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2 Perliminaries
2.1 Basic natation
The following notation will be used throughout this article. For 0 < p,q 6 ∞, we denote
1 1 1 1 σ(p,q) := d 0 ∧ ( − ) ∧ ( + − 1) ; q p q p 1 1 1 1 τ(p,q) := d 0 ∨ ( − ) ∨ ( + − 1) , q p q p where a ∧ b = min {a, b} , a ∨ b = max {a, b}. We write S (Rd) to denote the Schwartz space of all complex-valued rapidly decreasing infinity differentiable functions on Rd, and S ′(Rd) to denote the dual space of S (Rd), all called the space of all tempered distributions. For simplification, we omit Rd without causing ambiguity. The Fourier F ˆ −ixξ transform is defined by f(ξ) = f(ξ) = Rd f(x)e dξ, and the inverse Fourier transform by F −1 −d ixξ 6 p f(x) = (2π) Rd f(ξ)e dξ. For 1 p 2 p and kfk∞ = ess supx∈Rd |f(x)|. We also define the L Sobolev norm : s/2 kfkW s,p = (I −△) f . p s,p S ′ Recall that the Sobolev spaces is defined by W = {f ∈ : kfkW s,p < ∞}. And the Fourier Lp is defined by F Lp = f ∈ S ′ : fˆ < ∞ . p We use the notation I . J if there is an independently constant C such that I 6 CJ. Also we denote I ≈ J if I . J and J . I. For 1 6 p 6 ∞, we denote 1/p + 1/p′ = 1, for 0 2.2 Modulation spaces Let 0 < p,q 6 ∞, the short time Fourier tranform of f respect to a window function g ∈ S is defined as (see [4]): −itξ Vgf(x,ξ)= f(t)g(t − x)e dt. Rd Z And then for 1 6 p,q 6 ∞, we denote s kfk s = kVgf(x,ξ) hξi k q p , Mp,q LξLx s where hξi =1+ |ξ|. Modulation space Mp,q are defined as the space of all tempered distribution ′ f ∈ S for which kfk s is finite. Mp,q Also, we know another equivalent definition of modulation by uniform decomposition of fre- quency space (see [6]). Let σ be a smooth cut-off function adapted to the unit cube [−1/2, 1/2]d and σ = 0 outside the d cube [−3/4, 3/4] , we write σk = σ(·− k), and assume that d σk(ξ) ≡ 1, ∀ξ ∈ R . Zd kX∈ −1 Denote σk(ξ) = σ(ξ − k), and ✷k = F σkF , then we have the following equivalent norm of modulation space: s kfk s = hki k✷ fk p q . M k Lx Zd p,q ℓk( ) 0 s We simply write Mp,q instead of Mp,q. One can prove the Mp,q norm is independent of the s s choice of cut-off function σ. Also Mp,q is a quasi Banach space and when 1 6 p,q 6 ∞, Mp,q is a S s s Banach space. When p,q < ∞, then is dense in Mp,q. Also, Mp,q has some basic properties, we list them in the following lemma(see [2, 22]). Lemma 2.1. For s,s0,s1, ∈ R, 0 < p,p0,p1,q,q0,q1 6 ∞, s1 s0 ; if s0 6 s1,p1 6 p0,q1 6 q0, we have Mp1,q1 ֒→ Mp0,q0 (1) 3 s −s (2) when p,q < ∞, the dual space of Mp,q is Mp′,q′ ; s (3) the interpolation spaces theorem is true for Mp,q, i.e. for 0 <θ< 1 when 1 1 − θ θ 1 1 − θ θ s = (1 − θ)s0 + θs1, = + , = + , p p0 p1 q q0 q1 we have M s0 , M s1 = M s ; p0,q0 p1,q1 θ p,q s s1 . when q1 < q,s + d/q > s1 + d/q1, then we have Mp,q ֒→ Mp,q1 (4) 2.3 Besov-Triebel spaces Let 0 < p,q 6 ∞,s ∈ R, choose ψ : Rd :→ [0, 1] be a smooth radial bump function adapted to the ball B(0, 2): ψ(ξ)=1 as |ξ| 6 1 and ψ(ξ)=0as |ξ| > 