riesz transform characterization of hardy spaces associated with

Riesz transform characterization of Hardy spacesassociated with Schrodinger operators with

compactly supported potentials

Marcin Preisner (Wroc law University)

joint work with Jacek Dziubanski

Zakopane, 19.04.2010

Marcin Preisner (Wroc law University) Hardy spaces associated with Schrodinger operators

Classical Hardy spaces

Let Pt f (x) = Pt ∗ f (x), Pt(x) = (4πt)−d/2 exp(−|x |2/4t).

Theorem (Fefferman, Stein, Coifman, Latter, ...)

A function f ∈ L1(Rd) belongs to the Hardy space H1(Rd) when one of thefollowing equivalent conditions holds:

Mf (x) = supt>0 |Pt f (x)| ∈ L1(Rd),

Rj f (x) = cd limε→0

R|x−y|>ε

xj−yj

|x−y|d+1 f (y) dy ∈ L1(Rd)

for j = 1, 2, . . . , d ,

∃{λn, an}∞n=1 f (x) =P∞

n=1 λn an(x), whereP∞

n=1 |λn| <∞and an are atoms (i.e. supp an ⊂ Bn, ‖an‖∞ ≤ |Bn|−1,

Ra = 0)

Each of the above conditions define a complete norm of H1(Rd) and all ofthem are comparable, i.e.,

‖Mf ‖L1(Rd ) ∼ ‖f ‖L1(Rd ) +dX

‖Rj f ‖L1(Rd ) ∼ inff =

Pn λnan

|λn|.

Mf (x) = supt>0 |Pt f (x)| ∈ L1(Rd),

R|x−y|>ε

xj−yj

|x−y|d+1 f (y) dy ∈ L1(Rd)

for j = 1, 2, . . . , d ,

∃{λn, an}∞n=1 f (x) =P∞

Ra = 0)

‖Mf ‖L1(Rd ) ∼ ‖f ‖L1(Rd ) +dX

Pn λnan

|λn|.

Mf (x) = supt>0 |Pt f (x)| ∈ L1(Rd),

R|x−y|>ε

xj−yj

|x−y|d+1 f (y) dy ∈ L1(Rd)

for j = 1, 2, . . . , d ,

∃{λn, an}∞n=1 f (x) =P∞

Ra = 0)

‖Mf ‖L1(Rd ) ∼ ‖f ‖L1(Rd ) +dX

Pn λnan

|λn|.

Mf (x) = supt>0 |Pt f (x)| ∈ L1(Rd),

R|x−y|>ε

xj−yj

|x−y|d+1 f (y) dy ∈ L1(Rd)

for j = 1, 2, . . . , d ,

∃{λn, an}∞n=1 f (x) =P∞

Ra = 0)

‖Mf ‖L1(Rd ) ∼ ‖f ‖L1(Rd ) +dX

Pn λnan

|λn|.

Mf (x) = supt>0 |Pt f (x)| ∈ L1(Rd),

R|x−y|>ε

xj−yj

|x−y|d+1 f (y) dy ∈ L1(Rd)

for j = 1, 2, . . . , d ,

∃{λn, an}∞n=1 f (x) =P∞

Ra = 0)

‖Mf ‖L1(Rd ) ∼ ‖f ‖L1(Rd ) +dX

Pn λnan

|λn|.

Mf (x) = supt>0 |Pt f (x)| ∈ L1(Rd),

R|x−y|>ε

xj−yj

|x−y|d+1 f (y) dy ∈ L1(Rd)

for j = 1, 2, . . . , d ,

∃{λn, an}∞n=1 f (x) =P∞

Ra = 0)

‖Mf ‖L1(Rd ) ∼ ‖f ‖L1(Rd ) +dX

Pn λnan

|λn|.

The Schrodinger operator background

We consider a Schrodinger operator L = −∆ + V on Rd and assume

d ≥ 3,

V ≥ 0

V ∈ Lp(Rd) for some p > d/2,

supp V ⊆ B(0, 1) = {x ∈ R : |x | < 1}.

