Isotropic and dominating mixed Lizorkin–Triebel spaces

Isotropic and dominating mixed Lizorkin–Triebel spaces — a comparison

We shall compare isotropic Lizorkin–Triebel spaces with their counterparts of dominating mixed smoothness...

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Analysis Math., 43 (2) (2017), 303–337 DOI: 10.1007/s10476-017-0309-z

ISOTROPIC AND DOMINATING MIXED LIZORKIN–TRIEBEL SPACES — A COMPARISON V. K. NGUYEN1,2 and W. SICKEL1,∗ 1

Friedrich-Schiller-University Jena, Ernst-Abbe-Platz 2, 07737 Jena, Germany e-mails: [email protected], [email protected] 2

University of Transport and Communications, Dong Da, Hanoi, Vietnam e-mail: [email protected]

(Received September 30, 2016; revised January 10, 2017; accepted February 7, 2017)

Abstract. We shall compare isotropic Lizorkin–Triebel spaces with their counterparts of dominating mixed smoothness.

1. Introduction Let t ∈ N0 and 1 < p < ∞. The standard isotropic Sobolev space is deﬁned as Wpt (Rd ) = f ∈ Lp (Rd ) : f |Wpt (Rd ) = D α¯ f |Lp (Rd ) < ∞ . |α| ¯ 1 ≤t

The Sobolev space of dominating mixed smoothness is given by t d d t d α ¯ d D f |Lp (R ) < ∞ . Sp W (R ) = f ∈ Lp (R ) : f |Sp W (R ) = |α| ¯ ∞ ≤t

Here α ¯ = (α1 , . . . , αd ) ∈ Nd0 , |¯ α|1 = α1 + . . . + αd and |¯ α|∞ = maxi=1,...,d |αi |. Obviously we have the chain of continuous embeddings Wptd (Rd ) → Spt W (Rd ) → Wpt (Rd ). Also easy to see is the optimality of these embeddings in various directions. We will discuss this below. These two types of Sobolev spaces Wpt (Rd ) and ∗ Corresponding

author. Key words and phrases: (isotropic) Lizorkin–Triebel space, Lizorkin–Triebel space of dominating mixed smoothness, embedding. Mathematics Subject Classification: 46E35. c 2017 Akad´ 0133-3852/$ 20.00 emiai Kiad´ o, Budapest

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Spt W (Rd ) represent particular cases of corresponding scales of Bessel potential spaces (Sobolev spaces of fractional order t of smoothness), denoted by Hpt (Rd ) (isotropic smoothness) and Spt H(Rd ) (dominating mixed smoothness). Let t ∈ R and 1 < p < ∞. Then the space Hpt (Rd ) is the collection of all distributions f ∈ S (Rd ) such that f |Hpt (Rd ) = F −1 (1 + |ξ|2 )t/2 F f (·)|Lp (Rd ) < ∞ , whereas Spt H(Rd ) is the collection of all f ∈ S (Rd ) such that

d 2 t/2 d f |Spt H(Rd ) = F −1 (1 + ξ ) F f (·) (R ) L p i < ∞. i=1

Here ξ = (ξ1 , . . . , ξd ) ∈ Rd . Indeed, if t ∈ N0 we have Wpt (Rd) = Hpt (Rd )

and

Spt W (Rd ) = Spt H(Rd )

in the sense of equivalent norms. In case t > 0 Schmeisser [15] stated that (1.1)

Hptd (Rd ) → Spt H(Rd ) → Hpt (Rd).

In this paper we shall give a proof of (1.1) and we shall show the optimality of these assertions in the following directions: • Within all spaces Spt00 H(Rd ) satisfying Spt00 H(Rd ) → Hpt (Rd ) the class Spt H(Rd ) is the largest one. • Within all spaces Hpt00 (Rd ) satisfying Hpt00 (Rd ) → Spt H(Rd ) the class Hptd (Rd ) is the largest one. • Within all spaces Spt00 H(Rd ) satisfying Hptd (Rd ) → Spt00 H(Rd ) the class Spt H(Rd ) is the smallest one. In what follows we shall go one step further. Isotropic Sobolev spaces Hpt (Rd ) and Sobolev spaces of dominating mixed smoothness Spt H(Rd ) of fractional order t are contained as special cases in the scales of isotropic Lizorkin– t (Rd ) and Lizorkin–Triebel spaces of dominating mixed Triebel spaces Fp,q t F (Rd ). It is well-known that smoothness Sp,q t (Rd ) and Hpt (Rd ) = Fp,2

t Spt H(Rd ) = Sp,2 F (Rd),

1 < p < ∞ , t ∈ R,

in the sense of equivalent norm, see [23, Theorem 2.5.6] and [16, Theorem 2.3.1]. In this paper we address the question under which conditions on t, p, q the embeddings td t t (Rd ) → Sp,q F (Rd) → Fp,q (Rd ) Fp,q Analysis Mathematica 43, 2017

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hold true. In addition we shall discuss the optimality of these embeddings in various directions. Nowadays isotropic Lizorkin–Triebel spaces represent a well accepted regularity notion in various ﬁelds of mathematics. Lizorkin–Triebel spaces of dominating mixed smoothness, in particular the scale Spt H(Rd ), are of increasing importance in approximation theory and information-based comt F (Rd ) contains the tensor plexity. As special cases, the scale Spt H(Rd ) = Sp,2 t F (Rd ) contains products of the univariate spaces Hpt (R), and the scale Sp,p t (R), see [19,20]. It is the the tensor products of the univariate spaces Fp,p main aim of this paper to give a detailed comparison of these diﬀerent extensions of univariate Lizorkin–Triebel spaces into the multi-dimensional situation. The paper is organized as follows. Section 2 is devoted to the deﬁnition and some basic properties of the function spaces under consideration. Our main results are stated in Section 3. The proofs are concentrated in Section 4. In Subsection 4.1 we collect the required tools from Fourier analysis, especially some vector-valued Fourier multiplier assertions. The next Subsection 4.2 is devoted to complex interpolation. Dual spaces are discussed in Subsection 4.3. Finally, we collect families of test functions in Subsection 4.4. Notation. As usual, N denotes the natural numbers, N0 := N ∪ {0}, Z the integers and R the real numbers, C refers to the complex numbers. For a real number a we put a+ := max(a, 0). The letter d is always reserved for the underlying dimension in Rd , Zd . We denote by x, y or x · y the usual Euclidean inner product in Rd or Cd . By x y we mean x y = (x1 y1 , . . . , xd yd ) ∈ Rd . If k¯ = (k1, . . . , kd ) ∈ Nd0 , then we put ¯ 1 := k1 + . . . + kd |k|

¯ ∞ := max |kj |. and |k| j=1,...,d

¯

For k¯ ∈ Nd0 and a > 0 we write ak := (ak1 , . . . , akd ). By C, C1 , C2 , . . . we denote positive constants which are independent of the main parameters involved but whose values may diﬀer from line to line. The symbol A B means that there exist positive constants C1 and C2 such that C1 A ≤ B ≤ C2 A. Let X and Y be two quasi-Banach spaces. Then X → Y indicates that the embedding is continuous. Let Lp (Rd ), 0 < p ≤ ∞, be the space of all functions f : Rd → C such that f |Lp (R ) := d

1/p |f (x)| dx <∞ p

Rd

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with the usual modiﬁcation if p = ∞. By C0∞ (Rd ) the set of all compactly supported inﬁnitely diﬀerentiable functions f : Rd → C is denoted. Let S(Rd ) be the Schwartz space of all complex-valued rapidly decreasing inﬁnitely diﬀerentiable functions on Rd . The topological dual, the class of tempered distributions, is denoted by S (Rd ) (equipped with the weak topology). The Fourier transform on S(Rd ) is given by

−d/2 e−ixξ ϕ(x) dx , ξ ∈ Rd . F ϕ(ξ) = (2π) Rd

The inverse transformation is denoted by F −1 . We use both notations also for the transformations deﬁned on S (Rd). Let 0 < p, q ≤ ∞. For a sequence d of complex-valued functions {fk¯ }k∈ ¯ Nd0 on R , we put 1/q q d fk¯ |Lp (q ) = |fk¯ | Lp (R ). ¯ Nd0 k∈

2. Spaces of isotropic and dominating mixed smoothness t (Rd ) are invariant under rotations, the spaces The isotropic spaces Fp,q t F (Rd ) are not invariant under rotations. Both properties have been Sp,q known for a long time and are well reﬂected by trace assertions on hyperplanes, see, e.g., Triebel [23, 2.7] (isotropic spaces) and Triebel [24], Vyb´ıral [27], Vyb´ıral and Sickel [29] (dominating mixed smoothness).

2.1. Isotropic Besov–Lizorkin–Triebel spaces. For us it will be convenient to introduce Lizorkin–Triebel and Besov spaces simultaneously. Let φ0 ∈ S(Rd ) be a non-negative function such that φ0 (x) = 1 if |x| ≤ 1 and φ0 (x) = 0 if |x| ≥ 32 . For j ∈ N we deﬁne φj (x) := φ0 (2−j x) − φ0 (2−j+1x) ,

x ∈ Rd .

