Multiplication and Composition in Weighted Modulation Spaces

Abstract

We study the existence of the product of two weighted modulation spaces. For this purpose, we discuss two different strategies. The more simple one allows transparent proofs in various situations. However, our second method allows a closer look onto associated norm inequalities under restrictions in the Fourier image. This will give us the opportunity to treat the boundedness of composition operators.

1 Introduction

Since modulation spaces have been introduced by Feichtinger [7] they have become canonical for both time-frequency and phase-space analysis. However, in recent time modulation spaces have been found useful also in connection with linear and nonlinear partial differential equations, see e.g., Wang et al. [35–38], Ruzhansky et al. [26], or Bourdaud et al. [5]. Investigations of partial differential equations require partly different tools than used in time-frequency and phase-space analysis. In particular, Fourier multipliers, pointwise multiplication and composition of functions need to be studied. In our contribution, we will concentrate on pointwise multiplication and composition of functions. Already Feichtinger [7] was aware of the importance of pointwise multiplication in modulation spaces. In the meanwhile several authors have studied this problem, we refer, e.g., to [6, 13, 29, 30, 32]. In Sect. 3, we will give a survey about the known results. Therefore, we will discuss two different proof strategies. The more simple one, due to Toft [30, 32] and Sugimoto et al. [29], allows transparent proofs in various situations, in particular one can deal with those situations where the modulation spaces form algebras with respect to pointwise multiplication. As a consequence, Sugimoto et al. [29] are able to deal with composition operators on modulation spaces induced by analytic functions. Our second method, much more complicated, allows a closer look onto associated norm inequalities under restrictions in the Fourier image. This will give us the possibility to discuss the boundedness of composition operators on weighted modulation spaces based on a technique which goes back to Bourdaud [3], see also Bourdaud et al. [5] and Reich et al. [23]. Our approach will allow to deal with the boundedness of nonlinear operators $T_f:~g \mapsto f \circ g$ without assuming f to be analytic. However, as the case of $M^s_{2,2}$ shows, our sufficient conditions are not very close to the necessary conditions. There is still a certain gap.

The paper is organized as follows. In Sect. 2, we collect what is needed about the weighted modulation spaces we are interested in. The next section is devoted to the study of pointwise multiplication. In particular, we are interested in embeddings of the type

$$ M^{s_1}_{p,q} \, \cdot \, M^{s_2}_{p,q} \hookrightarrow M^{s_0}_{p,q}\, , $$

where $s_1,s_2, p$ and q are given and we are asking for an optimal $s_0$. These results will be applied to problems around the regularity of composition of functions in Sect. 4. For convenience of the reader we also recall what is known in the more general situation

$$ M^{s_1}_{p_1,q_1} \, \cdot \, M^{s_2}_{p_2,q_2} \hookrightarrow M^{s_0}_{p,q}\, . $$

Special attention will be paid to the algebra property. Here, the known sufficient conditions are supplemented by necessary conditions, see Theorem 3.5. Also only partly new is our main result in Sect. 3 stated in Theorem 3.22. Here we investigate multiplication of distributions (possibly singular) with regular functions (which are not assumed to be $C^\infty$). Partly we have found necessary and sufficient conditions also in this more general situation. Finally, Sect. 4 deals with composition operators. As direct consequences of the obtained results for pointwise multiplication we can deal with the mappings $g \mapsto g^\ell$, $\ell \ge 2$, see Sect. 4.1. In Sect. 4.4, we shall investigate $g \mapsto f \circ g$, where f is not assumed to be analytic. Sufficient conditions, either in terms of a decay for ${\mathcal F}f$ or in terms of regularity of f, are given.

Notation

We introduce some basic notation. As usual, $\mathbb {N}$ denotes the natural numbers, $\mathbb {N}_0 := \mathbb {N}\cup \{0\}$, $\mathbb {Z}$ the integers and $\mathbb {R}$ the real numbers, $\mathbb {C}$ refers to the complex numbers. For a real number a, we put $a_+ := \max (a,0)$. For $x \in {\mathbb {R}}^n$ we use $\Vert x\Vert _\infty := \max _{j=1, \ldots \,,n} \, |x_j|$. Many times we shall use the abbreviation $\langle \xi \rangle := (1+|\xi |^2)^{\frac{1}{2}}$, $\xi \in {\mathbb {R}}^n$.

The symbols $c,c_1, c_2, \, \ldots \, ,C, C_1,C_2, \, \ldots$ denote positive constants which are independent of the main parameters involved but whose values may differ from line to line. The notation $a \lesssim b$ is equivalent to $a \le Cb$ with a positive constant C. Moreover, by writing $a \asymp b$ we mean $a \lesssim b \lesssim a$.

Let X and Y be two Banach spaces. Then the symbol $X \hookrightarrow Y$ indicates that the embedding is continuous. By ${\mathcal L}(X,Y)$ we denote the collection of all linear and continuous operators which map X into Y. By $C_0^\infty ({\mathbb {R}}^n)$ the set of compactly supported infinitely differentiable functions $f:{\mathbb {R}}^n\rightarrow \mathbb {C}$ is denoted. Let ${\mathcal S}({\mathbb {R}}^n)$ be the Schwartz space of all complex-valued rapidly decreasing infinitely differentiable functions on ${\mathbb {R}}^n$. The topological dual, the class of tempered distributions, is denoted by ${\mathcal S}'({\mathbb {R}}^n)$ (equipped with the weak topology). The Fourier transform on ${\mathcal S}({\mathbb {R}}^n)$ is given by

$$ {\mathcal F}\varphi (\xi ) = (2\pi )^{-n/2} \int _{{\mathbb {R}}^n} \, e^{ix \cdot \xi }\, \varphi (x)\, dx \, , \qquad \xi \in {\mathbb {R}}^n\, . $$

The inverse transformation is denoted by ${{\mathcal F}}^{-1}$. We use both notations also for the transformations defined on ${\mathcal S}'({\mathbb {R}}^n)$.

Convention. If not otherwise stated all functions will be considered on the Euclidean n-space ${\mathbb {R}}^n$. Therefore ${\mathbb {R}}^n$ will be omitted in notation.

2 Basics on Modulation Spaces

2.1 Definitions

A general reference for definition and properties of weighted modulation spaces is Gröchenig’s monograph [10, Chap. 11].

Definition 2.1

Let $\phi \in {\mathcal S}$ be nontrivial. Then the short-time Fourier transform of a function f with respect to $\phi$ is defined as

$$ V_\phi f(x,\xi ) = (2\pi )^{-\frac{n}{2}} \int _{{\mathbb {R}}^n} f(s) \overline{\phi (s-x)} e^{-i s \cdot \xi } \, ds \qquad (x,\xi \in {\mathbb {R}}^n). $$

The function $\phi$ is usually called the window function. For $f \in {\mathcal S}'$ the short-time Fourier transform $V_\phi f$ is a continuous function of at most polynomial growth on $\mathbb {R}^{2n}$, see [10, Theorem 11.2.3].

Definition 2.2

Let $1\le p,q\le \infty$. Let $\phi \in {\mathcal S}$ be a fixed window and assume $s \in \mathbb {R}$. Then the weighted modulation space ${M}_{p,q}^{s}$ is the collection of all $f \in {\mathcal S}'$ such that

$$ \Vert f\Vert _{{M}_{p,q}^{s}} = \Big ( \int _{{\mathbb {R}}^n} \Big ( \int _{{\mathbb {R}}^n} |V_\phi f(x,\xi ) \, \langle \xi \rangle ^s|^p dx \Big )^{\frac{q}{p}} d\xi \Big )^{\frac{1}{q}} < \infty \, $$

(with obvious modifications if $p= \infty$ and/or $q=\infty$).

Formally these spaces ${M}_{p,q}^{s}$ depend on the window $\phi$. However, for different windows $\phi _1, \phi _2$ the resulting spaces coincide as sets and the norms are equivalent, see [10, Proposition 11.3.2]. For that reason we do not indicate the window in the notation (we do not distinguish spaces which differ only by an equivalent norm).

Remark 2.3

(i) General references with respect to weighted modulation spaces are Feichtinger [7], Gröchenig [10, Chap. 11], Gol’dman [9], Guo et al. [11], Toft [30–32], Triebel [34] and Wang et al. [38] to mention only a few.

(ii) There is an important special case. In case of $p=q=2$ we obtain $M^s_{2,2} = H^s$ in the sense of equivalent norms, see Feichtinger [7], Gröchenig [10, Proposition 11.3.1]. Here $H^s$ is nothing but the standard Sobolev space built on $L_2$, at least for $s\in \mathbb {N}$. In general $H^s$ is the collection of all $f \in {\mathcal S}'$ such that

$$ \Vert f\Vert _{H^s} := \Big (\int _{{\mathbb {R}}^n} (1+|\xi |^2)^{s}\, |{\mathcal F}f (\xi )|^2 \, d\xi \Big )^{1/2}<\infty . $$

For us of great use will be another alternative approach to the spaces $M^s_{p,q}$. This will be more close to the standard techniques used in connection with Besov spaces. We shall use the so-called frequency-uniform decomposition, see e.g., Wang [37]. Therefore, let $\rho : {\mathbb {R}}^n\mapsto [0,1]$ be a Schwartz function which is compactly supported in the cube

$$ Q_0 := \{ \xi \in {\mathbb {R}}^n: -1\le \xi _i \le 1,\, i=1,\ldots ,n \}\, . $$

Moreover, we assume

$$ \rho (\xi )=1 \qquad \text{ if } \quad |\xi _i|\le \frac{1}{2}\, , \qquad i=1, 2, \ldots , n. $$

With $\rho _k (\xi ) :=\rho (\xi -k)$, $\xi \in {\mathbb {R}}^n$, $k \in {\mathbb {Z}}^n$, it follows

$$ \sum _{k \in {\mathbb {Z}}^n} \rho _k (\xi )\ge 1 \qquad \text{ for } \text{ all }\quad \xi \in {\mathbb {R}}^n\, . $$

Finally, we define

$$ \sigma _k(\xi ) := \rho _k(\xi ) \Big (\sum _{k\in {\mathbb {Z}}^n} \rho _k(\xi )\Big )^{-1}, \qquad \xi \in {\mathbb {R}}^n\, , \quad k\in {\mathbb {Z}}^n\, . $$

The following properties are obvious:

• $0 \le \sigma _k(\xi ) \le 1$ for all $\xi \in {\mathbb {R}}^n$;

• $\mathrm{supp \, }\sigma _k \subset Q_k := \{ \xi \in {\mathbb {R}}^n: -1\le \xi _i -k_i \le 1, \, i=1,\ldots ,n \}$;

• $\displaystyle \sum _{k\in {\mathbb {Z}}^n} \sigma _k(\xi ) \equiv 1$ for all $\xi \in {\mathbb {R}}^n$;

• There exists a constant $C>0$ such that $\sigma _k (\xi ) \ge C$ if $\max _{i=1, \ldots , n}\, |\xi _i-k_i|\le \frac{1}{2}$;

• For all $m \in \mathbb {N}_0$ there exist positive constants $C_m$ such that for $|\alpha |\le m$

  $$ \sup _{k \in {\mathbb {Z}}^n}\, \sup _{\xi \in {\mathbb {R}}^n} \, |D^\alpha \sigma _k(\xi )|\le C_m\, . $$

We shall call the mapping

$$ \Box _k f := {{\mathcal F}}^{-1}\left[ \sigma _k (\xi ) \, {\mathcal F}f(\xi )\right] (\cdot ), \qquad k\in {\mathbb {Z}}^n, \quad f \in {\mathcal S}'\, , $$

frequency-uniform decomposition operator.

As it is well-known there is an equivalent description of the modulation spaces by means of the frequency-uniform decomposition operators.

Proposition 2.4

Let $1\le p,q \le \infty$ and assume $s \in \mathbb {R}$. Then the weighted modulation space $M_{p,q}^{s}$ consists of all tempered distributions $f\in {\mathcal S}'$ such that

$$ \Vert f\Vert ^*_{M_{p,q}^{s}} = \Big ( \sum _{k\in {\mathbb {Z}}^n} \langle k \rangle ^{sq} \Vert \Box _k f\Vert _{L^p}^q \Big )^{\frac{1}{q}} < \infty \, . $$

Furthermore, the norms $\Vert f\Vert _{M_{p,q}^{s}}$ and $\Vert f\Vert ^*_{M_{p,q}^{s}}$ are equivalent.

We refer to Feichtinger [7] or Wang and Hudzik [37]. In what follows, we shall work with both characterizations. In general, we shall use the same notation $\Vert \, \cdot \, \Vert _{M_{p,q}^{s}}$ for both norms.

Lemma 2.5

(i) The modulation space ${M}_{p,q}^{s}$ is a Banach space.

(ii) ${M}_{p,q}^{s}$ is independent of the choice of the window $\rho \in C_0^\infty$ in the sense of equivalent norms.

(iii) $M^s_{p,q}$ is continuously embedded into ${\mathcal S}'$.

(iv) ${M}_{p,q}^{s}$ has the Fatou property, i.e., if $(f_m)_{m=1}^\infty \subset {M}_{p,q}^{s}$ is a sequence such that $f_m \rightharpoonup f$ (weak convergence in ${\mathcal S}'$) and

$$ \sup _{m \in \mathbb {N}}\, \Vert \, f_m \, \Vert _{{M}_{p,q}^{s}} < \infty \, , $$

then $f \in {M}_{p,q}^{s}$ follows and

$$ \Vert \, f \, \Vert _{{M}_{p,q}^{s}} \le \sup _{m\in \mathbb {N}}\, \Vert \, f_m \, \Vert _{{M}_{p,q}^{s}} < \infty \, . $$

Proof

For (i), (ii), (iii) we refer to [10].

We comment on a proof of (iv). Therefore, we follow [8] and work with the norm $\Vert \, \cdot \, \Vert ^*_{M_{p,q}^{s}}$. From the assumption, we obtain that for all $k\in {\mathbb {Z}}^n$ and $x\in {\mathbb {R}}^n$,

$$\begin{aligned} {{\mathcal F}}^{-1}\, [\sigma _k \, {\mathcal F}f_m](x)= (2 \pi )^{-n/2} \, f_m(x-\cdot )(\sigma _k) \rightarrow f(x-\cdot )(\sigma _k)= {{\mathcal F}}^{-1}\, [\sigma _k \, {\mathcal F}f](x) \end{aligned}$$

as $m\rightarrow \infty$. Fatou’s lemma yields

$$\begin{aligned} \sum _{|k| \le N} \Big (\int _{{\mathbb {R}}^n} \,& |{{\mathcal F}}^{-1}\, [\sigma _k \, {\mathcal F}f] (x)|^p dx\, \Big )^{\frac{q}{p}} \\\le & {} \liminf _{m\rightarrow \infty } \sum _{|k|\le N} \, \Big (\int _{{\mathbb {R}}^n} \, |{{\mathcal F}}^{-1}\, [\sigma _k \, {\mathcal F}f_m](x)|^p dx \, \Big )^{\frac{q}{p}} \, . \end{aligned}$$

An obvious monotonicity argument completes the proof.$\blacksquare$

2.2 Embeddings

Obviously the spaces ${M}_{p,q}^{s}$ are monotone in s and q. But they are also monotone with respect to p. To show this we recall Nikol’skij’s inequality, see e.g., Nikol’skij [21, 3.4] or Triebel [33, 1.3.2].

Lemma 2.6

Let $1\le p \le q \le \infty$ and f be an integrable function with $\mathrm{supp \, }{\mathcal F}f \subset B(y,r)$, i.e., the support of the Fourier transform of f is contained in a ball with radius $r>0$ and center in $y \in {\mathbb {R}}^n$. Then it holds

$$ \Vert f\Vert _{L_q} \le C r^{n(\frac{1}{p}-\frac{1}{q})} \Vert f\Vert _{L_p} $$

with a constant $C>0$ independent of r and y.

This implies $\Vert \Box _k f\Vert _{L_q} \le c \, \Vert \Box _k f\Vert _{L_p}$ if $p \le q$ with c independent of k and f which results in the following corollary (by using the norm $\Vert \, \cdot \, \Vert ^*_{M_{p,q}^{s}}$).

Corollary 2.7

Let $s_0 > s$, $p_0 < p$ and $q_0 < q$. Then the following embeddings hold and are continuous:

$$ {M}_{p,q}^{s_0} \hookrightarrow {M}_{p,q}^{s}\, , \qquad M_{p_0,q}^{s} \hookrightarrow {M}_{p,q}^{s} $$

and

$$ {M}_{p,q_0}^{s} \hookrightarrow {M}_{p,q}^{s}\, ; $$

i.e., for all p, q, $1\le p,q\le \infty$, we have

$$ {M}_{1,1}^{s} \hookrightarrow {M}_{p,q}^{s} \hookrightarrow {M}_{\infty ,\infty }^{s}\, . $$

Of some importance are embeddings with respect to different metrics. To find sufficient conditions is not difficult when working with $\Vert \, \cdot \, \Vert ^*_{M_{p,q}^{s}}$. A bit more tricky are the necessity parts. We refer to the recent paper by Guo et al. [11].

Proposition 2.8

Let $s_0, s_1 \in \mathbb {R}$ and $1 \le p_0, p_1 \le \infty$. Then

$$ {M}_{p_0,q_0}^{s_0} \hookrightarrow {M}_{p_1,q_1}^{s_1} $$

holds if and only if either

• $p_0 \le p_1$ and $s_0 - s_1 > n \Big (\frac{1}{q_1} - \frac{1}{q_0}\Big )$

• or $p_0 \le p_1$, $s_0 = s_1$ and $q_0 = q_1$.

Remark 2.9

Embeddings of modulation spaces are treated at various places, we refer to Feichtinger [7], Wang and Hudzik [37], Cordero and Nicola [6], Iwabuchi [13] and Guo et al. [11].

The weighted modulation spaces $M^s_{p,q}$ cannot distinguish between boundedness and continuity (as Besov spaces). Let $C_{ub}$ denote the class of all uniformly continuous and bounded functions $f:~{\mathbb {R}}^n\rightarrow \mathbb {C}$ equipped with the supremum norm. If $f \in M^{s}_{p,q}$ is a regular distribution it is determined (as a function) almost everywhere. We shall say that f is a continuous function if there is one continuous function g which equals f almost everywhere.

Corollary 2.10

Let $s \in \mathbb {R}$ and $1 \le p,q\le \infty$. Then the following assertions are equivalent:

• ${M}_{p,q}^{s} \hookrightarrow L_\infty$;

• ${M}_{p,q}^{s} \hookrightarrow C_{ub}$;

• ${M}_{p,q}^{s} \hookrightarrow M^0_{\infty ,1}$;

• either $s\ge 0$ and $q=1$ or $s>n/q'$.

Proof

We shall work with $\Vert \, \cdot \, \Vert ^*_{M_{p,q}^{s}}$.

