c 2009 by Institut Mittag-Leffler. All rights reserved
On the L p -boundedness of pseudo-differential operators with non-regular symbols
Naohito Tomita
Abstract. In this paper, we consider the continuity property of pseudo-differential oper- ators with symbols whose Fourier transforms have compact support. As applications, we obtain theLp-boundedness for symbols in Besov spaces and in modulation spaces.
1. Introduction
The continuity property of pseudo-differential operators has been well inves- tigated. Among many works, Calder´on–Vaillancourt [3] first treated the bounded- ness for the classS0,00 , which means that the boundedness of all the derivatives of a symbol assures theL2-boundedness of the corresponding operator. It should be mentioned that the boundedness of all the derivatives of a symbol is not necessary in their proof.
Authors such as Coifman–Meyer [4], Cordes [5], Miyachi [11] and Muramatu [12], improved the result of [3] along this line, that is, investigated the minimal assumption on the regularity of a symbol. In fact, they proved that the boundedness of the derivatives of a symbol up to a certain order, which exceedsn/2, assures the boundedness onL2(Rn) of the corresponding operator. In particular, Sugimoto [16]
showed that symbols σ(x, ξ) in the Besov spaceB∞(n/2,n/2),1 (Rn×Rn) imply the L2- boundedness. This assumption means thatσ(x, ξ) belongs toBn/2∞,1(Rn) with respect to bothxandξ. [16, Theorem 2.1.2] also stated that the classB(n/2,n/2)∞,1 (Rn×Rn) can be replaced by a wider oneB(1/2,...,1/2)∞,1 (R×...×R), which means thatσ(x, ξ)=
σ(x1, ..., xn, ξ1, ..., ξn) belongs to B1/2∞,1(R) with respect to xj and ξk for all j, k=
1, ..., n. This is equivalent to the result of Boulkhemair [1, Corollary 6].
Theorem 1.1. There exists a constant C >0 such that σ(X, D)fL2≤C(R1...R2n)1/2σ(x, ξ)L∞x,ξfL2
for allσ(x, ξ) with supp ˆσ⊂2n
j=1[−Rj, Rj], Rj≥1,j=1, ...,2n,and all f∈L2(Rn).
Here σ=ˆ F1,2σ andC >0 is independent of Rj≥1, j=1, ...,2n.
The objective of this paper is to induce a similar estimate for theLp-bounded- ness, p=2. In this direction, Miyachi [11] proved that symbols σ(x, ξ) in the weighted Besov space B∞(s,∞
1,s2),m(p)(Rn×Rn), s1>min{n/p, n/2}, s2>max{n/p, n/2}, m(p)=n|1/p−1/2| imply the Lp-boundedness, 1<p<∞. This assumption means that fractional derivatives ofσ(x, ξ) inx(resp.ξ) higher than min{n/p, n/2} (resp. max{n/p, n/2}) are bounded byC ξ−m(p). Sugimoto [17] treated the critical cases1=n/2 ands2=n/pas in the argument for theL2-boundedness, and showed that symbolsσ(x, ξ) in the weighted Besov spaceB(n/2,n/p),m(p)∞,1 (Rn×Rn) imply the Lp-boundedness for 1<p≤2. The following main theorem of this paper extends this result to the casep>2. Furthermore, in the casep<2, it replaces the symbol class B(n/2,n/p),m(p)∞,1 (Rn×Rn) by the wider oneB∞(1/2,...,1/2,1/p,...,1/p),m(p),1 (R×...×R) as in the case ofL2-boundedness.
Theorem 1.2. Let 0<p<∞. Then there exists a constant C >0 such that σ(X, D)fLp≤C(R1...Rn)min{1/p,1/2}(Rn+1...R2n)max{1/p,1/2}
×σ(x, ξ) ξn|1/p−1/2|L∞
x,ξfHp
for allσ(x, ξ)with supp ˆσ⊂2n
j=1[−Rj, Rj], Rj≥1,j=1, ...,2n,and allf∈Hp(Rn).
Here Hp(Rn)is the Hardy space and C >0 is independent of Rj≥1, j=1, ...,2n.
