The inclusion relations between α-modulation spaces and Lp-Sobolev spaces or local Hardy spaces

Journal of Functional Analysis 272 (2017) 1340–1405

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Journal of Functional Analysis www.elsevier.com/locate/jfa

The inclusion relations between α-modulation spaces and Lp -Sobolev spaces or local Hardy spaces Tomoya Kato Graduate School of Mathematics, Nagoya University, Furocho, Chikusa-ku, Nagoya 464-8602, Japan

a r t i c l e

i n f o

Article history: Received 16 July 2015 Accepted 13 December 2016 Available online 19 December 2016 Communicated by L. Gross Keywords: Inclusion relations α-Modulation spaces Lp -Sobolev spaces Local Hardy spaces

a b s t r a c t In this paper, we first discuss equivalent norms for α-modulation spaces which are composed from decomposition with noncompact (frequency) support. Then, we determine sharp inclusion relations between α-modulation spaces and Lp -Sobolev spaces, and between α-modulation spaces and local Hardy spaces. © 2016 Elsevier Inc. All rights reserved.

1. Introduction and main theorems 1.1. Introduction In this paper, we determine optimal embedding theorems between α-modulation spaces and Lp -Sobolev spaces or local Hardy spaces. α-Modulation spaces were introduced by Gröbner in his PhD thesis [7] as intermediate spaces between Besov and modulation spaces. The “mixture” of these function spaces is controlled by the params s eter 0 ≤ α ≤ 1. Besov spaces Bp,q and modulation spaces Mp,q are constituted by E-mail address: [email protected]. http://dx.doi.org/10.1016/j.jfa.2016.12.002 0022-1236/© 2016 Elsevier Inc. All rights reserved.

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decomposition of the frequency components, which are called dyadic decomposition φj (j ∈ N ∪ {0}) and frequency uniform decomposition σk (k ∈ Zn ), respectively. Their norms are denoted by 



f B s = p,q

2

jsq

  −1 F φj Ff q p

1/q ,

L

j≥0

where supp φj ⊂ {ξ : 2j−1 ≤ |ξ| ≤ 2j+1 } and supp φ0 ⊂ {ξ : |ξ| ≤ 2}, and  f M s =



p,q

q  k F −1 σk Ff Lp

1/q

sq

,

k∈Zn

where the supports of σk are contained in k + [−1, 1]n and · = (1 + | · |2 )1/2 (see Feichtinger [4]). Here, F −1 φj Ff and F −1 σk Ff correspond with F −1 [φj · Ff ] and s,α F −1 [σk · Ff ], respectively. Now, α-modulation spaces Mp,q are built from decompo  α n α/(1−α) α/(1−α) k| ≤ Ck sition ηk (k ∈ Z ) supported in ξ : |ξ − k and their norms are denoted by  s,α = f Mp,q



k

sq/(1−α)

 −1 α q F ηk Ff  p

1/q ,

L

k∈Zn

where α ∈ [0, 1) and F −1 ηkα Ff = F −1 [ηkα · Ff ]. Modulation spaces are the special case when α = 0 and Besov spaces can be regarded as the limit case when α → 1. For the α-modulation spaces, the distance of the support of ηkα to the origin is essentially 1

dk,ηkα := k 1−α ∼

sup α ξ∈supp ηk

ξ ∼

inf

α ξ∈supp ηk

ξ.

Here, a ∼ b means that cb ≤ a ≤ Cb for some constants 0 < c, C < ∞. With the above convention, the volumes of the supports satisfy  

supp ηkα 1/n ∼ dk,ηα α , k i.e., they grow like the α-th power of the distance to the origin. Likewise, for modulation spaces, the distance of the support of σk to the origin is essentially dk,σk := k and  1/n

0 the volumes of the supports satisfy supp σk  ∼ 1 ∼ dk,σ . On the other hand, k

for Besov spaces, the distance dj,φj (= 2j ) of the support of φj to the origin satisfies  

supp φj 1/n ∼ dj,φj 1 . In this sense, the decomposition used to define the α-modulation spaces is intermediate between those used to define modulation and Besov spaces. Fi(s) nally, the weight wk,ηα := ks/(1−α) which controls the “regularity” of the space satisfies k

s (s) wk,ηα ∼ dk,ηkα . This is again the same relationship as for modulation and Besov spaces. k

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Thus, the definition of the α-modulation space norm is quite natural and is similar to those of the other two function spaces. More precise definitions of these three function spaces will be given in Section 2. α-Modulation spaces have been studied in recent works. Gröbner [7], Han and Wang [9], and Toft and Wahlberg [21] give us the embedding theorems between α1 -modulation spaces and α2 -modulation spaces. In [9] and [21], the inclusions between Besov and α-modulation spaces are also given. As applications of α-modulation spaces, Borup and Nielsen [2] showed that pseudo-differential operators in Hörmander class are bounded on α-modulation spaces. Moreover, Kobayashi, Sugimoto and Tomita [15,16] studied Lp boundedness of pseudo-differential operators with symbols in α-modulation spaces, where 1 ≤ p ≤ 2. For the other operators, Feichtinger, Huang, and Wang analyze the behavior of trace-operators on α-modulation spaces in [5]. α-Modulation spaces are quite recently applied to the field of partial differential equations. For instance, by Misiołek and Yoneda [17], the local ill-posedness of the Euler equations in the frame of α-modulation spaces is proved. 1.2. Main theorems Here, we state our main results. In order to prove inclusion relations, we will make full use of the following equivalent norm for α-modulation spaces. Let a function Ψ ∈ S satisfy that |Ψ(ξ)| ≥ c > 0 on |ξ| ≤ 2. Then  s,α ∼ f Mp,q



k

sq/(1−α)

 −1 α q F Ψk Ff  p L

1/q

 , where Ψα k (ξ) := Ψ

k∈Zn

ξ − kα/(1−α) k Ckα/(1−α)

.

−1 Here, the Fourier multiplier F −1 Ψα [Ψα k Ff = F k · Ff ]. This above fact is not trivial α because, as |k| gets bigger, the support of Ψk gets wider, so that the overlap between the different (frequency) supports may get larger. However, a close investigation of the frequency components arising from Ψα k enables us to show that the resulting norm is indeed equivalent to the usual norm (see Lemma 3.2). In this equivalent norm, we can take Ψ such that F −1 Ψ has a compact support since the assumption of the decomposition Ψ is only that Ψ is a Schwartz function and never vanishes at the origin. This fact plays important roles in proofs of our main theorems. For instance, if a function f has a compact support, then by the Hölder inequality (f Lr1 ≤ Cr1 ,r2 ,|supp f | · f Lr2 for r1 ≤ r2 ), we are able to change the index 0 < p ≤ ∞ to a bigger one (see the proof of Lemma 3.3). Now, we state our main theorems. Before that, we define the indexes ν1 (p, q) and ν2 (p, q) which we will use in the following argument. For 0 < p, q ≤ ∞,

⎧ ⎪ ⎨0 ν1 (p, q) = 1/p + 1/q − 1 ⎪ ⎩ −1/p + 1/q

if (1/p, 1/q) ∈ I1∗ , if (1/p, 1/q) ∈ I2∗ , if (1/p, 1/q) ∈ I3∗ ,

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⎧ ⎪ if (1/p, 1/q) ∈ I1 , ⎨0 ν2 (p, q) = 1/p + 1/q − 1 if (1/p, 1/q) ∈ I2 , ⎪ ⎩ −1/p + 1/q if (1/p, 1/q) ∈ I3 . Here,   I1∗ = (1/p, 1/q) ∈ [0, ∞)2 : 1/q ≤ 1/p and 1/q ≤ 1 − 1/p ,   I2∗ = (1/p, 1/q) ∈ [0, ∞)2 : 1/p ≥ 1/2 and 1/q ≥ 1 − 1/p ,   I3∗ = (1/p, 1/q) ∈ [0, ∞)2 : 1/p ≤ 1/2 and 1/q ≥ 1/p , and   I1 = (1/p, 1/q) ∈ [0, ∞)2 : 1/q ≥ 1/p and 1/q ≥ 1 − 1/p ,   I2 = (1/p, 1/q) ∈ [0, ∞)2 : 1/p ≤ 1/2 and 1/q ≤ 1 − 1/p ,   I3 = (1/p, 1/q) ∈ [0, ∞)2 : 1/p ≥ 1/2 and 1/q ≤ 1/p . We note that ν1 (p, q) = −ν2 (p , q  ) for all 1 ≤ p, q ≤ ∞, and ν1 (p, q) ≥ 0 and ν2 (p, q) ≤ 0 for all 0 < p, q ≤ ∞. Here, 1/p + 1/p = 1/q + 1/q  = 1 for 1 ≤ p, q ≤ ∞. 1/q

1/q

1

1

I3∗

I1

I2∗ (1/2,1/2)

(1/2,1/2)

I1∗

0

I2

1

0

1/p

I3

1

1/p

The index sets of ν1 (p, q) and ν2 (p, q) Then the following statements are already known. Theorem 1.1. (See [9, Theorem 4.2].) Let 0 < p, q ≤ ∞, s1 , s2 ∈ R, and 0 ≤ α < 1. Then s1 s2 ,α (1) Bp,q ⊂ Mp,q holds if and only if s1 ≥ s2 + n(1 − α)ν1 (p, q) is satisfied; s1 ,α s2 (2) Mp,q ⊂ Bp,q holds if and only if s1 ≥ s2 + n(α − 1)ν2 (p, q) is satisfied. This theorem coincides with the inclusion relations between Besov and modulation s spaces if α = 0 (see [7,18,20,24]). Since we have the inclusion relations Lps+ε → Bp,q →

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Lps−ε for ε > 0 (see [23, Remark 3 in Section 2.3.2] or Remark 2.7), the following statements are given immediately. Corollary 1.2. Let 1 ≤ p, q ≤ ∞, s1 , s2 ∈ R, and 0 ≤ α < 1. Then s2 ,α s2 ,α (1) Lps1 ⊂ Mp,q holds if s1 > s2 +n(1 −α)ν1 (p, q) is satisfied. Conversely, if Lps1 ⊂ Mp,q holds, then s1 ≥ s2 + n(1 − α)ν1 (p, q) is satisfied; s1 ,α s1 ,α ⊂ Lps2 holds if s1 > s2 +n(α−1)ν2 (p, q) is satisfied. Conversely, if Mp,q ⊂ Lps2 (2) Mp,q holds, then s1 ≥ s2 + n(α − 1)ν2 (p, q) is satisfied. In Corollary 1.2, there are differences between the sufficient conditions and the necessary conditions for these inclusion relations. That is, the critical cases s1 = s2 + n(1 − α)ν1 (p, q) and s1 = s2 + n(α − 1)ν2 (p, q) are not included in the sufficient conditions, although they are included in the necessary conditions. Thus, the goal of this paper is to determine whether these critical cases are needed or not. The exact answers are provided by the following theorems. Theorem 1.3. Let 1 ≤ p, q ≤ ∞, s1 , s2 ∈ R, and 0 ≤ α < 1. Then, Lps1 (Rn ) → s2 ,α (Rn ) holds if and only if one of the following conditions is satisfied. Mp,q (1) 1 < p ≤ q and s1 ≥ s2 + n(1 − α)ν1 (p, q); (2) p > q and s1 > s2 + n(1 − α)ν1 (p, q); (3) p = 1, q = ∞, and s1 ≥ s2 + n(1 − α)ν1 (1, ∞); (4) p = 1, q = ∞, and s1 > s2 + n(1 − α)ν1 (1, q). s1 ,α Theorem 1.4. Let 1 ≤ p, q ≤ ∞, s1 , s2 ∈ R, and 0 ≤ α < 1. Then, Mp,q (Rn ) → p n Ls2 (R ) holds if and only if one of the following conditions is satisfied. (1) ∞ > p ≥ q and s1 ≥ s2 + n(α − 1)ν2 (p, q); (2) p < q and s1 > s2 + n(α − 1)ν2 (p, q); (3) p = ∞, q = 1, and s1 ≥ s2 + n(α − 1)ν2 (∞, 1); (4) p = ∞, q = 1, and s1 > s2 + n(α − 1)ν2 (∞, q).

In Theorems 1.3 and 1.4, if we set α = 0, then these two theorems are exactly identical with the embedding relations between modulation and Lp -Sobolev spaces by Kobayashi and Sugimoto [14]. For 0 < p ≤ 1, we have the following inclusion relations between α-modulation spaces and local Hardy spaces. Theorem 1.5. Let 0 < p ≤ 1, 0 < q ≤ ∞, s ∈ R, and 0 ≤ α < 1. Then, hp (Rn ) → s,α Mp,q (Rn ) holds if and only if either of the following conditions is satisfied. (1) p > q and s < −n(1 − α)(1/p + 1/q − 1); (2) p ≤ q and s ≤ −n(1 − α)(1/p + 1/q − 1). s,α Theorem 1.6. Let 0 < p ≤ 1, 0 < q ≤ ∞, s ∈ R, and 0 ≤ α < 1. Then, Mp,q (Rn ) → p n h (R ) holds if and only if either of the following conditions is satisfied.

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(1) p ≥ q and s ≥ 0; (2) p < q and s > n(α − 1)(−1/p + 1/q). We remark that the conditions for s ∈ R given by Theorems 1.5 and 1.6 are precisely that s ≤ −n(1 − α)ν1 (p, q) (with a strict inequality for p > q) and s ≥ n(α − 1)ν2 (p, q) (with a strict inequality for p < q). If we substitute α = 0 into Theorems 1.5 and 1.6, then these theorems are also completely identical with the inclusion relations between modulation and local Hardy spaces by Kobayashi, Miyachi, and Tomita [12]. The organization of this paper is as follows. In Section 2, we will state the basic notations that we will use throughout this paper, and then introduce the definitions and some properties of α-modulation, Besov, Triebel Lizorkin and local Hardy spaces. After stating (in Section 3) some lemmas needed for the proof of the main theorems, we will actually prove the main theorems in Sections 4 and 5. Since we will use Theorems 1.5 and 1.6 to prove Theorems 1.3 and 1.4, we establish Theorems 1.5 and 1.6 in Section 4 ahead. Then, in Section 5, we prove Theorems 1.3 and 1.4. Finally, we give proofs of the lemmas for Theorems 1.3–1.4 and 1.5–1.6 in Section 6. 2. Preliminaries First of all, we remark that we denote A := α/(1 − α) unless otherwise noted in the following statements. 2.1. Basic notations In this first subsection, we collect the notations which will be used throughout this paper. R, N, and Z denote the sets of reals, positive integers, and integers, respectively. Moreover, we write Z+ = N ∪{0}. We assume that the constants c and C satisfy 0 < c < 1 and C > 1. a  b means that a ≤ Cb, and a ∼ b means that a  b and a  b. For 1 ≤ p ≤ ∞, we set p ∈ [1, ∞] as the dual number of p, i.e., 1/p + 1/p = 1. For 0 < p < 1, we set p = ∞. F and · denote the Fourier transform by  Ff (ξ) = f(ξ) := e−iξ·x f (x)dx, Rn

and F −1 and ˇ· denote the inverse Fourier transform by  1 F −1 f (x) = fˇ(x) := eix·ξ f (ξ)dξ. (2π)n Rn

Moreover, we write Fourier multipliers as F −1 gFf = F −1 [g · Ff ] .

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We use some function spaces. The Lebesgue spaces Lp := Lp (Rn ) are equipped with the (quasi)-norm ⎛ f Lp = ⎝



⎞1/p p  f (x) dx⎠

Rn

for 0 < p < ∞. If p = ∞, f ∞ := ess. supx∈Rn |f (x)|. We denote Lp -Sobolev spaces (or Bessel potential spaces) Lps := Lps (Rn ) by      Lps = f ∈ S  : f Lps = F −1 ξs Ff 

Lp

<∞



for 1 ≤ p ≤ ∞. Here, · = (1 + | · |2 )1/2 and S  := S  (Rn ) is the dual space of the Schwartz space S := S(Rn ). For 0 < q ≤ ∞, we denote q by the set of all complex number sequences {ak } such that  ak q =



1/q |ak |

q

< ∞,

k

 1/q q with modification for q = ∞. We will sometimes use the expression k |ak | without mentioning that this should be understood in the usual way for q = ∞. In the end of this subsection, we introduce the following inequality for convolution f ∗ g. Proposition 2.1. (See [22, Section 1.5.3].) Let 0 < p ≤ 1. We set LpΩ (Rn ) = {f ∈ Lp (Rn ) ∩ S  (Rn ) : supp (Ff ) ⊂ Ω} and Ω = {x ∈ Rn : |x − x0 | ≤ R}. Then there exists a constant C = Cp > 0 such that f ∗ gLp ≤ CRn(1/p−1) f Lp · gLp for any f, g ∈ LpΩ . 2.2. α-Modulation spaces We introduce the definition of α-modulation spaces and their properties. α-Modulation spaces were invented by Gröbner [7], as intermediate spaces between Besov spaces and modulation spaces, the mixture of which is controlled by the parameter 0 ≤ α ≤ 1. More precisely, α ∈ [0, 1] determines how the frequency space is decomposed. Before we state the definition of the α-modulation spaces, let us prepare some notations. We set B(x, r) as the open ball with center x ∈ Rn and radius r > 0, and x = (1 + |x|2 )1/2 . A countable set Q of subsets Q ⊂ Rn is called an admissible covering if they satisfy both of the following statements.

