Find $g$ such that $\operatorname{div}g=f$ in the whole space $\mathbb{R}^d$. (Bogovskii operator lemma)

Question

For a fixed $f \in L^2(Q)$ with $Q$ a bounded domain of $\mathbb{R}^d$, I know we can find a function $g \in H^1_0(Q)$ such that

$$\operatorname{div}g =f \enspace\text{ in } Q$$ in the sense of distribution. This is only possible if we assume the compatibility condition :

$$\int_Q f =0 \quad \quad \quad (\star)$$

This theorem is also known as the Bogovskii operator, and there exist milder versions (with $f \in L^p(Q)$ essentially) that are well referenced in the book of Galdi, in the dedicated chapter :

Galdi, G. (2011). An introduction to the mathematical theory of the Navier-Stokes equations: Steady-state problems.

I was wondering if the following is true :

Let $f \in L^2(\mathbb{R}^d)$, can we find $g$ (for exemple in $H^1(\mathbb{R}^d)$) such that $\operatorname{div}g =f$ in $\mathbb{R}$ ? Is there a compatibility condition like $(\star)$ ?

Any references or help is welcomed.

Answer 1

If you assume that $f \in L^{\frac{2d}{d+2}}(\mathbb{R}^d) \cap L^2((\mathbb{R}^d)$ then the answer is yes (for some $g \in L^2((\mathbb{R}^d)$ : see Lemma 1.6.2 Chapter II in the book of Sohr - The Navier-Stokes equations: An elementary functional analytic approach.