The statement for symbols in the class B(1/2,...,1/2,1/p,...,1/p),m(p)∞,1 with p<2 (resp. B∞(1/p,...,1/p,1/2,...,1/2),m(p),1 withp>2) can be obtained from Theorem1.2as a corollary (see Corollary5.2(2)). We remark that symbols in the weighted modula- tion spaceMm(p)∞,1 also induce theLp-boundedness. Such a result is just a corollary of Theorem 1.2 again (see Corollary 5.2(3)). The L2-boundedness with symbols in the modulation space M0∞,1 was studied by Sj¨ostrand [13], Boulkhemair [1], Gr¨ochenig–Heil [8] and Toft [18].
Finally, we explain the organization of this paper. Section 2 is for the pre- liminaries, including the notation and definitions of function spaces. Sections 3
and 4 are devoted to the proofs of Theorem 1.2 with p≤2 and p≥2, respectively.
In Section5, we state several results induced from Theorem 1.2.
2. Preliminaries
LetS(Rn) andS(Rn) be the Schwartz spaces of all rapidly decreasing smooth functions and tempered distributions, respectively. We define theFourier transform Ff and theinverse Fourier transform F−1f off∈S(Rn) by
Ff(ξ) = ˆf(ξ) =
Rn
e−ix·ξf(x)dx and F−1f(x) = 1 (2π)n
Rn
eix·ξf(ξ)dξ.
Let σ(x, ξ)∈S(R2n), where x, ξ∈Rn. We denote by F1σ(y, ξ) and F2σ(x, η) the partial Fourier transforms ofσwith respect to the first variable and the second vari- able, respectively. That is, F1σ(y, ξ)=F[σ(·, ξ)](y) and F2σ(x, η)=F[σ(x,·)](η).
We also denote byF1−1σandF2−1σthe partial inverse Fourier transforms ofσwith respect to the first variable and the second variable, respectively. We writeF1,2= F1F2andF1,2−1=F1−1F2−1, and note thatF1,2andF1,2−1 are the usual Fourier trans- form and inverse Fourier transform of distributions on Rn×Rn. For σ∈S(R2n), the pseudo-differential operatorσ(X, D) is defined by
σ(X, D)f(x) = 1 (2π)n
Rn
eix·ξσ(x, ξ) ˆf(ξ)dξ forf∈ S(Rn).
We use the notation ξ=(1+|ξ|2)1/2, whereξ∈Rn. For 1≤p≤∞,pis the conjugate exponent ofp(that is, 1/p+1/p=1).
We introduce Hardy spaces based on Stein [14, Chapter 3]. Let 0<p<∞, and let Φ∈S(Rn) be such that
RnΦ(x)dx=0. Then theHardy space Hp(Rn) consists of allf∈S(Rn) such that
fHp= sup
0<t<∞|Φt∗f|
Lp<∞,
where Φt(x)=t−nΦ(x/t). It is known thatHp(Rn)=Lp(Rn) if 1<p<∞([14, Chap- ter 3, Section 1.2.1]), and the definition ofHp(Rn) is independent of the choice of Φ∈S(Rn) with
RnΦ(x)dx=0 ([14, Chapter 3, Theorem 1]). Let 0<p≤1. We say thatais anHp-atom ifasatisfies
(2.1) suppa⊂B, aL∞≤ |B|−1/p and
Rn
xβa(x)dx= 0
for all|β|≤[n/p−n], whereB is an open ball and [n/p−n] stands for the largest integer ≤n/p−n. It is also known that everyf∈Hp(Rn) can be written as a sum ofHp-atoms:
f= ∞ j=1
λjaj inS(Rn),
where {aj}j≥1 is a collection of Hp-atoms and {λj}j≥1 is a sequence of complex numbers with∞
j=1|λj|p<∞. Moreover, 1
Cinf ∞
j=1
|λj|p
1/p
≤ fHp≤Cinf ∞
j=1
|λj|p
1/p
,
where the infimum is taken over all representations of f ([14, Chapter 3, Theo- rem 2]). By virtue of atomic decomposition of Hp, if an L2-bounded operator T satisfiesT aLp≤Cfor allHp-atomsa, thenT is bounded fromHp(Rn) toLp(Rn), where 0<p≤1.
