Integration by parts with $\Delta^{-1}$ and $\nabla^{-1}$ Let $u:\mathbb R^n \to \mathbb R$ be smooth and $A:\mathbb R \to \mathbb R^n$ be Lipschitz.
How can we estimate the quantity $$\int_{\mathbb R^n} \nabla \cdot (A(u)) \Delta^{-1}udx$$ from below
in terms of $\|\nabla^{-1}u\|_{2}$ and the Lipschitz norm of $A$?
Is an estimate from above also available?
In the simpler case when we replace $A(u)=a(x)u$ with a Lipschitz function $a:\mathbb R^n \to \mathbb R^n$ with $\mathrm{div}(a) = 0$,
I already know how to do it: let $\phi=\Delta^{-1} u$ and compute
\begin{align*}
\begin{aligned}
\int a \cdot \nabla u \Delta^{-1} u d x &=\int \nabla \cdot(a u) \Delta^{-1} u \\
&=\int \nabla \cdot(a \Delta \phi) \phi \\
&=-\int \sum_{i, j} \partial_{i}\left(a^{i} \partial_{j j} \phi\right) \phi \\
&=-\int \sum_{i, j} a^{i} \partial_{j j} \phi \partial_{i} \phi \\
&=\int \sum_{i, j} \partial_{j}\left(a^{i} \partial_{i} \phi\right) \partial_{j} \phi \\
&=\int \sum_{i, j} \partial_{j} a^{i}\left|\partial_{i} \phi\right|^{2}+\int u^{i} \partial_{i} \frac{\left|\partial_{j} \phi\right|^{2}}{2} d x \\
&=\int \nabla^{-1} u\cdot \nabla a \cdot \nabla^{-1} u d x
\end{aligned}
\end{align*}
which gives the required estimate in terms of $\|\nabla^{-1}u\|_{2}$ and $\|\nabla a\|_\infty$, that is $... \ge - \|\nabla^{-1}u\|_{2}^2\|\nabla a\|_\infty$ and $... \le \|\nabla^{-1}u\|_{2}^2\|\nabla a\|_\infty$
 A: You can't get this kind of estimate.

*

*One probably needs to work with inhomogeneous Sobolev spaces over the domain $\mathbb{R}^n$.


*Consider now the case $n = 1$. Let $w$ be such that $u = w_{xx}$, and take $A(z) = z$. Then formally $\Delta^{-1} u \sim w, \; \nabla^{-1} u \sim w_x$. You are now asking for an estimate of the form
$$
\int_\Omega w_{xxx} \cdot w \le C \|w_x\|_{L^2}^2
$$
and that clearly does not work.
Added:
To clarify the first remark, $\Delta^{-1} u$ is not defined as an $L^1$ function for general smooth $u \in C_0^\infty(\mathbb{R}^n)$. For example, if $w = \Delta^{-1} u$ for such a $u$, then $\Delta w = u$ and hence $\int_{\mathbb{R}^n} u = \int_{\mathbb{R}^n} \Delta w = 0$ by integration by parts. Rather, $u$ must satisfy additional conditions.
More specifically assume that $u$ is a Schwartz function and therefore $\hat u$ is also a Schwartz function. For $\Delta^{-1} u$ to exist as an $L^2$ function, $ \xi \mapsto |\xi|^{-2} \hat u(\xi)$ must be an $L^2$ function, which can only be true if $\hat u(\xi) = O(|\xi|^2)$ as $\xi \to 0$, at least for $n \le 2$. For Schwartz functions, the latter condition is equivalent to requiring $0 = \int u(x) dx  = \int x_j u(x) dx$ for all $j$.
Regarding remark 2: Consider still the case $n=1$ and now take $A$ to be a general Lipschitz function. Using otherwise the same notation and setup, and assuming the integration by parts is justified, you are asking for an estimate relating
$$
\int_\Omega \partial_x (A(w_{xx})) \cdot w  = - \int_\mathbb{R} A(w_{xx}) \cdot w_x  dx \quad \text{and} \quad  \|w_x\|_{L^2}^2
$$
which clearly cannot be expected.