2. We denote ϕ(ξ)= ψ(ξ) − ψ(2ξ), and −j Z F −1 F ϕj(ξ) = ϕ(2 ξ) for 1 6 j, j ∈ , ϕ0(ξ) = 1 − j>1 ϕj(ξ). We say that △j = ϕj are the dyadic decomposition operators. The Besov spaces Bs and the Triebel spaces F s are definied in P p,q p,q the following way : s ′ d js B = f ∈ S (R ) : kfk s = 2 k△ fk p < ∞ , p,q Bp,q j Lx q ℓ > j 0 s S ′ Rd js Fp,q = f ∈ ( ) : kfkF s = 2 △jf q < ∞ . p,q ℓj>0 p ( Lx ) One can prove that the Besov-Triebel norms defined by different dyadic decompositions are all equivalent(see [23]), so without loss of generality, we can assume that when 1 6 j, ϕj(ξ) = 1 Rd 3 j 5 j on Dj := ξ ∈ : 4 2 6 |ξ| 6 4 2 for convenience. Also, Besov-Triebel spaces have some basic properties known already(see [23, 7, 22]). Lemma 2.2. Let s,s1,s2 ∈ R, 0 < p,p1,p2,q,q1,q2 6 ∞, s s s s ; if q1 6 q2, we have Bp,q1 ֒→ Bp,q2 , Fp,q1 ֒→ Fp,q2 (1) s+ε s s+ε s ; ε> 0, we have Bp,q1 ֒→ Bp,q2 , Fp,q1 ֒→ Fp,q2∀ (2) s s s ;Bp,p∧q ֒→ Fp,q ֒→ Bp,p∨q (3) s1 s2 ;if p1 6 p2,s1 − d/p1 = s2 − d/p2, we have Bp1,q ֒→ Bp2,q (4) s1 s2 ; if p1 s −s s −s (6) when 1 6 p,q < ∞, the dual space of Bp,q is Bp′,q′ , the dual space of Fp,q is Fp′,q′ ; s s (7) the interpolation spaces theorem is true for Bp,q and Fp,q, i.e. for 0 <θ< 1 when 1 1 − θ θ 1 1 − θ θ s = (1 − θ)s0 + θs1, = + , = + , p p0 p1 q q0 q1 we have Bs0 ,Bs1 = Bs , F s0 , F s1 = F s ; p0,q0 p1,q1 θ p,q p0,q0 p1,q1 θ p,q s s,p s s,1 s,∞ s .when 1 4 3 Some lemmas In this section, we give some useful lemmas and propositions proved already, which will be used in our proof. The following Bernstein’s inequality is very useful in time-frequency analysis (see [22]) : Rd p Lemma 3.1. [Bernstein’s inequality] Let 0 < p 6 q 6 ∞,b > 0,ξ0 ∈ . Denote LB(ξ,b) = f ∈ Lp : supp fˆ ⊆ B(ξ, R) . Then there exists C(d,p,q) > 0, such that n o d(1/p−1/q) kfkq 6 C(d,p,q)R kfkp p Rd holds for all f ∈ LB(ξ,b) and C(d,p,q) is independent of b> 0 and ξ0 ∈ . Also, by using the Bernstein’s inequality, we can get the following Young type inequality for 0 d Lemma 3.2 ([7]). Let 0 d(1/p−1) k|f| ∗ |g|kp 6 C(d, p)(R1 + R2) kfk kgk holds for all f ∈ Lp , g ∈ Lp . B(ξ1,R1) B(ξ2,R2) By the definition of ✷k and △j operators, we know that they are all convolution type operators, so by the Lemma above and the usual Young’s inequality, we have: Corollary 3.3. For 0 0 independent of k, j, such that ✷ ✷ k△j kfk 6 C k kfkp , k△j△ℓfkp 6 C k△jfkp . Finally we recall a result known already, which is the special case of Theorem 4.1 and Theorem 4.2. Proposition 3.4 ([2]). For 0 < p,q 6 ∞,s ∈ R, we have s ;(Bp,q ֒→ Mp,q if and only if s > τ(p,q (1) s .(Mp,q ֒→ Bp,q if and only if s 6 σ(p,q (2) s,p Proposition 3.5. ([3]) Let 1 6 p,q 6 ∞,s ∈ R, then W ֒→ Mp,q if and only if one of the following cases is satisfied: (1) 1 τ(p,q); (2) p>q,s>τ(p,q); (3) p = 1,q = ∞,s > τ(1, ∞); (4) p = 1,q 6= ∞,s>τ(1,q). 