*Note:* I just realized that using $\omega$ and $w$ might not have been the smartest choice of notation -- Sorry about that.

Let $\renewcommand{\div}{\operatorname{\div}}Q_0, Q_1$ be two real numbers, $1<Q_0<2$ and $2<Q_1$. Let $$\omega\in L^{Q_0}(\mathbb R^2)\cap L^{Q_1}(\mathbb R^2).$$ In particular, since $$L^2(\mathbb R^2)\subset L^{Q_0}(\mathbb R^2)\cap L^{Q_1}(\mathbb R^2),$$ we have $\omega\in L^2(\mathbb R^2)$.

Let $K_2$ denote the *Biot-Savart kernel*, defined by (note that I am writing the coordinates of a vector in $x\in\mathbb R^2$ in the upper index: $x=(x^1,x^2)$)
\begin{align}K_2:\mathbb R^2\setminus\{(0,0)\}&\to\mathbb R^2,\\ (x^1,x^2)&\mapsto\frac{(-x^2,x^1)}{(x^1)^2+(x^2)^2}.\end{align}

Define $w:\mathbb R^2\to\mathbb R^2$ as the convolution $w=K_2*\omega$. More explicitly if we write $K_2=(K_2^1, K_2^2)$, then the first coordinate of $w$ is $w^1=K_2^1*\omega$ and the second coordinate of $w$ is $w^2=K_2^2*\omega$. I want to prove that, in the **weak sense**, we have $$\operatorname{div}(w)=0 \quad \text{and}\quad \operatorname{curl}(w)=\omega.$$

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My ideas for the "classical case": Assume that $\omega\in\mathcal C^1(\mathbb R^2)$ and that $\omega$ decays rapidly enough as $|x|\to\infty$, say $|\omega(x)|\lesssim\frac1{|x|}$ asymptotically. Then I would like to differentiate under the integral sign to show that $$\operatorname{div}(K_2*\omega)=(\operatorname{div} K_2)*\omega=0*\omega,$$ *however*, $K_2$ is not differentiable at $0$ so I don't know how to justify the differentiation under the integral sign.

To show that $\operatorname{curl}w=\omega$, I would introduce the real-valued function $$\psi(x)\overset{\text{Def.}}=\int_{\mathbb R^2}\ln\lvert x-y\rvert\omega(y)\,\mathrm dy$$ and use differentiation under the integral sign to show that $\frac{\partial}{\partial x^2}\psi = -w^1$ and $\frac{\partial}{\partial x^1}\psi=w^2$, (*once again we have the problem that $\ln$ is not differentiable at $0$.*)

If we have this, then by a classical result (solutions of the Poisson equation) we know that $\Delta\psi=\omega$ and using $\Delta\psi=\operatorname{curl}(\nabla^\bot w)$ we are done.