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 $0$.