5 4 Embedding between Besov spaces and modulation spaces s In this section, we consider the embedding with the form Bp0,q0 ֒→ Mp,q. If we got the sufficient s1 s and necessary condition of this embedding, then the embedding Bp1,q1 ֒→ Mp,q can be done in the same way. Our main result is: R s Theorem 4.1. For 0 < p,q,p0,q0 6 ∞,s ∈ , the embedding Bp0,q0 ֒→ Mp,q is true if and only if one of the following two conditions is satisfied: (1) p0 6 p,q0 6 q,s > τ(p0,q); (2) p0 6 p,q0 >q,s>τ(p0,q). As for the other side of the embedding between Besov spaces and modulation spaces, we also have: R s Theorem 4.2. For 0 < p,q,p1,q1 6 ∞,s ∈ , the embedding Mp,q ֒→ Bp1,q1 is true if and only if one of the following two conditions is satisfied: (1) p1 > p,q1 > q,s 6 σ(p1,q); (2) p1 > p,q1 s s For a special case when p0 = q0 = 2, which means that Bp0,q0 = H , the classical Sobolev spaces, then we have Corollary 4.4. Let 0 < p,q 6 ∞,s ∈ R, we have s :the embedding H ֒→ Mp,q is true if and only if one of the following two cases is true (1) 1. 2 6 p, 2 6 q,s > 0; 2. 2 6 p, 2 >q,s>d(1/q − 1/2). s :The embedding Mp,q ֒→ H is true if and only if the following two cases is true (2) 1. p 6 2,q 6 2,s 6 0; 2. p 6 2,q > 2,s 4.1 Proof of Theorem 4.1 In this subsection, we prove the Theorem 4.1. For sufficiency, by using the lemmas above, we can easily prove it. As for necessity, it may be much more difficult. Here, we mostly use the way in [[2]] to prove it. 6 4.1.1 Sufficiency s τ(p0,q) Use Lemma 2.2, we know that for any case in 4.1, we have Bp0,q0 ֒→ Bp0,q ; then by Proposition τ(p0,q) .we know that Bp0,q ֒→ Mp0,q; also by Lemma 2.1, we have Mp0,q ֒→ Mp,q ,3.4 4.1.2 Necessity s If we have Bp0,q0 ֒→ Mp,q, then we have s kfkM . kfkBs , ∀f ∈ Bp0,q0 . (4.1) p,q p0,q0 Then we can take some different kinds of f into (4.1) to get some restrictions of indexes. d −1 (1) Choose η ∈ S , such that supp η ⊂ [−1/8, 1/8] , denote f = F η, for ∀λ 6 1, take fλ(x)= d f(λx) into (4.1). By the choose of η, we know that supp F fλ ⊂ λ[−1/8, 1/8] , which means that fλ, k = 0; fλ, k = 0; ✷kfλ = △jfλ = (0, k 6= 0; (0, k 6= 0. So, take f into (4.1), we have kf k . kf k , which means that λ−d/p . λ−d/p0 . Take λ λ p λ p0 λ → 0, we have p0 6 p. d d −1 (2) Denote Aj := k ∈ Z : k + [−3/4, 3/4] ⊆ Dj , for ℓ > 10, choose kℓ ∈ Aℓ, take fℓ = F η(·− F d kℓ), η is just the function in (1). Then we know that supp fℓ ⊆ kℓ + [−1/8, 1/8] ⊆ Dℓ, so we have fℓ, j = ℓ; f , k = k ; ✷ ℓ ℓ kfℓ = △jfℓ = △jfℓ, 0 < kj − ℓk∞ 6 3; (0, k 6= kℓ; 0, kj − ℓk∞ > 3. Then by Corollary 3.3 ,we have kfℓkMp,q = kfℓkp ≈ 1 js kfℓkBs = 2 k△jfℓkp q p0,q0 0 ℓ 0 kj−ℓk∞63 . 2 ℓs kf k ≈ 2ℓs . ℓ p0 Then take the estimate above into (4.1), we have 1 . 2ℓs, ∀ℓ > 10, so we get s > 0. −1 (3) Take fℓ = F ϕℓ, then by Corollary 3.3, we have ✷ ✷ F −1 ℓd/q kfℓkM = k kfℓkp q > k kfℓkp q = σk q ≈ 2 ; p,q ℓ ℓ p ℓ k k∈Aℓ k∈Aℓ js js ℓs ℓ(s+d(1−1/p0)) kfℓkBs = 2 k△jf ℓk q = 2 k△jfℓk q 6 2 kfℓkp ≈ 2 . p0,q0 p0 0 p0 0 0 ℓj ℓ kj−ℓk∞63 Take the estimates above into (4.1), let ℓ →∞, we have s > d(1/p0 + 1/q − 1). 