Step 1. Sufficiency. By Proposition 2.8 it will be enough to show $M^0_{\infty ,1} \hookrightarrow C_{ub}$. From the definition of $M^0_{\infty ,1}$ it follows that

$$ \sum _{k \in {\mathbb {Z}}^n} \Box _k f (x) $$

is pointwise convergent (for all $x\in {\mathbb {R}}^n$). Furthermore, since $\Box _k f \in C^\infty$, there is a continuous representative in the equivalence class f, given by $\sum _{k \in {\mathbb {Z}}^n} \Box _k f (x)$. In what follows, we shall work with this representative. Boundedness of $f \in M^0_{\infty ,1}$ is obvious, we have

$$\begin{aligned} |f(x)| = |\sum _{k\in {\mathbb {Z}}^n} \Box _k f(x)| \le \Vert \, f \,\Vert _{M^0_{\infty ,1}} \, . \end{aligned}$$

It remains to prove uniform continuity. For fixed $\varepsilon >0$ we choose N such that

$$ \sum _{|k|>N} \Vert \, \Box _k f\, \Vert _{L_\infty } < \varepsilon /2\, . $$

In case $|k|\le N$ we observe that

$$ |\Box _k f(x)- \Box _k f(y)| \le \Vert \, \nabla (\Box _k f) \, \Vert _{L_\infty } |x-y|\, . $$

It follows from [33, Theorem 1.3.1] that

$$ \Vert \, \nabla (\Box _k f) \, \Vert _{L_\infty }\le c_1 \, \Vert \, (M \Box _k f) \, \Vert _{L_\infty } $$

with a constant $c_1$ independent of f and k. Here M denotes the Hardy–Littlewood maximal function. In the quoted reference, the assumption $\Box _k f \in {\mathcal S}$ is used. A closer look at the proof shows that $\Box _k f \in L_1^{\ell oc}$ satisfying

$$ \int _{Q_k} |\Box _k f(x)|\, dx \le c_2\, (1+|k|)^N\, , \qquad k \in {\mathbb {Z}}^n\, , $$

for some $N \in \mathbb {N}$ is sufficient. Since $\Box _k f \in L_\infty$ this is obvious. Consequently we obtain

$$\begin{aligned}|\Box _k f(x)- \Box _k f(y)|\le & {} c_1 \, \Vert \, (M \Box _k f) \, \Vert _{L_\infty }\, |x-y| \le c_1\, \Vert \, \Box _k f\, \Vert _{L_\infty } \, |x-y| \\\le & {} c_2\, \Vert \, f\, \Vert _{L_\infty } \, |x-y|\, , \end{aligned}$$

where in the last step we used the standard convolution inequality $\Vert \, g * h \, \Vert _{L_\infty } \le \, \Vert \, g\, \Vert _{L_1} \Vert \, f\, \Vert _{L_\infty }$. This implies uniform continuity of $\Box _k f$ and therefore of $\sum _{|k|\le N} \Box _k f$. In particular, we find

$$\begin{aligned}|f(x)-f(y)|= & {} \Big | \sum _{k\in {\mathbb {Z}}^n} (\Box _k f (x) - \Box _k f (y)) \Big | \\\le & {} \sum _{|k| > N} (|\Box _k f (x)| + | \Box _k f (y)|) + c_2\, \Vert \, f \, \Vert _{L_\infty } \, |x-y| \, \sum _{|k| \le N} 1 \\\le & {} \varepsilon + c_2\, \Vert \, f \, \Vert _{L_\infty } \, |x-y| \, (2N+1)^n \, . \end{aligned}$$

Choosing $\delta = (c_2 \, \Vert \, f \, \Vert _{L_\infty }\, (2N+ 1)^{n})^{-1} \, \varepsilon$ we arrive at

$$ |f(x)-f(y)|< 2 \, \varepsilon \qquad \text{ if }\qquad |x-y|< \delta \, . $$

Step 2. Necessity. Let $\psi \in {\mathcal S}$ be a real-valued function such that $\psi (0)=1$ and

$$ \mathrm{supp \, }{\mathcal F}\psi \subset \{\xi : \max _{j=1, \ldots , n}\, |\xi _j|< \varepsilon \}\, \qquad \text{ with }\quad \varepsilon < 1/2. $$

We define f by

$$ {\mathcal F}f (\xi ) := \sum _{k \in {\mathbb {Z}}^n} a_k \, {\mathcal F}\psi (\xi -k)\, . $$

Clearly,

$$ \Box _k f (x) = a_k \, e^{ikx}\, \psi (x)\, , \qquad k \in {\mathbb {Z}}^n\, . $$

Substep 2.1. Let $s=0$ and $1 \le p \le \infty$. The above arguments imply $f \in M^0_{p,q}$ if and only if $(a_k)_k \in \ell _q$. On the other hand,

$$\begin{aligned} f (x) = \psi (x)\, \sum _{k \in {\mathbb {Z}}^n} a_k \, e^{ikx} \end{aligned}$$

(2.1)

which implies that f is unbounded in 0 if $\sum _{k\in {\mathbb {Z}}^n} a_k = \infty$. Choosing

$$a_k := \left\{ \begin{array}{lll} (k_1 \, \log (2+ k_1))^{-1}\, &{}\qquad &{} \text{ if }\quad k_1 \in \mathbb {N}\, , \quad k=(k_1, 0, \ldots \, 0)\,; \\ 0 &{}&{} \text{ otherwise }; \end{array}\right. $$

then $f \in M^0_{p,q} \setminus L_\infty$, $q >1$, follows.

Substep 2.2. Let $1 \le p \le \infty$ and $q=\infty$. Then we choose $a_k := \langle k\rangle ^{-n}$. It follows $f \in M^n_{p,\infty }$ but $f(0)=+\infty$.

Substep 2.3. Let $1\le p \le \infty$, $1<q<\infty$ and $s=n/q'$. Then, with $\delta >0$, we choose

$$a_k := \left\{ \begin{array}{lll} \langle k \rangle ^{-n} \,\left( \log \langle k \rangle \right) ^{-(1+\delta )/q} \, &{}\qquad &{} \text{ if }\quad |k| >0\,; \\ 0 &{}&{} \text{ otherwise }. \end{array}\right. $$

It follows

$$\begin{aligned} \Vert \, f \, \Vert _{M^{n/q'}_{p,q}}= & {} \Vert \psi \Vert _{L_p}\, \left( \sum _{|k|>0} \langle k \rangle ^{-nq+ nq/q'} \, (\log \langle k \rangle )^{-(1+\delta )}\right) ^{\frac{1}{q}} \\= & {} \Vert \psi \Vert _{L_p}\, \left( \sum _{|k|>0} \langle k \rangle ^{-n} \, (\log \langle k \rangle )^{-(1+\delta )}\right) ^{\frac{1}{q}} <\infty \, . \end{aligned}$$

On the other hand, we have

$$ f(0) = \sum _{|k|>0} \langle k \rangle ^{-n} \, \left( \log \langle k \rangle \right) ^{-(1+\delta )/q}= \infty $$

if $(1+\delta )/q \le 1$. Hence, for choosing $\delta = q-1$ the claim follows.$\blacksquare$

Remark 2.11

Sufficient conditions for embeddings of modulation spaces into spaces of continuous functions can be found at several places, in particular in Feichtinger’s original paper [7]. We did not find references for the necessity.

3 Pointwise Multiplication in Modulation Spaces

We are interested in embeddings of the type

$$\begin{aligned} M^{s_1}_{p,q} \, \cdot \, M^{s_2}_{p,q} \hookrightarrow M^{s_0}_{p,q}\, , \end{aligned}$$

where $s_1,s_2, p$ and q are given and we are asking for an optimal $s_0$. These results will be applied in connection with our investigations on the regularity of compositions of functions in Sect. 4. However, several times we shall deal with the slightly more general problem

$$ M^{s_1}_{p_1,q} \, \cdot \, M^{s_2}_{p_2,q} \hookrightarrow M^{s_0}_{p,q}\, , \qquad \frac{1}{p} = \frac{1}{p_1} + \frac{1}{p_2}\, . $$

In view of Corollary 2.7 this always yields

$$ M^{s_1}_{p_1,q} \, \cdot \, M^{s_2}_{p_2,q} \hookrightarrow M^{s_0}_{p,q}\, , \qquad \frac{1}{p} \le \frac{1}{p_1} + \frac{1}{p_2}\, . $$

For convenience of the reader we also recall what is known in the more general situation

$$ M^{s_1}_{p_1,q_1} \, \cdot \, M^{s_2}_{p_2,q_2} \hookrightarrow M^{s_0}_{p,q}\, . $$

At first we shall deal with the algebra property. Afterwards we turn to the existence of the product in more general situations.

3.1 On the Algebra Property

The main aim consists in giving necessary and sufficient conditions for the embedding $M^{s}_{p,q} \, \cdot \, M^{s}_{p,q} \hookrightarrow M^{s}_{p,q}$. To prepare this we recall a nice identity due to Toft [30], see also Sugimoto et al. [29].

Lemma 3.1

Let $\varphi _1, \varphi _2 \in {\mathcal S}$ be nontrivial. Let $f,g \in L_2^{\ell oc}$ such that there exist $c>0$ and $M>0$ with

$$ \int _{Q_k} |f(x)|^2 + |g(x)|^2 \, dx \le c \, (1+|k|)^{M} \, , \qquad k \in {\mathbb {Z}}^n\, . $$

For all $x, \xi \in {\mathbb {R}}^n$ the following identity takes place

$$\begin{aligned} V_{\varphi _1 \, \cdot \, \varphi _2} (fg) (x,\xi ) = (2\pi )^{-n/2} \int V_{\varphi _1} (f) (x,\xi -\eta ) \, V_{\varphi _2} (g) (x,\eta )\, d\eta \, . \end{aligned}$$

(3.1)

Proof

The main tool will be the Plancherel identity. Observe, that for any fixed $x \in {\mathbb {R}}^n$ the functions $f(t)\,\overline{\varphi _1 (t-x)}$, $\overline{g(t)}\,{\varphi _2 (t-x)}$ belong to $L_2$ and therefore their Fourier transforms as well. For brevity we put

$$I := \int V_{\varphi _1} (f) (x,\xi -\eta ) \, V_{\varphi _2} (g) (x,\eta )\, d\eta \, . $$

Applying the Plancherel identity, we conclude

$$\begin{aligned}I= & {} \int {\mathcal F}( f(t)\,\overline{\varphi _1 (t-x)} \, e^{-i\xi t})(-\eta )\, \overline{{\mathcal F}(\overline{g(t)}\,{\varphi _2 (t-x)})(-\eta )} \, d\eta \\= & {} \, \int f(t)\,\overline{\varphi _1 (t-x)} \, e^{-i\xi t} \, \overline{\overline{g(t)}\,{\varphi _2 (t-x)})} \, dt \\= & {} \int f(t)\,g(t)\, \overline{\varphi _1 (t-x) \, \varphi _2 (t-x)}\, e^{-i\xi t}\, dt \\= & {} (2\pi )^{n/2}\, V_{\varphi _1 \cdot \varphi _2} (fg) (x,\xi ) \, . \end{aligned}$$

The proof is complete.$\blacksquare$

Remark 3.2

It is clear that the assertion does not extend very much. For example, if $f,g \in L_p^{\ell oc}$ for some $p< 2$ then the above claim is not true. We may take

$$ f(x) = g(x)= \psi (x) \, |x|^{-n/2}\, , \qquad x \in {\mathbb {R}}^n\, , $$

where $\psi$ is a smooth and compactly supported cut-off function s.t. $\psi (0)=1$. Then $f \, \cdot \, g$ is not longer a distribution, i.e., the integral

$$ V_{\varphi _1 \,\cdot \, \varphi _2} (fg)(x,\xi ) = (2\pi )^{-\frac{n}{2}} \int _{{\mathbb {R}}^n} f(s)\, g(s)\, \overline{\varphi _1(s-x) \, \varphi _2 (s-x)}\, e^{-i s\cdot \xi } \, ds $$

does not make sense in general.

In [29, 30], the identity (3.1) is applied either in case $f,g \in {\mathcal S}$ or $f,g \in L_\infty$. Here, we shall apply it in the wider context of Lemma 3.1.

Lemma 3.3

Let $1\le p,q\le \infty$ and assume $M^s_{p,q} \hookrightarrow M^0_{\infty ,1}$. Then there exists a constant c such that

$$ \Vert \, f \, \cdot \, g \, \Vert _{M^s_{p,q}} \le c\, \Big (\Vert \, f \Vert _{M^0_{\infty ,1}} \, \Vert \, g \Vert _{M^s_{p,q}} + \Vert \, f \Vert _{M^s_{p,q}} \, \Vert \, g \Vert _{M^0_{\infty ,1}} \Big ) $$

holds for all $f,g \in M^s_{p,q}$.

Proof

The main idea in the proof consists in the fact that the modulation space can be characterized by different window functions. Since $M^0_{\infty ,1} \hookrightarrow L_\infty$ we know that f, g satisfy the conditions in Lemma 3.1. Hence

$$\begin{aligned} \Vert \, f\, \cdot \, g\, \Vert _{M^s_{p,q}}= & {} \Big \{ \int \Big [ \int \Big | V_{\varphi ^2} (f\, \cdot \, g)(x,\xi )\, \Big |^p dx\Big ]^{q/p} \langle \xi \rangle ^{sq} d\xi \Big \}^{1/q} \\\le & {} \Big \{ \int \Big [ \int \Big | \int V_{\varphi } f(x,\xi -\eta ) V_\varphi g (x,\eta ) d\eta \, \Big |^p dx\Big ]^{q/p} \langle \xi \rangle ^{sq} d\xi \Big \}^{1/q}. \nonumber \end{aligned}$$

(3.2)

We split the integration with respect to $\eta$ into two parts

$$\begin{aligned} \Omega _\xi := \{ \eta : \, |\xi - \eta |\ge |\eta |\} \qquad \text{ and }\qquad \Gamma _\xi := \{\eta : \, |\xi - \eta |< |\eta |\}\, , \quad \xi \in {\mathbb {R}}^n\, . \end{aligned}$$

(3.3)

It follows

$$ \Vert \, f\, \cdot \, g\, \Vert _{M^s_{p,q}} \le 2^s \, (A+ B)\, , $$

where

$$ A:= \Big \{ \int \Big [ \int \Big | \int _{\Omega _\xi } V_{\varphi } f(x,\xi -\eta )\, (1+|\xi -\eta |^2)^{s/2} V_\varphi g (x,\eta )\, d\eta \, \Big |^p dx\Big ]^{q/p} \, d\xi \Big \}^{1/q} $$

and

$$ B:= \Big \{ \int \Big [ \int \Big | \int _{\Gamma _\xi } V_{\varphi } f(x,\xi -\eta )\, V_\varphi g (x,\eta )\, (1+|\eta |^2)^{s/2} d\eta \, \Big |^p dx\Big ]^{q/p} \, d\xi \Big \}^{1/q}\, . $$

We continue by applying the generalized Minkowski inequality, see [18, Theorem 2.4]. This yields

$$\begin{aligned}A\le & {} \int \Big \{ \int \Big [ \int | V_{\varphi } f(x,\xi -\eta )\, \langle \xi -\eta \rangle ^{s} \, V_\varphi g (x,\eta )\,|^p \, dx\Big ]^{q/p} \, d\xi \Big \}^{1/q}\, d\eta \\\le & {} \int \Big (\sup _{x \in {\mathbb {R}}^n} \, | V_\varphi g (x,\eta )|\Big )\, \Big \{ \int \Big [ \int | V_{\varphi } f(x,\xi -\eta )\, \langle \xi -\eta \rangle ^{s} \,|^p \, dx\Big ]^{q/p} \, d\xi \Big \}^{1/q}\, d\eta \\= & {} \Vert \, g\, \Vert _{M^0_{\infty ,1}} \, \Vert \, f\, \Vert _{M^s_{p,q}} \, . \end{aligned}$$

Analogously one can prove

$$ B \le \Vert \, f\, \Vert _{M^0_{\infty ,1}} \, \Vert \, g\, \Vert _{M^s_{p,q}} \, . $$

The proof is complete.$\blacksquare$

Remark 3.4

(i) We proved a bit more than stated. In fact, we have shown

$$ \Vert \, f \, \cdot \, g \, \Vert _{M^s_{p,q}} \le 2^{s}\, \Big (\Vert \, f \Vert _{M^0_{\infty ,1}} \, \Vert \, g \Vert _{M^s_{p,q}} + \Vert \, f \Vert _{M^s_{p,q}} \, \Vert \, g \Vert _{M^0_{\infty ,1}} \Big ) $$

But here one has to notice that the norm on the left-hand side is generated by the window $\varphi ^2$, whereas the norms on the right-hand side are generated by the window $\varphi$.

(ii) Lemma 3.3 has been proved by Sugimoto et al. [29]. For partial results with a different proof we refer to Feichtinger [7].

Next, we turn to necessary and sufficient conditions for the algebra property.

Theorem 3.5

Let $1\le p,q\le \infty$ and $s \in \mathbb {R}$. Then $M^s_{p.q}$ is an algebra with respect to pointwise multiplication if and only if either $s \ge 0$ and $q=1$ or $s>n/q'$.

Remark 3.6

(i) By Corollary 2.10 the Theorem 3.5 can be reformulated as

$$ M^s_{p.q}\quad \text{ is } \text{ an } \text{ algebra } \qquad \Longleftrightarrow \qquad M^s_{p.q} \hookrightarrow L_\infty \, . $$

This is in some sense natural because otherwise one could increase local singularities by pointwise multiplication.

(ii) Theorem 3.5 has a partial counterpart for Besov spaces. Here one knows that $B^s_{p,q}$ is an algebra if and only if $B^s_{p,q} \hookrightarrow L_\infty$ and $s>0$. We refer to Peetre [22, Theorem 11, p. 147], Triebel [33, Theorem 2.8.3] (sufficiency) and to [25, Theorem 4.6.4/1] (necessity).

To prepare the proof, we need the following lemma which is of interest for its own.

Lemma 3.7

Let $1\le p,q < \infty$ and $s \in \mathbb {R}$. Let $f \in {\mathcal S}'$ and let there exists a constant $c>0$ such that

$$ \Vert f \, \cdot \, g\Vert _{M^s_{p,q}} \le c \, \Vert g\Vert _{M^s_{p,q}} $$

holds for all $g \in {\mathcal S}$. Then $f\in L_\infty$ follows.

