Friedrich's second inequality (or Maxwell Estimate or Gaffney’s inequality in the literature) is referred as follows: For all $\mathbf{u} \in H^1(\Omega)^2$ satisfying either $\mathbf{n} \cdot \mathbf{u}=0$ or $\mathbf{n} \times \mathbf{u}=\mathbf{0}$ on $\partial \Omega$ where $\Omega$ is a simply connected domain with Lipchitz boundary, then $$ \|\mathbf{u}\|_{L^2(\Omega)} \leq C_1\left(\|\nabla \cdot \mathbf{u}\|_{L^2(\Omega)}+\|\nabla \times \mathbf{u}\|_{L^2(\Omega)}\right). $$
My question is that if the boundary condition is satisfied, we only know the mean value of $\mathbf{u}$ is $0$ or simply replace the LHS with $\|\mathbf{u}-\frac{1}{|\Omega|}\int_{\Omega}\mathbf{u}\|_{L^2(\Omega)}$. Does the inequality above hold?
Or at least, particularly, if $\mathbf{u}$ is divergence-free, do we have $$ \|\mathbf{u}-\frac{1}{|\Omega|}\int_{\Omega}\mathbf{u}\|_{L^2(\Omega)} \leq C_1\|\nabla \times \mathbf{u}\|_{L^2(\Omega)}. $$
Thank you in advance!
I believe this is false.
Let $u=\nabla F$, where $F:\mathbb{R}^n\to \mathbb{R}$ is non-constant and harmonic. Then $\nabla \cdot u=0$ and $\nabla \times u=0$, and so the left hand side of the inequality is positive, but the right hand side is 0.
The boundary restrictions rule out the above example by imposing some boundary conditions on $F$: $n\cdot u=0$, for example, is equivalent to $\partial_n F=0$ (the normal derivative). By uniqueness for the Neumann problem, the only such $F$ is constant, and so $u\equiv0$ which leads to no contradiction.