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We have an $L^2$ function $u$ defined on $\mathbb{R^2}$ with compact support such that $u \in H^{2/3}$ (H stands for Sobolev spaces, as always), $\partial_y u \in L^2$, and $(x\partial_y - y\partial_x)u \in L^2$. I want to conclude that $u$ lies in some higher order $L^p$ space, that is, for some $p > 2$. Ideally, I would like to use all three pieces of information so that it leads to a high enough $p$. Any ideas?

Thanks a lot!

Edit: Is there some book that contains exercises specifically of this nature?

Another edit: I am sorry for not pointing this out explicitly: I am familiar of the conclusion that can be drawn from Sobolev Embedding type theorems. My main concern is how to make use of the facts $\partial_y u \in L^2$ and $(x\partial_y - y\partial_x)u \in L^2$, particularly the latter one.

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2 Answers

This is not a full answer, but it shows you can do better than p=6. In the following, subscripts x and y refer to the x and y dependence. You have $$u \in H^{2/3}_x ( L^2_y )\cap L^2_x( H^1_y ).$$ By interpolation, you find $$u\in H^{2\alpha/3}_x(H^{1-\alpha}_y).$$ For $\alpha=3/5$, we find $$u\in H^{2/5}_x(H^{2/5}_y).$$ In one dimension $H^{2/5}$ embeds into $L^{10}$, so you have at least $p=10$. Since this does not use your last condition, it is probably not optimal.

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@Michael Renardy Professor Renardy, thanks a lot for your answer. Would you please explain the notation $H^{2/3}_x(L^2_y)$ once? I have never seen this notation before. Thanks a lot! –  mathenthusiast Mar 19 '13 at 18:14
    
It is supposed to mean $H^{2/3}$ (as a function of x) with values in $L^2$ (as a function of y). –  Michael Renardy Mar 19 '13 at 18:30
    
@Michael Renardy Professor Renardy, can you please point to a source where I can read more about these mixed type Sobolev spaces? Thanks a lot! –  mathenthusiast Mar 21 '13 at 15:10
    
The monograph of Lions and Magenes, for instance. –  Michael Renardy Mar 21 '13 at 15:54
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The power of 10 from Michael Renardy's answer is in fact optimal, which follows from the fact that for $y\approx 0$, the vector fields $\partial_y$ and $x\partial_y - y\partial_x $ are parallel.

We can also get at it using a scaling argument. Let $\phi(x,y) \in C^\infty_0(\mathbb{R}^2)$. Let $\phi_{\alpha\beta}^\lambda(x,y) = \lambda \phi(\lambda^\alpha x, \lambda^\beta y)$. The usual scaling analysis shows that for $\alpha = 6$ and $\beta = 4$, we have that

$$ \|\phi^{\lambda}_{\alpha\beta} \|_{H^{2/3}} \leq \|\phi\|_{H^{2/3}} $$

and

$$ \|\partial_y\phi^{\lambda}_{\alpha\beta} \|_{L^2} \leq \|\partial_y \phi\|_{L^2} $$

and

$$ \|y\partial_x \phi^{\lambda}_{\alpha\beta} \|_{L^2} \searrow 0 $$

as $$\lambda\nearrow \infty $$

On the other hand, for $\gamma > 10$, we have that

$$ \| \phi^{\lambda}_{\alpha\beta} \|_{L^\gamma}^{\gamma} = \lambda^{\gamma - \alpha - \beta} \|\phi\|_{L^\gamma}^\gamma \nearrow \infty $$

Now you can do the usual trick of summing a bunch of these guys with disjoint support to show that the embedding in to $\gamma > 10$ is not possible.

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@Willie Wong Hi Willie, if I understand this correctly, in the last step you mean that we will get a function $u$ with countably many components of its support, with $u = \frac{1}{2^n}\phi^{\lambda}_{\alpha\beta}$ on each support, with $\lambda$ increasing and $n$ chosen suitably for each $\lambda$. This implies $u$ remains in $H^{2/3}$ and $\partial_{y}u$ in $L^2$. However, the way I am thinking, to get these disjoint supports, you have to push (translate) the original $\phi^{\lambda}_{\alpha\beta}$ with their supports in the plane. But does it readily follow that we still have a control –  mathenthusiast Mar 20 '13 at 23:56
    
on the $L^2$-norm of $y\partial_{x}\phi^{\lambda}_{\alpha\beta}$? What I am missing here? –  mathenthusiast Mar 21 '13 at 0:00
    
@math: translate along the $x$ axis. the $L^2$ of $y\partial_x \phi^\lambda_{\alpha\beta}$ remains invariant if you translate along the $x$ axis. And by choosing sufficiently fast decreasing supports, you can fit all of them inside $|x| < R$ for some finite $R$, making $x\partial_y \phi^\lambda_{\alpha\beta}$ bounded. –  Willie Wong Mar 21 '13 at 7:29
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