Properties of function $f(x) = (1 + x^2)^{-\alpha/2}(\log(2+x^2))^{-1},\text{ }x \in \mathbb{R}$ with $0 < \alpha < 1$.

Question

Consider the function$$f(x) = (1 + x^2)^{-\alpha/2}(\log(2+x^2))^{-1},\text{ }x \in \mathbb{R},$$with $0 < \alpha < 1$. How do I see that $f \in W^{1, p}(\mathbb{R})$ for all $p \in [1/\alpha, \infty]$ and that $f \notin L^q(\mathbb{R})$ for all $q \in [1, 1/\alpha)$?

Answer 1

Let $x>0$. Because $\log x\le x$, we have that $\log x\le x^t/t$ for $t>0$. Therefore $$\frac{1}{\log (2+x^2)}\ge \frac{t}{(2+x^2)^t},\ x\in\mathbb{R}.$$

Then $$\int_{\mathbb{R}} (1+x^2)^{-\alpha q/2}(\log (2+x^2))^{-q}\ge \int_{\mathbb{R}}(1+x^2)^{-\alpha q/2}(2+x^2)^{-qt}. $$

From the last inequality, you can conclude that if $q\in [1,1/\alpha)$ then $f\notin L^q(\mathbb{R})$.

On the other hand, note that $$\int_{\mathbb{R}} (1+x^2)^{-\alpha p/2}(\log (2+x^2))^{-p}\le C\int_{\mathbb{R}} (1+x^2)^{-\alpha q/2},\tag{1}$$

for some positive constant $C$ then, $f\in L^p(\mathbb{R})$ for $p>1/\alpha$. To conclude, note that $$f'(x)=-(2+\alpha)x(1+x^2)^{-1-\alpha/2}\left(\frac{1}{(\log(1+x^2))^2}+\frac{1}{\log(1+x^2)}\right).$$

As in $(1)$, there exist $D>0$ such that $$\int_{\mathbb{R}} |f'|^p\le D\int_{\mathbb{R}}|x|^p(1+x^2)^{-p-p\alpha/2},$$

whence $f'\in L^p(\mathbb{R})$ if $p> 1/(\alpha+1)$, which implies that $f\in W^{1,p}(\mathbb{R})$ for $p> 1/\alpha$.

The case $p=\infty$ is immediate.

Remark: The above calculation does not includes the case where $p=1/\alpha$, however, after thinking for a while, I noted that the change of variables $2+x^2=e^u$ prove all cases, in fact, after making this change of variables, you will have to estimate the following integral $$\int_{N}^\infty \frac{e^u}{(e^u-1)^{\alpha p/2}(e^u-2)^{1/2}u^{p}},$$

where $N$ is bigger thatn $\log 2$. Once $$\frac{e^u}{(e^u-1)^{\alpha p/2}(e^u-2)^{1/2}},$$

is bounded for $p\ge 1/\alpha$ and $1/u^{p}$ is integrable in $[N,\infty)$ for $\alpha\in (0,1)$, we have proved the desired result.

Answer 2

While a sketch, I think the below is an intuitive way to view the situation.

Lemma: $\int_2^\infty x^r(\ln x)^s\,dx < \infty$ iff $r<-1$ or $r=-1,s<-1.$

Proof: If $r<-1,$ then because of the slow growth of $\ln x$ compared with any positive power of $x,$ there is nothing $(\ln x)^s$ can do to ruin things, no matter how awesomely large $s$ may be. If $r=-1,$ then the integral takes the form $\int u^s\,du.$ An antiderivative can be found explicity, and we get the result. If $r > -1,$ then because of the slow decay of $1/\ln x,$ there is nothing $(\ln x)^s$ can do to save the day, no matter how awesomely large negative $s$ may be.

Now note that on $[2,\infty),$ $(1+x^2)^{1/2}$ is on the order of $1/x,$ and $\ln (2+x^2)$ is on the order of $\ln x.$ The results for $f$ then fall out from this and the lemma. Same for $f'$ even though it's a messier function.