# Is $(u\cdot\nabla)v\in H^1$, if $u,v\in H^2$?

Let

• $d\in\left\{2,3\right\}$ with
• $\Lambda\subseteq\mathbb R^d$ be bounded and open with $\partial\Lambda\in C^1$

In Lemma 6.1 of Navier-Stokes Equations and Nonlinear Functional Analysis by Roger Temam, the author is stating that if $u,v\in H^2(\Lambda,\mathbb R^d)$, then $$\frac\partial{\partial x_i}(u\cdot\nabla)v=\left(\frac{\partial u}{\partial x_i}\cdot\nabla\right)v+(u\cdot\nabla)\frac{\partial v}{\partial x_i}\tag 1$$ would belong to $L^2(\Lambda,\mathbb R^d)$, because $u,v$ belong to $L^6(\Lambda,\mathbb R^d)$ and $u,v$ are continuous on $\overline\Lambda$ (which is clear by the Sobolev inequalities as they can be found, for example, in the book of Evans).

Can the same statement be proved, if $\Lambda$ is just bounded and open and $u,v\in H_0^2(\Lambda,\mathbb R^d)$?

• It also follows from Sobolev embedding. – Fan Zheng Dec 12 '16 at 18:26
• @FanZheng Of course, you're right (the crucial point is that $d\le 3$). I didn't saw this at the first place, cause I got a domain in mind which is just bounded and open. I've modified the question accordingly. – 0xbadf00d Dec 12 '16 at 18:38
• If a function is in $H^2_0$, it can be extended by zero outside the domain, and the extended function is in $H^2$. – Michael Renardy Dec 12 '16 at 19:25
• @0xbadf00d For the first term, $\nabla u$ and $\nabla v$ are in $H_0^1$, which embeds in $L^6$. For the second term, it is $L^\infty$ multiplied by $L^2$. – Fan Zheng Dec 12 '16 at 19:57
• Yes, that is what I mean. – Michael Renardy Dec 12 '16 at 20:26