# Commutators with Lipschitz Functions and Nonintegral Operators

Authors: Peizhu Xie; Ruming Gong
Journal: Journal of Applied Mathematics
 (2013)

---

## Abstract

LetT be a singular nonintegral operator; that is, it does not have an integral representation by a kernel with size estimates, even rough. In this paper, we consider the boundedness of commutators with T and Lipschitz functions. Applications include spectral multipliers of self-adjoint, positive operators, Riesz transforms of second-order divergence form operators, and fractional power of elliptic operators.


---

## Body

## 1. Introduction

LetT be a bounded operator on Lp(ℝn) for some p, 1<p<∞. A measurable function K(x,y) is called an associated kernel of T if
(1)Tf(x)=∫XK(x,y)f(y)dy
holds for each continuous function f with compact support and for almost all x not in the support of f.The kernelK(x,y) is said to satisfy the following.(i) The pointwise Hörmander condition onx variable if there exist 0<α≤1 and c,c1≥1 such that
(2)|K(x,y)-K(z,y)|≤c|x-z|α|x-y|n+α,
when |x-y|≥c1|x-z|, and B(x,r) denotes the ball with center x, radius r.(ii) The integral Hörmander condition ony variable if there exist constants C and c2≥1 such that
(3)∫|x-y|≥c2|z-y|‍|K(x,y)-K(x,z)|dx≤C,
for all y,z∈ℝn.It is well known that ifT is bounded on Lq(ℝn) for some q, 1<q<∞, and b∈BMO, the two Hörmander conditions (i) and (ii) above are sufficient to imply that the commutator [b,T] is bounded on Lp(ℝn) for all p, 1<p<∞, with norm
(4)∥[b,T](f)∥p≤C∥b∥*∥f∥p,
where the commutator [b,T] is defined by [b,T](f)=T(bf)-bT(f) and ∥b∥* is the BMO seminorm of b. See [1, 2] for BMO functions on Euclidean spaces ℝn and [3] for spaces of homogeneous type.A particular case of the result of Janson [2] states that [b,T]:Lp→Lq is bounded, 1<p<q<∞, if b∈Λ˙β, β=n(1/p-1/q). Here, Λ˙β is the homogeneous Lipschitz space determined by the first difference operator.In [4], Duong and Yan have replaced the two Hörmander conditions (2) and (3) by the following weaker conditions (5) and (6) below which previously appeared in [5] and still concluded that the commutator [b,T] is bounded on Lp(ℝn) for all p, 1<p<∞. And in [6], Hu and Yang obtained the weighted boundedness of maximal commutator when T satisfy (5) and (6). Roughly speaking, we assume the following.(iii) There exists a class of operatorsAt with kernels at(x,y), which satisfy the condition (23) in Section 2, so that the kernels kt(x,y) of the operators (T-AtT) satisfy the condition
(5)|kt(x,y)|≤ctγ/m|x-y|n+γ,
when |x-y|≥c2t1/m for some γ, m>0, where c is a positive constant.(iv) There exists a class of operatorsBt with kernels bt(x,y), which satisfy the condition (23), such that (T-TBt) have associated kernels Kt(x,y) and there exist positive constants c3, c4 such that
(6)∫|x-y|≥c3t1/m‍|Kt(x,y)|dx≤c4,∀y∈ℝn.Under conditions (5) and (6), if T is bounded on Lq(ℝn) for some q, 1<q<∞, then the commutator [b,T] is bounded on Lp(ℝn) for all p, 1<p<∞.In [7], Auscher and Martell have considered the commutators of singular nonintegral operators, where the implicit terminology has been introduced in [8]. By this we mean that they are still of order 0, but they do not have an integral representation by a kernel with size and/or smoothness estimates. Let 1≤p0<q0≤∞. Suppose that the singular nonintegral operator T is a sublinear operator bounded on Lp0(ℝn) and that {Ar}r>0 is a family of operators acting from Lc∞(ℝn) into Lp0(ℝn). Auscher and Martell assume the following.(v) For allf∈Lc∞(ℝn) and all balls B where r(B) denotes its radius,
(7)(1|B|∫B‍|T(I-Ar(B))f|p0dx)1/p0≤C∑j=1∞‍αj(1|2j+1B|∫2j+1B‍|f|p0dx)1/p0.(vi) For allf∈Lc∞(ℝn) and all balls B where r(B) denotes its radius,
(8)(1|B|∫B‍|TAr(B)f|q0dx)1/q0≤C∑j=1∞‍αj(1|2j+1B|∫2j+1B‍|Tf|p0dx)1/p0.Letp0<p<q0 and w∈Ap/p0∩RH(q0/p)′ (for the definitions of Ap/p0 and RH(q0/p)′ see Section 2). Under conditions (7) and (8), if ∑j=1∞αjj<∞, then the commutator [b,T] is bounded on Lp(w); that is, ∥[b,T]f∥Lp(w)≤C∥b∥*∥f∥Lp(w) for all f∈Lc∞(ℝn).The main object of this paper is the commutators of nonintegral operators[b,T]. Compared to the result in [7], we can obtain a more general result for b belongs to the Lipschitz spaces Λ˙βi(X). To be more specific, we can obtain the following.Theorem 1.
Let0≤α<1, 1≤p0≤s0<q0≤∞ such that 1/s0=1/p0-α/n. Suppose that T is a sublinear operator bounded from Lp0(ℝn) to Ls0(ℝn) and that {Ar}r>0 is a family of operators acting from Lc∞(ℝn) into Lp0(ℝn). Assume that
(9)(1|B|∫B‍|T(I-Ar(B))f|s0dx)1/s0≤C∑j=1∞‍αj|2j+1B|α/n(1|2j+1B|∫2j+1B‍|f|p0dx)1/p0,(10)(1|B|∫B‍|TAr(B)f|q0dx)1/q0≤C∑j=1∞‍αj(1|2j+1B|∫2j+1B‍|Tf|s0dx)1/s0,
for all f∈Lc∞(ℝn) and all balls B, where r(B) denotes its radius. Let 0<β<1 such that α+β<1. Let p0<p<q<q0 and 1/q=1/p-(α+β)/n. If ∑j=1∞αj<∞, then there is a constant C such that
(11)∥[b,T]f∥Lq≤C∥b∥Λ˙β∥f∥Lp,
for all f∈Lc∞(ℝn) and for all b∈Λ˙β.The caseq0=∞ is understood in the sense that the Lq0-average in (10) is indeed an essential supremum.Remark 2.
Let1≤p0<p<q<q0 be such that 1/q=1/p-α/n. Under the assumptions above, we know that if ∑j=1∞αj<∞, then T is bounded from Lp to Lq. See Theorem 2.2 in [9].
In the limiting caseα=0, from the assumptions (9) and (10), we deduce
(12)(1|B|∫B‍|T(I-Ar(B))f|p0)1/p0≤CM(|f|p0)1/p0(x),(1|B|∫B‍|TAr(B)f|q0)1/q0≤CM(|Tf|p0)1/p0(x).
Consequently, from the Theorem 3.7 in [7], we know that if ∑j=1∞αj<∞, then ∥Tf∥Lp(w)≤C∥f∥Lp(w) for p0<p<q0 and for all w∈Ap/p0∩RH(q0/p)′.Theorem 3.
Let1≤p0<q0≤∞. Suppose that T is a sublinear operator bounded on LP0(ℝn) and that {Ar}r>0 is a family of operators acting from Lc∞(ℝn) to LP0(ℝn). Assume that T satisfy (9) and (10) with α=0. Let 0<β<min{1,n/p0}, p0<p<q<q0, b∈Λ˙β and w,v∈Ap/p0∩RH(q0/p)′. Assume that there exists a constant 1<s<min{n/βp0,p/p0} such that (w,v)∈A(p/p0s,q/p0s,βp0s/n). If ∑j=1∞αj<∞, then there is a constant C such that
(13)∥[b,T]f∥Lq(v)≤C∥b∥Λ˙β∥f∥Lp(w),
for all f∈Lc∞.The classA(p,q,s) is defined in Section 2.