Let {Kt}t>0 be the semigroup of linear operators generated by −L and

MLf (x) = supt>0|Kt f (x)|.

be an associated maximal function.

Definition

We say that an L1(Rd)-function f belongs the Hardy space H1L if

‖f ‖H1L

= ‖MLf ‖L1(Rd ) <∞.

d ≥ 3,

V ≥ 0

supp V ⊆ B(0, 1) = {x ∈ R : |x | < 1}.

Definition

‖f ‖H1L

= ‖MLf ‖L1(Rd ) <∞.

d ≥ 3,

V ≥ 0

supp V ⊆ B(0, 1) = {x ∈ R : |x | < 1}.

Definition

‖f ‖H1L

= ‖MLf ‖L1(Rd ) <∞.

d ≥ 3,

V ≥ 0

supp V ⊆ B(0, 1) = {x ∈ R : |x | < 1}.

Definition

‖f ‖H1L

= ‖MLf ‖L1(Rd ) <∞.

d ≥ 3,

V ≥ 0

supp V ⊆ B(0, 1) = {x ∈ R : |x | < 1}.

Definition

‖f ‖H1L

= ‖MLf ‖L1(Rd ) <∞.

d ≥ 3,

V ≥ 0

supp V ⊆ B(0, 1) = {x ∈ R : |x | < 1}.

Definition

‖f ‖H1L

= ‖MLf ‖L1(Rd ) <∞.

d ≥ 3,

V ≥ 0

supp V ⊆ B(0, 1) = {x ∈ R : |x | < 1}.

Definition

‖f ‖H1L

= ‖MLf ‖L1(Rd ) <∞.

Riesz transforms in the Schrodinger case

Recall that in the classical case for f ∈ L1(Rd) we have

Rj f (x) =√π∂

∂xj(−∆)−1/2 = cd lim

ε→0

Z|x−y|>ε

xj − yj

|x − y |d+1f (y) dy

= limε→0

Z ε−1

∂xjPt(x − y)f (y) dy

dt√t,

Definition

Thus, for f ∈ L1(Rd), we define Riesz transforms Rj associated with L forj = 1, 2, . . . , d by setting

Rj f =√π∂

∂xjL−1/2f = lim

ε→0

Z 1/ε

∂xjKt f

dt√t,

where the limit is understood in the sense of distributions.

Riesz transforms in the Schrodinger case

Recall that in the classical case for f ∈ L1(Rd) we have

Rj f (x) =√π∂

∂xj(−∆)−1/2 = cd lim

ε→0

Z|x−y|>ε

xj − yj

|x − y |d+1f (y) dy

= limε→0

Z ε−1

∂xjPt(x − y)f (y) dy

dt√t,

Definition

Thus, for f ∈ L1(Rd), we define Riesz transforms Rj associated with L forj = 1, 2, . . . , d by setting

Rj f =√π∂

∂xjL−1/2f = lim

ε→0

Z 1/ε

∂xjKt f

dt√t,

where the limit is understood in the sense of distributions.

Main result

Having all this prepared we present the main theorem:

Theorem

Assume that f ∈ L1(Rd). Then f is in the Hardy space H1L if and only if

Rj f ∈ L1(Rd) for every j = 1, ..., d. Moreover, there exists C > 0 such that

C−1‖f ‖H1L≤ ‖f ‖L1(Rd ) +

‖Rj f ‖L1(Rd ) ≤ C‖f ‖H1L.

Main result

Having all this prepared we present the main theorem:

Theorem

Assume that f ∈ L1(Rd). Then f is in the Hardy space H1L if and only if

Rj f ∈ L1(Rd) for every j = 1, ..., d. Moreover, there exists C > 0 such that

C−1‖f ‖H1L≤ ‖f ‖L1(Rd ) +

‖Rj f ‖L1(Rd ) ≤ C‖f ‖H1L.