7 (4) When p > 2, choose η ∈ S , such that suppη ˆ ⊆ [−1/8, 1/8]d, take f (x)= eikxη(x−k), 0 ℓ k∈Aℓ −ik(ξ−k) d then fˆℓ(ξ)= e ηˆ(ξ−k), so we know that supp fˆℓ ⊆ k+[−1/8, 1/8] ⊆ Dℓ, k∈Aℓ k∈Aℓ P so by the orthogonality of △ , ✷ , we have: P j k S fℓ, j = ℓ; eikxη(x − k), k ∈ A ; ✷ ℓ kfℓ = △jfℓ = △jfℓ, 0 < kj − ℓk∞ 6 3; (0, else; 0, kj − ℓk∞ > 3. Then we have ✷ ℓd/q kfℓkM > k kfℓkp q ≈ 2 ; p,q ℓ k∈Aℓ js js ℓs ℓs 2/p0 1−2/p0 kfℓkBs = 2 k△j fℓk q = 2 k△jfℓk q 6 2 kfℓkp 6 2 kfℓk2 kfk∞ . p0,q0 p0 0 p0 0 0 ℓj ℓ kj−ℓk∞63 By the orthogonality of L2 space, we have 1/2 kf k = kη(x − k)k2 ≈ 2ℓd/2. ℓ 2 2 kX∈Aℓ Also by the fast decay of η, we have −N |fℓ(x)| 6 |η(x − k)| 6 (1 + |x − k|) . 1. Zd kX∈Aℓ kX∈ ℓ(s+d/p0) Combine the estimates above, we have kfkBs . 2 . Then take it into (4.1), and p0,q0 put ℓ →∞, we have s > d(1/q − 1/p0). (5) Combine case(2)-(4), we have s > τ(p0,q). In contrast with the necessary condition we need τ(p0,q) .in Theorem 4.1, we only need to prove that if Bp0,q0 ֒→ Mp,q, then we have q0 6 q (5.1) When τ(p0,q) = 0, take f = ℓ aℓfℓ, where fℓ in case (2), then by the same calculation, we have P kfkM ≈ kaℓkℓq ; kfkB0 . kaℓkℓq0 . p,q p0,q0 Take the estimates into (4.1), we have q0 6 q. (5.2) When τ(p0,q)= d(1/p0 + 1/q − 1), take f = ℓ aℓfℓ, where fℓ in case (3), then by the same calculation, we have P ✷ > ✷ jd/q kfkMp,q = aℓ kfℓ aℓ kfℓ = aj2 q ; ℓj ℓ q ℓ q p ℓ p ℓ X k X k∈Aj q ℓj jτ jτ kfk τ(p ,q) = 2 a △ f 6 a 2 k△ f k 0 ℓ j ℓ j+ℓ j ℓ+j p0 q0 Bp0,q0 ℓ q j ℓ p0 0 kℓk∞63 X ℓj X 8 jd(1−1/p0) jτ jd/q 6 aℓ+j2 2 q . aj2 q . ℓ 0 ℓ 0 6 j j kℓk∞ 3 X Take the estimates into (4.1), we have q0 6 q. (5.3) When τ(p0,q) = d(1/q − 1/p0), take f = ℓ aℓfℓ, where fℓ in case (4), then by the same calculation, we have P kfk = a ✷ f > a ✷ f Mp,q ℓ k ℓ ℓ k ℓ ℓ q ℓ q p ℓ p ℓ X k X k∈Aj q ℓj −1 jd/q = aj F σk ≈ aj2 ; p ℓq ℓq k∈Aj q j ℓj jτ jτ kfkBτ = 2 aℓ△jfℓ 6 aj+ℓ2 k△jfℓ+jkp q p0,q0 0 ℓ 0 ℓ q kℓk 63 j p0 ℓ 0 ∞ X j X jτ jτ jd/p0 jd/q 6 aℓ+j kfℓ+jk 2 q . aj2 2 q = aj2 q . p0 ℓ 0 ℓ 0 ℓ 0 6 j j j kℓk∞ 3 X Take the estimates into (4.1), we have q0 6 q. 4.2 Proof of Theorem 4.2 In this section, we prove the Theorem 4.2. Because we consider the case when 0 < p,q 6 ∞, we can not only use the duality to prove it. Thanks to the wonderful method in [[2]], also combining the construction in the proof of Theorem 4.1, we can prove this theorem as well. 4.2.1 Sufficiency σ(p1,q) By Lemma 2.1, we have Mp,q ֒→ Mp1,q; by Proposition 3.4, we have Mp1,q ֒→ Bp1,q ; finally by σ(p1,q) s .Lemma 2.2, we have Bp1,q ֒→ Bp1,q1 for any case in Theorem 4.2 4.2.2 Necessity s If we have Mp,q ֒→ Bp1,q1 , then we have kfkBs . kfkM . (4.2) p1,q1 p,q Just like the proof of Theorem 4.1, we can take some different kinds of f into (4.2) to get some restrictions of indexes, and in some cases, we can take the same f as in 4.1.2. (1) Take fλ in the case (1) in 4.1.2 into (4.2), we can get p 6 p1. ℓs (2) Take fℓ in the case (2) in 4.1.2, we have kfℓkM ≈ 1, and kfℓkBs > 2 k△ℓfℓkp = p,q p1,q1 1 2ℓs kf k ≈ 2ℓs. Then we have s 6 0. ℓ p1 9 d d (3) Denote Bj = k ∈ Z : k + [−3/4, 3/4] ∩ Dj 6= ∅ , choose g ∈ S , such that supp g ⊆ D0, −ℓ F −1 F then denote gℓ(·) = g(2 ·). Take fℓ = gℓ, so supp fℓ ⊆ Dℓ. So by Corollary 3.3, we have ✷ F −1 ℓd/q kfℓkM = k kfℓkp q 6 σk q . 