Proof

Let $T_f (g):= f \, \cdot \, g$, $g\in {\mathcal S}$. Let $\mathring{M}^s_{p,q}$ denote the closure of ${\mathcal S}$ in $M^s_{p,q}$. Hence, there is a unique extension of $T_f$ to a continuous operator belonging to ${\mathcal L}(\mathring{M}^s_{p,q}, {M}^s_{p,q})$. Next we employ duality. We fix p, q and s ($1\le p,q\le \infty$, $s\in \mathbb {R}$). Let (g, h) denote the standard dual pairing on ${\mathcal S}' \times {\mathcal S}$. Then

$$ \Vert g \Vert := \sup \Big \{ |( g,h)|~: \quad h \in {\mathcal S}, \, \Vert h\Vert _{M^{-s}_{p',q'}} \le 1 \Big \} $$

is an equivalent norm on $M^s_{p,q}$, see Feichtinger [7] or Toft [30]. In view of this equivalent norm our assumption on $T_f$ implies ${\mathcal L}({M}^{-s}_{p',q'}, {M}^{-s}_{p',q'})$. Next we continue by complex interpolation. Let $0< \Theta < 1$. It is known that

$$ {M}^s_{p,q} = [{M}^{s_1}_{p_1,q_1}, {M}^{s_2}_{p_2,q_2}]_\Theta $$

if $1 \le p_1,q_1<\infty$, $1 \le p_2, q_2\le \infty$, $s_1,s_2 \in \mathbb {R}$ and

$$ s:= (1-\Theta )s_1 + \Theta \, s_2\, , \quad \frac{1}{p} := \frac{1-\Theta }{p_1} + \frac{\Theta }{p_2}\, , \quad \frac{1}{q} := \frac{1-\Theta }{q_1} + \frac{\Theta }{q_2}\, , $$

see Feichtinger [7]. Thanks to the interpolation property of the complex method we conclude

$$ T_f \in {\mathcal L}\Big ( [\mathring{M}^{s}_{p,q}, {M}^{-s}_{p',q'}]_{1/2}, [\mathring{M}^{s}_{p,q}, {M}^{-s}_{p',q'}]_{1/2} \Big )\, . $$

Because of $\mathring{M}^{s}_{p,q} = {M}^{s}_{p,q}$ if $\max (p,q)<\infty$ we find

$$ T_f \in {\mathcal L}\Big ( {M}^{0}_{2,2}, {M}^{0}_{2,2}\Big ) = {\mathcal L}\Big ( L_{2}, L_{2}\Big ) \, . $$

But this implies $f\in L_\infty$.$\blacksquare$

Proof

of Theorem 3.5.

Step 1. Sufficiency is covered by Lemma 3.3.

Step 2. Necessity in case $1\le p,q<\infty$ and $s\in \mathbb {R}$. In view of Lemma 3.7 the embedding $M^s_{p,q}\, \cdot \, M^s_{p,q}\hookrightarrow M^s_{p,q}$ implies $M^s_{p,q} \subset L_\infty$.

Step 3. To treat the remaining cases $\max (p,q) = \infty$ we argue by using explicit counterexamples.

Substep 3.1. Let $1 \le p \le \infty$, $s=0$ and $1 < q \le \infty$. We assume that $M^0_{p ,q}$ is an algebra. This implies the existence of a constant $c>0$ such that

$$\begin{aligned} \Vert \, f\, \cdot \, g \, \Vert _{M^0_{p,q}} \le c \, \Vert \, f\, \Vert _{M^0_{p,q}}\, \Vert \, g \, \Vert _{M^0_{p,q}} \end{aligned}$$

(3.4)

holds for all $f,g \in M^0_{p,q}$. Let

$$ f (x) = \psi (x)\, \sum _{k=1}^\infty a_k \, e^{ikx_1}\, , \qquad x=(x_1, \ldots \, , x_n) \in {\mathbb {R}}^n\, , $$

be as in (2.1). Then, as shown above,

$$ \Vert \, f\, \Vert _{M^0_{p,q}} = \Vert \psi \Vert _{L_p}\, \Vert (a_k)_k\Vert _{\ell _q} $$

follows. Let

$$ f_N (x) := \psi (x)\, \sum _{k=1}^N a_k\, e^{ikx_1}\, , \qquad x=(x_1, \ldots \, , x_n) \in {\mathbb {R}}^n\, , \quad N \in \mathbb {N}\, . $$

Obviously $f_N \in {\mathcal S}$. We assume that

$$ \mathrm{supp \, }{\mathcal F}\psi \subset \{\xi : \max _{j=1, \ldots , n}\, |\xi _j|< \varepsilon \}\, \qquad \text{ with }\quad \varepsilon < 1/4. $$

Then, because of

$$ f_N (x)\, \cdot \, f_N (x) = \psi ^2 (x) \, \sum _{m=2}^{2N} \Big (\sum _{k = 1}^{m-1} a_k \, a_{m-k} \Big )\, e^{imx}\, , $$

we conclude

$$ \Vert f_N \, \cdot \, f_N \Vert _{M^0_{p,q}} = \Vert \psi ^2\Vert _{L_p} \, \Bigg ( \sum _{m=2}^{2N} \Big |\sum _{k=1}^{m-1} a_k \, a_{m-k} \Big |^q\Bigg )^{1/q}\, . $$

Inequality (3.4) implies

$$\begin{aligned}\Bigg ( \sum _{m=2}^{2N} \Big |\sum _{k=1}^{m-1} a_k \, a_{m-k} \Big |^q\Bigg )^{1/q} \le c \, \frac{\Vert \psi \Vert ^2_{L_p}}{\Vert \psi ^2\Vert _{L_p}} \, \Big (\sum _{k=1}^{N} |a_k|^q\Big )^{2/q}\, . \end{aligned}$$

Clearly, in case $q>1$ this is impossible in this generality. Explicit counterexamples are given by

$$ a_k := k^{-1/q} \qquad \text{ if }\quad 1< q < \infty $$

and

$$ a_k =1 \qquad \text{ if }\quad q =\infty \, . $$

In case $1<q< \infty$ (3.4) yields

$$ \Vert f_N \, \cdot \, f_N \Vert _{M^0_{p,q}} \asymp N^{1-1/q}\qquad \text{ and }\qquad \Vert \, f_N \, \Vert _{M^0_{p,q}}^2 \asymp (\log N)^{2/q}\, . $$

For $q= \infty$ we obtain

$$ \Vert f_N \, \cdot \, f_N \Vert _{M^0_{p,\infty }} \asymp N \qquad \text{ and }\qquad \Vert \, f_N \, \Vert _{M^0_{p,\infty }}^2 \asymp 1\, . $$

For $N \rightarrow \infty$ we find a contradiction in both situations.

Substep 3.2. Let $1 \le p \le \infty$, $q=\infty$ and $0 < s \le n$. We argue as in Substep 3.1 and assume $M^s_{p,\infty }$ is an algebra with respect to pointwise multiplication. This leads to the existence of a constant $c>0$ such that

$$\begin{aligned} \Vert \, f\, \cdot \, g \, \Vert _{M^s_{p,\infty }} \le c \, \Vert \, f\, \Vert _{M^s_{p,\infty }}\, \Vert \, g \, \Vert _{M^s_{p,\infty }} \end{aligned}$$

holds for all $f,g \in M^s_{p,\infty }$. We choose

$$\begin{aligned} f(x)=g(x)=f_N(x) := \psi (x)\, \sum _{\Vert k\Vert _\infty \le N} a_k\, e^{ikx}\,\, , \qquad x \in {\mathbb {R}}^n\, , \end{aligned}$$

and obtain

$$\begin{aligned}\Vert \, f_N\, \Vert _{M^s_{p,q}}= & {} \Vert \psi \Vert _{L_p}\, \Big ( \sum _{\Vert k\Vert _\infty \le N} |a_k\, \langle k\rangle ^s | ^q \Big )^{1/q} \, , \\ \Vert f_N \, \cdot \, f_N \Vert _{M^s_{p,q}}= & {} \Vert \psi ^2\Vert _{L_p} \, \Bigg ( \sum _{\Vert m\Vert _\infty \le 2N} \langle m \rangle ^{sq}\, \Big |\sum _{{k:~ \Vert k\Vert _\infty \le N \atop \Vert m -k\Vert _\infty \le N}} a_k \, a_{m-k} \Big |^q\Bigg )^{1/q} \, . \end{aligned}$$

In case $s<n$ we choose $a_k :=1$ for all k and obtain

$$ \Vert f_N \, \cdot \, f_N \Vert _{M^s_{p,\infty }} \asymp N^{n+s} \qquad \text{ and }\qquad \Vert \, f_N \, \Vert _{M^s_{p,\infty }}^2 \asymp N^{2s}\, . $$

This yields a contradiction if $s<n$. For $s=n$ we consider $a_k := \langle k\rangle ^{-n}$ for all k. This yields

$$ \log N\lesssim \, \Vert f_N \, \cdot \, f_N \Vert _{M^n_{p,\infty }} \qquad \text{ and }\qquad \Vert \, f_N \, \Vert _{M^n_{p,\infty }}^2 \asymp 1\, , $$

yielding a contradiction as well.

Substep 3.3. Let $s<0$ and $1\le p,q\le \infty$. We choose $a_k := \langle k\rangle ^{2|s|}$ for all k and obtain

$$ N^{3|s| + n+n/q}\lesssim \, \Vert f_N \, \cdot \, f_N \Vert _{M^s_{p,q}} \qquad \text{ and }\qquad \Vert \, f_N \, \Vert _{M^s_{p,q}}^2 \asymp N^{2|s| + 2n/q}\, . $$

For $N \rightarrow \infty$ this implies $|s|+n \le n/q$. Since $|s|>0$ this is impossible. The proof is complete.$\blacksquare$

Corollary 3.8

Let $1\le p,q\le \infty$ and $s \ge 0$. Then $M^s_{p,q}\cap M^0_{\infty ,1}$ is an algebra with respect to pointwise multiplication and there exist a constant c such that

$$ \Vert \, f \, \cdot \, g \, \Vert _{M^s_{p.q}} \le c\, \Big (\Vert \, f \Vert _{M^0_{\infty ,1}} \, \Vert \, g \Vert _{M^s_{p,q}} + \Vert \, f \Vert _{M^s_{p,q}} \, \Vert \, g \Vert _{M^0_{\infty ,1}} \Big ) $$

holds for all $f,g \in M^s_{p,q}\cap M^0_{\infty ,1}$.

Proof

The same arguments as in Lemma 3.3 apply.$\blacksquare$

Remark 3.9

Corollary 3.8 has a counterpart for Besov spaces. Here, one knows that $B^s_{p,q} \cap L_\infty$ is an algebra if $1 \le p,q\le \infty$ and $s>0$. We refer to Peetre [22, Theorem 11, p. 147] and to [25, Theorem 4.6.4/2].

3.2 More General Products of Functions

Here, we consider the problem

$$ M^{s_1}_{p_1,q_1} \, \cdot \, M^{s_2}_{p_2,q} \hookrightarrow M^{s}_{p,q}\, . $$

As a first result, we mention a generalization of Lemma 3.3.

Lemma 3.10

Let $1\le p_1,p_2,q \le \infty$ and $s \ge 0$. We put $1/p:= (1/p_1) + (1/p_2)$. If $p \in [1,\infty ]$, then there exists a constant c such that

$$ \Vert \, f \, \cdot \, g \, \Vert _{M^s_{p.q}} \le c\, \Big (\Vert \, f \Vert _{M^0_{p_1,1}} \, \Vert \, g \Vert _{M^s_{p_2,q}} + \Vert \, f \Vert _{M^s_{p_1,q}} \, \Vert \, g \Vert _{M^0_{p_2,1}} \Big ) $$

holds for all $f \in M^s_{p_1,q} \cap M^0_{p_1,1}$ and all $g \in M^s_{p_2,q} \cap M^0_{p_2,1}$.

Proof

We argue similar as above but using Hölder’s inequality with respect to p before applying the generalized Minkowski inequality. $\blacksquare$

Remark 3.11

Observe that $M^0_{p_1,1}, M^0_{p_2,1} \hookrightarrow M^0_{\infty ,1} \hookrightarrow L_\infty$.

Lemma 3.12

Let $1\le p_1,p_2,q \le \infty$ and $s \le 0$. We put $1/p:= (1/p_1) + (1/p_2)$. If $p \in [1,\infty ]$, then there exists a constant c such that

$$\begin{aligned} \Vert \, f \, \cdot \, g \, \Vert _{M^s_{p.q}}\le c\, \Vert \, f\, \Vert _{M^{|s|}_{p_1,1}} \, \Vert \, g\, \Vert _{M^s_{p_2, q}} \end{aligned}$$

(3.5)

holds for all $f \in M^{|s|}_{p_1,1}$, $g \in M^s_{p_2,q}$ such that g satisfies $g \in L_2^{\ell oc}$ and

$$\begin{aligned} \int _{Q_k} |g(x)|^2 \, dx \le C \, (1+|k|)^{M} \, , \end{aligned}$$

(3.6)

for some $C>0$ and $M>0$ independent of $k\in {\mathbb {Z}}^n$.

Proof

Point of departure is the formula (3.2). Instead of the splitting in (3.3) we use now the elementary inequality

$$ 1+|\eta |^2 \le 2 \, (1+|\xi |^2)\, (1+|\xi - \eta |^2) $$

which implies

$$ (1+|\xi |^2)^{s/2} \le 2^{|s|/2} \, (1+|\xi - \eta |^2)^{|s|/2}\, (1+|\eta |^2)^{s/2}\, . $$

This leads to the estimate

$$\begin{aligned}& \Vert \, f \, \cdot \, g \, \Vert _{M^s_{p.q}} \\\lesssim & {} \Big \{ \int \Big [ \int \Big | \int V_{\varphi } f(x,\xi -\eta )\, \langle \xi - \eta \rangle ^{|s|} \, V_\varphi g (x,\eta ) \, \langle \eta \rangle ^{s} \, d\eta \, \Big |^p dx\Big ]^{q/p} \, d\xi \Big \}^{1/q} \\= & {} \Big \{ \int \Big [ \int \Big | \int V_{\varphi } f(x,\tau )\, \langle \tau \rangle ^{|s|} \, V_\varphi g (x,\xi -\tau ) \, \langle \xi - \tau \rangle ^s \, d\tau \, \Big |^p dx\Big ]^{q/p} \, d\xi \Big \}^{1/q}\, . \end{aligned}$$

We continue by applying the generalized Minkowski inequality and Hölder’s inequality (with respect to p) and obtain

$$\begin{aligned}& \Vert \, f \, \cdot \, g \, \Vert _{M^s_{p,q}} \\\lesssim & {} \int \Big \{ \int \Big [ \Vert V_{\varphi } f(x,\tau )\, \langle \tau \rangle ^{|s|} \Vert _{L_{p_1}} \, \, \Vert V_\varphi g (x,\xi - \tau )\, \langle \xi - \tau \rangle ^s \Vert _{L_{p_2}} \Big ]^{q} \, d\xi \Big \}^{1/q} d\tau \\\lesssim & {} \int \Vert V_{\varphi } f(x,\tau )\, \langle \tau \rangle ^{|s|} \Vert _{L_{p_1}} d\tau \,\, \Vert g\Vert _{M_{p_2,q}^s} \\\lesssim & {} \Vert \, f\, \Vert _{M^{|s|}_{p_1,1}} \, \Vert \, g\, \Vert _{M^s_{p_2,q}} \, . \end{aligned}$$

$\blacksquare$

Remark 3.13

Observe that $M^{|s|}_{p_1,1} \hookrightarrow M^{|s|}_{\infty ,1} \hookrightarrow L_\infty$. In addition we would like to mention that the constant c in (3.5) does not depend on the constant C in (3.6).

We recall a final result of Cordero and Nicola [6] concentrating on $s=0$. These authors study $M^{0}_{p_1,q_1} \, \cdot \, M^{0}_{p_2,q_2} \hookrightarrow M^{0}_{p,q}$.

Proposition 3.14

Let $1\le p_1,p_2,q_1,q_2 \le \infty$. Then $M^0_{p_1,q_1} \, \cdot \, M^0_{p_2,q_2} \hookrightarrow M^0_{p,q}$ holds if and only if

$$ \frac{1}{p} \le \frac{1}{p_1} + \frac{1}{p_2} \qquad \text{ and }\qquad 1 + \frac{1}{q} \le \frac{1}{q_1} + \frac{1}{q_2}\, . $$

Remark 3.15

(i) Proposition 3.14 shows that in case $s=0$ in Lemma 3.12 we proved an optimal estimate.

(ii) Necessity of the restrictions in Proposition 3.14 is shown by studying products of Gaussian functions. For extensions of Proposition 3.14 to the case of products with more than two factors we refer to Guo et al. [11] and Toft [30].

3.3 Products of a Distribution with a Function

Up to now, we considered only products of either $L_\infty$-functions or $L_2 ^{\ell oc}$-functions with $L_\infty$-functions. But now we turn to the product of a distribution with a function which is not assumed to be $C^\infty$. This requires a definition.

The Definition of the Product in $\varvec{{\mathcal S}}'$

Let $\psi \in {\mathcal S}$ be a function in $C_0^\infty$ such that $\psi (\xi )=1$ in a neighbourhood of the origin. We define

$$ S^j f (x) = {{\mathcal F}}^{-1}[ \psi (2^{-j} \xi ) \, {\mathcal F}f(\xi ) ] ( x ), \quad j = 0, 1, \ldots \, . $$

The Paley-Wiener theorem tells us that $S^j f$ is anentire analytic function of exponential type. Hence, if $f, \, g \in {\mathcal S}'$ the products $S^j f \cdot S^j g$ makes sense for any j. Further,

$$ \lim _{j \rightarrow \infty } {{\mathcal F}}^{-1}[ \psi (2^{-j} \xi ) \, {\mathcal F}f(\xi ) ] ( \cdot ) = f \qquad (\text {convergence in } {\mathcal S}') $$

for any $f \in {\mathcal S}'$.

Definition 3.16

Let $f, \, g \in {{\mathcal S}}'.$ We define

$$\begin{aligned} f\, \cdot \, g = \lim _{j \rightarrow \infty } \, S^j f \cdot S^j g \end{aligned}$$

whenever the limit on the right-hand side exists in ${\mathcal S}'$. We call $f \cdot g$ the product of f and g.

Remark 3.17

In defining the product we followed a usual practice, see e.g., [22], [33, 2.8], [14, 15] and [25, 4.2]. For basic properties of this notion, we refer to [14, 15] and [25, 4.2].

Theorem 3.18

Let $1\le p_1,p_2,q \le \infty$ and $s \le 0$. We put $1/p:= (1/p_1) + (1/p_2)$. If $p \in [1,\infty ]$, then there exists a constant c such that

$$ \Vert \, f \, \cdot \, g \, \Vert _{M^s_{p,q}}\le c\, \Vert \, f\, \Vert _{M^{|s|}_{p_1,1}} \, \Vert \, g\, \Vert _{M^s_{p_2, q}}) $$

holds for all $f \in M^{|s|}_{p_1,1}$ and $g \in M^s_{p_2,q}$.

Proof

We have to show that the limit of $(S^j f \, \cdot \, S^jg)_j$ exists in ${\mathcal S}'$. The remaining assertions, $\lim _{j\rightarrow \infty } S^j f \, \cdot \, S^jg \in M^s_{p,q}$ and the norm estimates will follow by employing the Fatou property, see Lemmas 2.5 and 3.12.