## 2. Definitions and Preliminary Results

We use the notation(14)⨏Ef=1|E|∫E‍f(x)dx,
and we often ignore the Lebesgue measure and the variable of the integrand in writing integrals, unless this is needed to avoid confusions.A weightw is a nonnegative locally integrable function. We say that w∈Ap, 1<p<∞, if there exists a constant C such that for every ball B⊂X(15)(⨏Bw)(⨏Bw1-p′)p-1≤C.
For p=1, we say that w∈A1 if there is a constant C such that for every ball B⊂ℝn, ⨏Bw≤Cw(x), for a.e. x∈B, or, equivalently, M(w)≤Cw a.e., where M(w) denotes the classical Hardy-Littlewood maximal function of w. The reverse Hölder classes are defined in the following way: w∈RHq, 1<q<∞, if there is a constant C such that for every ball B⊂ℝn(16)(⨏Bwq)1/q≤⨏Bw.The endpointq=∞ is given by the condition: w∈RH∞ whenever, for any ball B,
(17)w(x)≤⨏Bw,fora.e.x∈B.The homogenous Lipschitz function spaceΛ˙β(ℝn) is the space of functions f such that
(18)∥f∥Λ˙β=supx,h∈ℝn,h≠0|Δh[β]+1f(x)||h|β<∞,
where Δhk denotes the kth difference operator (see [10]). That is, Δh1f(x)=Δhf(x)=f(x+h)-f(x), Δhk+1f(x)=Δhkf(x+h)-Δhkf(x), k≥1.We have the following lemmas.Lemma 4 (see [10]).
For0<β<1, 1≤q<∞, one has
(19)∥f∥Λ˙β(ℝn)≈supB1|B|1+β/n∫B‍|f-fB|dx∥f∥Λ˙β(ℝn)≈supB1|B|β/n(1|B|∫B‍|f-fB|q)1/qdx.
For q=∞, the last formula should be modified appropriately.Lemma 5 (see [10]).
LetB*⊂B⊂ℝn, and then |fB*-fB|≤C∥f∥Λ˙β(ℝn)|B|β/n.Lemma 6 (see [11]).
For1≤γ<∞ and β>0, let
(20)Mβ,γ(f)(x)=supB∋x(1|B|1-βγ/n∫B‍|f(y)|γdy)1/γ.
Suppose that γ<p<n/β and 1/q=1/p-β/n, and then
(21)∥Mβ,γ(f)∥Lq(ℝn)≤C∥f∥Lp(ℝn).Theorem A (see [7]).
Fix1<q≤∞, a≥1, and ω∈RHs′, 1≤s<∞. Then, there exist C=C(q,n,a,ω,s) and K0=K0(n,a)≥1 with the following property: assume that F, G, H1, and H2 are nonnegative measurable functions on ℝn such that for any cube Q there exist nonnegative functions GQ and HQ with F(x)≤GQ(x)+HQ(x) for a.e. x∈Q and
(22)(⨏QHQq)1/q≤a(MF(x)+MH1(x)+H2(x-)),.∀x,x-∈Q,⨏QGQ≤G(x),∀x∈Q.
Then for all λ>0, K≥K0 and 0<γ<1(23)ω{MF>Kλ,G+H2≤γλ}≤C(aqKq+γK)1/sω{MF+MH1>λ}.
As a consequence, for all 0<p<1/s, one has
(24)∥MF∥Lp(ω)≤C(∥G∥Lp(ω)+∥MH1∥Lp(ω)+∥H2∥Lp(ω)),
provided ∥MF∥Lp(ω)<∞, and
(25)∥MF∥Lp,∞(ω)≤C(∥G∥Lp,∞(ω)+∥MH1∥Lp,∞(ω)+∥H2∥Lp,∞(ω)),
provided ∥MF∥Lp,∞(ω)<∞. Furthermore, if p≥1, then (24) and (25)  hold, provided F∈L1(whether or not MF∈Lp(ω)).For0<s<1 and 1≤γ<∞, we denote
(26)ℳs,γ(f)(x)=supB∋x(1|B|1-s∫B‍|f(y)|γdy)1/γ,
where the supremum is taken with respect to all balls B of positive measure containing the point x.Theorem B.
Let1<p<q<∞, 0<s<1, and let v and w be the weight functions. For a constant C>0 to exist so that the inequality
(27)(∫ℝn‍(ℳs,1(f)(x))qv(x)dx)1/q≤C(∫ℝn‍|f(x)|pw(x)dx)1/p
would hold, it is necessary and sufficient that the condition
(28)supx∈ℝn,r>0(w1-p′B(x,6r))1/p′×(∫ℝn∖B(x,r)‍v(y)|x-y|(s-1)qndy)1/q<∞,
where 1/p+1/p′=1, be fulfilled.For the proof of this theorem, see [12].Definition 7.
(
w
,
v
) is said to belong to A(p,q,s)(1<p<q<∞, 0<s<1) if (28) holds.Lemma 8.
Let1≤γ<p<q<∞, 0<s<1. If (w,v)∈A(p/γ,q/γ,s), then
(29)∥ℳs,γf∥Lq(v)≤C∥f∥Lp(w).Proof.
Sinceℳs,γ(f)(x)=(ℳs,1(|f|γ)(x))1/γ, we have
(30)∥ℳs,γf∥Lq(v)=∥(ℳs,1(|f|γ))1/γ∥Lq(v)∥ℳs,γf∥Lq(v)=∥ℳs,1(|f|γ)∥Lq/γ(v)1/γ.
By Theorem B, we have
(31)∥ℳs,1(|f|γ)∥Lq/γ(v)≤C∥|f|γ∥Lp/γ(w)∥ℳs,1(|f|γ)∥Lq/γ(v)=C∥f∥Lp(w)γ.
Thus,
(32)∥ℳs,γf∥Lq(v)≤C∥f∥Lp(w).