Definition 2.1. Let 0 < p, q ≤ ∞ and t ∈ R. t (Rd ) is then the collection of all tempered dis(i) The Besov space Bp,q d tributions f ∈ S (R ) such that f

t (Rd )φ |Bp,q

:=

∞ j=0

is ﬁnite. Analysis Mathematica 43, 2017

2

−1 q F [φj F f ](·)Lp (Rd )

jtq

1/q

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t (Rd ) is then the collection (ii) Let p < ∞. The Triebel–Lizorkin space Fp,q of all tempered distributions f ∈ S (Rd ) such that ∞ t jtq −1 q 1/q d f Fp,q (Rd )φ := 2 F [φj F f ](·) Lp (R ) j=0

is ﬁnite. Remark 2.2. Lizorkin–Triebel spaces are discussed in various monographs, let us refer, e.g., to [23], [25], [26] and [2]. They are quasi-Banach spaces (Banach spaces if min(p, q) ≥ 1) and they do not depend on the chosen generator φ0 of the smooth dyadic decomposition (in the sense of equivalent quasi-norms). We call them isotropic because they are invariant under rotations. Characterizations in terms of diﬀerences can be found at various places, see, e.g., [23, 2.5], [25, 3.5] or [2, Section 28]. Many times we will work with the following equivalent quasi-norm. Let ψ0 ∈ S(Rd ) such that for x ∈ Rd ψ0 (x) = 1 if

sup |xi | ≤ 1

i=1,...,d

and

ψ0 (x) = 0 if

sup |xi | ≥

i=1,...,d

3 . 2

For j ∈ N, we deﬁne ψj (x) := ψ0 (2−j x) − ψ0 (2−j+1x). Then we have

supp ψj ⊂ x : sup |xi | ≤ 3 · 2j−1 \ x : sup |xi | ≤ 2j−1 . i=1,...,d

i=1,...,d

t (Rd ) Proposition 2.3. Let t ∈ R, 0 < p < ∞ and 0 < q ≤ ∞. Then Fp,q is the collection of all f ∈ S (Rd ) such that

∞ 1/q jtq −1 t d ψ q d f |Fp,q (R ) = 2 |F [ψj F f ](·)| Lp (R ) j=0 t (Rd )ψ and f |F t (Rd )φ are equivalent. is finite. The quasi-norms f |Fp,q p,q

The equivalence of these quasi-norms has been proved in [23, Proposition 2.3.2] (in a much more general framework). Proposition 2.3 also holds true for Besov spaces (with the respective quasi-norms). From now on we t (Rd ) instead of will work with the ψ-norm and therefore we write f |Fp,q t (Rd )ψ . f |Fp,q Analysis Mathematica 43, 2017

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2.2. Besov–Lizorkin–Triebel spaces of dominating mixed smoothness. Next we will give the deﬁnitions of Besov and Lizorkin–Triebel spaces of dominating mixed smoothness. We start with a smooth dyadic decomposition on R and afterwards we shall take its d-fold tensor product. More exactly, let ϕ0 ∈ C0∞ (R) satisfying ϕ0 (ξ) = 1 on [−1, 1] and supp ϕ ⊂ [− 32 , 32 ]. For j ∈ N we deﬁne ϕj (ξ) = ϕ0 (2−j ξ) − ϕ0 (2−j+1ξ) ,

(2.1)

ξ ∈ R.

Now we turn to tensor products. For k¯ = (k1, . . . , kd ) ∈ Nd0 we put ϕk¯ (x) := ϕk1 (x1 ) · . . . · ϕkd (xd ) ,

(2.2)

x ∈ Rd .

This construction results in a smooth dyadic decomposition of unity {ϕk¯ }k∈ ¯ Nd0 d on R . Definition 2.4. Let 0 < p, q ≤ ∞ and t ∈ R. t B(Rd ) is the (i) The Besov space of dominating mixed smoothness Sp,q collection of all tempered distributions f ∈ S (Rd ) such that t 1/q ¯ 1 tq |k| −1 d q f S B(Rd ) := F [ϕk¯ F f ](·)|Lp (R ) 2 p,q ¯ Nd0 k∈

is ﬁnite. (ii) Let 0 < p < ∞. The Lizorkin–Triebel space of dominating mixed t F (Rd ) is the collection of all tempered distributions f ∈ smoothness Sp,q S (Rd ) such that 1/q t ¯ 1 tq d |k| −1 q d (2.3) f Sp,q F (R ) := 2 | F [ϕk¯ F f ](·)| Lp (R ) ¯ Nd0 k∈

is ﬁnite. t A(R) = At (R), A ∈ {B, F }. Remark 2.5. (i) For d = 1 we have Sp,q p,q (ii) Lizorkin–Triebel spaces of dominating mixed smoothness have a t cross-quasi-norm, i.e., if fi ∈ Fp,q (R), i = 1, . . . , d, then it follows

f (x) =

d

t fi (xi ) ∈ Sp,q F (Rd)

d t t fi F (R). and f Sp,q F (Rd ) = p,q

i=1

i=1

3. The main results As mentioned in the Introduction we will split our considerations into two cases: Analysis Mathematica 43, 2017

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t F (Rd ) → F t (Rd ); • Sp,q p,q td t • Fp,q (Rd ) → Sp,q F (Rd).

3.1. The embedding of dominating mixed spaces into isotropic spaces. The ﬁrst of our main results reads as follows. Theorem 3.1. Let d ≥ 2, 0 0; • t = 0, 1 < p < ∞ and 0 < q ≤ 2; • t = 0, 0 < p ≤ 1 and 0 < q < 2. 0 F (Rd ) = F 0 (Rd ) = L (Rd ), 1 < p < ∞, Remark 3.2. We recall that Sp,2 p p,2 in the sense of equivalent norms. This is a consequence of certain LittlewoodPaley assertions, see Nikol’skij [14, 1.5.6]. This identity does not extend to p = 1. Here we conjecture 0 0 F (Rd) → F1,2 (Rd ) → L1 (Rd ) . S1,2

Theorem 3.3. Let d ≥ 2, 1 0 or t = 0 and 1 ≤ q ≤ 2.

In addition we have the following. Proposition 3.4. Let d ≥ 2 and t < 0. t t (i) If 1 < p < ∞ and 1 ≤ q ≤ ∞, then Fp,q (Rd ) → Sp,q F (Rd) . t (Rd ) and S t F (Rd ) are not com(ii) If 0 < p < 1, 0 < q ≤ ∞, then Fp,q p,q parable. t (Rd ) and S t F (Rd ) (1 ≤ q ≤ ∞) We summarize the relation between Fp,q p,q in Fig. 1. These above embeddings are optimal in the following sense.

Theorem 3.5. Let d ≥ 2, 0 < p0 , p < ∞, 0 < q0 , q ≤ ∞ and t0 , t ∈ R. Let p, q and t be fixed. Within all spaces Spt00 ,q0 F (Rd ) satisfying t Spt00 ,q0 F (Rd) → Fp,q (Rd ) t F (Rd ) is the largest. the class Sp,q

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t

critical line t t Sp,q F (Rd ) → Fp,q (Rd )

0

1

1 p

not comparable t t Fp,q (Rd ) → Sp,q F (Rd )

t t Fig. 1: Comparison of Sp,q F (Rd ) and Fp,q (Rd )

t F (Rd ) Remark 3.6. Note that, within all spaces Fpt00,q0 (Rd ) satisfying Sp,q t (Rd ) is not the smallest one. There is no smallest → Fpt00,q0 (Rd ) the class Fp,q space with this respect. We may follow the arguments we have used in the context of the analogous problem for Besov spaces, see [13]. From Theo2 F (Rd ) → F 2 (Rd ). On the other hand, a Sobolev-type rem 3.1 we have S1,2 1,2 embedding and Theorem 3.1 imply 3/2

3/2

2 S1,2 F (Rd) → S2,2 F (Rd ) → F2,2 (Rd ) . 3/2

2 (Rd ) and F d However, for d ≥ 2 the spaces F1,2 2,2 (R ) are not comparable.

3.2. The embedding of isotropic spaces into dominating mixed spaces. Theorem 3.7. Let d ≥ 2, 0 min(p,q) − 1 + and 0 < q < ∞;

• t = 0, 1 1 and t > 0 or 0 1/p. For further comments, see Remark 4.14. By using a similar argument as in proof of Theorem 3.3 we conclude the following. Theorem 3.9. Let d ≥ 2, 1 0 or t = 0 and q ≥ 2.

In addition we have the following supplement. Proposition 3.10. Let d ≥ 2. (i) Let 0 < p < 1, 0 < q ≤ ∞ and 0 < t ≤ td Fp,q (Rd ) are not comparable.

1 p

t − 1. Then Sp,q F (Rd ) and

t F (Rd ) → F td (Rd ) (ii) Let 0 < p < ∞, 0 < q ≤ ∞ and t < 0. Then Sp,q p,q follows. td (Rd ) and S t F (Rd ) (1 ≤ q ≤ ∞) We summarize the relation between Fp,q p,q in Fig. 2.

t=

t

td (Rd ) → S t F (Rd ) Fp,q p,q

0

1 p

critical line

−1 not comparable

1

1 p

t F (Rd ) → F td (Rd ) Sp,q p,q

t td Fig. 2: Comparison of Sp,q F (Rd ) and Fp,q (Rd )

These embeddings are optimal in the following sense. Theorem 3.11. Let d ≥ 2, 0 < p0 , p < ∞, 0 < q0 , q ≤ ∞ and t0 , t ∈ R. Let p, q and t be fixed. Within all spaces Fpt00,q0 (Rd ) satisfying t Fpt00,q0 (Rd ) → Sp,q F (Rd ) td (Rd ) is the largest. the class Fp,q

Theorem 3.12. Let d ≥ 2, 0 < p0 , p < ∞, 0 < q0 , q ≤ ∞ and t0 , t ∈ R. Let p, q and t be fixed. Within all spaces Spt00 ,q0 F (Rd ) satisfying td Fp,q (Rd ) → Spt00 ,q0 F (Rd ) t F (Rd ) is the smallest. the class Sp,q Analysis Mathematica 43, 2017

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4. Proofs of the main results To prove our main results we will apply essentially four diﬀerent tools: vector-valued Fourier multipliers; complex interpolation; assertions on dual spaces and some test functions. In what follows we collect what is needed. 4.1. Tools from Fourier analysis. In this section we collect the required tools from Fourier analysis. For a locally integrable function f : Rd → C we denote by M f (x) the Hardy–Littlewood maximal function deﬁned by

1 |f (y)| dy, x ∈ Rd , (4.1) (M f )(x) = sup x∈Q |Q| Q where the supremum is taken over all cubes with sides parallel to the coordinate axes containing x. A vector-valued generalization of the classical Hardy–Littlewood maximal inequality is due to Feﬀerman and Stein [5].] Theorem 4.1. For 1 0, such that M f¯ |Lp (q ) ≤ C f¯ |Lp (q ) k

k

holds for all sequences {fk¯ (x)}k∈ ¯ Nd0 of locally Lebesgue-integrable functions on Rd .