Step 1. Let $1\le q<\infty$. We have

$$ \lim _{j\rightarrow \infty } \Vert S^j f- f\Vert _{M^s_{p,q}}= 0 \qquad \text{ for } \text{ all }\quad f \in M^s_{p,q}\, . $$

In addition it is easily seen that

$$\begin{aligned} \sup _{j\in \mathbb {N}_0} \Vert S^j f\Vert _{M^s_{p,q}}\le \Vert {{\mathcal F}}^{-1}\psi \Vert _{L_1} \, \Vert \, f \, \Vert _{M^s_{p,q}} \end{aligned}$$

(3.7)

holds for all $f \in M^s_{p,q}$. Hence, we conclude by means of Lemma 3.12

$$\begin{aligned}& \Vert S^k f \, S^k g - S^j f \, S^j g \Vert _{M^s_{p,q}} \\\le & {} \Vert ( S^k f - S^j f) \, S^k g \Vert _{M^s_{p,q}} + \Vert S^j f \, (S^k g - S^j g) \Vert _{M^s_{p,q}} \\\le & {} c\, ( \Vert \, S^k g\, \Vert _{M^s_{p_2,q}} \, \Vert \, S^k f - S^j f\, \Vert _{M^{|s|}_{p_1,1}} + \Vert \, S^k g - S^j g\, \Vert _{M^s_{p_2,q}} \, \Vert \, S^j f\, \Vert _{M^{|s|}_{p_1,1}}) \end{aligned}$$

the convergence of $(S^kf \, \cdot S^k g)_k$ in $M^s_{p,q}$ and therefore in ${\mathcal S}'$, see Lemma 2.5.

Step 2. Let $q= \infty$ and suppose $p=1$. Let $\psi , \psi ^* \in C_0^\infty$ be functions such that $\psi (\xi ) = 1$, $|\xi |\le 1$, $\psi (\xi ) =0$ if $|\xi |>3/2$ and $\psi ^* (\xi ) = 1$, $|\xi |\le 6$. Then checking the Fourier support of the product $S^k f \, S^k g$ and using linearity of ${\mathcal F}$ we conclude

$$\begin{aligned}\Big \langle S^k f \, S^k g- & {} S^j f \, S^j g , \varphi \Big \rangle \\= & {} \Big \langle S^k f \, S^k g - S^j f \, S^j g , {{\mathcal F}}^{-1}[(\psi ^* (2^k\xi ) - \psi ^* (2^j \xi )){\mathcal F}\varphi (\xi )](\, \cdot \, ) \Big \rangle \, . \end{aligned}$$

For brevity we put

$$ h_1:= S^k f \, S^k g - S^j f \, S^j g \qquad \text{ and }\qquad h_2:= {{\mathcal F}}^{-1}[(\psi ^* (2^k\xi ) - \psi ^* (2^j \xi )){\mathcal F}\varphi (\xi )](\, \cdot \, )\, . $$

$h_1,h_2$ are smooth functions with compactly supported Fourier transform. Hence,

$$ h_1 = \sum _{k \in I_1} \Box _k h_1 \qquad \text{ and }\qquad h_2 = \sum _{k \in I_2} \Box _k h_2\, , $$

where $I_1,I_2$ are finite subsets of ${\mathbb {Z}}^n$. This allows us to rewrite $\Big \langle S^k f \, S^k g - S^j f \, S^j g ,$ $\varphi \Big \rangle$ as follows

$$\begin{aligned}\Big \langle S^k f \, S^k g - S^j f \, S^j g , \varphi \Big \rangle= & {} \sum _{k \in I_1}\sum _{\ell \in I_2} \int \Box _k h_1(x) \, \Box _\ell h_2 (x)\, dx \\= & {} \sum _{\ell \in I_2}\sum _{k \in I_1: \, Q_k\cap Q_\ell \ne \emptyset } \int \Box _k h_1(x) \, \Box _\ell h_2 (x)\, dx \, . \end{aligned}$$

Application of Hölder’s inequality yields

$$\begin{aligned} \Big | \Big \langle S^k f \, S^k g - S^j f \, S^j g , \varphi \Big \rangle \Big |\le & {} 2^n \sup _{k \in {\mathbb {Z}}^n} \langle k\rangle ^{s} \Vert \, \Box _k h_1\, \Vert _{L_{p_1}} \, \Big ( \sum _{\ell \in {\mathbb {Z}}^n} \langle \ell \rangle ^{-s}\Vert \, \Box _\ell h_2 \, \Vert _{L_{p_2}}\Big ) \nonumber \\\le & {} 2^n \, \Vert \, h_1\, \Vert _{M^s_{p_1,\infty }} \, \Vert \, h_2 \, \Vert _{M^{-s}_{p_2,1}} \, . \end{aligned}$$

(3.8)

By means of Lemma 3.12 and (3.7) we know that

$$\begin{aligned}\Vert \, h_1\, \Vert _{M^s_{p_1,\infty }}= & {} \Vert \, S^k f \, S^k g - S^j f \, S^j g \, \Vert _{M^s_{p_1,\infty }} \\\le & {} c_1\, \sup _{j\in \mathbb {N}_0} \Vert \, S^j g\, \Vert _{M^s_{p_2,q}} \, \Vert \, S^jf\, \Vert _{M^{|s|}_{p_1,1}} \\\le & {} c_2\, \Vert \, g\, \Vert _{M^s_{p_2,q}} \, \Vert \, f\, \Vert _{M^{|s|}_{p_1,1}} \, . \end{aligned}$$

On the other hand, if $j \le k$, a standard Fourier multiplier argument yields

$$\begin{aligned}\Vert \, h_2 \, \Vert _{M^{-s}_{p_2,1}}= & {} \Vert \, {{\mathcal F}}^{-1}[(\psi ^* (2^k\xi ) - \psi ^* (2^j \xi )){\mathcal F}\varphi (\xi )](\, \cdot \, ) \, \Vert _{M^{-s}_{p_2,1}} \\\le & {} C \, \sum _{A\, 2^j \le |\ell | \le B \, 2^k} \langle \ell \rangle ^{-s}\, \Vert \, \Box _\ell \varphi \, \Vert _{L_{p_2}} \end{aligned}$$

for appropriate positive constants A, B, C independent of j, k and $\varphi$. Since $\varphi \in {\mathcal S}\subset M^{-s}_{p_2,1}$ we conclude that the right-hand side tends to 0 if $j \rightarrow \infty$. This finally proves

$$ \Big |\Big \langle S^k f \, S^k g - S^j f \, S^j g , \varphi \Big \rangle \Big | < \varepsilon \qquad \text{ if }\quad j,k \ge j_0 (\varepsilon )\, . $$

Hence $(S^k f \, S^k g)_k$ is weakly convergent in ${\mathcal S}'$. Now, Lemma 3.12 yields the claim also for $q=\infty$.

Step 3. Let $q= \infty$ and suppose $1 < p \le \infty$. We employ (3.8) with $p_1 = \infty$ and $p_2 =1$ and afterwards Proposition 2.8. It follows

$$\begin{aligned}\Big | \Big \langle S^k f \, S^k g - S^j f \, S^j g , \varphi \Big \rangle \Big |\le & {} 2^n \, \Vert \, h_1\, \Vert _{M^s_{\infty ,\infty }} \, \Vert \, h_2 \, \Vert _{M^{-s}_{1,1}} \\\le & {} c_1 \, \Vert \, h_1\, \Vert _{M^s_{p,q}} \, \Vert \, h_2 \, \Vert _{M^{-s}_{1,1}} \, . \end{aligned}$$

Now we can argue as in Step 2.$\blacksquare$

Remark 3.19

For a partial result concerning Theorem 3.18 we refer to

Feichtinger [7].

3.4 One Example

We consider the Dirac $\delta$ distribution. Since

$$ {\mathcal F}\delta (\xi ) = (2\pi )^{-n/2}\, , \qquad \xi \in {\mathbb {R}}^n\, , $$

it is easily seen that $\delta \in M^0_{p,\infty }$ for all p. Also not difficult to see is that $M^0_{1,\infty }$ is the smallest space of type $M^s_{p,q}$ to which $\delta$ belongs to. Theorem 3.18 yields

$$ \Vert \, f \, \cdot \, \delta \, \Vert _{M^0_{p,\infty }}\le c\, \Vert \, \delta \, \Vert _{M^0_{p,\infty }} \, \Vert \, f\, \Vert _{M^0_{\infty ,1}} $$

with some c independent of $f \in M^0_{\infty ,1}$. With other words, we can multiply $\delta$ with a modulation space $M^s_{p,q}$ if this space is embedded into $C_{ub}$, see Corollary 2.10. This looks reasonable.

3.5 The Second Method

Finally, we would like to investigate also the cases $\min (s_1 , s_2) \le n/q'$. For dealing with this special situation we turn to a different method which will allow a better localization in the Fourier image. Therefore we shall work with the frequency-uniform decomposition $(\sigma _k)_k$. Recall that $\mathrm{supp \, }\sigma _k \subset Q_k := \{\xi \in {\mathbb {R}}^n: -1\le \xi _i-k_i\le 1, \, i=1,\ldots ,n\}$. For brevity we put

$$ f_k (x):= {{\mathcal F}}^{-1}[\sigma _k (\xi ) {\mathcal F}f (\xi )](x) \, , \qquad x \in {\mathbb {R}}^n\, , \quad k \in {\mathbb {Z}}^n\, . $$

Then, at least formally, we have the following representation of the product $f\cdot g$ as

$$ f\cdot g = \sum _{k,l\in {\mathbb {Z}}^n} f_k \cdot g_l. $$

In what follows we shall study bounds for related partial sums.

Lemma 3.20

Let $1\le p_1,p_2\le \infty$, $1 < q \le \infty$ and $s_0 , s_1,s_2 \in \mathbb {R}$. Define p by $\frac{1}{p}:= \frac{1}{p_1}+\frac{1}{p_2}$. If $p \in [1,\infty ]$, $0 \le s_0 \le \min (s_1, s_2)$ and $s_2 + s_1 - s_0 > n/q'$, then there exists a constant c such that

$$ \Vert \sum _{k,l\in {\mathbb {Z}}^n} f_k \cdot g_l\Vert _{{M}_{p,q}^{s_0}} \le c \, \Vert f \Vert _{{M}_{p_1,q}^{s_1}}\, \Vert g\Vert _{{M}_{p_2,q}^{s_2}} $$

holds for all $f, g\in {\mathcal S}'$ such that $\mathrm{supp \, }{\mathcal F}f$ and $\mathrm{supp \, }{\mathcal F}g$ are compact. The constant c is independent from $\mathrm{supp \, }{\mathcal F}f$ and $\mathrm{supp \, }{\mathcal F}g$, respectively.

Proof

Later on, we shall use the same strategy of proof as below in slightly different situations. For this reason and later use we shall take care of all constants showing up in our estimates below.

Step 1. Preparations. Determining the Fourier support of $f_j \cdot g_l$ we see that

$$\begin{aligned} \mathrm{supp \, }{\mathcal F}(f_j\cdot g_l)= & {} \mathrm{supp \, }({\mathcal F}f_j *{\mathcal F}g_l) \\\subset & {} \{\xi \in {\mathbb {R}}^n: j_i+l_i-2 \le \xi _i \le j_i+l_i+2, \, i=1,\ldots ,n\}. \end{aligned}$$

Hence, the term ${{\mathcal F}}^{-1}(\sigma _k {\mathcal F}(f_j \cdot g_l))$ vanishes if $\Vert k - (j+l)\Vert _\infty \ge 3$. In addition, since $\mathrm{supp \, }{\mathcal F}f$ and $\mathrm{supp \, }{\mathcal F}g$ are compact, the sum $\sum _{j,l \in {\mathbb {Z}}^n} f_j \cdot g_l$ is a finite sum. We obtain

$$\begin{aligned} \sigma _k {\mathcal F}(f\cdot g)&= \sigma _k {\mathcal F}\Big (\sum _{j,l\in {\mathbb {Z}}^n} f_j\cdot g_l \Big ) = \sigma _k {\mathcal F}\Big (\sum _{\begin{array}{c} j,l \in {\mathbb {Z}}^n, \\ k_i-3<j_i+l_i< k_i +3, \\ i=1,\ldots , n \end{array}} f_j\cdot g_l \Big ) \\&\mathop {=}\limits ^{[r=j+l]} \sum _{\begin{array}{c} r \in {\mathbb {Z}}^n, \\ k_i-3<r_i < k_i +3, \\ i=1,\ldots , n \end{array}} \sum _{l\in {\mathbb {Z}}^n}\, \sigma _k {\mathcal F}\big ( f_{r-l}\cdot g_l \big ) \, . \end{aligned}$$

Consequently

$$\begin{aligned} \left\| {{\mathcal F}}^{-1}\big (\sigma _k {\mathcal F}(f\cdot g)\big ) \right\| _{L_p}&\le \sum _{\begin{array}{c} r \in {\mathbb {Z}}^n, \\ k_i-3<r_i< k_i +3, \\ i=1,\ldots , n \end{array}} \sum _{l\in {\mathbb {Z}}^n} \Vert {{\mathcal F}}^{-1}\big ( \sigma _k {\mathcal F}( f_{r-l}\cdot g_l) \big )\Vert _{L_p} \\&\mathop {=}\limits ^{[t=r-k]} \sum _{\begin{array}{c} t \in {\mathbb {Z}}^n, \\ -3<t_i < 3, \\ i=1,\ldots , n \end{array}} \sum _{l\in {\mathbb {Z}}^n} \Vert {{\mathcal F}}^{-1}\big ( \sigma _k {\mathcal F}( f_{t-(l-k)}\cdot g_l) \big )\Vert _{L_p}. \end{aligned}$$

Step 2. Norm estimates. These preparations yield the following estimates

$$\begin{aligned} \Big ( \sum _{k\in {\mathbb {Z}}^n}& \langle k\rangle ^{s_0 q} \Vert {{\mathcal F}}^{-1}\big (\sigma _k {\mathcal F}(f\cdot g) \big )\Vert _{L_p}^q \Big )^{\frac{1}{q}} \\\le & {} \Bigg ( \sum _{k\in {\mathbb {Z}}^n} \langle k\rangle ^{s_0q} \Bigg [ \sum _{\begin{array}{c} t \in {\mathbb {Z}}^n, \\ -3<t_i< 3, \\ i=1,\ldots , n \end{array}} \sum _{l\in {\mathbb {Z}}^n} \Vert {{\mathcal F}}^{-1}\big (\sigma _k {\mathcal F}( f_{t-(l-k)} \cdot g_l) \big )\Vert _{L_p} \Bigg ]^q \Bigg )^{\frac{1}{q}} \\\le & {} \sum _{\begin{array}{c} t \in {\mathbb {Z}}^n, \\ -3<t_i < 3, \\ i=1,\ldots , n \end{array}} \left( \sum _{k\in {\mathbb {Z}}^n} \langle k\rangle ^{s_0q} \left[ \sum _{l\in {\mathbb {Z}}^n} \Vert {{\mathcal F}}^{-1}\big (\sigma _k {\mathcal F}( f_{t-(l-k)} \cdot g_l) \big )\Vert _{L_p} \right] ^q \right) ^{\frac{1}{q}} \, . \end{aligned}$$

Observe

$$\begin{aligned} \Vert {{\mathcal F}}^{-1}\big (\sigma _k {\mathcal F}( f_{t-(l-k)} \cdot g_l) \big )\Vert _{L^p}= & {} (2\pi )^{-n/2} \Vert ({{\mathcal F}}^{-1}\sigma _k) * (f_{t-(l-k)} \cdot g_l) \, \Vert _{L_p} \\\le & {} (2\pi )^{-n/2} \Vert {{\mathcal F}}^{-1}\sigma _k \, \Vert _{L^1} \, \Vert \, f_{t-(l-k)} \cdot g_l \, \Vert _{L_p} \\= & {} (2\pi )^{-n/2} \Vert {{\mathcal F}}^{-1}\sigma _0 \, \Vert _{L^1} \, \Vert \, f_{t-(l-k)} \cdot g_l \, \Vert _{L_p} \, , \end{aligned}$$

where we used Young’s inequality. We put $c_1:= (2\pi )^{-n/2} \Vert {{\mathcal F}}^{-1}\sigma _0 \, \Vert _{L_1}$. This implies

$$\begin{aligned} \Big ( \sum _{k\in {\mathbb {Z}}^n} \langle k\rangle ^{s_0 q}& \Vert {{\mathcal F}}^{-1}\big (\sigma _k {\mathcal F}(f\cdot g) \big )\Vert _{L_p}^q \Big )^{\frac{1}{q}} \\\le & {} c_1\, \sum _{\begin{array}{c} t \in {\mathbb {Z}}^n, \\ -3<t_i < 3, \\ i=1,\ldots , n \end{array}} \left( \sum _{k\in {\mathbb {Z}}^n} \langle k\rangle ^{s_0q} \left[ \sum _{l\in {\mathbb {Z}}^n} \Vert f_{t-(l-k)}\cdot g_l\Vert _{L_p} \right] ^q \right) ^{\frac{1}{q}}\, . \end{aligned}$$

We continue by using Hölder’s inequality to get

$$\begin{aligned} \Big ( \sum _{k\in {\mathbb {Z}}^n} \langle k\rangle ^{s_0q}& \Vert {{\mathcal F}}^{-1}\big (\sigma _k {\mathcal F}(f\cdot g) \big )\Vert _{L_p}^q \Big )^{\frac{1}{q}} \\\le & {} c_2 \, \max _{\begin{array}{c} t \in {\mathbb {Z}}^n, \\ -3<t_i < 3, \\ i=1,\ldots , n \end{array}} \left( \sum _{k\in {\mathbb {Z}}^n} \langle k\rangle ^{s_0q} \left[ \sum _{l\in {\mathbb {Z}}^n} \Vert f_{t-(l-k)}\Vert _{L_{p_1}} \Vert g_l\Vert _{L_{p_2}} \right] ^q \right) ^{\frac{1}{q}} \end{aligned}$$

with $c_2:= c_1 \, 5^n$. Since $s_0\ge 0$ elementary calculations yield

$$\begin{aligned} \langle k\rangle ^{s_0} \Big [ \sum _{l\in {\mathbb {Z}}^n}& \Vert f_{t-(l-k)}\Vert _{L_{p_1}}\, \Vert g_l\Vert _{L_{p_2}} \Big ] \\\le & {} 2^{s_0}\, \sum _{\begin{array}{c} l\in {\mathbb {Z}}^n, \\ |l|\le |l-k| \end{array}} \langle k-l\rangle ^{s_0} \Vert f_{t-(l-k)}\Vert _{L_{p_1}} \Vert g_l\Vert _{L_{p_2}} \\&\qquad + 2^{s_0} \sum _{\begin{array}{c} l\in {\mathbb {Z}}^n, \\ |l-k|\le |l| \end{array}} \Vert f_{t-(l-k)}\Vert _{L_{p_1}} \langle l\rangle ^{s_0} \Vert g_l\Vert _{L_{p_2}} \,. \end{aligned}$$