## 3. The Proof of the Main Theorems

In order to prove Theorem1, we need the following lemma.Lemma 9.
Let1≤p0≤s0, p0<p<q<∞, and w,v∈A∞. Let T be a sublinear operator bounded from Lp0 to Ls0.(i)
Ifb∈Λ˙β∩L∞ and f∈Lc∞, then [b,T]f∈Ls0.(ii)
Assume that for anyb∈Λ˙β∩L∞ and for any f∈Lc∞ one has that
(33)∥[b,T]f∥Lq(v)≤C∥b∥Λ˙β∥f∥Lp(w),where C does not depend on b and f. Then for all b∈Λ˙β, (33) holds.Proof.
The ideas of the following argument are taken from [7].
Fixf∈Lc∞. Note that (i) follows easily observing that
(34)[b,T]f(x)≤|b(x)||Tf(x)|+|T(bf)(x)|[b,T]f(x)≤∥b∥L∞|Tf(x)|+|T(bf)(x)|∈Ls0
since b∈L∞, f∈Lc∞ imply that f,bf∈Lc∞⊂Lp0 and hence, by assumption, T(f),T(bf)∈Ls0.
To obtain (ii), we fixb∈Λ˙β and f∈Lc∞. Let Q0 be a cube such that suppf⊂Q0. We may assume that bQ0=0 since otherwise we can work with b-=b-bQ0 and observe that
(35)[b,T]=[b-,T],∥b∥Λ˙β=∥b-∥Λ˙β.
Note that for m=0,1, we have that |bmf| and |T(bmf)| are finite almost everywhere since they belong to Lp0.
LetN>0 and define bN as follows:
(36)bN(x)={-N,b(x)<-N,b(x),-N≤b(x)≤N,N,b(x)>N.
Then, it is immediate to see that |bN(x)-bN(y)|≤|b(x)-b(y)| for all x,y. Thus, ∥bN∥Λ˙β≤∥b∥Λ˙β. As bN∈L∞, we can use (33) and
(37)∥[bN,T]f∥Lq(v)≤C∥bN∥Λ˙β∥f∥Lp(w)∥[bN,T]f∥Lq(v)≤C∥b∥Λ˙β∥f∥Lp(w)<∞.
To conclude, by Fatou’s lemma, it suffices to show that |[bNj,T]f(x)|→|[b,T]f(x)| for a.e. x∈ℝn and for some subsequence {Nj}j such that Nj→∞.
As|bN|≤|b|∈Lp(Q0), for any 1≤p<∞, the dominated convergence theorem yields that bNf→bf in Lp0 as N→∞. Therefore, T is bounded from Lp0 to Ls0. It follows that T(bNf-bf)→0 in Ls0. Thus, there exists a subsequence Nj→∞ such that T(bNjf-bf)→0 for a.e. x∈ℝn. In this way we obtain
(38)||[bNj,T]f(x)|-|[b,T]f(x)||≤|[bNj,T]f(x)-[b,T]f(x)|≤|T(bNjf-bf)(x)|+|bNj(x)-b(x)||Tf(x)|
as desired, and we get that |[bNj,T]f(x)|→|[b,T]f(x)| for a.e x∈ℝn.Proof of Theorem1.
We assume thatq0<∞, for q0=∞, and the main ideas are the same and details are left to the interested reader. Lemma 9 ensures that it suffices to consider the case b∈Λ˙β∩L∞. Let f∈Lc∞ and set F=|[b,T]f|s0. Note that F∈L1 by (i) of Lemma 9. Given a ball B, we set fB,b=(b4B-b)f and decompose [b,T]f as follows:
(39)|[b,T]f(x)|=|T((b(x)-b)f)(x)|≤|b(x)-b4B||Tf(x)|+|T((b4B-b)f)(x)|≤|b(x)-b4B||Tf(x)|+|T(I-Ar(B))fB,b(x)|+|TAr(B)fB,b(x)|.
We observe that F≤GB+HB, where
