Show that $f(x)= \frac{x^{-1/2}}{1+ | \log x |} $ is only $L_{p} ((0,\infty])$ for p=2

Question

Show that $ {\rm f}\left(x\right)= {x^{-1/2} \over 1 + \left\vert\,\log\left(x\right)\,\right\vert}$ is only $L_{p}\left(\vphantom{\large A}\left(0,\infty\right]\right)$ for $p = 2$. Where $x > 0$.

For the case $p = 2$, I think that I need to show that the function is bounded when raised to the pth power near 0. For the other cases I expect the function to be unbounded. When I look at graphs of the functions however, they all look pretty similar to me. I guess there are separate arguments for when $p = 1$ and $p > 2$. Any hints or help would be great.

Hmm, I think that my talk about boundedness being necessary for the Lebesgue integral to exist is off. So if there's another thought I should explore that would be great.Thanks

Answer 1

Forget boundedness and think of comparison. What can you compare $|f|^p$ to?

The logarithm grows slower than any positive power of $x$ at infinity, and slower than any negative power of $x$ at $0$. Think of it as $x$ raised to infinitesimally small power, if it helps.

• for every $\epsilon$ you have $|f(x)|^p \ge x^{-p/2-\epsilon}$ when $x$ is large enough. When $p<2$, this makes $|f|^p$ nonintegrable on $(0,\infty)$. (Nothing special is needed for $p=1$.)
• for every $\epsilon$ you have $|f(x)|^p \ge x^{-p/2+\epsilon}$ when $x$ is small enough. When $p>2$, this makes $|f|^p$ nonintegrable on $(0,\infty)$.

To show $f^2$ is integrable, do a direct computation involving the substitution $u=\log x$.