2 ; p,q ℓ p ℓ k∈Bℓ k∈Bℓ ℓs ℓs ℓ (s+d(1−1/p1)) kfℓkBs > 2 k△ℓf ℓkp = 2 kfℓkp ≈ 2 . p1,q1 1 1 Take the estimate into (4.2), let ℓ →∞, we can get s 6 d(1/p1 + 1/q − 1). (4) Take f (x) = eikxη( x−k ), we can choose a << 1 such that kf k > C2ℓd/p1 (see [2]), ℓ k∈Aℓ a ℓ p1 then by the same calculation of case (4) in 4.1.2, we have P ℓ(s+d/p1) ℓd/q kfℓkBs > C2 ; kfℓkM . 2 . p1,q1 p,q Take the estimates above into (4.2) and put ℓ →∞, we have s 6 d(1/q − 1/p1). (5) Combine case(2)-(4), we have s 6 σ(p1,q). In contrast with the necessary condition we need σ(p1,q) .in Theorem 4.2, we only need to prove that if Mp,q ֒→ Bp1,q1 , then we have q1 > q By the same way in the proof of Theorem 4.1, take f = aℓfℓ, fℓ in case (2)-(4), we can get ka k . ka k , which means that q1 > q. ℓ q1 ℓ q P 5 Embedding between Sobolev spaces and modulation spaces In this section, we can get the sufficient and necessary conditions of the embedding between Lr- Sobolev spaces and modulation spaces. We consider the most general case of this kind of embedding s,r ,with the form like W ֒→ Mp,q, which is the general case of Proposition 3.5 where r = p. Also we consider the case when 0 < q 6 ∞. Unlike the proof in that proposition, we mostly use the uniform decomposition of frequency space and the Theorem 4.1 we get in last section. Our main result is: s,r Theorem 5.1. Let 1 6 p, r 6 ∞, 0 < q 6 ∞,s ∈ R, then W ֒→ Mp,q if and only if r 6 p and one of the following conditions is satisfied: (1) r>q,s>τ(r, q); (2) 1 < r 6 q,s > τ(r, q); (3) r = 1,q = ∞,s > τ(r, q); (4) r = 1, 0 Remark 5.2. As the can see, the sufficient and necessary conditions we got in the theorem above is the same as the result in Proposition 3.4. The index p here only make sense in r 6 p. The reason of this may be that the modulation space have good embedding property of the index p like in Lemma 2.1. 10 1 q 1 (1) (4) (2) (3) 0 1 1 r Figure 1: the four cases of (1/r, 1/q) in Theorem 5.1 As for the other side of this kind of embedding, we also have: s,r Theorem 5.3. Let 1 6 p, r 6 ∞, 0 < q 6 ∞,s ∈ R, then Mp,q ֒→ W if and only if p 6 r and one of the following conditions is satisfied: (1) r (2) q 6 r< ∞,s 6 σ(r, q); (3) r = ∞, 0 (4) r = ∞, 1 5.1 Proof of Theorem 5.1 In this subsection, we prove Theorem 5.1. By the Proposition 3.5, we can easily prove the sufficiency. As for necessity, we mainly construct some funcions to the embedding estimates to get some restrictions of the indexes. 5.1.1 Sufficiency s,r ,When 1 6 q 6 ∞, by Proposition 3.5, in any case of (1)-(4), we have W ֒→ Mr,q; so, if r 6 p .by Lemma 2.1, we have Mr,q ֒→ Mp,q 11 When 0 5.1.2 Necessity s,r If we have W ֒→ Mp,q, then we have . s,r kfkMp,q kfkW . (5.1) d −1 (i) Choose η ∈ S , such that supp η ⊂ [−1/8, 1/8] , denote f = F η, for ∀λ 6 1, take fλ(x) = s,r . 