T. Kato / Journal of Functional Analysis 272 (2017) 1340–1405

• Rn =



1347

Q,

Q∈Q

• there exists a constant n0 < ∞ such that #{Q ∈ Q : Q ∩ Q = ∅} ≤ n0 for all Q ∈ Q. Moreover, an admissible covering Q is called α-covering (0 ≤ α ≤ 1) if • |Q| ∼ xαn for all Q ∈ Q and all x ∈ Q, • there exists a constant K ≥ 1 such that RQ /rQ ≤ K for all Q ∈ Q,    where RQ = inf R > 0 : Q ⊂ B(cR , R) for some cR ∈ Rn and rQ = sup r > 0 :  B(cr , r) ⊂ Q for some cr ∈ Rn . For the definition of α-modulation spaces, we will need certain partitions of unity, which we now define. Let 0 < p ≤ ∞ and Q be an α-covering of Rn , then we choose a family of functions {ψQ }Q∈Q satisfying • supp  ψQ ⊂ Q, ψQ ≡ 1, • Q∈Q

• sup |Q| Q∈Q

−1+1/ min(1,p)

 −1  F ψQ 

Lmin(1,p)

< ∞.

s,α (Rn ) For 0 < p, q ≤ ∞, s ∈ R, and α ∈ [0, 1], we denote the α-modulation spaces Mp,q by ⎧ ⎫ ⎛ ⎞1/q ⎪ ⎪ ⎨ ⎬    q s,α n  sq  −1  ⎝ ⎠ s,α F ψQ Ff Lp Mp,q (R ) := f ∈ S : f Mp,q = ξQ  <∞ ⎪ ⎪ ⎩ ⎭ Q∈Q

with modification for q = ∞, where the sequence {ξQ }Q∈Q satisfies ξQ ∈ Q. It is known that the cases α = 1 and α = 0 are equivalent to Besov and modulation spaces, respectively. This definition is based on Borup and Nielsen [1]. For the case 0 ≤ α < 1, the following equivalent norm of α-modulation spaces is given in [2, Section 2.1]. Let a Schwartz function sequence {ηkα }k∈Zn satisfy that • • • •



 inf |ηkα (ξ)| : ξ ∈ B kα/(1−α) k, ckα/(1−α) , k ∈ Zn > 0,

α α/(1−α) k, Ckα/(1−α) , supp  B k −|β|α/(1−α)  β αηk ⊂ ∂ η (ξ) ≤ Cβ k for every multi-index β ∈ Zn+ , k ηkα ≡ 1. k∈Zn

Then, for 0 < p, q ≤ ∞, s ∈ R, and α ∈ [0, 1),  s,α ∼ f Mp,q



k∈Zn

k

sq/(1−α)

 −1 α q F ηk Ff  p L

1/q .

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s,α We note that Mp,q are quasi-Banach spaces (Banach spaces for 1 ≤ p, q ≤ ∞) and S ⊂ s,α  s,α Mp,q ⊂ S . Especially, for 0 < p, q < ∞, S is dense in Mp,q (see Borup and Nielsen [3]). s,α Mp,q are independent of the choice of the partitions of unity {ηkα }k∈Zn . If 0 < p, q < ∞, s,α ∗  ,α s,α   the dual spaces of Mp,q are identified as Mp,q = Mp−s  ,q  , where p = [max(1, p)] ,    q = [max(1, q)] , and s = s + nα(1 − 1/ min(1, p)) (see [9, Theorem 2.1]). Moreover, the complex interpolation theorem is given by Han and Wang [9, Theorem 2.2] as follows. If 0 < θ < 1, s = (1 − θ)s1 + θs2 , 1/p = (1 − θ)/p1 + θ/p2 , 1/q = (1 − θ)/q1 + θ/q2 , then s ,α

,α s,α = Mp,q . We remark that one can not obtain α-modulation spaces by Mp11,q1 , Mps22,q 2 θ complex interpolation between modulation spaces (α = 0) and Besov spaces (α = 1), which is proved by Guo, Fan, Wu, and Zhao [8]. Next, we recall some embedding relations.

Proposition 2.2. Let 0 < p, q ≤ ∞, s, σ ∈ R, and 0 ≤ α < 1. Then the mapping s,α s−σ,α (I − Δ)σ/2 : Mp,q → Mp,q is isomorphic. The proof of Proposition 2.2 is similar to that for Besov spaces, which can be found in [22, Section 2.3.8]. See also [2, Corollary 3.3] or the appendix of this paper. Although this result seems to be folklore, it seems that the appendix of this paper provides the first published proof which is valid for all 0 < p, q ≤ ∞. s,α Proposition 2.3. (See [7].) Let s ∈ R and 0 ≤ α < 1. Then, M2,2 = L2s with equivalent norms.

Proposition 2.4. Let 1 ≤ p ≤ ∞, 1/p + 1/p = 1, s ∈ R and 0 ≤ α < 1. Then we have s,α s,α p Mp,min(p,p  ) ⊂ Ls ⊂ Mp,max(p,p ) .

Proof of Proposition 2.4. By the triangle inequality, we have for 1 ≤ p ≤ ∞ f Lp ≤

   F −1 ηkα Ff  p ∼ f  0,α , Mp,1 L k∈Zn

0,α 0,α which yields that Mp,1 → Lp . In particular, M1,1 → L1 . Interpolating this fact and 0,α 2 0,α p 0,α M2,2 ≈ L , then Mp,p → L for 1 ≤ p ≤ 2. By duality, we also have Lp → Mp,p for 2 ≤ p ≤ ∞. Next, from the definition of {ψQ }Q∈Q , we obtain for 1 ≤ p ≤ ∞

 −1  0,α ∼ sup F f Mp,∞ ψQ Ff Lp Q∈Q

  ≤ sup F −1 ψQ L1 · f Lp Q∈Q

 f Lp ,

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0,α 0,α which yields Lp → Mp,∞ . In particular, L1 → M1,∞ . Again, interpolating this embed0,α 0,α 0,α 2 p p ding and M2,2 ≈ L , then L → Mp,p for 1 ≤ p ≤ 2. By duality, we have Mp,p  → L 0,α 0,α ∞ for 2 ≤ p < ∞. For p = ∞, we saw above that Mp,p = M∞,1 → L . Therefore, we have established the claim for s = 0. Then, using Proposition 2.2, we obtain the general case. 2

Proposition 2.5. (See [9, Proposition 2.4].) Let 0 < p1 ≤ p2 ≤ ∞, 0 < q1 , q2 ≤ ∞, s1 , s2 ∈ R, and 0 ≤ α < 1. Then ,α ,α Mps11,q ⊂ Mps22,q 1 2

holds if either of the following conditions is satisfied. (1) q1 ≤ q2 and s1 ≥ s2 + nα(1/p1 − 1/p2 ); (2) q1 > q2 and s1 > s2 + nα(1/p1 − 1/p2 ) + n(1 − α)(1/q2 − 1/q1 ). Theorem 2.6. (See [9, Theorem 4.1].) Let 0 < p, q ≤ ∞, s1 , s2 ∈ R, and 0 ≤ α1 , α2 < 1. Then s1 ,α1 s2 ,α2 Mp,q ⊂ Mp,q

holds if and only if either of the following conditions is satisfied. (1) α1 ≥ α2 and s1 ≥ s2 + n(α1 − α2 )ν1 (p, q); (2) α1 < α2 and s1 ≥ s2 + n(α1 − α2 )ν2 (p, q). In this paper, we don’t mention modulation spaces explicitly, since the above facts about α-modulation spaces easily yield those for modulation spaces if we set α = 0. However, one can find some properties for modulation spaces in [4,10,11,25,26]. 2.3. Besov and Triebel Lizorkin spaces In this subsection, we recall Besov spaces and Triebel Lizorkin spaces. Let 0 < p, q ≤ ∞ and s ∈ R. Choose a family of functions {φj }j∈Z+ satisfying that for j ∈ N • supp φ0 ⊂ B(0, 2), • φj = φ(/2j ), where supp φ ⊂ {ξ : 1/2 ≤ |ξ| ≤ 2},  φj ≡ 1. • φ0 + j∈N s s and Triebel Lizorkin spaces Fp,q as follows. Then, we denote Besov spaces Bp,q

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  f ∈ S  : f B s := F −1 φ0 Ff Lp

s Bp,q (Rn ) :=

p,q

 

   +  2js · F −1 φj Ff (x)Lp (Rn ) 

q (Nj )

x

! < +∞ ;

  f ∈ S  : f Fp,q := F −1 φ0 Ff Lp s

s Fp,q (Rn ) :=

 

   +  2js · F −1 φj Ff (x)q (N )  j

Lp (Rn x)

! < +∞ .

s We note that Fp,2 = Lps for 1 < p < ∞ with equivalent norms (see [22, Theorem in Section 2.5.6]). More properties about Besov and Triebel Lizorkin spaces can be found in [22,23]. s Remark 2.7. We justify the inclusion relations Lps+ε → Bp,q → Lps−ε for 1 ≤ p, q ≤ ∞, s s ∈ R and ε > 0. For 1 < p < ∞, since we have Fp,2 ≈ Lps , we obtain from [22, Proposition 2 in Section 2.3.2] s+ε s+ε s−ε s−ε s Lps+ε ≈ Fp,2 → Bp,max(p,2) → Bp,q → Bp,min(p,2) → Fp,2 ≈ Lps−ε .

Next, we state the case p = 1. Recalling from [22, Proposition in Section 2.5.7] that 0 0 B1,1 → L1 → B1,∞ , the lifting properties (see [22, Theorem in Section 2.3.8]): for σ∈R (I − Δ)−σ/2 : Lp → Lpσ

0 σ and (I − Δ)−σ/2 : Bp,q → Bp,q ,

σ σ yield that B1,1 → L1σ → B1,∞ . Hence, we obtain from [22, Proposition 2 in Section 2.3.2] s+ε s−ε s L1s+ε → B1,∞ → B1,q → B1,1 → L1s−ε . 0 Finally, we consider the case p = ∞. As in Proposition 2.4, we have B∞,1 → L∞ → 0 B∞,∞ , since

f L∞ ≤

   F −1 φj Ff 

L∞

= f B∞,1 ; 0

j∈Z+

f B 0

∞,∞

  ≤ sup F −1 φj L1 · f L∞ ∼ f L∞ . j∈Z+

σ σ Moreover, the lifting properties imply that B∞,1 → L∞ σ → B∞,∞ for any σ ∈ R. Therefore, using application of [22, Proposition 2 in Section 2.3.2], it follows that s−ε s+ε s ∞ L∞ s+ε → B∞,∞ → B∞,q → B∞,1 → Ls−ε .

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Collecting all of the above cases, we have for 1 ≤ p, q ≤ ∞, s ∈ R and ε > 0 s Lps+ε → Bp,q → Lps−ε .

2.4. Local Hardy spaces Here, we recall the definition and some properties of local Hardy spaces as given by Goldberg [6]. Let 0 < p < ∞. Suppose that a Schwartz function Φ satisfies " Φ(x)dx = 0. Then we denote the local Hardy spaces hp as follows. Rn # hp (Rn ) :=

   1      sup f ∈ S  : f hp =  ∗ f Φ  0
$ <∞ .

Lp

We note that hp is independent of the choice of Φ, and satisfies that h1 → L1 and hp = Lp 0 with equivalent norms if 1 < p < ∞. Moreover, hp = Fp,2 with equivalent (quasi)-norms if 0 < p < ∞ (see [22, Theorem 1 in Section 2.5.8]). The complex interpolation theorem is given as follows. Let 0 < θ < 1 and 1/p = (1 − θ)/p1 + θ/p2 , then (hp1 , hp2 )θ = hp (see [22, Remark 1 in Section 2.4.7]). In the end of this subsection, we state the atomic decomposition of hp (see [6, Section 4]). An hp -atom a is called type I if • supp a ⊂ Q with |Q| < 1, • aL∞ ≤ |Q|−1/p , " • Rn xβ a(x)dx = 0 for all |β| ≤ [n(1/p − 1)], where Q is a cube, |Q| is its Lebesgue measure and [x] = max{n ∈ Z : n ≤ x}. On the other hand, an hp -atom a is called type II if • supp a ⊂ Q with |Q| ≥ 1, • aL∞ ≤ |Q|−1/p . We note that the set of hp -atoms of type I and type II is bounded in hp , and every  f ∈ hp (with 0 < p ≤ 1) can be expressed as f = j∈N λj aj , where {aj }j∈N is a family of hp -atoms of types I and II, and {λj }j∈N is a sequence of complex numbers which belongs to p . Moreover, f hp ∼ inf λj p (N) , where we take the infimum over  all representations f = j∈N λj aj as above (see [6, Lemma 5]). 3. Lemmas for main theorems In this section, we state lemmas which will be used in the proofs of our main theorems. Proofs of the lemmas are given in Section 6.

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3.1. Equivalent (quasi)-norms for α-modulation spaces We first give equivalent (quasi)-norms for α-modulation spaces with weaker restrictions regarding the “frequency localizing functions” than in the original definition of the α-modulation space norms from Subsection 2.2. Han and Wang [9] showed the following. Proposition 3.1. (See [9, Proposition 6.1].) Let 0 < p, q ≤ ∞, s ∈ R, and 0 ≤ α < 1. Let a smooth radial bump function ρ satisfy that ρ(ξ) = 1 on |ξ| < 1, and ρ(ξ) = 0 on s,α |ξ| ≥ 2. Then we have for all f ∈ Mp,q  s,α ∼ f Mp,q



k

sq/(1−α)

 −1 α q F ρk Ff  p

1/q ,

L

k∈Zn

where  ρα k (ξ) := ρ

ξ − kα/(1−α) k Ckα/(1−α)



for a suitable constant C ≥ 1 which depends on the space dimension. In the above proposition, the frequency localizing functions have compact supports. Even if the supports of the functions are not compact, however, we are able to have equivalent norms for those of the α-modulation spaces. Lemma 3.2. Let 0 < p, q ≤ ∞, s ∈ R, and 0 ≤ α < 1. Let a Schwartz function Ψ ∈ S s,α satisfy that |Ψ(ξ)| ≥ c > 0 on |ξ| ≤ 2. Then we have for all f ∈ Mp,q  s,α ∼ f Mp,q



q   p ksq/(1−α) F −1 Ψα k Ff L

1/q ,

k∈Zn

where  Ψα k (ξ)

:= Ψ

ξ − kα/(1−α) k Ckα/(1−α)



for a suitable constant C ≥ 1 which depends on the space dimension. Note that equivalence of the norms is only claimed under the assumption that we s,α already know f ∈ Mp,q . It is not claimed that finiteness of the alternative norm for  s,α some f ∈ S implies f ∈ Mp,q . The corresponding statement for modulation spaces has been already shown in [13, Theorem 2.5], where a specific property of modulation spaces, namely, that the supports of the partition of unity are uniform, was used. On the other hand, those for α-modulation spaces are not so. However, if we require the frequency

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localizing functions to be Schwartz functions, the preceding lemma shows that we can ignore the “overlapping” portions of the frequency components also for α-modulation spaces. 3.2. Lemmas for Theorems 1.5 and 1.6 In this subsection, we state the following three lemmas to prove Theorems 1.5 and 1.6. The first one will be used in the “IF” part of Theorem 1.5 on the critical regularity. Lemma 3.3. Let 0 < p ≤ 1, 0 < q ≤ 2, and 0 ≤ α < 1. Then there exists a universal constant K > 0 such that s,α ≤ K aMp,q

holds for all hp -atoms “a”. Here, K = K(n, p, q, α) > 0 and s = −n(1 −α)(1/p +1/q −1). The remaining two lemmas play important roles in the proofs for the “ONLY IF” parts of Theorems 1.5 and 1.6. s,α Lemma 3.4. Let 0 < p ≤ 1, 0 < q ≤ ∞, s ∈ R, and 0 ≤ α < 1. If Mp,q → hp , then we have





1/p |ck |p







k∈Zn

1/q ksq/(1−α) |ck |q

k∈Zn

for all finitely supported sequences {ck }k∈Zn (that is, for a finite number  ck = 0 except  of k’s). Here, the right hand side has to be read as  ks/(1−α) ck k∈Zn ∞ for q = ∞. s,α Lemma 3.5. Let 0 < p ≤ 1, 0 < q ≤ ∞, s ∈ R, and 0 ≤ α < 1. If hp → Mp,q , then we have

⎛

⎛

⎜ sq/(1−α)+nq(1/p−1) ⎝ |k| ⎝ k=0



k 1/(1−α) /2≤|m|≤2k 1/(1−α)

⎛ ⎞q/p ⎞1/q ⎞1/p  ⎟ |cm |p ⎠ ⎠ ⎝ |ck |p ⎠ k=0

for all finitely supported sequences {ck }k∈Zn \{0} . Here, the left hand side has to be read with obvious modification for q = ∞. 3.3. Lemmas for Theorems 1.3 and 1.4 In this subsection, we state the following three lemmas which are needed to prove Theorems 1.3 and 1.4. We will derive the “IF” parts from existing results via interpolation and duality, and we will prove the “ONLY IF” parts by using the following lemmas.

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s,α Lemma 3.6. Let 1 ≤ p, q ≤ ∞, s ∈ R, and 0 ≤ α < 1. If Mp,q → Lp , then we have







1/p |ck |





p

k∈Zn

1/q k

sq/(1−α)

|ck |

q

k∈Zn

for all finitely supported sequences {ck }k∈Zn (that is,  ck = 0 except number  for a finite     of k’s). Here, both sides have to be read as  ck k∈Zn ∞ and  ks/(1−α) ck k∈Zn ∞ for p = ∞ and q = ∞, respectively. s,α Lemma 3.7. Let 1 ≤ p, q < ∞, s ∈ R, and 0 ≤ α < 1. If Lp → Mp,q , then we have

⎛

⎛  ⎜ |k|sq/(1−α)+nq(1/p−1) ⎝ ⎝

⎛ ⎞q/p ⎞1/q ⎞1/p  ⎟ |cm |p ⎠ ⎠ ⎝ |ck |p ⎠



k 1/(1−α) /2≤|m|≤2k 1/(1−α)

k=0

k=0

for all finitely supported sequences {ck }k∈Zn \{0} . s,α , then we have Lemma 3.8. Let 1 ≤ q < ∞, s ∈ R, and 0 ≤ α < 1. If L1 → M1,q

⎛ ⎝



⎛

⎞q ⎞1/q



|k|sq/(1−α) ⎝

|cm |⎠ ⎠

k 1/(1−α) /2≤|m|

k=0





|ck |

k=0

for all finitely supported sequences {ck }k∈Zn \{0} . 4. Proofs of Theorems 1.5 and 1.6 By using the lemmas in Section 3, we prove our main theorems. This section is divided into four subsections, where the “IF” and “ONLY IF” parts of Theorem 1.5 or Theorem 1.6 are proved, respectively. 4.1. Proof of “IF” part in Theorem 1.6 In this subsection, we state the proof of the “IF” part in Theorem 1.6. Proof of “IF” part in Theorem 1.6. Let 0 < p ≤ 1, 0 < q ≤ ∞, s ∈ R, and 0 ≤ α < 1. Then we have (i)

(ii)

(iii)

(iv)

0 0 0,α s,α hp ≈ Fp,2 ← Bp,p ← Mp,p ← Mp,q

if either of the following conditions is satisfied.

