Let $B = B(0,4/3)$, $C = \{x \in \mathbb R^d : 3/4 \leq \|x\|_2 \leq 8/3\}$ and $\tilde{C} = \{x \in \mathbb R^d : 1/12 \leq \|x\|_2 \leq 10/3\}$.
For a fixed Littlewood-Paley decomposition $\chi \in \mathcal{D}(B)$, $\varphi \in \mathcal{D}(C)$, we denote by $\dot{S}_j := \chi(2^{-j}D)$ the homogeneous low-frequency cut-off operator and $\dot{\Delta}_j := \varphi(2^{-j}D)$ the homogeneous dyadic block operator. Moreover, we define the following subspace of tempered distributions:
$$S'_h = \{u \in S'(\mathbb R^d) : \lim \limits_{j \to -\infty}\|\dot{S_j}u\|_{\mathcal{L}^{\infty}(\mathbb R^d)} = 0\}$$
Here, I'm reading and using the notations from "Fourier Analysis and Nonlinear Partial Differential Equations", Chapter 2.2 p.59, by H. Bahouri.
In Chapter 2.6 p.86, the homogeneous paraproduct of two tempered distributions $u, v \in S'(\mathbb R^d)$ is formally defined as:
$$\dot{T}_uv = \sum_{j \in \mathbb Z}\dot{S}_{j-1}u \dot{\Delta}_j v$$
Remark 2.46 from p.86 claims that if $u,v \in S'_h$, then $\dot{T}_uv$ makes sense in $S'(\mathbb R^d)$, i.e. the sum $\langle \dot{T}_uv , f \rangle$ converges in $\mathbb R$ for any $f \in S(\mathbb R^d)$. However, I'm unable to prove such a claim or find a reference proving it.
I know that $\dot{S}_{j-1}u \dot{\Delta}_j v$ is a tempered distribution whose Fourier transform is supported in the annulus $2^j \tilde{C}$. In particular, it is a smooth function with at most polynomial growth (independent of $j$, I believe). Similar statements can be made about $\dot{S}_{j-1}u$ and $\dot{\Delta}_j v$.
I also know that for $u \in S'_h$, we have the decomposition $$u = \lim \limits_{j \to +\infty} \dot{S}_j u = \sum_{j \in \mathbb Z} \dot{\Delta}_ju$$ in the space $S'(\mathbb R^d)$.
However, I didn't succeed in combining any of these pieces of information together.
