Equivalent Characterization on Besov Space

In Sobolev spaces, it is known that ∥f ∥H2ðR2Þ ~ ∥f ∥L2ðR2Þ + ∑i=1∥∂ 2 f /∂xi ∥L2ðR2Þ, where ∥f ∥H2ðR2Þ ≕ ∥f ∥L2ðR2Þ + ∥∂x1∂x2 f ∥ L2ðR2Þ+∥∂x1 f ∥L2ðR2Þ+∥∂ 2 x2 f ∥L2ðR2Þ: Note that on the right hand side of the definition ∥f ∥H2ðR2Þ, it contains the mixed derivative norm ∥∂x1∂x2 f ∥L2ðR2Þ: 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 ∥f ∥Bsp,rðRnÞ ~∑ n j=1 ∥f ∥Bsp,r,x j ðRnÞ 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 ∥f ∥Bsp,rðRnÞ in solving partial differential equations, it is equivalent to estimate ∥f ∥Bsp,r,x j ðRnÞ individually. For the other equivalent characterizations for Besov spaces, refer to [4–7] and the references therein.

In this paper, we aim to prove ∥f ∥ B s p,r ðℝ n Þ~∑ n j=1 ∥f ∥ B s p,r,x j ðℝ n Þ 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 ∥f ∥ B s p,r ðℝ n Þ in solving partial differential equations, it is equivalent to estimate ∥f ∥ B s p,r,x j ðℝ n Þ individually. For the other equivalent characterizations for Besov spaces, refer to [4][5][6][7] and the references therein.

Preliminaries
We first recall definitions on Besov spaces, see [8]. Given f ∈ S which is the Schwartz function, its Fourier transform and its inverse Fourier transform is defined by F −1 f ðxÞ = f ð−xÞ: We consider φ ∈ S satisfying supp φ ⊂ fξ ∈ ℝ n : 1/2≤|ξ| ≤2g: Setting φ j ðξÞ = φð2 −j ξÞ with j = f1, 2,⋯,g, we can adjust the normalization constant in front of φ and choose φ 0 ∈ S satisfying supp φ 0 ⊂ fξ ∈ ℝ n : |ξ|≤2g, such that We observe with the usual interpretation for p = ∞ or r = ∞: Throughout this paper, all the function spaces are defined on Euclidean space ℝ n ; 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 n ≥ 2, and take φ as in the definition of Besov space. Then, there exist functions χ j ∈ Sðℝ n Þðj = 1,⋯,nÞ, such that Next, we recall the real interpolation characterization for Besov spaces.
Remark 3. We also have Its proof can be repeated the process of Lemma 2 completely.

Equivalent Characterization
Now, we are in the position to state and prove our theorems. Firstly, we apply the Fourier multiplier [9] to prove that H s p ðℝ n Þ = T n j=1 H s p,x j ðℝ n Þ directly; H s p space has an advantage that the factor ð1 + jξj 2 Þ s/2 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. where Proof. On the one hand, if f ∈ H s p , i.e., ∥F −1 hξi s F f ∥ p < ∞ where hξi = ð1 + jξj 2 Þ 1/2 : Note that, for any j = 1, ⋯, n, we have Next, we just need to show that m 1 ðξÞ = hξ j i s /hξi s is an L p multiplier. To prove the assertion, we introduce an auxiliary function on ℝ n+1 defined bỹ It is easy to verify thatm 1 is homogeneous of degree 0 and smooth on ℝ n+1 \ f0g: The derivatives ∂ βm 1 are homogeneous of degree −|β | and satisfy whenever ðξ, tÞ ≠ 0 and β is a multiindex of n + 1 variables. In particular, taking β = ðα, 0Þ, we obtain and setting t = 1, we deduce that |∂ α m 1 ðξÞ | ≤C α ð1+jξj 2 Þ −|α|/2 ≤ C α jξj −|α| , which implies that m 1 ðξÞ is an L p Fourier multiplier by the Mihlin-H€ omander theorem [9] (page 446).
On the other hand, assume f ∈ T n j=1 H s p,x j , that is, ∥F −1 hξ j i s F f ∥ p < ∞, for any j = 1, ⋯, n: Note that, Similarly, we can verify that m 2 ðξÞ = hξi s /∑ n j=1 hξ j i s is an L p 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 and φ j k is the dyadic block of the unit decomposition for the jth variable as in the definition of Besov spaces.
Proof. We split the proof into the following two steps: Step I. To prove