(40)GB=4s0-1(GB,1+GB,2)GB=4s0-1(|b-b4B|s0|Tf|s0+|T(I-Ar(B))fB,b|s0)
and HB=2s0-1|TAr(B)fB,b|s0.
We first estimate the average ofGB on B. Fix any x∈B. Let 1<s<∞. Using Lemma 4,
(41)(⨏BGB,1)1/s0=(1|B|∫B‍|b-b4B|s0|Tf|s0)1/s0(⨏BGB,1)1/s0≤(1|B|∫B‍|b-b4B|s0s′)1/(s0s′)(⨏BGB,1)1/s0=×(1|B|∫B‍|Tf|s0s)1/(s0s)(⨏BGB,1)1/s0=1|B|β/n(1|B|∫B‍|b-b4B|s0s′)1/(s0s′)(⨏BGB,1)1/s0=×(1|B|1-s0sβ/n∫B‍|Tf|s0s)1/(s0s)(⨏BGB,1)1/s0≤C∥b∥Λ˙βMβ,s0s(Tf)(x).
Using (9) and Lemmas 4 and 5,
(42)(⨏BGB,2)1/s0=(⨏B|T(I-Ar(B))fB,b|s0)1/s0(⨏BGB,2)1/s0≤C∑j=1∞‍αj|2j+1B|α/n(⨏2j+1B|fB,b|p0)1/p0(⨏BGB,2)1/s0≤C∑j=1∞‍αj|2j+1B|α/n(⨏BGB,2)1/s0=×(1|2j+1B|∫2j+1B‍|b-b2j+1B|p0|f|p0)1/p0(⨏BGB,2)1/s0=+C∑j=1∞‍αj|2j+1B|α/n(⨏BGB,2)1/s0=×(1|2j+1B||b2j+1B-b4B|p0∫2j+1B|f|p0‍)1/p0(⨏BGB,2)1/s0≤C∑j=1∞‍αj∥b∥Λ˙βMα+β,p0s(f)(x)(⨏BGB,2)1/s0=+C∑j=1∞‍αj∥b∥Λ˙β|2j+1B|(α+β)/n(⨏BGB,2)1/s0=×(1|2j+1B|∫2j+1B|f|p0s‍)1/(p0s)(⨏BGB,2)1/s0≤C∥b∥Λ˙βMα+β,p0s(f)(x)
since ∑j=1∞‍αj<∞. Hence, for any x∈B,
(43)⨏BGB≤C(∥b∥Λ˙βs0Mβ,s0s(Tf)s0(x)z+∥b∥Λ˙βs0Mα+β,p0s(f)s0(x))≡G(x).
We next estimate the average of HBq′ on B with q′=q0/s0. Using (10) and proceeding as before, we see that
(44)(⨏BHBq′)1/q0=2(s0-1)/s0(⨏B|TAr(B)fB,b|q0)1/q0≤C∑j=1∞‍αj(⨏2j+1B|TfB,b|s0)1/s0≤C∑j=1∞‍αj(⨏2j+1B|Tbf|s0)1/s0+C∑j=1∞‍αj(⨏2j+1B|b-b4B|s0|Tf|s0)1/s0≤C(MF)1/s0(x)+C∥b∥Λ˙βMβ,s0s(Tf)(x-),
for any x,x-∈B. Thus we have obtained
(45)(⨏BHBq′)1/q′≤C(MF(x)+∥b∥Λ˙βs0Mβ,s0s(Tf)s0(x-))(⨏BHBq′)1/q′≡C(MF(x)+H2(x-)).
Forp0<p<q<q0 and 1/q=1/p-(α+β)/n, we can find a 1<s<∞ such that s0s<1/(1/p-α/n) and p0s<p. As mentioned before F∈L1. Applying Theorem A and Remark 2 with q/s0 in place of p, we obtain
(46)∥[b,T]f∥qs0≤∥MF∥q/s0≤C(∥G∥q/s0+∥H2∥q/s0)≤C∥b∥Λ˙βs0(∥Mβ,s0s(Tf)∥qs0+∥Mα+β,p0s(f)∥qs0)≤C∥b∥Λ˙βs0(∥Tf∥1/(1/p-α/n)s0+∥f∥ps0)≤C∥b∥Λ˙βs0∥f∥ps0,
where we have used Lemma 6. This implies that
(47)∥[b,T]f∥q≤C∥b∥Λ˙β∥f∥p.Proof of Theorem3.
LetF, G, and H2 be the same as those used in the proof of Theorem 1. As mentioned before F∈L1. Since v∈Ap/p0∩RH(q0/p)′, applying Theorem A with p/p0 in place of p and s=q0/p, we obtain
(48)∥[b,T]f∥Lq(v)p0≤∥MF∥Lq/p0(v)≤C(∥G∥Lq/p0(v)+∥H2∥Lq/p0(v))≤C∥b∥Λ˙βp0(∥Mβ,p0s(Tf)∥Lq(v)p0+∥Mβ,p0s(f)∥Lq(v)p0)=C∥b∥Λ˙βp0(∥ℳβp0s/n,p0s(Tf)∥Lq(v)p0+∥ℳβp0s/n,p0s(f)∥Lq(v)p0).
Noting that (w,v)∈A(p/p0s,q/p0s,βp0s/n), Lemma 8 and Remark 2 give us that
(49)∥ℳβp0s/n,p0s(Tf)∥Lq(v)≤C∥Tf∥Lp(w)∥ℳ(βp0s/n),p0s(Tf)∥Lq(v)≤C∥f∥Lp(w).
This implies that
(50)∥[b,T]f∥Lq(v)≤C∥b∥Λ˙β∥f∥Lp(w).