Both parts of this right-hand side will be estimated separately. We put

$$\begin{aligned} S_{1,t,k}:= & {} \sum _{\begin{array}{c} l\in {\mathbb {Z}}^n, \\ |l|\le |l-k| \end{array}} \langle k-l\rangle ^{s_1} \Vert f_{t-(l-k)}\Vert _{L_{p_1}} \langle l\rangle ^{s_2} \Vert g_l\Vert _{L_{p_2}} \, \langle k-l\rangle ^{s_0 - s_1} \, \langle l\rangle ^{-s_2} \, ; \\ S_{2,t,k}:= & {} \sum _{\begin{array}{c} l\in {\mathbb {Z}}^n, \\ |l-k|\le |l| \end{array}} \langle k-l\rangle ^{s_1} \, \Vert f_{t-(l-k)}\Vert _{L_{p_1}} \langle l\rangle ^{s_2} \, \Vert g_l\Vert _{L_{p_2}} \, \langle k-l\rangle ^{- s_1} \, \langle l\rangle ^{s_0-s_2} \, . \end{aligned}$$

With $\frac{1}{q}+\frac{1}{q'}=1$ we find

$$\begin{aligned} S_{1,t,k}&\mathop {=}\limits ^{[j=l-k]} \sum _{\begin{array}{c} j\in {\mathbb {Z}}^n, \\ |j+k|\le |j| \end{array}} \langle j\rangle ^{s_0} \Vert f_{t-j}\Vert _{L_{p_1}} \Vert g_{j+k}\Vert _{L_{p_2}} \\&\le \Big ( \sum _{\begin{array}{c} j\in {\mathbb {Z}}^n, \\ |j+k|\le |j| \end{array}} ( \langle j\rangle ^{s_1} \Vert f_{t-j}\Vert _{L_{p_1}} \langle j+k\rangle ^{s_2} \Vert g_{j+k}\Vert _{L_{p_2}} )^{q}\Big )^{1/q} \\&\qquad \qquad \qquad \times \quad \Big ( \sum _{\begin{array}{c} j\in {\mathbb {Z}}^n, \\ |j+k|\le |j| \end{array}} (\langle j \rangle ^{s_0 - s_1} \, \langle j+k \rangle ^{-s_2})^{q'} \Big )^{\frac{1}{q'}} \, . \end{aligned}$$

Substep 2.1. Our assumptions $s_0 \le s_1$, $s_2 \ge 0$ and $s_1 + s_2 -s_0 >n/q'$ imply

$$\begin{aligned} \Big ( \sum _{\begin{array}{c} j\in {\mathbb {Z}}^n, \\ |j+k|\le |j| \end{array}} \Big | \langle j \rangle ^{s_0 - s_1} \, \langle j+k \rangle ^{-s_2} \Big |^{q'} \Big )^{\frac{1}{q'}} \le \Big ( \sum _{m \in {\mathbb {Z}}^n} \langle m \rangle ^{(s_0 -s_1 -s_2) q'} \Big )^{\frac{1}{q'}} =: c_3 <\infty \, . \end{aligned}$$

Inserting this in our previous estimate we obtain

$$\begin{aligned} \Big (\sum _{k \in {\mathbb {Z}}^n} S_{1,t,k}^q\Big )^{1/q}\le & {} c_3\, \Bigg ( \sum _{k\in {\mathbb {Z}}^n} \sum _{\begin{array}{c} j\in {\mathbb {Z}}^n, \\ |j+k|\le |j| \end{array}} \langle j \rangle ^{s_1q} \Vert f_{t-j}\Vert _{L^{p_1}}^q \langle j+k \rangle ^{s_2q} \Vert g_{j+k}\Vert _{L^{p_2}}^q \Bigg )^{1/q} \\\le & {} c_3 \, \Bigg ( \sum _{j\in {\mathbb {Z}}^n} \langle j \rangle ^{s_1q} \Vert f_{t-j}\Vert _{L^{p_1}}^q \sum _{k\in {\mathbb {Z}}^n} \langle j+k \rangle ^{s_2q} \Vert g_{j+k}\Vert _{L^{p_2}}^q \Bigg )^{\frac{1}{q}} \, . \end{aligned}$$

Because of $1+|j|^{2} \le 1+ 8n + |j-t|^{2}$ we know

$$\begin{aligned} \max _{\begin{array}{c} t \in {\mathbb {Z}}^n, \\ -3<t_i< 3, \\ i=1,\ldots , n \end{array}} \sup _{j\in {\mathbb {Z}}^n} \frac{\langle j \rangle ^{s_1}}{\langle j-t \rangle ^{s_1}} \le (1+ 8n)^{s_1/2} =: c_4 < \infty \, . \end{aligned}$$

This implies

$$\begin{aligned} \Big (\sum _{k \in {\mathbb {Z}}^n} S_{1,t,k}^q\Big )^{1/q} \le c_3 \, c_4 \, \Vert g\Vert _{{M}^{s_2}_{p_2,q}} \Vert f\Vert _{{M}^{s_1}_{p_1,q}}, \end{aligned}$$

(3.9)

where $c_3$, $c_4$ are independent of f, g and t.

Substep 2.2. Because of $0 \le s_0 \le s_1$, $s_0 \le s_2$ and $s_1 + s_2 -s_0 >n/q'$ we conclude

$$\begin{aligned} \Big ( \sum _{\begin{array}{c} l\in {\mathbb {Z}}^n, \\ |l-k|\le |l| \end{array}} \Big | \langle k-l \rangle ^{- s_1} \, \langle l \rangle ^{s_0-s_2} \Big |^{q'} \Big )^{\frac{1}{q'}} \le \Big ( \sum _{m \in {\mathbb {Z}}^n} \langle m \rangle ^{(s_0 -s_1 -s_2) q'} \Big )^{\frac{1}{q'}} =: c_5 <\infty \, . \end{aligned}$$

This leads to the estimate

$$\begin{aligned} \Big (\sum _{k \in {\mathbb {Z}}^n} S_{2,t,k}^q\Big )^{1/q} \le c_5 c_6\, \Vert g\Vert _{{M}^{s_2}_{p_2,q}} \Vert f\Vert _{{M}^{s_1}_{p_1,q}} \end{aligned}$$

(3.10)

with some constants $c_6$ independent from f and g. Combining the inequalities (3.9) and (3.10) we have proved the claim.$\blacksquare$

Remark 3.21

Some basic ideas of the above proof are taken over from [5], see also [23].

Of course the above method of proof works as well for $q=1$. But all spaces $M^s_{p,1}$, $s\ge 0$, are algebras.

Theorem 3.22

Let $1\le p, p_1,p_2\le \infty$ and $s_0, s_1, s_2 \in \mathbb {R}$. Let $1/p \le (1/p_1) + (1/p_2)$, $1 < q \le \infty$, $0 \le s_0 \le \min (s_1, s_2)$ and $s_1 + s_2-s_0>n/q'$. There exists a constant c such that

$$ \Vert \, f \cdot g \, \Vert _{{M}_{p,q}^{s_0}} \le c \, \Vert f \Vert _{{M}_{p_1,q}^{s_1}}\, \Vert g\Vert _{{M}_{p_2,q}^{s_2}} $$

holds for all $f\in {M}_{p_1,q}^{s_1}$ and all $g \in {M}_{p_2,q}^{s_2}$.

Proof

We only comment on the case $1/p = (1/p_1) + (1/p_2)$, see Corollary 2.7. It will be enough to prove the weak convergence of $(S^kf \cdot S^k g)_k$ in ${\mathcal S}'$. The claimed estimate will then follow from Lemma 3.20. We employ the method and the notation used in proof of Theorem 3.18 (Steps 2 and 3). There we have proved

$$ \Big | \Big \langle S^k f \, S^k g - S^j f \, S^j g , \varphi \Big \rangle \Big | \le c_1 \, \Vert \, h_1\, \Vert _{M^{s_0}_{p,q}} \, \Vert \, h_2 \, \Vert _{M^{-s_0}_{1,1}} $$

with $c_1$ independent of f, g, k and j. By means of Lemma 3.20 we know the uniform boundedness of $\Vert \, h_1\, \Vert _{M^{s_0}_{p,q}}$ in k and j. The estimate of $\Vert \, h_2 \, \Vert _{M^{-s_0}_{1,1}}$ can be done as above. It follows

$$ \Vert \, h_2 \, \Vert _{M^{-s_0}_{1,1}} \le \varepsilon $$

if $j,k\ge j_0(\varepsilon )$. This guarantees the weak convergence of $(S^kf \cdot S^k g)_k$ in ${\mathcal S}'$. $\blacksquare$

Our sufficient conditions are not far away from necessary conditions.

Lemma 3.23

Let $1\le p_1,p_2, p, q\le \infty$ and $s_0, s_1, s_2 \in \mathbb {R}$. Suppose that there exists a constant c such that

$$\begin{aligned} \Vert \, f \cdot g \, \Vert _{{M}_{p,q}^{s_0}} \le c \, \Vert f \Vert _{{M}_{p_1,q}^{s_1}}\, \Vert g\Vert _{{M}_{p_2,q}^{s_2}} \end{aligned}$$

(3.11)

holds for all $f, g \in {\mathcal S}$.

(i) It follows $s_0 \le \min (s_1,s_2)$, $s_1 + s_ 2 \ge 0$ and $s_1 + s_2 -s_0 \ge n/q'$.

(ii) If $1 \le p_2 =p < \infty$ and $1\le q < \infty$, then either $q=1$ and $s_1 \ge 0$ or $1< q < \infty$ and $s_1 >n/q'$.

Proof

Part (ii) is an immediate consequence of Lemma 3.7. Concerning the proof of (i) we shall work with the same test functions as used in Step 2 of the proof of Corollary 2.10, see (2.1).

Step 1. We choose $a_k := \delta _{k,\ell }$, $k \in {\mathbb {Z}}^n$, for a fixed given $\ell \in {\mathbb {Z}}^n$ and put $b_k:= \delta _{k,0}$, $k \in {\mathbb {Z}}^n$. Then we define

$$ f(x) := \psi (x) \, e^{i \ell x} \qquad \text{ and } \qquad g(x):= \psi (x)\, . $$

We obtain

$$ \Vert f\Vert _{M^{s_1}_{p_1,q}} \, \cdot \, \Vert g\Vert _{M^{s_2}_{p_2,q}} = \Vert \psi \Vert _{L_{p_1}}\, \Vert \psi \Vert _{L_{p_2}} \, \langle \ell \rangle ^{s_1} $$

as well as

$$ \Vert f\, \cdot \, g \Vert _{M^{s_0}_{p,q}} = \Vert \psi ^2\Vert _{L_{p}} \, \langle \ell \rangle ^{s_0} \, . $$

Hence, (3.11) implies $s_0 \le s_1$. Interchanging the roles of f and g leads to the conclusion $s_0 \le s_2$.

Step 2. Let $\ell \in {\mathbb {Z}}^n$ be fixed. We choose $a_k := \delta _{k,\ell }$, $k \in {\mathbb {Z}}^n$, and $b_k:= \delta _{k,-\ell }$, $k \in {\mathbb {Z}}^n$. Then we define

$$ f(x) := \psi (x) \, e^{i \ell x} \qquad \text{ and } \qquad g(x):= \psi (x)\, e^{-i \ell x} \, . $$

It follows

$$ \Vert f\Vert _{M^{s_1}_{p_1,q}} \, \cdot \, \Vert g\Vert _{M^{s_2}_{p_2,q}} = \Vert \psi \Vert _{L_{p_1}}\, \Vert \psi \Vert _{L_{p_2}} \, \langle \ell \rangle ^{s_1+s_2} $$

as well as

$$ \Vert f\, \cdot \, g \Vert _{M^{s_0}_{p,q}} = \Vert \psi ^2\Vert _{L_{p}} \, . $$

Hence, (3.11) implies $s_1+ s_2\ge 0$.

Step 3. Let $\varepsilon _1, \varepsilon _2 \ge 0$. These two numbers will be chosen such that

$$ \min (s_1 + \varepsilon _1 + n/q, s_2 + \varepsilon _2+ n/q)>0 \qquad \text{ and }\qquad s_0 + \varepsilon _2 + \varepsilon _1 + n>0\, . $$

We choose $a_k := \langle k \rangle ^{\varepsilon _1}$, $k \in {\mathbb {Z}}^n$, and $b_k:= \langle k \rangle ^{\varepsilon _2}$, $k \in {\mathbb {Z}}^n$. Then we define

$$ f(x) := \psi (x) \, \sum _{\Vert k\Vert _\infty \le N} a_k \, e^{i k x} \qquad \text{ and } \qquad g(x):= \psi (x) \, \sum _{\Vert k\Vert _\infty \le N} b_k \, e^{i k x} \, . $$

By means of the same arguments as used in Substep 3.1 of the proof of Theorem 3.5, we conclude

$$ \Vert f\Vert _{M^{s_1}_{p_1,q}} \asymp N^{s_1 + \varepsilon _1 + n/q} \qquad \text{ and }\qquad \, \Vert g\Vert _{M^{s_2}_{p_2,q}} \asymp N^{s_2 + \varepsilon _2 + n/q} \, . $$

In addition, we have

$$\begin{aligned} \Vert f \, \cdot \, g \Vert _{M^{s_0}_{p,q}}&\asymp \Bigg ( \sum _{\Vert m\Vert _\infty \le 2N} \langle m \rangle ^{s_0q}\, \Big |\sum _{{k: ~\Vert k\Vert _\infty \le N \atop \Vert m -k\Vert _\infty \le N}} a_k \, b_{m-k} \Big |^q\Bigg )^{1/q} \\&\ge \frac{1}{2n}\, \Bigg (\sum _{\Vert m\Vert _\infty \le N} \langle m \rangle ^{(s_0 + \varepsilon _2)q}\, \Big |\sum _{{k: ~\Vert k\Vert _\infty \le \Vert m\Vert _\infty /2}} \langle k \rangle ^{\varepsilon _1} \Big |^q\Bigg )^{1/q} \\&\ge C_1 \, \Bigg (\sum _{\Vert m\Vert _\infty \le N} \langle m \rangle ^{(s_0 + \varepsilon _2 + \varepsilon _1 + n)q}\, \Bigg )^{1/q} \\&\ge C_2 \, N^{s_0 + \varepsilon _2 + \varepsilon _1 + n + n/q} \end{aligned}$$

for some $C_1,C_2$ independent of N, see Substep 3.2 of the proof of Theorem 3.5. The inequality (3.11) yields

$$ s_0 + \varepsilon _2 + \varepsilon _1 + n + n/q \le s_1 + \varepsilon _1 + n/q + s_2 + \varepsilon _2 + n/q $$

which proves the claim.$\blacksquare$

The duality argument used in the proof of Lemma 3.7 allows to treat the case $s_0 < 0$.

Theorem 3.24

Let $1\le p, p_1,p_2\le \infty$ and $s_0, s_1, s_2 \in \mathbb {R}$. Let $1/p \le (1/p_1) + (1/p_2)$, $1 \le q < \infty$, $s_0 \le s_2 \le 0$, $0 \le s_1 + s_2$ and $s_1 + s_2-s_0>n/q$. There exists a constant c such that

$$ \Vert \, f \cdot g \, \Vert _{{M}_{p,q}^{s_0}} \le c \, \Vert f \Vert _{{M}_{p_1,q'}^{s_1}}\, \Vert g\Vert _{{M}_{p_2,q}^{s_2}} $$

holds for all $f\in {M}_{p_1,q'}^{s_1}$ and all $g \in {M}_{p_2,q}^{s_2}$.

Remark 3.25

Theorems 3.18 and 3.24 have some overlap.

3.6 Some Further Remarks to the Literature

Here, we recall results of Iwabuchi [13] and Toft et al. [32]. As Cordero and Nicola [6] also Iwabuchi considered the more general situation ${M}_{p_1,q_1}^{s_1} \, \cdot \, {M}_{p_2,q_2}^{s_2} \hookrightarrow M^{s_0}_{p,q}$. This greater flexibility with respect to the tripel $q,q_1,q_2$ allows to treat cases not covered by Theorems 3.22, 3.24.

Proposition 3.26

(Iwabuchi [13])

Let $1\le p, p_1,p_2 \le \infty$, $1<q,q_1,q_2< \infty$ and $0< s_0 < n/q$.

(i) If $q\ge q_1$,

$$\begin{aligned} \frac{1}{p} \le \frac{1}{p_1} + \frac{1}{p_2} \qquad \text{ and }\qquad 1 + \frac{1}{q} - \Big (\frac{1}{q_1} + \frac{1}{q_2}\Big ) = \frac{s_0}{n}\, , \end{aligned}$$

(3.12)

then there exists a constant c such that

$$ \Vert \, f \cdot g \, \Vert _{{M}_{p,q}^{-s_0}} \le c \, \Vert f \Vert _{{M}_{p_1,q_1}^{0}}\, \Vert g\Vert _{{M}_{p_2,q_2}^{0}} $$

holds for all $f \in {M}_{p_1,q_1}^{0}$ and all $g \in {M}_{p_2,q_2}^{0}$.

(ii) Assume $q\ge \max (q_1,q_2)$ and (3.12). Then, there exists a constant c such that

$$ \Vert \, f \cdot g \, \Vert _{{M}_{p,q}^{s_0}} \le c \, \Vert f \Vert _{{M}_{p_1,q_1}^{s_0}}\, \Vert g\Vert _{{M}_{p_2,q_2}^{s_0}} $$

holds for all $f \in {M}_{p_1,q_1}^{s_0}$ and all $g \in {M}_{p_2,q_2}^{s_0}$.

Remark 3.27

Let us take $q=q_1 = q_2$. Then (3.12) reads as $s_0=n/q'$. In combination with $0<s_0<n/q$ this yields $1< q< 2$. Hence, (i) reads as

$$ \Vert \, f \cdot g \, \Vert _{{M}_{p,q}^{-n/q'}} \le c \, \Vert f \Vert _{{M}_{p_1,q}^{0}}\, \Vert g\Vert _{{M}_{p_2,q}^{0}} \, , $$

whereas (ii) gives

$$ \Vert \, f \cdot g \, \Vert _{{M}_{p,q}^{n/q'}} \le c \, \Vert f \Vert _{{M}_{p_1,q_1}^{n/q'}}\, \Vert g\Vert _{{M}_{p_2,q_2}^{n/q'}} \, . $$

Toft et al. [32] also consider the situation ${M}_{p_1,q_1}^{s_1} \, \cdot \, {M}_{p_2,q_2}^{s_2} \hookrightarrow M^{s_0}_{p,q}$. Recall, $\mathring{M}^{s_0}_{p,q}$ denotes the closure of ${\mathcal S}$ in $M^{s_0}_{p,q}$.

Proposition 3.28

(Toft et al. [32])

Let $1\le p,p_1,p_2,q,q_1,q_2\le \infty$ and $s_0,s_1,s_2 \in \mathbb {R}$.

