ON THE BOUNDEDNESS OF PSEUDO-DIFFERENTIAL OPERATORS ...
arXiv:1602.08811v1 [math.AP] 29 Feb 2016
ON THE BOUNDEDNESS OF PSEUDO-DIFFERENTIAL OPERATORS ON TRIEBEL-LIZORKIN AND BESOV SPACES BAE JUN PARK Abstract. In this work we study the boundedness of pseudo-differential operators corm responding to a ∈ Sρ,δ , when 0 ≤ δ ≤ ρ < 1, on Triebel-Lizorkin spaces Fps,q and Besov s,q Bp spaces. We also discuss the sharpness of our estimates in a certain sense and give an application.
1. Introduction For 0 ≤ ρ < 1 and m ∈ R, the multiplier operator 1−ρ
(1.1)
ei(−△) 2 cm,ρ (D) = m (1 − △)− 2
is one example of translation invariant Pseudo-differential operator and there are several boundedness results for the operators cm,ρ (D). Fefferman [5], Hirschman [13], Stein [20], and Wainger [24] prove that for 1 < p < ∞ and 0 < ρ < 1, cm,ρ (D) extends to a bounded operator on Lp if 1 1 (1.2) m ≤ −d(1 − ρ) − 2 p and it is unbounded otherwise.
The boundedness result also holds for general Pseudo-differential operators of type (ρ, δ) and order m when 0 ≤ δ ≤ ρ < 1. Calder¨on and Vaillancourt [3] proved the L2 boundedness if m = 0 by using the almost orthogonality technique in a Hilbert space. Fefferman [6] generalized this result to Lp boundedness(1 < p < ∞) with the condition (1.2), but 0 ≤ δ < ρ < 1, by using the interpolation theorem in [7]. P¨ aiv¨arinta and Somersalo [18] proved that these operators map hp into itself for 0 < p < ∞ if (1.2) holds. When 0 < ρ < 1, the example (1.1) gives the sharpness in the sense that for m > −d(1−ρ) 21 − 1p , the boundedness does not hold.(See [6], [13], [17], [24]). A natural question is whether we can extend these results to Besov spaces and TriebelLizorkin spaces. We shall answer this question in this paper. Christ and Seeger [4] proved that for 0 < q < p ≤ 2, (1.1) is unbounded in Fps,q when 1 1 m = −d(1 − ρ) − 2 p
and this partially establishes the sharpness of our result.
The author would like to thank his advisor Andreas Seeger for the guideline and helpful discussions. Supported in part by NSF grant DMS 1500162. 1
2
BAE JUN PARK
m, Recall that by definition, the symbol a(x, ξ) of type (ρ, δ) of order m ∈ R, saying a ∈ Sρ,δ satisfies
|∂ξα ∂xβ a(x, ξ)| ≤ cα,β (1 + |ξ|)m−ρ|α|+δ|β| for (x, ξ) ∈ Rd × Rd
and the corresponding pseudo-differential operator is given, for f ∈ S, by Z a(x, ξ)fb(ξ)e2πi<x,ξ> dξ. Ta f (x) = Rd
Denote by
m OpSρ,δ
m. the class of pseudo-differential operators with symbols in Sρ,δ
Let S(Rd ) denote Schwartz space and S ′ (Rd ) the space of tempered distributions. Let 0 < p < ∞, 0 < q ≤ ∞, and s ∈ R. According to [22], Besov space Bps,q and Triebeld Lizorkin space Fps,q are defined in the following way. Let Φ be a system {ϕj (x)}∞ j=0 ⊂ S(R ) satisfying ∞ X 1= ϕk (x) k=0
Supp(ϕ0 ) ⊂ {x : |x| ≤ 2}
Supp(ϕj ) ⊂ {x : 2k−1 ≤ |x| ≤ 2k+1 } if k = 1, 2, 3, . . .
2k|α| |∂ α ϕk (x)| < ∞ for every multi-index α
sup x∈Rd
k=0,1,2,...
If we choose such a system Φ, define (1.3) and (1.4)
Bps,q = {f ∈ S ′ (Rd ) : kf kBps,q = {2ks (ϕk fb)∨ } q
l (Lp )
Fps,q = {f ∈ S ′ (Rd ) : kf kFps,q = {2ks (ϕk fb)∨ }
Lp (lq )
< ∞}
< ∞} if p < ∞
By definition, the norms in (1.3) and (1.4) depend on the choice of Φ, but these are equivalent to each other for different systems.(the proof is base on Lemma 2.3 below) Thus, both spaces are independent of the chosen system Φ.(See [22, 2.3.2]) In this paper, we use ck for convenience. a Littlewood-Paley partition of unity for Φ and use a notation ϕk = φ ck }∞ satisfies So our system Φ = {φ k=0 1=
∞ X k=0
where
ck φ
φ, φ0 ∈ S b ⊂ {ξ : 1 ≤ |ξ| ≤ 2} Supp(φ) 2 c Supp(φ0 ) ⊂ {ξ : |ξ| ≤ 2} c0 (ξ) = 1 on {ξ : |ξ| ≤ 1} φ ck (ξ) = φb ξ for 1 ≤ k. φ 2k Then the norms on these spaces are kf kBps,q =
∞ X k=0
!1 q
(2sk kφk ∗ f kLp )q
FUNCTION SPACES
and kf kFps,q
3
!1
∞ q
X sk q
= (2 |φ ∗ f |) k
k=0
.
Lp
Let’s denote Besov space and Triebel-Lizorkin space by B-space and F -space, respectively. According to those norms, B- and F -spaces are quasi-Banach space (Banach space if p ≥ 1, q ≥ 1). Note that those are a generalization of many standard function spaces such as Lp spaces, Sobolev spaces, and Hardy spaces in the following sense([10],[22]); Lp ≈ Fp0,2 , p
h
≈
Lps ≈
Λp,q s
≈
Fp0,2 , Fps,2 , Bps,q ,
We recall that hp ≈ Lp for 1 < p < ∞.
1 0 and σ > 0, set Mσ,r u(x) = sup
y∈Rd
|u(x + y)| . (1 + r|y|)σ
As shown in [19], one has the majorization Mσ,r u(x) . Mt u(x)
for all σ ≥ dt , provided that u ∈ S ′ whose Fourier transform is supported in {ξ : |ξ| ≤ r}. These estimates imply the following maximal inequality. Lemma 2.3. (Maximal inequality) Suppose 0 < p < ∞ and 0 < q ≤ ∞. Then for any sequence of positive numbers {rk } and uk ∈ S whose Fourier transform is supported in {|ξ| ≤ rk },
!1 !1
X
X q q
d d q q
|uk | |Mσ,rk uk |
.
for σ > max p , q .
p k
p
k L
L
By using Lemma 2.3 we can easily derive the following estimate.
Lemma 2.4. Suppose 0 < p < ∞, 0 < q ≤ ∞, and c > 0. Let µ0 , µ ∈ S satisfy µ c0 ⊂ {|ξ| . 1} and µ b ⊂ { 1c ≤ |ξ| ≤ c} and set µk (x) = 2kd µ(2k x) for k ≥ 1. Then
!1
∞
q
X sk
q
. kuk s,q . (2 |µ ∗ u|) k Fp
k=0
p L
Another property of F -spaces is the decomposition by using wavelet transforms, so called ϕ-transform. Let D be the set of all dyadic cubes in Rd and Dk be the subset of D containing cubes with side length 2−k . For Q ∈ D, denote the side length of Q by l(Q), lower left corner of Q by xQ , center of Q by cQ , and the characteristic function of Q by χQ . For a sequence of complex numbers b = {bQ } Q∈D , define l(Q)≤1
1
q
X q − ds − 21 |b |χ (x)) gs,q (b)(x) = (|Q| Q Q Q∈D l(Q)≤1
and
f0 , ϑe ∈ S satisfy Let ϑ0 , ϑ, ϑ
kbkfps,q = kgs,q (b)kLp .
c c0 ), Supp(ϑ f0 ) ⊂ {ξ : |ξ| ≤ 2}, Supp(ϑ
6
BAE JUN PARK
1 be b Supp(ϑ) Supp(ϑ), ⊂ {ξ : ≤ |ξ| ≤ 2} 2 c c0 (ξ)|, |ϑ f0 (ξ)| ≥ c > 0 for |ξ| ≤ 5 |ϑ 3
5 3 be b |ϑ(ξ)|, |ϑ(ξ)| ≥ c > 0 for ≤ |ξ| ≤ 4 3
and
∞ X k=0
fk (ξ)ϑk (ξ) = 1 if ξ 6= 0 ϑ
e k x) for k ≥ 1. where ϑk (x) = 2kd ϑ(2k x) and ϑek (x) = 2kd ϑ(2 s,q Then the norms in Fp can be characterized by the discrete fps,q norms.
Lemma 2.5. [8], [9], [10] Suppose 0 < p < ∞, 0 < q ≤ ∞, s ∈ R. (a) Every f ∈ Fps,q can be decomposed as X vQ ϑQ (x) f (x) = Q∈D l(Q)≤1
1
where ϑQ (x) = |Q| 2 ϑk (x − xQ ) for l(Q) = 2−k and vQ =< f, ϑeQ >. Moreover, kvkfps,q . kf kFps,q
(b) Let v = {vQ }Q∈D be a sequence of complex numbers satisfying kvkfps,q < ∞ Then f (x) =
X
vQ ϑQ (x)
Q∈D l(Q)≤1
is in Fps,q and kf kFps,q . kvkfps,q In addition to ϕ-transform, we have an atomic decomposition of B- and F -spaces([9],[10]). It has the same characterization as in Lemma 2.5, but the benefit of ϕ-transform we will use is that when l(Q) = 2−k , the Fourier transforms of ϑQ and φk have comparable compact supports, which is the key to prove (4.6). 3. Paradifferential technique for Pseudo-differential operators Pick a smooth function γ, compactly supported, with γ(0) = 1. Then note that aǫ (x, ξ) = a(x, ξ)γ(ǫx), 0 < ǫ ≤ 1
m uniformly in ǫ and that, for g ∈ S, belongs to Sρ,δ
Taǫ g → Ta g, as ǫ → 0 in the topology of S. Since all of the estimates later will be made independent of ǫ, without m has a compact support in x. Indeed, we need loss of generality, assume our symbol a ∈ Sρ,δ this assumption just to do integration by part in x-variable.
FUNCTION SPACES
Let
7
( ck (ξ) φj ∗ a(·, ξ)(x)φ aj,k (x, ξ) = 0
j, k ≥ 0 otherwise.
Using this, we decompose the symbol a(x, ξ) as XX a(x, ξ) = aj,k (x, ξ) k
=
j
j−3 ∞ X X
aj,k (x, ξ) +
j=3 k=0
k+2 ∞ X X
aj,k (x, ξ) +
k−3 ∞ X X
aj,k (x, ξ)
k=3 j=0
k=0 j=k−2
≡ a(1) (x, ξ) + a(2) (x, ξ) + a(3) (x, ξ)
and proceed by estimating each term separately.
Let T (j) be the pseudo-differential operators corresponding to a(j) for j = 1, 2, 3. Then our claim is that for any s, t, m ∈ R, T (1) and T (2) satisfy the estimates
(1) (3.1)
T s,q .s,t kf kFpt,q Fp
and
(2)
T
(3.2)
Fps,q
.s,t kf kFpt,q .
(3.1) follows exactly in the same way as in [15], which is based on the following two m instead of a ∈ S m . lemmas. The only difference is that we have a ∈ Sρ,δ 1,1
Lemma 3.1. [23, Theorem 3.6]Let s ∈ R, 0 < p < ∞, 0 < q ≤ ∞ and {uj }∞ j=0 be a ′ d sequence in S (R ) satisfying Supp(c u0 ) ⊂ {ξ : |ξ| ≤ 2}
and
Supp(ubj ) ⊂ {ξ : 2j−1 ≤ |ξ| ≤ 2j+1 }
for j ≥ 1.Then ∞ X uj (x) converges in S ′ (Rd ) to some u ∈ Bps,q and (1) j=0
(2)
∞ X j=0
kukBps,q ≤ c
∞ X j=0
1 q
sj
q
(2 kuj kp )
.
uj (x) converges in S ′ (Rd ) to some u ∈ Fps,q and
kukFps,q
1
q ∞
X
sj q (2 |uj |) ≤ c
.
j=0 p
Lemma 3.2. [15],[16] Let v ∈ and
[
S ′ (Rd )
x∈Rd
and b(x, ξ) be a function on Rd × Rd such that
Supp(b v ) ⊂ {ξ : |ξ| ≤ 2k A}
supp(b(x, ·)) ⊂ {ξ : |ξ| ≤ 2k A}.
8
BAE JUN PARK
Then for all 0 < r ≤ 1,
|Tb v(x)| . b(x, 2k ·)
Observe that the Fourier transform of uj ≡
L1d+1
j−3 X
Mr v(x).
Taj,k f is supported in the annulus
k=0
{η : 2j−2 ≤ |η| ≤ 2j+2 } and Taj,k f (x) = Taj,k fk (x) where fk = f ∗ φek for φek = φk−1 + φk + φk+1 . Then by using Lemma 3.1 we have
∞ q 1q j−3
X X
2sjq Taj,k fk (x) kT (1) f kFps,q . (3.3)
k=0
j=3
and then apply lemma 3.2 to get
k Ta fk (x) . (x, 2 ·)
a
j,k j,k Since
aj,k (x, 2k ·)
L1d+1
.
X
X
|α|≤d+1 α1 +α2 =α
and
Mr fk (x).
X
α b
∂ξ [φj ∗ a(·, 2k ξ)(x)φ(ξ)]
|α|≤d+1
.
L1d+1
p
L1 (ξ)
b 2k|α1 | φj ∗ ∂ξα1 a(·, 2k ξ)(x) ∂ α2 φ(ξ)
L1 (ξ)
Z α α1 k k 1 φj (y)∂ξ a(x − y, 2 ξ)dy φj ∗ ∂ξ a(·, 2 ξ)(x) = d R Z X (−y)β α β k 1 φj (y) = ∂x ∂ξ a(x − ηy , 2 ξ)dy β! Rd |β|=N Z X |φj (y)||y|N |∂xβ ∂ξα1 a(x − ηy , 2k ξ)|dy . |β|=N
.
Rd
X Z
|β|=N
Rd
|φj (y)||y|N (1 + 2k |ξ|)m−ρ|α1 |+δ|β| dy
. 2k(m−ρ|α1 |+δN )−jN for |ξ| ∼ 1 and N sufficiently large, we obtain
and this gives
k
aj,k (x, 2 ·)
L1d+1
. 2k(m+(d+1)(1−ρ))+δN k−jN
|Taj,k fk (x)| . 2k(m+(d+1)(1−ρ))+δN k−jN Mr fk (x).
