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Suppose that we have a function $u$ on $\mathbb{R}^2$ such that $r\frac{\partial}{\partial\theta}u \in L^2(\mathbb{R}^2)$. Suppose it is also given that $u$ lies in some fractional Sobolev space $H^s(\mathbb{R}^2)$ for $1/3 < s < 1$. We also add the strong assumption about the differentiability of $u$ about a second vector field: $\partial_x u \in L^2$. My question is:

Can we use the first information somehow to get $u$ to lie in a greater Sobolev space, that is, $u \in H^{s'}(\mathbb{R}^2)$ where $s' > s$?

For specificity, we can maybe use $s = 2/5$ or $s = 3/5$, etc. We can also, if need be, assume that $u$ is in Schwartz space, or even compactly supported. Any help will be highly appreciated.

Thanks.

Edit: After @Bazin's comment. My intention is to understand how to use the condition $r\frac{\partial}{\partial\theta}u \in L^2(\mathbb{R}^2)$ because it seems a bit "weak" in the sense that it somehow cannot control the behaviour near $0$.

More edit: The way in which I say take $s = 2/5$ or $s = 3/5$ or say one can add assumptions in $u$ might seem a bit lax, but just for clarification: I am more interested in learning some generic approaches/methods as opposed to solving this particular problem. I have observed that there are frequently situations where you have some conditions which are not enough to say $H^1$, but very nearly so. So if you have some reference in mind from where I can learn problems like these (please give the chapter number if you can, it is sometimes very difficult to find something from an entire book), that will also be a very acceptable and welcome answer. I just hope the motivation is clear :)

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  • $\begingroup$ You need to specify what $r$, $\partial/\partial\theta$, and $\partial_x$ mean. Partial derivatives only makes sense when you have a full coordinate system. $\endgroup$ Commented Apr 29, 2013 at 9:33
  • $\begingroup$ See mathoverflow.net/questions/124942/… This question is quite similar. $\endgroup$
    – gerw
    Commented May 28, 2013 at 18:48

1 Answer 1

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No: take a radial function in $H^s$ and not in $H^{s+\epsilon}$ for any $\epsilon >0$.

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