Let
- $d\in\mathbb N$
- $\Lambda$ be open
- $\mathcal V:=\left\{\phi\in C_c^\infty(\Lambda,\mathbb R^d):\nabla\cdot\phi=0\right\}$, $$V:=\overline{\mathcal V}^{\left\|\;\cdot\;\right\|_{H^1(\Lambda,\:\mathbb R^d)}}$$ and $$H:=\overline{\mathcal V}^{\left\|\;\cdot\;\right\|_{L^2(\Lambda,\:\mathbb R^d)}}$$
- $\operatorname P_H$ denote the orthogonal projection from $L^2(\Lambda,\mathbb R^d)$ onto $H$
Note that
- $H$ is a closed subspace of $L^2(\Lambda,\mathbb R^d)$
- $V$ is a closed subspace of $H_0^1(\Lambda,\mathbb R^d)$
- $V=H\cap H_0^1(\Lambda,\mathbb R^d)$
If we assume that $\Lambda$ is bounded and $\partial\Lambda$ is Lipschitz, then $$V=\left\{u\in H_0^1(\Lambda,\mathbb R^d):\nabla\cdot u=0\right\}\;.\tag 1$$ Moreover, if $d\le 4$, then $$u\cdot\nabla v\in L^2(\Lambda,\mathbb R^d)\;\;\;\text{for all }u\in H_0^1(\Lambda,\mathbb R^d)\text{ and }v\in H_0^2(\Lambda,\mathbb R^d)\;.\tag 2$$
Question:$\;\;\;$Let $u\in V\cap H^2(\Lambda,\mathbb R^d)$. I want to show that $$\text P_H(u\cdot\nabla u)\in V\;.\tag 3$$ Is that possible?
It's clear that $u\cdot\nabla u$ doesn't need to belong to $V$ and hence $\text P_H(u\cdot\nabla u)$ doesn't need to equal $u\cdot\nabla u$.
However, if I'm not terribly wrong, the conclusion should be trivial: We know that $u\cdot\nabla u\in H_0^1(\Lambda,\mathbb R^d)$ and we should be able to conclude (this is the only part where I'm not sure) $\text P_H(u\cdot\nabla u)\in H_0^1(\Lambda,\mathbb R^d)$. The conclusion would now follow by (3.).