(i) We suppose

• (a) $1+ \frac{1}{p} - \frac{1}{p_1} - \frac{1}{p_2}\le 1$;

• (b) $0 \le 1+ \frac{1}{q} - \frac{1}{q_1} - \frac{1}{q_2}\le 1/2$;

• (c) $s_0 \le \min (s_1,s_2)$;

• (d) $s_1 + s_2 \ge 0$;

• (e) $s_1 + s_2 -s_0 - n \, \Big (1+ \frac{1}{q} - \frac{1}{q_1} - \frac{1}{q_2}\Big ) \ge 0$;

• (f) $s_1 + s_2 -s_0 - n \, \Big (1+ \frac{1}{q} - \frac{1}{q_1} - \frac{1}{q_2}\Big ) > 0$ if $1+ \frac{1}{q} - \frac{1}{q_1} - \frac{1}{q_2}>0$ and either $s_1$ or $s_2$ or $-s_0$ equals $n \,\Big ( 1+ \frac{1}{q} - \frac{1}{q_1} - \frac{1}{q_2}\Big )$.

Then there exists a constant c such that

$$\begin{aligned} \Vert \, f \cdot g \, \Vert _{{M}_{p,q}^{s_0}} \le c \, \Vert f \Vert _{{M}_{p_1,q_1}^{s_1}}\, \Vert g\Vert _{{M}_{p_2,q_2}^{s_2}} \end{aligned}$$

(3.13)

holds for all $f \in \mathring{M}_{p_1,q_1}^{s_1}$ and all $g \in \mathring{M}_{p_2,q_2}^{s_2}$.

(ii) If (3.13) holds for all $f,g \in {\mathcal S}$, then (c), (d) and (e) follow.

Remark 3.29

Again we consider the case $q=q_1=q_2$. Then (b) implies $1\le q \le 2$ and (e) reads as $s_1 + s_2 -s_0 - n/q' \ge 0$. Hence, if we restrict us to $1 < q \le 2$, Proposition 3.28 is slightly more general than Theorem 3.22 and Theorem 3.24. However, for our purpose, see the next section on composition of functions, Theorem 3.22 is already sufficient. Let us mention that Proposition 4.5 below, which is nothing but a modification of Lemma 3.20, is of central importance for the applications to composition operators we have in mind.

3.7 An Important Special Case

We consider $M^s_{2,2}$. A simple argument, based on the frequency-uniform decomposition yields $M^s_{2,2} = H^s$ in the sense of equivalent norms, see Remark 2.3. For these Sobolev spaces $H^s$ almost all is known.

• $H^s$ is an algebra with respect to pointwise multiplication if and only if $s>n/2$, see Strichartz [28], Triebel [33, 2.8] or [25, Theorem 4.6.4/1]. This coincides with Theorem 3.5.

• Let E be a Banach space of functions. By M(E) we denote the set of all pointwise multipliers of E, i.e., the set of all f such that $T_f$, defined as $T_f (g) = f \cdot g$, maps E into E. We equip M(E) with the norm $\Vert f\Vert _{M(E)}:= \Vert T_f\Vert _{{\mathcal L}(E)}$. For a description of $M(H^s)$ one needs the classes $H^{s,\ell oc}$. Here $H^{s,\ell oc}$ denotes the collection of all distributions $f\in {\mathcal S}'$ such that $f \, \cdot \, \varphi \in H^s$ for all $\varphi \in C_0^\infty$. In case $s>n/2$ it holds

  $$ M(H^s) = \Big \{f \in H^{s,\ell oc}: \quad \Vert f\Vert _{M(H^s)}^*:= \sup _{\lambda \in {\mathbb {R}}^n} \Vert \psi (\, \cdot -\lambda )\, f\, \Vert _{H^s} < \infty \Big \} $$

  in the sense of equivalent norms. Here $\psi$ is a smooth nontrivial cut-off function supported around the origin. For all this we refer to Strichartz [28].

• In case $0 \le s < n/2$ also characterizations of $M(H^s)$ are known, this time more complicated, based on capacities. For all details we refer to the monograph of Maz’ya and Shaposnikova [20, Theorem 3.2.2, pp. 86].

• Now we concentrate on the situation described in Theorem 3.22 in case $0< s < \frac{n}{2}$. As it is well-known, there exists a constant c such that

  $$ \Vert \, f \cdot g \, \Vert _{{H}^{2s-n/2}} \le c \, \Vert f \Vert _{H^{s}}\, \Vert g\Vert _{H^{s}} $$

  holds for all $f, g \in {H}^{s}$, see e.g., [25, Theorem 4.5.2]. In Theorem 3.22 we proved that for any $\varepsilon >0$ there exists a constant $c_\varepsilon$ such that

  $$ \Vert \, f \cdot g \, \Vert _{{M}^{2s-n/2-\varepsilon }_{1,2}} \le c_\varepsilon \, \Vert f \Vert _{H^{s}}\, \Vert g\Vert _{H^{s}} $$

  holds for all $f, g \in {H}^{s}$. We conjecture that ${M}^{2s-n/2-\varepsilon }_{1,2}$ and ${H}^{2s-n/2}$ are incomparable.

4 Composition of Functions

There are some attempts to investigate composition of functions in the framework of modulation spaces, i.e., we consider the operator

$$\begin{aligned} T_f : \quad g \mapsto f \circ g, \qquad g \in M^s_{p,q}, \end{aligned}$$

(4.1)

and ask for mapping properties. Of course, we used the symbol $T_f$ before with a different meaning, but we hope that will not cause problems. Within Sect. 4 $T_f$ will have the meaning as in (4.1). Based on pointwise multiplication one can treat f to be a polynomial or even the more general case of f being an entire function.

4.1 Polynomials

We consider the case

$$ f(z):= \sum _{\ell =1}^m \, a_\ell \, z^\ell \, , \qquad z \in \mathbb {C}\, , $$

where $m \in \mathbb {N}$, $m \ge 2$, and $a_\ell \in \mathbb {C}$, $\ell =1, \ldots \, , m$. For brevity we denote the associated composition operator by $T_m$. In addition we need the abbreviation

$$ t_m (s):= s + (m-1)(s-n/q')\, , \qquad m=2,3,\ldots \, . $$

Theorem 4.1

Let $1\le p,q\le \infty$ and $m \in \mathbb {N}$, $m \ge 2$.

(i) Let either $s\ge 0$ and $q=1$ or $s>n/q'$. Then $T_m$ maps $M^s_{p,q}$ into itself. There exists a constant c such that

$$ \Vert \, T_m g\, \Vert _{M^s_{p,q}} \le c \, \Vert \, g\, \Vert _{M^s_{p,q}}\, \sum _{\ell =1}^m \, |a_\ell | \, \Vert \, g\, \Vert _{M^0_{\infty ,1}}^{\ell -1} $$

holds for all $g \in M^s_{p,q}$.

(ii) Let $1 < q \le \infty$, $0 < s \le n/q'$ and $t_{m}(s)> 0$. If $p \in [m,\infty ]$ and $t<t_m(s)$, then there exists a constant c such that

$$ \Vert \, T_m g \, \Vert _{{M}_{p/m,q}^{t}} \le c \, \sum _{\ell =1}^m \, |a_\ell | \, \Vert g\Vert _{{M}_{p,q}^{s}}^{\ell } $$

holds for all $g \in {M}_{p,q}^{s}$.

(iii) Let $q=1$ and $s \ge 0$. If $p \in [m,\infty ]$, then there exists a constant c such that

$$ \Vert \, T_m g \, \Vert _{{M}_{p/m,1}^{s}} \le c \, \sum _{\ell =1}^m \, |a_\ell | \, \Vert g\Vert _{{M}_{p,1}^{s}}^{\ell } $$

holds for all $g \in {M}_{p,1}^{s}$.

Proof

Step 1. Both parts, (i) and (ii), can be proved by induction based on Theorem 3.5 or Theorem 3.22. We concentrate on the proof of (ii). Let $m=2$. Then by assumption $t_2 (s)= 2s-n/q' > 0$. Hence, we may apply Theorem 3.22 with $p_1=p_2 =p$ and $s_1=s_2$ and obtain

$$ \Vert \, g^2 \, \Vert _{{M}_{p/2,q}^{t}} \le c \, \Vert g\Vert _{{M}_{p,q}^{s}}^2 $$

for any $t< 2s-n/q'= t_2(s)$. Now we assume that part (ii) is correct for all natural numbers in the interval [2, m]. We split the product $g^{m+1}$ into the two factors $g^{m}$ and g. By assumption $g^{m} \in {M}_{p/m,q}^{t}$ for any $t< t_{m} (s)$. We put $s_1 = t = t_{m} (s) -\varepsilon$, $s_2 =s$, $p_1 = p/m$ and $p_2 = p$, where we assume that $\varepsilon >0$ is sufficiently small. This guarantees

$$ s_1 + s_2- \frac{n}{q'} = s+ (m-1) \Big (s-\frac{n}{q'}\Big ) -\varepsilon + s - \frac{n}{q'} = t_{m+1}(s)-\varepsilon >0 \,. $$

Hence, we may choose $s_0$ by

$$ s_0 < \min (s_1,s_2, t_{m+1}(s)-\varepsilon ) = t_{m+1}(s)-\varepsilon \, . $$

Since $\varepsilon >0$ is arbitrary, any value $< t_{m+1} (s)$ becomes admissible for $s_0$. An application of Theorem 3.22 yields

$$ \Vert \, g^{m} \cdot g \, \Vert _{{M}_{p/(m+1),q}^{s_0}} \le c \, \Vert \, g_m \, \Vert _{{M}_{p/m,q}^{t_m-\varepsilon }}\, \Vert g\Vert _{{M}_{p,q}^{s}} \, . $$

Step 2. Part (iii) is an immediate consequence of Lemma 3.10. $\blacksquare$

Remark 4.2

For the case $s=0$ we refer to Cordero, Nicola [6], Toft [30] and Guo et al. [11].

4.2 Entire Functions

We consider the case of f being an entire analytic function on $\mathbb {C}$, i.e.,

$$ f(z):= \sum _{\ell =0}^\infty \, a_\ell \, z^\ell \, , \qquad z \in \mathbb {C}\, , $$

where $a_\ell \in \mathbb {C}$, $\ell \in \mathbb {N}_{0}$. Clearly, we need to assume $f(0)=a_0 = 0$. Otherwise $T_f g$ will not have global integrability properties. Let

$$ f_0 (r):= \sum _{\ell =1}^\infty \, |a_\ell | \, r^\ell \, , \qquad r >0\, . $$

Theorem 4.3

Let $1\le p,q\le \infty$ and let either $s\ge 0$ and $q=1$ or $s>n/q'$. Let f be an entire function satisfying $f(0)=0$. Then $T_f$ maps $M^s_{p,q}$ into itself. There exist two constants a, b, independent of f, such that

$$ \Vert \, T_f g\, \Vert _{M^s_{p,q}} \le a \, f_0(b\, \Vert \, g\, \Vert _{M^s_{p,q}}) $$

holds for all $g \in M^s_{p,q}$.

Proof

The constant c in Theorem 4.1 (i) depends on m. To clarify the dependence on m we proceed by induction. Let $c_1$ be the best constant in the inequality

$$\begin{aligned} \Vert \, g_1\cdot g_2\, \Vert _{M^s_{p,q}} \le c_1 \, (\Vert \, g_1\, \Vert _{M^s_{p,q}}\, \Vert \, g_2\, \Vert _{M^0_{\infty ,1}} + \Vert \, g_2\, \Vert _{M^s_{p,q}}\, \Vert \, g_1\, \Vert _{M^0_{\infty ,1}} ) \, , \end{aligned}$$

(4.2)

see Lemma 3.3. Further, let $c_2$ be the best constant in the inequality

$$\begin{aligned} \Vert \, g_1\cdot g_2\, \Vert _{M^0_{\infty ,1}} \le c_2 \, \Vert \, g_1\, \Vert _{M^0_{\infty ,1}}\, \Vert \, g_2\, \Vert _{M^0_{\infty ,1}}\, , \end{aligned}$$

(4.3)

see also Lemma 3.3. By $c_3$ we denote $\max (1,c_1,c_2)$. Our induction hypothesis consists in: the inequality

$$ \Vert \, g^m\, \Vert _{M^s_{p,q}} \le c_3^{m-1} \, m \, \Vert \, g\, \Vert _{M^s_{p,q}} \Vert \, g\, \Vert _{M^0_{\infty ,1}}^{m-1} $$

holds for all $g \in {M^s_{p,q}}$ and all $m \ge 2$. This follows easily from (4.2) and (4.3). Next we need the best constant, denoted by $c_4$, in the inequality

$$ \Vert \, g\, \Vert _{M^0_{\infty ,1}} \le c_4\, \Vert \, g\, \Vert _{M^s_{p,q}}\, , \qquad g \in M^s_{p,q}\, . $$

This proves that

$$\begin{aligned} \Vert \, g^m\, \Vert _{M^s_{p,q}} \le c_3^{m-1} \, m\, c_4^{m-1} \, \Vert \, g\, \Vert ^m_{M^s_{p,q}} \end{aligned}$$

(4.4)

holds for all $g \in {M^s_{p,q}}$ and all $m \ge 2$. Hence

$$\begin{aligned}\Vert \, T_f g\, \Vert _{M^s_{p,q}}\le & {} \sum _{m=1}^\infty |a_m| c_3^{m-1} \, m\, c_4^{m-1} \, \Vert \, g\, \Vert ^m_{M^s_{p,q}} \\= & {} \frac{1}{c_3 \, c_4}\, \sum _{m=1}^\infty |a_m| \, m\, (c_3 \, c_4 \, \Vert \, g\, \Vert _{M^s_{p,q}} )^m \, . \end{aligned}$$

Since

$$ \sup _{m \in \mathbb {N}} m^{1/m} = 3^{1/3} $$

the claimed estimate follows.$\blacksquare$

Remark 4.4

Theorem 4.3 is essentially known, see e.g., Sugimoto et al. [29] or Bhimani [1].

4.3 One Example

The following example has been considered at various places. Let $f(z):= e^z-1$, $z \in \mathbb {C}$. For appropriate constants $a,b>0$ it follows that

$$\begin{aligned} \Vert \, e^g - 1\, \Vert _{M^s_{p,q}} \le a \, e^{b\, \Vert \, g\, \Vert _{M^s_{p,q}}} \end{aligned}$$

(4.5)

holds for all $g \in M^s_{p,q}$.

It will be essential for our approach to non-analytic composition results that we can improve this estimate.

4.4 Non-analytic Superposition Operators

There is a famous classical result by Katznelson [17] (in the periodic case) and by Helson, Kahane, Katznelson, Rudin [12] (nonperiodic case) which says that only analytic functions operate on the Wiener algebra $\mathcal {A}$. More exactly, the operator $T_f : ~u \mapsto f(u)$ maps $\mathcal {A}$ into $\mathcal {A}$ if and only if $f(0)=0$ and f is analytic. Here, $\mathcal {A}$ is the collection of all $u \in C$ such that ${\mathcal F}u \in L_1$. Moreover, a similar result is obtained for particular standard modulation spaces. Bhimani and Ratnakumar [2], see also Bhimani [1], proved that $T_f$ maps $M_{1,1}$ into $M_{1,1}$ if and only if $f(0)=0$ and f is analytic. Therefore, the existence of non-analytic superposition results for weighted modulation spaces is a priori not so clear.

We shall concentrate on the algebra case. Our first aim consists in deriving a better estimate than (4.5).

To proceed we need some preparations. An essential tool in proving our main result will be a certain subalgebra property. Therefore, we consider the following decomposition of the phase space. Let $R>0$ and $\epsilon =(\epsilon _1,\ldots , \epsilon _n)$ be fixed with $\epsilon _j \in \{0,1\}$, $j=1,\ldots ,n$. Then a decomposition of ${\mathbb {R}}^n$ into $(2^n+1)$ parts is given by

$$\begin{aligned} P_R := \{ \xi \in {\mathbb {R}}^n:\, |\xi _j|\le R, j=1,\ldots , n \} \end{aligned}$$

and

$$ P_R(\epsilon ) := \{ \xi \in {\mathbb {R}}^n: \, \mathrm{sign \,}(\xi _j) = (-1)^{\epsilon _j}, \, j=1,\ldots ,n \} \setminus P_R. $$

For given p, q, s, $\epsilon =(\epsilon _1,\ldots , \epsilon _n)$ and $R>0$ we introduce the spaces

$$ {M}_{p,q}^{s}(\epsilon , R) := \{ f\in {M}_{p,q}^{s}: \, \mathrm{supp \, }{\mathcal F}f \subset P_R(\epsilon ) \}\, . $$

Proposition 4.5

Let $1\le p_1,p_2\le \infty$, $1 < q \le \infty$ and $s_0 , s_1,s_2 \in \mathbb {R}$. Define p by $\frac{1}{p}:= \frac{1}{p_1}+\frac{1}{p_2}$. Let $R > 2$. If $p \in [1,\infty ]$, $s_0 \le \min (s_1,s_2)$, $s_1, s_2 \ge 0$ and $s_1 + s_2-s_0 > n/q'$, then there exists a constant c such that

$$ \Vert \, f \cdot g \, \Vert _{{M}_{p,q}^{s_0}} \le c \, (R-2)^{-[(s_1 +s_2 -s_0 ) - n/q']} \, \Vert f \Vert _{{M}_{p_1,q}^{s_1}}\, \Vert g\Vert _{{M}_{p_2,q}^{s_2}} $$

holds for all $f \in {M}_{p_1,q}^{s_1}(\epsilon ,R)$ and all $g \in {M}_{p_2,q}^{s_2}(\epsilon ,R)$. The constant c is independent from $R>2$ and $\epsilon$.

