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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

(2)

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 . . .

(3)

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 . . .

(4)

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 . . .

(5)

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 . . .

(6)

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 . . .

(7)

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

(8)

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

(9)

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

(10)

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

w

Σw |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

(11)

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

w

Σw |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

(12)

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

w

Σw |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

(13)

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 .

(14)

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 .

(15)

Ricci versus Riemann

Riemann tensor R(X , Y ) = ∇XY− ∇YX− ∇[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.

(16)

Ricci versus Riemann

Riemann tensor R(X , Y ) = ∇XY− ∇YX− ∇[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.

(17)

Ricci versus Riemann

Riemann tensor R(X , Y ) = ∇XY− ∇YX− ∇[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.

(18)

Lower Ricci bounded vs C

bounded geometry.

Def.

M has lower Ricci bounded geometry if

I rinj(M) > 0

I RicM≥ −κ2.

Consequences:

I V Br(p) ≤ C(1 + r )αeβr

I M islocallydoubling

I covering theorem: M =SNj=1Sm=1B1(xjm) B2(xjm) ∩B2(xj`) = /0 if xjm6= xi`.

(19)

Lower Ricci bounded vs C

bounded geometry.

Def.

M has lower Ricci bounded geometry if

I rinj(M) > 0

I RicM≥ −κ2.

Consequences:

I V Br(p) ≤ C(1 + r )αeβr

I M islocallydoubling

I covering theorem: M =SNj=1Sm=1B1(xjm) B2(xjm) ∩B2(xj`) = /0 if xjm6= xi`.

(20)

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.

(21)

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.

(22)

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.

(23)

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

(24)

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

(25)

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

(26)

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

(27)

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

(28)

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

(29)

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

(30)

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

(31)

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.

(32)

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.

(33)

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

Iyk

L2 Br(y )+ kψyk

L2 Br(y )≤ C

If M has only lower Ricci bounded geometry we use ultracontractivityof theheat semigroupto obtain aSobolev inequality.

(34)

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

Iyk

L2 Br(y )+ kψyk

L2 Br(y )≤ C

If M has only lower Ricci bounded geometry we use ultracontractivityof theheat semigroupto obtain aSobolev inequality.

(35)

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.

(36)

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.

(37)

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.

(38)

The estimates of k

]

est

Divide M into shells:

Aj(y ) = {x ∈ M : j ≤ d (x , y ) < j + 1}

M = [

j≥0

Aj(y )

(39)

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 sup

y

kk](·,y )k

L2 Aj(y )≤ C`(1 + j)−`e−jw ∀j ≥ 0, ` ∈ N.

(40)

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 sup

y

kk](·,y )k

L2 Aj(y )≤ C`(1 + j)−`e−jw ∀j ≥ 0, ` ∈ N.

(41)

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 sup

y

kk](·,y )k

L2 Aj(y )≤ C`(1 + j)−`e−jw ∀j ≥ 0, ` ∈ N.

(42)

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 sup

y

kk](·,y )k

L2 Aj(y )≤ C`(1 + j)−`e−jw ∀j ≥ 0, ` ∈ N.

(43)

Proof of the Lemma

Chose a cutoff ψj: ψj(t) = 1 for |t| ≥ j ,

(44)

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

(45)

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

(46)

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

(47)

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

(48)

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

(49)

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

(50)

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.

(51)

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.

(52)

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.

(53)

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.

(54)

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.

(55)

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.

(56)

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.

(57)

Recall that

L = dd and L = d d+dd 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.

(58)

Recall that

L = dd and L = d d+dd 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.

(59)

Recall that

L = dd and L = d d+dd 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.

(60)

Recall that

L = dd and L = d d+dd 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.

(61)

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

(62)

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

(63)

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

(64)

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

(65)

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

(66)

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

(67)

m

[

(H) d

= Z

−∞

P

t

b 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.

(68)

m

[

(H) d

= Z

−∞

P

t

b 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.

(69)

m

[

(H) d

= Z

−∞

P

t

b 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.

(70)

m

[

(H) d

= Z

−∞

P

t

b 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.

(71)

m

[

(H) d

= Z

−∞

P

t

b 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.

(72)

The proof of the lemma is based on

I the intertwining identity

L d=dL

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

(73)

The proof of the lemma is based on

I the intertwining identity

L d=dL

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

(74)

Proof of the lemma

Choose σ > n/4 . Then for 0 < t < 1 kF (tH1)dk1,2= kdF (tH1)k1,2

≤ k(I + t2H21)−σk1,2k dF (tH1) (I + t2H21)σk2,2

≤ C t−n/2k dG(tH1)k2,2 Next observe that

k dG(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.

(75)

Proof of the lemma

Choose σ > n/4 . Then for 0 < t < 1 kF (tH1)dk1,2= kdF (tH1)k1,2

≤ k(I + t2H21)−σk1,2k dF (tH1) (I + t2H21)σk2,2

≤ C t−n/2k dG(tH1)k2,2 Next observe that

k dG(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.

(76)

Proof of the lemma

Choose σ > n/4 . Then for 0 < t < 1 kF (tH1)dk1,2= kdF (tH1)k1,2

≤ k(I + t2H21)−σk1,2k dF (tH1) (I + t2H21)σk2,2

≤ C t−n/2k dG(tH1)k2,2 Next observe that

k dG(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.

(77)

Proof of the lemma

Choose σ > n/4 . Then for 0 < t < 1 kF (tH1)dk1,2= kdF (tH1)k1,2

≤ k(I + t2H21)−σk1,2k dF (tH1) (I + t2H21)σk2,2

≤ C t−n/2k dG(tH1)k2,2 Next observe that

k dG(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.

(78)

Proof of the lemma

Choose σ > n/4 . Then for 0 < t < 1 kF (tH1)dk1,2= kdF (tH1)k1,2

≤ k(I + t2H21)−σk1,2k dF (tH1) (I + t2H21)σk2,2

≤ C t−n/2k dG(tH1)k2,2 Next observe that

k dG(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.

(79)

Proof of the lemma

Choose σ > n/4 . Then for 0 < t < 1 kF (tH1)dk1,2= kdF (tH1)k1,2

≤ k(I + t2H21)−σk1,2k dF (tH1) (I + t2H21)σk2,2

≤ C t−n/2k dG(tH1)k2,2 Next observe that

k dG(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.

(80)

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 .

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