Thus the Fourier transform of $\phi_i$ satisfies for all ${|\alpha|}=d+1$, $${|k^\alpha \hat\phi_i(k)|}
    \sim
    {\Big|\int \partial_z^\alpha\phi_i(z) e^{-i2\pi k\cdot z}{\,\mathrm{d}}z\Big|} \\
    \leq\int{|\partial_z^\alpha\phi_i(z)|}{\,\mathrm{d}}z<\infty,$$ and hence ${|\hat\phi_i(k)|}\lesssim{|k|}^{-1-d}$. If $\Phi_i$ is the $1$-periodic extension of $\phi_i$, its Fourier coefficients satisfy this same estimate. In particular, these Fourier coefficient are in $\ell^1({\mathbb{Z}}^d)$. Recalling that $\phi_i(z)$ agrees with $z_i{|z|}^{d-1}$ in $Q(0,\frac14)$, we hence have shown that $$\label{eq:FourierExp}
    z_i{|z|}^{d-1}=\sum_{k\in{\mathbb{Z}}^d} a_i(k)e^{i2\pi k\cdot z},\qquad\forall z\in Q(0,\frac14),$$ where $\sum_{k\in{\mathbb{Z}}^d}{|a_i(k)|}<\infty$.

And observe that we only need to apply the formula when $x,y\in Q$, a given cube. By composing back and forth with dilations in addition to translations, we may assume that $Q=Q(0,\tfrac18)$. Then if $x,y\in Q$, we see that $x-y\in Q(0,\tfrac14)$, where is valid. Substituting with $z=x-y$ into , we obtain $$1=\sum_{i=1}^d\frac{(x_i-y_i)}{{|x-y|}^{d+1}}\sum_{k\in{\mathbb{Z}}^d}a_i(k)e^{i2\pi k\cdot(x-y)}
    =\sum_{k\in{\mathbb{Z}}^d}a_i(k)\sum_{i=1}^d\frac{(x_i-y_i)}{{|x-y|}^{d+1}}e^{i2\pi k\cdot x} e^{-i2\pi k\cdot y},$$ which is a convergent series of expressions of the desired form, namely the Riesz transform kernel multiplied by (bounded) functions that depend only on $x$ or only on $y$. After this, the argument can be concluded in the same way as before.

This Fourier series idea is based on Svante Janson [@JANS].

We gather two more basic estimates.

\[magiclemma\] Let $Q$ be a cube and $b_i\in  L^3_{loc}, i =1,2,$ be such that $\dashint_Qb_i=0$. 