FUNCTION SPACES
9
Finally,
∞ !q q1 j−3
X X
2k(m+(d+1)(1−ρ))+δN k−jN Mr fk . 2sjq
k=0
j=3 p
1
∞ ! q q j−3
X X
(s−N )jq k(m+(d+1)(1−ρ))+δN k = 2 2 Mr fk
.
k=0
j=3
kT (1) f kFps,q
p
Case1. 1 < q ≤ ∞ Choose M such that M > 0 and N − s − M > 0. By using H¨older’s inequality, j−3 X
k(m+(d+1)(1−ρ))+δN k
2
k=0
j−3 X
=
k(m+(d+1)(1−ρ))+δN k−kM
2
k=0
j−3 X
≤ . 2
kM
Mr fk (x)2
2kq(m+(d+1)(1−ρ)+δN −M ) (Mr fk (x))q
k=0
jM q
Mr fk (x)
!q
j−3 X
!q
!
j−2 X k=0
kM q ′
2
! qq′
2kq(m+(d+1)(1−ρ)+δN −M ) (Mr fk (x))q .
k=0
Then
kT (1) f kFps,q
.
=
.
.
∞ ! q1 j−3
X X
2kq(m+(d+1)(1−ρ)+δN −M ) (Mr fk )q 2(s−N +M )jq
k=0
j=3 p
1
∞ q ∞
X X
2kq(m+(d+1)(1−ρ)+δN −M ) (Mr fk (x))q 2−jq(N −s−M )
j=k+3
k=0 p
" 1 #
X q
∞ kq(m+s+(d+1)(1−ρ)+δN −N ) q
2 (M f (x)) r k
k=0 p
" 1 #
X q
∞ kq(m+s+(d+1)(1−ρ)−N (1−δ)) q
2 (f (x)) k
k=0 p
. kf kF m+s+(d+1)(1−ρ)−N(1−δ),q . p
by the boundedness of maximal operator Mr with the choice r > p, q and by using Lemma 2.4. Case2. 0 < q ≤ 1
10
BAE JUN PARK
In this case, just use lq ⊂ l1 .
! q1 j−3 ∞
X
X
2kq(m+(d+1)(1−ρ)+δN ) (Mr fk )q 2(s−N )jq kT (1) f kFps,q .
k=0
j=3
p
1
∞
q ∞
X
X
= 2kq(m+(d+1)(1−ρ)+δN ) (Mr fk )q 2(s−N )jq
j=k+3
k=0
p
" 1 #
X q
∞ kq(m+s+(d+1)(1−ρ)−N (1−δ)) q
. 2 (M f ) r k
k=0 p
" 1 #
∞ q
X kq(m+s+(d+1)(1−ρ)−N (1−δ)) q
. 2 (fk (x))
k=0 p
. kf kF m+s+(d+1)(1−ρ)−N(1−δ),q p
For all t ∈ R, choose N sufficiently large so that m + s + (d + 1)(1 − ρ) − N (1 − δ) −
d d dξ Z Z Z b k (η)φbk (ξ)e−2πi e2πi<x,η> e2πi<x−y,ξ> dzdηdξ. = a(z, ξ)Φ
Kk (x, y) = (3.4)
Z
Rd
One observation in the integral is that the variables ξ and η live in |ξ| ∼ 2k and |η| ∼ 2k . To be precise, this is true for all but a finite number of terms (for k = 0, |ξ| ≤ 2 and for 0 ≤ k ≤ 2, |η| ≤ 32), but it is not really concern because our argument also covers those cases. Then the kernel Kk (x, y) has the size estimate (3.5)
|Kk (x, y)| .J,N 2−Jk
for any J > 0 and N > 0.
1 (1 + |x − y|)N
FUNCTION SPACES
11
The idea to get (3.5) is to apply the technique of oscillatory integrals by integrating by parts with respect to each variable. First, we perform this with respect to the z-variable by writing e−2πi =
(I − △z )M e−2πi (1 + 4π 2 |η|2 )M
and then carry out a similar process on the η-variable with e2πi<x−z,η> =
(I − △η )d e2πi<x−z,η> . (1 + 4π 2 |x − z|2 )d
Then (3.4) is dominated by Z Z Z M a(z, ξ) (I − △ ) z 2πi<x−z,η> 2πi<x−y,ξ> e e dηdξdz Rd |ξ|∼2k |η|∼2k (1 + 4π 2 |η|2 )M Z Z Z 1 1 M 2πi<x−y,ξ> (I − △z ) a(z, ξ)e dξ .M dηdz 2d (1 + |η|)2M (1 + |x − z|) k d k |ξ|∼2 R |η|∼2 Z Z 1 dz. (I − △z )M a(z, ξ)e2πi<x−y,ξ> dξ .M 2−k(2M −d) (1 + |x − z|)2d Rd |ξ|∼2k If |x − y| ≤ 1, then
Z M 2πi<x−y,ξ> (I − △ ) a(z, ξ)e dξ z |ξ|∼2k Z (1 + |ξ|)m+2δM dξ . |ξ|∼2k
k(m+2δM +d)
. 2
.
If |x − y| > 1, then do integration by part in ξ-variable as we did, beginning with e2πi<x−y,ξ> = , for |α| = N , to get
.N,M .N,M
1 ∂ α e2πi<x−y,ξ> (2πi(x − y))α ξ
Z (I − △z )M a(z, ξ)e2πi<x−y,ξ> dξ |ξ|∼2k Z 1 (1 + |ξ|)m−ρN +2δM dξ |x − y|N |ξ|∼2k 1 2k(m−ρN +2δM +d) . |x − y|N
These yield (3.5) by choosing M sufficiently large. Another observation is that since the Fourier transform of Kk (x, y) in x-variable has a k+4 }. By the support compact support in {|η| ≤ 2k+4 }, T[ ak f (η) is supported in {|η| ≤ 2
12
BAE JUN PARK
rule, we have
(2)
T f
Fps,q
(3.6)
d d p, q
1
∞
q
X
sjq (2) q = 2 |φj ∗ T f |
j=0
p L
∞ !q q1 ∞
X X
2sjq ≤ |φj ∗ Tak fk |
k=0
p
j=0 L
q 1
∞ q ∞
X
X
sjq ≤ 2 |φj ∗ Tak fk |
k=j−4
j=4
. Lp
and choose N > σ. Then Z |φj (x − z)||Kk (z, y)||fk (y)|dydz |φj ∗ Tak fk (x)| ≤ Rd Rd Z Z (1 + 2k |x − y|)σ |φj (x − z)| |Kk (z, y)|dydz . Mσ,2k fk (x)
Now let σ > max
Z
Rd
−k(J−σ)
. 2
Rd
Mσ,2k fk (x)
by the size estimate (3.5). So (3.6) is controlled by
q 1
∞ q ∞
X
X
sjq −k(J−σ) 2 2 Mσ,2k fk (3.7)
k=j−4
j=4
. Lp
We can easily check that q ∞ ∞ X X q −jqL −k(J−σ) 2−kq(J−σ−L) Mσ,2k fk (x) .2 2 Mσ,2k fk (x) k=0
k=j−4
for any L > 0. (if q > 1, use H¨older’s inequality, if 0 < q ≤ 1, use lq ⊂ l1 ). Pick L > |s| to get
!1
∞ q X
q −kq(J−σ−L)
(3.7) . 2 (Mσ,2k fk (x))
p
k=0 L
. kf kF −(J −σ−L),q p
by Lemma 2.3 and Lemma 2.4. (3.2) follows from the choice of J satisfying −(J − σ − L) < t. Now it remains to establish the boundedness of T (3) . m as We express a symbol a(3) ∈ Sρ,δ (3)
a
(x, ξ) =
∞ X k=3
bk (x, ξ)
FUNCTION SPACES
where bk (x, ξ) =
k−3 X
13
aj,k (x, ξ).
j=0
Observe that
k−3 X φj ∗ (a(·, ξ)) (x) j=0
is a u∈
m symbol Sρ,δ ′ S (Rd ),
(3.8)
with a constant which is independent of k, and thus for 0 < r < ∞ and kTbk ukLr
i h k s1 −s2 +d(1−ρ) | r1 − 21 |− p1 − 21
. 2
by Corollary 1.2 and the fact that
kφk ∗ ukLr
hr ֒→ Lr and kφk ∗ ukhr ≤ kφk ∗ ukLr . By the same reasoning as (3.3), we have
!1
∞ q
X
(3) skq q
2 |Tbk f |
T f s,q .
Fp
k=3
Lp
k because Td bk f (η) has a compact support in {|η| ∼ 2 }. Thus, our goal is to prove that
!1
∞ q
X skq q
.1
(3.9) 2 |Tbk f |
p
k=3 L
and we will prove it in the next section.
4. The proof of (3.9) We start with the case 0 < p ≤ 1. Assume
p≤q≤∞
and
1 1 m − s1 + s2 = −d(1 − ρ) − . p 2 The idea of the proof of Theorem 1.1 when 0 < p ≤ 1 is the atomic decomposition for hp . It asserts that every v ∈ hp can be decomposed as ∞ X λj vj , v= j=1
where λj are complex numbers such that (4.1)
∞ X j=1
and vj are
hp
|λj |p . kvkphp
atoms. By the virtue of the atomic decomposition, it is enough to prove
uniformly in j.
kT vj khp . 1
14
BAE JUN PARK
To extend q = 2 to 0 < q ≤ ∞, we shall use a tool which substitutes the atomic decomposition for hp . Definition 1. Let 0 < p ≤ 1, 0 < q ≤ ∞, and s ∈ R. A sequence of complex numbers r = {rQ } Q∈D is called an ∞-atom for fps,q if there exists a dyadic cube Q0 such that l(Q)≤1
rQ = 0
if
Q 6⊂ Q0
and 1
kg s,q (r)kL∞ ≤ |Q0 |− p .
(4.2)
Then the following lemma can replace (4.1). Lemma 4.1. [10], [12] Suppose 0 < p ≤ 1, p ≤ q ≤ ∞, and b = {bQ }Q∈D,l(Q)≤1 ∈ fps,q . Then there exist Cd,p,q > 0, a sequence of scalars {λj }, and a sequence of ∞-atoms rj = {rj,Q } Q∈D for fps,q such that l(Q)≤1
b = {bQ }Q =
∞ X j=1
λj {rj,Q }Q =
∞ X
λj rj
j=1
and such that 1 p ∞ X p |λj | ≤ Cd,p,q kbkfps,q . j=1
Moreoever,
kbkfps,q
1 p ∞ ∞ X X s,q p λj rj , rj is an ∞-atom for fp . |λj | : s = ≈ inf j=1 j=1
By Lemma 2.5 and Lemma 4.1, f ∈ Fps1 ,q can be decomposed with {bQ } and there exist a sequence of scalars {λj } and a sequence of ∞-atoms that f (x) =
X
Q∈D l(Q)≤1
bQ ϑQ (x) =
∞ X j=1
λj
X
Q∈D l(Q)≤1
rj,Q ϑQ (x).
Q∈D ∈ l(Q)≤1 {rj,Q } for fps1,q
fps1,q such
FUNCTION SPACES
15
Then
(3)
T f
s ,q
Fp 2
!1
∞ q X
s2 k q
. (2 |Tbk f |)
p
k=3 L
q 1
q
∞ ∞
X X X
2s2 k λ T r ϑ = j j,Q Q b k
j=1 Q∈D
k=3
l(Q)≤1
Lp
1 p p q pq p Z ∞ ∞ X X X s2 kp p 2 |λ | T r ϑ ≤ j bk j,Q Q dx d R k=3 j=1 Q∈D l(Q)≤1
(4.3)
q p p1 q Z ∞ ∞ X X X p s2 kq |λj | ≤ T dx 2 r ϑ j,Q Q bk d R j=1 Q∈D k=3 l(Q)≤1
q 1
q 1
p ∞ ∞
X X X
2s2 k Tb |λj |p sup . r ϑ j,Q Q k
j j=1 Q∈D
k=3
l(Q)≤1
Lp
by using lp ⊂ l1 and Minkowski’s inequality for
q p
≥ 1.
Since
∞ X j=1
1
p
p
|λj |
. kf kFps1 ,q ,
it is sufficient to show the supremum in (4.3) is bounded by a constant. Let Q0 be any dyadic cubes with side length 2−µ and rQ be an ∞-atome for fps1,q with Q0 and define RQ0 (x) =
X
rQ ϑQ (x).
Q∈D,Q⊂Q0 l(Q)≤1
Then we obtain the desired result by showing
(4.4)
!1
∞ q X
s2 k q
(2 |Tbk RQ0 |)
k=3
Lp
. 1 uniformly in Q0 .
16
BAE JUN PARK
, but this is true when p ≤ q < ∞ because
!1 !1
X
∞ q p X
∞ s k
p q s2 kp 2
(2 |T R |) ≤ 2 kT R k bk Q0 bk Q0 Lp
k=3
p k=3 L !1 ∞ p X p s1 kp 2 kφk ∗ RQ0 kLp . k=3
= kRQ0 kFps1 ,p
1
p
X
s1 1
(|rQ ||Q|− 2 − d χQ )p .
Q∈D,Q⊂Q0
p l(Q)≤1 L 1 p ∞ X X s1 1 p − − −kd (|rQ ||Q| 2 d ) 2 ≤ Q∈Dk ,Q⊂Q0
k=µ
− dq
. l(Q0 ) . 1
∞ X
1
p
− pdk q
2
k=µ
by lp ⊂ lq , (3.8), Lemma 2.5, H¨older’s inequality, and (4.2). Now assume q = ∞. Then by (4.2), we have 1
s1
1
|rQ | ≤ |Q| 2 + d |Q0 |− p .
(4.5)
Since l2 ⊂ l∞ , we need to show (4.6)
!1
X
2
∞ sk
2 2
(2 |T R |) Q b 0 k
k=2
. 1. Lp
with (4.5), but when we define
RQ0 ,k (x) =
X
rQ ϑQ (x),
Q∈Dk ,Q⊂Q0 l(Q)≤1
by the support rule it suffices to prove (4.6) with Tbk RQ0 replaced by Tbk RQ0 ,k . We consider first the case l(Q0 ) < 1. Let (4.7)
ν =ρ−
1 . µ
fν with side For Q ∈ D, denote by Qν a dilate of Q with side 2l(Q)ν and by Q
√
d24 l(Q)ν .
FUNCTION SPACES
Since RQ0 ,k = 0 for k < µ,
!1
∞ 2
X sk 2 2
(2 |T R |) b Q ,k 0 k
p
k=3 L
1
∞
2
X
s2 k 2 . (2 |Tbk RQ0 ,k |)
k=µ
p
17
1
∞ 2
X
+ (2s2 k |Tbk RQ0 ,k |)2
k=µ
fν ) L (Q 0
The former part is ,by H¨older’s inequality, less than 1 2 ∞ X 1 − 21 2 s2 k ν f p |Q0 | (2 kTbk RQ0 ,k kL2 ) k=µ
and by (3.8), it is controlled by 1 2 ∞ X 1 1 1 1 k(s1 −d(1−ρ)( p − 2 )) 2 ν p−2 (2 kRQ0 ,k kL2 ) |Q0 | . k=µ
However, by Lemma 2.5 and (4.5),
1
2
X 1
kRQ0 ,k kL2 . (|rQ ||Q|− 2 χQ )2
Q∈Dk ,Q⊂Q0 µd p1 − 12 −s1 k
and this gives
≤ 2
L2
2
1
2 ∞
X
(2s2 k |Tbk RQ0 ,k |)2
k=µ
fν ) Lp (Q 0
µd(ρ−ν) p1 − 12
. 2
. 1
due to (4.7). For the latter one we split it into
1
∞
2
X
(2s2 k |Tbk (χQν0 RQ0 ,k )|)2 (4.8)
k=µ
c
fν ) Lp (Q 0
and
1
2 ∞
X
s2 k 2 c (2 |Tbk (χ(Qν0 ) RQ0 ,k )|)
k=µ
(4.9)
fν c ) Lp (Q 0
and our claim is that each part can be bounded by a constant.