We recall the definitions of Besov and modulation spaces. Assume that 0<q≤∞andρ∈R. Letψ0, ψ∈S(RN) be such that
suppψ0⊂ {ξ∈RN:|ξ| ≤2}, suppψ⊂ {ξ∈RN: 2−1≤ |ξ| ≤2}, (2.2)
ψ0(ξ)+
∞ j=1
ψ(2−jξ) = 1 for allξ∈RN, and setψj(ξ)=ψ(2−jξ) if j≥1. Theweighted Besov space B(s∞,q
1,s2),ρ(Rn×Rn) con- sists of all σ∈S(R2n) such that
σB∞,q
(s1,s2 )
= ∞
j=0
∞ k=0
(2js12ks2ψj(Dx)ψk(Dξ)σ(x, ξ) ξρL∞x,ξ)q
1/q
<∞,
wheres1, s2∈R,{ψj}j≥0,{ψk}k≥0 are as in (2.2) withN=n, ψj(Dx)ψk(Dξ)σ(x, ξ) =F(y,η)−1 →(x,ξ)[ψj(y)ψk(η)ˆσ(y, η)]
andx, y, ξ, η∈Rn. Similarly, B(s∞,q
1,...,s2n),ρ(R×...×R) is defined by the norm σB∞,q
(s1,...,s2n),ρ=
∞
j1,...,jn=0
∞ k1,...,kn=0
(2j1s1+...+jnsn2k1sn+1+...+kns2n
×ψj1(Dx1)...ψjn(Dxn)ψk1(Dξ1)...ψkn(Dξn)σ(x, ξ) ξρL∞x,ξ)q
1/q
,
wheres1, ..., s2n∈R,{ψj1}j1≥0, ...,{ψkn}kn≥0 are as in (2.2) withN=1, ψj1(Dx1)...ψjn(Dxn)ψk1(Dξ1)...ψkn(Dξn)σ(x, ξ)
=F(y,η)−1 →(x,ξ)[ψj1(y1)...ψjn(yn)ψk1(η1)...ψkn(ηn)ˆσ(y, η)], y=(y1, ..., yn)∈Rn andη=(η1, ..., ηn)∈Rn. Letϕ∈S(Rn) be such that
(2.3) suppϕ⊂[−1,1]n and
k∈Zn
ϕ(ξ−k) = 1 for allξ∈Rn.
Then theweighted modulation spaceMρ∞,q(Rn×Rn) consists of allσ∈S(R2n) such that
σMρ∞,q=
k∈Zn
l∈Zn
ϕ(Dx−k)ϕ(Dξ−l)σ(x, ξ)ξρqL∞ x,ξ
1/q
<∞,
where
ϕ(Dx−k)ϕ(Dξ−l)σ(x, ξ) =F(y,η)−1 →(x,ξ)[ϕ(y−k)ϕ(η−l)ˆσ(y, η)].
3. Proof of Theorem1.2 withp≤2
In this section we prove Theorem1.2 withp≤2.
Lemma 3.1. Let s∈R. Then there exists a constant C >0 such that σ(X, D)DsfL2≤C(R1...R2n)1/2σ(x, ξ)ξsL∞
x,ξfL2
for allσ(x, ξ)with supp ˆσ⊂2n
k=1[−Rk, Rk], Rk≥1,k=1, ...,2n,and allf∈L2(Rn).
Here C >0 is independent ofRk≥1,k=1, ...,2n.
Proof. Letωs(ξ)= ξs, and let{ψj}j≥0 be as in (2.2) withN=n. Then (3.1) σ(x, ξ)ξs=
∞ j=0
σ(x, ξ)ψj(D)ωs(ξ) = ∞ j=0
σj(x, ξ).
Since supp ˆσ⊂2n
k=1[−Rk, Rk] and suppψj(D)ωs⊂{ξ∈R:|ξ|≤2j+1}, we have suppσj⊂
n k=1
[−Rk, Rk] × 2n
k=n+1
[−(Rk+2j+1), Rk+2j+1] for allj≥0.
