Integrability of $g(x)=\frac{f(x)}{|x|^\alpha}$

Question

How can I show for which $\alpha > 0$ the function $g(x)=\frac{f(x)}{|x|^\alpha}$ is in $L^1([1,\infty))$ for every $f \in L^3([1, \infty))$?

My approach was to apply the Hölder-inequality: $\frac{1}{|x|^\alpha}$ is in $L^p([1,\infty))$ if and only if $\alpha p >1$. Since $f \in L^3([1, \infty))$, we need $p= \frac{3}{2}$. Hence if $\alpha > \frac{2}{3}$ the function $g$ is in $ L^1([1, \infty))$.

How do I exclude $ \alpha \leq \frac{2}{3}$?

Answer 1

Suppose $\alpha<2/3$. For $\epsilon>0$ small, pick $f_\epsilon(x)=x^{-\frac{1+\epsilon}{3}}$. Notice that $f_\epsilon \in L^3([1,\infty))$. We have $g(x)=x^{-(\frac{1+\epsilon}{3}+\alpha)}$. Since $\alpha<2/3$, we can say $\alpha+1/3<1$. Choose $\epsilon$ small enough so that $\alpha+1/3+\epsilon/3<1$. Then $g$ is not integrable.

Also, in your proof, you really covered the $\alpha \geq 2/3$ case. Hence, you are done. I don't really understand why you use the notation $|x|^\alpha$ since $x>0 \implies |x|^\alpha=x^\alpha$.