By integration by parts, the kernel of Tbk has the estimate, for any L > 0, Z 1 2πi<x−y,ξ> k(d+m−ρL) . b (x, ξ)e dξ k d .L 2 |x − y|L R
fν c ) Lp (Q 0
18
BAE JUN PARK
fν c , Thus, for x ∈ Q 0
X
|Tbk (χQν0 RQ0 ,k )(x)| ≤
Q∈Dk ,Q⊂Q0 k(d+m−ρL)
. 2
|rQ | Tbk (χQν0 ϑQ )(x) X
Q∈Dk ,Q⊂Q0 1
|rQ |
. 2k(m−s1 +2d−ρL) 2µd( p −1−ν)
Z
Qν0
|ϑQ (y)| dy |x − y|L
1 . |x − cQ0 |L
The last inequality is from (4.5) and the fact that |x − y| & |x − cQ0 | in the integral. Now, by using H¨older’s inequality,
(4.8) =
Z
fν c Q 0
∞ X
k=µ
2
1
p
22s2 k |Tbk (χQν0 RQ0 ,k )(x)|2 dx
µνN −µνd p1 − 21
. 2
p
∞ X
2s2 k
2
Z
2N
fν c Q 0
k=µ
|x − cQ0 |
2
|Tbk (χQν0 RQ0 ,k )(x)| dx 2
1
2 ∞ X 1 1 µd p1 −1−ν µνN −µνd p − 2 2k(m−s1 +s2 +2d−ρL)
. 2
2
2
2
k=µ
. 2−µL(ρ−ν) 2 . 1
µd (ρ−ν) p
2
3µd 2
2−µνd 2−
Z
µρd 2
for sufficiently large N and L. Moreover, by using lp ⊂ l2 and (3.8),
(4.9) ≤
∞ X
k=µ
1
1 p
p
s1 kp . 2
φk ∗ (χ(Qν0 )c RQ0 ,k ) p
Since l(Q0 ) = 2−µ < 1, Q0 ⊂ Qν0 and therefore |y − xQ | & |y − cQ0 |
L
fν c Q 0
1 |x − cQ0 |2(L−N )
!1 2
FUNCTION SPACES
19
for y ∈ (Qν0 )c and Q ⊂ Q0 . So we have
p
φk ∗ χ(Qν0 )c RQ0 ,k p L p Z Z X |φ (x − y)| |rQ ||ϑQ (y)|dy ≤ k dx Rd
y∈(Qν0 )c
d
. 2−kp(N − 2 )
−kp(N − d2 )
≤ 2
Z
R
d
Q∈Dk Q⊂Q0
Z
y∈(Qν0 )c
p
|φk (x − y)|
Z X |rQ | Q∈Dk Q⊂Q0
Rd
X
|rQ | dy N dx |y − xQ |
Q∈Dk Q⊂Q0
Z
y∈(Qν0 )c
p
|φk (x − y)| dy |y − cQ0 |N
!p
dx
!p |φk (x − y)| ≤ 2 2 dy dx N y∈(Qν0 )c |y − cQ0 | Rd !p Z Z 1 L d 1 (1 + 2k |x − cQ0 |) p |φk (x − y)|dxdy . 2−kp(N +s1 −2d+ p ) 2µdp( p −1) N y∈(Qν0 )c |y − cQ0 | Rd !p Z 1 −kp(N +s1 −2d+ dp ) µdp( 1p −1) kL 2 2 . 2 L dy y∈(Qν0 )c |y − cQ |N − p 0 1 −kp(N +s1 −d) µdp( p −1)
Z
Z
−kp(N +s1 −2d+ pd ) µdp( 1p −1) kL µνp(N − L −d) p
. 2 for
L 1−p
2
> d and N −
L p
2 2
> d and so L
(4.9) . 2−µ(1−ν)(N − p −d) . 1. When l(Q0 ) ≥ 1, it still holds in a similar way, but is simpler. However, in this case, we cannot use the above argument to get (4.9) . 1 because Q0 may not contained in Qν0 . Instead of the usage of Qν0 in the decomposition, we use a dilation of Q by a factor of some constants independent of µ. Moreover, note that since µ ≤ 0, we may use the range k ≥ 3 in our summations. Since the proof is derived in the almost same way and easier way, we omit it here. This completes the proof of (2) in Theorem 1.3. We now assume 1 < p < ∞, q is between p and p′ , and 1 m − s1 + s2 = −d(1 − ρ) − 2 We proved for any 1 ≤ r ≤ ∞,
Ta : F1s1 ,r → F1s2 ,r
if
1 . p
d m − s1 + s2 ≤ − (1 − ρ), 2
and already know Ta : L2s1 → L2s2
if
m − s1 + s2 ≤ 0.
20
BAE JUN PARK
By applying the complex interpolation theorem in Section 7, we have 1 1 s1 ,q s2 ,q − (4.10) Ta : Fp → Fp if m − s1 + s2 ≤ −d(1 − ρ) p 2 for 1 ≤ p ≤ 2 and p ≤ q ≤ p′ .
m , the case 2 < p < ∞ can be derived Since the adjoint operator (Ta )∗ is also in OpSρ,δ via duality (Lemma 2.2) and this completes the proof of Theorem 1.3.
Theorem 1.4 is proved in a similar way. By repeating the process in Section 3, (3.1) and (3.2) hold if F-spaces are replaced by B-spaces, and the boundedness of T (3) on B-spaces follows just from (3.8). We omit the details. 5. Sharpness of Theorem 1.3 In this section, when 0 < ρ < 1, we shall prove the necessity of 1 1 m − s1 + s2 ≤ −d(1 − ρ) − 2 p
and when the equality holds, the necessity of the condition on q in Theorem 1.3. Recall Theorem 1.1 does not hold without the assumption (1.2). When 0 < ρ < 1, (1.1) is one example of Pseudo-differential operator that is not bounded in hp for 0 < p < ∞ if 1 1 m > −d(1 − ρ) − . 2 p
See [6], [13], [24] for 1 < p < ∞ and [17] for 0 < p ≤ 1. Once we have a counter example for the necessity of the condition (1.2), it is easy to show the necessity of (1.5) for Corollary 1.2. Suppose (5.1)
1 1 m − s1 + s2 > −d(1 − ρ) − . 2 p
m−s1 +s2 Then there exist a ∈ Sρ,δ and f ∈ Fp0,2 such that
kf kFp0,2 . 1, but kTa f kFp0,2 = ∞. Now put Tb = n−s2 (D)Ta ns1 (D) and g(x) = n−s1 (D)f (x). Then it is clear that m , b ∈ Sρ,δ
kgkF s1 ,2 = kf kFp0,2 . 1, p
and kTb gkF s2 ,2 = kTa f kFp0,2 = ∞. p
FUNCTION SPACES
21
This is also true for Theorem 1.3. If (5.1) holds, there exists ǫ > 0 such that 1 1 m − s1 + s2 − ǫ > −d(1 − ρ) − . 2 p
m and f ∈ F s1 ,2 such that If 2 < q, choose a ∈ Sρ,δ p
kTa f kF s2 −ǫ,2 = ∞. p
Fps1 ,2
Then by ֒→ If q ≤ 2, choose a
Fps1 ,q and Fps2 ,q m and f ∈ ∈ Sρ,δ
֒→ Fps2 −ǫ,2 , it is done. Fps1 +ǫ,2 such that kTa f kF s2 ,2 = ∞. p
Then by
Fps1 +ǫ,2
֒→
Fps1 ,q
and
Fps2 ,q
֒→
Fps2 ,2 ,
it is done.
Now assume
1 1 m − s1 + s2 = −d(1 − ρ) − 2 p and consider the condition of q. We may assume s1 = s2 . Christ and Seeger [4] show that (1.1) is unbounded in Fps,q if 0 < q < p ≤ 2 with 1 1 m = −d(1 − ρ) − . p 2 Theorem 5.1. [4] Let 0 < q < p ≤ 2, s ∈ R, 0 < ρ < 1, m = −d(1 − ρ) p1 − 21 . Then, for r ≥ 2, 1
sup {kcm,ρ (D)f kFps,q : kf kFps,q ≤ 1, f ∈ E(r)} ≈ (log r) q
− p1
.
where E(r) is the space of all (smooth) distributions on Rd whose Fourier transforms are supported in {ξ : |ξ| ≤ r}. Theorem 5.1 also yields ,via duality, the necessity of the condition q ≤ p for the boundedness of (1.1) when p ≥ 2. Suppose 2 ≤ p < q and 1 1 1 1 − − = −d(1 − ρ) . m = −d(1 − ρ) 2 p p′ 2 The upper boundedness
1
sup {kcm,ρ (D)f kFps,q : kf kFps,q ≤ 1, f ∈ E(r)} . (log r) p
− q1
follows immediately from the duality. ′ such that f ∈ E(r), kf kF −s,q′ = 1, According to Theorem 5.1, there exists f ∈ Fp−s,q ′ p′
and
1
kCm,ρ (D)f kF −s,q′ & (logr) q′
− p1′
.
p′
There exists g ∈ Fps,q such that (5.2)
1
< Cm,ρ (D)f, g >& (log r) q′
− p1′
Choose K ∈ E(2r) such that K = 1 in the support of fb. Then
Cm,ρ (D)(g ∗ K) s,q ≥ | < Cmρ (D)(g ∗ K), f > | F p
= < Cm,ρ (D)f, g > 1
1
& (log r) p − q
22
BAE JUN PARK
and it is clear that kg ∗ KkFps,q . 1.
Thus, we conclude that
1
sup {kCm,ρ (D)f kFps,q : kf kFps,q ≤ 1, f ∈ E(r)} ≈ (log r) p
− 1q
for 2 ≤ p < q and large r > 0. Remark. It remains to show the necessity of p′ ≤ q when 2 ≤ p (and via duality, q ≤ p′ when p ≤ 2), however we do not know this works with the same exmaple (1.1). Another part we need to show in order to complete the sharpness is to show it is also true for the case ρ = 0. Unfortunately, Miyachi [17] pointed out that cm,0 (D) is bounded in hp if and only if 1 1 m ≤ −(d − 1) − . 2 p Thus, we cannot use cm,0 (D) for the sharpness.
6. An application Finally we shall give a short discussion on the continuity of Ta mapping Fps11 ,q1 into Fps22 ,q2 . H¨ormander [14] gave a question that for p1 ≤ p2 , the estimate kTa f kLp2 . kf kLp1 holds if m ≤ −d
1 1 1 1 1 1 − + (1 − ρ) max ( − , − , 0) p1 p2 2 p1 p2 2
and Alvarez and Hounie [1] answered the question.. Theorem 6.1. [1] Let 0 ≤ δ ≤ ρ < 1, 0 < ρ, 1 < p1 ≤ p2 < ∞, and m ∈ R. Suppose m . Then T maps Lp1 into Lp2 in the following cases: a ∈ Sρ,δ a (1) if 1 < p1 ≤ 2 ≤ p2 and
m ≤ −d (2) if 2 ≤ p1 ≤ p2 and m ≤ −d (3) if p1 ≤ p2 ≤ 2 and m ≤ −d
1 1 − p1 p2
;
1 1 1 1 − + (1 − ρ)( − ) ; p1 p2 2 p1
1 1 1 1 − + (1 − ρ)( − ) . p1 p2 p2 2
Our results give the generalization of the above results. Theorem 6.2. Let 0 ≤ δ ≤ ρ < 1, 0 < p1 ≤ p2 < ∞, 0 < q1 , q2 ≤ ∞, and s1 , s2 , m ∈ R. m . Then T maps F s1 ,q1 into F s2 ,q2 in the following cases: Suppose a ∈ Sρ,δ p1 p2 a (1) if 0 < p1 = p2 = p < ∞, 0 < q1 = q2 = q ≤ ∞ satisfies the assumptions of q in Theorem 1.3, and 1 1 m − s1 + s2 ≤ −d(1 − ρ) − ; 2 p
FUNCTION SPACES
(2) if p1 < 2 < p2 , 0 < q1 , q2 ≤ ∞, and
m − s1 + s2 ≤ −d (3) if p1 < p2 = 2, 0 < q1 ≤ ∞, q2 = 2, and
m − s1 + s2 ≤ −d (4) if p1 = 2 < p2 , q1 = 2, 0 < q2 ≤ ∞, and m − s1 + s2 ≤ −d
23
1 1 − p1 p2
;
1 1 − p1 2
;
1 1 − 2 p2
;
(5) if 2 < p1 < p2 , p′1 ≤ q1 ≤ p1 , 0 < q2 ≤ ∞, and 1 1 1 1 − + (1 − ρ) − m − s1 + s2 ≤ −d ; p1 p2 2 p1 (6) if p1 < p2 < 2, 0 < q1 ≤ ∞, p2 ≤ q2 ≤ p′2 , and 1 1 1 1 m − s1 + s2 ≤ −d − + (1 − ρ) − . p1 p2 p2 2
Proof. (1) is from Theorem 1.3 and let us prove other cases. It suffices to prove the endpoint results. (2) Suppose 0 < p1 < 2 < p2 . Then kTa f kFps2 ,q2 2
. kTa f k . kf k
)
(
1 − 1 ,2 s2 +d 2 p2
F2
(
)
s1 −d p1 − 1 ,2 1 2
F2
. kf kFps1 ,q1 1
by Lemma 2.1 and Theorem 1.3. (3) Suppose p1 < p2 = 2 and q2 = 2. Then kTa f kF s2 ,2 . kf k 2
s1 −d( p1 − 1 ),2 1 2
F2
. kf kFps1 ,q1 1
by Lemma 2.1 and Theorem 1.3. (4) Suppose p1 = 2 < p2 , q1 = 2. Then kTa f kFps2 ,q2 2
. kTa f k
1 − 1 ),2 s2 +d( 2 p2
F2
. kf kF s1 ,2 2
by Lemma 2.1 and Theorem 1.3. (5) Suppose 2 < p1 < p2 and p′1 ≤ q1 ≤ p′1 . Then kTa f kFps2 ,q2 2
. kTa f k
s2 +d( p1 − p1 ),q1 1 2
F p1
. kf kFps1 ,q1 1
by Lemma 2.1 and Theorem 1.3.