Hence, by Theorem1.1, σj(X, D)fL2
(3.2)
≤C(R1...Rn)1/2((Rn+1+2j)...(R2n+2j))1/2σj(x, ξ)L∞x,ξfL2
for allf∈L2(Rn) andj≥0. If j=0 then
|σ0(x, ξ)|= σ(x, ξ)
Rn
Ψ0(η) ξ−ηsdη
≤C|σ(x, ξ)|
Rn|Ψ0(η)| ξs η|s|dη=C|σ(x, ξ)| ξs, where Ψ0=F−1ψ0, and consequently
(3.3) σ0(x, ξ)L∞x,ξ≤Cσ(x, ξ) ξsL∞x,ξ. We consider the case j≥1, and note that
RnηβΨ(η)dη=i|β|∂βψ(0)=0 for allβ, where Ψ=F−1ψ. LetN=[n/2]. Since
ψj(D)ωs(ξ) =
Rn
2jnΨ(2j(ξ−η))
ωs(η)−
|β|≤N
(η−ξ)β
β! ∂βωs(ξ) dη
=
Rn
2jnΨ(2j(ξ−η))
×
(N+1)
|β|=N+1
(η−ξ)β β!
1 0
(1−t)N∂βωs(ξ+t(η−ξ))dt dη
andωs∈Ss1,0 (that is,|∂βωs(ξ)|≤Cβ ξs−|β|), we see that
|ψj(D)ωs(ξ)| ≤C
Rn|Ψ(η)| |2−jη|N+1 1
0
ξ−2−jtηs−(N+1)dt dη
≤C
Rn|Ψ(η)| |2−jη|N+1 1
0
ξs−(N+1) 2−jtη|s−(N+1)|dt dη
≤C2−j(N+1) ξs−(N+1)
Rn|Ψ(η)| |η|N+1 η|s−(N+1)|dη for allj≥1. This gives
(3.4) σj(x, ξ)L∞x,ξ=σ(x, ξ)ψj(D)ωs(ξ)L∞x,ξ≤C2−j(N+1)σ(x, ξ) ξsL∞x,ξ
for allj≥1. Note that ifa, b≥1 thena+b≤2ab. Therefore, since 2j≥1,j≥0,Rk≥1, 1≤k≤2n, andN+1>n/2, we have by (3.1)–(3.4),
σ(X, D) DsfL2≤ ∞ j=0
σj(X, D)fL2
≤C ∞ j=0
(R1...Rn)1/2((Rn+1+2j)...(R2n+2j))1/2
×σj(x, ξ)L∞x,ξfL2
≤C ∞ j=0
(R1...Rn)1/2((2jRn+1)...(2jR2n))1/2
×σj(x, ξ)L∞x,ξfL2
≤C(R1...R2n)1/2 ∞ j=0
2−j((N+1)−n/2)σ(x, ξ) ξsL∞x,ξfL2
=C(R1...R2n)1/2σ(x, ξ)ξsL∞x,ξfL2. The proof is complete.
We are now ready to prove Theorem1.2withp≤2.
Proof of Theorem 1.2 withp≤2. Letσ(x, ξ) be such that supp ˆσ⊂
2n j=1
[−Rj, Rj], Rj≥1, j= 1, ...,2n.
We first consider the case 0<p<1. By virtue of atomic decomposition of Hp, it is enough to prove that
(3.5) σ(X, D)fLp≤C(R1...Rn)1/2(Rn+1...R2n)1/pσ(x, ξ)ξn(1/p−1/2)L∞
x,ξ
for allf∈Hp(Rn) satisfying
(3.6) suppf⊂ {x∈Rn:|x| ≤r}, fL∞≤r−n/p and
Rn
xβf(x)dx= 0 for all |β|≤[n/p−n], where C >0 is independent of r. In fact, since · Lp and · Hp are translation invariant,σ(X, D)(f(· −x0))(x)=σ(X+x0, D)f(x−x0) and suppF(x,ξ)→(y,η)[σ(x+x0, ξ)]⊂2n
j=1[−Rj, Rj], if (3.5) holds for allf satisfying (3.6)
then (3.5) holds for allf satisfying (2.1) (see Remark 5.1). Note that iff satisfies (3.6) thenfL2≤Cr−n(1/p−1/2).
Letf be as in (3.6) withr>1, and splitσ(X, D)fpLp as σ(X, D)pLp
=
n
j=1[−2rRn+j,2rRn+j]
+
cn
j=1[−2rRn+j,2rRn+j] |σ(X, D)f(x)|pdx.