Previous results

Hardy spaces H1L (with the same assumptions on potential V ) were studied in

the paper:

J. Dziubanski and J. Zienkiewicz, ”Hardy space H1 for Schrodingeroperators with compactly supported potentials”,Ann. Mat. Pura Appl., 184 (2005), 315–326.

It was proved there that the elements of the space H1L admit special atomic

decompositions. In order to define the atoms the authors introduced thefunction

ω(x) = limt→∞

ZKt(x , y) dy

which turns out to be L-harmonic, Lipschitz, and 0 < ε < ω ≤ 1.

The atoms of the Hardy space for the Schrodinger operator satisfy the sameconditions as in the classical case, the only difference is that the cancellationcontition holds with respect to ω, i.e.,

Ra(x)ω(x)dx = 0.

In other words the mapping

H1L 3 f 7→ f · ω ∈ H1(Rd)

is an isomorphism of the Hardy space for the Schrodinger operator and theclassical one.

Previous results

the paper:

ω(x) = limt→∞

ZKt(x , y) dy

Ra(x)ω(x)dx = 0.

H1L 3 f 7→ f · ω ∈ H1(Rd)

Previous results

the paper:

ω(x) = limt→∞

ZKt(x , y) dy

Ra(x)ω(x)dx = 0.

H1L 3 f 7→ f · ω ∈ H1(Rd)

Previous results

the paper:

ω(x) = limt→∞

ZKt(x , y) dy

Ra(x)ω(x)dx = 0.

H1L 3 f 7→ f · ω ∈ H1(Rd)

An isomorphism with classical Hardy space

In order to prove this result Dziubanski and Zienkiewicz first obtained anotherisomorphism between H1

L and the classical Hardy space H1(Rd).

Namely, theyproved that the operator

I − VL−1 : H1L → H1(Rd)

is bounded and invertible (whose inverse is I − V (−∆)−1).

The operator I − VL−1 is also an isomorphism on L1(Rd).

Also, corresponding norms are equivalent, i.e.,

‖(I − VL−1)f ‖H1(Rd ) ' ‖f ‖H1L

for f ∈ H1L .

RemarkHaving these results, in order to prove the Riesz transform characterization ofH1

L , it suffices to show that Rj(I − VL−1)− Rj , j = 1, 2..., d, are boundedoperators on L1(Rd).

L and the classical Hardy space H1(Rd). Namely, theyproved that the operator

I − VL−1 : H1L → H1(Rd)

‖(I − VL−1)f ‖H1(Rd ) ' ‖f ‖H1L

for f ∈ H1L .

I − VL−1 : H1L → H1(Rd)

‖(I − VL−1)f ‖H1(Rd ) ' ‖f ‖H1L

for f ∈ H1L .

I − VL−1 : H1L → H1(Rd)

‖(I − VL−1)f ‖H1(Rd ) ' ‖f ‖H1L

for f ∈ H1L .

I − VL−1 : H1L → H1(Rd)

‖(I − VL−1)f ‖H1(Rd ) ' ‖f ‖H1L

for f ∈ H1L .

Perturbation formula

Similarly to Dziubanski and Zienkiewicz paper we take adventage of theperturbation formula

Pt = Kt +

Pt−sVKs ds

from which the following two identities hold

Pt(I − VL−1)− Kt =

Pt−sVKsds − PtVL−1 = fWt −fQt .

Pt(I − VL−1)− Kt =

(Pt−s − Pt)VKs ds −Z ∞

PtVKs ds = Wt − Qt .

Derivating both formulas with respect to xj , integrating the first one from ε to1 and the second one from 1 to ε−1 (with respect to the weight dt/

√t) we get:

Pt = Kt +

Pt−sVKs ds

Pt(I − VL−1)− Kt =

√t) we get:

Pt = Kt +

Pt−sVKs ds

Pt(I − VL−1)− Kt =

√t) we get:

Crucial difference

Z ε−1

∂xjPt

dt√t

!“I − VL−1

”−Z ε−1

∂xjKt

dt√t

fWtdt√

t−Z 1

eQtdt√

∂xjWt

dt√t−Z ε

∂xjQt

dt√t

= fWεj − eQεj +Wε

j −Qεj .

