Showing that $u(x)=(1+x^2)^{-k/2} (\ln(2+x^2))^{-1}$ belongs to $\mathcal W^{1,p}(\Bbb R)$.

Question

Consider the function $$u(x)=(1+x^2)^{-\frac{k}{2}} (\ln(2+x^2))^{-1}$$ $x\in\Bbb R$ with $k\in(0,1)$. Prove that $u\in \mathcal W^{1,p}(\Bbb R)$ for every $p\in[1/k,+\infty]$ and that $u\notin L^q(\Bbb R)$ for $q\in[1,1/k)$.

By definition, I have to prove that $u(x)\in L^p(\Bbb R)$ and that the weak derivative $Du$ exists and it belongs to $L^p(\Bbb R)$.

I think I computed $Du$ correctly, but I am having problems showing that $u$ and $Du$ belong to $L^p(\Bbb R).$

How can I show that $$\int_{\Bbb R}|u(x)|^p<+\infty $$ $$\int_{\Bbb R}|Du(x)|^p<+\infty. $$

Thanks in advance.

Answer 1

I'll show you how to prove $u\in\ L^p$. It is clear that $u$ is smooth. We have to look at the behavior of $u$ at $\infty$. For $x$ large, $$ u(x)^p=(1+x^2)^{-\frac{k\,p}{2}} (\ln(2+x^2))^{-p}\le 2^{-p}|x|^{-k\,p}(\ln|x|)^{-p}. $$ If $k\,p>1$ this is integrable. If $k\,p=1$, since $p\ge1/k>1$, it is also integrable.

Similar arguments will show that $Du\in L^p$.