We require a direction-wise version of (4.1)

xi +s 1 (Mif )(x) = sup |f (x1 , . . . , xi−1 , t, xi+1 , . . . , xd )| dt, s>0 2s xi −s

i = 1, . . . , d .

The following version of the Feﬀerman–Stein maximal inequality is due to Bagby [1], see also St¨ockert [21]. Theorem 4.2. For 1 0, such that for any i = 1, . . . , d Mi f¯ |Lp (q ) ≤ C f¯ |Lp (q ) k k holds for all sequences {fk¯ (x)}k∈ ¯ Nd0 of locally Lebesgue-integrable functions d on R .

Iterative application of this theorem yields a similar boundedness property for the operator M = Md ◦ · · · ◦ M1 . The following proposition will be a consequence of Theorem 4.2. In it’s proof we will follow the arguments in the isotropic case, see [31] . Analysis Mathematica 43, 2017

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Proposition 4.3. Suppose 1 < p < ∞, 1 ≤ q ≤ ∞ and let φ ∈ S(Rd ). Then there exists a constant C such that −1 F [φ(2−k¯ ξ)F f¯ (ξ)](·)Lp (q ) ≤ Cf¯ Lp (q ) k k for all {fk¯ }k∈ ¯ Nd0 ∈ Lp (q ). ¯

Proof. Step 1. The case 1 < q ≤ ∞. Recall the notation 2k := (2k1 , . . . , 2kd ). Observe that for k¯ ∈ Nd0 we have (4.2)

¯

¯

¯

F −1 [φ(2−k ·)F fk¯ (·)](x) = (2π)−d/2 2|k|1

Rd

(F −1 φ)(2k y)fk¯ (x − y)dy .

Let α > 1. The assumption φ ∈ S(Rd ) implies ¯ −1 k (4.3) (F φ)(2 y)f (x − y) dy ¯ k Rd

d

≤ sup

y∈Rd

d

×

Rd

≤ c1

(1 + |2 yi | )

i=1

Rd

ki

2 α/2

−1 (F φ)(2k¯ y)

(1 + |2ki yi |2 )−α/2 |fk¯ (x − y)| dy

i=1

d ki 2 −α/2 (1 + |2 yi | ) |fk¯ (x − y)| dy i=1

with a constant c1 depending on φ, but not on k¯ and fk¯ . For ¯ ∈ Zd and k¯ ∈ Nd0 we put ¯ ) ¯ := {x ∈ Rd : 2−ki 2i ≤ |xi | < 2−ki 2i +1 , i = 1, . . . , d}. P (k, Observe Rd \{0} =

¯ ) ¯ P (k,

for all k¯ ∈ Nd0 .

¯ Zd ∈

Then we obtain from (4.3) ¯ k −1 (4.4) (F φ)(2 y)f (x − y)dy ¯ k ≤ c1

¯ Zd ∈

Rd

sup

d

¯ ) ¯ y∈P (k, i=1

2 −α/2

(1 + |2 yi | ) ki

¯ ) ¯ P (k,

|fk¯ (x − y)| dy.

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By applying M to the integral on the right-hand side of (4.4) we derive ¯ k −1 (F φ)(2 y)f (x − y)dy ¯ k Rd

≤ c1(Mfk¯ )(x)

2−|k|1 ¯

¯ Zd ∈

≤ c1 2−|k|1 (Mfk¯ )(x) ¯

d ¯ Zd i=1 ∈

d

2 i α (1 + |2ki yi |2 ) 2 ¯ ) ¯ y∈P (k, i=1 sup

2 i ¯ ≤ c2 2−|k|1 (Mfk¯ )(x) . α i (1 + 2 )

Inserting this into (4.2) we arrive at −1 F [φ(2−k¯ ·)F f¯ (·)](x) ≤ c3 (Mf¯ )(x) k k with some c3 independent of k¯ and fk¯ . Now the desired estimate follows from Theorem 4.2. Step 2. The case q = 1. From Step 1 we derive that the linear operator ¯ T : {fk¯ }k¯ → F −1 [φ(2−k y)F fk¯ (y)] k¯ is bounded from Lp (∞ ) into itself. By a duality argument we conclude that the adjoint operator T of T is bounded from [Lp (∞ )] into itself. That is −1 F [φ(2−k¯ y)F f¯ (y)](·)[Lp (∞ )] ≤ C f¯ [Lp (∞ )] k k for all {fk¯ }k∈ ¯ Nd0 ∈ [Lp (∞ )] . Of course the same inequality follows with φ replaced by φ. The canonical embedding

Lp (1 ) → [Lp (∞ )] is a linear isometry, see, e.g., [30, Satz III.3.1]. We put d ¯} . A := {{fk¯ }k∈ ¯ Nd0 : fk ¯ ∈ S(R ), fk ¯ ≡ 0 for all but a ﬁnite number of k

It is obvious that

¯ F −1 [φ(2−k y)F fk¯ (y)] k∈ ¯ Nd ∈ A ⊂ Lp (1 ) 0

if {fk¯ }k∈ ¯ Nd0 ∈ A. Because A is dense in Lp (1 ) we conclude that −1 F [φ(2−k¯ y)F f¯ (y)](·)Lp (1 ) ≤ C f¯ Lp (1 ) k k

holds for all {fk¯ }k∈ ¯ Nd0 ∈ Lp (1 ). The proof is complete. Analysis Mathematica 43, 2017

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Definition 4.4. Let 0 < p, q ≤ ∞. Let Ω = {Ωk¯ }k∈ ¯ Nd0 be a sequence of d compact subsets in R . Then we deﬁne d LΩ ¯ }k∈ ¯ Nd0 : fk ¯ ∈ S (R ), p (q ) = {{fk

supp F fk¯ ⊂ Ωk¯ if k¯ ∈ Nd0 , fk¯ Lp (q ) < ∞}. We specify the sequence of compact subsets of Rd by choosing (4.5) Ωk¯ := x ∈ Rd : |xki | ≤ aki , i = 1, . . . , d , with ak¯ = (ak1 , . . . , akd ), k¯ ∈ Nd0 , aki > 0, i = 1, . . . , d. The following proposition has been proved in [16, Theorem 1.10.2] for d = 2. A proof for general d can be found in [7, Proposition 2.3.4]. Proposition 4.5. Let 0 < p < ∞, 0 < q ≤ ∞ and Ω = {Ωk¯ }k∈ ¯ Nd0 be the sequence given in (4.5). Let 0 < s < min(p, q). Then there exists a positive constant C , independent of the sequence {ak¯ }k¯ , such that |f (· − z)| ¯ k sup ¯ |Lp (q ) 1 Lp (q ) ≤ Cfk d s z∈Rd i=1 (1 + |aki zi | ) holds for all systems {fk¯ } ∈ LΩ p (q ).

Next we recall a Fourier multiplier assertion for the spaces LΩ p (q ). We refer to [16, Theorem 1.10.3], see also [28, Theorem 1.12] or [7, Proposition 2.3.5]. Lemma 4.6. Let 0 min(p,q) + 12 . Then there exists a constant C , independent of the sequence {ak¯ }k¯ , such that −1 F M¯ F f¯ | Lp (q ) ≤ C sup M ¯(a ¯ ·)|S r H(Rd ) · f¯ |Lp (q ) 2 k k k ¯ Nd0 ∈

r d holds for all systems {fk¯ }k¯ ∈ LΩ ¯ }k ¯ ∈ S2 H(R ). p (q ) and all systems {Mk

Of certain use for us will be the following Nikol’skij representation for Lizorkin–Triebel spaces of dominating mixed smoothness. Proposition 4.7. Let 1 < p < ∞, 1 ≤ q ≤ ∞ and t ∈ R. Let further t d {ϕk¯ }k∈ ¯ Nd0 be the above system. Then the space Sp,q F (R ) is a collection of d d all f ∈ S (R ) such that there exists a sequence {fk¯ }k∈ ¯ Nd0 ⊂ Lp (R ) satisfying ¯ (4.6) f= F −1 ϕk¯ F fk¯ in S (Rd ) and 2t|k|1 fk¯ |Lp (q ) < ∞ . ¯ Nd0 k∈ Analysis Mathematica 43, 2017

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The norm ¯

t f |Sp,q F (Rd)∗ := inf 2t|k|1 fk¯ |Lp (q )

is equivalent to the norm in (2.3). Here the infimum is taken over all admissible representations in (4.6).

Proof. Step 1. Let {ϕj }∞ j=0 be the system given in (2.1). We put j ∈ N0 ,

ϕ ˜j := ϕj−1 + ϕj + ϕj+1 ,

(4.7)

t F (Rd ) ˜k¯ := ϕ˜k1 ⊗ · · · ⊗ ϕ ˜kd . For f ∈ Sp,q with ϕ−1 ≡ 0. If k¯ ∈ Nd0 we deﬁne ϕ d we choose fk¯ = F −1 ϕ˜k¯ F f . It follows from k∈ ¯ (x) = 1 for all x ∈ R ¯ Nd0 ϕk and ϕ˜j (x) = 1 if x ∈ supp ϕj that F −1 [ϕk¯ F fk¯ ] = F −1 [ϕk¯ ϕ ˜k¯ F f ] = f . ¯ Nd0 k∈

Hence

¯ Nd0 k∈

∗ ¯ t f |Sp,q F (Rd) ≤ 2t|k|1 fk¯ |Lp (q ) ¯ ¯ = 2t|k|1 F −1 ϕ˜k¯ F f |Lp (q ) ≤ 3d 2t|k|1 F −1 ϕk¯ F f |Lp (q ).

