Integration by parts with $\Delta^{-1}$ and $\nabla^{-1}$

Solution 1:

You can't get this kind of estimate.

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

2. 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.