#### Abstract

In this paper, we give an equivalent characterization of the Besov space. This reveals the equivalent relation between the mixed derivative norm and single-variable norm. Fourier multiplier, real interpolation, and Littlewood-Paley decomposition are applied.

#### 1. Introduction

In Sobolev spaces, it is known that where Note that on the right hand side of the definition , it contains the mixed derivative norm This mixed derivative norm would make the calculation more complicated or even infeasible to estimate partial differential equations with some anisotropy property, like Vlasov-Poisson equation [1, 2], in fractional Sobolev space [3]. So, separating variables becomes necessary and meaningful.

In this paper, we aim to prove which realizes the separation, i.e., the right hand side does not contain the “mixed derivative” term, it only contains fractional derivative with respect to a single variable for each term. Thus, when it comes to estimate in solving partial differential equations, it is equivalent to estimate individually. For the other equivalent characterizations for Besov spaces, refer to [4–7] and the references therein.

#### 2. Preliminaries

We first recall definitions on Besov spaces, see [8]. Given which is the Schwartz function, its Fourier transform is defined by and its inverse Fourier transform is defined by

We consider satisfying Setting with , we can adjust the normalization constant in front of and choose satisfying such that

We observe

Given , we denote For , then we define the inhomogeneous Besov space by with the usual interpretation for or Throughout this paper, all the function spaces are defined on Euclidean space ; we will omit it whenever there is no confusion.

Next, we would like to present some known results which will be used later. The first one is the unit decomposition.

Lemma 1 (see [8], page 145). *Assume that and take as in the definition of Besov space. Then, there exist functions , such that
**Next, we recall the real interpolation characterization for Besov spaces.*

Lemma 2 (see [8], page 142). *Suppose where We have
*

*Remark 3. *We also have
Its proof can be repeated the process of Lemma 2 completely.

#### 3. Equivalent Characterization

Now, we are in the position to state and prove our theorems. Firstly, we apply the Fourier multiplier [9] to prove that directly; space has an advantage that the factor is positive everywhere, which is fundamentally important when applying the Fourier multiplier theorem.

For the sake of brevity, we denote

We have the following equivalent norm theorem in Sobolev spaces.

Theorem 4. *Suppose We have
where
*

*Proof. *On the one hand, if , i.e., where Note that, for any , we have
Next, we just need to show that is an multiplier. To prove the assertion, we introduce an auxiliary function on defined by
It is easy to verify that is homogeneous of degree 0 and smooth on The derivatives are homogeneous of degree and satisfy
whenever and is a multiindex of variables. In particular, taking , we obtain
and setting , we deduce that , which implies that is an Fourier multiplier by the Mihlin-Hmander theorem [9] (page 446).

On the other hand, assume , that is, Note that,
Similarly, we can verify that is an Fourier multiplier which finishes the proof of Theorem 4.

We return to prove the equivalent characterization on Besov spaces. However, we cannot do the same trick as in space since is not positive everywhere as Fortunately, we have , see Lemma 2. This observation is favourable to prove the equivalent relation in one direction; however, for the other direction, we need a more delicate technique, in fact, we establish an identity by applying the Littlewood-Paley decomposition [10], which is very important in our proof. In what follows, means there exists a constant independent of the main parameters such that means and

Theorem 5. *Suppose We have
where and is the dyadic block of the unit decomposition for the th variable as in the definition of Besov spaces.*

*Proof. *We split the proof into the following two steps:

Step I. To prove
Assume , by the real interpolation Lemma 2, we have
where , and we applied the equivalent norm for the interpolation space , see [8] ((3) page 39 and (5) page 40).

By Remark 3, we obtain, for any combining (18) and (19), it follows that
the arbitrariness of implies that (17) holds.

Step II. To prove
For it is trival.

For , we need the following key claim.

*Claim. *There exists a positive integer depending on only such that
where
which is the dyadic block for variable, is the usual dyadic block as in the definition of Besov spaces, and is the same as in Lemma 1.

*Proof of Claim. *By Lemma 1, we have

Note
In order to get , for any chosen and , we must have
which implies that with , ending the proof of the claim. With this claim in mind, we get
where we used the fact that is the Fourier multiplier.

With Young’s inequality [11], taking the norm on both sides of (26) yields that
which implies (21) holds; thus, we complete the proof our main theorem.

*Remark 6. *The methods could be adapted to the weighted Sobolev spaces and weighted Besov space, or even in the anisotropic function space.

#### Data Availability

The data in this paper is available on request. Please contact Jingchun Chen at [email protected].

#### Conflicts of Interest

The authors declare that they have no conflicts of interest.