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(1) p ≥ q and s ≥ nα(1/p − 1/p) = 0; (2) p < q and s > nα(1/p − 1/p) + n(1 − α)(1/p − 1/q) = n(1 − α)(1/p − 1/q). The embedding (i) is given by [22, Theorem 1 in Section 2.5.8]. At (ii), we used the s s fact that Bp,u → Fp,q holds if u ≤ min(p, q) by [22, Proposition 2 in Section 2.3.2]. Theorem 1.1 and Proposition 2.5 provide (iii) and (iv), respectively. 2 4.2. Proof of “ONLY IF” part in Theorem 1.6 Next, we prove the “ONLY IF” part in Theorem 1.6. s,α → hp . Proof of “ONLY IF” part in Theorem 1.6. We assume that Mp,q By Lemma 3.4, we have for all finitely supported sequences {ck }k∈Zn ,





1/p |ck |







p

k∈Zn

1/q k

sq/(1−α)

|ck |

q

.

k∈Zn

We use this with ck = k−s/(1−α) |dk |1/p for a given finitely supported sequence {dk }k∈Zn . Then 

 −sp/(1−α)

k

|dk | 

k∈Zn



p/q |dk |

q/p

.

(1)

k∈Zn

Here, we take {dk }k∈Zn satisfying dk =

1 0

if k = (K, 0, · · · , 0), otherwise,

where K ∈ N. Substituting this sequence {dk }k∈Zn into the inequality (1), we have K−sp/(1−α)  1. This implies that necessarily s ≥ 0. Next, we assume that p < q and q = ∞ (⇒ 1 < q/p < ∞). Taking the supremum over {dk } such that dk q/p = 1, then by the inequality (1)          −sp/(1−α)  sup  k−sp/(1−α) dk   1. k  (q/p) =    dk  q/p =1 n 

k∈Z

This yields that (q/p) sp/(1 − α) > n, namely, s > n(1 − α)(1/p − 1/q). Finally, we assume that p < q = ∞. Note that in this case, the inequality (1) has to be read as    k−sp/(1−α) |dk |  {dk }k∈Zn ∞ , k∈Zn

  which easily yields k−sp/(1−α) 1  1. Thus, s > n(1 − α)/p. 2

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4.3. Proof of “IF” part in Theorem 1.5 Here, we give the proof of the “IF” part in Theorem 1.5. Proof of “IF” part in Theorem 1.5. Let 0 < p ≤ 1, 0 < q ≤ ∞, s ∈ R, and 0 ≤ α < 1. We first assume that 2 ≤ q ≤ ∞. Then (i)

(ii)

(iii)

0 0 s,α hp ≈ Fp,2 → Bp,q → Mp,q

holds if s ≤ −n(1 −α)(1/p +1/q −1). The inclusion relation (i) is given by [22, Theorem 1 s s in Section 2.5.8]. At (ii), we used the fact that Fp,q → Bp,v holds if v ≥ max(p, q). One can find this in [22, Proposition 2 in Section 2.3.2]. Theorem 1.1 provides (iii). Next, we consider the case when 0 < q ≤ 2. We first assume that p ≤ q (⇒ p/q ≤ 1). We set f=

∞ 

λ i a i ∈ hp ,

i=1

 where ai are hp -atoms and λi are complex numbers with |λi |p < ∞. Since p p p f + gMp,q s,α ≤ f  s,α + g s,α for p ≤ q and p ≤ 1, we have by Lemma 3.3 Mp,q Mp,q p

f Mp,q s0 ,α ≤

∞ 

p

|λi |p ai Mp,q s0 ,α 

i=1

∞ 

|λi |p < ∞,

i=1

where s0 = −n(1 − α)(1/p + 1/q − 1), 0 < p ≤ 1, and 0 < q ≤ 2. This means that for 0 < p ≤ 1 and 0 < q ≤ 2, s0 ,α s,α hp → Mp,q → Mp,q

holds if p ≤ q and s ≤ s0 = −n(1 − α)(1/p + 1/q − 1). Finally, we consider the case p > q and 0 < q ≤ 2. In view of the embedding relation just above, we see that s,α hp → Mp,p

holds if s = −n(1 − α)(2/p − 1). Thus, by Proposition 2.5, −n(1−α)(2/p−1),α s,α hp → Mp,p → Mp,q

holds if p > q and −n(1 − α)(2/p − 1) > s + nα(1/p − 1/p) + n(1 − α)(1/q − 1/p), that is, if s < −n(1 − α)(1/p + 1/q − 1). 2

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4.4. Proof of “ONLY IF” part in Theorem 1.5 Finally, we show the “ONLY IF” part in Theorem 1.5. s,α . Then LemProof of “ONLY IF” part in Theorem 1.5. We assume that hp → Mp,q ma 3.5 gives us the fact

⎛

⎛  ⎜ |k|sq/(1−α)+nq(1/p−1) ⎝ ⎝

⎛ ⎞q/p ⎞1/q ⎞1/p  ⎟ |cm |p ⎠ ⎠ ⎝ |ck |p ⎠



k 1/(1−α) /2≤|m|≤2k 1/(1−α)

k=0

k=0

(2) for all finitely supported sequences {ck }k∈Zn \{0} . Here, we take {ck }k∈Zn \{0} satisfying 1

for some sufficiently large integer K ≥ 21+ 2(1−α) that ck =

if |k| ≤ K, if |k| > K.

1 0

If we substitute this sequence into (2), we have ⎛ ⎝



⎞1/p |ck |p ⎠

k=0

⎛ =⎝



⎞1/p 1⎠

 K n/p

0<|k|≤K

and ⎛

⎛  ⎜ |k|sq/(1−α)+nq(1/p−1) ⎝ ⎝

k 1/(1−α) /2≤|m|≤2k 1/(1−α)

k=0

⎛ ⎜ ≥⎝

⎛



|k|sq/(1−α)+nq(1/p−1) ⎝

⎛

⎛



|k|sq/(1−α)+nq(1/p−1) ⎝ ⎞1/q

⎛ ⎜ ⎝

sq q⎟ 1 n |k| 1−α +nq( p −1)+ 1−α · p ⎠

1−α 0<|k|≤ √12 ·( K 2 )

 K s+n(1−α)( p −1)+ p +(1−α) q 1

n

n

= K s+n(1−α)( p + q −1)+ p . 1

1



k 1/(1−α) /2≤|m|≤2k 1/(1−α)

1−α 0<|k|≤ √12 ·( K 2 )





⎞q/p ⎞1/q ⎟ |cm |p ⎠ ⎠

k 1/(1−α) /2≤|m|≤2k 1/(1−α)

1−α 0<|k|≤ √12 ·( K 2 )

⎜ =⎝

⎞q/p ⎞1/q ⎟ |cm |p ⎠ ⎠



n

⎞q/p ⎞1/q ⎟ 1⎠ ⎠

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In the third line, we used that 1

|m| ≤ 2k 1−α

1 = 2 1 + |k|2 2(1−α) 1 · 2

≤2



K 2

2(1−α)

1 + · 2



K 2

1 2(1−α) ! 2(1−α)

= K, 1−α 2(1−α) 1 since we have 0 < |k| ≤ √12 · K and K ≥ 21+ 2(1−α) ⇔ 1 ≤ 12 · K . In view of 2 2 these results, (2) connotes the inequality s + n(1 − α)(1/p + 1/q − 1) + n/p ≤ n/p, that is, s ≤ −n(1 − α)(1/p + 1/q − 1). The case p ≤ q is proven. Next, we examine p > q. We assume towards a contradiction that s ≥ −n(1 −α)(1/p + s,α 1/q − 1). We can take ε > 0 such that (1 + ε)q/p < 1. By the assumption hp → Mp,q and Lemma 3.5, we have ⎛

⎛

⎛ ⎞q/p ⎞1/q ⎞1/p  ⎟ |cm |p ⎠ ⎠ ⎝ |ck |p ⎠



⎜ sq/(1−α)+nq(1/p−1) ⎝ |k| ⎝

k 1/(1−α) /2≤|m|≤2k 1/(1−α)

k=0

k=0

(3) for all {ck }k∈Zn \{0} ∈ p , where we used a limit argument to pass from finitely supported sequences to arbitrary p sequences. We choose the sequence {ck }k∈Zn \{0} as ck =

|k|−n/p (log |k|) 0

−(1+ε)/p

if |k| ≥ K, if |k| < K,

where K > 0 is a sufficiently large integer. Here, we use the following fact. Fact 4.1. (See [19, Remark 4.3].) Let 0 < γ < ∞ and K > 0 be a sufficiently large number. Then we have   −β/γ ∈ γ (Zn ) if β > 1; |k|−n/γ (log |k|) 

|k|≥K

−n/γ

|k|

(log |k|)

−β/γ



|k|≥K

∈ / γ (Zn ) if β ≤ 1.

From Fact 4.1, the right hand side of (3) satisfies ⎛ ⎝



k=0

⎞1/p |ck |p ⎠

⎛ =⎝

 

|k|≥K

|k|−n/p (log |k|)

−(1+ε)/p

p

⎞1/p ⎠

< ∞.

(4)

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On the other hand, if K = Kα > 0 is a sufficient large number, the left hand side of (3) satisfies ⎛

⎛ ⎜  sq/(1−α)+nq(1/p−1) ⎝ k ⎝

k1/(1−α) /2≤|m|≤2k1/(1−α)

k=0

⎛

⎛



≥⎜ ⎝

ksq/(1−α)+nq(1/p−1) ⎝

|k|≥(2K)1−α

⎛



ksq/(1−α)+nq(1/p−1) ⎝

|k|≥(2K)1−α

⎛

⎞q/p ⎞1/q ⎟ |m|−n (log |m|)−(1+ε) ⎠ ⎠



k1/(1−α) /2≤|m|≤2k1/(1−α)

⎞1/q



⎝

⎞q/p ⎞1/q ⎟ |cm |p ⎠ ⎠



k1/(1−α) /2≤|m|≤2k1/(1−α)

⎛

=⎜ ⎝

⎞q/p ⎞1/q ⎟ |cm |p ⎠ ⎠



|k|

−n

(log |k|)

−(1+ε)q/p ⎠

|k|≥(2K)1−α

⎛

⎞1/q  q −n/q −{(1+ε)q/p}/q ⎠ (log |k|) = ∞. |k|



=⎝

(5)

|k|≥(2K)1−α

Here, in the third line, we used that |m| ≥ k1/(1−α) /2 ≥ |k|1/(1−α) /2 ≥ K from |k| ≥ (2K)1−α . Moreover, in the fourth line, we used the assumption s ≥ −n(1 − α)(1/p + 1/q − 1) for the outer summand and the following argument for the inner summation. That is,  −(1+ε) |m|−n (log |k|) k 1/(1−α) /2≤|m|≤2k 1/(1−α)

⎛   −(1+ε) ≥ 2−n k−n/(1−α) log 2k1/(1−α) ·⎝ 



∼ log 2k1/(1−α)  =

log 2 +

 (logk) ∼ (log |k|)

−(1+ε)



⎞ 1⎠

k 1/(1−α) /2≤|m|≤2k 1/(1−α)

−(1+ε) 1 logk 1−α

−(1+ε)

−(1+ε)

.

Here, in the fifth line, we used the fact log 2 ≤ logk if |k| ≥ (2K)1−α , with a sufficiently large K = Kα > 0. In the last line, we used the fact logk ∼ log |k| if |k| ≥ (2K)1−α , with K sufficiently large.

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However, these two estimates (4) and (5) contradict (3). Therefore, we obtain s < −n(1 − α)(1/p + 1/q − 1) for p > q. 2 5. Proofs of Theorems 1.3 and 1.4 Next, we prove Theorems 1.3 and 1.4. As a matter of fact, Theorems 1.3 and 1.4 are closely related with each other by duality. Hence, essentially, it suffices to show one theorem. Moreover, in view of Corollary 1.2, the remaining question is to get the answer whether the critical cases, that is, s1 = s2 + n(1 − α)ν1 (p, q) or s1 = s2 + n(α − 1)ν2 (p, q), are sufficient or not for the inclusion relations. This section is divided into four subsections. The first and second subsections are devoted to proving the “IF” and the “ONLY IF” parts of Theorem 1.4. In the third and fourth subsections, the “IF” and the “ONLY IF” parts of Theorem 1.3 are proved. 5.1. Proof of “IF” part in Theorem 1.4 We begin this subsection by introducing the following lemma. Lemma 5.1. Let 1 < p ≤ 2, p ≤ q ≤ p , and s ≤ −n(1 − α)(1/p + 1/q − 1). Then s,α Lp → Mp,q holds. s

,α

1,q Proof of Lemma 5.1. By Theorem 1.5, we have h1 → M1,q with s1,q = −n(1−α)/q for arbitrary 1 ≤ q ≤ ∞. Here, sp,q = −n(1 −α)(1/p +1/q−1). Moreover, by Proposition 2.3, 0,α we have M2,2 ≈ L2 ≈ h2 . Interpolating between these two inclusion relations, we obtain

sp,q ,α hp → Mp,q

for 1 ≤ p ≤ 2, p ≤ q ≤ p . Indeed, the gray area in the following figure indicates the area given by interpolation, which corresponds to 1 ≤ p ≤ 2 and p ≤ q ≤ p . 1/q 1

(1/2,1/2)

0

1

1/p

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0,α Here, the point (1/2, 1/2) and the line 1/p = 1 (0 ≤ 1/q ≤ 1) indicate h2 ≈ M2,2 and s ,α 1,q 1 p p h → M1,q , respectively. Therefore, since we have h ≈ L for 1 < p < ∞, sp,q ,α s,α Lp ≈ hp → Mp,q → Mp,q

holds if 1 < p ≤ 2, p ≤ q ≤ p , and s ≤ sp,q = −n(1 − α)(1/p + 1/q − 1). 2 Now, we start the proof of the “IF” part in Theorem 1.4. Proof of the “IF” part in Theorem 1.4. We assume that p ≥ q. If q ≤ min(p, p ) and s ≤ 0, then we have by Proposition 2.4 s,α 0,α s,α p Mp,q → Mp,q → Mp,min(p,p  ) → Ls .

Therefore, recalling that ν2 (p, q) = 0 for q ≤ min(p, p ), we obtain from Proposition 2.2 s1 ,α that if q ≤ min(p, p ) and s1 ≥ s2 = s2 + n(α − 1)ν2 (p, q), then Mp,q → Lps2 . In  particular, note that q ≤ min(p, p ) holds for q = 1 and arbitrary 1 ≤ p ≤ ∞. Next, by the dual statement of Lemma 5.1, we have 

p Mp−s,α  ,q  → L

for 1 < p ≤ 2, p ≤ q ≤ p , and s ≤ −n(1 − α)(1/p + 1/q − 1). By setting p = p0 , q  = q0 , and −s = s1 − s2 , we have −s2 ,α Mps01,q → Lp0 0

for 1 < p0 ≤ 2, p0 ≤ q0 ≤ p0 , and −(s1 − s2 ) ≤ −n(1 − α)(1/p0 + 1/q0 − 1). This yields from Proposition 2.2 that ,α Mps01,q → Lps20 0

for 2 ≤ p0 < ∞, p0 ≤ q0 ≤ p0 , and

⇐⇒

−(s1 − s2 ) ≤ −n(1 − α)(1/p0 + 1/q0 − 1)  1 1 +1− −1 s1 ≥ s2 + n(1 − α) 1 − p0 q0

⇐⇒

s1 ≥ s2 + n(α − 1)(1/p0 + 1/q0 − 1)

⇐⇒

s1 ≥ s2 + n(α − 1)ν2 (p0 , q0 ).

s1 ,α Therefore, Mp,q → Lps2 holds if 2 ≤ p < ∞, p ≤ q ≤ p, and s1 ≥ s2 + n(α − 1)ν2 (p, q). The remaining cases of (p, q) are sufficient for the existence of the embedding relations by Corollary 1.2. 2

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5.2. Proof of “ONLY IF” part in Theorem 1.4 We start this subsection with preparations for the proof of the “ONLY IF” part of Theorem 1.4. The arguments of the proofs are very similar to those of Theorems 1.5 and 1.6, and the following statements are almost repetition of the proofs of the theorems for the embedding relations between α-modulation and local Hardy spaces. Hence, we only sketch the proofs. s,α → Lp holds, then Lemma 5.2. Let 1 ≤ p, q ≤ ∞, p < q, and s ∈ R. Then if Mp,q s > n(1 − α)(1/p − 1/q). s,α → Lp for p < q and q = ∞ (⇒ 1 < Proof of Lemma 5.2. We first assume that Mp,q q/p < ∞). By Lemma 3.6, we have for all finitely supported sequences {ck }k∈Zn ,





1/p



|ck |

p



k∈Zn



1/q k

sq/(1−α)

|ck |

q

.

k∈Zn

We use this with ck = k−s/(1−α) |dk |1/p for a given finitely supported sequence {dk }k∈Zn . Then 



−sp/(1−α)

k

|dk | 

k∈Zn



p/q |dk |

q/p

.

(6)

k∈Zn

Taking the supremum over {dk } such that dk q/p = 1, then by the inequality (6)    −sp/(1−α)  k 

(q/p)

      = sup  k−sp/(1−α) dk   1.   dk  q/p =1 n 

k∈Z

This yields that (q/p) sp/(1 − α) > n, namely, s> n(1 − α)(1/p  − 1/q). Next, we assume that p < q = ∞. We have k−sp/(1−α) 1  1. Thus, s > n(1 − α)/p. 2 s,α holds, then s < Lemma 5.3. Let 1 ≤ q < p < ∞ and s ∈ R. Then if Lp → Mp,q −n(1 − α)(1/p + 1/q − 1). s,α . Then Lemma 3.7 gives us the inProof of Lemma 5.3. We assume that Lp → Mp,q equality

⎛

⎛

⎜ sq/(1−α)+nq(1/p−1) ⎝ |k| ⎝ k=0



k 1/(1−α) /2≤|m|≤2k 1/(1−α)

⎛ ⎞q/p ⎞1/q ⎞1/p  ⎟ |cm |p ⎠ ⎠ ⎝ |ck |p ⎠ k=0

(7)

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for all {ck }k∈Zn \{0} ∈ p , where we used a limit argument to pass from finitely supported sequences to arbitrary p sequences. We assume towards a contraction that s ≥ −n(1 − α)(1/p + 1/q − 1). Since q < p, we can take ε > 0 such that (1 + ε)q/p < 1. We set the sequence {ck }k∈Zn \{0} as ck =

|k|−n/p (log |k|) 0

−(1+ε)/p

if |k| ≥ K, if |k| < K,

where K > 0 is a sufficiently large integer. Recalling Fact 4.1 and the proof of “ONLY IF” part of Theorem 1.5, we have ⎛ ⎝

⎞1/p



|ck |p ⎠

⎛ =⎝

 

|k|−n/p (log |k|)

−(1+ε)/p

p

⎞1/p ⎠

<∞

|k|≥K

k=0

and ⎛

⎛

⎞q/p ⎞1/q ⎟ |cm |p ⎠ ⎠ = ∞.



