Let $K(x):=|x|^{-\alpha}$ be a function on $\mathbb{R}^{n}\setminus\{0\}$ with $0<\alpha<n$. Suppose $\phi$ is a $C^{\infty}_{c}(\mathbb{R}^n)$ function such that
$$\text{(i)}\quad \text{supp}\phi \subset \left\{\frac{1}{2}<|x|<2\right\},$$ $$\text{(ii)}\quad\sum_{k\in \mathbb{Z}} \phi(2^{-k}x)=1.$$
Let $$g(\xi):=\mathcal{F}\left(\sum_{k=1}^{\infty}\phi(2^{-k}x)K \right)(\xi),$$ where $\mathcal{F}$ denotes Fourier transform.
Problem: show that $$g(\xi)=\sum_{k=1}^{\infty} \mathcal{F}\left(\phi(2^{-k}x)K \right)(\xi)$$ with convergence in the space of tempered distributions.
Given a Schwartz function $\psi$, we need to show that $$ \sum_{k=1}^{m}\langle \mathcal{F}\left(\phi(2^{-k} x)|x|^{-\alpha}\right),\psi\rangle \to\langle \mathcal{F}\left(\sum_{k=1}^{\infty}\phi(2^{-k} x)|x|^{-\alpha}\right), \psi\rangle.$$ Equivalently, we need to prove that
$$ \sum_{k=1}^{m}\langle \phi(2^{-k} x)|x|^{-\alpha},\widehat{\psi}\rangle=\langle \sum_{k=1}^{m}\phi(2^{-k} x)|x|^{-\alpha},\widehat{\psi}\rangle\\ \to \langle \sum_{k=1}^{\infty}\phi(2^{-k} x)|x|^{-\alpha}, \widehat{\psi}\rangle.$$
My argument is simple: $$\sum_{k=1}^{m}\phi(2^{-k} x)\to \sum_{k=1}^{\infty}\phi(2^{-k} x)$$ which is a bounded continuous function (Recall that $\phi$ is smooth and $\sum_{k=-\infty}^{\infty}\phi(2^{-k} x)=1$). Also, $|\cdot|^{-\alpha} \widehat{\psi} $ is an $L^1$ function. So the passing of the limit is now justified by the dominated convergence theorem.
Did I get the question correctly ? And if so is the solution correct ?