24
BAE JUN PARK
(6) Suppose p1 < p2 < 2 and p2 ≤ q2 ≤ p′2 . Then kTa f kFps2 ,q2
. kf k
2
s1 −d( p1 − p1 ),q2 1 2
F p2
. kf kFps1 ,q1 1
by Lemma 2.1 and Theorem 1.3. Theorem 6.3. Let 0 ≤ δ ≤ ρ < 1, 0 < p1 < p2 < ∞, 0 < q1 , q2 ≤ ∞, and s, m ∈ R. m . Then T maps F s1 ,q1 into F s2 ,q2 in the following cases: Suppose a ∈ Sρ,δ p1 p2 a (1) if p1 ≤ 2 ≤ p2 and
1 1 − p1 p2
;
1 1 − + (1 − ρ) p1 p2
1 1 − 2 p1
m − s1 + s2 < −d (2) if 2 ≤ p1 ≤ p2 and m − s1 + s2 < −d (3) if p1 ≤ p2 ≤ 2 and
;
1 1 1 1 m − s1 + s2 < −d − + (1 − ρ) − . p1 p2 p2 2 1 1 − . Proof. (1) Assume p1 ≤ 2 ≤ p2 and m − s1 + s2 < −d p1 p2 If 0 < q2 < 2, then for any ǫ > 0, kTa f kFps2 ,q2 .ǫ kTa f kLp2
ǫ+s2
2
by H¨older’s inequality.
1 1 − . Then we have p1 p 1 1 − . m + s2 − s < −d p p2
Pick p ∈ (p1 , p2 ) and let s = s1 − d
Now choose ǫ > 0 so small that
m + s2 + ǫ − s < −d and apply Theorem 6.2 to get kTa f kLp2
ǫ+s2
Since s −
1 1 − p p2
. kf kLps .
d d = s1 − and p1 < p, we have p p1 kf kLps . kf kFps1 ,q1 1
by Lemma 2.1. If 2 < q2 < ∞, then
kTa f kFps2 ,q2 .ǫ kTa f kLps 2 2
by
l2
2
l q2 .
⊂ Choose ǫ > 0 so small that m − s1 + s2 + ǫ < −d
1 1 − p1 p2
.
FUNCTION SPACES
25
1 1 Pick p ∈ (p1 , p2 ) and let s = s1 − ǫ − d − . Then we have p1 p 1 1 − . m + s2 − s < −d p p2 By Theorem 6.2, kTa f kLps 2 . kf |Lps . 2
Since s −
d d = s1 − ǫ − and p1 < p, we have p p1 kf kLps . kf kF s1 −ǫ,q1 ≤ kf kFps1 ,q1 p1
1
by Lemma 2.1. By using the same technique, (2) and (3) can be proved. By using Theorem 1.4 and Lemma 2.1, we can get analogous conclusions on B-spaces. Theorem 6.4. Let 0 ≤ δ ≤ ρ < 1, 0 < p1 ≤ p2 < ∞, 0 < q ≤ ∞, and s, m ∈ R. Suppose m . Then T maps B s1 ,q into B s2 ,q in the following cases: a ∈ Sρ,δ p1 p2 a (1) if p1 ≤ 2 ≤ p2 and
1 1 − p1 p2
:
1 1 − + (1 − ρ) p1 p2
1 1 − 2 p1
:
1 1 − + (1 − ρ) p1 p2
1 1 − p2 2
.
m − s1 + s2 ≤ −d (2) if 2 ≤ p1 ≤ p2 and m − s1 + s2 ≤ −d (3) if p1 ≤ p2 ≤ 2 and m − s1 + s2 ≤ −d
Theorem 6.5. Let 0 ≤ δ ≤ ρ < 1, 0 < p1 < p2 < ∞, 0 < q1 , q2 ≤ ∞, and s, m ∈ R. m . Then T maps B s1 ,q1 into B s2 ,q2 in the following cases: Suppose a ∈ Sρ,δ p1 p2 a (1) if p1 ≤ 2 ≤ p2 and
1 1 − p1 p2
;
1 1 − + (1 − ρ) p1 p2
1 1 − 2 p1
;
1 1 − + (1 − ρ) p1 p2
1 1 − p2 2
.
m − s1 + s2 < −d (2) if 2 ≤ p1 < p2 and m − s1 + s2 < −d (3) if p1 < p2 ≤ 2 and m − s1 + s2 < −d
26
BAE JUN PARK
7. Appendix - Complex Interpolation theorem for F- spaces This section deals with a complex interpolation method for F-spaces. Let S = {z ∈ C : 0 < Re(z) < 1} be a strip in the complex plane and S = {z ∈ C : 0 ≤ Re(z) ≤ 1} be its closure. We say that f is an S ′ -analytic function in S if the following properties are satisfied: (a) For any fixed z ∈ S, we have f (z) ∈ S ′ ; (b) For any ϕ ∈ S with compact support in Rd , h(x, z) = (ϕfd (z))∨ (x) is uniformly d continuous and bounded function in R × S; (c) For any ϕ ∈ S with compact support in Rd and any fixed x ∈ Rd , h(x, ·) = (ϕfd (·))∨ (x) is an analytic function in S. Let s0 , s1 ∈ R, 0 < p0 , p1 < ∞, and 0 < q0 , q1 ≤ ∞. Suppose that for any t ∈ R, f (it) ∈ Fps00 ,q0 and f (1 + it) ∈ Fps11 ,q1 . We introduce the norm kf kF (Fps0 ,q0 ,Fps1 ,q1 ) = max {sup kf (it)kFps0 ,q0 , sup kf (1 + it)kFps1 ,q1 } 0
1
0
t
1
t
and then F(Fps00 ,q0 , Fps11 ,q1 ) becomes a quasi-Banach space(Banach space if 1 ≤ p0 , p1 , q0 , q1 ). And define a subspace Fps00 ,q0 + Fps11 ,q1 of S ′ with the norm kvkFps0 ,q0 +Fps1 ,q1 = 0
1
inf
v=v0 +v1
(kv0 kFps0 ,q0 + kv1 kFps1 ,q1 ). 0
1
Then F(Fps11 ,q1 , Fps11 ,q1 ) ֒→ Fps00 ,q0 + Fps11 ,q1 ֒→ S ′ .
Definition 2. Fix 0 ≤ θ ≤ 1. The complex interpolation space of order θ between Fps00 ,q0 and Fps11 ,q1 is the normed linear subspace (Fps00 ,q0 , Fps11 ,q1 )θ = {v ∈ Fps00 ,q0 + Fps11 ,q1 : v = f (θ) for some f ∈ F(Fps00 ,q0 , Fps11 ,q1 )}
of Fps00 ,q0 + Fps11 ,q1 , with the norm
kvk(Fps0 ,q0 ,Fps1 ,q1 )θ = 0
1
inf
s ,q
s ,q1 )
f ∈F (Fp00 0 ,Fp11 f (θ)=v
kf kF (Fps0 ,q0 ,Fps1 ,q1 ) . 0
1
Theorem 7.1. [22] Let s0 , s1 ∈ R, 0 < q0 , q1 ≤ ∞, and 0 < p0 , p1 < ∞. For fixed 0 < θ < 1, assume s = (1 − θ)s0 + θs1 ,
1 1−θ 1−θ θ 1 θ = = + , + . p p0 p1 q q0 q1
Then (Fps00 ,q0 , Fps11 ,q1 )θ = Fps,q . Lemma 7.2. [21], [22] Let s ∈ R, 0 < p < ∞, 0 < q ≤ ∞, and τ > d2 + all a ∈ L∞ and f ∈ Fps,q ,
ka(D)f kFps,q ≤ c(kγ1 akL2τ + sup γ2 a(2k ·) 2 ) kf kFps,q 0≤k
where γ1 , γ2 ∈ S with
Lτ
0 ≤ γ1 (ξ), γ2 (ξ) ≤ 1 for all ξ,
Supp(γ1 ) ⊂ {ξ : |ξ| ≤ 4}, γ1 (ξ) = 1 if |ξ| ≤ 2,
Supp(γ2 ) ⊂ {ξ : γ2 (ξ) = 1 if
d min (p,q) .
1 ≤ |ξ| ≤ 4}, 4
1 ≤ |ξ| ≤ 2. 2
Then for
FUNCTION SPACES
27
Theorem 7.3. (Complex interpolation theorem for F -spaces) Let m, s1 , s2 ∈ R, 0 ≤ δ ≤ ρ < 1, ρ > 0, 1 ≤ q0 , q1 ≤ ∞, and 1 ≤ p0 , p1 ≤ 2. For fixed 0 < θ < 1, let 1−θ 1−θ θ 1 θ 1 = = + , + q q0 q1 p p0 p1 m and assume a ∈ Sρ,δ . Suppose that 1 1 s ,q s ,q Ta : Fpj1 j → Fpj2 j if m − s1 + s2 = −d(1 − ρ) − pj 2 for each j = 0, 1. Then Ta :
Fps1 ,q
→
Fps2 ,q
if
m − s1 + s2 = −d(1 − ρ)
Proof. Assume m − s1 + s2 = −d(1 − ρ) By Theorem 7.1, it suffices to show
1 p
1 2
−
1 1 − p 2
.
m. and a ∈ Sρ,δ
Ta : (Fps01 ,q0 , Fps11 q1 )θ → (Fps02 ,q0 , Fps12 ,q1 )θ . For z ∈ S, define and
d(1−ρ)(( p1 − p1 )z+( p1 − p1 ))
az (x, ξ) = a(x, ξ)(1 + |ξ|)
0
1
0
Tz = Taz . Then Tθ = Ta . By the assumption, T0 : Fps01 ,q0 → Fps02 ,q0
and
T1 : Fps11 ,q1 → Fps12 ,q1 . Let k0 and k1 be the constants for the boundedness of T0 and T1 , respectively. Let v ∈ (Fps01 ,q0 , Fps11 ,q1 )θ and ǫ > 0. By the definition of norm of interpolation space, there exists f ∈ F(Fps01 ,q0 , Fps11 ,q1 ) such that f (θ) = v
and kf kF (Fps1 ,q0 ,Fps1 ,q1 ) . kvk(Fps1 ,q0 ,Fps1,q1 )θ + ǫ. 0
Put
1
0
g(z) =
1
k0z−1 k1−z Tz (f (z)). (1 + z)2d
Then our claim is that g ∈ F(Fps02 ,q0 , Fps12 ,q1 )
and (7.1)
kgkF (Fps2 ,q0 ,Fps2 ,q1 ) . kvk(Fps1 ,q0 ,Fps1 ,q1 )θ + ǫ. 0
1
0
1
Observe that kTit kFps1 ,q0 →Fps2 ,q0 . |t|2d k0 0
and
0
kT1+it kFps1 ,q1 →Fps2 ,q1 . |t|2d k1 1
1
28
BAE JUN PARK
by Lemma 7.2 because
itd(1−ρ)( p1 − p1 )
0 1
(1 + |D|)
s ,qj
Fpj1
s ,qj
→Fpj1
. |t|2d
for each j = 0, 1. Another observation is that g is a S ′ -analytic function in S. Thus, we shall prove (7.1). k0−1 kTit (f (it))kFps2 ,q0 0 |1 + it|2d . kf (it)kFps1 ,q0
kg(it)kFps2 ,q0
=
0
0
and similarly, kg(1 + it)kFps2 ,q1
. kf (1 + it)kFps1 ,q1 .
1
1
Thus, kgkF (Fps2 ,q0 ,Fps2 ,q1 ) = max {sup kg(it)kFps2 ,q0 , sup kg(1 + it)kFps2 ,q1 } 0
1
0
t
1
t
. kf kF (Fps1 ,q0 ,Fps1 ,q1 ) 0
1
≤ kvk(Fps1 ,q0 ,Fps1 ,q1 )θ + ǫ 0
1
and this implies that = (1 + θ)2d k01−θ k1θ kg(θ)k(Fps2 ,q0 ,Fps2 ,q1 )θ
kTa (v)k(Fps2 ,q0 ,Fps2 ,q1 )θ 0
1
0
.
k01−θ k1θ
kgk
1
s ,q s ,q F (Fp02 0 ,Fp12 1 )
. k01−θ k1θ kvk(Fps1 ,q0 ,Fps1 ,q1 )θ + ǫ. 0
1
Since ǫ > 0 is arbitrary, we have the inequality kTa (v)k(Fps2 ,q0 ,Fps2 ,q1 )θ . k01−θ k1θ kvk(Fps1 ,q0 ,Fps1 ,q1 )θ . 0
1
0
1
References [1] J. Alvarez and J. Hounie, Estimates for the kernel and continuity properties of pseudo-differential operators, Ark. Mat. 28 (1990) 1-22. [2] J. M. Bony, Calcul symbolique et propagation des singularit´es pour les ´equations aux d´eriv´ees partielles non lin´eaires, Ann.Sc.E.N.S. Paris 14 (1981) 209-246. [3] A. P. Calder¨ on and R. Vaillancourt, A class of bounded pseudo-differential operators, Proc. Nat. Acad. Sci. U.S.A. 69 (1972) 1185-1187. [4] M. Christ and A. Seeger, Necessary conditions for vector-valued operator inequalities in harmonic analysis, London Math. Soc. (3)93 (2006) 447-473. [5] C. Fefferman, Inequalities for strongly singular convolution operators, Acta Math. 123 (1969) 9-36. [6] C. Fefferman, Lp bounds for pseudo-differential operators, Israel J. Math. 14 (1973) 413-417. [7] C. Fefferman and E. M. Stein, Some maximal inequalities, Amer. J. Math. 93 (1971) 107-115. [8] M. Frazier and B. Jawerth, The ϕ-transform and applications to distribution spaces, in ”Function Spaces and Applications”, Lecture Notes in Math. Vol. 1302, Springer-Verlag, New York/Berlin, (1988) 223246. [9] M. Frazier and B. Jawerth, A discrete transform and decomposition of distribution spaces, J. Functional Anal. 93 (1990) 34-170. [10] M. Frazier and B. Jawerth, Applications of the φ and wavelet transforms to the theory of function spaces, Wavelets and Their Applications, pp.377-417, Jones and Bartlett, Boston, MA, 1992 [11] D. Goldberg, A local version of real Hardy spaces, Duke Math. J. 46 (1979) 27-42. [12] L. Grafakos, Modern Fourier Analysis, Springer (2008) [13] I. I. Hirschman, On multiplier transformations, Duke Math. J., 26 (1959) 221-242.