⎜ sq/(1−α)+nq(1/p−1) ⎝ |k| ⎝

k 1/(1−α) /2≤|m|≤2k 1/(1−α)

k=0

However, these two estimates contradict (7). Therefore, we obtain s < −n(1 − α)(1/p + 1/q − 1). 2 s,α Lemma 5.4. Let 1 ≤ q < ∞. Then L1 → M1,q holds only if s < −n(1 − α)/q. s,α Proof of Lemma 5.4. Let 1 ≤ q < ∞. The assumption L1 → M1,q gives us from Lemma 3.8

⎛ ⎝



⎛ |k|sq/(1−α) ⎝

⎞q ⎞1/q



|cm |⎠ ⎠



k 1/(1−α) /2≤|m|

k=0



|ck |

(8)

k=0

for all {ck }k∈Zn \{0} ∈ 1 , where we used a limit argument to pass from finitely supported sequences to arbitrary 1 sequences. We assume towards a contradiction that s ≥ −n(1 − α)/q. We can take ε > 0 such that εq < 1. For this ε > 0, we set the sequence {ck }k∈Zn \{0} as ck =

|k|−n (log |k|) 0

−(1+ε)

if |k| ≥ K, if |k| < K,

where K > 0 is a sufficiently large integer. Recall Fact 4.1. Then, the right hand side of (8) satisfies

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|ck | =



|k|−n (log |k|)

−(1+ε)

< ∞.

(9)

|k|≥K

k=0

On the other hand, the left hand side of (8) satisfies ⎛ ⎝



⎛ |k|sq/(1−α) ⎝

⎛



≥⎝

−(1+ε) ⎠

⎠

|m|−n (log |m|)

k 1/(1−α) /2≤|m|

⎞1/q



⎝

⎞q ⎞1/q



|k|sq/(1−α) ⎝

|k|≥(2K)1−α

⎛

|cm |⎠ ⎠

k 1/(1−α) /2≤|m|

k=0

⎛

⎞q ⎞1/q



|k|−n (log |k|)

−εq ⎠

|k|≥(2K)1−α

⎛





=⎝

|k|−n/q (log |k|)

(εq)/q

q

⎞1/q ⎠

= ∞,

(10)

|k|≥(2K)1−α

where, in the third line, we used the assumption which s ≥ −n(1 − α)/q for the outer summand and the following argument for the inner summation. Set B = k1/(1−α) /2 and assume |k| ≥ (2K)1−α with a sufficient large K = Kα > 0. Then, we have 

−(1+ε)

|m|−n (log |m|)

B≤|m|



−(1+ε)

|x|−n (log |x|)



dx

˜ |x|≥CB

for some C˜ = C˜n > 0. By the spherical coordinate transform and integration by parts, we have 

−n

|x|

−(1+ε)

(log |x|)

˜ |x|≥CB

∞ dx =

r−1 (log r)

−(1+ε)

dr

˜ CB



−ε ˜ + (1 + ε) = − log CB

∞

˜ CB

which yields that 

|x|−n (log |x|)

−(1+ε)

−(1+ε)

r−1 (log r)



−ε ˜ dx = ε−1 log CB .

˜ |x|≥CB

Thus, if |k| ≥ (2K)1−α with a sufficient large K = Kα > 0, we obtain

dr,

T. Kato / Journal of Functional Analysis 272 (2017) 1340–1405



|m|−n (log |m|)

−(1+ε)



k 1/(1−α) /2≤|m|

  −ε ˜ 1/(1−α) /2 log Ck #

=

1365

$

−ε 1 ˜ logk + log C/2 1−α

 (logk) ∼ (log |k|)

−ε

−ε

,

where we used the fact logk ∼ log |k| for sufficiently large |k| > 0. However, these two estimates (9) and (10) contradict (8). Thus, we obtain s < −n(1 − α)/q. 2 Now, we begin with the proof of the “ONLY IF” part in Theorem 1.4. 0,α → Lps . By CorolProof of the “ONLY IF” part in Theorem 1.4. We assume that Mp,q lary 1.2, we have s ≤ n(1 − α)ν2 (p, q). Therefore, we only consider the case p < q and the case p = ∞ and 1 < q ≤ ∞. −s,α We first investigate the case p < q. Proposition 2.2 yields that Mp,q → Lp , so that Lemma 5.2 implies that −s > n(1 − α)(1/p − 1/q). This means that s < n(1 − α)(−1/p + 1/q) = n(1 − α)ν2 (p, q) for p ≤ 2 and p < q. Next, we consider the case  2 < p < q. Since we have Lp−s → Mp0,α  ,q  by the dual of the assumption if p < q < ∞, s,α p we obtain L → Mp ,q from Proposition 2.2. Hence, since 1 < q  < p < 2, Lemma 5.3 yields that s < −n(1 − α)(1/p + 1/q  − 1) = n(1 − α)(1/p + 1/q − 1). This means that s < n(1 − α)ν2 (p, q) for 2 < p < q < ∞. We next state the case 2 < p < q = ∞, that is, 2 < p < ∞ and q = ∞. Note that we can’t use a duality argument in this case. However, p 0,α  we can see Mp,∞ and Lps (2 < p < ∞) as the dual spaces of Mp0,α  ,1 and L−s (1 < p < 2),  ∗   ∗ that is, Mp0,α and Lp−s , respectively. This yields from the assumption that  ,1



Mp0,α  ,1

∗

  ∗ 0,α = Mp,∞ → Lps = Lp−s .

(11)

Applying the Hahn–Banach theorem, we see that for g ∈ S ⊂ Mp0,α  ,1 g

Mp0,α  ,1

# $  ∗ 0,α   ∗ = sup |F (g)| : F ∈ Mp ,1 and F  M 0,α ≤ 1 . p ,1

Then, we obtain from (11) #   ∗ gM 0,α  sup |F (g)| : F ∈ Lp−s and F Lp  p ,1

Thus, we get

−s

$ ∗

≤1

= gLp . −s

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S,  · Lp



−s

→ Mp0,α  ,1 ,

and thus 

S,  · Lp



→ Mps,α  ,1 .

Investigating the proofs of Lemmas 3.7 and 5.3, it is easy to see that this suffices to get s < −n(1 − α)/p , since 1 < p < 2. Hence, we have s < −n(1 − α)(1 − 1/p) = n(1 − α)ν2 (p, ∞) for 2 < p < ∞. Next, we consider the case p = ∞ and 1 < q ≤ ∞. In this case, we also can not use a 0,α duality argument. But we can regard M∞,q (1 < q ≤ ∞) and L∞ s as the dual spaces of  ∗ 1 ∗ 0,α 0,α  1 M1,q (1 ≤ q < ∞) and L−s , that is, M1,q and L−s , respectively. This means that  ∗ 1 ∗ 0,α 0,α = M∞,q → L∞ from the assumption. Applying the Hahn–Banach M1,q  s = L−s 0,α theorem, then we have for g ∈ S ⊂ M1,q 

#



0,α M1,q 

$

∗

F 

∗

gM 0,α = sup |F (g)| : F ∈ and ≤1 0,α M1,q 1,q   

∗  sup |F (g)| : F ∈ L1−s and F  L1 ∗ ≤ 1 −s

= gL1 . −s

Thus, we get 

 0,α S,  · L1−s → M1,q 

 and hence

 s,α S,  · L1 → M1,q .

Investigating the proofs of Lemmas 3.8 and 5.4, it is easy to see that this suffices to get s < −n(1 − α)/q  , since 1 ≤ q  < ∞. Hence, we have s < −n(1 − α)(1 − 1/q) = n(1 − α)ν2 (∞, q) for 1 < q ≤ ∞. s1 ,α Finally, using Proposition 2.2, we get the desired result for Mp,q → Lps2 . 2 Remark 5.5. When we proved the case 2 < p < ∞ and q = ∞, we investigated  the proofs of Lemmas 3.7 and 5.3 and then obtained the desired results from S,  · Lp → Mps,α  ,1 . However, the following statement might be more helpful and clearer, the idea of which is given by the reviewer.   Here, we use that S ⊂ Lp (1 < p < 2) is dense. Hence, given f ∈ Lp , there exists a sequence ofSchwartz functions {fn }n∈N satisfying fn −f Lp → 0. Using the embedding  S,  · Lp

s,α s,α → Mps,α  ,1 , we see that {fn }n∈N is a Cauchy sequence in Mp ,1 . Since Mp ,1

(1 < p < 2) is complete, we obtain fn − gM s,α → 0 for some g ∈ Mps,α  ,1 . Recalling that  p ,1



Lp → S  and Mps,α → S  , we have fn → f in S  and fn → g in S  . This yields that  ,1  f = g. Hence, by S,  · Lp → Mps,α  ,1 , we have f M s,α = gM s,α = lim fn M s,α  lim fn Lp = f Lp .    p ,1

p ,1

n→∞

p ,1

n→∞

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 This means that Lp → Mps,α  ,1 . Then, applying Lemma 5.3, we obtain s < −n(1 − α)/p , since 1 < p < 2. Thus, we have s < −n(1 − α)(1 − 1/p) = n(1 − α)ν2 (p, ∞).

Remark 5.6. As in the previous remark, we can obtain the desired result for the case p = ∞ and 1 < q ≤ ∞ without the investigations of the proofs of Lemmas 3.8 and 5.4. S ⊂ L1 is dense. Hence, given f ∈ L1 , there exists a sequenceof Schwartz functions s,α {fn }n∈N satisfying fn −f L1 → 0. Using the embedding S, ·L1 → M1,q  , we see that s,α s,α  {fn }n∈N is a Cauchy sequence in M1,q  . Since M1,q  (1 ≤ q < ∞) is complete, we obtain s,α s,α  fn − gM s,α → 0 for some g ∈ M1,q . Recalling that L1 → S  and M1,q  → S , we have 1,q   s,α fn → f in S  and fn → g in S  . This yields that f = g. Hence, by S,  · L1 → M1,q , we have

f M s,α = gM s,α = lim fn M s,α  lim fn L1 = f L1 . 1,q

1,q

n→∞

1,q

n→∞

s,α  This means that L1 → M1,q  . Then, applying Lemma 5.4, we obtain s < −n(1 − α)/q , since 1 ≤ q  < ∞. Thus, we have s < −n(1 − α)(1 − 1/q) = n(1 − α)ν2 (∞, q).

5.3. Proof of “IF” part in Theorem 1.3 In this subsection, we sketch the proof of the “IF” part in Theorem 1.3. Proof of “IF” part in Theorem 1.3. In view of Corollary 1.2, we already have that s2 ,α Lps1 → Mp,q holds if s1 > s2 + n(1 − α)ν1 (p, q). Hence, it suffices to consider the following two cases 1 < p ≤ q ≤ ∞ and (p, q) = (1, ∞). By interpolation between 0,α 0,α 0,α M2,2 ≈ L2 and Lp → Mp,∞ (1 ≤ p ≤ ∞) from Proposition 2.4, we get Lp → Mp,q for 1/q ≤ min(1/p, 1/p ) (see the following figure). This yields from Proposition 2.2 that s1 ,α s2 ,α Lps1 → Mp,q → Mp,q for all s1 ≥ s2 = s2 + n(1 − α)ν1 (p, q), since ν1 (p, q) = 0 for  1/q ≤ min(1/p, 1/p ). 1/q 1

(1/2,1/2)

0

1

1/p

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The inclusion relations on the remaining region, i.e., 1 < p ≤ 2 and p ≤ q ≤ p , can be easily derived from Lemma 5.1 by similar arguments. 2 Remark 5.7. When we proved the inclusion relation for 1/q ≤ min(1/p, 1/p ), we used the interpolation theorem. However, this result can be directly given from Proposition 2.4, the idea of which is a gift from the reviewer. Also, his/her proof is simpler and might be clearer. s,α s,α  By Proposition 2.4, we have Lps → Mp,max(p,p  ) → Mp,q for q ≥ max(p, p ). It is

easy to see that q ≥ max(p, p ) is equivalent to 1/q ≤ min 1/p, 1/p . Thus, using s1 ,α s2 ,α Proposition 2.2, then we obtain Lps1 → Mp,q → Mp,q for all s1 ≥ s2 = s2 + n(1 −  α)ν1 (p, q), since ν1 (p, q) = 0 for 1/q ≤ min(1/p, 1/p ). 5.4. Proof of “ONLY IF” part in Theorem 1.3 We sketch the proof of the “ONLY IF” part in Theorem 1.3. s2 ,α Proof of “ONLY IF” part in Theorem 1.3. We assume that Lps1 → Mp,q . 2 If (p, q) ∈ [1, ∞) , by duality, we have 

s2 ,α 2 ,α Lps1 → Mp,q =⇒ Mp−s → Lp−s1 .  ,q 

Then, by Theorem 1.4, one of the following condition is satisfied; (1) ∞ > p ≥ q  > 1 and −s2 ≥ −s1 + n(α − 1)ν2 (p , q  ); (2) 1 < p < q  ≤ ∞ and −s2 > −s1 + n(α − 1)ν2 (p , q  ); (4) p = ∞, q  = 1, and −s2 > −s1 + n(α − 1)ν2 (∞, q  ), and thus we have (1) 1 < p ≤ q < ∞ and s1 ≥ s2 + n(1 − α)ν1 (p, q); (2) ∞ > p > q ≥ 1 and s1 > s2 + n(1 − α)ν1 (p, q); (4) p = 1, q = ∞, and s1 > s2 + n(1 − α)ν1 (1, q). Here, we used the fact that ν1 (p, q) = −ν2 (p , q  ) for all 1 ≤ p, q ≤ ∞ as stated in Section 1. For the case when 1 ≤ p ≤ ∞ and q = ∞, the desired result is already given by Corollary 1.2. Next, we investigate the case p = ∞ and 1 < q < ∞. In this case, we can see L∞ s1 and  ∗ 1 ∗ −s2 ,α s2 ,α ∞ s2 ,α M∞,q as L−s1 and M1,q , respectively. Thus, by the assumption Ls1 → M∞,q and the Hahn–Banach theorem, we have 

   S,  · M −s2 ,α → L1−s1 ⇐⇒ S,  · M s1 −s2 ,α → L1 , 1,q 

1,q 

as in the proof of the “ONLY IF” part in Theorem 1.4. Here, 1 < q  < ∞. But investigating the proof of Lemmas 3.6 and 5.2, it is easy to see that this suffices to get s1 − s2 > n(1 − α)(1 − 1/q  ) = n(1 − α)/q = n(1 − α)ν1 (∞, q) for 1 < q < ∞.

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Finally, we consider p = ∞ and q = 1. In this case, ν1 (p, q) = ν1 (∞, 1) = 1. In view of Corollary 1.2, we assume towards a contradiction that s1 = s2 + n(1 − α). We note s2 ,α s2 ,α 2 that we have L∞ by Proposition 2.3. s1 → M∞,1 by the assumption and Ls2 ≈ M2,2 Interpolation yields for arbitrary p ∈ (2, ∞) and θ = 2/p ∈ (0, 1) that s2 ,α Lpθs2 +(1−θ)s1 → Mp, 1

θ + 1−θ 2 1

where we used

1 1−θ θ 2+ 1

=

1 1 1− p

s2 ,α = Mp,p  ,

= p . Now, we remark that

s˜1 := θs2 + (1 − θ)s1  2 2 (s2 + n(1 − α)) = s2 + 1 − p p  2 = s2 + n(1 − α) 1 − p = s2 + n(1 − α)ν1 (p, p ). s2 ,α However, this is a contradiction, since we already showed above that Lps˜1 → Mp,p  holds  only if s˜1 > s2 + n(1 − α)ν1 (p, p ). Here, we used that p ∈ (2, ∞), so that ∞ > p > s2 ,α 2 > p > 1. Thus, we obtain that L∞ s1 → M∞,1 holds only if s1 > s2 + n(1 − α) = s2 + n(1 − α)ν1 (∞, 1). Therefore, we complete the proof of Theorem 1.3 on the whole region (p, q) ∈ [1, ∞]2 . 2

Remark 5.8. In the proof for p = ∞ and 1 < q < ∞, recalling Remarks 5.5 and 5.6, s1 −s2 ,α we obtain the following argument. Since 1 < q < ∞, S ⊂ M1,q is dense. Thus,  s1 −s2 ,α given f ∈ M1,q , there exists a sequence of Schwartz functions {fn }n∈N satisfying   fn −f M s1 −s2 ,α → 0. Using the embedding S,  ·M s1 −s2 ,α → L1 , we see that {fn }n∈N 1,q 

1,q 

is a Cauchy sequence in L1 . Since L1 is complete, we obtain fn − gL1 → 0 for some s1 −s2 ,α g ∈ L1 . Recalling that M1,q → S  and L1 → S  , we have fn → f in S  and fn → g    in S  . This yields that f = g. Hence, by S,  · M s1 −s2 ,α → L1 , we have 1,q 

f L1 = gL1 = lim fn L1  lim fn M s1 −s2 ,α = f M s1 −s2 ,α . n→∞

n→∞

1,q 

1,q 

s1 −s2 ,α This means that M1,q → L1 . Then, applying Lemma 5.2, we obtain s1 − s2 >   n(1 − α)(1 − 1/q ) = n(1 − α)/q = n(1 − α)ν1 (∞, q).

6. Proofs of lemmas In this section, we prove the lemmas from Section 3, which we needed to prove our main theorems. This section is divided into three subsections. The first one is the proof for the equivalent norms, and the second and third ones are the proofs of the lemmas needed to establish Theorems 1.5–1.6 and 1.3–1.4, respectively.

