤ C t−n/2−1 t ∈ (0, 1).
Remark:Note that |BN(λ )| ≤ C (1 + λ )−N.
m
[(H) d
∗= Z
∞−∞
P
tb f (t) B
N(tH
1) d
∗dt
Pass to kernels, use the finite speed of propagation and apply Schwarz. Matters reduce to proving that for N large enough
BN(tH1)d∗
1,2≤ C t−n/2−1 t ∈ (0, 1).
Lemma
Suppose that for some N > n/2 + 1|F (λ )| ≤ C (1 + λ )−N ∀λ > 0.
Then
F (tH1)d∗
1,2 ≤ C t−n/2−1 t ∈ (0, 1).
Remark:Note that |BN(λ )| ≤ C (1 + λ )−N.
The proof of the lemma is based on
I the intertwining identity
L d∗=d∗L
I the ultracontractive estimate for the Hodge semigroup ke−tLk1,2≤ C (1 + t−n/2) e(k2−b)t ∀t > 0 which implies the Sobolev estimate
k(I + t2H2)−σk1,2≤ C t−n/2 ∀t ∈ (0, 1), σ > n/4 for
H = p
L − b + k2.
Skip proof
The proof of the lemma is based on
I the intertwining identity
L d∗=d∗L
I the ultracontractive estimate for the Hodge semigroup ke−tLk1,2≤ C (1 + t−n/2) e(k2−b)t ∀t > 0 which implies the Sobolev estimate
k(I + t2H2)−σk1,2≤ C t−n/2 ∀t ∈ (0, 1), σ > n/4 for
H = p
L − b + k2.
Skip proof
Proof of the lemma
Choose σ > n/4 . Then for 0 < t < 1 kF (tH1)d∗k1,2= kd∗F (tH1)k1,2
≤ k(I + t2H21)−σk1,2k d∗F (tH1) (I + t2H21)σk2,2
≤ C t−n/2k d∗G(tH1)k2,2 Next observe that
k d∗G(tH1)k2,2≤ C
kH1G(tH1)k2,2+ kG(tH1)k2,2
≤ C t−1 sup
λ
|λ G(λ )| + sup
λ
|G(λ )|
≤ C t−1.
Proof of the lemma
Choose σ > n/4 . Then for 0 < t < 1 kF (tH1)d∗k1,2= kd∗F (tH1)k1,2
≤ k(I + t2H21)−σk1,2k d∗F (tH1) (I + t2H21)σk2,2
≤ C t−n/2k d∗G(tH1)k2,2 Next observe that
k d∗G(tH1)k2,2≤ C
kH1G(tH1)k2,2+ kG(tH1)k2,2
≤ C t−1 sup
λ
|λ G(λ )| + sup
λ
|G(λ )|
≤ C t−1.
Proof of the lemma
Choose σ > n/4 . Then for 0 < t < 1 kF (tH1)d∗k1,2= kd∗F (tH1)k1,2
≤ k(I + t2H21)−σk1,2k d∗F (tH1) (I + t2H21)σk2,2
≤ C t−n/2k d∗G(tH1)k2,2 Next observe that
k d∗G(tH1)k2,2≤ C
kH1G(tH1)k2,2+ kG(tH1)k2,2
≤ C t−1 sup
λ
|λ G(λ )| + sup
λ
|G(λ )|
≤ C t−1.
Proof of the lemma
Choose σ > n/4 . Then for 0 < t < 1 kF (tH1)d∗k1,2= kd∗F (tH1)k1,2
≤ k(I + t2H21)−σk1,2k d∗F (tH1) (I + t2H21)σk2,2
≤ C t−n/2k d∗G(tH1)k2,2 Next observe that
k d∗G(tH1)k2,2≤ C
kH1G(tH1)k2,2+ kG(tH1)k2,2
≤ C t−1 sup
λ
|λ G(λ )| + sup
λ
|G(λ )|
≤ C t−1.
Proof of the lemma
Choose σ > n/4 . Then for 0 < t < 1 kF (tH1)d∗k1,2= kd∗F (tH1)k1,2
≤ k(I + t2H21)−σk1,2k d∗F (tH1) (I + t2H21)σk2,2
≤ C t−n/2k d∗G(tH1)k2,2 Next observe that
k d∗G(tH1)k2,2≤ C
kH1G(tH1)k2,2+ kG(tH1)k2,2
≤ C t−1 sup
λ
|λ G(λ )| + sup
λ
|G(λ )|
≤ C t−1.
Proof of the lemma
Choose σ > n/4 . Then for 0 < t < 1 kF (tH1)d∗k1,2= kd∗F (tH1)k1,2
≤ k(I + t2H21)−σk1,2k d∗F (tH1) (I + t2H21)σk2,2
≤ C t−n/2k d∗G(tH1)k2,2 Next observe that
k d∗G(tH1)k2,2≤ C
kH1G(tH1)k2,2+ kG(tH1)k2,2
≤ C t−1 sup
λ
|λ G(λ )| + sup
λ
|G(λ )|
≤ C t−1.
Further developments
How large is the scope of Taylor’s thm?
Meda-Vallarino: on Riemannian noncpct symmetric spaces of arbitrary rankthe operators
Liu= (H2+b)iu u ∈ R are of weak type (1, 1) .
Taylor’s theorem does not apply to Liu because the function m(λ ) = (λ2+b)iu∈/Sβ /2 since
√
b = β /2.
M-Meda-Vallarino: Endpoint estimates on Hardy type spaces for m(H) even when m is singular at ±iβ /2 .