Step 2. Assume that f can be represented as in (4.6). We put ϕk¯ ≡ 0 if mini=1,...,d ki < 0. Then we obtain −1 −1 F ϕk¯ F f = F ϕk+ ϕk¯ ¯ ¯F fk+ ¯ ¯ . d ¯ ∈{−1,0,1}

Applying Lemma 4.6 we get

¯ −1 t|k| 1 2 ¯ 1 F −1 ϕ¯ F f |Lp (q ) ≤ c1 2t|k| F k

d ¯ ∈{−1,0,1}

ϕk+ ¯ ¯F fk+ ¯ ¯ Lp (q )

¯ ≤ c2 2t|k|1 F −1 ϕk¯ F fk¯ |Lp (q ). To continue we split k¯ into several parts. Observe that ¯ ¯ 2t|k|1 F −1 [ϕ¯ F f¯ ]q = 2t|k|1 F −1 [ϕ¯ F f¯ ]q . (4.8) k k k k ¯ Nd0 k∈

e⊂{1,...,d}

ki ≥1,i∈e kj =0,j∈e

Proposition 4.3 can be applied to each subsum. This yields t|k| ¯ 1 2 ¯ 1 F −1 ϕ¯ F f¯ Lp (q ) ≤ c3 2t|k| fk¯ |Lp (q ) k k with a constant c3 independent of f . The proof is complete. Analysis Mathematica 43, 2017

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4.2. Complex interpolation. For the basics of Calder´on’s complex interpolation method we refer to the monographs [3,11,22]. It is well-known that this complex interpolation method can be extended to a special class of quasi-Banach spaces, called analytically convex, see [10]. Note that any Banach space is analytically convex. The following proposition was wellknown in the classical context of Banach spaces, see [11, Theorem 2.1.6], [3, Theorem 4.1.2] or [22, Theorem 1.10.3.1]. The extension to quasi-Banach spaces can be found in Kalton, Mayboroda and Mitrea [10]. Proposition 4.8. Let 0 < Θ < 1. Let (X1, Y1 ) and (X2 , Y2 ) be two compatible couples of quasi-Banach spaces. In addition, let X1 + Y1 , X2 + Y2 be analytically convex. If T is in L (X1 , X2 ) and in L (Y1 , Y2 ), then the restriction of T to [X1 , Y1 ]Θ is in L ([X1 , Y1 ]Θ , [X2 , Y2 ]Θ ) for every Θ. Moreover, T : [X1 , Y1 ]Θ → [X2 , Y2 ]Θ ≤ T : X1 → X2 1−Θ T : Y1 → Y2 Θ . Complex interpolation of isotropic Lizorkin–Triebel spaces has been studied, e.g., in [6,10,22,23]. For the case of dominating mixed smoothness one can ﬁnd the proof for associated sequence spaces in [28, Theorem 4.6]. However, these results can be shifted to the level of function spaces by some wavelet isomorphisms, see [28, Theorem 2.12]. Proposition 4.9. Let ti ∈ R, 0 < pi < ∞, 0 < qi ≤ ∞, i = 1, 2, and min(q1 , q2 ) < ∞. Let 0 < Θ < 1. If t0 , p0 and q0 are given by 1−Θ Θ 1 = + , p0 p1 p2 then

and

1 1−Θ Θ = + , q0 q1 q2

t0 = (1 − Θ)t1 + Θt2 ,

Fpt00,q0 (Rd ) = Fpt11,q1 (Rd ), Fpt22,q2 (Rd ) Θ Spt00 ,q0 F (Rd) = Spt11 ,q1 F (Rd), Spt22 ,q2 F (Rd) Θ .

4.3. Dual spaces. Next we will recall some results about the dual t t spaces of Fp,q (Rd ) and Sp,q F (Rd). Note that S(Rd ) is dense either in t (Rd ) or in S t F (Rd ) if and only if max(p, q) < ∞. By F t (Rd ) we de˚p,q Fp,q p,q t (Rd ) and by S ˚t F (Rd ) the closure of S(Rd ) note the closure of S(Rd ) in Fp,q p,q t F (Rd ). The dual space of F t (Rd ) must be understood in the followin Sp,q p,q t (Rd )] of F t (Rd ) if and ing sense: f ∈ S (Rd ) belongs to the dual space [Fp,q p,q only if there exists a positive constant C such that t (Rd) |f (ϕ)| ≤ Cϕ|Fp,q

for all ϕ ∈ S(Rd ). Analysis Mathematica 43, 2017

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t F (Rd ). For 1 < p < ∞ the conjugate exponent Similarly for the space Sp,q p is determined by 1p + p1 = 1. If 0 < p ≤ 1 we put p := ∞ and if p = ∞ we put p := 1. Let c0 be the space of all sequences converging to zero. Let Lp (c0 ) denote the space of all sequences {ψk¯ }k¯ of measurable functions such that

lim |ψk¯ (x)| = 0

¯ 1 →∞ |k|

a.e.

equipped with the norm ψk¯ |Lp (c0) := sup |ψk¯ (·)| Lp (Rd ). ¯ Nd0 k∈

The following lemma is well-known, see, e.g., [23, Proposition 2.11.1] and [4, Theorems 8.18.2, 8.20.3] . Lemma 4.10. (i) Let 1 ≤ p < ∞ and 0 < q < ∞. Then g ∈ (Lp (q )) if and only if it can be represented uniquely as g(f ) = gk¯ (x)fk¯ (x) dx for every f = {fk¯ (x)}k∈ ¯ Nd0 ∈ Lp (q ), ¯ Nd0 k∈

Rd

where

g = {gk¯ (x)}k∈ ¯ Nd0 ∈ Lp (q )

and

g = gk¯ |Lp (q ).

(ii) Let 1 < p < ∞. Then we have (Lp (c0 )) = Lp (1 ) . Proposition 4.11. Let t ∈ R. (i) If 1 < p < ∞ and 1 ≤ q ≤ ∞, then d ˚t (Rd )] = F −t [F p,q p ,q (R )

and

d ˚t F (Rd )] = S −t [S p,q p ,q F (R ).

(ii) If 0 < p < 1 and 0 < q ≤ ∞ then −t+d( p1 −1)

t ˚p,q [F (Rd )] = B∞,∞

(Rd )

and

−t+ 1 −1

t ˚p,q [S F (Rd)] = S∞,∞p

B(Rd).

Proof. The proof in the isotropic setting can be found in [23, Section 2.11] and [12]. Duality of spaces of dominating mixed smoothness has been considered in [7, Section 5.5]. But there only partial results with respect to sequence spaces associated to Lizorkin–Triebel spaces of dominating mixed smoothness can be found. Analysis Mathematica 43, 2017

ISOTROPIC AND DOMINATING MIXED LIZORKIN–TRIEBEL SPACES

319

Step 1. The case 1 < p < ∞ and 1 ≤ q < ∞. d t d Let f ∈ Substep 1.1. We shall prove that Sp−t ,q F (R ) → [Sp,q F (R )] . d Sp−t ,q F (R ). We shall employ Proposition 4.7. Since 1 < p < ∞ and 1 < q ≤ ∞ we can ﬁnd {fk¯ }k∈ ¯ Nd0 such that ¯ d ∗ F −1 ϕk¯ F fk¯ in S (Rd ) and 2−|k|1 t fk¯ |Lp (q ) ≤ 2 f |Sp−t f= ,q F (R ) . ¯ Nd0 k∈

With ∈ S(Rd ) we conclude −1 −1 (F ϕk¯ F fk¯ )( ) = fk¯ (F ϕk¯ F ) |f ( )| = ¯ Nd0 k∈

¯ Nd0 k∈

¯ ¯ ≤ c1 2−|k|1 t fk¯ Lp (q ) · 2|k|1 t F ϕk¯ F −1 Lp (q ) d ∗ t ≤ c2 f Sp−t · Sp,q F (Rd ). ,q F (R ) Substep 1.2. Now we prove the reverse direction. Here we assume that the underlying decomposition of unity, see (2.1) and (2.2), is generated by an even function ϕ0 . Then all elements of the sequence {ϕk¯ }k∈ ¯ Nd0 are even t d functions as well. Let f ∈ Sp,q F (R ). Then the operator ¯ f → 2|k|1 t F −1 ϕk¯ F f k∈ ¯ Nd 0

t F (Rd ) onto a subspace Y of L ( ). Hence, is one-to-one mapping from Sp,q p q t F (Rd )] can be interpreted as a functional on that every functional g ∈ [Sp,q

subspace. From the Hahn–Banach theorem we derive that g can be extended to a continuous linear functional on Lp (q ) with preservation of the norm. We still denote this extension by g. Now if ∈ S(Rd ), then Lemma 4.10 yields the existence of a sequence {gk¯ }k¯ such that ¯ gk¯ (x) 2|k|1 t F −1 [ϕk¯ F ](x) dx g( ) = ¯ Nd0 k∈

Rd

and g = gk¯ |Lp (q ) . Next we continue with a simple observations. Since ϕk¯ is even we obtain ϕk¯ (ξ) (F )(ξ) = ϕk¯ (−ξ) (F −1 )(−ξ) ,

ξ ∈ Rd .