FUNCTION SPACES
29
[14] L. H¨ ormander, Pseudo-Differential Operators and hypoelliptic equations, Proc. Sympos. Pure Math. 10 (1967) 138-183. [15] J. Johnsen, Domains of pseudo-differential operators: a case for the Triebel-Lizorkin spaces, J. Funct. Spaces Appl. 3 (2005) 263-286. [16] J. Marschall, Nonregular pseudo-differential operators, J. Anal. Anwendungen 15(1)(1996) 109-148. [17] A. Miyachi, On some Fourier multipliers for H p (Rd ), J. Fac. Sci. Univ. Tokyo Sect. IA Math. 27(1980) 157-179. [18] L. P¨ aiv¨ arinta and E. Somersalo, A generalization of the Calder´ on-Vaillancourt theorem to Lp and hp , Math. Nachr. 138 (1988) 145-156. [19] J. Peetre, On spaces of Triebel-Lizorkin type, Ark. Mat. 13 (1975) 123-130. [20] E. M. Stein, Singular integrals, harmonic functions, and differentiability properties of functions of several variables, Proc. Sympos. Pure Math., vol. 10 Amer. Math. Soc., Providence, R.I. (1967) 316355. [21] H. Triebel, Complex Interpolation and Fourier Multipliers for the Space Bps,q and Fps,q of Besov-HardySobolev Type:The Case 0 < p ≤ ∞, 0 < q ≤ ∞, Math. Z. 176 (1981) 495-510. [22] H. Triebel, Theory of Function Spaces, Birkhauser, Basel-Boston-Stuttgart (1983). [23] M. Yamazaki, A quasi-homogeneous version of paradifferential operators, I. Boundedness on spaces of Besov type, J. Fac. Sci. Univ. Tokyo Sect. IA, Math., 33(1986) 131-174. [24] S. Wainger, Special trigonometric series in k dimensions, Memoirs of the AMS., 59(1965). Department of Mathematics, University of Wisconsin - Madison, , U.S.A. E-mail address: [email protected]
ON THE BOUNDEDNESS OF PSEUDO-DIFFERENTIAL OPERATORS ON TRIEBEL-LIZORKIN AND BESOV SPACES BAE JUN PARK Abstract. In this work we study the boundedness of pseudo-differential operators corm responding to a ∈ Sρ,δ , when 0 ≤ δ ≤ ρ < 1, on Triebel-Lizorkin spaces Fps,q and Besov s,q Bp spaces. We also discuss the sharpness of our estimates in a certain sense and give an application.
1. Introduction For 0 ≤ ρ < 1 and m ∈ R, the multiplier operator 1−ρ
(1.1)
ei(−△) 2 cm,ρ (D) = m (1 − △)− 2
is one example of translation invariant Pseudo-differential operator and there are several boundedness results for the operators cm,ρ (D). Fefferman [5], Hirschman [13], Stein [20], and Wainger [24] prove that for 1 < p < ∞ and 0 < ρ < 1, cm,ρ (D) extends to a bounded operator on Lp if 1 1 (1.2) m ≤ −d(1 − ρ) − 2 p and it is unbounded otherwise.
The boundedness result also holds for general Pseudo-differential operators of type (ρ, δ) and order m when 0 ≤ δ ≤ ρ < 1. Calder¨on and Vaillancourt [3] proved the L2 boundedness if m = 0 by using the almost orthogonality technique in a Hilbert space. Fefferman [6] generalized this result to Lp boundedness(1 < p < ∞) with the condition (1.2), but 0 ≤ δ < ρ < 1, by using the interpolation theorem in [7]. P¨ aiv¨arinta and Somersalo [18] proved that these operators map hp into itself for 0 < p < ∞ if (1.2) holds. When 0 < ρ < 1, the example (1.1) gives the sharpness in the sense that for m > −d(1−ρ) 21 − 1p , the boundedness does not hold.(See [6], [13], [17], [24]). A natural question is whether we can extend these results to Besov spaces and TriebelLizorkin spaces. We shall answer this question in this paper. Christ and Seeger [4] proved that for 0 < q < p ≤ 2, (1.1) is unbounded in Fps,q when 1 1 m = −d(1 − ρ) − 2 p
and this partially establishes the sharpness of our result.
The author would like to thank his advisor Andreas Seeger for the guideline and helpful discussions. Supported in part by NSF grant DMS 1500162. 1
2
BAE JUN PARK
m, Recall that by definition, the symbol a(x, ξ) of type (ρ, δ) of order m ∈ R, saying a ∈ Sρ,δ satisfies
|∂ξα ∂xβ a(x, ξ)| ≤ cα,β (1 + |ξ|)m−ρ|α|+δ|β| for (x, ξ) ∈ Rd × Rd
and the corresponding pseudo-differential operator is given, for f ∈ S, by Z a(x, ξ)fb(ξ)e2πi<x,ξ> dξ. Ta f (x) = Rd
Denote by
m OpSρ,δ
m. the class of pseudo-differential operators with symbols in Sρ,δ
Let S(Rd ) denote Schwartz space and S ′ (Rd ) the space of tempered distributions. Let 0 < p < ∞, 0 < q ≤ ∞, and s ∈ R. According to [22], Besov space Bps,q and Triebeld Lizorkin space Fps,q are defined in the following way. Let Φ be a system {ϕj (x)}∞ j=0 ⊂ S(R ) satisfying ∞ X 1= ϕk (x) k=0
Supp(ϕ0 ) ⊂ {x : |x| ≤ 2}
Supp(ϕj ) ⊂ {x : 2k−1 ≤ |x| ≤ 2k+1 } if k = 1, 2, 3, . . .
2k|α| |∂ α ϕk (x)| < ∞ for every multi-index α
sup x∈Rd
k=0,1,2,...
If we choose such a system Φ, define (1.3) and (1.4)
Bps,q = {f ∈ S ′ (Rd ) : kf kBps,q = {2ks (ϕk fb)∨ } q
l (Lp )
Fps,q = {f ∈ S ′ (Rd ) : kf kFps,q = {2ks (ϕk fb)∨ }
Lp (lq )
< ∞}
< ∞} if p < ∞
By definition, the norms in (1.3) and (1.4) depend on the choice of Φ, but these are equivalent to each other for different systems.(the proof is base on Lemma 2.3 below) Thus, both spaces are independent of the chosen system Φ.(See [22, 2.3.2]) In this paper, we use ck for convenience. a Littlewood-Paley partition of unity for Φ and use a notation ϕk = φ ck }∞ satisfies So our system Φ = {φ k=0 1=
∞ X k=0
where
ck φ
φ, φ0 ∈ S b ⊂ {ξ : 1 ≤ |ξ| ≤ 2} Supp(φ) 2 c Supp(φ0 ) ⊂ {ξ : |ξ| ≤ 2} c0 (ξ) = 1 on {ξ : |ξ| ≤ 1} φ ck (ξ) = φb ξ for 1 ≤ k. φ 2k Then the norms on these spaces are kf kBps,q =
∞ X k=0
!1 q
(2sk kφk ∗ f kLp )q
FUNCTION SPACES
and kf kFps,q
3
!1
∞ q
X sk q
= (2 |φ ∗ f |) k
k=0
.
Lp
Let’s denote Besov space and Triebel-Lizorkin space by B-space and F -space, respectively. According to those norms, B- and F -spaces are quasi-Banach space (Banach space if p ≥ 1, q ≥ 1). Note that those are a generalization of many standard function spaces such as Lp spaces, Sobolev spaces, and Hardy spaces in the following sense([10],[22]); Lp ≈ Fp0,2 , p
h
≈
Lps ≈
Λp,q s
≈
Fp0,2 , Fps,2 , Bps,q ,
We recall that hp ≈ Lp for 1 < p < ∞.
1 0 and σ > 0, set Mσ,r u(x) = sup
y∈Rd
|u(x + y)| . (1 + r|y|)σ
As shown in [19], one has the majorization Mσ,r u(x) . Mt u(x)
for all σ ≥ dt , provided that u ∈ S ′ whose Fourier transform is supported in {ξ : |ξ| ≤ r}. These estimates imply the following maximal inequality. Lemma 2.3. (Maximal inequality) Suppose 0 < p < ∞ and 0 < q ≤ ∞. Then for any sequence of positive numbers {rk } and uk ∈ S whose Fourier transform is supported in {|ξ| ≤ rk },
!1 !1
X
X q q
d d q q
|uk | |Mσ,rk uk |
.
for σ > max p , q .
p k
p
k L
L
By using Lemma 2.3 we can easily derive the following estimate.
Lemma 2.4. Suppose 0 < p < ∞, 0 < q ≤ ∞, and c > 0. Let µ0 , µ ∈ S satisfy µ c0 ⊂ {|ξ| . 1} and µ b ⊂ { 1c ≤ |ξ| ≤ c} and set µk (x) = 2kd µ(2k x) for k ≥ 1. Then
!1
∞
q
X sk
q
. kuk s,q . (2 |µ ∗ u|) k Fp
k=0
p L
Another property of F -spaces is the decomposition by using wavelet transforms, so called ϕ-transform. Let D be the set of all dyadic cubes in Rd and Dk be the subset of D containing cubes with side length 2−k . For Q ∈ D, denote the side length of Q by l(Q), lower left corner of Q by xQ , center of Q by cQ , and the characteristic function of Q by χQ . For a sequence of complex numbers b = {bQ } Q∈D , define l(Q)≤1
1
q
X q − ds − 21 |b |χ (x)) gs,q (b)(x) = (|Q| Q Q Q∈D l(Q)≤1
and
f0 , ϑe ∈ S satisfy Let ϑ0 , ϑ, ϑ
kbkfps,q = kgs,q (b)kLp .
c c0 ), Supp(ϑ f0 ) ⊂ {ξ : |ξ| ≤ 2}, Supp(ϑ
6
BAE JUN PARK
1 be b Supp(ϑ) Supp(ϑ), ⊂ {ξ : ≤ |ξ| ≤ 2} 2 c c0 (ξ)|, |ϑ f0 (ξ)| ≥ c > 0 for |ξ| ≤ 5 |ϑ 3
5 3 be b |ϑ(ξ)|, |ϑ(ξ)| ≥ c > 0 for ≤ |ξ| ≤ 4 3
and
∞ X k=0
fk (ξ)ϑk (ξ) = 1 if ξ 6= 0 ϑ
e k x) for k ≥ 1. where ϑk (x) = 2kd ϑ(2k x) and ϑek (x) = 2kd ϑ(2 s,q Then the norms in Fp can be characterized by the discrete fps,q norms.
Lemma 2.5. [8], [9], [10] Suppose 0 < p < ∞, 0 < q ≤ ∞, s ∈ R. (a) Every f ∈ Fps,q can be decomposed as X vQ ϑQ (x) f (x) = Q∈D l(Q)≤1
1
where ϑQ (x) = |Q| 2 ϑk (x − xQ ) for l(Q) = 2−k and vQ =< f, ϑeQ >. Moreover, kvkfps,q . kf kFps,q
(b) Let v = {vQ }Q∈D be a sequence of complex numbers satisfying kvkfps,q < ∞ Then f (x) =
X
vQ ϑQ (x)
Q∈D l(Q)≤1
is in Fps,q and kf kFps,q . kvkfps,q In addition to ϕ-transform, we have an atomic decomposition of B- and F -spaces([9],[10]). It has the same characterization as in Lemma 2.5, but the benefit of ϕ-transform we will use is that when l(Q) = 2−k , the Fourier transforms of ϑQ and φk have comparable compact supports, which is the key to prove (4.6). 3. Paradifferential technique for Pseudo-differential operators Pick a smooth function γ, compactly supported, with γ(0) = 1. Then note that aǫ (x, ξ) = a(x, ξ)γ(ǫx), 0 < ǫ ≤ 1
m uniformly in ǫ and that, for g ∈ S, belongs to Sρ,δ
Taǫ g → Ta g, as ǫ → 0 in the topology of S. Since all of the estimates later will be made independent of ǫ, without m has a compact support in x. Indeed, we need loss of generality, assume our symbol a ∈ Sρ,δ this assumption just to do integration by part in x-variable.
FUNCTION SPACES
Let
7
( ck (ξ) φj ∗ a(·, ξ)(x)φ aj,k (x, ξ) = 0
j, k ≥ 0 otherwise.
Using this, we decompose the symbol a(x, ξ) as XX a(x, ξ) = aj,k (x, ξ) k
=
j
j−3 ∞ X X
aj,k (x, ξ) +
j=3 k=0
k+2 ∞ X X
aj,k (x, ξ) +
k−3 ∞ X X
aj,k (x, ξ)
k=3 j=0
k=0 j=k−2
≡ a(1) (x, ξ) + a(2) (x, ξ) + a(3) (x, ξ)
and proceed by estimating each term separately.
Let T (j) be the pseudo-differential operators corresponding to a(j) for j = 1, 2, 3. Then our claim is that for any s, t, m ∈ R, T (1) and T (2) satisfy the estimates
(1) (3.1)
T s,q .s,t kf kFpt,q Fp
and
(2)
T
(3.2)
Fps,q
.s,t kf kFpt,q .
(3.1) follows exactly in the same way as in [15], which is based on the following two m instead of a ∈ S m . lemmas. The only difference is that we have a ∈ Sρ,δ 1,1
Lemma 3.1. [23, Theorem 3.6]Let s ∈ R, 0 < p < ∞, 0 < q ≤ ∞ and {uj }∞ j=0 be a ′ d sequence in S (R ) satisfying Supp(c u0 ) ⊂ {ξ : |ξ| ≤ 2}
and
Supp(ubj ) ⊂ {ξ : 2j−1 ≤ |ξ| ≤ 2j+1 }
for j ≥ 1.Then ∞ X uj (x) converges in S ′ (Rd ) to some u ∈ Bps,q and (1) j=0
(2)
∞ X j=0
kukBps,q ≤ c
∞ X j=0
1 q
sj
q
(2 kuj kp )
.
uj (x) converges in S ′ (Rd ) to some u ∈ Fps,q and
kukFps,q
1
q ∞
X
sj q (2 |uj |) ≤ c
.
j=0 p
Lemma 3.2. [15],[16] Let v ∈ and
[
S ′ (Rd )
x∈Rd
and b(x, ξ) be a function on Rd × Rd such that
Supp(b v ) ⊂ {ξ : |ξ| ≤ 2k A}
supp(b(x, ·)) ⊂ {ξ : |ξ| ≤ 2k A}.
8
BAE JUN PARK
Then for all 0 < r ≤ 1,
|Tb v(x)| . b(x, 2k ·)
Observe that the Fourier transform of uj ≡
L1d+1
j−3 X
Mr v(x).
Taj,k f is supported in the annulus
k=0
{η : 2j−2 ≤ |η| ≤ 2j+2 } and Taj,k f (x) = Taj,k fk (x) where fk = f ∗ φek for φek = φk−1 + φk + φk+1 . Then by using Lemma 3.1 we have
∞ q 1q j−3
X X
2sjq Taj,k fk (x) kT (1) f kFps,q . (3.3)
k=0
j=3
and then apply lemma 3.2 to get
k Ta fk (x) . (x, 2 ·)
a
j,k j,k Since
aj,k (x, 2k ·)
L1d+1
.
X
X
|α|≤d+1 α1 +α2 =α
and
Mr fk (x).
X
α b
∂ξ [φj ∗ a(·, 2k ξ)(x)φ(ξ)]
|α|≤d+1
.
L1d+1
p
L1 (ξ)
b 2k|α1 | φj ∗ ∂ξα1 a(·, 2k ξ)(x) ∂ α2 φ(ξ)
L1 (ξ)
Z α α1 k k 1 φj (y)∂ξ a(x − y, 2 ξ)dy φj ∗ ∂ξ a(·, 2 ξ)(x) = d R Z X (−y)β α β k 1 φj (y) = ∂x ∂ξ a(x − ηy , 2 ξ)dy β! Rd |β|=N Z X |φj (y)||y|N |∂xβ ∂ξα1 a(x − ηy , 2k ξ)|dy . |β|=N
.