## 4. Applications

### 4.1.  Spectral Multipliers: Off-Diagonal Estimates

Suppose thatL is a self-adjoint nonnegative definite operator on L2(ℝn). Let E(λ) be the spectral resolution of L. For any bounded Borel function m:[0,∞)→ℂ, by using the spectral theorem, we can define the operator
(51)m(L)=∫0∞‍m(λ)dE(λ).
This is of course bounded on L2(ℝn).The following will be assumed throughout this subsection.(H1)
L is a nonnegative self-adjoint operator on L2(ℝn).(H2)
The operatorL generates an analytic semigroup {e-tL}t>0 which satisfies the Davies-Gaffney condition. That is, there exist constants C,c>0 such that for any open subsets U1,U2⊂ℝn,
(52)|〈e-tLf1,f2〉|≤Cexp(-dist(U1,U2)2ct)z×∥f1∥L2(ℝn)∥f2∥L2(ℝn),∀t>0,for everyfi∈L2(ℝn) with suppfi⊂Ui, i=1,2, where dist(U1,U2):=infx∈U1,y∈U2d(x,y).(H3)
Suppose2<q0≤∞. Assume that the analytic semigroup e-tL generated by L satisfies “L2-Lq0 off-diagonal” estimates: there exist coefficients {aj}j≥0 satisfying ∑j=0∞aj<∞ such that for all balls B and for all functions f∈L2(ℝn)(53)(1|B|∫B‍|e-rB2Lf|q0dx)1/q0≤∑j=0∞‍aj(1|2jB|∫2jB‍|f|2dx)1/2.Letϕ be a nonnegative C0∞ function such that
(54)suppϕ⊂(14,1),∑l∈ℤ‍ϕ(2-lλ)=1,∀λ>0.
For s≥0, let [s] denote the integer part of s. Recall that Cs is the space of functions m on ℝ for which
(55)∥m∥Cs={∑k=0ssupλ∈ℝ|m(k)(λ)|ifs∈ℤ,∥m([s])∥Lip(s-[s])+∑k=0[s]supλ∈ℝ|m(k)(λ)|ifs∉ℤ
is finite.Then the following result holds.Theorem 10.
LetL satisfy assumptions (H1)–(H3). Let ϕ be a nonnegative C0∞ function satisfying (54), and suppose that the bounded measurable function m:[0,∞)→ℂ satisfies
(56)Cϕ,s=supt>0∥ϕ(·)m(t·)∥Cs+|m(0)|<∞
for some s>n/2. Then(i)
let0<β<1. If 2<p<1/(1/q0+β/n) and 1/q=1/p-β/n, then there is a constant C such that
(57)∥[b,m(L)]f∥Lq≤C∥b∥Λ˙β∥f∥Lp,for allf∈Lc∞ and for all b∈Λ˙β.(ii)
Let0<β<min{1,n/2}, 2<p<q<q0, and w,v∈Ap/p0∩RH(q0/p)′. If there exists a constant 1<s<min{n/β2,p/2} such that (w,v)∈A(p/2s,q/2s,β2s/n), then there is a constant C such that
(58)∥[b,m(L)]f∥Lq(v)≤C∥b∥Λ˙β∥f∥Lp(w),for allf∈Lc∞ and for all b∈Λ˙β.Proof.
Estimate (57) follows from Theorem 1 with α=0 and estimate (58) follows from Theorem 3, applied to Tf=m(L)f and Ar=I-(I-e-r2L)M with M∈ℕ and M>s/2. It suffices to show that there exist coefficients {aj}j≥0 satisfying ∑j=1∞aj<∞ such that (9) and (10) hold for all f∈Lc∞(ℝn).
Fix1≤k≤M. From (53), we deduce that
(59)(1|B|∫B‍|e-krB2Lf|q0dx)1/q0≤∑j=0∞‍Caj(1|2jB|∫2jB‍|f|2dx)1/2.
This estimate with m(L)f in place of f yields (10). Since, by functional calculus, m(L)e-kr2Lf=e-kr2Lm(L)f, (9) was proved in [13].