Applying the inverse Fourier transform to both sides of this identity it follows F −1 [ϕk¯ (ξ) (F )(ξ)](x) = F [ϕk¯ (ξ) (F −1 )(ξ)](x) . Analysis Mathematica 43, 2017

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¯ We put hk¯ := 2|k|1 t gk¯ , k¯ ∈ Nd0 , By ·, · we denote the scalar product in L2 (Rd ). Because of ϕk¯ is real-valued, we ﬁnd

hk¯ (x)F −1 [ϕk¯ F ](x) dx = hk¯ , F −1 [ϕk¯ F ] = hk¯ , F [ϕk¯ F −1 ] Rd

= hk¯ , F −1 [ϕk¯ F ] = F [ϕk¯ F −1 hk¯ ] ( ) .

Altogether this proves the identity (F −1[ϕk¯ F hk¯ ])( ) g( ) = ¯ Nd0 k∈

for all ∈ S(Rd ). This means g=

F −1 [ϕk¯ F hk¯ ] in S (Rd )

¯ Nd0 k∈ ¯

and g = 2−|k|1 t hk¯ |Lp (q ) < ∞. In view of Proposition 4.7 we conclude d that g ∈ Sp−t ,q F (R ). Step 2. The case 1 < p < ∞ and q = ∞. Substep 2.1. We prove that (4.9)

¯ lim 2|k|1 t F −1 [ϕk¯ F f ](x) = 0 a.e.

¯ 1 →∞ |k|

˚t F (Rd). Indeed, let g ∈ S(Rd ) ⊂ S t F (Rd ). This imholds for all f ∈ S p,∞ p,p plies 1/p ¯ t d | k| tp −1 p d 1 2 |F [ϕk¯ F g](·)| g|Sp,p F (R ) = Lp (R ) <∞ ¯ Nd0 k∈

and hence the function

¯

2|k|1 tp |F −1 [ϕk¯ F g](x)|p

¯ Nd0 k∈

has to be ﬁnite a.e. in Rd . Consequently (4.9) holds for g ∈ S(Rd ) ⊂ ˚t F (Rd ). S p,∞ ˚t F (Rd) and let (gN )N Now we turn to the general situation. Let f ∈ S p,∞ t ⊂ S(Rd ) be a sequence which approximates f in Sp,∞ F (Rd), i.e., we assume Analysis Mathematica 43, 2017

ISOTROPIC AND DOMINATING MIXED LIZORKIN–TRIEBEL SPACES

that for all ε > 0 there exists some N0 such that f − gN S t F (Rd ) (4.10) p,∞ ¯ = sup 2|k|1 t F −1 [ϕk¯ F (f − gN )]Lp (Rd ) < ε ¯ Nd0 k∈

321

for all N ≥ N0 .

Next we assume that there exists a set Ω ⊂ Rd with positive Lebesgue mea ¯ ¯ 1 → ∞ in Ω. For x ∈ Ω it sure such that 2|k|1 t F −1 [ϕk¯ F f ](x) → 0 as |k| follows that there exist a positive number δ and a subsequence (k¯ ) such that ¯ 2|k |1 t F −1 [ϕk¯ F f ](x) ≥ δ for all ∈ N . Here δ = δ(x), x ∈ Ω. For all N this implies ¯ sup 2|k|1 t F −1 [ϕk¯ F (f − gN )](x) ≥ δ(x) a.e. in Ω

¯ Nd0 k∈

and therefore t f − gN Sp,∞ F (Rd ) ¯ = sup 2|k|1 t F −1 [ϕk¯ F (f − gN )]Lp (Rd) ≥ δ |Lp (Ω) > 0. ¯ Nd0 k∈

But this is in contradiction with (4.10). Substep 2.2. From Substep 2.1 we have the mapping ¯

f → {2|k|1 t F −1 ϕk¯ F f }k∈ ¯ Nd0 ˚t F (Rd) to Lp (c0). Now by the same arguis an isometric mapping from S p,∞ ment as in Step 1, the assertion follows from Lemma 4.10(ii). Step 3. Proof of (ii). We have 1

1

d d ˚t ˚t ˚t− p +1 F (Rd) = S ˚t− p +1 B(Rd ) , S 1,1 p,min(p,q) B(R ) → Sp,q F (R ) → S1,1

see [16, Subsections 2.2.3 and 2.4.1] and [8]. The known duality relations of t B(Rd ), see [13] and references given there, yields the space Sp,q −t+ 1 −1

S∞,∞p

−t+ 1 −1

˚t F (Rd )] → S∞,∞p B(Rd ) → [S p,q

B(Rd ) .

This ﬁnishes the proof. Analysis Mathematica 43, 2017

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4.4. Test function. Let us give some few more properties of our smooth decompositions of unity in Section 2. As a consequence of the deﬁnitions we obtain for j ∈ N ϕj (ξ) = 1 and ∈ N ψ (x) = 1

on the set

if

3 4

2j ≤ ξ ≤ 2j ,

x : sup |xj | ≤ 2 \ x : sup |xj | ≤ j=1,...,d

j=1,...,d

3 4

2 .

Let ∈ N. In the following examples we assume that aj , j = 1, . . . , , are complex numbers which will be chosen later on. Example 1. Let us ﬁx ∈ N and let η ∈ S(R) such that supp F η ⊂ {ξ ∈ R : 0 < ξ < 14 }. We deﬁne the function g by its Fourier transform F g (ξ) =

aj (F η) ξ − 78 2j .

j=1

Then we arrive at

7 j F −1 ϕj F g (ξ) = aj e 8 2 iξ η(ξ) ,

j ≤ .

Consequently we obtain 1/q 0 q g F (R) = η Lp (R) |aj | . p,q j=1

Now we turn to the multi-dimensional case and introduce a new family of test functions f : Rd → C as follows: F f (x) = θ (x1 ) · · · θ (xd−1 )(F g)(xd ) ,

x = (x1 , . . . , xd ) ∈ Rd ,

where θ1 ∈ S(R) is a function satisfying supp θ1 ⊂ ξ : ϕ1 (ξ) = 1 and θ (ξ) = θ1 (2−+1ξ). Clearly,

supp θ ⊂ ξ : ϕ (ξ) = 1 and

supp(F f ) ⊂ {x : ψ (x) = 1}.

By means of the cross norm property we obtain (4.11)

0 f Sp,q F (Rd) =

d−1 j=1

Analysis Mathematica 43, 2017

−1 0 0 F θ |Fp,q (R) g |Fp,q (R)

ISOTROPIC AND DOMINATING MIXED LIZORKIN–TRIEBEL SPACES

=

d−1

323

−1 0 F θ |Lp (R) g |Fp,q (R)

j=1

d−1 0 1 (R) = 2(−1)(1− p )(d−1) F −1 θ1 Lp (Rd ) g Fp,q = C1 2

(1− p1 )(d−1)

|aj |

q

1/q .

j=1 0 (Rd ), the construction of f and basic properties From the deﬁnition of Fp,q of the Fourier transform we conclude that ∞ 1/q −1 0 d q d |F ψj F f | f |Fp,q (R ) = Lp (R ) j=0

= F −1 θ (x1 ) · · · θ (xd−1 )(F g)(xd ) Lp (Rd ) =

d−1

1 F −1 θ |Lp (R) g |Lp (R) = C1 2(1− p )(d−1) g |Lp (R)

j=1

holds for an appropriate positive constant C1 . Using the Littlewood-Paley characterization of Lp (R), 1 < p < ∞, it follows 0 f |Fp,q (Rd )

(4.12)

2

(1− p1 )(d−1)

1/2 |aj |

2

,

∈ N.

j=1

Example 2. Let us consider a function g ∈ C0∞ (R) such that supp g ⊂ {t ∈ R : 3/4 ≤ |t| ≤ 1}. For ∈ N0 we deﬁne g (t) := g(2− t) and f (x) := F −1 g (ξ1 )g0 (ξ2 ) · · · g0 (ξd ) (x) . Then we ﬁnd t f |Fp,q (Rd ) = 2t F −1 g (ξ1 )g0 (ξ2 ) · . . . · g0 (ξd ) Lp (Rd ) 1 = F −1 g0 (ξ1 )g0 (ξ2) · . . . · g0 (ξd ) Lp (Rd ) 2(t+1− p ) and t f |Sp,q F (Rd) = 2t F −1 g (ξ1)g0 (ξ2 ) · · · g0 (ξd ) Lp (Rd ) 1 = F −1 g0 (ξ1)g0 (ξ2 ) · . . . · g0 (ξd ) Lp (Rd ) 2(t+1− p ) . Analysis Mathematica 43, 2017

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Example 3. We consider the same functions g as in Example 2. This time we deﬁne f (x) := F −1 g (ξ1 )g (ξ2 ) · · · g (ξd ) (x). Let G (ξ) := g (ξ1 )g (ξ2) · . . . · g (ξd ). As above we conclude t 1 f Sp,q F (Rd) = 2td F −1 G Lp (Rd ) = C3 2d(t+1− p ) ,

∈ N,

and t t 1 f F (Rd ) = 2t F −1 G Lp (Rd) = C3 2d( d +1− p ) , p,q

∈ N.

Example 4. Let g ∈ S(Rd ) with supp F g ⊂ [0, 14 ]d . We deﬁne f (x) :=

7

j

aj ei 8 2

x1

x = (x1 , . . . , xd ) ∈ Rd .

g(x),

j=1

Then we have F f (ξ) =

aj (F g) ξ1 − 78 2j , ξ2 , . . . , ξd .

j=1

We obtain F −1 [ϕk¯ F f ](x) =

7

j

δk,(j,0,...,0) aj ei 8 2 ¯

x1

g(x)

j=1

and 7

F −1 [ψj F f ](x) = aj ei 8 2

j

x1

g(x) ,

j ≤ .