Rd
X Z
|β|=N
Rd
|φj (y)||y|N (1 + 2k |ξ|)m−ρ|α1 |+δ|β| dy
. 2k(m−ρ|α1 |+δN )−jN for |ξ| ∼ 1 and N sufficiently large, we obtain
and this gives
k
aj,k (x, 2 ·)
L1d+1
. 2k(m+(d+1)(1−ρ))+δN k−jN
|Taj,k fk (x)| . 2k(m+(d+1)(1−ρ))+δN k−jN Mr fk (x).
FUNCTION SPACES
9
Finally,
∞ !q q1 j−3
X X
2k(m+(d+1)(1−ρ))+δN k−jN Mr fk . 2sjq
k=0
j=3 p
1
∞ ! q q j−3
X X
(s−N )jq k(m+(d+1)(1−ρ))+δN k = 2 2 Mr fk
.
k=0
j=3
kT (1) f kFps,q
p
Case1. 1 < q ≤ ∞ Choose M such that M > 0 and N − s − M > 0. By using H¨older’s inequality, j−3 X
k(m+(d+1)(1−ρ))+δN k
2
k=0
j−3 X
=
k(m+(d+1)(1−ρ))+δN k−kM
2
k=0
j−3 X
≤ . 2
kM
Mr fk (x)2
2kq(m+(d+1)(1−ρ)+δN −M ) (Mr fk (x))q
k=0
jM q
Mr fk (x)
!q
j−3 X
!q
!
j−2 X k=0
kM q ′
2
! qq′
2kq(m+(d+1)(1−ρ)+δN −M ) (Mr fk (x))q .
k=0
Then
kT (1) f kFps,q
.
=
.
.
∞ ! q1 j−3
X X
2kq(m+(d+1)(1−ρ)+δN −M ) (Mr fk )q 2(s−N +M )jq
k=0
j=3 p
1
∞ q ∞
X X
2kq(m+(d+1)(1−ρ)+δN −M ) (Mr fk (x))q 2−jq(N −s−M )
j=k+3
k=0 p
" 1 #
X q
∞ kq(m+s+(d+1)(1−ρ)+δN −N ) q
2 (M f (x)) r k
k=0 p
" 1 #
X q
∞ kq(m+s+(d+1)(1−ρ)−N (1−δ)) q
2 (f (x)) k
k=0 p
. kf kF m+s+(d+1)(1−ρ)−N(1−δ),q . p
by the boundedness of maximal operator Mr with the choice r > p, q and by using Lemma 2.4. Case2. 0 < q ≤ 1
10
BAE JUN PARK
In this case, just use lq ⊂ l1 .
! q1 j−3 ∞
X
X
2kq(m+(d+1)(1−ρ)+δN ) (Mr fk )q 2(s−N )jq kT (1) f kFps,q .
k=0
j=3
p
1
∞
q ∞
X
X
= 2kq(m+(d+1)(1−ρ)+δN ) (Mr fk )q 2(s−N )jq
j=k+3
k=0
p
" 1 #
X q
∞ kq(m+s+(d+1)(1−ρ)−N (1−δ)) q
. 2 (M f ) r k
k=0 p
" 1 #
∞ q
X kq(m+s+(d+1)(1−ρ)−N (1−δ)) q
. 2 (fk (x))
k=0 p
. kf kF m+s+(d+1)(1−ρ)−N(1−δ),q p
For all t ∈ R, choose N sufficiently large so that m + s + (d + 1)(1 − ρ) − N (1 − δ) −
d d dξ Z Z Z b k (η)φbk (ξ)e−2πi e2πi<x,η> e2πi<x−y,ξ> dzdηdξ. = a(z, ξ)Φ
Kk (x, y) = (3.4)
Z
Rd
One observation in the integral is that the variables ξ and η live in |ξ| ∼ 2k and |η| ∼ 2k . To be precise, this is true for all but a finite number of terms (for k = 0, |ξ| ≤ 2 and for 0 ≤ k ≤ 2, |η| ≤ 32), but it is not really concern because our argument also covers those cases. Then the kernel Kk (x, y) has the size estimate (3.5)
|Kk (x, y)| .J,N 2−Jk
for any J > 0 and N > 0.
1 (1 + |x − y|)N
FUNCTION SPACES
11
The idea to get (3.5) is to apply the technique of oscillatory integrals by integrating by parts with respect to each variable. First, we perform this with respect to the z-variable by writing e−2πi =
(I − △z )M e−2πi (1 + 4π 2 |η|2 )M
and then carry out a similar process on the η-variable with e2πi<x−z,η> =
(I − △η )d e2πi<x−z,η> . (1 + 4π 2 |x − z|2 )d
Then (3.4) is dominated by Z Z Z M a(z, ξ) (I − △ ) z 2πi<x−z,η> 2πi<x−y,ξ> e e dηdξdz Rd |ξ|∼2k |η|∼2k (1 + 4π 2 |η|2 )M Z Z Z 1 1 M 2πi<x−y,ξ> (I − △z ) a(z, ξ)e dξ .M dηdz 2d (1 + |η|)2M (1 + |x − z|) k d k |ξ|∼2 R |η|∼2 Z Z 1 dz. (I − △z )M a(z, ξ)e2πi<x−y,ξ> dξ .M 2−k(2M −d) (1 + |x − z|)2d Rd |ξ|∼2k If |x − y| ≤ 1, then
Z M 2πi<x−y,ξ> (I − △ ) a(z, ξ)e dξ z |ξ|∼2k Z (1 + |ξ|)m+2δM dξ . |ξ|∼2k
k(m+2δM +d)
. 2
.
If |x − y| > 1, then do integration by part in ξ-variable as we did, beginning with e2πi<x−y,ξ> = , for |α| = N , to get
.N,M .N,M
1 ∂ α e2πi<x−y,ξ> (2πi(x − y))α ξ
Z (I − △z )M a(z, ξ)e2πi<x−y,ξ> dξ |ξ|∼2k Z 1 (1 + |ξ|)m−ρN +2δM dξ |x − y|N |ξ|∼2k 1 2k(m−ρN +2δM +d) . |x − y|N
These yield (3.5) by choosing M sufficiently large. Another observation is that since the Fourier transform of Kk (x, y) in x-variable has a k+4 }. By the support compact support in {|η| ≤ 2k+4 }, T[ ak f (η) is supported in {|η| ≤ 2
12
BAE JUN PARK
rule, we have
(2)
T f
Fps,q
(3.6)
d d p, q
1
∞
q
X
sjq (2) q = 2 |φj ∗ T f |
j=0
p L
∞ !q q1 ∞
X X
2sjq ≤ |φj ∗ Tak fk |
k=0
p
j=0 L
q 1
∞ q ∞
X
X
sjq ≤ 2 |φj ∗ Tak fk |
k=j−4
j=4
. Lp
and choose N > σ. Then Z |φj (x − z)||Kk (z, y)||fk (y)|dydz |φj ∗ Tak fk (x)| ≤ Rd Rd Z Z (1 + 2k |x − y|)σ |φj (x − z)| |Kk (z, y)|dydz . Mσ,2k fk (x)
Now let σ > max
Z
Rd
−k(J−σ)
. 2
Rd
Mσ,2k fk (x)
by the size estimate (3.5). So (3.6) is controlled by
q 1
∞ q ∞
X
X
sjq −k(J−σ) 2 2 Mσ,2k fk (3.7)
k=j−4
j=4
. Lp
We can easily check that q ∞ ∞ X X q −jqL −k(J−σ) 2−kq(J−σ−L) Mσ,2k fk (x) .2 2 Mσ,2k fk (x) k=0
k=j−4
for any L > 0. (if q > 1, use H¨older’s inequality, if 0 < q ≤ 1, use lq ⊂ l1 ). Pick L > |s| to get
!1
∞ q X
q −kq(J−σ−L)
(3.7) . 2 (Mσ,2k fk (x))
p
k=0 L
. kf kF −(J −σ−L),q p
by Lemma 2.3 and Lemma 2.4. (3.2) follows from the choice of J satisfying −(J − σ − L) < t. Now it remains to establish the boundedness of T (3) . m as We express a symbol a(3) ∈ Sρ,δ (3)
a
(x, ξ) =
∞ X k=3
bk (x, ξ)
FUNCTION SPACES
where bk (x, ξ) =
k−3 X
13
aj,k (x, ξ).
j=0
Observe that
k−3 X φj ∗ (a(·, ξ)) (x) j=0
is a u∈
m symbol Sρ,δ ′ S (Rd ),
(3.8)
with a constant which is independent of k, and thus for 0 < r < ∞ and kTbk ukLr
i h k s1 −s2 +d(1−ρ) | r1 − 21 |− p1 − 21
. 2
by Corollary 1.2 and the fact that
kφk ∗ ukLr
hr ֒→ Lr and kφk ∗ ukhr ≤ kφk ∗ ukLr . By the same reasoning as (3.3), we have
!1
∞ q
X
(3) skq q
2 |Tbk f |
T f s,q .
Fp
k=3
Lp
k because Td bk f (η) has a compact support in {|η| ∼ 2 }. Thus, our goal is to prove that
!1
∞ q
X skq q
.1
(3.9) 2 |Tbk f |
p
k=3 L
and we will prove it in the next section.
4. The proof of (3.9) We start with the case 0 < p ≤ 1. Assume
p≤q≤∞
and
1 1 m − s1 + s2 = −d(1 − ρ) − . p 2 The idea of the proof of Theorem 1.1 when 0 < p ≤ 1 is the atomic decomposition for hp . It asserts that every v ∈ hp can be decomposed as ∞ X λj vj , v= j=1
where λj are complex numbers such that (4.1)
∞ X j=1
and vj are
hp
|λj |p . kvkphp
atoms. By the virtue of the atomic decomposition, it is enough to prove
uniformly in j.
kT vj khp . 1
14
BAE JUN PARK
To extend q = 2 to 0 < q ≤ ∞, we shall use a tool which substitutes the atomic decomposition for hp . Definition 1. Let 0 < p ≤ 1, 0 < q ≤ ∞, and s ∈ R. A sequence of complex numbers r = {rQ } Q∈D is called an ∞-atom for fps,q if there exists a dyadic cube Q0 such that l(Q)≤1
rQ = 0
if
Q 6⊂ Q0
and 1
kg s,q (r)kL∞ ≤ |Q0 |− p .
(4.2)
Then the following lemma can replace (4.1). Lemma 4.1. [10], [12] Suppose 0 < p ≤ 1, p ≤ q ≤ ∞, and b = {bQ }Q∈D,l(Q)≤1 ∈ fps,q . Then there exist Cd,p,q > 0, a sequence of scalars {λj }, and a sequence of ∞-atoms rj = {rj,Q } Q∈D for fps,q such that l(Q)≤1
b = {bQ }Q =
∞ X j=1
λj {rj,Q }Q =
∞ X
λj rj
j=1
and such that 1 p ∞ X p |λj | ≤ Cd,p,q kbkfps,q . j=1
Moreoever,
kbkfps,q
1 p ∞ ∞ X X s,q p λj rj , rj is an ∞-atom for fp . |λj | : s = ≈ inf j=1 j=1
By Lemma 2.5 and Lemma 4.1, f ∈ Fps1 ,q can be decomposed with {bQ } and there exist a sequence of scalars {λj } and a sequence of ∞-atoms that f (x) =
X
Q∈D l(Q)≤1
bQ ϑQ (x) =
∞ X j=1
λj
X
Q∈D l(Q)≤1
rj,Q ϑQ (x).
Q∈D ∈ l(Q)≤1 {rj,Q } for fps1,q
fps1,q such
FUNCTION SPACES
15
Then
(3)
T f
s ,q
Fp 2
!1
∞ q X
s2 k q
. (2 |Tbk f |)
p
k=3 L
q 1
q
∞ ∞
X X X
2s2 k λ T r ϑ = j j,Q Q b k
j=1 Q∈D
k=3
l(Q)≤1
Lp
1 p p q pq p Z ∞ ∞ X X X s2 kp p 2 |λ | T r ϑ ≤ j bk j,Q Q dx d R k=3 j=1 Q∈D l(Q)≤1
(4.3)
q p p1 q Z ∞ ∞ X X X p s2 kq |λj | ≤ T dx 2 r ϑ j,Q Q bk d R j=1 Q∈D k=3 l(Q)≤1
q 1
q 1
p ∞ ∞
X X X
2s2 k Tb |λj |p sup . r ϑ j,Q Q k
j j=1 Q∈D
k=3
l(Q)≤1
Lp
by using lp ⊂ l1 and Minkowski’s inequality for
q p
≥ 1.
Since
∞ X j=1
1
p
p
|λj |
. kf kFps1 ,q ,
it is sufficient to show the supremum in (4.3) is bounded by a constant. Let Q0 be any dyadic cubes with side length 2−µ and rQ be an ∞-atome for fps1,q with Q0 and define RQ0 (x) =
X
rQ ϑQ (x).
Q∈D,Q⊂Q0 l(Q)≤1
Then we obtain the desired result by showing
(4.4)
!1
∞ q X
s2 k q
(2 |Tbk RQ0 |)
k=3
Lp
. 1 uniformly in Q0 .
16
BAE JUN PARK
, but this is true when p ≤ q < ∞ because
!1 !1
X
∞ q p X
∞ s k
p q s2 kp 2
(2 |T R |) ≤ 2 kT R k bk Q0 bk Q0 Lp
k=3
p k=3 L !1 ∞ p X p s1 kp 2 kφk ∗ RQ0 kLp . k=3
= kRQ0 kFps1 ,p
1
p
X
s1 1
(|rQ ||Q|− 2 − d χQ )p .
Q∈D,Q⊂Q0
p l(Q)≤1 L 1 p ∞ X X s1 1 p − − −kd (|rQ ||Q| 2 d ) 2 ≤ Q∈Dk ,Q⊂Q0
k=µ
− dq
. l(Q0 ) . 1
∞ X
1
p
− pdk q
2
k=µ
by lp ⊂ lq , (3.8), Lemma 2.5, H¨older’s inequality, and (4.2). Now assume q = ∞. Then by (4.2), we have 1
s1
1
|rQ | ≤ |Q| 2 + d |Q0 |− p .
(4.5)
Since l2 ⊂ l∞ , we need to show (4.6)
!1
X
2
∞ sk
2 2
(2 |T R |) Q b 0 k
k=2
. 1. Lp
with (4.5), but when we define
RQ0 ,k (x) =
X
rQ ϑQ (x),
Q∈Dk ,Q⊂Q0 l(Q)≤1
by the support rule it suffices to prove (4.6) with Tbk RQ0 replaced by Tbk RQ0 ,k . We consider first the case l(Q0 ) < 1. Let (4.7)
ν =ρ−
1 . µ
fν with side For Q ∈ D, denote by Qν a dilate of Q with side 2l(Q)ν and by Q
√
d24 l(Q)ν .