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6.1. Proof of equivalent (quasi)-norms for α-modulation spaces Proof of Lemma 3.2. We divide the proof into four steps, in which we show the estimates “  ” and “  ” for 0 < p ≤ 1 or for 1 < p ≤ ∞, respectively. Since the proof techniques for 0 < p ≤ 1 are very similar to those for 1 < p ≤ ∞, we handle the case 0 < p ≤ 1 in detail and only sketch the case 1 < p ≤ ∞. Step 1. Let 0 < p ≤ 1. Then, we first prove 



s,α  f Mp,q

q   p ksq/(1−α) F −1 Ψα k Ff L

1/q .

k∈Zn

Choose an auxiliary function κ ∈ S satisfying 1 on |ξ| ≤ 1, 0 on |ξ| ≥ 2,

κ(ξ) =

(12)

and  κα  (ξ) := κ

ξ − α/(1−α)  Cα/(1−α)

,

α α then κα  = 1 on the support of η , with η and C as in Section 2.2. By Proposition 2.1, we have  p  −1 α p α α  p F Ψk Ff  p ≤ F −1 [(Ψα k η κ ) · Ff ] L

L

∈Zn





   −1 α p α p  An(1−p) F −1 [Ψα η Ff Lp . k κ ] Lp · F

∈Zn

Thus, we consider   α p An(1−p) F −1 [Ψα k κ ] Lp

 p      −n An(1−p)  ix·ξ α α = (2π)   e Ψk (ξ)κ (ξ)dξ    Rnξ 

Lp (Rn x)

for the following 6 cases. (i) (ii) (iii)

|k| ≤ C  and || ≤ |k|/2, |k| ≥ C  and || ≤ |k|/2, |k| ≤ C  and |k|/2 ≤ || ≤ 2|k|,

where C  = 6C.

(iv) |k| ≥ C  and |k|/2 ≤ || ≤ 2|k|, (v) |k| ≤ C  and || ≥ 2|k|, (vi) |k| ≥ C  and || ≥ 2|k|,

(13)

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Cases (i) and (iii). Since |k|, ||  1 for the cases (i) and (iii), we clearly have for ˜ >0 sufficiently large N   ˜p α p −N An(1−p) F −1 [Ψα . k κ ] Lp  k −  Case (ii). (⇒  ≤ k). Changing variables by ξ → kA ξ and kA x → x,   α p An(1−p) F −1 [Ψα k κ ] Lp p  An(1−p)   A       A  k ξ −   ξ−k   κ dξ = (2π)−n   eix·ξ Ψ   k C CA  Rnξ

.

(14)

Lp (Rn x)

From the definition of the function κ, its support satisfies the following.  supp κ

kA  −A  CA



# ⊂

  $  A  2CA  ≤ ξ∈R :ξ− .  kA  kA n

This implies that the domain of integration in (14) is contained in the set {ξ ∈ Rn : |ξ| ≤ 5|k|/6}. Thus, since the function Ψ is a Schwartz function, we have on this domain of integration for all β ∈ Zn+ and N ∈ Z+       −N  β  ξ−k    1 +  ξ − k   ∂ Ψ   ξ C C

−N  1 + |k| ≤ k−N  k−N/2 k − −N/2 .

(15)



1/2 In the third line, we used that x = 1 + |x|2 ≤ 1 + |x|. In the last line, we used that

2 9 9 k − 2 ≤ 1 + |k| + || ≤ 1 + |k|2 ≤ k2 4 4 from the assumption || ≤ |k|/2. On the other hand, we have n 1  ∂eix·ξ xj = i eix·ξ for |x| = 0. |x|2 j=1 ∂ξj

Thus, by dividing the Lp -integral into the cases |x| ≤ 1 and |x| ≥ 1, and by integrating ˜ > 0 that by parts sufficiently often for |x| ≥ 1, it follows for sufficiently large N   ˜p α p −N An(1−p) F −1 [Ψα . k κ ] Lp  k − 

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On each time when we use integration by parts,

k A  A

≤ kA appears from the function

κ in (14). However, we can cancel it by k−N/2 in the estimate (15). Case (iv). Changing variables by ξ → A ξ, ξ −  → ξ, and A x → x,   α p An(1−p) F −1 [Ψα k κ ] L p p      A   A A ξ  ξ +   − k k   κ dξ  = (2π)−n  eix·ξ Ψ A   Ck C  Rnξ

.

(16)

Lp (Rn x)

Note that κ(ξ/C) = 0 unless |ξ| ≤ 2C. Thus, the domain of integration in (16) is contained in {ξ : |ξ| ≤ 2C}. For these ξ, using Ψ ∈ S, we have for all β ∈ Zn+ and N ∈ Z+ and a suitable large constant Kα ≥ 1, depending on α,  A      A A  −N   β  A ξ + A  − kA k   ∂ Ψ   Kα +   ξ +   − k k    ξ   CkA CkA  A     − kA k  A |ξ| −N   ≤ Kα +   − CkA CkA   −N   1  A  1 +  A  − k  (17) C k   −N   1  kA ∼ 1 +  A k −   . (18) C  

More precisely, we can choose Kα = 1 + 2A+1 . We estimate (18) and (17) by considering the two cases |k|/2 ≤ || ≤ |k| and |k| < || ≤ 2|k|, respectively. First, we consider the estimate (18) in the case |k|/2 ≤ || ≤ |k| (⇒ 1 ≤ k/ ≤ 2). The point (k/)A k moves between the points k and 2A k on the line “L” and the point    moves in ξ : |k|/2 ≤ |ξ| ≤ |k| (see the following figure).

Here, ξ = (ξ  , ξn ) ∈ Rn−1 × R. Hence, we have

T. Kato / Journal of Functional Analysis 272 (2017) 1340–1405

1373

 A   k     A k −   ≥ |k − |. We can also get this fact by the following direct estimates. Set M = (k/)A . Since it is clear when M = 1 (i.e., |k| = ||), we assume that 1 < M ≤ 2A . Then we have |M k −  | ≥ |k − | ⇐⇒ (M 2 − 1)|k|2 − 2(M − 1)k ·  ≥ 0 ⇐⇒ (M + 1)|k|2 /2 ≥ k · . Now, since we assume that M > 1 and || ≤ |k|, the last inequality holds true. Thus, it follows that     β  A ξ + A  − kA k 

  1 + |k − | −N ≤ k − −N ,  ∂ Ψ   ξ A Ck

1/2 where we used x = 1 + |x|2 ≤ 1 + |x|. For the case |k| < || ≤ 2|k|, we get the same result by using a similar argument on the estimate (17). Analogously to Case (ii), using integration by parts sufficiently often, then we have ˜ >0 for sufficiently large N   ˜p α p −N An(1−p) F −1 [Ψα . k κ ] Lp  k −  Case (v). (⇒ k ≤ .) We estimate the right hand side of the expression (16) under this case. Since the case 2|k| ≤ || ≤ 2C  is clear as in Cases (i) and (iii), we only state the case 2|k| ≤ 2C  ≤ ||. We also note that the domain of integration in (16) is contained in {ξ : |ξ| ≤ 2C}. Using Ψ ∈ S, we have for all β ∈ Zn+ and N ∈ Z+  A      A A  −N   β  A ξ + A  − kA k    1 +   ξ +   − k k   ∂ Ψ    ξ  CkA CkA  # $−N  1 A  1+ || − |ξ| − |k| C kA

−N  1 + || ≤

≤ −N  −N/2 k − −N/2 .

(19)

Here, in the third line, we used |ξ| ≤ 2C = C  /3 ≤ ||/6, |k| ≤ ||/2, and /k ≥ 1. In the fourth line, we used that x ≤ 1 + |x|. In the last line, we used that

2 9 9 k − 2 ≤ 1 + |k| + || ≤ 1 + ||2 ≤ 2 4 4

T. Kato / Journal of Functional Analysis 272 (2017) 1340–1405

1374

from the assumption |k| ≤ ||/2. By the same argument as in the previous cases, we have ˜ >0 for sufficiently large N   ˜p α p −N An(1−p) F −1 [Ψα . k κ ] Lp  k −  As mentioned at the end of Case (ii), some parts sufficiently often. However, since

 A k A

 A -factors k A A

appear when we integrate by

≤  , we can cancel all these A -factors

by −N/2 in the inequality (19). Case (vi). In this case, || ≥ 2|k| ≥ 2C  . By the same argument as in Case (v), we ˜ >0 have for sufficiently large N   ˜p α p −N An(1−p) F −1 [Ψα . k κ ] Lp  k −  Taking Cases (i)–(vi) together and recalling (13), we get 



 q  p ksq/(1−α) F −1 Ψα k Ff

1/q

L

k∈Zn

⎛ ⎝



 ksq/(1−α)

k∈Zn

⎛ ⎝





   −1 α p α p  An(1−p) F −1 [Ψα η Ff Lp k κ ] L p · F

q/p ⎞1/q ⎠

∈Zn



k∈Zn



p  ˜ k − (|s|/(1−α)−N )p sp/(1−α) F −1 ηα Ff Lp

q/p ⎞1/q ⎠

,

(20)

∈Zn

with the usual modification for q = ∞. In the last inequality, we used the following inequalities. kx ≤ 2x/2 x k − x kx = (k−|x| k − −|x| )k − |x| ≤ 2|x|/2 −|x| k − |x|

if x ≥ 0, if x < 0.

(21)

˜ > 0, we have If 0 < q/p < 1, by the Fubini–Tonelli theorem and sufficiently large N  (20) ≤

 

˜ )q (|s|/(1−α)−N

k − 



sq/(1−α)

 −1 α q F η Ff  p L

1/q

k∈Zn ∈Zn

=





∈Zn s,α .  f Mp,q

sq/(1−α)

 −1 α q F η Ff  p L





k∈Zn

!1/q ˜ )q (|s|/(1−α)−N

k − 

T. Kato / Journal of Functional Analysis 272 (2017) 1340–1405

1375

In the first line, we used (a + b)t ≤ at + bt for a, b > 0 and 0 < t ≤ 1. On the other hand, ˜ > 0, we if q/p ≥ 1, by the convolution relation 1 ∗ q/p → q/p and sufficiently large N have ⎧   ⎫1/p ⎨  ⎬ q/p p/q    p ˜ (20) ≤ · k(|s|/(1−α)−N )p sp/(1−α) F −1 ηα Ff Lp ⎩ ⎭ n n ∈Z

k∈Z

s,α .  f Mp,q

Hence, we have  s,α  f Mp,q



k

sq/(1−α)

 −1 α q F Ψk Ff  p

1/q .

L

k∈Zn

Finally, we remark that we only used Ψ ∈ S in this step. In the next step, namely, the estimate “  ” for 0 < p ≤ 1, we will use the assumption which |Ψ| ≥ c > 0 around the origin. Step 2. Next, we show for 0 < p ≤ 1 that  s,α  f Mp,q



k

sq/(1−α)

 −1 α q F Ψk Ff  p L

1/q .

k∈Zn

In view of the equivalent norm given in Proposition 3.1, it suffices to show that 



k

sq/(1−α)

 −1 α q F ρk Ff  p



1/q 

L

k∈Zn



k

sq/(1−α)

 −1 α q F Ψk Ff  p

1/q ,

L

k∈Zn

α,2 where ρα and k is the smooth radial bump function from Proposition 3.1. We set κk κα,M as k

 κα,2 k (ξ)

=κ

ξ − kα/(1−α) k 2Ckα/(1−α)



 and

κα,M (ξ) k

=κ

ξ − kα/(1−α) k M Ckα/(1−α)



with sufficiently large M = Mα,n,p > 0. Here κ is the same auxiliary function as in Step 1 and C > 1 is the constant taken from Proposition 3.1. Then κα,2 = κα,M = 1 on k k α α supp ρα . Moreover, since |Ψ| ≥ c > 0 on |ξ| ≤ 2, Ψ never vanishes on supp ρ . Thus we k k k have  ' ( p α    −1 α p α,2 α,M Ψk  F ρk Ff  p =  F −1 ρα · Ff κ κ k k k α  p  L Ψk L ' ( p  )  * p    −1 α,2 ρα

n(1−p)  −1  α k   F κ  M kA Ψ · κα,M · Ff F  k k k α  p Ψ  p L k

L

T. Kato / Journal of Functional Analysis 272 (2017) 1340–1405

1376

= M

n(1−p)

 )  * p   −1 Ψα κα,M F k · Ff  p k L

 ' ( p  −1 ρ(·/C)    κ(·/2C) ·F Ψ(·/C)  p L

 )  * p   ∼ M n(1−p)  F −1 κα,M Ψα k · Ff  p . k

(22)

L

In the second line, we applied Proposition 2.1 with     A A supp κα,M Ψα , k · Ff ⊂ ξ : |ξ − k k| ≤ M Ck k      ρα k ⊂ ξ : |ξ − kA k| ≤ 2CkA ⊂ ξ : |ξ − kA k| ≤ M CkA . supp κα,2 k α Ψk Now, using the fact that F −1

)

 * )  * α,M α α −1 α −1 α Ψ Ψ Ff − F − κ Ψ κα,M · Ff = F Ψ · Ff , k k k k k k

we have by Proposition 2.1   )   * p  p   −1 α,M  (22) ≤ M n(1−p) F −1 Ψα Ψα · Ff  p k Ff Lp +  F k 1 − κk L  p n(1−p)  −1 α F Ψk Ff Lp M  )   * p   p   α,M + M n(1−p) An(1−p)  F −1 Ψα · F −1 ηα Ff Lp . κα k 1 − κk   ∈Zn

Lp

α Here, κα  is the auxiliary function from Step 1 which satisfies that κ = 1 on the support of ηα . Changing variables in the first factor of the summand of the series above, we have

 )   * p   α,M An(1−p)  F −1 Ψα κα k 1 − κk   p L     ξ A ξ + A  − kA k −n  ix·ξ = (2π)  e κ Ψ  C CkA Rnξ p  $ #  A   ξ + A  − kA k  × 1−κ dξ   M CkA 

.

(23)

Lp (Rn x)

We consider this expression (23) in the following argument. On the other hand, since we have        α,M supp Ψα 1 − κ ⊂ ξ : ξ − kA k  ≥ M CkA , k k     A  A  supp κα ,  ⊂ ξ : ξ −   ≤ 2C the term (23) always vanishes unless the condition

T. Kato / Journal of Functional Analysis 272 (2017) 1340–1405

1377

  A k k − A   ≥ M CkA − 2CA

(24)

is satisfied. Now, we estimate the expression (23) for k,  ∈ Zn by considering the following three cases. (i)

|| ≤ |k|/2,

(ii)

|k|/2 ≤ || ≤ 2|k|,

|| ≥ 2|k|.

(iii)

Case (i). Note that we can always assume that (24) holds, since otherwise (23) = 0. We also remark that κ(ξ/C) = 0 implies |ξ| ≤ 2C, so that we can restrict the domain   of integration in (23) to {ξ : |ξ| ≤ 2C}. On this set, A ξ/(CkA ) ≤ 2. Because of Ψ ∈ S, we thus have for all β ∈ Zn+ and N ∈ Z+ , and for sufficiently large M > 0,  A      A  −N   β  A ξ + A  − kA k    2 +    − k k   ∂ Ψ     CkA CkA  ≤

2+M −2

A kA

−N

≤ M −N , where we used the condition (24) and /k ≤ 1. On the other hand, we also have  A      A  −N   β  A ξ + A  − kA k   ∂ Ψ   1 +    − k k      CkA CkA  (1 + |k| − ||)  (1 + |k|)

−N

−N

≤ k−N  k−N/2 k − −N/2 .

1/2 Here, in the fourth and last lines, we used that x = 1 + |x|2 ≤ 1 + |x| and

2 9 2 2 k− ≤ 1+ |k|+|| ≤ 4 k from the assumption || ≤ |k|/2, respectively. Integrating by parts sufficiently often as in Step 1, we obtain ˜

˜

(23)  M −N p k − −N p ˜ > 0. Although A /kA appears on each time when we integrate for sufficiently large N by parts, we can see it as a constant since A /kA ≤ 1. Case (ii). Note that we can always assume that (24) holds, since otherwise (23) = 0.   As above, on the relevant domain of integration in (23), we have A ξ/(CkA ) ≤ 2A+1 . Since Ψ ∈ S, we get for β ∈ Zn+ and N ∈ Z+ that

1378

T. Kato / Journal of Functional Analysis 272 (2017) 1340–1405

 A   −N      β  A ξ + A  − kA k     1 + 1    − k   ∂ Ψ    ξ  A A Ck C k   −N   1  kA ∼ 1 +  A k −   C 

(25)

(26)

by the same argument as in Case (iv) of Step 1. We estimate (26) and (25) by considering the two cases |k|/2 ≤ || ≤ |k| and |k| < || ≤ 2|k|, respectively. We first consider (26) in the case |k|/2 ≤ || ≤ |k| (⇒ k/2 ≤  ≤ k). As we proved in Case (iv) of Step 1, we have  A   −N      β  A ξ + A  − kA k     1 +  k k −   ∂ Ψ     A A Ck  −N

≤ (1 + |k − |) ≤ k − −N ,

where we used that x ≤ 1 + |x|. On the other hand, as in Case (i), applying the condition (24),     β  A ξ + A  − kA k    M −N .  ∂ Ψ   CkA ˜ >0 Thus we obtain for sufficiently large N ˜

˜

(23)  M −N p k − −N p . Since the proof for the case |k| < || ≤ 2|k| is same as above, we omit it. Case (iii). We divide this case into the following three cases. (a) 2|k| ≤ || ≤

1 2

M 1−α 2

, (b) 2|k| ≤

1 2

M 1−α 2

≤ ||, (c)

1 2

M 1−α 2

≤ 2|k| ≤ ||.

We first consider (23) in the case (a). In this case, it follows that 1 ≤ /k ≤ (M/2)1−α and from the condition (24)   A     Mα    kA  − k  ≥ C M − 2 · 2α .

(27)

Since M = Mα,n,p > 0 is a sufficiently large number, we can take M > 0 such that

1 M 1−α ≥ 6C (⇔ M/24 ≥ C(M/2)α ), where C > 1 is the constant from the equivalent 2 2 norm for α-modulation space from Proposition 3.1. In particular, this choice of M implies

T. Kato / Journal of Functional Analysis 272 (2017) 1340–1405

that 2 ·

Mα 2α

≤

2 C

·

M 24

≤

M 12

≤

M 2 ,

1379

so that (27) yields that

 A        kA  − k ≥ C (M − M/2) = M C/2. Hence, since Ψ ∈ S, we have for all β ∈ Zn+ and N ∈ Z+      β  A ξ + A  − kA k   ∂ Ψ  1+ 1   CkA C

 A     − kA k  |ξ| A −N  −   C kA kA −N

≤ (1 + M/2 − M/12)  M −N ≤ −N

 −N/2 k − −N/2 .