This leads to 1/q t t d d jtq q f Fp,q (Rd ) = f Sp,q F (R ) = g Lp (R ) 2 |aj | . j=1

Example 5. We shall modify Example 4. This time we deﬁne the function f (x) :=

7

j

aj ei 8 2

j=1 Analysis Mathematica 43, 2017

(x1 +...+xd )

g(x) ,

x = (x1 , . . . , xd ) ∈ Rd .

ISOTROPIC AND DOMINATING MIXED LIZORKIN–TRIEBEL SPACES

325

As above we conclude F −1 [ϕk¯ F f ](x) =

7

j

δk,(j,...,j) aj ei 8 2 ¯

(x1 +...+xd )

g(x)

j=1

and 7

F −1 [ψj F f ](x) = aj ei 8 2

j

(x1 +...+xd )

g(x) ,

j ≤ .

Hence, we obtain 1/q t d d jtq q f Fp,q (R ) = g Lp (R ) 2 |aj | j=1

and

1/q t d d djtq q f S F (R ) = g Lp (R ) 2 |aj | . p,q j=1

Example 6. This example is taken from [23, 2.3.9], see also [13]. Let

∈ S(Rd ) be a function such that supp F ⊂ {ξ :: |ξ| ≤ 1}. We deﬁne hj (x) := (2−j x) ,

x ∈ Rd , j ∈ N.

For all admissible p, q, t we conclude t t F (Rd) = hj |Fp,q (Rd ) = hj |Lp (Rd ) = 2jd/p |Lp (Rd ), j ∈ N . hj |Sp,q

As an immediate conclusion of this example we obtain the following result. Lemma 4.12. Let 0 < p0 , p1 < ∞, 0 < q0 , q1 ≤ ∞ and t0 , t1 ∈ R. (i) An embedding Spt00 ,q0 F (Rd) → Fpt11,q1 (Rd ) implies p0 ≤ p1 . (ii) An embedding Fpt00,q0 (Rd ) → Spt11 ,q1 F (Rd) implies p0 ≤ p1 . 4.5. Proof of results in Section 3.1. Proof of Theorem 3.1. Step 1. Preparations. For k¯ ∈ Nd0 we deﬁne k¯ := j ∈ N0 : supp ψj ∩ supp ϕk¯ = ∅ and j ∈ N0

j := k¯ ∈ Nd0 : supp ψj ∩ supp ϕk¯ = ∅ .

The condition supp ψj ∩ supp ϕk¯ = ∅ implies (4.13)

max ki − 1 ≤ j ≤ max ki + 1.

i=1,...,d

i=1,...,d

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V. K. NGUYEN and W. SICKEL

Consequently we obtain | k¯ | 1, k¯ ∈ Nd0

and | j | (1 + j)d−1 , j ∈ N0 .

By deﬁnition we have

ψj (x) =

(4.14)

ϕk¯ (x)ψj (x),

x ∈ Rd .

¯ k∈Δ j

Step 2. The case t > 0. From (4.14) we have (4.15) q 1/q ∞ tj−t|k| ¯ 1 t|k| ¯ 1 −1 t d L (Rd ) = 2 2 F [ϕ ψ F f ] (R ) f |Fp,q p ¯ j k . ¯ j=0 k∈

j

If q ≤ 1, then (4.16) t f |Fp,q (Rd )

∞ t(j−|k| q 1/q ¯ 1 ) t|k| ¯ 1 −1 d 2 ≤ 2 F [ϕk¯ ψj F f ] Lp (R ) ¯ j=0 k∈

j

∞ t|k| q 1/q ¯ 1 −1 d ≤ c1 2 F [ϕk¯ ψj F f ] Lp (R ). ¯ j=0 k∈Δ j

The last inequality is due to ¯

sup sup 2t(j−|k|1 ) ≤ c1 < ∞ . ¯ j≥0 k∈Δ j

Now we turn to q > 1. Using H¨older’s inequality we obtain from (4.15) ¯ 1 t|k| ¯ 1 −1 tj−t|k| 2 2 F [ϕk¯ ψj F f ] ¯ k∈

j

≤

|2

¯1 t|k|

F

−1

q [ϕk¯ ψj F f ]

¯ k∈

j

1/q

2

¯ 1 )q t(j−|k|

1/q ,

¯ k∈

j

where 1q + q1 = 1. Because of t > 0 and (4.13) the second sum on the righthand side is uniformly bounded, i.e., 1/q ¯ 1 )q t(j−|k| sup 2 ≤ c2 < ∞. j≥0 Analysis Mathematica 43, 2017

¯ k∈Δ j

ISOTROPIC AND DOMINATING MIXED LIZORKIN–TRIEBEL SPACES

327

Consequently, we obtain for all q t ∞ t|k| 1/q f F (Rd ) ≤ c3 Lp (Rd ). 2 ¯ 1 F −1 [ϕ¯ ψj F f ]q p,q k ¯ j=0 k∈Δ j

Let τ := min(1, p, q). Interchanging the order of summation we ﬁnd τ t t|k| τ q 1/q ¯ d τ −1 d 1 f Fp,q (R ) ≤ c3 2 F [ϕk¯ ψj F f ] Lp (R ) ¯ Nd0 j∈ k¯ k∈

≤

cτ3

τ 1 t|k| q 1/q ¯ −1 d 1 2 F [ϕk¯ ψj+i F f ] Lp (R ) , ¯ Nd0 k∈

i=−1

¯ ∞ , see (4.13). We estimate the term with i = 0. The terms where j := |k| with i = ±1 can be treated in a similar way. Let {ϕ˜k¯ }k∈ ¯ Nd0 be the system deﬁned in the proof of Proposition 4.7, i.e., ϕ ˜k¯ = (ϕk1 −1 + ϕk1 + ϕk1 +1 ) ⊗ · · · ⊗ (ϕkd −1 + ϕkd + ϕkd +1 ) with {ϕj }∞ ˜k¯ ≡ 1 on supp ϕk¯ for all j=0 given in (2.1) and ϕ−1 ≡ 0. Since ϕ k¯ ∈ Nd0 we have t|k| 1/q d 2 ¯ 1 F −1 [ϕ¯ ψj F f ]q Lp (R ) k ¯ Nd0 k∈

−1 t|k| q 1/q ¯ 1 −1 d F ϕ = ˜k¯ ψj F 2 F ϕk¯ F f Lp (R ). ¯ Nd0 k∈

¯ ¯ Applying Lemma 4.6 with Mk¯ = ϕ˜k¯ ψj and ak¯ = 2k+1 , k¯ ∈ Nd0 , we obtain t|k| q 1/q ¯ 1 −1 d L 2 (4.17) F ϕ ψ F f (R ) ¯ j p k ¯ Nd0 k∈

≤ c4 sup (ϕ˜k¯ ψj )(2

¯ ¯ k+ 1

¯ Nd0 k∈

·)|S2r W (Rd )

t|k| q 1/q ¯ 1 −1 d 2 F ϕk¯ F f |Lp (R ), ¯ Nd0 k∈

where we have chosen r ∈ N such that r >

1 min(p,q)

+ 12 . To estimate the

factor · · · |S2r W (Rd ) we consider several cases. First, we assume that mini=1,...,d ki ≥ 1. Then it follows ¯ ¯

(ϕ˜k¯ ψj )(2k+1 x) = ϕ ˜¯1 (4x) ψ1 (2k1 −j+2 x1 , . . . , 2kd −j+2 xd ) . Analysis Mathematica 43, 2017

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For any α ¯ ∈ Nd0 , since k1 − j + 2 = k1 − |k|∞ + 2 ≤ 2, we conclude the existence of a positive constant Cα¯ such that ¯ sup D α¯ (ψ1 (4 2k−|k|∞ x)) ≤ Cα¯ < ∞ . x∈Rd

Furthermore r r ¯ ¯ 1 ϕ˜¯ (2k+ S W (Rd ) = ϕ S W (Rd ) = Cr < ∞ , ·) ˜ (4·) ¯ 1 2 2 k which implies

¯ ¯ sup (ϕ˜k¯ ψj )(2k+1 ·)S2r W (Rd) ≤ c5 .

¯ Nd k∈

Now we turn to the cases mini=1,...,d ki = 0. Let us assume that 0 = k1 = · · · = km < km+1 ≤ · · · ≤ kd for some m < d. Recall the notation from (4.7). Then ¯ ¯

˜0 (2x1 ) · . . . · ϕ˜0 (2xm ) · ϕ ˜1 (4xm+1 ) · . . . · ϕ ˜1 (4xd ) (ϕ˜k¯ ψj )(2k+1 x) = ϕ × ψ1 (2−j+2x1 , . . . , 2−j+2 xm , 2km+1 −j+2 xd , . . . , 2kd −j+2xd ) . Now we proceed as above and ﬁnd also in this case ¯ ¯ 1 (ϕ˜¯ ψj )(2k+ ·)S2r W (Rd ) ≤ c6 . k Similarly we can treat all cases caused by a diﬀerent ordering of the compo¯ The case k¯ = ¯0, j = 0 can be handled in the same way. Summanents of k. rizing, we get ¯ ¯ (4.18) sup (ϕ˜k¯ ψj )(2k+1 ·)S2r W (Rd ) ≤ c7 < ∞ . ¯ Nd0 k∈

From (4.18) and (4.17) we derive t|k| 1/q d t 2 ¯ 1 F −1 ϕ¯ ψj F f q F (Rd) . Lp (R ) ≤ c8 f |Sp,q k ¯ Nd0 k∈

t F (Rd ) → F t (Rd ). We conclude that Sp,q p,q Step 3. The case t = 0. If 0 < q ≤ 1 we can argue as in (4.16) with t = 0. This implies ∞ −1 q 1/q 0 d d F [ϕk¯ ψj F f ] f |Fp,q (R ) ≤ Lp (R ). ¯ j=0 k∈Δ j