FUNCTION SPACES
Since RQ0 ,k = 0 for k < µ,
!1
∞ 2
X sk 2 2
(2 |T R |) b Q ,k 0 k
p
k=3 L
1
∞
2
X
s2 k 2 . (2 |Tbk RQ0 ,k |)
k=µ
p
17
1
∞ 2
X
+ (2s2 k |Tbk RQ0 ,k |)2
k=µ
fν ) L (Q 0
The former part is ,by H¨older’s inequality, less than 1 2 ∞ X 1 − 21 2 s2 k ν f p |Q0 | (2 kTbk RQ0 ,k kL2 ) k=µ
and by (3.8), it is controlled by 1 2 ∞ X 1 1 1 1 k(s1 −d(1−ρ)( p − 2 )) 2 ν p−2 (2 kRQ0 ,k kL2 ) |Q0 | . k=µ
However, by Lemma 2.5 and (4.5),
1
2
X 1
kRQ0 ,k kL2 . (|rQ ||Q|− 2 χQ )2
Q∈Dk ,Q⊂Q0 µd p1 − 12 −s1 k
and this gives
≤ 2
L2
2
1
2 ∞
X
(2s2 k |Tbk RQ0 ,k |)2
k=µ
fν ) Lp (Q 0
µd(ρ−ν) p1 − 12
. 2
. 1
due to (4.7). For the latter one we split it into
1
∞
2
X
(2s2 k |Tbk (χQν0 RQ0 ,k )|)2 (4.8)
k=µ
c
fν ) Lp (Q 0
and
1
2 ∞
X
s2 k 2 c (2 |Tbk (χ(Qν0 ) RQ0 ,k )|)
k=µ
(4.9)
fν c ) Lp (Q 0
and our claim is that each part can be bounded by a constant.
By integration by parts, the kernel of Tbk has the estimate, for any L > 0, Z 1 2πi<x−y,ξ> k(d+m−ρL) . b (x, ξ)e dξ k d .L 2 |x − y|L R
fν c ) Lp (Q 0
18
BAE JUN PARK
fν c , Thus, for x ∈ Q 0
X
|Tbk (χQν0 RQ0 ,k )(x)| ≤
Q∈Dk ,Q⊂Q0 k(d+m−ρL)
. 2
|rQ | Tbk (χQν0 ϑQ )(x) X
Q∈Dk ,Q⊂Q0 1
|rQ |
. 2k(m−s1 +2d−ρL) 2µd( p −1−ν)
Z
Qν0
|ϑQ (y)| dy |x − y|L
1 . |x − cQ0 |L
The last inequality is from (4.5) and the fact that |x − y| & |x − cQ0 | in the integral. Now, by using H¨older’s inequality,
(4.8) =
Z
fν c Q 0
∞ X
k=µ
2
1
p
22s2 k |Tbk (χQν0 RQ0 ,k )(x)|2 dx
µνN −µνd p1 − 21
. 2
p
∞ X
2s2 k
2
Z
2N
fν c Q 0
k=µ
|x − cQ0 |
2
|Tbk (χQν0 RQ0 ,k )(x)| dx 2
1
2 ∞ X 1 1 µd p1 −1−ν µνN −µνd p − 2 2k(m−s1 +s2 +2d−ρL)
. 2
2
2
2
k=µ
. 2−µL(ρ−ν) 2 . 1
µd (ρ−ν) p
2
3µd 2
2−µνd 2−
Z
µρd 2
for sufficiently large N and L. Moreover, by using lp ⊂ l2 and (3.8),
(4.9) ≤
∞ X
k=µ
1
1 p
p
s1 kp . 2
φk ∗ (χ(Qν0 )c RQ0 ,k ) p
Since l(Q0 ) = 2−µ < 1, Q0 ⊂ Qν0 and therefore |y − xQ | & |y − cQ0 |
L
fν c Q 0
1 |x − cQ0 |2(L−N )
!1 2
FUNCTION SPACES
19
for y ∈ (Qν0 )c and Q ⊂ Q0 . So we have
p
φk ∗ χ(Qν0 )c RQ0 ,k p L p Z Z X |φ (x − y)| |rQ ||ϑQ (y)|dy ≤ k dx Rd
y∈(Qν0 )c
d
. 2−kp(N − 2 )
−kp(N − d2 )
≤ 2
Z
R
d
Q∈Dk Q⊂Q0
Z
y∈(Qν0 )c
p
|φk (x − y)|
Z X |rQ | Q∈Dk Q⊂Q0
Rd
X
|rQ | dy N dx |y − xQ |
Q∈Dk Q⊂Q0
Z
y∈(Qν0 )c
p
|φk (x − y)| dy |y − cQ0 |N
!p
dx
!p |φk (x − y)| ≤ 2 2 dy dx N y∈(Qν0 )c |y − cQ0 | Rd !p Z Z 1 L d 1 (1 + 2k |x − cQ0 |) p |φk (x − y)|dxdy . 2−kp(N +s1 −2d+ p ) 2µdp( p −1) N y∈(Qν0 )c |y − cQ0 | Rd !p Z 1 −kp(N +s1 −2d+ dp ) µdp( 1p −1) kL 2 2 . 2 L dy y∈(Qν0 )c |y − cQ |N − p 0 1 −kp(N +s1 −d) µdp( p −1)
Z
Z
−kp(N +s1 −2d+ pd ) µdp( 1p −1) kL µνp(N − L −d) p
. 2 for
L 1−p
2
> d and N −
L p
2 2
> d and so L
(4.9) . 2−µ(1−ν)(N − p −d) . 1. When l(Q0 ) ≥ 1, it still holds in a similar way, but is simpler. However, in this case, we cannot use the above argument to get (4.9) . 1 because Q0 may not contained in Qν0 . Instead of the usage of Qν0 in the decomposition, we use a dilation of Q by a factor of some constants independent of µ. Moreover, note that since µ ≤ 0, we may use the range k ≥ 3 in our summations. Since the proof is derived in the almost same way and easier way, we omit it here. This completes the proof of (2) in Theorem 1.3. We now assume 1 < p < ∞, q is between p and p′ , and 1 m − s1 + s2 = −d(1 − ρ) − 2 We proved for any 1 ≤ r ≤ ∞,
Ta : F1s1 ,r → F1s2 ,r
if
1 . p
d m − s1 + s2 ≤ − (1 − ρ), 2
and already know Ta : L2s1 → L2s2
if
m − s1 + s2 ≤ 0.
20
BAE JUN PARK
By applying the complex interpolation theorem in Section 7, we have 1 1 s1 ,q s2 ,q − (4.10) Ta : Fp → Fp if m − s1 + s2 ≤ −d(1 − ρ) p 2 for 1 ≤ p ≤ 2 and p ≤ q ≤ p′ .
m , the case 2 < p < ∞ can be derived Since the adjoint operator (Ta )∗ is also in OpSρ,δ via duality (Lemma 2.2) and this completes the proof of Theorem 1.3.
Theorem 1.4 is proved in a similar way. By repeating the process in Section 3, (3.1) and (3.2) hold if F-spaces are replaced by B-spaces, and the boundedness of T (3) on B-spaces follows just from (3.8). We omit the details. 5. Sharpness of Theorem 1.3 In this section, when 0 < ρ < 1, we shall prove the necessity of 1 1 m − s1 + s2 ≤ −d(1 − ρ) − 2 p
and when the equality holds, the necessity of the condition on q in Theorem 1.3. Recall Theorem 1.1 does not hold without the assumption (1.2). When 0 < ρ < 1, (1.1) is one example of Pseudo-differential operator that is not bounded in hp for 0 < p < ∞ if 1 1 m > −d(1 − ρ) − . 2 p
See [6], [13], [24] for 1 < p < ∞ and [17] for 0 < p ≤ 1. Once we have a counter example for the necessity of the condition (1.2), it is easy to show the necessity of (1.5) for Corollary 1.2. Suppose (5.1)
1 1 m − s1 + s2 > −d(1 − ρ) − . 2 p
m−s1 +s2 Then there exist a ∈ Sρ,δ and f ∈ Fp0,2 such that
kf kFp0,2 . 1, but kTa f kFp0,2 = ∞. Now put Tb = n−s2 (D)Ta ns1 (D) and g(x) = n−s1 (D)f (x). Then it is clear that m , b ∈ Sρ,δ
kgkF s1 ,2 = kf kFp0,2 . 1, p
and kTb gkF s2 ,2 = kTa f kFp0,2 = ∞. p
FUNCTION SPACES
21
This is also true for Theorem 1.3. If (5.1) holds, there exists ǫ > 0 such that 1 1 m − s1 + s2 − ǫ > −d(1 − ρ) − . 2 p
m and f ∈ F s1 ,2 such that If 2 < q, choose a ∈ Sρ,δ p
kTa f kF s2 −ǫ,2 = ∞. p
Fps1 ,2
Then by ֒→ If q ≤ 2, choose a
Fps1 ,q and Fps2 ,q m and f ∈ ∈ Sρ,δ
֒→ Fps2 −ǫ,2 , it is done. Fps1 +ǫ,2 such that kTa f kF s2 ,2 = ∞. p
Then by
Fps1 +ǫ,2
֒→
Fps1 ,q
and
Fps2 ,q
֒→
Fps2 ,2 ,
it is done.
Now assume
1 1 m − s1 + s2 = −d(1 − ρ) − 2 p and consider the condition of q. We may assume s1 = s2 . Christ and Seeger [4] show that (1.1) is unbounded in Fps,q if 0 < q < p ≤ 2 with 1 1 m = −d(1 − ρ) − . p 2 Theorem 5.1. [4] Let 0 < q < p ≤ 2, s ∈ R, 0 < ρ < 1, m = −d(1 − ρ) p1 − 21 . Then, for r ≥ 2, 1
sup {kcm,ρ (D)f kFps,q : kf kFps,q ≤ 1, f ∈ E(r)} ≈ (log r) q
− p1
.
where E(r) is the space of all (smooth) distributions on Rd whose Fourier transforms are supported in {ξ : |ξ| ≤ r}. Theorem 5.1 also yields ,via duality, the necessity of the condition q ≤ p for the boundedness of (1.1) when p ≥ 2. Suppose 2 ≤ p < q and 1 1 1 1 − − = −d(1 − ρ) . m = −d(1 − ρ) 2 p p′ 2 The upper boundedness
1
sup {kcm,ρ (D)f kFps,q : kf kFps,q ≤ 1, f ∈ E(r)} . (log r) p
− q1
follows immediately from the duality. ′ such that f ∈ E(r), kf kF −s,q′ = 1, According to Theorem 5.1, there exists f ∈ Fp−s,q ′ p′
and
1
kCm,ρ (D)f kF −s,q′ & (logr) q′
− p1′
.
p′
There exists g ∈ Fps,q such that (5.2)
1
< Cm,ρ (D)f, g >& (log r) q′
− p1′
Choose K ∈ E(2r) such that K = 1 in the support of fb. Then
Cm,ρ (D)(g ∗ K) s,q ≥ | < Cmρ (D)(g ∗ K), f > | F p
= < Cm,ρ (D)f, g > 1
1
& (log r) p − q
22
BAE JUN PARK
and it is clear that kg ∗ KkFps,q . 1.
Thus, we conclude that
1
sup {kCm,ρ (D)f kFps,q : kf kFps,q ≤ 1, f ∈ E(r)} ≈ (log r) p
− 1q
for 2 ≤ p < q and large r > 0. Remark. It remains to show the necessity of p′ ≤ q when 2 ≤ p (and via duality, q ≤ p′ when p ≤ 2), however we do not know this works with the same exmaple (1.1). Another part we need to show in order to complete the sharpness is to show it is also true for the case ρ = 0. Unfortunately, Miyachi [17] pointed out that cm,0 (D) is bounded in hp if and only if 1 1 m ≤ −(d − 1) − . 2 p Thus, we cannot use cm,0 (D) for the sharpness.
6. An application Finally we shall give a short discussion on the continuity of Ta mapping Fps11 ,q1 into Fps22 ,q2 . H¨ormander [14] gave a question that for p1 ≤ p2 , the estimate kTa f kLp2 . kf kLp1 holds if m ≤ −d
1 1 1 1 1 1 − + (1 − ρ) max ( − , − , 0) p1 p2 2 p1 p2 2
and Alvarez and Hounie [1] answered the question.. Theorem 6.1. [1] Let 0 ≤ δ ≤ ρ < 1, 0 < ρ, 1 < p1 ≤ p2 < ∞, and m ∈ R. Suppose m . Then T maps Lp1 into Lp2 in the following cases: a ∈ Sρ,δ a (1) if 1 < p1 ≤ 2 ≤ p2 and
m ≤ −d (2) if 2 ≤ p1 ≤ p2 and m ≤ −d (3) if p1 ≤ p2 ≤ 2 and m ≤ −d
1 1 − p1 p2
;
1 1 1 1 − + (1 − ρ)( − ) ; p1 p2 2 p1
1 1 1 1 − + (1 − ρ)( − ) . p1 p2 p2 2
Our results give the generalization of the above results. Theorem 6.2. Let 0 ≤ δ ≤ ρ < 1, 0 < p1 ≤ p2 < ∞, 0 < q1 , q2 ≤ ∞, and s1 , s2 , m ∈ R. m . Then T maps F s1 ,q1 into F s2 ,q2 in the following cases: Suppose a ∈ Sρ,δ p1 p2 a (1) if 0 < p1 = p2 = p < ∞, 0 < q1 = q2 = q ≤ ∞ satisfies the assumptions of q in Theorem 1.3, and 1 1 m − s1 + s2 ≤ −d(1 − ρ) − ; 2 p
FUNCTION SPACES
(2) if p1 < 2 < p2 , 0 < q1 , q2 ≤ ∞, and
m − s1 + s2 ≤ −d (3) if p1 < p2 = 2, 0 < q1 ≤ ∞, q2 = 2, and
m − s1 + s2 ≤ −d (4) if p1 = 2 < p2 , q1 = 2, 0 < q2 ≤ ∞, and m − s1 + s2 ≤ −d
23
1 1 − p1 p2
;
1 1 − p1 2
;
1 1 − 2 p2
;
(5) if 2 < p1 < p2 , p′1 ≤ q1 ≤ p1 , 0 < q2 ≤ ∞, and 1 1 1 1 − + (1 − ρ) − m − s1 + s2 ≤ −d ; p1 p2 2 p1 (6) if p1 < p2 < 2, 0 < q1 ≤ ∞, p2 ≤ q2 ≤ p′2 , and 1 1 1 1 m − s1 + s2 ≤ −d − + (1 − ρ) − . p1 p2 p2 2
Proof. (1) is from Theorem 1.3 and let us prove other cases. It suffices to prove the endpoint results. (2) Suppose 0 < p1 < 2 < p2 . Then kTa f kFps2 ,q2 2
. kTa f k . kf k
)
(
1 − 1 ,2 s2 +d 2 p2
F2
(
)
s1 −d p1 − 1 ,2 1 2
F2
. kf kFps1 ,q1 1
by Lemma 2.1 and Theorem 1.3. (3) Suppose p1 < p2 = 2 and q2 = 2. Then kTa f kF s2 ,2 . kf k 2
s1 −d( p1 − 1 ),2 1 2
F2
. kf kFps1 ,q1 1
by Lemma 2.1 and Theorem 1.3. (4) Suppose p1 = 2 < p2 , q1 = 2. Then kTa f kFps2 ,q2 2
. kTa f k
1 − 1 ),2 s2 +d( 2 p2
F2
. kf kF s1 ,2 2
by Lemma 2.1 and Theorem 1.3. (5) Suppose 2 < p1 < p2 and p′1 ≤ q1 ≤ p′1 . Then kTa f kFps2 ,q2 2
. kTa f k
s2 +d( p1 − p1 ),q1 1 2
F p1
. kf kFps1 ,q1 1
by Lemma 2.1 and Theorem 1.3.