### 4.2. Riesz Transforms

LetA be an n×n matrix of complex and L∞-valued coefficients on ℝn. We assume that this matrix satisfies the following ellipticity (or “accretivity”) condition: there exist 0<λ≤Λ<∞ such that
(60)λ|ξ|2≤ReA(x)ξ·ξ-,|A(x)ξ·ζ-|≤Λ|ξ||ζ|,
for all ξ,ζ∈ℂn and almost every x∈ℝn. Associated with this matrix we define the second-order divergence form operator
(61)L=-div(A∇).The Riesz transforms associated toL are ∂jL-1/2, 1≤j≤n. Set ∇L-1/2=(∂1L-1/2,…,∂nL-1/2). The solution of the Kato conjecture [14] implies that this operator extends boundedly to L2. This allows the representation
(62)∇L-1/2f=1π∫0∞∇e-tLfdtt
in which the integral converges strongly in L2 both at 0 and ∞ when f∈L2.Defineϑ∈[0,π/2) by
(63)ϑ=sup{|arg〈Lf,f〉|:f∈𝒟(L)}.We write for0<θ<∞,  Σθ={z∈ℂ∖{0}:|argz|<θ}.We extract from [15] some definitions and results on unweighted off-diagonal estimates.Definition 11.
Let1≤p≤q≤∞. One says that a family {Tt}t>0 of sublinear operators satisfies Lp-Lq full off-diagonal estimates, in short Tt∈ℱ(Lp-Lq), if for some c>0, for all closed sets E and F, all f, and all t>0, we have
(64)(∫F‍|Tt(χEf)|qdx)1/q≤Ct-(1/2)(n/p-n/q)e-cd2(E,F)/2(∫E‍|f|pdx)1/p.IfI is a subinterval of [1,∞], Int I denotes the interior in ℝ of I∩ℝ.Proposition 12 (see [15]).
Fixm∈ℕ and 0<μ<π/2-ϑ.(a)
There exists a nonempty maximal interval in[1,∞], denoted by𝒥(L), such that ifp,q∈𝒥(L)withp≤q, then{(zL)me-zL}z∈ΣμsatisfiesLp-Lqfull off-diagonal estimates and is a bounded set inℒ(Lp).(b)
There exists a nonempty maximal interval in[1,∞], denoted by𝒦(L), such that ifp,q∈𝒦(L)withp≤q, then{z∇(zL)me-zL}z∈ΣμsatisfiesLp-Lqfull off-diagonal estimates and is a bounded set inℒ(Lp).(c)
𝒦(L)⊂𝒥(L)and, forp<2, we havep∈𝒦(L)if and only ifp∈𝒥(L).(d)
Denote byp-(L),p+(L)the lower and upper bounds of𝒥(L)and byq-(L),q+(L)those of𝒦(L). We havep-(L)=q-(L)and(q-(L))*≤p+(L). (We have setq*=(qn/(n-q)), the Sobolev exponent ofqwhenq<nandq*=∞, otherwise.)(e)
Ifn=1,𝒥(L)=𝒦(L)=[1,∞]. 											Ifn=2,𝒥(L)=[1,∞]and𝒦(L)⊃[1,q+(L))withq+(L)>2.(f)
Ifn≥3,p-(L)<2n/(n+2),p+(L)>2n/(n-2), andq+(L)>2.Then forq-<p0<q0<q+, T=∇L-1/2 satisfy (9) and (10) with α=0 and Ar=I-(I-e-r2L)M, where M is a large enough integer. For the proof of this argument, see [15]. So Theorem 1 with α=0 and Theorem 3 can be applied to T=∇L-1/2.


### 4.3. Fractional Operators

LetL=-div(A∇). The fractional power of an elliptic operator L on Rn is given formally by
(65)L-α/2=1Γ(α/2)∫0∞tα/2e-tLdtt,
with α>0. There exist p-=p-(L) and p+=p+(L), 1≤p-<2<p+≤∞ such that the semigroup {e-tL}t>0 is uniformly bounded on Lp for every p-<p<p+ (see Proposition 12). We have the following results.Lemma 13 (see [9]).
Letp-<p0<s0<q0<p+ so that 1/p0-1/s0=α/n. Fix a ball B with radius r. For f∈Lc∞ and M large enough, one has
(66)(⨏B|L-α/2(I-e-r2L)Mf|s0)1/s0≤C∑j=1∞‍g1(j)|2j+1B|α/n(⨏2j+1B|f|p0)1/p0,
and for 1≤l≤M(67)(⨏B|L-α/2e-lr2Lf|q0)1/q0≤C∑j=1∞‍g2(j)(⨏2j+1B|L-α/2f|s0)1/s0,
where gj=C2-j(2M-n/s0) and g2(j)=Ce-c4j.Theorem 14.
Letp-<p<q<p+, 0<α,β,α+β<1, and 1/q=1/p-(α+β)/n. Given b∈Λ˙β, one has
(68)∥[b,L-α/2]f∥q≤C∥b∥Λ˙β∥f∥p.Proof.
We are going to apply Theorem1 to the linear operator T=L-α/2. We fix p-<p<q<p+, α, and β so that 1/q=1/p-(α+β)/n. Then we can find p0, q0, s0 such that 1/p0-1/s0=α/n, p-<p0<s0<q0<p+, and p0<p<q<q0. Notice that as 1≤p-<p+≤∞, we have that 1<p0<s0<q0<∞. By Theorem 1.2 in [9], we know that T=L-α/2 is bounded from Lp0 to Ls0.
We takeAr=I-(I-e-r2L)m, where m≥1 is an integer to be chosen. We apply Lemma 13. Note that (66) is (9). Also, (10) follows from (67) after expanding Ar=I-(I-e-r2L)m. Then, we have that ∑j≤1‍gi(j) for i=1,2 by choosing 2m>n/s0. Consequently applying Theorem 1, we conclude that ∥[b,L-α/2]f∥q≤C∥b∥Λ˙β∥f∥p.