The operator on the LHS when we apply to f ∈ L1(Rd) tends toRj(I − VL−1)f − Rj f as ε→ 0, so it is enough to study the limits of the fourterms on the RHS.

Crucial difference

Z ε−1

∂xjPt

dt√t

!“I − VL−1

”−Z ε−1

∂xjKt

dt√t

fWtdt√

t−Z 1

eQtdt√

∂xjWt

dt√t−Z ε

∂xjQt

dt√t

= fWεj − eQεj +Wε

j −Qεj .

The operator on the LHS when we apply to f ∈ L1(Rd) tends toRj(I − VL−1)f − Rj f as ε→ 0, so it is enough to study the limits of the fourterms on the RHS.

The limits of Wεj , Qε

j , Wεj

The operators Qεj , fWεj , and Wε

j converge in the norm-operator topology tobounded operators on L1(Rd) as ε→ 0.

Proof. In this talk we will only concentrate on limε→0Wεj .

The integral kernels of these operators are

Wεj (x , y) =

Z ε−1

∂xjWt(x , y)

dt√t,

Wt(x , y) =

Z(Pt−s(x − z)− Pt(x − z))V (z)Ks(z , y) dz ds.

j , Wεj

Wεj (x , y) =

Z ε−1

∂xjWt(x , y)

dt√t,

Wt(x , y) =

j , Wεj

Wεj (x , y) =

Z ε−1

∂xjWt(x , y)

dt√t,

Wt(x , y) =

Let 0 < ε2 < ε1 < 1/2. Then

Z ˛Wε1

j (x , y)−Wε2j (x , y)

˛dx ≤ W ′(y) +W ′′(y), where

W ′(y) =

Z Z ε−12

ε−11

Z t8/9

“Pt(x − z)− Pt−s(x − z)

”˛V (z)Ks(z, y) dz ds

Observe that for 0 < s < t8/9, there exists 0 ≤ φ ∈ S(R) such that˛ ∂∂xj

“Pt(x − z)− Pt−s(x − z)

”˛≤ s t−3/2φt(x − z) (φt(x) = t−d/2φ(x/

√t))

Moreover, sKs(z, y) ≤ Cs(2−d)/2 exp`−c|z − y |2/s

´≤ C |z − y |2−d . Thus

W ′(y) ≤Z Z ε−1

ε−11

Z t8/9

Zs t−2φt(x − z)V (z)Ks(z, y) dz ds dt dx

≤Z ε−1

ε−11

t−10/9dt ·Z

V (z)|z − y |2−ddz ≤ Cε1/91

uniformly in y . The last inequality is a simple consequence of the Holder inequalityand the assumption p > d/2.

A similar analysis leads to W ′′(y) ≤ Cε4d/9−11 .

Let 0 < ε2 < ε1 < 1/2. ThenZ ˛Wε1

j (x , y)−Wε2j (x , y)

˛dx ≤ W ′(y) +W ′′(y), where

W ′(y) =

Z Z ε−12

ε−11

Z t8/9

“Pt(x − z)− Pt−s(x − z)

”˛≤ s t−3/2φt(x − z) (φt(x) = t−d/2φ(x/

√t))

´≤ C |z − y |2−d . Thus

W ′(y) ≤Z Z ε−1

ε−11

Z t8/9

≤Z ε−1

ε−11

t−10/9dt ·Z

V (z)|z − y |2−ddz ≤ Cε1/91

Let 0 < ε2 < ε1 < 1/2. ThenZ ˛Wε1

j (x , y)−Wε2j (x , y)

˛dx ≤ W ′(y) +W ′′(y), where

W ′(y) =

Z Z ε−12

ε−11

Z t8/9

“Pt(x − z)− Pt−s(x − z)

”˛≤ s t−3/2φt(x − z) (φt(x) = t−d/2φ(x/

√t))

