Let
- $\Lambda\subseteq\mathbb R^d$ be open
- $\mathcal V:=\left\{v\in C_c^\infty(\Lambda)^d:\nabla\cdot v=0\right\}$, $$V:=\overline{\mathcal V}^{\left\|\;\cdot\;\right\|_{H^1(\Lambda,\:\mathbb R^d)}}$$ and $$H:=\overline{\mathcal V}^{\left\|\;\cdot\;\right\|_{L^2(\Lambda,\:\mathbb R^d)}}=\overline V^{\left\|\;\cdot\;\right\|_{L^2(\Lambda,\:\mathbb R^d)}}\tag 1$$
Note that $H$ is a closed subspace of $L^2(\Lambda,\mathbb R^d)$ and hence there is a unique orthogonal projection $\operatorname P_H$ from $L^2(\Lambda,\mathbb R^d)$ onto $H$. Moreover, if $\Lambda$ is bounded and $\partial\Lambda$ is Lipschitz, then $$V=\left\{u\in H_0^1(\Lambda,\mathbb R^d):\nabla\cdot u=0\right\}\;.\tag 2$$
Now, I've seen some authors claiming that $$B(u,v):=\operatorname P_H\left[(u\cdot\nabla)v\right]\;\;\;\text{for }u,v\in V$$ is a well-defined bounded bilinear operator from $V\times V$ to $H$ (see page 2). However, I don't think that $B$ is well-defined, since I doubt that $(u\cdot\nabla)v\in L^2(\Lambda,\mathbb R^d)$ for all $u,v\in V$.
We would need to show that $$u_j\frac{\partial v_i}{\partial x_j}\in L^2(\Lambda)\tag 3$$ for all $i,j\in\left\{1,\ldots,d\right\}$, but while the Sobolev inequalities yield $$W_0^{1,\:p}(\Lambda)\subseteq L^q(\Lambda)\;\;\;\text{for all }q\in[1,p^\ast]\;.\tag 4$$ for all $p\in[1,d)$ and $p^\ast:=dp/(d-p)$, we only know that the weak partial derivatives of $w\in W_0^{1,\:p}(\Lambda)$ are in $L^p(\Lambda)$.
So, what am I missing?