## 4.1.  Spectral Multipliers: Off-Diagonal Estimates

Suppose thatL is a self-adjoint nonnegative definite operator on L2(ℝn). Let E(λ) be the spectral resolution of L. For any bounded Borel function m:[0,∞)→ℂ, by using the spectral theorem, we can define the operator
(51)m(L)=∫0∞‍m(λ)dE(λ).
This is of course bounded on L2(ℝn).The following will be assumed throughout this subsection.(H1)
L is a nonnegative self-adjoint operator on L2(ℝn).(H2)
The operatorL generates an analytic semigroup {e-tL}t>0 which satisfies the Davies-Gaffney condition. That is, there exist constants C,c>0 such that for any open subsets U1,U2⊂ℝn,
(52)|〈e-tLf1,f2〉|≤Cexp(-dist(U1,U2)2ct)z×∥f1∥L2(ℝn)∥f2∥L2(ℝn),∀t>0,for everyfi∈L2(ℝn) with suppfi⊂Ui, i=1,2, where dist(U1,U2):=infx∈U1,y∈U2d(x,y).(H3)
Suppose2<q0≤∞. Assume that the analytic semigroup e-tL generated by L satisfies “L2-Lq0 off-diagonal” estimates: there exist coefficients {aj}j≥0 satisfying ∑j=0∞aj<∞ such that for all balls B and for all functions f∈L2(ℝn)(53)(1|B|∫B‍|e-rB2Lf|q0dx)1/q0≤∑j=0∞‍aj(1|2jB|∫2jB‍|f|2dx)1/2.Letϕ be a nonnegative C0∞ function such that
(54)suppϕ⊂(14,1),∑l∈ℤ‍ϕ(2-lλ)=1,∀λ>0.
For s≥0, let [s] denote the integer part of s. Recall that Cs is the space of functions m on ℝ for which
(55)∥m∥Cs={∑k=0ssupλ∈ℝ|m(k)(λ)|ifs∈ℤ,∥m([s])∥Lip(s-[s])+∑k=0[s]supλ∈ℝ|m(k)(λ)|ifs∉ℤ
is finite.Then the following result holds.Theorem 10.
LetL satisfy assumptions (H1)–(H3). Let ϕ be a nonnegative C0∞ function satisfying (54), and suppose that the bounded measurable function m:[0,∞)→ℂ satisfies
(56)Cϕ,s=supt>0∥ϕ(·)m(t·)∥Cs+|m(0)|<∞
for some s>n/2. Then(i)
let0<β<1. If 2<p<1/(1/q0+β/n) and 1/q=1/p-β/n, then there is a constant C such that
(57)∥[b,m(L)]f∥Lq≤C∥b∥Λ˙β∥f∥Lp,for allf∈Lc∞ and for all b∈Λ˙β.(ii)
Let0<β<min{1,n/2}, 2<p<q<q0, and w,v∈Ap/p0∩RH(q0/p)′. If there exists a constant 1<s<min{n/β2,p/2} such that (w,v)∈A(p/2s,q/2s,β2s/n), then there is a constant C such that
(58)∥[b,m(L)]f∥Lq(v)≤C∥b∥Λ˙β∥f∥Lp(w),for allf∈Lc∞ and for all b∈Λ˙β.Proof.
Estimate (57) follows from Theorem 1 with α=0 and estimate (58) follows from Theorem 3, applied to Tf=m(L)f and Ar=I-(I-e-r2L)M with M∈ℕ and M>s/2. It suffices to show that there exist coefficients {aj}j≥0 satisfying ∑j=1∞aj<∞ such that (9) and (10) hold for all f∈Lc∞(ℝn).
Fix1≤k≤M. From (53), we deduce that
(59)(1|B|∫B‍|e-krB2Lf|q0dx)1/q0≤∑j=0∞‍Caj(1|2jB|∫2jB‍|f|2dx)1/2.
This estimate with m(L)f in place of f yields (10). Since, by functional calculus, m(L)e-kr2Lf=e-kr2Lm(L)f, (9) was proved in [13].