´≤ C |z − y |2−d . Thus

W ′(y) ≤Z Z ε−1

ε−11

Z t8/9

≤Z ε−1

ε−11

t−10/9dt ·Z

V (z)|z − y |2−ddz ≤ Cε1/91

Let 0 < ε2 < ε1 < 1/2. ThenZ ˛Wε1

j (x , y)−Wε2j (x , y)

˛dx ≤ W ′(y) +W ′′(y), where

W ′(y) =

Z Z ε−12

ε−11

Z t8/9

“Pt(x − z)− Pt−s(x − z)

”˛≤ s t−3/2φt(x − z) (φt(x) = t−d/2φ(x/

√t))

´≤ C |z − y |2−d . Thus

W ′(y) ≤Z Z ε−1

ε−11

Z t8/9

≤Z ε−1

ε−11

t−10/9dt ·Z

V (z)|z − y |2−ddz ≤ Cε1/91

Let 0 < ε2 < ε1 < 1/2. ThenZ ˛Wε1

j (x , y)−Wε2j (x , y)

˛dx ≤ W ′(y) +W ′′(y), where

W ′(y) =

Z Z ε−12

ε−11

Z t8/9

“Pt(x − z)− Pt−s(x − z)

”˛≤ s t−3/2φt(x − z) (φt(x) = t−d/2φ(x/

√t))

´≤ C |z − y |2−d . Thus

W ′(y) ≤Z Z ε−1

ε−11

Z t8/9

≤Z ε−1

ε−11

t−10/9dt ·Z

V (z)|z − y |2−ddz ≤ Cε1/91

Let 0 < ε2 < ε1 < 1/2. ThenZ ˛Wε1

j (x , y)−Wε2j (x , y)

˛dx ≤ W ′(y) +W ′′(y), where

W ′(y) =

Z Z ε−12

ε−11

Z t8/9

“Pt(x − z)− Pt−s(x − z)

”˛≤ s t−3/2φt(x − z) (φt(x) = t−d/2φ(x/

√t))

´≤ C |z − y |2−d . Thus

W ′(y) ≤Z Z ε−1

ε−11

Z t8/9

≤Z ε−1

ε−11

t−10/9dt ·Z

V (z)|z − y |2−ddz ≤ Cε1/91

The limit of Qεj f for f ∈ L1(Rd)

Assume that f ∈ L1(Rd). Then the limit F = limε→0fQεj f exists in the

L1-norm. Moreover, ‖F‖L1(Rd ) ≤ C‖f ‖L1(Rd ) with C independent of f .

Proof. Of course, for any fixed y ∈ Rd , the functionz 7→ U(z , y) = V (z)Γ(z , y) is supported in the unit ball and

‖U(z , y)‖Lr (dz) ≤ Cr for fixed r ∈h1, dp

dp+d−2p

”with Cr independent of y . Let

Hεj (x , z) =

∂xjPt(x − z)

dt√t, H∗j g(x) = sup

0<ε<1|Hε

j g(x)|.

It follows from the theory of singular integral convolution operators that for1 < r <∞ there exists Cr such that

‖H∗j g‖Lr (Rd ) ≤ Cr‖g‖Lr (Rd ) for g ∈ Lr (Rd)

and limε→0 Hεj g(x) = Hjg(x) a.e. and in Lr (Rd)-norm.

dp+d−2p

”with Cr independent of y .

Hεj (x , z) =

∂xjPt(x − z)

0<ε<1|Hε

j g(x)|.

dp+d−2p

Hεj (x , z) =

∂xjPt(x − z)

0<ε<1|Hε

j g(x)|.

dp+d−2p

Hεj (x , z) =

∂xjPt(x − z)

0<ε<1|Hε

j g(x)|.

Note that fQεj (x , y) = Hεj U( · , y)(x).

Hence, there exists a function fQj(x , y)

such that limε→0fQεj (x , y) = fQj(x , y) a.e. and

sup0<ε<1

˛fQεj (x , y)˛r

dx ≤ C ′r for 1 < r <dp

dp + d − 2p.

