Let $\phi$ be a $C^{\infty}$ function on $\mathbb R^{n}$ with
$ \operatorname{supp} \phi \subset \{\xi \in \mathbb R^{n}: |\xi|\leq 2, \phi(\xi)=1$ if $|\xi|\leq 1.$
Let $j\in \mathbb N$ and $$\phi_{j}(\xi)= \phi(2^{-j}\xi)- \phi(2^{-j+1}\xi), (\xi \in \mathbb R^{n})$$
Then we have $$\operatorname{supp} \phi_{j} \subset \{\xi\in \mathbb R^{n}: 2^{j-1}\leq |\xi| \leq 2^{j+1} \}, j\in \mathbb N $$ and, with $\phi_{0}=\phi,$ $$\sum_{k=0}^{\infty} \phi_{k}(\xi)=1, \text{if} \ \xi\in \mathbb R^{n}.$$ In other words, $\{\phi_{k}\}_{k\in\mathbb N_{0}}$ is a resolution of unity. Let $f\in \mathcal{S'}$ then $\phi_{k}(D)f(x)= (\phi_{k}\hat{f})^{\vee}(x), k\in \mathbb N_{0}, x \in \mathbb R^{n}$ is an entire analytic function.
My Question is: (1) How to show $ \sum_{k=0}^{\infty} \phi_{k}(D)f;$ convergence in $\mathcal{S'},$ and $f=\sum_{k=0}^{\infty} \phi_{k}(D)f,$ that is, $f$ can be decompose by entire analytic functions.
Motivation: The sequences $(\phi_{k}\hat{f})^{\vee}$ generats Besove spaces if it in $L^{p}(\ell^{q}).$
My attempt: [A sequence $\{f_{j}\}_{j\in \mathbb N}$ in $\mathcal{S'}$ is said to converge to $f\in \mathcal{S'}$ if $\langle f_{j}, \psi \rangle \to \langle f, \psi\rangle$ for all $\psi \in \mathcal{S}$ as $j\to \infty.$]
Formally, we put $F=\sum_{k=0}^{\infty} \phi_{k}^{\vee}\ast f.$ I guess, first I should show: $F\in \mathcal{S'}$, that is, $\langle F, \psi_{1}+\psi_{2}\rangle =\langle F, \psi_{1}\rangle + \langle F, \psi_{2}\rangle$ for $\psi_{1}, \psi_{2} \in \mathcal{S}$ and for any sequence $\{\psi_{j}\}_{j\in \mathbb N}$ in $\mathcal{S}$ that converges in $\mathcal{S}$ to zero, the sequence of numbers $\langle F, \psi_{j}\rangle$ converges to 0.
For any $\psi \in \mathcal{S},$ we note that, $\langle \sum_{k=0}^{n}\phi_{k}^{\vee}\ast f, \psi \rangle = \sum_{k=0}^{n}\langle \phi_{k}^{\vee}\ast f, \psi \rangle$