I have the following elliptic problem: $$ \Delta u = \operatorname{div}\operatorname{div}S, $$ where $S=(S_{i,j})\colon \mathbb{T}^n\to \mathbb{R}^{n\times n} $ is bounded and $\mathbb{T}^n$ is the $n$-dimensional torus. My goal is to prove that $u\in \operatorname{BMO}(\mathbb{T}^n)$.
Now if I worked in $\mathbb{R}^n$ instead of the torus, it would be a direct consequence of Calderon-Zygmund theorem. I would have $$ u(x) = \int_{\mathbb{R}^n} \frac{\operatorname{div}\operatorname{div}S(y)}{|x-y|^{n-2}} dy$$ and after integration by parts I get $u = K\ast S :=\sum_{i,j=1}^n K_{i,j}\ast S_{i,j}$, where $K_{i,j}$ are singular kernels. Therefore $\|K_{i,j}\ast S_{i,j}\|_\mathrm{BMO}\leq C\|S_{i,j}\|_{L^\infty}$.
Now the question: how to obtain the analogous result on the torus?
I tried to use the Fourier transform to solve the equation and in the end I got the formula $$ u(x) = -\sum_{k\in\mathbb{Z}^d}\frac{k\otimes k}{|k|^2}:\hat{S}(k)e^{-k\cdot x}. $$ This means that:
a) $ u = K\ast S $ for $K(x)=\sum_{k\in\mathbb{Z}^d}\frac{k\otimes k}{|k|^2}e^{ikx}$. I tried to show that $K$ is a singular kernel and apply Calderon-Zygmund theorem straight away, but I don't know how to show that $|K(x)|\leq \frac{C}{|x|^n}$.
b) in terms of Fourier multipliers, $u$ corresponds to a multiplier $m(k)=\frac{k\otimes k}{|k|^2}$ (i.e. $\hat{u}(k) = m(k)\hat{S}(k)$). Are there some $L^\infty\text{-}\operatorname{BMO}$ estimates in this setting? There are plenty of results concerning $L^p$ estimates, but I had problem in finding anything more.
I would be very grateful for any suggestions or references where I could find the answer.