6 f(λx) into (5.1), we have kfλkr ≈ kfλkW kfλkMp,q ≈ kfλkp, so we have r p. Therefore, when r = ∞, we know that p = ∞, this is the case in Proposition 3.5. s s s,r ii) When 1 6 r < ∞, by Lemma 2.2, we have Br,r∧2 ֒→ Fr,2 ֒→ W ֒→ Mp,q, so when r 6 2, we) s have Br,r ֒→ Mp,q, by Theorem 4.1, we know that if r 6 q, then s > τ(r, q); if r>q, then s>τ(r, q). s When r > 2, we have Br,2 ֒→ Mp,q, also by Theorem 4.1, we know that if 2 6 q, then s > τ(r, q); if 2 >q, then s>τ(r, q). τ(r,q),r = (iii) We prove that W ֒→ Mp,q is not true for any 2 6 q < r 6 p, in which case τ(r, q) d(1/q − 1/r), for convenience we also denote it by τ. τ,r r −τ If not, we have W ֒→ Mp,q which is equivalent to L ֒→ Mp,q . So we have kfk −τ . kfk . (5.2) Mp,q r S d ikx Choose η ∈ , such that suppη ˆ ⊂ [−1/8, 1/8] , take f(x) = k∈Zd ake η(x − k), then ik(ξ−k) fˆ(ξ)= d a e ηˆ(ξ − k). Then by the orthogonality of ✷ , we have k∈Z k Pk P ikx ′ ake η(x − k); k = k ; ✷k′ f = (0; k 6= k′. −τ So, we have kfk −τ ≈ hki ak q . Mp,q ℓk 2 By the orthogonality of L space, we know that 1/2 2 ikx kfk2 = ake η(x − k) ≈ kakkℓ2 . 2 k∈Zd X By the rapidly decrease of η, we know that −N |f(x)| 6 |ak| |η(x − k)| 6 kakk ∞ (1 + |x − k|) . kakk ∞ . ℓk ℓk Zd Zd kX∈ kX∈ So we have kfk . kakk ∞ , then by interpolation, we have kfk . kakk r . ∞ ℓk r ℓk 12 −τ Take the two estimates into (5.2), we have hki ak . kakk r , which is equivalent to ℓk q/r hki−τq b . br/q . k k Zd Zd kX∈ kX∈ ′ So we know that hki−τq ∈ ℓ(r/q) , which is equivalent to τq(r/q)′ > d, so we have τ > d(1/q − 1/r), which is a contradiction. s,1 .iv) We prove that when r = 1 6 p, 0 kfk −d/q . kfk . (5.3) M∞,q 1 Choose η ∈ S such thatη ˆ = 1 on [−1, 1]d, for 0 −d/q ✷ −d/q ✷ −d/q kfk −d/q = hki k kfk∞ q > hki k kfk∞ q ≈ hki q . M∞,q ℓ ℓ ℓ k k∈Kt k∈Kt − d/q Also, kfk1 = kηk1 ≈ 1. So, take the two estimates into (5.3), we have hki q . 1, ℓ k∈Kt take t → 0, we have hki−d/q . 1, which is a contraction. ℓq k∈Zd In fact, we already get the result as desired. To be specific,from (i,ii) we can get the necessity of the condition (2)(3); the necessity of condition (4) can be got by (ii,iv); as for condition (1), when q < r 6 2 or r > 2 > q, the necessity is proved in (ii), when 2 6 q < r, the necessity is proved in (iii). 5.2 Proof of Theorem 5.3 When 1 6 q 6 ∞, we can get the result as desired just by duality of Theorem 5.1, so we only need to consider the case 0 kfk . kfk −s . r Mp,q For any k ∈ Zd, take f(x) = eikxη(x) into the equation above, where η ∈ S and suppη ˆ ⊆ [−1/8, 1/8]d. Then we have 1 . hki−s, which means that s 6 0. 