Since |Hεj (x , z)| ≤ CN |x − z |−N for |x − z | > 1,

|fQεj (x , y)| =

˛˛Z|z|≤1

Hεj (x , z)U(z , y) dz

˛˛ ≤ CN |x |−N for |x | > 2, y ∈ Rd .

Applying the Holder inequality we get

sup0<ε<1

˛fQεj (x , y)˛

dx ≤ C ,

limε→0

˛fQεj (x , y)−fQj(x , y)˛

dx = 0 for every y .

Now the lemma can be easily concluded from the Lebesgue dominatedconvergence theorem.

Note that fQεj (x , y) = Hεj U( · , y)(x). Hence, there exists a function fQj(x , y)

sup0<ε<1

˛fQεj (x , y)˛r

dp + d − 2p.

|fQεj (x , y)| =

˛˛Z|z|≤1

˛˛ ≤ CN |x |−N for |x | > 2, y ∈ Rd .

sup0<ε<1

˛fQεj (x , y)˛

dx ≤ C ,

limε→0

sup0<ε<1

˛fQεj (x , y)˛r

dp + d − 2p.

|fQεj (x , y)| =

˛˛Z|z|≤1

˛˛ ≤ CN |x |−N for |x | > 2, y ∈ Rd .

sup0<ε<1

˛fQεj (x , y)˛

dx ≤ C ,

limε→0

sup0<ε<1

˛fQεj (x , y)˛r

dp + d − 2p.

|fQεj (x , y)| =

˛˛Z|z|≤1

˛˛ ≤ CN |x |−N for |x | > 2, y ∈ Rd .

sup0<ε<1

˛fQεj (x , y)˛

dx ≤ C ,

limε→0

Proof of the main theorem

Consider f ∈ L1(Rd).

Using above lemmas we get that Rj f ∈ L1(Rd) if and only ifRj(I − VL−1)f ∈ L1(Rd) and

‖f ‖L1(Rd ) +dX

‖Rj f ‖L1(Rd ) ∼ ‖(I − VL−1)f ‖L1(Rd ) +dX

‖Rj (I − VL−1)f ‖L1(Rd )

Classical characterization implies that (I −VL−1)f ∈ H1(Rd) if and only if(I − VL−1)f , Rj(I − VL−1)f ∈ L1(Rd) for j = 1, ..., d . Moreover,

‖(I − VL−1)f ‖L1(Rd ) +dX

‖Rj (I − VL−1)f ‖L1(Rd ) ∼ ‖(I − VL−1)f ‖H1(Rd ).

Finally, recall that (I − VL−1) is the isomorphism of Hardy spaces H1L and

H1(Rd). Thus (I − VL−1)f ∈ H1(Rd) if and only if f ∈ H1L and

‖(I − VL−1)f ‖H1(Rd ) ∼ ‖f ‖H1L.

Putting these facts together we obtain the proof of the main theorem.

‖f ‖L1(Rd ) +dX

‖Rj (I − VL−1)f ‖L1(Rd )

‖(I − VL−1)f ‖L1(Rd ) +dX

‖(I − VL−1)f ‖H1(Rd ) ∼ ‖f ‖H1L.

‖f ‖L1(Rd ) +dX

‖Rj (I − VL−1)f ‖L1(Rd )

‖(I − VL−1)f ‖L1(Rd ) +dX

‖(I − VL−1)f ‖H1(Rd ) ∼ ‖f ‖H1L.

‖f ‖L1(Rd ) +dX

‖Rj (I − VL−1)f ‖L1(Rd )

‖(I − VL−1)f ‖L1(Rd ) +dX

‖(I − VL−1)f ‖H1(Rd ) ∼ ‖f ‖H1L.

‖f ‖L1(Rd ) +dX

‖Rj (I − VL−1)f ‖L1(Rd )

‖(I − VL−1)f ‖L1(Rd ) +dX

‖(I − VL−1)f ‖H1(Rd ) ∼ ‖f ‖H1L.

riesz transform characterization of hardy spaces associated with

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