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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 $u,v\in H_0^2(\Lambda,\mathbb R^d)$?

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    $\begingroup$ It also follows from Sobolev embedding. $\endgroup$
    – Fan Zheng
    Commented Dec 12, 2016 at 18:26
  • $\begingroup$ @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. $\endgroup$
    – 0xbadf00d
    Commented Dec 12, 2016 at 18:38
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    $\begingroup$ If a function is in $H^2_0$, it can be extended by zero outside the domain, and the extended function is in $H^2$. $\endgroup$
    – Michael Renardy
    Commented Dec 12, 2016 at 19:25
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    $\begingroup$ @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 $L^2$. $\endgroup$
    – Fan Zheng
    Commented Dec 12, 2016 at 19:57
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    $\begingroup$ Yes, that is what I mean. $\endgroup$
    – Michael Renardy
    Commented Dec 12, 2016 at 20:26

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