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Let $f(x,y)=\frac{x}{\sqrt{x^2+y^2}}$. Does $f \in W^{1,p}(B)$ for some $p \ge 1$, where $B$ is the open unit disk in $\mathbb{R}^2$?

(I guess we can replace $B$ with a disk with arbitrarily small radius; the singularity is centered at the origin).

Here is what I know:

$f \le 1$ is bounded, so it is in $L^p(B)$ for any $p \ge 1$. Let us consider its derivatives:

$f_x=\frac{1}{\sqrt{x^2+y^2}}(\frac{y^2}{x^2+y^2})\le \frac{1}{\sqrt{x^2+y^2}}$, so for sure $f_x \in L^p(B)$ for $p<2$.

(In fact $f_x \in L^p(B) \iff p<2$).

$f_y=-\frac{yx}{x^2+y^2}(\frac{1}{\sqrt{x^2+y^2}})$, so $|f_y|\le \frac{1}{\sqrt{x^2+y^2}}$, hence $f_y \in L^p(B)$ for $p<2$.

So, is it true that $f \in W^{1,p}(B)$ for some $ 1 \le p <2$?

Asaf Shachar
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1 Answers1

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As shown in the answer to this question of mine, since $f$ is singular at a point only its classical derivatives coincide with the distributional ones. Therefore, by definition, $f\in W^{1,p}(B)$ if and only if $f$ and $\nabla f$ belong to $L^p(B)$. The computations in the question show that this is the case for all $p\in[1, 2)$.

Giuseppe Negro
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  • Thank you. By the way: Do you know if there is any "systematic way" to say something intelligent on the difference between the distributional derivative and the classical derivative, when they do not coincide? (let's say that I know that my function is smooth on an open set of full measure). My motivation comes from this question here: https://mathoverflow.net/questions/328119/the-distributional-gradient-of-the-closest-isometry-to-the-differential-of-a-smo – Asaf Shachar Apr 21 '19 at 14:39
  • Dear @AsafShachar, I have been thinking a bit at your question, but I am afraid I cannot say anything "intelligent". Of course, your function $q$ (cfr. MathOverflow question) is Sobolev if you assume that $U$ is not only full measure, but also that its complementary is codimension 2 or higher; this is the linked Fubini argument. With the only assumption that $U$ is full measure, the distributional derivative may differ from the classical one by a singular measure, supported in the complementary of $U$. As I said, this is not very intelligent, I am afraid. – Giuseppe Negro Apr 23 '19 at 09:10