Since suppF2−1σ(x,·)⊂n
j=1[−Rn+j, Rn+j] for all x∈Rn, suppf⊂{x∈Rn:|x|≤r} and
(3.7) σ(X, D)f(x) =
Rn
1 (2π)n
Rn
ei(x−y)·ξσ(x, ξ)dξ f(y)dy
=
RnF2−1σ(x, x−y)f(y)dy, we see that
suppσ(X, D)f⊂ n j=1
[−Rn+j, Rn+j]+{x∈Rn:|x| ≤r} ⊂ n j=1
[−(Rn+j+r), Rn+j+r].
By H¨older’s inequality and Theorem1.1,
n
j=1[−2rRn+j,2rRn+j]
|σ(X, D)f(x)|pdx
1/p
(3.8)
≤ n
j=1
4rRn+j
1/p−1/2
σ(X, D)fL2
≤Crn(1/p−1/2)(Rn+1...R2n)1/p−1/2(R1...R2n)1/2σ(x, ξ)L∞x,ξfL2
≤C(R1...Rn)1/2(Rn+1...R2n)1/pσ(x, ξ) ξn(1/p−1/2)L∞
x,ξ, where we have used the fact that 1/p−1/2>0. On the other hand, (3.9)
cn
j=1[−2rRn+j,2rRn+j]|σ(X, D)f(x)|pdx= 0.
In fact, sincer>1 andRn+j≥1, 1≤j≤n, suppσ(X, D)f⊂
n j=1
[−(Rn+j+r), Rn+j+r]⊂ n j=1
[−2rRn+j,2rRn+j].
Hence, (3.8) and (3.9) give (3.5) forf satisfying (3.6) withr>1.
Letf be as in (3.6) withr≤1. In this case, our proof is similar to that of [17], but we give it for the reader’s convenience. Since
suppσ(X, D)f⊂ n j=1
[−(Rn+j+r), Rn+j+r]
⊂ n j=1
[−(Rn+j+1), Rn+j+1]⊂ n j=1
[−2Rn+j,2Rn+j], we have by H¨older’s inequality
(3.10) σ(X, D)fLp≤C(Rn+1...R2n)1/p−1/2σ(X, D)fL2.
Letφ∈S(Rn) be such thatφ=1 on{ξ∈Rn:|ξ|≤1}and suppφ⊂{ξ∈Rn:|ξ|≤2}, and splitσ(x, ξ) as
σ(x, ξ) =σ(x, ξ)φ(rξ)+σ(x, ξ)(1−φ(rξ)) =σ(1)(x, ξ)+σ(2)(x, ξ).
SetK(l)(x, y)=F2−1σ(l)(x, y), l=1,2, andN=[n(1/p−1)]. By (3.6) and (3.7), σ(1)(X, D)f(x) =
|y|≤r
K(1)(x, x−y)−
|β|≤N
(−y)β
β! ∂2βK(1)(x, x) f(y)dy
=
|y|≤r
(N+1)
|β|=N+1
(−y)β β!
× 1
0
(1−t)N∂2βK(1)(x, x−ty)dt f(y)dy and
σ(2)(X, D)f(x) =
|y|≤r
K(2)(x, x−y)f(y)dy,
where∂2βK(1)(x, y)=∂βyK(1)(x, y). Then, by Minkowski’s inequality for integrals,
(3.11)
σ(1)(X, D)fL2≤CrN+1−n(1/p−1) sup
0≤t≤1
|y|≤r
|β|=N+1
∂β2K(1)(x, x−ty)L2x,
σ(2)(X, D)fL2≤Cr−n(1/p−1) sup
|y|≤r
K(2)(x, x−y)L2x.
SettingFg(1)t,y,β,r(ξ)=φ(rξ)(iξ)βe−ity·ξ ξ−n(1/p−1/2), we obtain
∂2βK(1)(x, x−ty) = 1 (2π)n
Rn
ei(x−ty)·ξσ(x, ξ)φ(rξ)(iξ)βdξ
= 1
(2π)n
Rn
eix·ξσ(x, ξ)ξn(1/p−1/2)
×φ(rξ)(iξ)βe−ity·ξ ξ−n(1/p−1/2)dξ
=σ(X, D)Dn(1/p−1/2)gt,y,β,r(1) (x).
