An inequality about Besov norm: transfer the index $s$ to the multiplier $\langle \xi \rangle^s$

Question

I try to prove the following inequality about Besov norm:

Let $s,\sigma > 0$ and $r \geq 1$, show that there is a constant $C = C(d,s,\sigma,r)$ such that for any $u \in \mathcal{S}(\mathbb{R}^d)$ (the space of Schwartz functions)

$\| u \|_{B^{s-\sigma}_{r,2}} \leq C \| \mathcal{F}^{-1}(\langle \xi \rangle^s \mathcal{F} u)\|_{B^{-\sigma}_{r,2}}$ or simply $\| u \|_{B^{s-\sigma}_{r,2}} \lesssim \| \mathcal{F}^{-1}(\langle \xi \rangle^s \mathcal{F}u)\|_{B^{-\sigma}_{r,2}}$,

where $\langle \xi \rangle = (1 + |\xi|^2)^{\frac{1}{2}}$, $\mathcal{F}$ means the Fourier transform and $\| \cdot \|_{B^k_{p,q}}$ is the Besov norm defined by Littlewood-Paley multiplier.

More precisely, let $\eta$ be a smooth bump function that equals to 1 on unit ball $B(0,1)$ and vanish outside $B(0,2)$. Let $\phi_k(\xi) = \eta(\frac{\xi}{2^k}) - \eta(\frac{\xi}{2^{k-1}})$. I want to prove

$\| \mathcal{F}^{-1}(\eta \mathcal{F}u) \|_{L^r(\mathbb{R}^d)} + \Big( \sum_{j=1}^\infty \big( 2^{(s-\sigma)j} \| \mathcal{F}^{-1} ( \phi_j \mathcal{F}u)\|_{L^r} \big)^2 \Big)^{\frac{1}{2}} \\ \leq C \Big\{ \| \mathcal{F}^{-1}(\eta(\xi) \langle \xi \rangle^s \mathcal{F}u(\xi)) \|_{L^r(\mathbb{R}^d)} + \Big( \sum_{j=1}^\infty \big( 2^{-\sigma j} \| \mathcal{F}^{-1} ( \phi_j(\xi) \langle \xi \rangle^s \mathcal{F}u(\xi))\|_{L^r} \big)^2 \Big)^{\frac{1}{2}} \Big\}$

(Note that the Besov norm can be proved to be independent of the choice of $\eta$.)

This problem comes from Machihara-Nakanishi-Ozawa's paper in 2003. In (2.24), I think they apply their Strichartz estimates (2.3)(2.4) in Lemma 3. If the inequality I proposed is true, then part of (2.24) is proved, for example, in part of $X^s$ norm of $U(t)\psi_0$,

$\| K_\pm(t) \psi_0 \|_{L^2(\mathbb{R};B^{s-\sigma}_{r,2})} \lesssim \| \mathcal{F}^{-1} \langle \xi \rangle^s \mathcal{F}(K_\pm(t) \psi_0 )\|_{L^2(\mathbb{R};B^{-\sigma}_{r,2})} \\\;\;\;\;\;\; = \|K_\pm(t) (\mathcal{F}^{-1} \langle \xi \rangle^s \mathcal{F}\psi_0 )\|_{L^2(\mathbb{R};B^{-\sigma}_{r,2})} \lesssim \| \mathcal{F}^{-1} \langle \xi \rangle^s \mathcal{F}\psi_0 \|_{L^2(\mathbb{R}^3)} = \| \psi_0 \|_{H^s(\mathbb{R}^3)}.$

Many thanks for any discussion or hint.

Answer 1

Solved. It's an easy consequence of the following inequality.

(See Bergh-Löfström's book:Interpolation Spaces, Lemma 6.2.1)

Let $s \in \mathbb{R}$ and $1 \leq p \leq \infty$. Then $\| \mathcal{F}^{-1}(\langle \xi \rangle^s \phi_k \mathcal{F} u) \|_{L^p(\mathbb{R}^d)} \leq C(s,d) 2^{sk} \| \mathcal{F}^{-1}(\phi_k \mathcal{F} u)\|_{L^p(\mathbb{R}^d)}$ for all tempered distributions $u$ with $\mathcal{F}^{-1}(\phi_k \mathcal{F} u) \in L^p(\mathbb{R}^d)$ and $k \geq 0$ (where $\phi_0 := \eta$).