# reference needed for sobolev type estimates

I'm reading a paper and the authors applied the following sobolev type estimates $$||(Dv)^{2}||_{H^{3k-2}(\Omega)}\leq C||v||_{H^{3k-1+\alpha}(\Omega)}^{2}$$ for $\alpha>\frac{1}{4}$, where $v$ is a $n$ velocity vector field, $\Omega\in \mathbb{R^{n}}$ and $k$ is chosen such that $3k\geq \frac{n+3}{2}$.

I was wondering if anyone has seen this inequality before and can provide me one of its reference? I couldn't find the exact inequality applies to it. The closest one should be general sobolev embedding but its condition involves dimension $n$ so doesn't work.

• Link to the paper, please? As gerw has written there seems to be a problem. Maybe there is a hidden condition which you didn't quote or perhaps a typo (since there is a lowerbound condition on $k$, which would suggest some type of Sobolev). Apr 3 '14 at 10:49
• The paper is "Geometry and a priori estimates for the free boundary problem of the Euler's equation" by Shatah and Zeng. The link is as follows math.nyu.edu/faculty/shatah/preprints/sz06a.pdf Apr 3 '14 at 21:59
• This estimate appears on page 12, definition 4.2, formula (4.6). Apr 3 '14 at 22:12

Okay, the estimate actually holds. I had a "moment" yesterday when I first read gerw's argument, which is actually subtly flawed. Let me quickly illustrate:

To estimate $\| u^2\|_{H^1}$ you need to estimate $\|u^2\|_2$ and $\|\nabla (u^2)\|_2$. The first term is bounded by $\|u\|_4^2$ as claimed, but the second term is not $\|\nabla u\|_4^2$! What you have is in fact $\|u \nabla u\|_2 \leq \|u \|_4 \|\nabla u\|_4^2$. This behaves a bit better than $\|\nabla u\|_4^2$ since the "total number of derivatives is fewer".

By examining the binomial formula, we have that

$$\| \nabla^k(vw)\|_{2} \lesssim \sum_{j = 0}^k \| \nabla^j v \nabla^{k-j}w\|_2$$

Only one of the two terms can receive more than half the derivatives. We put that term in $L^2$ and the other in $L^\infty$, and we get very roughly

$$\|\nabla^k(vw)\|_2 \lesssim \| v\|_{W^{\alpha,\infty}}\| w\|_{H^{k,2}} + \| w\|_{W^{\alpha,\infty}}\| v\|_{H^{k,2}}$$

where $\alpha = \lfloor k/2\rfloor$. Sobolev embedding gives

$$W^{\alpha + n/2 + 1,2} \to W^{\alpha,\infty}$$

So for $k >n + 2$ we have shown that $W^{k,2}$ is an algebra.

The bound $k > n+2$ is not sharp. If you do better interpolation between the terms you can get it all the way down to almost the boundary for Sobolev embedding to $L^\infty$ to hold.

Basically, you need to apply the product rule to distribute the derivatives on the two terms and estimate each of the two terms in the optimal Sobolev spaces given by the Sobolev inequalities. I leave the details as an exercise for you to check.

You can manipulate the left-hand side to \begin{equation*} \lVert (Dv)^2 \rVert_{H^{3k-2}} = \lVert D v \rVert_{W^{3k-2,4}}^2 \le \lVert v \rVert_{W^{3k-1,4}}^2. \end{equation*} (Note that what we have lost in the inequality is just $\| v\|_{L^4}$ which is not critical).

Now, in order that your inequality holds, we need that $H^{3k-1+\alpha}$ embeds into $W^{3k-1,4}$. With Sobolev's embedding, you would get this for small $n$ but not for all, as you already mentioned. Hence, I would say that this inequality does not hold.

Maybe you could try to construct a counterexample.

• Yes, that's exactly what I'm confusing of. Because the dimension n is arbitrary. Apr 3 '14 at 22:03
• The first equality doesn't hold, strictly speaking. Apr 4 '14 at 8:21