Proof

In order to show the subalgebra property we follow the same steps as in the proof of Lemma 3.20. We start with some almost trivial observations. Let $f\in M_{p_1,q}^{s}(\epsilon , R)$ and $g\in M_{p_2,q}^{s}(\epsilon , R)$. By

$$ \mathrm{supp \, }({\mathcal F}f *{\mathcal F}g) \subset \{ \xi +\eta : \xi \in \mathrm{supp \, }{\mathcal F}f, \eta \in \mathrm{supp \, }{\mathcal F}g \} $$

we have $\mathrm{supp \, }{\mathcal F}(fg) \subset P_R(\epsilon )$. Let

$$ P_R^* (\epsilon ):= \Big \{k \in {\mathbb {Z}}^n: \quad \Vert k\Vert _\infty >R-1\, , \quad \mathrm{sign \,}(k_j) = (-1)^{\epsilon _j}, \, j=1,\ldots ,n\Big \}. $$

Hence, if $\mathrm{supp \, }\sigma _k \cap P_R(\epsilon ) \ne \emptyset$, then $k \in P_R^*(\epsilon )$ follows. Now we continue as in proof of Lemma 3.20, Step 2, and obtain

$$\begin{aligned}& \Big ( \sum _{k\in P_R^* (\epsilon )} \langle k\rangle ^{s_0 q} \Vert {{\mathcal F}}^{-1}\big (\sigma _k {\mathcal F}(f\cdot g) \big )\Vert _{L_p}^q \Big )^{\frac{1}{q}} \\\le & {} \sum _{\begin{array}{c} t \in {\mathbb {Z}}^n, \\ -3<t_i < 3, \\ i=1,\ldots , n \end{array}} \Bigg ( \sum _{k\in P_R^* (\epsilon )} \langle k\rangle ^{s_0q} \Big [ \sum _{{l \in {\mathbb {Z}}^n: \atop t-l+k, l\in P_R^* (\epsilon )}} \Vert {{\mathcal F}}^{-1}\big (\sigma _k {\mathcal F}( f_{t-(l-k)} \cdot g_l) \big )\Vert _{L_p} \Big ]^q \Bigg )^{\frac{1}{q}} \, . \end{aligned}$$

This implies

$$\begin{aligned}& \Big ( \sum _{k\in P_R^* (\epsilon )} \langle k\rangle ^{s_0 q} \Vert {{\mathcal F}}^{-1}\big (\sigma _k {\mathcal F}(f\cdot g) \big )\Vert _{L_p}^q \Big )^{\frac{1}{q}} \\\le & {} c_1\, \sum _{\begin{array}{c} t \in {\mathbb {Z}}^n, \\ -3<t_i< 3, \\ i=1,\ldots , n \end{array}} \Bigg ( \sum _{k\in P_R^* (\epsilon )} \langle k\rangle ^{s_0q} \Big [ \sum _{{l \in {\mathbb {Z}}^n: \atop t-l+k, l\in P_R^* (\epsilon )}} \Vert f_{t-(l-k)}\cdot g_l\Vert _{L_p} \Big ]^q \Bigg )^{\frac{1}{q}} \\\le & {} c_2 \, \max _{\begin{array}{c} t \in {\mathbb {Z}}^n, \\ -3<t_i < 3, \\ i=1,\ldots , n \end{array}} \Bigg ( \sum _{k\in P_R^* (\epsilon )} \langle k\rangle ^{s_0q} \Big [ \sum _{{l \in {\mathbb {Z}}^n: \atop t-l+k, l\in P_R^* (\epsilon )}} \Vert f_{t-(l-k)}\Vert _{L_{p_1}} \Vert g_l\Vert _{L_{p_2}} \Big ]^q \Bigg )^{\frac{1}{q}} \end{aligned}$$

with $c_2$ and $c_1$ as above. We put

$$\begin{aligned} S_{1,t,k}:= & {} \sum _{\begin{array}{c} l \in {\mathbb {Z}}^n: ~t-l+k, l\in P_R^* (\epsilon ), \\ |l|\le |l-k| \end{array}} \langle k-l\rangle ^{s_1} \Vert f_{t-(l-k)}\Vert _{L_{p_1}} \langle l\rangle ^{s_2} \Vert g_l\Vert _{L_{p_2}} \, \langle k-l\rangle ^{s_0 - s_1} \, \langle l\rangle ^{-s_2} \, ; \\ S_{2,t,k}:= & {} \sum _{\begin{array}{c} l \in {\mathbb {Z}}^n: ~t-l+k, l\in P_R^* (\epsilon ), \\ |l-k|\le |l| \end{array}} \langle k-l\rangle ^{s_1} \, \Vert f_{t-(l-k)}\Vert _{L_{p_1}} \langle l\rangle ^{s_2} \, \Vert g_l\Vert _{L_{p_2}} \, \langle k-l\rangle ^{- s_1} \, \langle l\rangle ^{s_0-s_2} \, . \end{aligned}$$

Hölder’s inequality leads to

$$\begin{aligned} S_{1,t,k}\le & {} \Big ( \sum _{\begin{array}{c} j\in {\mathbb {Z}}^n, \\ |j+k|\le |j| \end{array}} ( \langle j\rangle ^{s_1} \Vert f_{t-j}\Vert _{L_{p_1}} \langle j+k\rangle ^{s_2} \Vert g_{j+k}\Vert _{L_{p_2}} )^{q}\Big )^{1/q} \\&\qquad \qquad \times \quad \Big ( \sum _{\begin{array}{c} j\in {\mathbb {Z}}^n: t-j, j+k \in P_R^* (\epsilon )\\ |j+k|\le |j| \end{array}} (\langle j \rangle ^{s_0 - s_1} \, \langle j+k \rangle ^{-s_2})^{q'} \Big )^{\frac{1}{q'}} \, . \end{aligned}$$

Our assumptions $s_0 \le s_1$, $s_2 \ge 0$ and $s_1 + s_2 -s_0 >n/q'$ and $j+k \in P_R^* (\epsilon )$ imply

$$\begin{aligned} \Big ( \sum _{\begin{array}{c} j\in {\mathbb {Z}}^n, \\ |j+k|\le |j| \end{array}} \Big |&(\langle j \rangle ^{s_0 - s_1} \, \langle j+k \rangle ^{-s_2}) \Big |^{q'} \Big )^{\frac{1}{q'}} \le \Big ( \sum _{m \in P_R^* (\epsilon )} \langle m \rangle ^{(s_0 -s_1 -s_2) q'} \Big )^{\frac{1}{q'}} \\\le & {} \Big ( 2^{-n} \int _{\Vert x\Vert _\infty>R-2} (1+|x|^2)^{(s_0 -s_1 -s_2) q'/2}\, dx\Big )^{1/q'} \\\le & {} \Big ( 2^{-n} \int _{|x| > R-2} |x|^{(s_0 -s_1 -s_2) q'}\, dx\Big )^{1/q'} \\\le & {} \Big (\frac{2^{-n}}{(s_1 + s_2 - s_0)q'-n}\Big )^{1/q'} \, (R-2)^{-[(s_1 +s_2 -s_0 ) - n/q']} \, . \end{aligned}$$

With $c_3 := \Big (\frac{2^{-n}}{(s_1 + s_2 - s_0)q'-n}\Big )^{1/q'}$ we insert this in our previous estimate and obtain

$$\begin{aligned} \Big (\sum _{k \in {\mathbb {Z}}^n} S_{1,t,k}^q\Big )^{1/q}\le & {} c_3\, (R-2)^{-[(s_1 +s_2 -s_0 ) - n/q']} \\& \times \, \Bigg ( \sum _{k\in {\mathbb {Z}}^n} \sum _{\begin{array}{c} j\in {\mathbb {Z}}^n, \\ |j+k|\le |j| \end{array}} \langle j \rangle ^{s_1q} \Vert f_{t-j}\Vert _{L^{p_1}}^q \langle j+k \rangle ^{s_2q} \Vert g_{j+k}\Vert _{L^{p_2}}^q \Bigg )^{1/q} \\\le & {} c_3 \, c_4\, (R-2)^{-[(s_1 +s_2 -s_0 ) - n/q']} \, \Vert g\Vert _{{M}^{s_2}_{p_2,q}} \Vert f\Vert _{{M}^{s_1}_{p_1,q}} \, . \end{aligned}$$

Here, $c_3,c_4$ are independent of $f,g, \epsilon$ and R. For the second sum the estimate

$$ \Big (\sum _{k \in {\mathbb {Z}}^n} S_{2,t,k}^q\Big )^{1/q} \le c_5\, (R-2)^{-[(s_1 +s_2 -s_0 ) - n/q']} \, \Vert g\Vert _{{M}^{s_2}_{p_2,q}} \Vert f\Vert _{{M}^{s_1}_{p_1,q}} $$

follows by analogous computations. The proof is complete.$\blacksquare$

Of course, the above arguments have a counterpart in case $q' = \infty$.

Proposition 4.6

Let $1\le p_1,p_2\le \infty$, $q=1$ and $s_1,s_2 \in \mathbb {R}$. Define p by $\frac{1}{p}:= \frac{1}{p_1}+\frac{1}{p_2}$. Let $R > 2$. If $p \in [1,\infty ]$, $s_1, s_2 \ge 0$ and $s_0 := \min (s_1, s_2)$, then there exists a constant c such that

$$ \Vert \, f \cdot g \, \Vert _{{M}_{p,1}^{s_0}} \le c \, (R-2)^{-(s_1 +s_2 -s_0 )} \, \Vert f \Vert _{{M}_{p_1,1}^{s_1}}\, \Vert g\Vert _{{M}_{p_2,1}^{s_2}} $$

holds for all $f \in {M}_{p_1,1}^{s_1}(\epsilon ,R)$ and all $g \in {M}_{p_2,1}^{s_2}(\epsilon ,R)$. The constant c is independent from $R>2$ and $\epsilon$.

As a consequence of Nikol’kij’s inequality, see Lemma 2.6, Proposotion 4.5 (with $s_0=s_1=s_2$ and $p_1=p$, $p_2 = \infty$) and Corollary 2.7 we obtain the following.

Proposition 4.7

Let $1\le p \le \infty$ and $R > 2$.

(i) Let $1 < q \le \infty$ and $s>n/q'$. Then there exists a constant c such that

$$ \Vert \, f \cdot g \, \Vert _{{M}_{p,q}^{s}} \le c \, (R-2)^{-(s- n/q')} \, \Vert f \Vert _{{M}_{p,q}^{s}}\, \Vert g\Vert _{{M}_{p,q}^{s}} $$

holds for all $f, g \in {M}_{p,q}^{s}(\epsilon ,R)$. The constant c is independent from $R>2$ and $\epsilon$.

(ii) Let $q=1$ and $s \ge 0$. Then there exists a constant c such that

$$ \Vert \, f \cdot g \, \Vert _{{M}_{p,1}^{s}} \le c \, (R-2)^{-s} \, \Vert f \Vert _{{M}_{p,1}^{s}}\, \Vert g\Vert _{{M}_{p,1}^{s}} $$

holds for all $f, g \in {M}_{p,1}^{s}(\epsilon ,R)$. The constant c is independent from $R>2$ and $\epsilon$.

Note that in the following, we assume every function to be real-valued unless it is explicitly stated that complex-valued functions are allowed. To make this more clear we switch from $g \in M^s_{p,q}$ to $u \in M^s_{p,q}$.

Next we have to recall some assertions from harmonic analysis. The first one concerns a standard estimate of Fourier multipliers, see e.g., [33, Theorem1.5.2].

Lemma 4.8

Let $1 \le r \le \infty$ and assume that $s>n/2$. Then there exists a constant $c>0$ such that

$$ \Vert {{\mathcal F}}^{-1}[ \phi \, {\mathcal F}g](\,\cdot \, )\Vert _{L_r} \le c\, \Vert \phi \Vert _{H^s} \, \Vert g\Vert _{L_r} $$

holds for all $g \in L_r$ and all $\phi \in H^s$.

The next lemma is taken from [5].

Lemma 4.9

Let $N\in \mathbb {N}$ and suppose $a_1, a_2,\ldots , a_N$ to be complex numbers. Then, it holds

$$ a_1 \cdot a_2 \cdot \ldots \cdot a_N -1 = \sum _{l=1}^{N} \sum _{\begin{array}{c} j=(j_1,\ldots , j_l), \\ 0\le j_1< \ldots < j_l \le N \end{array}} (a_{j_1} -1) \cdot \ldots \cdot (a_{j_l}-1). $$

In our approach the next estimate will be fundamental.

Proposition 4.10

Let $1< p< \infty$, $1 \le q \le \infty$ and $s>n/q'$. Then there exists a positive constant C such that

$$ \Vert e^{i u} -1\Vert _{M_{p,q}^{s}} \le C \, \Vert u\Vert _{M_{p,q}^{s}} \, \left( 1 + \Vert u\Vert _{M_{p,q}^s}\right) ^{(s+n/q)(1+ \frac{1}{s-n/q'})} $$

holds for all real-valued $u \in M^s_{p,q}$.

Proof

This proof follows ideas developed in [5], but see also [23].

Step 1. Let u be a nontrivial function in ${M}_{p,q}^s$ satisfying $\mathrm{supp \, }{\mathcal F}u \subset P_R$ for some $R\ge 2$.

First we consider the Taylor expansion

$$ e^{i u}-1 = \sum _{l=1}^r \frac{(i u)^l}{l!} + \sum _{l=r+1}^{\infty } \frac{(i u)^l}{l!} $$

resulting in the norm estimate

$$ \Vert e^{i u}-1\Vert _{{M}_{p,q}^s} \le \Big \Vert \sum _{l=1}^r \frac{(i u)^l}{l!} \Big \Vert _{{M}_{p,q}^s} + \Big \Vert \sum _{l=r+1}^{\infty } \frac{(i u)^l}{l!} \Big \Vert _{{M}_{p,q}^s} \, . $$

For brevity we put

$$ S_1:= \Big \Vert \sum _{l=1}^r \frac{(i u)^l}{l!} \Big \Vert _{{M}_{p,q}^s} \qquad \text{ and }\qquad S_2 := \Big \Vert \sum _{l=r+1}^{\infty } \frac{(i u)^l}{l!} \Big \Vert _{{M}_{p,q}^s} \, . $$

The natural number r will be chosen later on. Next we employ the algebra property, in particular the estimate (4.4) with $C_1 := 2\, c_3 \, c_4$. We obtain

$$ S_2 \le \sum _{l=r+1}^\infty \frac{1}{l!} \Vert u^l\Vert _{M_{p,q}^s} \le \frac{1}{C_1} \sum _{l=r+1}^{\infty } \frac{ (C_1\, \Vert u\Vert _{M_{p,q}^s})^l}{l!}. $$

Now we choose r as a function of $\Vert u\Vert _{M_{p,q}^s}$ and distinguish two cases:

• 1. $C_1\, \Vert u\Vert _{{M}_{p,q}^s}>1$. Assume that

  $$\begin{aligned} 3\, C_1\, \Vert u\Vert _{{M}_{p,q}^s} \le r \le 3\, C_1 \, \Vert u\Vert _{{M}_{p,q}^s} +1 \end{aligned}$$

  (4.6)

  and recall Stirling’s formula $l! = \Gamma (l+1) \ge l^l e^{-l} \sqrt{2\pi l}$. Thus, we get

  $$\begin{aligned} \sum _{l=r+1}^{\infty } \frac{(C_1\Vert u\Vert _{{M}_{p,q}^s})^l}{l!}\le & {} \sum _{l=r+1}^{\infty } \left( \frac{r}{l} \right) ^l \left( \frac{e}{3}\right) ^l \frac{1}{\sqrt{2\pi l}} \\\le & {} \sum _{l=r+1}^\infty \left( \frac{e}{3} \right) ^l \le \frac{3}{3-e}. \end{aligned}$$

• 2. $C_1 \, \Vert u\Vert _{{M}_{p,q}^s}\le 1$. It follows

  $$\begin{aligned} \sum _{l=r+1}^{\infty } \frac{(C_1\, \Vert u\Vert _{{M}_{p,q}^s})^l}{l!} \le C_1\, \Vert u\Vert _{{M}_{p,q}^s} \, \sum _{l=1}^{\infty } \frac{1}{l!} \le C_1\, e\, \Vert u\Vert _{{M}_{p,q}^s}. \end{aligned}$$

Both together can be summarized as

$$\begin{aligned} S_2 \le C_2 \, \Vert u\Vert _{M_{p,q}^s}\, , \qquad C_2 := \max \Big (e, \frac{3}{C_1(3-e)}\Big ). \end{aligned}$$

To estimate $S_1$ we check the support of ${\mathcal F}u^\ell$ and find

$$\begin{aligned} S_1 = \Big \Vert \sum _{l=1}^r \frac{(i u)^l}{l!} \Big \Vert _{M_{p,q}^{s}}= & {} \Big ( \sum _{k\in {\mathbb {Z}}^n} \langle k \rangle ^{sq} \Big \Vert \Box _k \Big (\sum _{l=1}^{r} \frac{(i u)^l}{l!} \Big ) \Big \Vert _{L_p}^q \Big )^{\frac{1}{q}} \\= & {} \Big ( \sum _{\begin{array}{c} k\in {\mathbb {Z}}^n, \\ -Rr-1<k_i< Rr+1, \\ i=1,\ldots ,n \end{array}} \langle k \rangle ^{sq} \Big \Vert \Box _k \Big (\sum _{l=1}^{r} \frac{(i u)^l}{l!} \Big ) \Big \Vert _{L_p}^q \Big )^{\frac{1}{q}} \\\le & {} \Big ( \sum _{\begin{array}{c} k\in {\mathbb {Z}}^n, \\ -Rr-1<k_i< Rr+1, \\ i=1,\ldots ,n \end{array}} \langle k \rangle ^{sq} \Vert \Box _k (e^{i u}-1) \Vert _{L_p}^q \Big )^{\frac{1}{q}} + S_2\, . \end{aligned}$$

Concerning $S_2$ we proceed as above. To estimate the first part we observe that

$$ C_3 := \sup _{k \in {\mathbb {Z}}^n} \, \Vert \, \sigma _k \, \Vert _{H^t} = \Vert \, \sigma _0 \, \Vert _{H^t} <\infty \, , $$

see Lemma 4.8. Furthermore, $\cos , \sin$ are Lipschitz continuous and consequently we get

$$\begin{aligned} \Vert \Box _k (e^{i u}-1) \Vert _{L_p}\le & {} C_3 \, \Vert e^{i u}-1 \Vert _{L_p} \\\le & {} C_3 \, (\Vert \cos u - \cos 0 \Vert _{L_p} + \Vert \sin u - \sin 0 \Vert _{L_p}) \\\le & {} 2\, C_3 \, \Vert u - 0 \Vert _{L_p} \, . \end{aligned}$$

This implies

$$\begin{aligned} \Big ( \sum _{\begin{array}{c} k\in {\mathbb {Z}}^n, \\ -Rr-1<k_i< Rr+1, \\ i=1,\ldots ,n \end{array}} \langle k \rangle ^{sq}& \Vert \Box _k (e^{i u}-1) \Vert _{L_p}^q \Big )^{\frac{1}{q}} \\\le & {} 2\, C_3 \, \Vert \, u \, \Vert _{L_p}\, \Big ( \sum _{\begin{array}{c} k\in {\mathbb {Z}}^n, \\ -Rr-1<k_i< Rr+1, \\ i=1,\ldots ,n \end{array}} \langle k \rangle ^{sq} \Big )^{\frac{1}{q}} \, . \end{aligned}$$

Clearly,

$$\begin{aligned} \sum _{\begin{array}{c} k\in {\mathbb {Z}}^n, \\ -Rr-1<k_i< Rr+1, \\ i=1,\ldots ,n \end{array}} \langle k \rangle ^{sq}\le & {} \int _{\Vert \, x\, \Vert _\infty< Rr+1}\, \langle x \rangle ^{sq} \, dx \nonumber \\\le & {} \int _{|x|< \sqrt{n}(Rr+1)}\, \langle x \rangle ^{sq} \, dx \nonumber \\\le & {} 2\, \frac{\pi ^{n/2}}{\Gamma (n/2)}\, \int _{0}^{\sqrt{n} (Rr+1)} (1+\tau )^{n-1+sq} \, d\tau \nonumber \\\le & {} 2\, \frac{\pi ^{n/2}}{\Gamma (n/2)}\, \frac{1}{n+sq}\, (\sqrt{n}(Rr+2))^{n+ sq}\, \, . \end{aligned}$$

To simplify notation we define

$$ C_4 := \Big (2\, \frac{\pi ^{n/2}}{\Gamma (n/2)}\, \frac{1}{n+sq}\, \sqrt{n}^{n+sq}\Big )^{1/q}\, . $$

In addition we shall use in case $1 < q \le \infty$

$$ \Vert u \Vert _{L^p}\le C_5 \, \Vert u\Vert _{{M}_{p,q}^{s}} \, , \qquad C_5 := \Big (\sum _{k\in {\mathbb {Z}}^n} \langle k \rangle ^{-sq'} \Big )^{1/q'} $$

which follows from Hölder’s inequality and in case $q=1$

$$ \Vert u \Vert _{L^p}\le \, \Vert u\Vert _{{M}_{p,1}^{s}} $$

as a consequence of triangle inequality. Summarizing we have found

$$ \Vert e^{i u}-1\Vert _{{M}_{p,q}^s} \le \Big (2 \, C_2 + 2\, \max (C_5,1) \, C_4\,C_3\, (Rr+2)^{s+n/q}\Big )\, \Vert u\Vert _{{M}_{p,q}^{s}} \, . $$

Next we apply (4.6) which results in

$$\begin{aligned} \Vert e^{i u}-1 \Vert _{{M}_{p,q}^{s}} \le C_6 \, \Vert u\Vert _{{M}_{p,q}^{s}} \left( 1 + R \, \Vert u\Vert _{M_{p,q}^s}\right) ^{s+n/q} \, , \end{aligned}$$

(4.7)

valid for all $u\in {M}_{p,q}^{s}$ satisfying $\mathrm{supp \, }{\mathcal F}u \subset P_R$ and with positive constant $C_6$ not depending on u and $R\ge 2$.