(28)

A

Here, we used |ξ| ≤ 2C and (/k) ≤ (M/2)α ≤ M/(24C) ≤ M/24 in the second line, and  ≤ (M/2)1−α ≤ M in the fourth line. Moreover, in the last line, we used

2 that k − 2 ≤ 1 + |k| + || ≤ 94 2 from the assumption 2|k| ≤ ||. Integrating by ˜ >0 parts sufficiently often, we have for sufficiently large N ˜

˜

(23)  M −N p k − −N p . Although A /kA appears on each time when we use integration by parts, we can cancel them by −N/2 in the inequality (28). Next, we consider (23) in the case (b). Since we take M > 0 satisfying that

1−α 1 M 1−α ≥ 6C, it follows that || ≥ 12 M ≥ 6C ≥ 3|ξ| on the relevant domain of 2 2 2 n integration. Thus, we obtain for all β ∈ Z+ and N ∈ Z+     −N  A  β  A ξ + A  − kA k    1 + 1  || − |ξ| − |k|  ∂ Ψ   CkA C kA C  −N  M −(1−α)N/2 −N/4 k − −N/4 from k −  ≤ 3/2 and  ≥ ˜ >0 sufficiently large N

1 2

M 1−α 2

. Thus, by integration by parts, we have for

˜

˜

(23)  M −N p k − −N p . For the case (c), we can apply the same argument as is used in the case (b), so that we omit it.

T. Kato / Journal of Functional Analysis 272 (2017) 1340–1405

1380

Taking all cases together and recalling estimates (21), (22) and (23), 



q   p ksq/(1−α) F −1 ρα k Ff

1/q

L

k∈Zn







 −1 α p F Ψk Ff  p

ksq/(1−α) M n(1/p−1)q

L

k∈Zn



+

 )   * p     α,M  −1 ηα Ff p p An(1−p)  F −1 Ψα κα k 1 − κk   p · F L

⎠

L

∈Zn

 M

!q/p ⎞1/q



n(1/p−1)

k

sq/(1−α)

 −1 α q F Ψk Ff  p

1/q

L

k∈Zn

⎛ + M n(1/p−1)−N ⎝ ˜







p ˜  sp/(1−α) k − |s|p/(1−α)−N p F −1 ηα Ff  p L

k∈Zn

q/p ⎞1/q ⎠

.

∈Zn

Using the convolution relation 1 ∗q/p → q/p if q/p > 1 and the Fubini–Tonelli theorem if 0 < q/p ≤ 1, we have 



s,α ∼ f Mp,q

k

sq/(1−α)

 −1 α q F ρk Ff  p L

1/q

k∈Zn

  M

n(1/p−1)



k

sq/(1−α)

 −1 α q F Ψk Ff  p L

1/q ˜

s,α . + M n(1/p−1)−N f Mp,q

k∈Zn

˜ > 0 as sufficiently large numbers, we see that Since we take M > 0 and N  s,α  f Mp,q



q   ksq/(1−α) F −1 Ψα k Ff Lp

1/q .

k∈Zn

Hence, combining Step 1 and Step 2, we obtain the desired results for 0 < p ≤ 1. Step 3. Here, we prove for 1 < p ≤ ∞ that  s,α  f Mp,q



k

sq/(1−α)

 −1 α q F Ψk Ff  p L

1/q .

k∈Zn

With the same choice of the function κ as in Step 1, it follows from the Young inequality that

T. Kato / Journal of Functional Analysis 272 (2017) 1340–1405

 −1 α  F Ψk Ff 

Lp

≤

  α α   F −1 [(Ψα k η κ ) · Ff ]

Lp

∈Zn

≤

1381

  α   F −1 [Ψα k κ ]

L1

  · F −1 ηα Ff Lp .

∈Zn

  ˜ α  −N ˜ ∈N Then, recalling from Step 1 the estimate  F −1 [Ψα for all N k κ ] L1  k −  ˜ and (21), we have for sufficiently large N > 0 



k

sq/(1−α)

 −1 α q F Ψk Ff  p

1/q

L

k∈Zn

 





k∈Zn

∈Zn

˜ |s|/(1−α)−N

k − 



s/(1−α)

 −1 α  F η Ff 

!q 1/q Lp

⎧ 1/q ⎪    −1 α q ⎪ ˜q ⎪ |s|q/(1−α)− N sq/(1−α) F η Ff  p ⎪ k −   ⎪ L ⎨ k∈Zn ∈Zn ≤  !q 1/q ⎪   ⎪   ˜ ⎪ |s|/(1−α)−N s/(1−α)  −1 α ⎪  F η Ff Lp k −   ⎪ ⎩ k∈Zn

if 0 < q ≤ 1, if 1 < q ≤ ∞

∈Zn

⎧ 1/q ⎪   −1 α q  ⎪ ˜q ⎪ sq/(1−α) |s|q/(1−α)− N   ⎪ F η Ff Lp  k −  ⎪ ⎨ ∈Zn k∈Zn ≤  1/q   ⎪    −1 α q ⎪ ˜ ⎪ sq/(1−α) |s|/(1−α)− N F η Ff  p ⎪  · k ⎪ ⎩ L ∈Zn

if 0 < q ≤ 1, if 1 < q ≤ ∞

k∈Zn

s,α .  f Mp,q

Here, we used the Fubini–Tonelli theorem for 0 < q ≤ 1 and the convolution relation 1 ∗ q → q for 1 < q ≤ ∞ in the third and fourth inequality, respectively. Step 4. For 1 < p ≤ ∞, we prove that 



s,α  f Mp,q

k

sq/(1−α)

 −1 α q F Ψk Ff  p

1/q

L

.

k∈Zn

By Proposition 3.1, we have  s,α ∼ f Mp,q



k

sq/(1−α)

 −1 α q F ρk Ff  p L

1/q .

k∈Zn

The Young inequality and the suitable changes of variables yield that

1382

T. Kato / Journal of Functional Analysis 272 (2017) 1340–1405

 −1 α  F ρk Ff 

Lp

 ' (  −1 ρα  α k  F = · Ψ Ff k α   p Ψk L  ' (   −1 ρα   k  · F −1 Ψα  ≤ k Ff Lp α  F Ψk L1  )ρ*      = F −1  1 · F −1 Ψα k Ff Lp Ψ L    ∼ F −1 Ψα k Ff Lp .

α In the above estimate, we remark that |Ψα k | ≥ c > 0 on the support of ρk and |Ψ| ≥ c > 0 on the support of ρ. Therefore, the proof is complete. 2

6.2. Proofs of lemmas for Theorems 1.5 and 1.6 Here, we prove Lemmas 3.3–3.5 for the inclusion relations between α-modulation and local Hardy spaces. The first lemma is to give the “IF” part of Theorem 1.5 for 0 < p ≤ 1 and 0 < q ≤ 2.

Proof of Lemma 3.3. Let Ψ ∈ S satisfy that supp F −1 Ψ ⊂ {x : |x| ≤ 1} and |Ψ| ≥ c > 0 on |ξ| ≤ 2 (the existence of such a function is proven in [12, Lemma 4.3]). We set the " Friedrichs mollifier as {ε }ε>0 , that is, ε := ε−n (x/ε) for  ∈ S with Rn (x)dx = 1 and supp  ⊂ [−1, 1]n . Then, (ε ∗ a)/(2n L1 ) is an hp -atom of type I (resp. type II) if a is an hp -atom of type I (resp. type II) and ε > 0 is sufficiently small (see [12]). In the following statement, we denote aε := (ε ∗ a)/(2n L1 ). By the Fatou lemma, we have  a

s,α Mp,q

=



k

k∈Zn

sq/(1−α)





≤ lim inf ε→0

ε→0

k

sq/(1−α)

1/q

 −1 α  F ηk Faε q p

1/q

L

k∈Zn

  lim inf

 −1 α q F ηk Fa p L



k

sq/(1−α)

 −1 α  F Ψk Faε q p L

1/q ,

k∈Zn

where Ψα k is the same function as is set in Proposition 3.2. Here, we should note that s,α ε ∗ a ∈ S ⊂ Mp,q and limε→0 ε ∗ a = a in S  (also in Lp ). Thus, since lim inf ε→0 bε ≤ lim supε→0 bε , it is enough to prove that  lim sup ε→0



k

sq/(1−α)

 −1 α  F Ψk Faε q p L

1/q (29)

k∈Zn

 is bounded independently of the specific hp -atom a. We split the sum k into two parts, so that we changed “lim inf” to “lim sup”. We note that, in general, lim sup X +



Y ≤ lim sup X + lim sup Y is true, but this is not true for “lim inf” (lim inf X + Y ≥

T. Kato / Journal of Functional Analysis 272 (2017) 1340–1405

1383

lim inf X + lim inf Y is true, in general). Now, we assume that a is an hp -atom with the cube Q = [−r, r]n , without loss of generality, since  · Lp is invariant under translations. Then, supp aε ⊂ [−(r + ε), r + ε]n . In the following, we divide this proof into two steps for type I and type II atoms, respectively. Step 1. First, we assume that a is an hp -atom of type I with Q = [−r, r]n . By the definition of an hp -atom of type I, we have |Q| < 1, i.e., r < 1/2. We consider (29) in the following two cases. (i)

k < |Q|−(1−α)/n ,

(ii)

k ≥ |Q|−(1−α)/n .

Case (i). We write N = [n(1/p − 1)], where [x] = max{n ∈ Z : n ≤ x}. By the " definition of an hp -atom of type I, we have xβ aε (x)dx = 0 for |β| ≤ N , which we will now apply to a suitable Taylor expansion. Indeed, by the Fubini–Tonelli theorem, suitable changes of variables and the Taylor expansion, we have  −1 α  F Ψk Faε p p  L p     + ,

 ˇ CkA (x − y) dy  = C np kAnp  aε (y) exp ikA k · (x − y) Ψ    Rny  Lp (Rn x)    + ,

 ˇ CkA (x − y) = C np kAnp  aε (y) exp ikA k · (x − y) Ψ  Rny  ! p

β )   * A  −Ck y

k  ˇ () CkA x dy  − ∂ β ei C · Ψ  β! |β|≤N  Lp (Rn x)  1

 β  A  −Ck y  = C np kAnp  aε (y)(N + 1) (1 − θ)N  β! |β|=N +1

Rn y

)

· ∂

β



k iC ·

e

0

ˇ () Ψ

*

p   Ck (x − θy) dθ dy   p A



!

.

L (Rn x)

In the second equality, we used that aε is an hp -atom of type I. Hence, we see that y ∈ supp aε () ⊂ [−(r + ε), r + ε]n ,  + ,

 ˇ CkA  + supp aε () x ∈ supp exp ikA k ·  · Ψ (n ' 1 1 ⊂ − , + [−(r + ε), r + ε]n CkA CkA '  ˜  ˜ (n C C ⊂ − + ε , + ε . kA kA

(30)

T. Kato / Journal of Functional Analysis 272 (2017) 1340–1405

1384

Here, we used r < 2r < k−1/(1−α) ≤ k−A , which follows from k < |Q|−(1−α)/n = (2r)−(1−α) . We denote the sets V0 and V1 by V0 := [−(r + ε), r + ε]n , '  ˜  ˜ (n C C V1 := − + ε , + ε . kA kA



−n/p Then, since we have ∂ β exp[i Ck · ] ∞  k|β| and aε L∞ ≤ 2(r + ε) , we get  −np/p kAnp 2(r + ε)



(30)



(N +1)p

× kA(N +1)p r + ε

⎧ ⎛ ⎞p ⎫  ⎨ ⎬ ⎝ 1dy ⎠ dx k(N +1)p · ⎩ ⎭ V1

V0

 −n  (N +1)p kAnp 2(r + ε) kA(N +1)p 2r + ε  np

n × k(N +1)p 2(r + ε) k−A + ε



−−−→ kAp{(N +1)−n(1/p−1)} k(N +1)p |Q|p{(N +1)/n−(1/p−1)} . ε→0

In order to estimate (29), summing (30) on k < |Q|−(1−α)/n and using s = −n(1 − α) (1/p + 1/q − 1), we have ⎛ ⎝lim sup ε→0

⎞1/q   q ⎠ k−n(1/p−1)q−n F −1 Ψα k Faε Lp

 k <|Q|−(1−α)/n

⎛  |Q|

(N +1)/n−(1/p−1)

⎝



⎞1/q −n

k

{(N +1)−n(1/p−1)}q/(1−α) ⎠

k

k <|Q|−(1−α)/n

 |Q|(N +1)/n−(1/p−1) · |Q|−{(N +1)−n(1/p−1)}/n = 1,   where we used the fact (N + 1) − n(1/p − 1) q/(1 − α) > 0 in the last inequality. Case (ii). By the Fubini–Tonelli theorem and suitable changes of variables, we have  −1 α  F Ψk F aε  p L

     + ,

  n An  A A ˇ = C k  aε (y) exp ik k · (x − y) Ψ Ck (x − y) dy     Rny

, Lp (Rn x)

(31) and the integrand in (31) vanishes unless we have

T. Kato / Journal of Functional Analysis 272 (2017) 1340–1405

1385

y ∈ supp aε () ⊂ [−(r + ε), r + ε]n ,  ,

 + ˇ CkA  + supp aε () x ∈ supp exp ikA k ·  · Ψ (n ' 1 1 , + [−(r + ε), r + ε]n ⊂ − CkA CkA + , ˜ α + ε), Cr ˜ α+ε n. ⊂ −(Cr Here, we used the facts that k−A ≤ (2r)α from the assumption k ≥ |Q|−(1−α)/n = (2r)−(1−α) , and r < rα from 0 < r < 1/2. Thus, using the Hölder inequality, we have n(1/p−1/2)  ˜ α + ε) (31) ≤ C n kAn 2(Cr       + ,

 ˇ CkA (x − y) dy  ×  aε (y) exp ikA k · (x − y) Ψ    Rny  

n(1/p−1/2)   F −1 Ψα  ∼ (2r)α + ε k Faε L2 .

L2 (Rn x)

(32)

Again, in order to estimate (29), summing (31) on k ≥ |Q|−(1−α)/n and using s = −n(1 − α) (1/p + 1/q − 1), we have 

 q  k−n(1/p+1/q−1)q F −1 Ψα k Faε Lp

k ≥|Q|−(1−α)/n

⎛

≤⎝



⎞1/u ⎛ k−n(1/p+1/q−1)qu ⎠

k ≥|Q|−(1−α)/n

·⎝

⎞q/2  −1 α 2 F Ψk Faε  p ⎠ . L



k ≥|Q|−(1−α)/n

(33) In this inequality, we used the Hölder inequality with 1/u + q/2 = 1, i.e., 1 < u ≤ ∞ from 0 < q ≤ 2. This implies that n(1/p + 1/q − 1)qu > n. Therefore, since we have (32), 0,α M2,2 = L2 , and aε L∞ ≤ (2(r + ε))−n/p (from the definition of the type I hp -atom), we obtain

(33)



⎛  −n(1/p+1/q−1)qu+n 1/u ·⎝ |Q|−(1−α)/n 

nq(1/p−1/2)



⎞q/2  −1 α 2 F Ψk Faε  2⎠ L

k ≥|Q|−(1−α)/n

· (2r)α + ε  ≤

 nq(1/p−1/2) q |Q|(1−α){(1/p+1/q−1)q−1/u} · aε L2 · (2r)α + ε  nq(1/2−1/p)  nq(1/p−1/2) |Q|(1−α){(1/p+1/q−1)q−1+q/2} · 2(r + ε) · (2r)α + ε

−−−→ |Q|(1−α)(1/p−1/2)q · |Q|−q(1/p−1/2) · |Q|αq(1/p−1/2) = 1. ε→0

T. Kato / Journal of Functional Analysis 272 (2017) 1340–1405

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Hence, if a is an hp -atom of type I, then we have s,α ≤ K aMp,q

for s = −n(1 − α)(1/p + 1/q − 1), 0 < p ≤ 1, and 0 < q ≤ 2. Here, the constant K = K(n, p, q, α) is universal and independent of the type I hp -atom a. Step 2. Next, we assume that a is an hp -atom of type II with Q = [−r, r]. By the definition of a type II hp -atom, we have 1 ≤ |Q| = (2r)n , i.e., r ≥ 1/2. In particular, k−A ≤ 1 ≤ 2r. Now, we note that  −1 α  F Ψk F aε 

Lp

      + ,

n An  A A ˇ Ck (x − y) dy  = C k  aε (y) exp ik k · (x − y) Ψ     Rny

, Lp (Rn x)

(34) where the integrand in (34) vanishes unless we have y ∈ supp aε () ⊂ [−(r + ε), r + ε]n ,  ,

 + ˇ CkA  + supp aε () x ∈ supp exp ikA k ·  · Ψ ' (n 1 1 ⊂ − , + [−(r + ε), r + ε]n CkA CkA + , ˜ + ε), Cr ˜ +ε n. ⊂ −(Cr Thus, using the Hölder inequality, we have  n(1/p−1/2) ˜ + ε) (34) ≤ C n kAn 2(Cr      + ,

  A A ˇ Ck (x − y) dy  ×  aε (y) exp ik k · (x − y) Ψ    Rny  

n(1/p−1/2)   F −1 Ψα  ∼ 2r + ε k Faε L2 .

