Exercise 8.1 in Brezis' Functional Analysis

Question

Consider the function $$u(x) = \frac{1}{(1+x^2)^{\frac{\alpha}{2}}} \frac{1}{\ln(2+ x^2)} \qquad\; x\in \mathbb{R}$$ with $0<\alpha<1$. Check that $u\in W^{1,p}(\mathbb{R})$ for all $p\in \left[\frac{1}{\alpha},+\infty\right)$ and that $u\notin L^q(\mathbb{R})$ for all $q\in \left[1,\frac{1}{\alpha}\right)$

For $p> \frac{1}{\alpha}$, using AM-GM's inequality I have $x^2 +1 \geq \frac{(x+1)^2}{2}$, thus $$ u(x)^p \leq \frac{2^{p\alpha/2}}{\ln(2)^p} \frac{1}{(x+1)^{p\alpha}}$$ Combine with the fact $$\int_1^\infty \frac{dx}{x^k} <\infty \qquad\text{if and only if}\qquad k>1$$ I obtain $u\in L^p(\mathbb{R})$. But incase $p = \frac{1}{\alpha}$, I cannot find any useful estimate for $$ u(x) = \frac{1}{\sqrt{1+x^2}}\frac{1}{\ln(x^2+2)^p}$$ where $p > \frac{1}{\alpha} > 1$. Could anyone help me?

Answer 1

Split into small $x$ and large $x$:

• $|x|\le 2$. Nothing to discuss here: the function is continuous.
• $|x|>2$. Then $u(x)\le \dfrac{1}{|x| \ln^p |x|}$. By symmetry, it's enough to consider positive $x$. Integrate using substitution $y=1/x$. This is one of standard improper integrals that just barely converge (thanks to $p>1$).