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0 F (Rd ) → F 0 (Rd ) if 0 < q Then we continue as in Step 2 resulting in Sp,q p,q ≤ 1. Next we consider the case 1 < q < 2. Here we employ complex interpolation. For 0 < p < ∞ and 1 < q < 2 there exist Θ ∈ (0, 1), 0 < p0 < ∞ and 1 < p1 < ∞ such that

1−Θ Θ 1 = + p p0 p1

and

1−Θ Θ 1 = + . q 1 2

Proposition 4.9 yields 0 Sp,q F (Rd) = Sp00 ,1 F (Rd ), Sp01 ,2 F (Rd) Θ and

0 (Rd ) = Fp00 ,1 (Rd ), Fp01 ,2 (Rd ) Θ . Fp,q

As a consequence of the Littlewood–Paley assertion we have Sp01 ,2 F (Rd) = Fp01 ,2 (Rd ), see Remark 3.2. Now, the claim follows from Proposition 4.8. Lemma 4.13. Let d ≥ 2, 0 < p < ∞ and 0 < q ≤ ∞. Then the embedding 0 0 Sp,q F (Rd) → Fp,q (Rd )

q ≤ 2.

implies

Proof. Step 1. The case 1 0 such that 2

(1− p1 )(d−1)

1/2 |aj |

2

≤ c2

j=1

(1− p1 )(d−1)

1/q |aj |

q

j=1

holds for all ∈ N and all sequences {aj }j , see (4.11) and (4.12). This requires q ≤ 2. 0 F (Rd ) → F 0 (Rd ) with Step 2. The case 0 < p ≤ 1. Assume that Sp,q p,q 0 < p ≤ 1 and 2 < q ≤ ∞. Then we can ﬁnd a triple (p1 , q1 , Θ) such that Θ ∈ (0, 1), 1 < p1 < 2 < q1 < ∞, Θ 1−Θ 1 + = p1 p 2

and

1 Θ 1−Θ + . = q1 q 2

Complex interpolation, see Propositions 4.8 and 4.9, yields Sp01 ,q1 F (Rd ) → Fp01 ,q1 (Rd). But this is in contradiction with Step 1 since p1 > 1 and q1 > 2. Analysis Mathematica 43, 2017

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Proof of Theorem 3.3. As a consequence of Theorem 3.1 and Lemma t F (Rd ) 4.13 it will be enough to consider the case t < 0. We assume Sp,q t d t d t d ˚ F (R ) → F ˚ (R ). Then, → Fp,q (R ) if t < 0. Observe that this implies S p,q p,q −t −t d by duality, see Proposition 4.11, we obtain Fp ,q (R ) → Sp ,q F (Rd ). Using the test function from Example 5 with aj := δj, we can disprove this embedding. Proof of Proposition 3.4. Proof of (i). Theorem 3.1 implies ˚t (Rd ) ˚t F (Rd ) → F S p,q p,q

if t > 0.

−t d d Proposition 4.11 yields Fp−t ,q (R ) → Sp ,q F (R ), if 1 < p < ∞ and 1 ≤ q ≤ ∞. Proof of (ii). Since −t + 1p − 1 < −t + d( 1p − 1) and t < 0 we can use −t+ 1 −1

¯

−t+d( p1 −1)

gk¯ = eikx as test functions to prove that S∞,∞p B(Rd ) and B∞,∞ are not comparable. Here one can use s ikx ¯ 2 ) s , k ∈ Nd e ¯ B∞,∞ (Rd) (1 + |k| 0

(Rd )

and d s ikx d e ¯ S∞,∞ B(R )

(1 + |ki |)s ,

k ∈ Nd0 .

i=1

˚t F (Rd ) and F ˚t (Rd ) Then, from Proposition 4.11, we can conclude that S p,q p,q t d t d are incomparable and therefore Sp,q F (R ) and Fp,q (R ) as well. This ﬁnishes the proof. 4.6. Proof of the results in Section 3.2. Proof of Theorem 3.7. The claim for the case t = 0 is a consequence of Theorem and a duality argument, see Proposition 4.11. The proof in 3.1 1 case t > min(p,q) − 1 + will be divided into several steps. Step 1. We shall prove the embedding under the assumptions 0 min(p,q) and (4.13) we obtain q 1/q τ |k| ¯ 1 t −1 t d τ d (4.19) f |Sp,q F (R ) = 2 F ϕ ψ F f (R ) L ¯ j p k ¯ Nd0 j∈ k¯ k∈

τ 1 |k| q 1/q ¯ 1 t −1 d L 2 F ϕ ψ F f (R ) ≤ ¯ j+i p k , i=−1

¯ Nd0 k∈

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¯ ∞ . It will be enough to deal with the term for i = 0. where again j := |k| 1 we can write The others terms can be treated similarly. Since t > min(p,q) 1 t = a + ε with a > min(p,q) and ε > 0. We put ¯ gk¯ (x) := F −1 2(|k|1 −jd)ε 2jtd ψj F f (x) ,

¯ ∞. k¯ ∈ Nd0 , j = |k|

It follows |k| q 1/q ¯ t −1 d 1 2 F [ϕk¯ ψj F f ] Lp (R )

(4.20)

¯ Nd0 k∈

(|k| q 1/q ¯ 1 −jd)a −1 d 2 = F [ϕk¯ F gk¯ ] Lp (R ) . ¯ Nd0 k∈

Next we need the related Peetre maximal function. We deﬁne P2¯j+¯1 ,a gk¯ (x) := sup d

|gk¯ (x − z)|

j+1 z |a ) i i=1 (1 + |2

z∈Rd

¯ ∞, x ∈ Rd , k ∈ Nd0 , j = |k|

,

compare with Proposition 4.5. A standard convolution argument, given by

−1 −1 F [ϕ¯ F g¯ ] (x − z) ≤ (2π)−d/2 (F ϕ¯ )(x − z − y) · |g¯ (y)| dy k k k k ≤ (2π)

−d/2

P2¯j+¯1 ,a gk¯ (x)

Rd

Rd

d −1 (F ϕ¯ )(x − z − y) (1 + |2j+1 (xi − yi )|a ) dy, k

i=1

the elementary inequality (1 + |2j+1(xi − yi )|a ) ≤ 2a 1 + |2j+1 zi |a 1 + |2j+1(xi − zi − yi )|a , i = 1, . . . , d, and a change of variable lead to −1

d F ϕ¯ F g¯ (x − z) −1 k k (F ϕk¯ )(y) (1 + |2j+1 yi |a ) dy. ≤c1 P2¯j+¯1 ,a gk¯ (x) d j+1 z |a ) d (1 + |2 R i i=1 i=1 Temporarily we assume mini=1,...,d ki ≥ 1. Then

Rd

d d −1 (F ϕ¯ )(y) (1 + |2j+1 yi |a ) dy = k i=1

i=1

R

−1 F ϕ1 (t)(1 + 2j+2−ki |t|)a dt Analysis Mathematica 43, 2017

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¯ ∞ and F −1 ϕ1 ∈ S(R) we have follows. Since ki ≤ j = |k|

−1 F ϕ1 (t)(1 + 2j+2−ki |t|)a dt

R

= 2(j−ki )a R

−1 F ϕ1 (t)(2ki −j + 4|t|)a dt ≤ c2 2(j−ki )a .

This estimate carries over to the situation mini=1,...,d ki = 0 by obvious modiﬁcations. Consequently −1 F [ϕ¯ F g¯ ] (x − z) ¯ 1 −jd)a k k (|k| 2 ≤ c3 P2¯j+¯1 ,a gk¯ (x) d j+1 |z |)a (1 + 2 i i=1 with a constant c3 independent of x and {gk¯ }k¯ . Obviously, this implies ¯ 2(|k|1 −jd)a F −1 [ϕk¯ F gk¯ ] (x) ¯ 2(|k|1 −jd)a F −1 [ϕk¯ F gk¯ ] (x − z) ≤ c3 P2¯j+¯1 ,a gk¯ (x), ≤ sup d j+1 z |a ) z∈Rd i i=1 (1 + |2 which results in the estimate |k| 1/q d 2 ¯ 1 t F −1 [ϕ¯ ψj F f ] q Lp (R ) k ¯ Nd0 k∈

q 1/q d ¯ ¯ P2j+1 ,a gk¯ (·) ≤ c3 Lp (R ), ¯ Nd0 k∈

see (4.20). Now, applying Proposition 4.5 with respect to {gk¯ }k∈ ¯ Nd0 and with j+1 ¯ aki chosen to be 2 , i = 1, . . . , d, j = |k|∞ , we obtain |k| q 1/q ¯ t −1 d 2 1 F [ϕ¯ ψj F f ] Lp (R ) k ¯ Nd0 k∈

−1 (|k| q 1/q ¯ −jd)ε jtd d 1 F [2 ≤ c4 2 ψj F f ] Lp (R ) ¯ Nd0 k∈

−1 (|k| q 1/q ¯ 1 −jd)ε jtd d F [2 ≤ c4 2 ψj F f ] Lp (R ) ¯ Nd0 j∈ k¯ k∈

∞ 1/q jtd −1 q (|k| ¯ 1 −jd)εq d 2 F [ψj F f ] = c4 2 Lp (R ). j=0 Analysis Mathematica 43, 2017

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Our assumption ε > 0 guarantees |k| 1/q d 2 ¯ 1 t F −1 [ϕ¯ ψj F f ] q Lp (R ) k ¯ Nd0 k∈

∞ jtd −1 q 1/q d td 2 F [ψj F f ] ≤ c5 (Rd ) , Lp (R ) = c5 f |Fp,q j=0

see (4.13). Inserting this into (4.19) and carrying out the estimates of the other terms in the same way, the claim follows. Step 2. We shall prove the embedding under the assumptions 1 0. This time we use Proposition 4.3. Starting point is ¯ ∞ . The inequality (4.19). As above it will be enough to deal with j := |k| remaining terms can be treated in a similar way. Using the same decomposition as in (4.8) we obtain q 1/q |k| ¯ t −1 d 2 1 F [ϕ¯ ψj F f ] Lp (R ) k =

≤ c6

¯ Nd0 k∈

−1 ¯ 1 t −1 q F ϕ¯ F 2|k| F ψj F f k

e⊂{1,...,d} ki ≥1,i∈e kj =0,j ∈e

e⊂{1,...,d}

1/q d Lp (R )

−1 ¯ 1 t −1 q F ϕ¯ F 2|k| F ψj F f k

ki ≥1,i∈e kj =0,j ∈e

1/q Lp (Rd ).