L2 (Rn x)

(35)

Thus, by (35), we have 

 ≤



−n(1/p+1/q−1)q

k

 −1 α  F Ψk Faε q p L

k∈Zn

 k∈Zn

1/u  −n(1/p+1/q−1)qu

k

·

1/q

  2 F −1 Ψα  p k Faε L

k∈Zn

q/2

T. Kato / Journal of Functional Analysis 272 (2017) 1340–1405





n(1/p−1/2)q



1 · 2r + ε

  2 F −1 Ψα  2 k Faε

1387

q/2

L

k∈Zn



∼

n(1/p−1/2)q 2r + ε



≤

q

· aε L2

n(1/p−1/2)q  n(1/2−1/p)q · 2(r + ε)

2r + ε

−−−→ |Q|(1/p−1/2)q · |Q|−(1/p−1/2)q = 1, ε→0

where u is the same exponent as used in the case of type I hp -atoms, which yields n(1/p + 1/q − 1)qu > n. Hence, if a is an hp -atom of type II, then we have for s = −n(1 − α)(1/p + 1/q − 1), 0 < p ≤ 1, and 0 < q ≤ 2 s,α ≤ K, aMp,q

where the universal constant K = K(n, p, q, α) is independent of the type II hp -atom a. 2 Next we prove the lemma needed to show the “ONLY IF” part of Theorem 1.6. Proof of Lemma 3.4. We choose ϕ ∈ S \ {0} satisfying that supp ϕ ⊂ {ξ : |ξ| ≤ 1/2} and define 

f (x) =

+ ,

c An/p exp iA  · x ϕ CA (x − )

∈Zn

for some finitely supported sequence {c }∈Zn . Here, the constant C > 1 is that one from Proposition 3.1. Then, changing variables by x −  → x and CA x → x, we obtain f(ξ) =



 c An/p

∈Zn

=C

−n

+ ,

exp −iξ · x + iA  · x ϕ CA (x − ) dx

Rn x

 ∈Zn

c 

An(1/p−1)

 + , ξ − A  A 2 . exp −iξ ·  + i || ϕ  CA

Then, for the above function f, we have  −1 α p F ρk Ff  p L

p        ξ − A   −n p An(1−p)  α dξ ≤ (2πC) |c |     exp [ix · ξ − iξ · ] ρk (ξ)ϕ A   C ∈Zn  Rnξ

Lp (Rn x)

T. Kato / Journal of Functional Analysis 272 (2017) 1340–1405

1388

p        A A ξ −   ξ − k k   ϕ  dξ = (2πC)−n |c |p An(1−p)  eix·ξ ρ  A A   Ck C ∈Zn  Rnξ 

Lp (Rn x)

p  An(1−p)      A    A A  k ξ + k k −    ξ   −n ϕ  dξ |c |p   eix·ξ ρ = (2πC) A   k C C ∈Zn  Rnξ

Lp (Rn x)

(36)

p      A   A  ξ−  ξ − k k   ϕ  dξ = (2πC)−n |c |p  eix·ξ ρ  A   Ck C ∈Zn  Rnξ

,

(37)

Lp (Rn x)

where the functions ρα k are as in Proposition 3.1. In the second line, we used the fact p p that g + hLp ≤ gLp + hpLp for 0 < p ≤ 1. In the third line, we changed variables by x −  → x. In order to get (36), we changed variables by ξ → kA ξ, ξ − k → ξ, and kA x → x in the third expression. On the other hand, changing variables by ξ → A ξ and A x → x in the third expression, we have (37). We estimate (36) or (37) for k,  ∈ Zn by considering the following six cases. (i) (ii) (iii)

|k| ≤ C  and || ≤ |k|/2, |k| ≥ C  and || ≤ |k|/2, |k| ≤ C  and |k|/2 ≤ || ≤ 2|k|,

(iv) |k| ≥ C  and |k|/2 ≤ || ≤ 2|k|, (v) |k| ≤ C  and || ≥ 2|k|, (vi) |k| ≥ C  and || ≥ 2|k|,

where C  = 6C and C > 1 is the constant taken from the definition of ρα k from Proposition 3.1. ˜ >0 Cases (i) and (iii). From (36), we clearly have for sufficiently large N 

p   An(1−p)    A     k ξ + kA k − A  ξ   ix·ξ dξ ϕ    e ρ   k C CA  Rnξ

˜

 k − −N p .

Lp (Rn x)

Case (ii). (⇒  ≤ k.) In this case, we also consider (36). Note that ρ(ξ/C) = 0 unless |ξ| ≤ 2C (⇒ |ξ| ≤ |k|/3). Thus, the effective domain of integration in (36) is over ξ ∈ {|ξ| ≤ 2C}. For these ξ, using ϕ  ∈ S, we get for all β ∈ Zn+ and N ∈ Z+ that  A       k ξ + kA k − A   −N  β  kA ξ + kA k − A      ∂ ϕ   1+  ξ   CA CA  # $−N  1 kA  1+ |k| − |ξ| − || C A

−N  1 + |k| ≤

T. Kato / Journal of Functional Analysis 272 (2017) 1340–1405

1389

≤ k−N  k−N/2 k − −N/2 .

(38)

Here, in the fourth and last lines, we used that x ≤ 1 + |x| and k −  ≤ 3k/2 from || ≤ |k|/2, respectively. Thus, by dividing the Lp -integral (36) into the cases |x| ≤ 1 and |x| ≥ 1, and by integrating by parts sufficiently often for |x| ≥ 1, it follows for sufficiently ˜ > 0 that large N 

p  An(1−p)     A    A A  k ξ + k k −   ξ   ϕ  dξ   eix·ξ ρ A   k C C  Rnξ

˜

 k − −N p .

Lp (Rn x)

On each time when we use integration by parts, (k/)A appears from the function ϕ  in (36). However, since (k/)A ≤ kA , we can cancel it by k−N/2 in the estimate (38). One can find the details in Case (ii) of Step 1 in the proof of Lemma 3.2. Case (iv). (⇒  ∼ k.) Since ρ(ξ/C) = 0 unless |ξ| ≤ 2C, the effective domain of integration in (36) is over ξ ∈ {|ξ| ≤ 2C}. For these ξ, using ϕ  ∈ S, we get for all β ∈ Zn+ and N ∈ Z+ that  A   −N      β  kA ξ + kA k − A      1 + 1  k k −    ∂ ϕ     ξ  A A C C    −N   1  A ∼ 1 +  A  − k  . C k

(39) (40)

We estimate (39) and (40) by considering the two cases |k|/2 ≤ || ≤ |k| and |k| < || ≤ 2|k|, respectively. Then we have ⎧ ⎪ ⎨ 1+  ⎪ ⎩ 1+

1 C 1 C

 A −N  k   k − −N   A k −    A −N     k − −N  k A  − k 

for |k|/2 ≤ || ≤ |k|, for |k| < || ≤ 2|k|,

as we showed in Case (iv) of Step 1 in the proof of Lemma 3.2. Analogously to the above case, using integration by parts sufficiently often, then we have for sufficiently ˜ >0 large N 

p   An(1−p)    A    A A  k ξ + k k −   ξ   dξ ϕ    eix·ξ ρ   k C CA  Rnξ

˜

 k − −N p .

Lp (Rn x)

Case (v). (⇒ k ≤ .) In this case, we estimate (37) and only consider the case ˜ >0 2|k| ≤ 2C  ≤ ||, since otherwise, we have |k| ≤ C  and || ≤ 2C  , so that for any N

1390

T. Kato / Journal of Functional Analysis 272 (2017) 1340–1405

p      A   A ξ−  ξ − k k   ϕ  dξ  eix·ξ ρ  A   Ck C  Rnξ

˜

 k − −N p

Lp (Rn x)

is trivial. From the definition of the function ρ in Proposition 3.1,   # $  A  kA  2CkA   −kA k n  ξ − ≤ ⊂ ξ ∈ R . supp ρ : k  CkA A  A This yields that the domain of integration is included in {ξ ∈ Rn : |ξ| ≤ 2||/3}. Thus, by ϕ  ∈ S, we have for all β ∈ Zn+ and N ∈ Z+        ξ −   −N  β  ξ−      ∂ ϕ  1+ C   ξ  C

−N  1 + || ≤ −N  −N/2  − k−N/2 .

(41)

Here, in the third and last lines, we used that x ≤ 1 + |x| and k −  ≤ 3/2 from 2|k| ≤ ||, respectively. Integrating by parts sufficiently often, we have for sufficiently ˜ >0 large N p      A   A ξ−  ξ − k k ˜   ϕ  dξ  k − −N p .  eix·ξ ρ  A   Ck C  Rnξ Lp (Rn x)

In this case, each integration by parts yields a factor

 A k A

≤ A , which can be absorbed

by the factor −N/2 in the estimate (41). Case (vi). Since the proof of this case is the same as Case (v), we omit it. Combining Cases (i)–(vi) and recalling estimates (36) and (37), we have for all k ∈ Zn ˜ >0 and sufficiently large N   −1 α p ˜ F ρk Ff  p  |c |p k − −N p . L ∈Zn

˜ >0 Thus, recalling (21), we have for sufficiently large N 



k

sq/(1−α)

⎝

 k∈Zn



1/q

L

k∈Zn

⎛

 −1 α q F ρk Ff  p

 ∈Zn

q/p ⎞1/q ˜ ⎠ . k − (|s|/(1−α)−N )p sp/(1−α) |c |p

T. Kato / Journal of Functional Analysis 272 (2017) 1340–1405

1391

Using the convolution relation 1 ∗q/p → q/p if q/p ≥ 1 and the Fubini–Tonelli theorem if 0 < q/p ≤ 1, we obtain 



s,α ∼ f Mp,q

k

sq/(1−α)

 −1 α q F ρk Ff  p L

k∈Zn

 



1/q

1/q k

sq/(1−α)

|ck |

q

.

k∈Zn

As usual, this notation has to be interpreted suitably for q = ∞. Next, we want the* hp norm of f from below. Since f ∈ S, it follows that ) to estimate

" f (x) = limt→0 t−n Φ t ∗ f (x) for any x ∈ Rn , where Φ ∈ S and Rn Φ(x)dx = 1. Then, we have    1      sup f hp =  ∗ f Φ  0


supp ϕ CA (· − ) ∩ supp ϕ CkA (· − k) = ∅ if  = k. Thus, f pLp

p    + ,

  An/p A A =  c  exp i  · x ϕ C (x − )  dx   n ∈Z

Rn

=

 



p |c |p An ϕ CA (x − )  dx

n Rn ∈Z

∼



∈Zn

∼



 p

|c |p

|ϕ(x)| dx

Rn

|c |p .

∈Zn s,α Hence, from the assumption Mp,q → hp , we obtain (with the usual modification for q = ∞)







1/p |ck |

p

k∈Zn





1/q k

sq/(1−α)

|ck |

q

,

k∈Zn

which is the desired estimate. 2 Lastly, we prove the following lemma used in the proof of the “ONLY IF” part of Theorem 1.5.

T. Kato / Journal of Functional Analysis 272 (2017) 1340–1405

1392

Proof of Lemma 3.5. We set f (x) =



c C n/p ||n/p a (C|| (x − )) ,

=0

where a ∈ S satisfies that ⎧ supp a ⊂ [−δ/8, δ/8]n , ⎪ ⎪ ⎪ ⎪ a∞ ≤ 1, ⎪ ⎨ xβ a(x)dx = 0 for all |β| ≤ [n(1/p − 1)], ⎪ ⎪ ⎪ ⎪ n ⎪ √ ⎩R | a(ξ)| ≥ c > 0 for any 1/(2 2C) ≤ |ξ| ≤ 2/C,

(42)

where C > 1 is the constant taken from Lemma 3.2. Existence of such a function a (for arbitrary δ > 0) is already proved in [12, Lemma 4.3]. Then (δ/4)−n/p C n/p ||n/p × a (C|| (x − )) is an hp -atom of type I. In fact, we have (n δ δ , , 8C|| 8C||   (δ/4)−n/p C n/p ||n/p a (C|| ( − )) ≤ (δ/4)−n/p C n/p ||n/p ∞  ' (n −1/p   δ δ  , =  + − , 8C|| 8C||  

−n " −1 −1

β xβ a (C|| (x − )) dx = C|| C || x +  a(x)dx Rn ' supp a (C|| ( − )) ⊂  + −

Rn

=0 for all |β| ≤ [n(1/p − 1)] (see also [12, Proof of Lemma 4.4]). Hence, we have f php ≤

 =0

 p    |c |p C n/p ||n/p a (C|| (x − ))  |c |p . hp

(43)

=0

Next, we estimate the α-modulation space norm of f from below. To this end, let Ψ be a Schwartz function satisfying that + 3 3 ,n ⎧ ˇ ⎪ ⎨ supp Ψ ⊂ )− 8 δ, 8√δ , √ *n ˇ = 1 on − 1+ 2 δ, 1+ 2 δ , Ψ 8 8 ⎪ ⎩ |Ψ(ξ)| ≥ c > 0 for any |ξ| ≤ 2.

(44)

Existence of such a function Ψ (for a small δ > 0) is proved by Kobayashi, Miyachi, and Tomita [12, Lemma 4.3]. Since f ∈ S, using the equivalent norm for the α-modulation spaces from Lemma 3.2, we get

T. Kato / Journal of Functional Analysis 272 (2017) 1340–1405

⎛ s,α  ⎝ f Mp,q



1393

⎞1/q   q  ⎠ , ksq/(1−α) F −1 Ψα k Ff Lp

k=0

where Ψα k is as in Lemma 3.2. By the Fubini–Tonelli theorem, and the changes of variables by ξ → kA ξ, ξ − k → ξ, ξ → Cξ, and CkA x → x, we have  −1 α  F Ψk Ff 

Lp

     + ,

  n(1−1/p) An(1−1/p)  A A ˇ =C k  f (y) exp −ik k · y Ψ x − Ck y dy     Rny

.

(45)

Lp (Rn x)

Considering the supports of the functions in the above integral (45), we have (n ' δ δ , supp a (C|| ( − )) ⊂  + − , 8C|| 8C|| (n '

x 3δ 3δ A ˇ supp Ψ x − Ck  ⊂ + − , CkA 8CkA 8CkA ⊂ m + [−δ/2, δ/2]n for all x ∈ CkA m + [−δ/8, δ/8]n and m ∈ Zn . We see from the above statement that if we choose δ > 0 such that δ/2 + δ/8 < 1, then we have

ˇ x − CkA  = ∅ if m =  supp a (C|| ( − )) ∩ supp Ψ for all x ∈ CkA m + [−δ/8, δ/8]n and m ∈ Zn . Hence, we have ⎛ ⎜ (45)  kAn(1−1/p) ⎝



m∈Znx∈Ω

k,m

   + ,

 ˇ x − CkA y  exp −ikA k · y Ψ  Rny

p ⎞1/p    ⎟ n/p c || a (C|| (y − )) dy  dx⎠ ·  =0      + ,  p n |cm | |m|  exp −ikA k · y  1/(1−α) 1/(1−α) 

 ≥ k

An(1−1/p) k

/2≤|m|≤2k

x∈Ωk,m Rn y

p 1/p 

A ˇ x − Ck y a (C|m| (y − m)) dy  dx ·Ψ , 

(46)

where we set Ωk,m := CkA m + [−δ/8, δ/8]n . Moreover, since k1/(1−α) /2 ≤ |m| ≤ 2k1/(1−α) , if x ∈ Ωk,m and y ∈ supp a (C|m| ( − m)), then

T. Kato / Journal of Functional Analysis 272 (2017) 1340–1405

1394

(n  ' (n ' δ δ δ δ A − Ck m+ − , x − Ck y ∈ Ck m + − , 8 8 8C|m| 8C|m| (n ' (n ' δ kA δ kA δ δ + − , ⊂ − , 8 8 4 k1/(1−α) 4 k1/(1−α) (n ' (n ' δ δ δ δ , ⊂ − , + − 8 8 4k 4k √ (n √ ' 1+ 2 1+ 2 δ, δ , ⊂ − 8 8 

A

A



ˇ x − CkA y = 1 by the second property in assumpwhere |k| ≥ 1. This implies Ψ tion (44). Using the changes of variables by y − m → y and C|m|y → y in (46), we obtain   An(1−1/p) (46) ∼ k |cm |p |m|n(1−p) k 1/(1−α) /2≤|m|≤2k 1/(1−α)

 ·

 p 1/p ( '   kA   k · y a (y) dy  dx  exp −i   C|m|

x∈Ωk,m Rn y

∼ kAn(1−1/p) kn(1/p−1)/(1−α) ⎛  ×⎝ |cm |p k 1/(1−α) /2≤|m|≤2k 1/(1−α)

⎞   p 1/p  A k   a k  ⎠ .    C|m|

Since we have 2kA 2 2|k| kA |k| ≤ ≤ , |k| = C|m| Ck C Ck1/(1−α) kA kA 1 |k| |k| ≥ ≥ √ , |k| = C|m| 2Ck 2Ck1/(1−α) 2 2C and thus     kA   a k ≥c>0    C|m| from the fourth property in (42), we obtain ⎛ (47)  kn(1/p−1) ⎝



k 1/(1−α) /2≤|m|≤2k 1/(1−α)

Hence, with the usual modification for q = ∞,

⎞1/p |cm |p ⎠

.

(47)

T. Kato / Journal of Functional Analysis 272 (2017) 1340–1405

⎞q/p ⎞1/q ⎟ |cm |p ⎠ ⎠ .

⎛

s,α f Mp,q

⎛  ⎜  ⎝ ksq/(1−α) kn(1/p−1)q ⎝

1395

 k 1/(1−α) /2≤|m|≤2k 1/(1−α)

k=0

(48) s,α Therefore, using the assumption hp → Mp,q and combining (43) with (48), we obtain the desired result. 2

6.3. Proofs of lemmas for Theorems 1.3 and 1.4 In this subsection, we prove Lemmas 3.6–3.8. Since the arguments of the proofs are very similar to those of Lemmas 3.4–3.5, we just sketch the proofs. However one can find the precise statement in each related lemma. Proof of Lemma 3.6. Let ϕ ∈ S \ {0} satisfy that supp ϕ ⊂ {ξ : |ξ| ≤ 1/2}. We set f (x) =



+ ,

c An/p exp iA  · x ϕ CA (x − )

∈Zn

for an arbitrary finitely supported sequence {c }∈Zn . Then, using exactly the same ˜ >0 arguments as in the proof of Lemma 3.4 (for p = 1), we have for sufficiently large N   −1 α  ˜ F ρk Ff  p  |c |k − −N . L ∈Zn

Multiplying by ks/(1−α) , taking q-th power and summing these inequalities over k ∈ Zn , we obtain  s,α  f Mp,q





k∈Zn



q 1/q ˜ |s|/(1−α)−N

k − 



s/(1−α)

|c |

∈Zn

from the estimate (21). As usual, this notation has to be interpreted suitably for q = ∞. ˜ >0 By the convolution relation 1 ∗ q → q , we have for sufficiently large N  s,α  f Mp,q



1/q  

sq/(1−α)

|c |

q

·

∈Zn

 





 ˜ |s|/(1−α)−N

k

k∈Zn

1/q k

sq/(1−α)

|ck |

q

.

k∈Zn

Next, we consider the Lp norm of f . By the choice of ϕ, we have supp ϕ ⊂ {ξ : |ξ| ≤ 1/2} and thus

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T. Kato / Journal of Functional Analysis 272 (2017) 1340–1405





supp ϕ CA (· − ) ∩ supp ϕ CkA (· − k) = ∅, if  = k. Thus, we get for p < ∞ that     + ,

p c An/p exp iA  · x ϕ CA (x − )  dx f pLp =   ∈ZnRn

∼





|c |p

∈Zn

∼



|ϕ(x)|p dx

Rn

|c |p .

∈Zn

Likewise, for p = ∞, it is not hard to see f L∞ ∼ {ck }k∈Zn ∞ . Hence, from the s,α assumption Mp,q → Lp , we obtain 



1/p



|ck |



p

k∈Zn

1/q



k

sq/(1−α)

|ck |

q

,

k∈Zn

with the usual modification for p = ∞ or q = ∞. This completes the proof. 2 Proof of Lemma 3.7. Choose a function a ∈ S which satisfies for some δ ∈ (0, 1) that ⎧ n ⎪ ⎨ supp a ⊂ [−δ/8, δ/8] , a∞ ≤ 1, ⎪ ⎩ | a(ξ)| ≥ c > 0 on |ξ| ≤ 2/C.

