The Besov spaces can be defined using the dyadic decomposition. For this purpose, let $\varphi \in C^\infty_c(\mathbb R^n)$ be such $ {\rm supp}(\varphi) \subseteq \left\{x; \, \frac 12< |x| < 2\right\}$ and denote $$\varphi_j(x) = \varphi\left(2^{-j}x\right)\quad\text{for} \quad j\in \mathbb N^\ast\quad \text{and}\quad \varphi_0 = 1-\sum_{j=1}^\infty \varphi_j.$$
In this paper. For any $1\leq p, q \leq \infty,$ the Besov space $B^0_{pq}$ is defined as the closure of $C^\infty_c(\mathbb R^n)$ with respect to the norm $$\|u\|_{B_{pq}^0} := \left(\sum_{j=0}^\infty \left\| \mathfrak{F}^{-1} (\varphi_j \hat u)\right\|_p^q \right)^{\frac 1q},$$ where $\mathfrak{F}^{-1}$ is the inverse Fourier transform. Can this definition be extended for $B^s_{pq}$ with $s\in \mathbb R,$ so that $B^s_{pq}$ is the closure of $C^\infty_c(\mathbb R^n)$ with respect to the norm $$\|u\|_{B_{pq}^s} := \left(\sum_{j=0}^\infty 2^{jsq} \left\| \mathfrak{F}^{-1} (\varphi_j \hat u)\right\|_p^q \right)^{\frac 1q}?$$
Or there is a condition on $s$? Thank you for any hint.