Applying Proposition 4.3 for each term under the outer sum we arrive at

e⊂{1,...,d}

...

|k| 1/q d 2 ¯ 1 t F −1 [ϕ¯ ψj F f ] q Lp (R ) k ¯ Nd0 k∈

|k| q 1/q ¯ 1 t −1 d 2 ≤ c7 F [ψj F f ] Lp (R ) ¯ Nd0 k∈

1/q ∞ jtd −1 q (|k| ¯ 1 −jd)tq d 2 F [ψj F f ] ≤ c7 2 Lp (R ) . j=0

¯ k∈Δ j

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Because of t > 0 we conclude that td d |k| 1/q d 2 ¯ 1 t F −1 [ϕ¯ ψj F f ] q (R ). Lp (R ) ≤ c7 f Fp,q k ¯ Nd0 k∈

From this the claim follows. 1 − 1 + . We shall proceed by Step 3. Let 0 < p, q < ∞ and t > min(p,q) interpolation. Substep 3.1. Assume that min(p, q) ≤ 1 and p ≤ q. Since t > 1p − 1 we choose p0 > 1, 0 < Θ < 1 and ε > 0 such that t = ε + 1p − p10 + pΘ0 . Next we Θ deﬁne (p0 , q0 ), (p1 , q1 ) by 1p = 1−Θ and pq = pq00 = pq11 . Now we put p0 + p1 t0 := ε and t1 := min(p11 ,q1 ) + ε = p11 + ε since p1 ≤ q1 . Hence we obtain t = (1 − Θ)t0 + Θt1

and

1−Θ Θ 1 = + . q q0 q1

Proposition 4.9 yields td d d Fp,q (Rd ) = [Fpt00,q (Rd ), Fpt11,q (Rd )]Θ 0 1

and t F (Rd) = [Spt00 ,q0 F (Rd ), Spt11 ,q1 F (Rd )]Θ . Sp,q td (Rd ) → S t F (Rd ). In view of Proposition 4.8, Steps 1 and 2 we ﬁnd Fp,q p,q Substep 3.2. Assume that min(p, q) ≤ 1 and q < p. It is enough to interchanges the roles of p and q in Substep 3.1.

Remark 4.14. The interpolation argument in Substep 3.1 does not extend to the case q0 = q1 = ∞. It is known that d d td [Fpt00,∞ (Rd ), Fpt11,∞ (Rd )]Θ = Fp,∞ (Rd ) d (Rd ) = F t1 d (Rd ), see [32]. However, one could apply the ± method if Fpt00,∞ p1 ,∞ of Gustavsson and Peetre, denoted by ·, ·, Θ, to obtain t0 d d td Fp0 ,∞ (Rd ), Fpt11,∞ (Rd ), Θ = Fp,∞ (Rd ),

see [32]. However, there is no proof of the assertion t0 t F (Rd) Sp0 ,∞ F (Rd ), Spt11 ,∞ F (Rd), Θ = Sp,∞ available in the literature. Analysis Mathematica 43, 2017

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Proof of Theorem 3.9. By Theorem 3.7 and Lemma 4.13 it will be td (Rd ) → S t F (Rd ) if t < 0. enough to deal with t < 0. We assume that Fp,q p,q td d t d d ˚ ˚ This implies Fp,q (R ) → Sp,q F (R ) and therefore, by duality, Sp−t ,q F (R ) → −td d Fp ,q (R ). Applying Example 4 with aj := δj, we come to a contradiction. Proof of Proposition 3.10. Proof of (i). Since 0 < p < 1 we know −t+ 1 −1

˚t F (Rd)] = S∞,∞p [S p,q see Proposition 4.11.

−td+d( p1 −1)

˚td (Rd )] = B∞,∞ B(Rd ) and [F p,q

(Rd ) ,

td (Rd ) → S t F (Rd ) we get F ˚td (Rd ) Assuming Fp,q p,q p,q −t+ 1 −1

−td+d( 1 −1)

t ˚p,q F (Rd) and therefore S∞,∞p B(Rd ) → B∞,∞ p (Rd ). This is im→ S possible. If t = 1p − 1 we refer to [13] where we have discussed the ret B(Rd ) to B td (Rd ) and B t (Rd ) in great detail. In case lations of Sp,q p,q p,q ¯ −td + d( 1p − 1) > −t + 1p − 1 > 0 it is enough to use the test functions eikx to td (Rd ) → S t F (Rd ). By employing the disprove this embedding. Hence Fp,q p,q test function in Example 4 with aj := δj, we can disprove the embedding t F (Rd ) → F td (Rd ) as well. Sp,q p,q t (Rd ) Proof of (ii). We argue as in the proof of Theorem 3.1 replacing Fp,q td d by Fp,q (R ) and taking into account that t < 0. The proof is complete.

4.7. Proofs of the optimality assertions. Let us recall some results about embeddings of Lizorkin–Triebel spaces. Lemma 4.15. Let 0 < p < p0 < ∞ and 0 < q, q0 ≤ ∞. t (Rd ) → F t0 (Rd ) holds if and only if t − (i) The embedding Fp,q 0 p0 ,q0 d t − p.

d p0

≤

t F (Rd ) → S t0 F (Rd ) holds if and only if t − (ii) The embedding Sp,q 0 p0 ,q0 ≤ t − 1p .

1 p0

t (Rd ) Note that in case p = p0 and t = t0 , that is the embedding Fp,q t (Rd ), holds true if and only if q ≤ q . A similar statement is true → Fp,q 0 0 Lizorkin–Triebel spaces of dominating mixed smoothness. The assertion (i) in Lemma 4.15 can be found in [9], [23, 2.7.1] (suﬃciency) and in [18] (necessity). In case of Triebel–Lizorkin spaces of dominating mixed smoothness we refer to [17] and [8] (suﬃciency). Necessity can be traced back to the isotropic case by standard arguments (one considers tensor products of appropriate test functions). Now we are in position to prove the optimality assertions. Analysis Mathematica 43, 2017

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t (Rd ), Lemma Proof of Theorem 3.5. Assuming Spt00 ,q0 F (Rd) → Fp,q 4.12 yields p0 ≤ p. Applying Example 2 we derive

t+1−

1 1 1 1 ≤ t0 + 1 − ⇐⇒ t − ≤ t0 − . p p0 p p0

If p0 = p and t0 = t we use in addition Example 4. With aj := 2−jt , the emt F (Rd ) → F t (Rd ) implies that q ≤ q. Altogether Lemma 4.15 bedding Sp,q 0 p,q 0 t d t d 0 implies that Sp0 ,q0 F (R ) → Sp,q F (R ). t F (Rd ) Lemma Proof of Theorem 3.11. Assuming Fpt00,q0 (Rd ) → Sp,q 4.12 implies p0 ≤ p. Next we employ Example 3. Then the embedding t F (Rd) yields Fpt00,q0 (Rd ) → Sp,q t 1 d d 1 0 +1− ⇐⇒ t0 − ≥ dt − . ≥d t+1− d d p0 p p0 p

In case p0 = p and t0 = td we use Example 5, again with aj := 2−jt to obtain q0 ≤ q. As a consequence of Lemma 4.15 we arrive at Fpt00,q0 (Rd ) td F (Rd ) . → Fp,q td (Rd ) → Spt00 ,q0 F (Rd) Lemma Proof of Theorem 3.12. Assuming Fp,q 4.12 implies p ≤ p0 . Next we apply Example 3 and get 1 1 1 1 ⇐⇒ t0 − ≤ t− . ≤d t+1− d t0 + 1 − p0 p p0 p

Working with Example 4 with aj := 2−jt we obtain q ≤ q0 in case p = p0 and t F (Rd ) → S t0 F (Rd ). t = t0 . In a view of Lemma 4.15 we conclude that Sp,q p0 ,q0 Acknowledgement. The authors are grateful to the referee for a careful reading of the manuscript and many detailed hints to improve the paper. References [1] R. J. Bagby, An extended inequality for the maximal function, Proc. Amer. Math. Soc., 48 (1975), 419–422. [2] O. V. Besov, V. P. Il’in and S. M. Nikol’skij, Integralnye predstavleniya funktsii i teoremy vlozheniya, Second ed., Fizmatlit “Nauka” (Moscow, 1996) (in Russian). [3] J. Bergh and J. L¨ ofstr¨ om, Interpolation Spaces. An Introduction, Springer (New York, 1976). [4] R. E. Edwards, Functional Analysis, Theory and Applications, Holt, Rinehart and Winston (New York, 1965). [5] C. Feﬀerman and E. M. Stein, Some maximal inequalities, Amer. J. Math., 93 (1971), 107–115. [6] M. Frazier and B. Jawerth, A discrete transform and decompositions of distribution spaces, J. Funct. Anal., 93 (1990), 34–170. Analysis Mathematica 43, 2017

ISOTROPIC AND DOMINATING MIXED LIZORKIN–TRIEBEL SPACES

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Isotropic and dominating mixed Lizorkin–Triebel spaces — a comparison

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