Similarly,
K(2)(x, x−y) =σ(X, D) Dn(1/p−1/2)g(2)y,r(x), whereFg(2)y,r(ξ)=(1−φ(rξ))e−iy·ξ ξ−n(1/p−1/2). Then, by Lemma3.1,
∂β2K(1)(x, x−ty)L2x=σ(X, D)Dn(1/p−1/2)g(1)t,y,β,rL2
(3.12)
≤C(R1...R2n)1/2σ(x, ξ)ξn(1/p−1/2)L∞x,ξg(1)t,y,β,rL2
and
K(2)(x, x−y)L2x=σ(X, D) Dn(1/p−1/2)g(2)y,rL2
(3.13)
≤C(R1...R2n)1/2σ(x, ξ)ξn(1/p−1/2)L∞x,ξg(2)y,rL2. Since N+1−n(1/p−1/2)>−n/2 and −n(1/p−1/2)<−n/2, by Plancherel’s theo- rem,
sup
0≤t≤1
|y|≤r
|β|=N+1
g(1)t,y,β,rL2= sup
0≤t≤1
|y|≤r
|β|=N+1
(2π)−n/2Fgt,y,β,r(1) L2
(3.14)
≤(2π)−n/2φ(rξ)|ξ|N+1−n(1/p−1/2)L2ξ
=Cr−(N+1−n(1/p−1)) and
(3.15) sup
|y|≤r
gy,r(2)L2x≤(2π)−n/2(1−φ(rξ))|ξ|−n(1/p−1/2)L2ξ=Crn(1/p−1). Combining (3.10)–(3.15), we have (3.5) forf satisfying (3.6) with r≤1. Thus, we obtain Theorem1.2with 0<p<1.
We next consider the case 1≤p<2 (since we already have the case p=2 as Theorem1.1). We use the interpolation theorem for analytic families of operators (see Stein–Weiss [15] and Calder´on–Torchinsky [2]). Since our proof is very similar to that of Sugimoto [17, Section 5] (or Miyachi [11, p. 150]), we shall only indicate the necessary modifications.
Let 0<p0<1 and S={z∈C:0≤Rez≤1}. For 1≤p<2, we take 0<θ<1 such that 1/p=(1−θ)/p0+θ/2. Set
σz(x, ξ) =e(z−θ)2σ(x, ξ) ξn(1/p0−1/2)(z−θ) forz∈S, and note that
(3.16) σθ(X, D) =σ(X, D).
Letωn(1/p0−1/2)(ξ)= ξn(1/p0−1/2), and let{ψj}j≥0be as in (2.2) withN=n. Then σz(x, ξ) =
∞ j=0
e(z−θ)2σ(x, ξ)ψj(D)ωzn(1/p−θ
0−1/2)(ξ) = ∞ j=0
σz,j(x, ξ).
In the same way as in the proof of Lemma3.1, we can prove that
|ψj(D)ωn(1/pit−θ
0−1/2)(ξ)| ≤Cmax{1, t}N+12−j(N+1) ξ−nθ(1/p0−1/2) for allt∈Rand j≥0, whereN=[n/p0]. Hence, since
supp ˆσit,j⊂ n
k=1
[−Rk, Rk] × 2n
k=n+1
[−(Rk+2j+1), Rk+2j+1]
for all t∈Rand j≥0 (see the proof of Lemma 3.1), we have by Theorem1.2 with 0<p0<1,
σit(X, D)fpL0p0≤∞
j=0
σit,j(X, D)fpL0p0
≤C ∞ j=0
(R1...Rn)p0/2((Rn+1+2j)...(R2n+2j))p0/p0
×σit,j(x, ξ)ξn(1/p0−1/2)pL0∞
x,ξfpH0p0
≤C(R1...Rn)p0/2(Rn+1...R2n)p0/p0
× ∞ j=0
(e−t2+θ2max{1, t}N+12−j(N+1−n/p0))p0
×σ(x, ξ)ξ−nθ(1/p0−1/2) ξn(1/p0−1/2)pL0∞
x,ξfpH0p0
≤C(R1...Rn)p0/2(Rn+1...R2n)p0/p0
×σ(x, ξ)ξn(1/p−1/2)pL0∞x,ξfpH0p0,
that is,
σit(X, D)fLp0≤C(R1...Rn)1/2(Rn+1...R2n)1/p0 (3.17)
×σ(x, ξ) ξn(1/p−1/2)L∞x,ξfHp0
for allf∈S(Rn)∩Hp0(Rn) andt∈R. Similarly,
σ1+it(X, D)fL2≤C(R1...Rn)1/2(Rn+1...R2n)1/2 (3.18)
×σ(x, ξ) ξn(1/p−1/2)L∞x,ξfL2
for all f∈S(Rn) and t∈R. Therefore, by the interpolation theorem for analytic families of operators with (3.16)–(3.18), we obtain Theorem1.2with 1≤p<2.
