Powers of a BMO function

Question

We know that $f(x) = |\log{|x|}|$ is in BMO (bounded mean oscillation). Using this, how can we prove that $f(x) = |\log{|x|}|^p$ for $0 < p < 1$ is also in BMO? I tried proving this using really cumbersome computational arguments but I feel like I am missing something here.

Answer 1

I have finally figured out how to prove this. There is a standard result for BMO functions: $$f \in BMO \iff \exists c > 0 \text{ for all balls } B \text{ such that } \frac{1}{|B|^2}\iint |f(x)-f(y)| dx dy \le c.$$ We already know that $f(x) = |\log |x||$ is in BMO and hence will satisfy the given result. So we can compare $||\log |x||^p - |\log |y||^p|$ with $||\log |x|| - |\log |y|||$ and if $$||\log |x||^p - |\log |y||^p| \le ||\log |x|| - |\log |y|||,$$ then we are done. Consider $|x| \ge |y|$ and $z = \log |y|$ if $|y| \ge 1$ ($\Rightarrow z \ge 0$) . We can say $|x| = e^h |y|$ for some $h \ge 0$. Then proving the given inequality is reduced to proving, that for $0 < p < 1$, $$(h+z)^p - z^p \le h.$$ This can be verified using basic calculus. If $|y| < 1$, one can assume $z = \log \frac{1}{|y|}$ and proceed in a similar manner. In this case, one would have to check whether $|x|$ is still bigger than $\frac{1}{|y|}$. If $|x| < 1$ as well, then one can replace $|x|$ with $\frac{1}{|x|}$ in this entire procedure.