Find $p$ that a function will be in $L^p_\text{loc}$

Question

Let the function $G(x)=\ln|x|$ defined in $\mathbb{R}^2 \setminus \{0\}$. How can we prove that $\nabla G \in L^p_\text{loc} ((]0,1[)^2)$ when $p<2$?

I calculate $\nabla G = \dfrac{x_1+x_2}{x_1^2+x_2^2}$ when $x=(x_1,x_2)$, but I can't find $p$.

Thanks in advance.

Answer 1

$$ \nabla G=\frac{x}{\left|x\right|^2}\implies\left|\nabla G\right|=\frac1{\left|x\right|} $$ For $p\lt2$, the $L^p$ norm in the unit ball is $$ \left(\int_0^12\pi r\frac1{r^p}\,\mathrm{d}r\right)^{1/p} =\left(\frac{2\pi}{2-p}\right)^{1/p} $$ For $p\ge2$, $r^{1-p}$ is not integrable on $[0,1]$.

Answer 2

It's important to note that we are talking about locally integrable functions. In the following I will only consider $p>0$.

Let $A$ be a compact subset of the open set $(0,1)^2$. Then $A$ has an empty intersection with the boundary of $(0,1)^2$.

In $A$, the function $\nabla G = \frac{x_1 + x_2}{x_1^2 + x_2^2}$ is continuous (note that the origin cannot be in $A$). Let $$f(x_1,x_2)= \max(1,|(\nabla G)(x_1,x_2)|).$$ Then, $f$ is bounded and continuous in $A$ and for $p>0$ we have $|\nabla G(x_1,x_2)|^p \leq f^p(x_1,x_2)$.

$$\int_A|\nabla G(x_1,x_2)|^p dx_1 dx_2 \leq \int_A f^p(x_1,x_2)dx_1 dx_2 \leq \sup_{x\in A}f^p(x) \int_A dx_1 dx_2 \leq \sup_{x\in A}f^p(x) < \infty.$$

For $p< 0$, the same argument can be applied to $\frac{x_1^2 +x_2^2 }{x_1 + x_2}$ (note that our domain of integration does not contain $0$ and both $x_1,x_2$ are positive).

For $p=0$, the integral is precisely the Lebesgue measure of $A$.