13 6 Embedding between Triebel spaces and modulation spaces s In this section, we mainly study the embedding Fp0,q ֒→ Mp,q. We first consider a special case s like Fp,2 ֒→ Mp,q, which is just the same as Proposition 3.5 in some cases. And using this embedding and interpolation method we can get the result we need. As for special case, we have: s Theorem 6.1. Let 0 < p,q 6 ∞, then Fp,2 ֒→ Mp,q if and only if one of the following conditions is satisfied: (1) q > p,s > τ(p,q); (2) q s Theorem 6.2. Let 0 < p,q 6 ∞, then Mp,q ֒→ Fp,2 if and only if one of the following conditions is satisfied : (1) q 6 p,s 6 σ(p,q); (2) q>p,s<σ(p,q). Then, our main result is: s Theorem 6.3. Let 0 < p,p0,q 6 ∞, then Fp0,q ֒→ Mp,q if and only if p0 6 p and one of the following conditions is satisfied: (1) p0 6 q,s > τ(p0,q); (2) p0 >q,s>τ(p0,q). As for the other side of this kind of embedding, we also have: s Theorem 6.4. Let 0 < p,p1,q 6 ∞, then Mp,q ֒→ Fp1,q if and only if p1 > p and one of the following conditions is satisfied: (1) p1 > q,s 6 σ(p1,q); (2) p1 6.1 Proof of Theorem 6.1 s s,p For 1 14 1 q τ = d( 1 − 1 ) q p0 τ = d( 1 + 1 − 1) p0 q (1) (2) τ = 0 0 1 1 p0 Figure 2: the two cases of (1/p0, 1/q) in Theorem 6.3 6.2 Proof of Theorem 6.2 By the same discussion in the proof of Theorem 6.1, we only need to consider the case when p = ∞. Also, when 1 6 q 6 ∞, by duality, we can get the result as desired. When 0 ,∞,Sufficiency: by Lemma 2.1, we have M∞,q ֒→ M∞,1. By Theorem 6.1, we have F1,2 ֒→ M1 s .by duality, we have M∞,1 ֒→ F∞,2, so we have M∞,q ֒→ F∞,2 ֒→ F∞,2 for s 6 0 s s s Necessity: if we have M∞,q ֒→ F∞,2, then by Lemma 2.2, we have F∞,2 ֒→ B∞,∞. By Theorem 4.2, we have s 6 σ(∞,q). 6.3 Proof of Theorem 6.3 6.3.1 Sufficiency s s For condition (1): by Lemma 2.2, we have Fp0,q ֒→ Bp0,q, then by Theorem 4.1, we have s .Bp0,q ֒→ Mp,q s s τ(p0,q)+ε ,For condition (2): by Lemma 2.2, we have Fp0,q ֒→ Bp0,p0 ֒→ Bp0,q for some 0 < ε << 1 τ(p0,q)+ε .then by Theorem 4.1, we have Bp0,q ֒→ Mp,q 6.3.2 Necessity s If we have Fp0,q ֒→ Mp,q, then we have kfkM . kfkF s . (6.1) p,q p0,q 15 (i) By the same method as in subsection 5.1.2, we can get p0 6 p. s s .(ii) By Lemma 2.2, we have Bp0,p0∧q ֒→ Fp0,q ֒→ Mp,q, then by Theorem 4.1, we have s > τ(p0,q) > s,p0 s s iii) When p0 >q 2, by Lemma 2.2, we have W ≈ Fp0,2 ֒→ Fp0,q ֒→ Mp,q, then by Theorem) 5.1, we have s>τ(p0,q). jd/q (iv) When 2 > p0 > q, by a similar proof of Proposition 3.2 in [21], we have 2 aj q . ℓj j(s+d(1−1/p0)) 2 aj , then we have s > d(1/q + 1/p0 − 1). p0 τ (p0,q) 0,q ֒→ Mp,q is not true for any p0 > q,q < 2,p0 > 2, in which case prove that Fp v) We) τ(p0,q)= d(1/q − 1/p0). d(1/q−1/p0) 0 , If not, we have Fp0,q ֒→ Mp,q, also by sufficiency, we have Fp0,p0 ֒→ Mp0,p0 ֒→ Mp,p0 d(1/2−1/p0) -then by interpolation in Lemma 2.2 and 2.1, we have Fp0,2 ֒→ Mp,2 which is contra diction with case (iii). Combine the case(i),(ii), we get the condition (1); combine the case (i),(iii),(iv),(v), we get the condition (2) as desired. Remark 6.5. 1. As for case (v) in the proof above, we prove it by contradiction using the method of interpola- tion. Also, we can prove it directly by take some good function into the norm inequality. One can modify the proof of Proposition 3.1 in [21] to get the result. 