4. Proof of Theorem 1.2with p≥2
Hwang–Lee [9] essentially proved the following lemma, and obtained Miyachi’s result (see the introduction) as a corollary. In order to prove Theorem 1.2 with p≥2, we modify their estimate as follows:
Lemma 4.1. Let 2≤p<∞and m=−n(1/2−1/p). Forϕ1, ..., ϕn, ψ1, ..., ψn∈ S(R)andf, g∈S(Rn), set
ϕ(y) = [ϕ1⊗...⊗ϕn](y), ψ(η) = [ψ1⊗...⊗ψn](η), V(f, g)(y, η) =
Rn
eiy·tf(η+t)g(t)dt, W(ϕ, ψ, f, g)(x, ξ) =
R2n
e−i(x·y+ξ·η)ϕ(y)ψ(η)V(f, g)(y, η)dy dη,
where [ϕ1⊗...⊗ϕn](y)=ϕ1(y1)...ϕn(yn) and y=(y1, ..., yn)∈Rn. Then there exists a constant C >0 such that
R2n
ξm|W(ϕ, ψ, f, g)(x, ξ)|dx dξ
≤C n
j=1
ϕjLp+ϕ(1)j
Lp+ϕ(2)j
Lp
× n
k=1
ψˆkL2+ψ(1)k
L1
+(ψk(1))(1)
L1 fLpgLp, where ϕ(1)j (yj)=dϕj(yj)/dyj, ϕ(2)j (yj)=d2ϕj(yj)/dy2j and C >0 is independent of ϕ1, ..., ϕn, ψ1, ..., ψn∈S(R)and f, g∈S(Rn).
To prove Lemma4.1, we use the following result.
Lemma 4.2.[9, Lemma 2.4] Let 2≤p<∞ andp≤q≤p. Then there exists a constantC >0 such that
Rn|fˆ(ξ)|q|ξ|−n(1−q/p)dξ
1/q
≤CfLp.
Proof of Lemma 4.1. By the Fourier inversion formula, we have W(ϕ, ψ, f, g)(x, ξ) =
R2n
e−i(x·y+ξ·η)ϕ(y)ψ(η)V(f, g)(y, η)dy dη
=
R2n
e−i(x·y+ξ·η)ϕ(y)ψ(η)
× Rn
eiy·tf(η+t) 1
(2π)n
Rn
eit·ζˆg(ζ)dζ dt dy dη
= 1
(2π)n
R3n
e−i(x·y+ξ·η)ϕ(y)ψ(η)ˆg(ζ)
× Rn
ei(y−ζ)·(t−η)f(t)dt dy dη dζ
= 1
(2π)n
R3n
e−ix·yei(y−ζ)·tϕ(y)f(t)ˆg(ζ)
× Rn
e−i(y+ξ−ζ)·ηψ(η)dη dt dy dζ
= 1
(2π)n
R2ne−iζ·tf(t)ˆg(ζ)
× Rn
ei(t−x)·yϕ(y) ˆψ(y+ξ−ζ)dy dt dζ
= 1
(2π)n
R2n
eix·(ξ−ζ)e−it·ξ
× Rn
ei(t−x)·yϕ(y+ζ−ξ) ˆψ(y)dy f(t)ˆg(ζ)dt dζ.
For the sake of simplicity, we only consider the case n=2 (the case n=2 can be proved in the same way). This means that ϕ(y)=ϕ1(y1)ϕ2(y2) and ψ(η)=