24
BAE JUN PARK
(6) Suppose p1 < p2 < 2 and p2 ≤ q2 ≤ p′2 . Then kTa f kFps2 ,q2
. kf k
2
s1 −d( p1 − p1 ),q2 1 2
F p2
. kf kFps1 ,q1 1
by Lemma 2.1 and Theorem 1.3. Theorem 6.3. Let 0 ≤ δ ≤ ρ < 1, 0 < p1 < p2 < ∞, 0 < q1 , q2 ≤ ∞, and s, m ∈ R. m . Then T maps F s1 ,q1 into F s2 ,q2 in the following cases: Suppose a ∈ Sρ,δ p1 p2 a (1) if p1 ≤ 2 ≤ p2 and
1 1 − p1 p2
;
1 1 − + (1 − ρ) p1 p2
1 1 − 2 p1
m − s1 + s2 < −d (2) if 2 ≤ p1 ≤ p2 and m − s1 + s2 < −d (3) if p1 ≤ p2 ≤ 2 and
;
1 1 1 1 m − s1 + s2 < −d − + (1 − ρ) − . p1 p2 p2 2 1 1 − . Proof. (1) Assume p1 ≤ 2 ≤ p2 and m − s1 + s2 < −d p1 p2 If 0 < q2 < 2, then for any ǫ > 0, kTa f kFps2 ,q2 .ǫ kTa f kLp2
ǫ+s2
2
by H¨older’s inequality.
1 1 − . Then we have p1 p 1 1 − . m + s2 − s < −d p p2
Pick p ∈ (p1 , p2 ) and let s = s1 − d
Now choose ǫ > 0 so small that
m + s2 + ǫ − s < −d and apply Theorem 6.2 to get kTa f kLp2
ǫ+s2
Since s −
1 1 − p p2
. kf kLps .
d d = s1 − and p1 < p, we have p p1 kf kLps . kf kFps1 ,q1 1
by Lemma 2.1. If 2 < q2 < ∞, then
kTa f kFps2 ,q2 .ǫ kTa f kLps 2 2
by
l2
2
l q2 .
⊂ Choose ǫ > 0 so small that m − s1 + s2 + ǫ < −d
1 1 − p1 p2
.
FUNCTION SPACES
25
1 1 Pick p ∈ (p1 , p2 ) and let s = s1 − ǫ − d − . Then we have p1 p 1 1 − . m + s2 − s < −d p p2 By Theorem 6.2, kTa f kLps 2 . kf |Lps . 2
Since s −
d d = s1 − ǫ − and p1 < p, we have p p1 kf kLps . kf kF s1 −ǫ,q1 ≤ kf kFps1 ,q1 p1
1
by Lemma 2.1. By using the same technique, (2) and (3) can be proved. By using Theorem 1.4 and Lemma 2.1, we can get analogous conclusions on B-spaces. Theorem 6.4. Let 0 ≤ δ ≤ ρ < 1, 0 < p1 ≤ p2 < ∞, 0 < q ≤ ∞, and s, m ∈ R. Suppose m . Then T maps B s1 ,q into B s2 ,q in the following cases: a ∈ Sρ,δ p1 p2 a (1) if p1 ≤ 2 ≤ p2 and
1 1 − p1 p2
:
1 1 − + (1 − ρ) p1 p2
1 1 − 2 p1
:
1 1 − + (1 − ρ) p1 p2
1 1 − p2 2
.
m − s1 + s2 ≤ −d (2) if 2 ≤ p1 ≤ p2 and m − s1 + s2 ≤ −d (3) if p1 ≤ p2 ≤ 2 and m − s1 + s2 ≤ −d
Theorem 6.5. Let 0 ≤ δ ≤ ρ < 1, 0 < p1 < p2 < ∞, 0 < q1 , q2 ≤ ∞, and s, m ∈ R. m . Then T maps B s1 ,q1 into B s2 ,q2 in the following cases: Suppose a ∈ Sρ,δ p1 p2 a (1) if p1 ≤ 2 ≤ p2 and
1 1 − p1 p2
;
1 1 − + (1 − ρ) p1 p2
1 1 − 2 p1
;
1 1 − + (1 − ρ) p1 p2
1 1 − p2 2
.
m − s1 + s2 < −d (2) if 2 ≤ p1 < p2 and m − s1 + s2 < −d (3) if p1 < p2 ≤ 2 and m − s1 + s2 < −d
26
BAE JUN PARK
7. Appendix - Complex Interpolation theorem for F- spaces This section deals with a complex interpolation method for F-spaces. Let S = {z ∈ C : 0 < Re(z) < 1} be a strip in the complex plane and S = {z ∈ C : 0 ≤ Re(z) ≤ 1} be its closure. We say that f is an S ′ -analytic function in S if the following properties are satisfied: (a) For any fixed z ∈ S, we have f (z) ∈ S ′ ; (b) For any ϕ ∈ S with compact support in Rd , h(x, z) = (ϕfd (z))∨ (x) is uniformly d continuous and bounded function in R × S; (c) For any ϕ ∈ S with compact support in Rd and any fixed x ∈ Rd , h(x, ·) = (ϕfd (·))∨ (x) is an analytic function in S. Let s0 , s1 ∈ R, 0 < p0 , p1 < ∞, and 0 < q0 , q1 ≤ ∞. Suppose that for any t ∈ R, f (it) ∈ Fps00 ,q0 and f (1 + it) ∈ Fps11 ,q1 . We introduce the norm kf kF (Fps0 ,q0 ,Fps1 ,q1 ) = max {sup kf (it)kFps0 ,q0 , sup kf (1 + it)kFps1 ,q1 } 0
1
0
t
1
t
and then F(Fps00 ,q0 , Fps11 ,q1 ) becomes a quasi-Banach space(Banach space if 1 ≤ p0 , p1 , q0 , q1 ). And define a subspace Fps00 ,q0 + Fps11 ,q1 of S ′ with the norm kvkFps0 ,q0 +Fps1 ,q1 = 0
1
inf
v=v0 +v1
(kv0 kFps0 ,q0 + kv1 kFps1 ,q1 ). 0
1
Then F(Fps11 ,q1 , Fps11 ,q1 ) ֒→ Fps00 ,q0 + Fps11 ,q1 ֒→ S ′ .
Definition 2. Fix 0 ≤ θ ≤ 1. The complex interpolation space of order θ between Fps00 ,q0 and Fps11 ,q1 is the normed linear subspace (Fps00 ,q0 , Fps11 ,q1 )θ = {v ∈ Fps00 ,q0 + Fps11 ,q1 : v = f (θ) for some f ∈ F(Fps00 ,q0 , Fps11 ,q1 )}
of Fps00 ,q0 + Fps11 ,q1 , with the norm
kvk(Fps0 ,q0 ,Fps1 ,q1 )θ = 0
1
inf
s ,q
s ,q1 )
f ∈F (Fp00 0 ,Fp11 f (θ)=v
kf kF (Fps0 ,q0 ,Fps1 ,q1 ) . 0
1
Theorem 7.1. [22] Let s0 , s1 ∈ R, 0 < q0 , q1 ≤ ∞, and 0 < p0 , p1 < ∞. For fixed 0 < θ < 1, assume s = (1 − θ)s0 + θs1 ,
1 1−θ 1−θ θ 1 θ = = + , + . p p0 p1 q q0 q1
Then (Fps00 ,q0 , Fps11 ,q1 )θ = Fps,q . Lemma 7.2. [21], [22] Let s ∈ R, 0 < p < ∞, 0 < q ≤ ∞, and τ > d2 + all a ∈ L∞ and f ∈ Fps,q ,
ka(D)f kFps,q ≤ c(kγ1 akL2τ + sup γ2 a(2k ·) 2 ) kf kFps,q 0≤k
where γ1 , γ2 ∈ S with
Lτ
0 ≤ γ1 (ξ), γ2 (ξ) ≤ 1 for all ξ,
Supp(γ1 ) ⊂ {ξ : |ξ| ≤ 4}, γ1 (ξ) = 1 if |ξ| ≤ 2,
Supp(γ2 ) ⊂ {ξ : γ2 (ξ) = 1 if
d min (p,q) .
1 ≤ |ξ| ≤ 4}, 4
1 ≤ |ξ| ≤ 2. 2
Then for
FUNCTION SPACES
27
Theorem 7.3. (Complex interpolation theorem for F -spaces) Let m, s1 , s2 ∈ R, 0 ≤ δ ≤ ρ < 1, ρ > 0, 1 ≤ q0 , q1 ≤ ∞, and 1 ≤ p0 , p1 ≤ 2. For fixed 0 < θ < 1, let 1−θ 1−θ θ 1 θ 1 = = + , + q q0 q1 p p0 p1 m and assume a ∈ Sρ,δ . Suppose that 1 1 s ,q s ,q Ta : Fpj1 j → Fpj2 j if m − s1 + s2 = −d(1 − ρ) − pj 2 for each j = 0, 1. Then Ta :
Fps1 ,q
→
Fps2 ,q
if
m − s1 + s2 = −d(1 − ρ)
Proof. Assume m − s1 + s2 = −d(1 − ρ) By Theorem 7.1, it suffices to show
1 p
1 2
−
1 1 − p 2
.
m. and a ∈ Sρ,δ
Ta : (Fps01 ,q0 , Fps11 q1 )θ → (Fps02 ,q0 , Fps12 ,q1 )θ . For z ∈ S, define and
d(1−ρ)(( p1 − p1 )z+( p1 − p1 ))
az (x, ξ) = a(x, ξ)(1 + |ξ|)
0
1
0
Tz = Taz . Then Tθ = Ta . By the assumption, T0 : Fps01 ,q0 → Fps02 ,q0
and
T1 : Fps11 ,q1 → Fps12 ,q1 . Let k0 and k1 be the constants for the boundedness of T0 and T1 , respectively. Let v ∈ (Fps01 ,q0 , Fps11 ,q1 )θ and ǫ > 0. By the definition of norm of interpolation space, there exists f ∈ F(Fps01 ,q0 , Fps11 ,q1 ) such that f (θ) = v
and kf kF (Fps1 ,q0 ,Fps1 ,q1 ) . kvk(Fps1 ,q0 ,Fps1,q1 )θ + ǫ. 0
Put
1
0
g(z) =
1
k0z−1 k1−z Tz (f (z)). (1 + z)2d
Then our claim is that g ∈ F(Fps02 ,q0 , Fps12 ,q1 )
and (7.1)
kgkF (Fps2 ,q0 ,Fps2 ,q1 ) . kvk(Fps1 ,q0 ,Fps1 ,q1 )θ + ǫ. 0
1
0
1
Observe that kTit kFps1 ,q0 →Fps2 ,q0 . |t|2d k0 0
and
0
kT1+it kFps1 ,q1 →Fps2 ,q1 . |t|2d k1 1
1
28
BAE JUN PARK
by Lemma 7.2 because
itd(1−ρ)( p1 − p1 )
0 1
(1 + |D|)
s ,qj
Fpj1
s ,qj
→Fpj1
. |t|2d
for each j = 0, 1. Another observation is that g is a S ′ -analytic function in S. Thus, we shall prove (7.1). k0−1 kTit (f (it))kFps2 ,q0 0 |1 + it|2d . kf (it)kFps1 ,q0
kg(it)kFps2 ,q0
=
0
0
and similarly, kg(1 + it)kFps2 ,q1
. kf (1 + it)kFps1 ,q1 .
1
1
Thus, kgkF (Fps2 ,q0 ,Fps2 ,q1 ) = max {sup kg(it)kFps2 ,q0 , sup kg(1 + it)kFps2 ,q1 } 0
1
0
t
1
t
. kf kF (Fps1 ,q0 ,Fps1 ,q1 ) 0
1
≤ kvk(Fps1 ,q0 ,Fps1 ,q1 )θ + ǫ 0
1
and this implies that = (1 + θ)2d k01−θ k1θ kg(θ)k(Fps2 ,q0 ,Fps2 ,q1 )θ
kTa (v)k(Fps2 ,q0 ,Fps2 ,q1 )θ 0
1
0
.
k01−θ k1θ
kgk
1
s ,q s ,q F (Fp02 0 ,Fp12 1 )
. k01−θ k1θ kvk(Fps1 ,q0 ,Fps1 ,q1 )θ + ǫ. 0
1
Since ǫ > 0 is arbitrary, we have the inequality kTa (v)k(Fps2 ,q0 ,Fps2 ,q1 )θ . k01−θ k1θ kvk(Fps1 ,q0 ,Fps1 ,q1 )θ . 0
1
0
1
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[14] L. H¨ ormander, Pseudo-Differential Operators and hypoelliptic equations, Proc. Sympos. Pure Math. 10 (1967) 138-183. [15] J. Johnsen, Domains of pseudo-differential operators: a case for the Triebel-Lizorkin spaces, J. Funct. Spaces Appl. 3 (2005) 263-286. [16] J. Marschall, Nonregular pseudo-differential operators, J. Anal. Anwendungen 15(1)(1996) 109-148. [17] A. Miyachi, On some Fourier multipliers for H p (Rd ), J. Fac. Sci. Univ. Tokyo Sect. IA Math. 27(1980) 157-179. [18] L. P¨ aiv¨ arinta and E. Somersalo, A generalization of the Calder´ on-Vaillancourt theorem to Lp and hp , Math. Nachr. 138 (1988) 145-156. [19] J. Peetre, On spaces of Triebel-Lizorkin type, Ark. Mat. 13 (1975) 123-130. [20] E. M. Stein, Singular integrals, harmonic functions, and differentiability properties of functions of several variables, Proc. Sympos. Pure Math., vol. 10 Amer. Math. Soc., Providence, R.I. (1967) 316355. [21] H. Triebel, Complex Interpolation and Fourier Multipliers for the Space Bps,q and Fps,q of Besov-HardySobolev Type:The Case 0 < p ≤ ∞, 0 < q ≤ ∞, Math. Z. 176 (1981) 495-510. [22] H. Triebel, Theory of Function Spaces, Birkhauser, Basel-Boston-Stuttgart (1983). [23] M. Yamazaki, A quasi-homogeneous version of paradifferential operators, I. Boundedness on spaces of Besov type, J. Fac. Sci. Univ. Tokyo Sect. IA, Math., 33(1986) 131-174. [24] S. Wainger, Special trigonometric series in k dimensions, Memoirs of the AMS., 59(1965). Department of Mathematics, University of Wisconsin - Madison, , U.S.A. E-mail address: [email protected]