(49)

Existence of such a function a (for arbitrary δ > 0) is shown in [12, Lemma 4.3]. We set f (x) =



c C n/p ||n/p a (C|| (x − )) .

=0

Since supp a (C|| (· − )) ⊂  + [−δ/(8C||), δ/(8C||)]n , the support of the summands is pairwise disjoint. Thus, we have   p p f pLp = C n |c | ||n |a (C|| (x − ))| dx =0

=

 =0

∼





|c |

p

Rn p

|a(x)| dx

Rn

|c | . p

(50)

=0

We next consider the α-modulation space norm of f . However, this is exactly the same as is preformed in the proof of Lemma 3.5. Thus, we omit the details and note

T. Kato / Journal of Functional Analysis 272 (2017) 1340–1405

⎞q/p ⎞1/q ⎟ |cm |p ⎠ ⎠ .

⎛

s,α f Mp,q

⎛  ⎜  ⎝ ksq/(1−α) kn(1/p−1)q ⎝

1397

 k 1/(1−α) /2≤|m|≤2k 1/(1−α)

k=0

(51) s,α Therefore, combining (50) with (51) and using the assumption Lp → Mp,q , we obtain

⎛

⎛  ⎜ |k|sq/(1−α)+nq(1/p−1) ⎝ ⎝

⎞q/p ⎞1/q ⎛ ⎞1/p  ⎟ |cm |p ⎠ ⎠  ⎝ |ck |p ⎠ ,



k 1/(1−α) /2≤|m|≤2k 1/(1−α)

k=0

k=0

which end this proof. 2 Proof of Lemma 3.8. Choose functions Ψ ∈ S and a ∈ S satisfying the properties of (44) and (49), respectively. We set f (x) =



c C n ||n a (C|| (x − ))

=0

for a finitely supported sequence {c }k∈Zn \{0} . Then, as in (50), we have f L1 ∼



|c |.

(52)

=0 s,α Next, we estimate the M1,q norm of f from below. Since f ∈ S, by Lemma 3.2,

⎛ s,α  ⎝ f M1,q



⎞1/q   q  1⎠ . ksq/(1−α) F −1 Ψα k Ff L

k=0

Analogously to the proof of Lemma 3.5 (cf. equation (45) for p = 1), we have  −1 α  F Ψk Ff 

L1

      + ,

 A A ˇ x − Ck y dy  =  f (y) exp −ik k · y Ψ    Rny 

L1 (Rn x)

and (n δ δ , , 8C|| 8C|| (n '

x 3δ 3δ A ˇ supp Ψ x − Ck  ⊂ + − , CkA 8CkA 8CkA ' supp a (C|| ( − )) ⊂  + −

⊂ m + [−δ/2, δ/2]n

,

(53)

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T. Kato / Journal of Functional Analysis 272 (2017) 1340–1405

for all x ∈ CkA m + [−δ/8, δ/8]n and m ∈ Zn . We see from the above statement that if we choose δ > 0 such that δ/2 + δ/8 < 1, then we have

ˇ x − CkA  = ∅ if m =  supp a (C|| ( − )) ∩ supp Ψ for all x ∈ CkA m + [−δ/8, δ/8]n and m ∈ Zn . From these properties, we obtain

(53) 





m∈Znx∈Ω

k,m



≥

   + ,

 ˇ x − CkA y  exp −ikA k · y Ψ  Rny      n c || a (C|| (y − )) dy  dx ·  =0     + ,  n |cm ||m|  exp −ikA k · y 

k 1/(1−α) /2≤|m|

x∈Ωk,m Rn y

  ˇ x − CkA y a (C|m| (y − m)) dy dx, ·Ψ 



(54)

where we set Ωk,m := CkA m + [−δ/8, δ/8]n . Moreover, since k1/(1−α) /2 ≤ |m|, if x ∈ Ωk,m and y ∈ supp a (C|m| ( − m)), then  ' (n (n ' δ δ δ δ − CkA m + − , CkA m + − , 8 8 8C|m| 8C|m| ( (n ' ' n δ δ δ kA δ kA ⊂ − , + − , 1/(1−α) 8 8 4 k 4 k1/(1−α) (n ' (n ' δ δ δ δ , ⊂ − , + − 8 8 4k 4k √ (n √ ' 1+ 2 1+ 2 δ, δ , ⊂ − 8 8 

x − CkA y ∈



ˇ x − CkA y = 1. where |k| ≥ 1. This implies from the second property in (44) that Ψ By suitable changes of variables, we have  ( '      kA   k · y a (y) dy  dx (54) ∼ |cm |  exp −i   C|m| 1/(1−α) k

∼

/2≤|m|

 k 1/(1−α) /2≤|m|

x∈Ωk,m Rn y

    kA   a k . |cm |     C|m|

(55)

T. Kato / Journal of Functional Analysis 272 (2017) 1340–1405

1399

Since we have 2kA 2 2|k| kA |k| ≤ ≤ |k| = C|m| Ck C Ck1/(1−α) and thus     kA   a k ≥c>0    C|m| from the third property in (49), we obtain (55) 



|cm |.

k 1/(1−α) /2≤|m|

Hence, ⎛ s,α  ⎝ f M1,q

 k=0

⎛ ksq/(1−α) ⎝



⎞q ⎞1/q |cm |⎠ ⎠

.

(56)

k 1/(1−α) /2≤|m|

s,α Therefore, combining (52) with (56) by using the assumption L1 → M1,q , we obtain the desired result. 2

Acknowledgments The author would like to express his deepest thanks to the anonymous referee. His/Her careful reading and valuable suggestions on this paper really helped the author. He expresses his sincerest gratitude to Professor Mitsuru Sugimoto for his helpful advice and comments. They spared a lot of their precious time for this manuscript. Without the referee’s and Prof. Sugimoto’s help, he couldn’t complete this paper. He also thanks Dr. J. Cunanan, Dr. Y. Nagata and Dr. Y. Wakasugi for giving him useful papers and encouragement. This work is supported by Grant-in-aid for Scientific Research of JSPS Fellow No. 15J07897. Appendix A We sketch the proof of the lifting property in Proposition 2.2, i.e., the differential s,α s−σ,α operator (I − Δ)σ/2 maps Mp,q (Rn ) isomorphically onto Mp,q (Rn ). In the following σ/2 argument, we will write this operator as Iσ = (I − Δ) for simplicity. Now, the full statement is as follows. Proposition A.1. Let 0 < p, q ≤ ∞, s, σ ∈ R, and 0 ≤ α ≤ 1. Then the mapping s,α s−σ,α Iσ : Mp,q → Mp,q is isomorphic and

T. Kato / Journal of Functional Analysis 272 (2017) 1340–1405

1400

s,α . s−σ,α ∼ f  Iσ f Mp,q Mp,q

Before we begin with its proof, we prepare the following two lemmas. Lemma A.2. Let σ ∈ R, β ∈ Zn+ . Then we have ∂ξβ



1 + |ξ|2

σ/2 



σ/2−|β| = Pβ,σ (ξ) · 1 + |ξ|2

with a suitable polynomial Pβ,σ (ξ) of degree at most |β|. Proof of Lemma A.2. We prove this lemma by mathematical induction. The case |β| = 0 is clear. For the case |β| = 1, since we have ∂ξj



1 + |ξ|2

σ/2 



σ/2−1 for j = 1, · · · , n, = σξj · 1 + |ξ|2

we see that the above expression holds. Next, we assume that ∂ξβ



1 + |ξ|2

σ/2 



σ/2−|β| = Pβ,σ (ξ) · 1 + |ξ|2

holds true when |β| = m, and then we prove that the expression holds when |β| = m + 1. For all j = 1, · · · , n, we have 

σ/2  1 + |ξ|2  

σ/2  = ∂ξj ∂ξγ 1 + |ξ|2 

σ/2−|γ|  = ∂ξj Pγ,σ (ξ) 1 + |ξ|2 ∂ξβ



σ/2−|γ|

σ/2−|γ|−1  + Pγ,σ (ξ) · 1 + |ξ|2 (σ − 2|γ|) · ξj = ∂ξj (Pγ,σ (ξ)) · 1 + |ξ|2 ' (  



σ/2−|γ|−1 = ∂ξj (Pγ,σ (ξ)) · 1 + |ξ|2 + (σ − 2|γ|) · ξj · Pγ,σ (ξ) 1 + |ξ|2 ,  

where γ ∈ Zn+ with |γ| = m. Setting Pβ,σ (ξ) = ∂ξj (Pγ,σ (ξ)) · 1 + |ξ|2 + (σ − 2|γ|) · ξj · Pγ,σ (ξ), then Pβ,σ is a polynomial of degree at most |β|. Thus, we obtain the desired result by mathematical induction. 2 Lemma A.3. Let A ≥ 0, σ ∈ R, β ∈ Zn+ . Then, we have    2 σ/2   β  A   kσ(A+1)−|β|  ∂  1 + k (ξ + k)   ξ for any k ∈ Zn and |ξ| ≤ C. Here, the invisible constant depends on σ, β and C.

T. Kato / Journal of Functional Analysis 272 (2017) 1340–1405

1401

Proof of Lemma A.3. By Lemma A.2, we have ∂ξβ



2 σ/2  1 + kA (ξ + k)



 2 σ/2−|β|

 , = k|β|A Pβ,σ kA (ξ + k) · 1 + kA (ξ + k)

(57)

where Pβ,σ is a polynomial of degree at most |β|. If |k|  1, then we clearly have    2 σ/2   β  A  ∂    1 ∼ kσ(A+1)−|β| . 1 + k (ξ + k)  ξ  We next consider the case |k|  1. From |ξ| ≤ C, we obtain for any s ∈ R   2 s    1 + kA (ξ + k)   k2s(A+1) , 

 Pβ,σ kA (ξ + k)   k|β|(A+1) . Thus, |(57)|  k|β|A k|β|(A+1) k(σ−2|β|)(A+1) = kσ(A+1)−|β| , which is the desired estimate. 2 We are now in the position to prove Proposition A.1. Proof of Proposition A.1. For α = 1, namely, Besov spaces, this statement is already given by Triebel [22, Theorem in Section 2.3.8]. Thus, we only consider the case 0 ≤ α < 1. We divide the proof into three steps. We prove the estimates “  ” for 0 < p ≤ 1 and for 1 < p ≤ ∞, and then prove “  ” for 0 < p ≤ ∞. Step 1. Let 0 < p ≤ 1. Then, we first show that s,α . s−σ,α  f  Iσ f Mp,q Mp,q

In view of Subsection 2.2, we consider 



k

(s−σ)q/(1−α)

 −1 α  F [ηk · F (Iσ f )]q p L

k∈Zn

Choosing an auxiliary function κ ∈ S satisfying κ(ξ) =

1 on |ξ| ≤ 1, 0 on |ξ| ≥ 2,

1/q .

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T. Kato / Journal of Functional Analysis 272 (2017) 1340–1405

and  κα k (ξ) := κ

ξ − kα/(1−α) k Ckα/(1−α)

,

α then κα k = 1 on the support of ηk . Here, the constant C > 1 is that one taken from the support of ηkα in Subsection 2.2. Then, by Proposition 2.1, we have

 −1 α  F [ηk · F (Iσ f )]

Lp

 ) *

σ/2 α α    = F −1 κk ηk · Ff  p 1 + | · |2 L  ) *

An(1/p−1)  −1 2 σ/2 α   k κk  1+|·| F

Lp

  · F −1 [ηkα · Ff ]Lp ,

where A = α/(1 − α). Changing variables by ξ → kA ξ, ξ − k → ξ, and kA x → x,  )

σ/2 α *    kAn(1/p−1) F −1 1 + | · |2 κk  Lp    

ξ − kα/(1−α) k −n An(1/p−1)  ix·ξ 2 σ/2 dξ = (2π) k 1 + |ξ| κ  e  Ckα/(1−α) Rnξ      2 σ/2  A ξ −n  ix·ξ   dξ = (2π)  e κ 1 + k (ξ + k)  C Rnξ

      

      

Lp (Rn x)

. Lp (Rn x)

On the other hand, we have n 1  ∂eix·ξ xj = i eix·ξ for |x| = 0. |x|2 j=1 ∂ξj

Thus, by dividing the Lp -integral into the cases |x| ≤ 1 and |x| ≥ 1, and by integrating by parts sufficiently often for |x| ≥ 1, we easily see from Lemma A.3 that      2 σ/2  A ξ  ix·ξ   dξ κ 1 + k (ξ + k)  e  C Rnξ

      

 kσ(A+1) = kσ/(1−α) . Lp (Rn x)

Therefore, we have   −1 α F [ηk · F (Iσ f )]

Lp

   kσ/(1−α) F −1 [ηkα · Ff ]Lp ,

and thus, with the usual modification for q = ∞,

T. Kato / Journal of Functional Analysis 272 (2017) 1340–1405





k

(s−σ)q/(1−α)

 −1 α  F [ηk · F (Iσ f )]q p L

1403

1/q

k∈Zn

 



k

(s−σ)q/(1−α)

k

σq/(1−α)

 −1 α  F [ηk · Ff ]q p

1/q

L

k∈Zn

 =



k

sq/(1−α)

 −1 α  F [ηk · Ff ]q p

1/q .

L

k∈Zn

This yields that s,α s−σ,α  f  Iσ f Mp,q Mp,q

holds for 0 < p ≤ 1, 0 < q ≤ ∞, s, σ ∈ R, and 0 ≤ α ≤ 1. Step 2. Let 1 < p ≤ ∞. Then, we show that s,α . s−σ,α  f  Iσ f Mp,q Mp,q

As in Step 1, choosing an auxiliary function κ ∈ S, we have by the Young inequality and suitable changes of variables  −1 α  F [ηk · F (Iσ f )] p L  ) *  

 −1 2 σ/2 α  1+|·| ≤ F κk  · F −1 [ηkα · Ff ]Lp 1 L          σ/2   ξ 2   = (2π)−n  eix·ξ 1 + kA (ξ + k) κ dξ    C Rnξ 

  · F −1 [ηkα · Ff ]Lp . L1 (Rn x)

Then, recalling from Step 1 the estimate           2 σ/2 ξ   dξ  κ  eix·ξ 1 + kA (ξ + k)   C Rnξ 

 kσ/(1−α) ,

L1 (Rn x)

we obtain   −1 α F [ηk · F (Iσ f )]

Lp

   kσ/(1−α) F −1 [ηkα · Ff ]Lp ,

and thus s,α . s−σ,α  f  Iσ f Mp,q Mp,q

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T. Kato / Journal of Functional Analysis 272 (2017) 1340–1405

s−σ,α s,α Step 3. Let 0 < p ≤ ∞. If g ∈ Mp,q , then we have by Steps 1–2 f = I−σ g ∈ Mp,q and s,α  g s−σ,α = Iσ f  s−σ,α f Mp,q Mp,q Mp,q

(see [22, Section 2.3.8] in detail). Finally, taking all steps together, we have s,α s−σ,α ∼ f  Iσ f Mp,q Mp,q

for s, σ ∈ R, 0 ≤ α ≤ 1, and 0 < p, q ≤ ∞. 2 References [1] L. Borup, M. Nielsen, Banach frames for multivariate α-modulation spaces, J. Math. Anal. Appl. 321 (2006) 880–895. [2] L. Borup, M. Nielsen, Boundedness for pseudodifferential operators on multivariate α-modulation spaces, Ark. Mat. 44 (2006) 241–259. [3] L. Borup, M. Nielsen, Nonlinear approximation in α-modulation spaces, Math. Nachr. 279 (2006) 101–120. [4] H.G. Feichtinger, Modulation Spaces on Locally Compact Abelian Groups, Technical Report, University of Vienna, 1983. [5] H.G. Feichtinger, C. Huang, B. Wang, Trace operators for modulation, α-modulation and Besov spaces, Appl. Comput. Harmon. Anal. 30 (2011) 110–127. [6] D. Goldberg, A local version of real Hardy spaces, Duke Math. J. 46 (1979) 27–42. [7] P. Gröbner, Banachräume glatter Funktionen und Zerlegungsmethoden, Ph.D. thesis, University of Vienna, 1992. [8] W. Guo, D. Fan, H. Wu, G. Zhao, Sharpness of complex interpolation on α-modulation spaces, J. Fourier Anal. Appl. 22 (2016) 427–461. [9] J. Han, B. Wang, α-Modulation spaces (I) scaling, embedding and algebraic properties, J. Math. Soc. Japan 66 (2014) 1315–1373. [10] M. Kobayashi, Modulation spaces M p,q for 0 < p, q ≤ ∞, J. Funct. Spaces Appl. 4 (2006) 329–341. [11] M. Kobayashi, Dual of modulation spaces, J. Funct. Spaces Appl. 5 (2007) 1–8. [12] M. Kobayashi, A. Miyachi, N. Tomita, Embedding relations between local Hardy and modulation spaces, Studia Math. 192 (2009) 79–96. [13] M. Kobayashi, Y. Sawano, Molecular decomposition of the modulation spaces, Osaka J. Math. 47 (2010) 1029–1053. [14] M. Kobayashi, M. Sugimoto, The inclusion relation between Sobolev and modulation spaces, J. Funct. Anal. 260 (2011) 3189–3208. [15] M. Kobayashi, M. Sugimoto, N. Tomita, On the L2 -boundedness of pseudo-differential operators and their commutators with symbols in α-modulation spaces, J. Math. Anal. Appl. 350 (2009) 157–169. [16] M. Kobayashi, M. Sugimoto, N. Tomita, Trace ideals for pseudo-differential operators and their commutators with symbols in α-modulation spaces, J. Anal. Math. 107 (2009) 141–160. [17] G. Misiołek, T. Yoneda, Loss of continuity of the solution map for the Euler equations in α-modulation and Hölder spaces, arXiv:1412.4619v1, 2014. [18] M. Sugimoto, N. Tomita, The dilation property of modulation spaces and their inclusion relation with Besov spaces, J. Funct. Anal. 248 (2007) 79–106. [19] M. Sugimoto, N. Tomita, A remark on fractional integrals on modulation spaces, Math. Nachr. 281 (2008) 1372–1379. [20] J. Toft, Continuity properties for modulation spaces with applications to pseudo-differential calculus—I, J. Funct. Anal. 207 (2004) 399–429. [21] J. Toft, P. Wahlberg, Embeddings of α-modulation spaces, Pliska Stud. Math. Bulgar. 21 (2012) 25–46.

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