# Completeness of Besov spaces

Problem: Let us recall that: $$\dot{B}^{-\sigma}_{\infty,\infty}=\{u \in S'(\mathbb{R}^d): \|u\|_{\dot{B}^{-\sigma}_{\infty,\infty}} < \infty\}$$ where $$\|u\|_{\dot{B}^{-\sigma}_{\infty,\infty}}=\sup\limits_{A>0}\{A^{d-\sigma}\| \theta(A\cdot)* u\|_{L^{\infty}}\}$$ and $$S'(\mathbb{R}^d)$$ is the set of tempered distributions. I want to understand two things:

• why $$\dot{B}^{-\sigma}_{\infty,\infty}$$ is complete (i.e. a Banach space)
• why if I choose another $$\theta'$$ I obtain an equivalent norm over $$\dot{B}^{-\sigma}_{\infty,\infty}$$.

Attempt. I tried exploiting the Fourier transorm of a tempered distribution: $$\theta(A\cdot)* u= \mathcal{F}^{-1}\mathcal{F}(\theta(A\cdot)* u)=\mathcal{F}^{-1}\bigg(A^{-d}\mathcal{F}\Big(\theta\Big(\frac{\cdot}{A}\Big)\Big) \mathcal{F}(u)\bigg)$$ but I cannot go on. Any help or reference will be appreciate. Please notice that I would like a self-contained proof instead one for general Besov spaces as $$\dot{B}^{\sigma}_{p,q}$$.

Edit: we assume $$\mathcal{F}(\theta)$$ is such that $$\mathcal{F}(\theta) \in C^{\infty}_c(\mathbb{R}^d)$$, $$0 \leq \mathcal{F}(\theta) \leq 1$$ and $$\mathcal{F}(\theta)=1$$ in a neighborhood of $$0$$.

• What are the assumptions on $\theta$? Dec 1, 2022 at 21:24
• @Jose27 I edited the quesiton Dec 1, 2022 at 21:41
• I posted an answer about the completeness of this space in this post: math.stackexchange.com/questions/4127104/… Nov 12, 2023 at 18:12