## 4.2. Riesz Transforms

LetA be an n×n matrix of complex and L∞-valued coefficients on ℝn. We assume that this matrix satisfies the following ellipticity (or “accretivity”) condition: there exist 0<λ≤Λ<∞ such that
(60)λ|ξ|2≤ReA(x)ξ·ξ-,|A(x)ξ·ζ-|≤Λ|ξ||ζ|,
for all ξ,ζ∈ℂn and almost every x∈ℝn. Associated with this matrix we define the second-order divergence form operator
(61)L=-div(A∇).The Riesz transforms associated toL are ∂jL-1/2, 1≤j≤n. Set ∇L-1/2=(∂1L-1/2,…,∂nL-1/2). The solution of the Kato conjecture [14] implies that this operator extends boundedly to L2. This allows the representation
(62)∇L-1/2f=1π∫0∞∇e-tLfdtt
in which the integral converges strongly in L2 both at 0 and ∞ when f∈L2.Defineϑ∈[0,π/2) by
(63)ϑ=sup{|arg〈Lf,f〉|:f∈𝒟(L)}.We write for0<θ<∞,  Σθ={z∈ℂ∖{0}:|argz|<θ}.We extract from [15] some definitions and results on unweighted off-diagonal estimates.Definition 11.
Let1≤p≤q≤∞. One says that a family {Tt}t>0 of sublinear operators satisfies Lp-Lq full off-diagonal estimates, in short Tt∈ℱ(Lp-Lq), if for some c>0, for all closed sets E and F, all f, and all t>0, we have
(64)(∫F‍|Tt(χEf)|qdx)1/q≤Ct-(1/2)(n/p-n/q)e-cd2(E,F)/2(∫E‍|f|pdx)1/p.IfI is a subinterval of [1,∞], Int I denotes the interior in ℝ of I∩ℝ.Proposition 12 (see [15]).
Fixm∈ℕ and 0<μ<π/2-ϑ.(a)
There exists a nonempty maximal interval in[1,∞], denoted by𝒥(L), such that ifp,q∈𝒥(L)withp≤q, then{(zL)me-zL}z∈ΣμsatisfiesLp-Lqfull off-diagonal estimates and is a bounded set inℒ(Lp).(b)
There exists a nonempty maximal interval in[1,∞], denoted by𝒦(L), such that ifp,q∈𝒦(L)withp≤q, then{z∇(zL)me-zL}z∈ΣμsatisfiesLp-Lqfull off-diagonal estimates and is a bounded set inℒ(Lp).(c)
𝒦(L)⊂𝒥(L)and, forp<2, we havep∈𝒦(L)if and only ifp∈𝒥(L).(d)
Denote byp-(L),p+(L)the lower and upper bounds of𝒥(L)and byq-(L),q+(L)those of𝒦(L). We havep-(L)=q-(L)and(q-(L))*≤p+(L). (We have setq*=(qn/(n-q)), the Sobolev exponent ofqwhenq<nandq*=∞, otherwise.)(e)
Ifn=1,𝒥(L)=𝒦(L)=[1,∞]. 											Ifn=2,𝒥(L)=[1,∞]and𝒦(L)⊃[1,q+(L))withq+(L)>2.(f)
Ifn≥3,p-(L)<2n/(n+2),p+(L)>2n/(n-2), andq+(L)>2.Then forq-<p0<q0<q+, T=∇L-1/2 satisfy (9) and (10) with α=0 and Ar=I-(I-e-r2L)M, where M is a large enough integer. For the proof of this argument, see [15]. So Theorem 1 with α=0 and Theorem 3 can be applied to T=∇L-1/2.


## 4.3. Fractional Operators

LetL=-div(A∇). The fractional power of an elliptic operator L on Rn is given formally by
(65)L-α/2=1Γ(α/2)∫0∞tα/2e-tLdtt,
with α>0. There exist p-=p-(L) and p+=p+(L), 1≤p-<2<p+≤∞ such that the semigroup {e-tL}t>0 is uniformly bounded on Lp for every p-<p<p+ (see Proposition 12). We have the following results.Lemma 13 (see [9]).
Letp-<p0<s0<q0<p+ so that 1/p0-1/s0=α/n. Fix a ball B with radius r. For f∈Lc∞ and M large enough, one has
(66)(⨏B|L-α/2(I-e-r2L)Mf|s0)1/s0≤C∑j=1∞‍g1(j)|2j+1B|α/n(⨏2j+1B|f|p0)1/p0,
and for 1≤l≤M(67)(⨏B|L-α/2e-lr2Lf|q0)1/q0≤C∑j=1∞‍g2(j)(⨏2j+1B|L-α/2f|s0)1/s0,
where gj=C2-j(2M-n/s0) and g2(j)=Ce-c4j.Theorem 14.
Letp-<p<q<p+, 0<α,β,α+β<1, and 1/q=1/p-(α+β)/n. Given b∈Λ˙β, one has
(68)∥[b,L-α/2]f∥q≤C∥b∥Λ˙β∥f∥p.Proof.
We are going to apply Theorem1 to the linear operator T=L-α/2. We fix p-<p<q<p+, α, and β so that 1/q=1/p-(α+β)/n. Then we can find p0, q0, s0 such that 1/p0-1/s0=α/n, p-<p0<s0<q0<p+, and p0<p<q<q0. Notice that as 1≤p-<p+≤∞, we have that 1<p0<s0<q0<∞. By Theorem 1.2 in [9], we know that T=L-α/2 is bounded from Lp0 to Ls0.
We takeAr=I-(I-e-r2L)m, where m≥1 is an integer to be chosen. We apply Lemma 13. Note that (66) is (9). Also, (10) follows from (67) after expanding Ar=I-(I-e-r2L)m. Then, we have that ∑j≤1‍gi(j) for i=1,2 by choosing 2m>n/s0. Consequently applying Theorem 1, we conclude that ∥[b,L-α/2]f∥q≤C∥b∥Λ˙β∥f∥p.



---

*Source: 178961-2013-06-22.xml*