Step 2. This time we consider $u\in {M}_{p,q}^{s}$ without any restriction on the Fourier support. Here we need the restriction $1< p< \infty$. For those p the characteristic functions $\chi$ of cubes are Fourier multipliers in $L^p$ by the famous Riesz Theorem and therefore also in ${M}_{p,q}^{s}$. In addition we shall make use of the fact that the norm of the operator $f \mapsto {{\mathcal F}}^{-1}\chi \, {\mathcal F}f$ does not depend on the size of the cube. Below we shall denote this norm by $C_7 = C_7 (p)$. We refer to Lizorkin [19] for all details. For decomposing u on the phase space we introduce functions $\chi _{R,\epsilon }$ and $\chi _R$, that is, the characteristic functions of the sets $P_R(\epsilon )$ and $P_R$, respectively. By defining

$$\begin{aligned} u_\epsilon (x)= & {} {{\mathcal F}}^{-1}[\chi _{R,\epsilon } (\xi ) \, {\mathcal F}u(\xi ) ] (x) , \qquad x \in {\mathbb {R}}^n\, , \\ u_0 (x)= & {} {{\mathcal F}}^{-1}[\chi _R (\xi ) \, {\mathcal F}u (\xi ) ] (x)\, , \qquad x \in {\mathbb {R}}^n\, , \end{aligned}$$

we can rewrite u as

$$\begin{aligned} u(x) = u_0(x) + \sum _{\epsilon \in I} u_\epsilon (x) , \end{aligned}$$

(4.8)

where I is the set of all $\epsilon =(\epsilon _1,\ldots , \epsilon _n)$ with $\epsilon _j \in \{0,1\}$, $j=1,\ldots ,n$. Hence

$$\begin{aligned} \Vert u\Vert _{{M}_{p,q}^{s}} \le \Vert u_0\Vert _{{M}_{p,q}^{s}} + \sum _{\epsilon \in I} \Vert u_\epsilon \Vert _{M_{p,q}^{s}} \end{aligned}$$

and

$$ \max \Big (\Vert u_0\Vert _{M_{p,q}^{s}}, \Vert u_\epsilon \Vert _{M_{p,q}^{s}}\Big ) \le C_7 \, \Vert \, u\, \Vert _{M_{p,q}^{s}}\, . $$

Due to the representation (4.8) and using an appropriate enumeration Lemma 4.9 leads to

$$ e^{i u} -1 = \sum _{l=1}^{2^n +1} \sum _{0\le j_1<\ldots < j_l \le 2^n} (e^{i u_{j_1}} -1)\cdot \ldots \cdot (e^{i u_{j_l}} -1) \, . $$

The algebra property, in particular the estimate (4.4) with $C_1 := 2\, c_3 \, c_4$, yields

$$\begin{aligned} \Vert e^{i u} -1\Vert _{M_{p,q}^{s}} \le \sum _{l=1}^{2^n +1} C^{l-1}_1\, \sum _{0\le j_1<\ldots < j_l \le 2^n} \Vert e^{i u_{j_1}} -1 \Vert _{M_{p,q}^{s}} \cdot \ldots \cdot \Vert e^{i u_{j_l}} -1 \Vert _{M_{p,q}^{s}}. \end{aligned}$$

(4.9)

By Proposition 4.7 and (4.7) it follows

$$\begin{aligned} \Vert e^{i u_{j_k}} -1\Vert _{M_{p,q}^{s}}= & {} \Big \Vert \sum _{l=1}^\infty \frac{(i u_{j_k})^l}{l!} \, \Big \Vert _{M_{p,q}^{s}} \le \frac{R^{s-n/q'}}{c} \, \Big (e^{c\, \Vert u_{j_k}\Vert _{M_{p,q}^{s}}/R^{s-n/q'} } -1 \Big ) \nonumber \\\le & {} \frac{(R-2)^{s-n/q'}}{c} \, \Big (e^{c\, C_7 \, \Vert u\Vert _{M_{p,q}^{s}}/(R-2)^{s-n/q'} } -1 \Big ) , \end{aligned}$$

(4.10)

as well as

$$\begin{aligned} \Vert e^{i u_0}-1 \Vert _{{M}_{p,q}^{s}} \le C_6 \, C_7\, \Vert u\Vert _{{M}_{p,q}^{s}} \left( 1 + R \, C_7 \Vert u\Vert _{M_{p,q}^s}\right) ^{s+n/q} \, , \end{aligned}$$

(4.11)

where we used the Fourier multiplier assertion mentioned at the beginning of this step. The final step in our proof is to choose the number R as a function of $\Vert u\Vert _{M_{p,q}^{s}}$ such that (4.10) and (4.11) will be approximately of the same size.

Substep 2.1. Let $\Vert u\Vert _{M_{p,q}^{s}} \le 1$. We choose $R=3$. Then (4.9) combined with (4.10) and (4.11) results in the estimate

$$\begin{aligned} \Vert e^{i u} -1\Vert _{M_{p,q}^{s}} \le C_8 \, \Vert u\Vert _{M_{p,q}^{s}} , \end{aligned}$$

where $C_8$ does not depend on u.

Substep 2.2. Let $\Vert u\Vert _{M_{p,q}^{s}} >1$. We choose $R\ge 3$ such that

$$\begin{aligned} (R-2)^{s-n/q'} = \Vert u\Vert _{M_{p,q}^{s}} \, . \end{aligned}$$

Now (4.9), combined with (4.10) and (4.11), results in

$$\begin{aligned} \Vert e^{i u} -1\Vert _{M_{p,q}^{s}}\le & {} C_{9} \, \Vert u\Vert _{M_{p,q}^{s}} \, \left( 1 + \Vert u\Vert _{M_{p,q}^s}\right) ^{(s+n/q)(1+ \frac{1}{s-n/q'})} \, , \, \end{aligned}$$

(4.12)

with a constant $C_{9}$ independent of u.$\blacksquare$

Remark 4.11

The restriction of p to the interval $(1,\infty )$ is caused by our decomposition technique, see Step 2 of the preceding proof. We do not know whether Proposition 4.10 extends to $p=1$ and/or $p = \infty$.

Next, we need again a technical lemma.

Lemma 4.12

Let $1< p< \infty$, $1 \le q \le \infty$ and $s>n/q'$.

(i) The mapping $u \mapsto e^{iu}-1$ is locally Lipschitz continuous (considered as a mapping of $M_{p,q}^s$ into $M_{p,q}^s$).

(ii) Assume $u\in M_{p,q}^s$ to be fixed and define a function $g: \mathbb {R}\mapsto M_{p,q}^s$ by $g(\xi ) = e^{i u(x) \xi }-1$. Then the function g is continuous.

Proof

Local Lipschitz continuity follows from the identity

$$\begin{aligned} e^{iu}- e^{iv} = (e^{iv}-1)\, (e^{i(u-v)}-1) + (e^{i(u-v)}-1)\, , \end{aligned}$$

(4.13)

the algebra property of $M_{p,q}^s$ and Proposition 4.10.

To prove the continuity of g we also employ the identity (4.13). The claim follows by using the algebra property and Proposition 4.10. $\blacksquare$

Now we are in position to prove the main result of this section.

Theorem 4.13

Let $1< p < \infty$, $1\le q \le \infty$ and $s>n/q'$. Let $\mu$ be a complex measure on $\mathbb {R}$ such that

$$\begin{aligned} L := \int _{-\infty }^\infty \, (1+|\xi |)^{1+(s+n/q)(1+ \frac{1}{s-n/q'})} \, d|\mu | (\xi ) < \infty \end{aligned}$$

(4.14)

and such that $\mu (\mathbb {R}) = 0$. Furthermore, assume that the function f is the inverse Fourier transform of $\mu$. Then f is a continuous function and the composition operator $T_f: u \mapsto f \circ u$ maps $M_{p,q}^s$ into $M_{p,q}^s$.

Proof

Equation (4.14) yields $\int _{{\mathbb {R}}^n} d|\mu |(\xi ) < \infty$. Thus, $\mu$ is a finite measure and $\mu (\mathbb {R}) =0$ makes sense. Now we define the inverse Fourier transform of $\mu$

$$\begin{aligned} f(t) = \frac{1}{\sqrt{2\pi }} \int _{{\mathbb {R}}^n} e^{i \xi t} \, d \mu (\xi ). \end{aligned}$$

Moreover, since

$$ (s+n/q)\Big (1+ \frac{1}{s-n/q'}\Big ) >n $$

we conclude that $\int _{\mathbb {R}} |(i \xi )^j| \, d|\mu |(\xi ) <\infty$, $j=1, \ldots \, , n+1$, which implies $f \in C^{n+1}$. Due to $\mu (\mathbb {R})=0$ we can also write f as follows:

$$ f(t) = \frac{1}{\sqrt{2\pi }} \int _{\mathbb {R}} (e^{i \xi t}-1) \, d \mu (\xi ). $$

Since $\mu$ is a complex measure we can split it up into real part $\mu _r$ and imaginary part $\mu _i$, where each of them is a signed measure. Without loss of generality we proceed our computations only with the positive real measure $\mu _r^+$. For all measurable sets E we have $\mu _r^+(E) \le |\mu |(E)$.

Let $u\in M_{p,q}^s$ and define the function $g(\xi ) = e^{i u(x) \xi } -1$ analogously to Lemma 4.12. Then g is Bochner integrable because of its continuity and taking into account that the measure $\mu _r^+$ is finite. Therefore we obtain the Bochner integral

$$ \int _{-\infty }^\infty \big (e^{i u(x) \xi } -1 \big )\, d\mu _r^+(\xi ) = \int _{-\infty }^\infty g(\xi ) \, d\mu _r^+(\xi ) $$

with values in $M_{p,q}^s$. By applying Minkowski inequality it follows

$$ \Big \Vert \int _{-\infty }^\infty \big (e^{i u(\cdot ) \xi } -1 \big )\, d\mu _r^+(\xi ) \Big \Vert _{M_{p,q}^s} \le \int _{-\infty }^\infty \big \Vert e^{i u(\cdot ) \xi } -1 \big \Vert _{M_{p,q}^s} \, d|\mu |(\xi ). $$

Using the abbreviation $\Vert u\Vert := \Vert u\Vert _{M_{p,q}^s}$, Proposition 4.10 together with equation (4.14) yields

$$\begin{aligned} \int _{|\xi | \Vert u\Vert \ge 1}& \big \Vert e^{i u(\cdot ) \xi } -1 \big \Vert _{M_{p,q}^s} \, d|\mu |(\xi ) \\\le & {} C'\, \Vert \, u\, \Vert _{M_{p,q}^s}^{1+(s+n/q)(1+ \frac{1}{s-n/q'})} \, \int _{|\xi | \Vert u\Vert \ge 1} | \xi |^{1+ (s+n/q)(1+ \frac{1}{s-n/q'})} \, d|\mu |(\xi ) \\< & {} \infty . \end{aligned}$$

In a similar way the remaining part $|\xi | \le 1/ \Vert u\Vert$ of the integral can be treated.

The same estimates also hold for the measures $\mu _r^-$, $\mu _i^+$ and $\mu _i^-$. Thus, the result is obtained by

$$\begin{aligned}& \Vert \sqrt{2\pi } f(u(x))\Vert _{M_{p,q}^s} \\= & {} \Big \Vert \int _{-\infty }^\infty g(\xi ) \, d\mu _r^+ - \int _{-\infty }^\infty g(\xi ) \, d\mu _r^- + i \int _{-\infty }^\infty g(\xi ) \, d\mu _i^+ - i \int _{-\infty }^\infty g(\xi ) \, d\mu _i^- \Big \Vert _{M_{p,q}^s} \\\le & {} \int _{-\infty }^\infty \Vert g(\xi ) \Vert _{M_{p,q}^s} \, d|\mu _r^+| + \int _{-\infty }^\infty \Vert g(\xi ) \Vert _{M_{p,q}^s}\, d|\mu _r^-| \\&\qquad + \int _{-\infty }^\infty \Vert g(\xi )\Vert _{M_{p,q}^s} \, d|\mu _i^+| + \int _{-\infty }^\infty \Vert g(\xi ) \Vert _{M_{p,q}^s} \, d|\mu _i^-| \, , \end{aligned}$$

where every integral on the right-hand side is finite. Thus, the statement is proved.$\blacksquare$

A bit more transparent sufficient conditions can be obtained by using Szasz theorem, see Peetre [22, pp. 9–11] and [27, Proposition 1.7.5]. By $B^{s}_{p,q}(\mathbb {R})$ we denote the Besov spaces on $\mathbb {R}$, see e.g., [33] or [25] for details.

Lemma 4.14

Let $t\ge 0$ and suppose $f \in B^{t+1/2}_{2,1}(\mathbb {R})$. Then the Fourier transform of f is a regular distribution and

$$ \int _{-\infty }^\infty (1+|\xi |^2)^{t/2} |{\mathcal F}f (\xi )|\, d\xi \le c\, \Vert f\Vert _{B^{t+1/2}_{2,1}(\mathbb {R})} $$

follows with some c independent of f.

Based on Lemma 4.14 and Theorem 4.13 one obtains the next result.

Corollary 4.15

Let $1< p < \infty$, $1\le q \le \infty$ and $s>n/q'$. Let $f \in B^{t}_{2,1}(\mathbb {R})$ for some

$$ t\ge \frac{3}{2} + (s+n/q)\, \Big (1+ \frac{1}{s-n/q'}\Big ) $$

and suppose $f(0)=0$. Then the composition operator $T_f: u \mapsto f \circ u$ maps real-valued functions in $M_{p,q}^s$ boundedly into $M_{p,q}^s$.

Proof

Boundedness of $T_f$ follows from Proposition 4.10, the proof of Theorem 4.13 and Lemma 4.14. $\blacksquare$

Remark 4.16

Let $t>0$ be given. A function $f:~\mathbb {R}\rightarrow \mathbb {R}$, m-times continuously differentiable, compactly supported and satisfying $f^m \in \mathrm{Lip}\, \alpha$ for some $\alpha \in (0,1]$, belongs to $B^{t}_{2,1}(\mathbb {R})$ if $t< m+\alpha$.

4.5 One Example

Ruzhansky, Sugimoto, and Wang [26] suggested to study the operator $T_\alpha$ associated to $f_\alpha (t):= t\, |t|^\alpha$, $t\in \mathbb {R}$, with $\alpha >0$. This function belongs locally to the Besov space $B^{\alpha +1+1/p}_{p,\infty }(\mathbb {R})$, $1 \le p \le \infty$, see [25, Lemma 2.3.1/1] for a related case. Let $\psi \in C_0^\infty (\mathbb {R})$ be a smooth cut-off function such that $\psi (x)=1$ if $|x|\le 1$. Then the function

$$ \tilde{f}_{\alpha , \lambda } (t) := \psi (t/\lambda ) \cdot f_\alpha (t), \qquad t \in \mathbb {R}\, , $$

belongs to $B^{\alpha +1+1/p}_{p,\infty }$ for any p, $1 \le p \le \infty$, and any $\lambda >0$. Applying Corollary 4.15 and

$$ u(x)\, |u(x)|^\alpha = \tilde{f}_{\alpha , \lambda } (u(x))\, , \qquad x \in {\mathbb {R}}^n\, , \quad \lambda := \Vert u\Vert _{L_\infty }\, , $$

we find the following.

Corollary 4.17

Let $1< p < \infty$, $1\le q \le \infty$ and $s>n/q'$. Let $\alpha$ be a positive real number such that

$$ (s+n/q)\Big (1+ \frac{1}{s-n/q'}\Big ) < \alpha \, . $$

Then the composition operator $T_\alpha :~u \mapsto u \, |u|^\alpha$ maps real-valued functions in $M_{p,q}^s$ boundedly into $M_{p,q}^s$.

4.6 The Special Case $p=q=2$

Finally, we will have a look onto the special case $M^s_{2,2} = H^s$, $s>n/2$. In Bourdaud, Moussai, S. [4] the set of functions f such that $T_f:~g \mapsto f \circ g$ maps $H^s$ into itself has been characterized.

Proposition 4.18

Let $s> \frac{1}{2} \, \max (n,3)$. For a Borel measurable function $f:\, \mathbb {R}\rightarrow \mathbb {R}$ the composition operator $T_f$ acts on $H^{s}$ if and only if $f(0)=0$ and $f\in H^{s,\ell oc} (\mathbb {R})$.

Concerning our example $T_\alpha$ treated above this yields the following: $T_\alpha$ maps $H^s$ into itself if and only if $\alpha > s-3/2$ (instead of $\alpha > s+\frac{n}{2} + \frac{s+ n/2}{s-n/2}$ as required in Corollary 4.17).

Corollary 4.15 and Corollary 4.17 may be understood as first results about sufficient conditions, not more.

4.7 A Final Remark

The method employed here has been used before in connection with composition operators on Gevrey-modulation spaces and modulation spaces of ultradifferentiable functions, see Bourdaud [3], Bourdaud et al. [5], Reich et al. [5], and Reich [24], for Hörmander-type spaces $B_{p,k}$ we refer to Jornet and Oliaro [16]. It would be desirable to develop this method more systematically.