2. The proof of Theorem 6.4 is similar to the proof above. For the reader’s convenience, we give the outline of the proof here. The sufficiency can get by the relation of Triebel spaces and Besov spaces, then use Theorem 4.2 can get the result. As for necessity, we can use a revised version of Proposition 3.1,3.2 in [21] directly to get the result as desired. s →֒ As we can see, the embedding we consider above is not the most general case like Fp0,q0 .3 Mp,q. I believe that for the general case the sufficient and necessary is the similar to Theorem 6.3. But I have not proven it yet. The most difficult case left is the same as the open problem in [21]. 7 Embedding between Fourier Lp spaces and modulation spaces s F r In this section, we consider the embedding Mp,q ֒→ L . Because of the orthogonality of r ✷k(σk), we can calculate the F L norm easily. Then, we can get the result as follow. Recall that when 0 R s F r Theorem 7.1. Let 0 < p, q, r 6 ∞,s ∈ , then we have Mp,q ֒→ L if and only if one of the following conditions is satisfied: (1) p 6 2,q 6 r 6 p′,s > 0; 16 (2) p 6 2, r 6 p′,r As for the other side of this kind of embedding, we also have: R F r s Theorem 7.2. Let Let 0 < p, q, r 6 ∞,s ∈ , then we have L ֒→ Mp,q if and only if one of the following conditions is satisfied: (1) p > 2,p′ 6 r 6 q,s 6 0; (2) p > 2,p′ 6 r,r>q,s 7.1 Proof of Theorem 7.1 7.1.1 Sufficiency If we have condition (1), when 1 6 p 6 ∞, by the orthogonality of σk, H¨older’s inequality of Lp spaces and Hausdorff-Young’s inequality of Fourier transform, we have fˆ ≈ σ fˆ 6 σ fˆ 6 k✷ fk 6 k✷ fk = kfk 6 kfk s . k k k p r k p q Mp,q M r r r p′ r ℓ ℓ p,q ℓk ℓ k k k ˆ ✷ ✷ When 0 s If we have condition (2), by Lemma 2.1, we have Mp,q ֒→ Mp,r, then by the sufficiency of r . condition(1), we have Mp,r ֒→ F L 7.1.2 Necessity s F r If we have Mp,q ֒→ L , then we have ˆ f . kfkM s . (7.1) r p,q The conditions we need are contained the following cases. d (i) For any f ∈ S with supp fˆ ⊆ [−1/8, 1/8] , in this case we have kfk s ≈ kfk . So, (7.1) Mp,q p means that ˆ f . kfkp . r Then by Hausdorff-Young’s inequality on compact set (see Tao’s note of Math 254B, one can extend the result to the case of 0 (ii) For any k ∈ Zd, take f(x)= eikxη(x), where η ∈ S ,suppη ˆ ⊆ [−1/8, 1/8]d, then we know that supp fˆ ⊆ k + [−1/8, 1/8]d. So, we have s s kfk s = hki kηk ≈ hki ; kfkF r = kηˆk ≈ 1. Mp,q p L r Take the estimates into (7.1), we have 1 . hkis, which means that s > 0, which is the result in condition (1). 17 ikx (iii) When r fˆ(ξ)= d a ηˆ(ξ − k). So, we have k∈Z k P P s ˆ kfkM s ≈ khki akkℓq ; f ≈ kakkℓr . p,q k r k s −sr Take the estimates above into (7.1), we have kakkℓr . kh ki akkℓq , then we have hki bk 1 . k k ℓk −sr (q/r)′ ′ kb k q/r , which means that hki ∈ ℓ . So, we have sr(q/r) > d, i.e. s > d(1/r − 1/q), k ℓ k which is the result in condition (2). Remark 7.3. The proof of Theorem 7.2 is similar to the proof above, we omit it here. References [1] M. Sugimoto and N. 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